How to plot celestial equator in galactic coordinates? Why does my plot appear "backwards"?

Ok, community helped me to figure out that I faced the “mapping from the inside” issue (explained here). I'll use this answer to show my final code.
import numpy as np
from astropy.coordinates import SkyCoord
import astropy.units as u
import matplotlib.pyplot as plt
import seaborn as sns

sns.set()

def eq2gal(ra, dec):

    '''
    Transforms equatorial coordinates to galactic ones.
    Then prepares them for matplotlib aitoff projection.
    '''

    eq = SkyCoord(ra, dec, unit=u.deg)
    gal = eq.galactic

    # Minus appears because of “mapping from the inside” issue
    l_gal, b_gal = -gal.l.wrap_at('180d').radian, gal.b.radian

    return l_gal, b_gal

def ecl2gal(lon_ecl, lat_ecl):

    '''
    Transforms ecliptic coordinates to galactic ones.
    Then prepares them for matplotlib aitoff projection.
    '''

    ecl = SkyCoord(lon_ecl, lat_ecl, unit=u.deg, frame='barycentricmeanecliptic')
    gal = ecl.transform_to('galactic')

    # Minus appears because of “mapping from the inside” issue
    l_gal, b_gal = -gal.l.wrap_at('180d').radian, gal.b.radian

    return l_gal, b_gal

# Equatorial plane
ra_all = np.linspace(-180, 180, 100)
dec_0 = np.zeros(100)

l_eq_gal, b_eq_gal = eq2gal(ra_all, dec_0)

# Ecliptic plane
lon_ecl = np.linspace(0, 360, 100)
lat_ecl = np.zeros(100)

l_ecl_gal, b_ecl_gal = ecl2gal(lon_ecl, lat_ecl)

plt.figure(figsize=(14,7))
plt.subplot(111, projection='aitoff')

plt.scatter(l_eq_gal, b_eq_gal, s=4, marker='v', label='Celestial Equator')
plt.scatter(l_ecl_gal, b_ecl_gal, s=4, marker='^', label='Eclptic')

# Essential thing is to rename RA axis ticks to transform them to conventional format
plt.xticks(ticks=np.radians([-150, -120, -90, -60, -30, 0, \
                             30, 60, 90, 120, 150]),
           labels=['150°', '120°', '90°', '60°', '30°', '0°', \
                   '330°', '300°', '270°', '240°', '210°'])

plt.grid(True)
plt.legend(fontsize=16, loc='lower center')

plt.title('Galactic', fontsize='16')

plt.show()


Why is time considered a (fourth) dimension?

(The speed of light) times (time) has the properties of a linear dimension.  Just like length, width, or height, it can be used to measure how far you can go.  The Lorentz contraction is consistent with applying the Pythagorean theorem to it and the other dimensions.


Can moon set after being up when due north? (pseudo circumpolarity)

Yes - here is a good explanation of the situation. Just as it's possible to have continuous daylight for days or weeks on end, you can have continuous moonlight too.

The Moon's orbit is tilted at (a maximum of) just over 5 degrees to the ecliptic, however this tilt has an oscillatory precession with a period of 18.6 degrees. This precession means that our "lunar arctic circle" will vary in latitude over time.

For example, if the moon is sitting exactly on the ecliptic (latitudinally speaking), it would remain above the horizon until it is at a sufficiently low phase of illumination that its current ecliptic longitude is below the horizon. Of course in reality the moon only crosses the ecliptic twice during each lunar cycle (it's these two points that precess). Stellarium is a cool program to visualize this on if you turn on the ecliptic grids.

One important thing to note is that this phenomenon is dependent on both the time of year and the phase of the moon. In the summer months, new and near-new moons are the ones that can remain visible for days or weeks on end. Conversely, at the North pole, you might not see a full moon at all for a few months!


Assuming that you have a phaser array sufficiently powerful to pulverize the NEO into a harmless meteor storm,
I read this as a question of when an object with those geocentric orbital elements arrives at perigee.
However, I get a different value for $r$ if I treat that $\nu$ value as a true anomaly; it could be a mean anomaly or hyperbolic anomaly instead.

I suppose your textbook covers this stuff in detail, but for other readers
these lecture notes
by M. Peet at ASU, using $f$ instead of $\nu$, may help.


If F is the focal length of the objective, and f the focal length of the ocular, then the magnification of the instrument is:

M = F / f

So you'll get 10x with the first combination. This is enough to see craters, mountains and valleys on the Moon. You will probably not see planetary disks or the rings of Saturn.

The large diameter of the ocular might be awkward, but it should work. Make sure both lenses are installed on a rigid tube - plastic tubes from water pipes should work. Make sure the lenses are perpendicular to the axis of the tube - small errors there reduce overall performance.

You will need a way to finely adjust the distance between objective and ocular - the ideal distance is F + f but don't just glue them in place; instead, see if you can make two concentric sliding tubes, you know, like the spyglasses from pirate movies. Then you'll be able to look through the instrument and adjust the distance until the image is sharp.

If there's any way you could find a used ocular from an old microscope or binoculars, use that as the ocular in your instrument. It will provide higher magnification (smaller focal length) and its performance should be much better than a simple biconvex lens - fewer aberrations, since true oculars have internal corrections (well, most of them). With the higher magnification you will probably see the rings of Saturn, the moons of Jupiter, or even some equatorial belts on Jupiter.

Take it slowly and do it right. This is totally doable with simple materials. This is the design of the first telescope I've built many years ago. You'll be very proud when something you made with your own hands actually works.

EDIT: Yes, you can stack lenses for extra power in the ocular. If you stack two convergent lenses, so close that they touch each other, having the same focal length, the resulting system has about half the focal length of one lens. If there's some distance in between them, things are a bit more complex.

You will notice that power is not everything, and sometimes more power is beneficial while other times it creates problems. It depends on many factors, which you can discover through experimentation.

Also read about the various eyepiece designs, which are made for specific purposes - not just magnifying power, but also lack of aberrations and a high quality image. But those require special components to make.

https://en.wikipedia.org/wiki/Eyepiece

By all means go ahead and try various combinations yourself. There's a lot to discover. It's a lot of fun.

EDIT2:

Telescopes for astronomy cannot be used hand-held. You must use some kind of mount. You could make one, doesn't matter how simple, as long as it's stable and it can turn in both directions (up/down and left/right). E.g. you could make a mount out of a cardboard box:

https://aquillam.wordpress.com/2010/01/04/a-cheap-tripod-alternative/

Google "diy refractor mount" or "galileoscope mount" for more ideas.

EDIT3:

When the eyepiece is convergent, the image is reversed. This is normal for usage in astronomy, where the orientation of the image doesn't matter.

There are special devices called prism diagonals that can be inserted between objective and ocular, that fix the reverse image problem. Star diagonals show the image correct-side-up, but left and right are reversed. Correct-image diagonals show the image correct-side-up and also left and right are correct. They achieve this by doing one or multiple reflections in prisms.


If the Sun got larger, but maintained its luminosity, would the Earth get hotter or colder?

To a first approximation if the Sun got bigger (by a relatively small factor, say a few times) but maintained its present luminosity, as the Earth would still be intercepting the same total energy per unit time the temperature would stay the same. Constant luminosity gives a constant intensity at the Earth's distance from the Sun (intensity is the energy crossing unit area in unit time) which would mean that the total solar energy intercepted by the Earth would remain the same.

However in this scenario the colour of the sun would change and we would get secondary effects due to the reflectivity of the Earth being frequency dependent so the fraction of the incident energy reflected rather than absorbed would change, which would result in a change in temperature. Which direction this would go in is difficult to say as an increase in temperature should result in more cloud which would increase reflectivity ...


Matter/mass exist into the space. Where does space exist?

There is still no way to solve this question for all people.
You also need to keep in mind that time is directly linked to space. So before the universe there was no space and no time. However, if you want describe processes or even states you should also be able to talk about time.
Time is the perception of a sequence of events. So before the big bang no events could happen. To be more detailed time is what physicsists call the increase of entropy. Entropy itself is the randomness of a closed physical system. Each physical systems wants to have maximum entropy or randomness. The big bang at point of time -zero- can be seen as a point with the lowest possible value of entropy, because any matter/energy was located at a single point - at a singularity. There was ne place for particles to get at a random place.

With the expansion of this point space and time expanded into "nothing". If you think about it: how could it expand if there was something?

Next: If there is something outside, we still have a neverending question: What's outside the outside? Is it infinite?
Our physical laws don't work outside - so how should we talk about things nobody could ever describe?

Of course, there are theories that try to explain why or how the big bang happened - the ethereal expansion by Alan H. Guth and Alexander Vilenkin, for example. Universes are "created" like holes in a cheese by massive discharges of energy (like lightnings during a storm), but those are very hypothetical theories without any prove or even evidence. And all those theories need parallel universes / multiverses.

So, for my opinion, I keep at this point: There was nothing. As human beings we cannot deal with nothing. For us there is always something. If we try to think about nothing, we still think on how much time has passed not thinking about anything - for example. Nothing is mathematically described as 0. But also 0 is a number. It's more than nothing. It still consumes place in a formula. So don't think about nothing as "nothing". Think about nothing as everything. Because or universe was created out of nothing, this nothing is also everything for any or every time.


What does it mean 'black holes is created when center of a very massive star collapses in upon itself'?

This Wikipedia page on Stellar Evolution is a good starting point with lots of links to more info.

A star is a huge sphere of mostly hydrogen.

Like any large mass it has a gravitational field and that field acts on star itself.

So it would always want to collapse to as dense a sphere as possible under it's own gravity.

Now as the density increases, particularly at the center, the pressure also increases and so does the temperature.  When these are high enough the core of the star will be able to fuse hydrogen (and later other materials) into heavier elements.  Doing so releases huge amounts of energy and this is what creates all that light we see.

But the energy released also prevents the star from collapsing more.

And it can stay that way (more or less) for billions of years.  It is this energy produced by fusion at the core that balances the gravitational force (weight) of the rest of the star trying to crush the core.  So fusion that generates more energy than it requires to happen is essential to keep a star "alive".

After a long time, the hydrogen at the core runs out and the star starts to burn heavier elements at the core (helium, carbon, etc. ).  It does that until those run out and (roughly) we reach elements like Iron remaining.  At that point fusing those elements no longer generates enough energy to stop the star collapsing and it starts to collapse again, but very, very rapidly.

Exactly what happens next depends mostly on the mass of the star.  It can end up as a white dwarf, neutron star or black hole.

But the final collapse is a very rapid process and quite violent.

Once the useful fuel at the core runs out, the stars dies a spectacular death one way or another.

What is a black hole (in simple terms) ?

Gravity is a result of the way mass (or even energy) can distort space.

Beyond a certain point this distortion becomes so great that everything on one side becomes trapped inside a "surface" called the event horizon.  An event horizon is what a black hole actually is - no event horizon, no black hole.

Within the black hole, space is distorted in a way that's basically impossible to describe without some mathematics.  The effect is that there is no direction you can move in and no speed you can travel at which will not lead you to the center of the black hole - the singularity.  Don't even try and visualize it, it just isn't a common sense idea.

This even works with light.

Now an event horizon is not a surface you can feel.  You can actually travel across an event horizon (but only inward, not outward !).  So this change in the way space-time is distorted is not something that is sudden or obvious to the person doing the falling.

Back to our star collapsing.

As it collapses (without energy from fusion to stop that), it becomes denser and denser.

Now if it is large enough it will collapse to a black hole, because the core will become so dense from the collapse that it will be enough to form an event horizon.

Our Sun is not large enough to do that.


Given that the methane variations are seasonal, the diagram is labeled in Martian years - the amount of time it take Mars to travel around the Sun once (about 1.88 Earth years). This makes it much easier to see the cycles rather than using Earth years. In particular, the x-axis has a length of one Martian year; the data point show seasonal methane measurements for that year. So, for instance, the red data point was take early in the spring of Mars year 34, the first blue point was taken at the start of summer of Mars year 32 (earlier!), and so on and so forth.

The specific year numbering is an arbitrary convention begun on April 11, 1955 (the northern spring equinox). The solar longitude system is applied to Mars; it's a geometric way of describing where in its orbit a planet is. The solar longitude is denoted $L_2$; on April 11, 1955, $L_s=0^\circ$; it increases to $L_s=360^\circ$, at which point a new Martian year begins.


"Juxtapositions" of moon,stars and planets in the sky,

There are several words to describe two celestial bodies being in the same place on the celestial sphere.

A conjunction is when two bodies seem to pass close to one another, as seen from the Earth.

A transit is when a celestial  body passes between the observer (us) and a bigger body. For instance, Venus can transit in front of the Sun, when it passes between the Earth and the Sun.

An occultation is when a celestial body in the foreground hides a celestial body in the background. For instance, the Moon occults the stars it passes in front of.

An eclipse designates what happens when one body passes in front of the other for an observer, as a result of a transit or an occultation.

A syzygy is simply when three celestial bodies are roughly aligned. One of those three bodies can be the Earth, in which case, the two other bodies will either be in the same place on the celestial sphere (a conjunction) or at the antipodes.


Black dot visible when looking into the telescope

What we see in your image is an out of focus object (the white moon shape) with the shadow of the secondary mirror (causing the out of focus object to have a crescent moon shape) and the struts (or "spider vanes") that hold the secondary mirror in place (the vertical and horizontal lines).
Focus on the moon and get that focus sharp.  Then try looking at stars (they should appear as pin-points) then try Jupiter and Saturn.  Don't touch the focus knob, when looking at the planets, since the focus point for the moon, star and planet is the same (they are all at "infinity" as far as focusing is concerned).
Jupiter will be "small" with that telescope.  Perhaps much smaller than you expect. Look a tiny circle of light with four dots (the moon) around it. You may see some banding.  Saturn is often better looking, as the rings should be clearly visible.
As a_donda suggests, the telescope may benefit from collimation (getting the mirrors and eyepiece perfectly lined up) The manual should have details (if not, search the internet) It's hard to give specific advice.  Look through the manual perhaps there's a faceslapper that you're missing, or get a friend to go through it with you (four eyes are better than two). If you got it from a specialist shop, you could try talking to the shop people. You could try the website of the company that made the telescope, they may have a contact number.  You could search for a local astronomy club.


Astronomical animations of deep space objects

Yes there are! Here are some examples:

A giant flare in the XZ Tauri system
The light echo from the outburst of V838 Monocerotis
The planetary system at HR 8799

The first two are obviously sequences of still images taken a while apart, the third one uses interpolation to smooth out the motion.


How do I align my finder scope National Geographic 76/700 EQ

The moon is generally big enough to not need a finder scope. In general, The easiest way to aim a telescope directly at something is to look parallel to the telescope, right above the edge. If you can just see the object you are aiming at, it should be at least visible in the finder scope (if properly configured, see below). In the case of the moon, try to make just the top half visible when looking parallel to your telescope.

If you are having trouble using a finder scope, it might not be properly aligned. A finder scope has two wires forming a cross. A properly aligned finder scope will have an object in the middle of the telescope's view if it is exactly on the cross of the wires. If this is not the case, the position of the wires can be adjusted using the screws. This is best done at daytime by aiming the telescope at a recognizable landmark (somewhat) nearby, such as a tower.


If an Asteroid was to strike the Earth, would it affect the Earth's rotation?

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.


If the Sun's trajectory were altered, would the solar system follow?

The motion of bodies orbiting anything, is described by orbital elements - their distance from the barycenter (central mass), direction, velocity - and so two bodies of mass significantly lower than the central mass, with identical orbital elements, will move along the same orbit; with very similar elements their orbits will be very similar.
For example, an astronaut on ISS, will circle Earth in nearly perfectly same trajectory as the station itself; both the astronaut and the station are vastly lighter than Earth, they are at the same altitude, the same speed, and so the astronaut is in freefall inside the station, not pushed against the walls or otherwise accelerating relative to the station - they both orbit Earth, their trajectory mostly unaffected by the minuscule differences in location between center of mass of the station and the astronaut - that difference being of order on ~30 meters at most, while the station orbits Earth on an orbit of radius of about 6700km (Earth radius + 400km) - with such small difference in orbital parameters they are both in the same orbit.
Similarly, Sun is about 25,000 light years from the center of the galaxy, and as you can guess the galaxy is much, much heavier than the Sun. Pluto is about 4.6 light hours away from Sun, so it moves 'in' and 'out' relative to the galaxy by maybe 9.2 light hours. So the influence of the galaxy on it practically doesn't differ between these positions - the galaxy influences it just the same as it does the Sun, so they both orbit the center of Galaxy on practically the same orbit.
Now, there is a certain rather significant difference between the orbit described by the Sun, and, say, Mercury. The Sun moves in the orbit around the galaxy at 230km per second. Mercury circles the Sun at 47.36 km/s. Now, when it goes "in the opposite direction" relative to the Sun than the motion around galaxy, that's 230-47=183km/s; when it goes "in the same direction", it's 230+47=277km/s; a body moving at either of these velocities at the same 25,000km from the center of the galaxy would have a much different orbit than the Sun. But as Sun's gravity reverses Mercury's direction relative to the center of the galaxy every 44 days (88 days is its orbital period), it all averages out over the 226 million years to complete the orbit around Milky Way.
So, in short, yes, the entire solar system is affected just the same as the Sun and orbits the center of the galaxy just the same, all the differences either averaging out or too small to be noticeable. And if you tried to change Sun's trajectory by a gravitational influence that is sufficiently strong and sufficiently distant, you'd redirect the entire solar system just fine. Of course if you used a source of gravity that varies more across the width of the solar system - say, another star or a regular (not supermassive) black hole - you'd completely break the precarious balance of orbital motion of our system, planets closer to the new source of gravity captured by it and entering its orbit, leaving the orbit of the Sun, planets farther out possibly escaping into interstellar space.
Gravitational field from a central mass (like a star, or the center of galaxy) varies with square of the distance from the central mass, so you need it acting over relatively short distance in at a very long distance from that mass, to have it "locally" behave like uniform field and act on all the affected objects nearly the same.


I don't think there are any in the Solar system. We do have around 250 asteroids with moons. Rhea's ring seems to be the only exception.

Edit: Originally I said "a moon with a moon would be an unstable system, due to the gravitational influence of the planet." @Florian disagrees with this. However, the answer is more complex than the Hill sphere alone.

At first approximation, the Hill sphere gives a radius in which orbits around a moon could be stable. Our Moon's Hill radius is 64000 km.

For our own Moon, we know that most low orbits are unstable due to mascons: mass concentrations below the surface which make the Moon's gravitational field noticeably uneven. There are only four inclinations where an object orbiting the Moon avoids all mascons and would be stable: 27º, 50º, 76º, and 86º.

High orbits above the Moon aren't all safe either: above 1200 km and inclinations of more than 39.6º, Earth's gravity disrupts the satellite's orbit. Note that these orbits are comfortably within the Moon's Hill sphere.

There are stable orbits at high inclinations and high eccentricity:

As for other moons in the solar system: most of them are smaller and orbit around larger planets, so their Hill spheres are small, and the planet's gravity will disrupt much of the volume inside the Hill sphere too.

Moons below the limit where their gravity is strong enough to make them spherical, will have problems with uneven gravitational fields. Mascons may also be present.


Will gravitational waves too far away ever reach us?

The answer here is very similar to if you were asking about light.

In principle gravitational waves might allow us to fractions of a second after the big bang. Electromagnetic waves can see back to where the cosmic background radiation formed, about 400,000 years after the big bang.

You are right, the universe has expanded. At the present epoch it is estimated that the observable universe, containing objects that have emitted light or GWs that may reach us now, is around 46 billion light years.

However, it seems quite likely that the universe continues well beyond this horizon, and sources beyond this horizon can never have emitted light or GWs that will reach us.

As the relevant section of wikipedia (https://en.m.wikipedia.org/wiki/Observable_universe) points out, the detection of GWs extends our view very slightly. We cannot "see" with electromagnetic waves beyond 45.7 billion light years because of the "fog" of the cosmic microwave background,  but GWs can penetrate this fog allowing us (in principle) to see signals from objects currently 46.6 billion light years away.


What is left of a universe where no matter and energy exist and is there a term for this?

However, due to the Law of Conservation of Energy, I suspect that energy will be all that is left, only in equilibrium meaning no ability for energy to form matter.

The law of conservation of energy states that energy is conserved in a closed system. This means the Universe, which is likely an open system, does not necessarily have to obey that conservation law. Not to mention, scientific laws are not absolute, despite their names; they are entirely empirical, and only apply as far as we've observed. As Edwin Hubble put it, "Beyond the limit of the observations, the form of [a] function, or relation, is speculative. For this reason, [a] law is empirical and it must remain empirical until it is explained by an accepted theory."

Anyway, let's get on with the answer. You've just described the Black Hole Era in the heat death scenario. If Hawking radiation occurs, then the black hole will continue to emit particles. During the last stages of its evaporation, a black hole will emit not only massless particles, but also heavier particles, such as electrons, positrons, protons, and antiprotons.

Once all black holes have evaporated, the Universe will be nearly empty, sans some particles that will almost never interract. Eventually, even the protons and antiprotons emitted by the black holes will decay into photons and leptons. Thus begins the Dark Era, when the Universe will be dominated by dark matter, electrons and positrons.

Electrons and positrons will rarely come together and form positronium atoms (in which the electron and positron are electromagnetically bound). Positronium decays very within nanoseconds, since the two particles annihilate. Thus, over time, the energy levels in the Universe will become lower and lower.

What happens next is a matter of speculation. According to Carroll & Chen (2004), quantum tunnelling and quantum fluctuations could hypothetically create another Big Bang, approximately $10^{10^{10^{56}}}$ years from now.

Adams & Laughlin (1997) speculates that quantum effects will dominate once all macroscopic matter is gone. Essentially, the laws of "macro-physics" will break down, while quantum mechanical phenomena prevail. As a result, the smallest perturbations would make the biggest difference in this era, so there is no telling what may happen to space or time.

Furthermore, they speculate that if the Universe is a false vacuum, the vacuum may decay into a lower-energy state, creating a new "child universe":

  It is also possible for the universe to spontaneously create “child
  universes” through a quantum tunneling process roughly analogous to
  that considered above (e.g., Sato et al., 1982; Hawking, 1987; Blau,
  Guendelman, & Guth, 1987). In this situation, a bubble of false vacuum
  energy nucleates in an otherwise empty space-time. If this bubble is
  sufficiently large, it will grow exponentially and will eventually
  become causally disconnected from the original space-time. In this
  sense, the newly created bubble becomes a separate “child universe”.

So yes, something as small as quantum tunnelling will potentially have an enormous effect on the future of the Universe.

In the end, we really have no idea what will happen after enough time, so there's no definitive answer to your question in the long run.


The solar wind is just a stream of charged particles, mostly protons and electrons, some alpha particles and a tiny trace amount of other stuff.

If the solar wind approached an uncharged Neutron star, then the particles would simply fall onto the neutron star or circle around it in a hyperbolic orbit.  The Neutron star's gravity would add considerable velocity to the particles due to it's strong gravity.   Any particle that falls within about 2 radii of the neutron star would inevitably fall towards it due to the space-time properties of relativistic orbits, so there would be some spiraling in and probably some kind of light-show in the process.  The impacts of particles onto the surface would likely be in the x-ray or gamma ray spectrums and might not be that visible.   Ultimately it wouldn't be that interesting, but perhaps there would be some kind of light-show.

If we give the Neutron star a strong magnetic field, then the solar wind would simply get redirected, similar to how it gets redirected by Earth's magnetic field, except in this case, the magnetic field is much stronger and the central object, much smaller.  Probably very few of the solar wind particles would hit the Neutron star and most would just circle around it's magnetic field, which would produce (I think) less of a light show.   There wouldn't be much effect that I can think of.  The Magnetar star is very massive with a very strong magnetic field.  The solar wind is comparatively, especially after spreading out for 93 million miles, quite diffuse and weak.

If a person got hit directly by the solar wind, it might be deadly.   We're not very tough, but a magnetar is one of the most indestructible objects in the universe.   The effect would be negligible.

It might be more interesting if you moved the Magnetar much closer to the star.   Magnetic field lines can't cross each other, but they can push each other and flatten out, and in the case of solar flares they can get very twisted (but still never cross).

A magnetar is several orders of magnitude smaller than a sunspot.   The interaction between the two magnetic fields, if the magnetar passed close to a sunspot would be interesting to see. (though the gravitational effect would still probably be the more impressive of the two)  But if you move a Magnetar as distant as the Earth, the solar wind would have lost much of it's strength and it would just be ejected material from a solar magnetic storm.  It wouldn't have much effect on the magnetar.

That's my hobbyist understanding anyway.   If I missed any key points, please clarify your question or ask a new question.


Yes it does. If you are at a high latitude the crescent Moon will set looking like a reversed C orientation but in the tropics it will set more like a U orientation. Though the time of the year will also have some effect due to the angle of the ecliptic to the horizon.


How can I use Polar Alignment to find objects in the sky?

If a mount is polar aligned, it becomes very easy to find objects that have the same Right Ascension or Declination as a bright star. It works like this

1) you align the mount to the pole

2) using a star map, or an app, you try to find an object that has the same R.A. or declination as a bright star. As an example, use Messier 4, which has almost the same declination as Antares.

3) you point the telescope at Antares. Lock the declination axis.

4) Look through the finder, or through the scope using the eyepiece that gives you the biggest field of view. This most often this is the eyepiece that gives you the lowest magnification.

5) move the scope to the west, it rotates around the RA. axis. Messier 4 should now enter the field. Move slowly, and move back to Antares when you get lost.

6) When you see Messier 4, lock the RA. axis too, and center Messier 4 in the field using the fine adjustment controls.

7) There are a couple of other bright objects you can try. I'm not familiar with the Southern sky, but for the Northern sky, Messier 13 can be found starting at eta Her(cules), and moving in declination.

8) a bit harder is Messier 57 in Lyra, you need to start a bit to the north of gamma Lyr, or a bit to the south of beta. Another easy target is IC 4665, which is very close to Beta Ophiuchi (Cebalrai), almost but not quite to the north.

9) Other objects to try are Messier 5 starting with Beta Librae (Zubeneschamali), and Messier 10 starting at epsilon Ophiuch (Yed Posterior).

10) The next step is to try and find objects by using multiple steps, alternating between moving in RA and in declination. You would move to a star or better, to a recognizable star pattern, and continue from there.

11) This way of star hopping works best if the angles moved are not huge. Think 10, 15 degrees. As you get more experienced you will start using the stars in between as waypoints so you can move bigger distances.


One would not, under normal conditions, die within seconds. This is a common trope in science fiction, but there has been quite a bit of real research into the subject.

Let's speak on each point.

The extreme cold

This isn't really a huge problem. Space isn't necessarily cold. The extraordinarily low density of matter in space means that the movement of energy through induction is extremely limited. If one were to step into 'space' one wouldn't feel cold, or even cool.  For engineers, often the greater problem is not how to deal with cold, but how to deal with heat, as even though one can't easily release heat through induction, one can certainly gain it through radiation or internal chemical processes.

A person, however, is likely to experience cooling due to the process of liquids in the body evaporating and escaping, taking some of their heat with them. It's unlikely that this would prove fatal though, or at least it's unlikely to be the first thing to prove fatal.

The Intense Radiation

Radiation is a problem in space, but it's not going to kill you so quickly.

  During a 360-day round trip, an astronaut would receive a dose of about 662 millisieverts (mSv), according to RAD measurements. National space agencies limit exposure to about 1000 mSv or less during an astronaut's entire career; NASA's limit corresponds to a 3% risk of exposure-induced death from cancer. (Kerr, 1031)

Currently, orbital missions get some protection from radiation as they are within the Earth's magnetic field, but for those travelling beyond this there isn't a lot of protection. Part of the reason is that things that provide good protection are also very heavy, which can be quite difficult in the economies of reaching escape velocity. As such, one wouldn't necessarily be exposed to that much more radiation outside of a shuttle or their suit in space than they would otherwise.

The Vacuum

This is where things can get messy, but there has been some research on the subject. Obviously one can't breath in a vacuum, in fact the lungs actually begin to dump oxygen, essentially working in reverse due to the lack of air pressure. Our victim is likely to lose consciousness within 10-15 seconds. If they happen to be holding their breath when they become exposed, the pressure difference could cause some real damage to their lungs.

People often theorize that blood would boil in space. Geoffrey Landis disagrees with this on his blog, in which he speaks on this subject. I highly recommend reading through it as I would consider him both thorough and authoritative on the subject.

I couldn't find the official report, but there are frequent references to a 1965 (Earth based) pressurization incident, in which a technician's suit depressurized within a chamber. He remarked afterwords that the last thing he recalled before blacking out was feeling his saliva begin to boil on his tongue.

While your blood arguably won't boil, you would experience the boiling of other liquids in your body, water particularly.  This can result in rapid swelling of the body. While not pretty, this isn't likely to kill you. However, nitrogen can also become gaseous due to this, potentially resulting the bends, a condition more closely associated with pressure differences in diving. This condition could prove fatal, but AFAIK there hasn't been a lot of research into this in reference to vacuum exposure (This condition tends to become dangerous during the return to normal pressures).

All of this can be survived, as long as pressure is returned relatively quickly. Within a minute is probably enough to save a person's life with little lasting damage. After 2-3 minutes, it's likely too late. The difficulty is that the unfortunate victim may be able to survive 1-2 minutes of exposure, they really only get 10-15 seconds to do something about the problem. Of those who have survived vacuum pressurization incidents, their survival is mostly credited to the quick thinking of those around them.

To respond to your more specific questions:

Exposure for half of what would generally be fatal

This could vary a lot, like depending on whether they were holding their breath. However, generally it's believed that a person could recover from this with little to no permanent damage. Probably the biggest risks are lung damage due to pressure differences, the bends and other related complications, and brain damage due to oxygen deprivation.

Exposure to space that wouldn't result in imminent or eventual death

A little exposure without other complications will generally be survivable. We don't have a ton of data on the subject, so it's possible that unpredictable complications could arise, and frankly it's not exactly advisable, especially considering the person will lose consciousness quite quickly.

Partial exposure

I can't really rely on any specific source for this, but presumably exposure of an extremity would be perfectly survivable. I'm not quite sure how this would work as it would be very difficult to partially expose someone to a vacuum. The human body isn't exactly leak proof. In theory, you would begin to exhibit symptoms in the exposed body part, but you would remain conscious (assuming your respiratory system isn't exposed) and generally be in control of what's going on. Eventually other complications would likely set in and need to be managed carefully.

Exposed to space from inside a room with a single open door

As noted the radiation and temperature is not terribly dangerous, but with no way to repressurize the room this can quickly turn fatal. The Russian Soyuz II mission ended catastrophically when the capsule carrying the cosmonauts depressurized on way for reentry, killing all three crew members on board.  A valve malfunctioned during the reentry, causing depressurization. The craft was recovered after reentry and had not taken any significant damage. It was believed that Patsayev had made attempts to fix the valve in order to save the crew, but unfortunately there was simply too little time to accomplish this. Initial reports suggested that the crew had died of asphyxiation, however autopsies showed that the cause was hemorrhaging in the brain - reportedly caused by oxygen and nitrogen in the blood boiling. Not sure how this works out with Landis's claim that blood does not boil.

Radiation Will Make Astronauts' Trip to Mars Even Riskier
Richard A. Kerr
Science 31 May 2013: 340 (6136), 1031. [DOI:10.1126/science.340.6136.1031]


No, the sun won't ever become a black hole.
The choice between the three fates of stars (white dwarf, neutron star, black hole) is entirely determined by the star's mass.
A star on the main sequence (like most stars, including our sun) is constantly in a balance between the inward pressure of gravity and the outward pressure of the energy generated by the hydrogen fusion that makes it "burn".1  This balance stays relatively stable until the star runs out of whatever its current fuel is - at that point, it stops burning, which means there's no longer outward pressure, which means it starts collapsing.  Depending on how much mass there is, it might get hot enough as it collapses to start fusing helium together. (If it's really massive, it might continue on to burn carbon, neon, oxygen, silicon, and finally iron, which can't be usefully fused.)
Regardless of what its final fuel is, eventually the star will reach a point where the collapse from gravity is insufficient to start burning the next fuel in line.  This is when the star "dies".
White dwarfs
If the star's remains2 mass less than 1.44 solar masses (the Chandrasekhar limit3), eventually gravity will collapse the star to the point where each atom is pushed right up against the next.  They can't collapse further, because the electrons can't overlap.  While white dwarfs do shed light, they do so because they are extremely hot and slowly cooling off, not because they're generating new energy.  Theoretically, a white dwarf will eventually dim until it becomes a black dwarf, although the universe isn't old enough for this to have happened yet.
Neutron stars
If the collapsing star is above the Chandraskhar limit, gravity is so strong that it can overcome the "electrons can't overlap" restriction.  At that point, all the electrons in the star will be pushed into combining with protons to form neutrons.  Eventually, the entire star will composed primarily of neutrons pushed right up next to each other.  The neutrons can't be pushed into occupying the same space, so the star eventually settles into being a single ball of pure neutrons.
Black holes
Black holes are the step beyond neutron stars, although they're worth discussing in a bit more detail.  Everything, in theory, has a Schwarzschild radius.  That's the radius where a ball of that mass would be so dense that light can't escape.  For example, the Schwarzschild radius for Earth is about 9mm.  However, for all masses smaller than somewhere between 2-3 times the mass of the sun, it's impossible to squeeze the matter small enough to get it inside that radius.  Even a neutron star isn't massive enough.
But a star that becomes a black hole is.  We don't actually know what happens to a star once it's become a black hole - the edges of the "hole" itself is simply the Schwarzschild radius - the point light can't escape.  From outside, it doesn't matter whether the matter collapsed to the point that the neutrons started overlapping, whether it stopped just inside the radius, or whether it continued collapsing until it broke all known physical laws.  The edges are still the same, because they're just a cutoff based on the escape velocity.

1  I'm ignoring the red giant phase here, since it's just a delay in the "run out of fuel" step.  Basically, the core is helium "ash", while the hydrogen fusion process takes place further and further out.  Once that runs out, you get a nova and the collapse continues.
2 Likewise, I'm ignoring the mass that stars shed in their various nova phases.  All given masses are based on the remnants left behind.
3 Every source I've found for Chandrasekhar mass, except Wikipedia, gives 1.44 or 1.4 solar masses (which are compatible).  Wikipedia gives 1.39, and gives at least one source to back that number.


How can I id this celestial without data?

My proposal in case of real time simulation is Stellarium software or an excellent android app like Sky safari.


What are some examples of routine human activity times for certain cultures/peopes that are based on the position of the sun?

We are now dependent for timekeeping on clocks, most clocks are synced (directly or indirectly) to atomic clocks that keep time without direct reference to the sun.

However, the principal definition of clock time, UTC, is coordinated to the position of the sun, through the insertion of leap seconds every 18 months or so. So anything that happens at a particular time is coordinated to the position of the sun, though not directly.


What does the Sun's orbit within the Solar System look like?

In case anyone can't follow the links in my comment (above), here are the two pictures I mentioned. From: here and here.

The first claims to show the track of the solar system barycentre in the heliocentric reference frame. The outer yellow circle marks the photosphere of the Sun. The second plot claims to show the track of the centre of the Sun in the barycentric reference frame. The yellow circle shows the photosphere of the Sun to scale. As you can see, the plots are actually (almost) the same! Given that to go from one frame to the other is just a translation, I suppose they can both be right providing the x and y axes are defined appropriately.

To answer the questions posed: "What does it look like" - it looks like these two pictures. "How large is it?" As you can see, the maximum separation between the barycentre and the solar centre appears to be about 2 solar radii over the timescale covered by these plots, but is as small as a tenth of a solar radius (e.g. in 1950). "How elliptic?" Not at all really, it is a complicated superposition caused by the orbits mainly of Jupiter and Saturn, but all the planets contribute to a greater or lesser extent.

The barycenter is calculated from the instantaneous positions of all the discrete masses in the solar system. I do not know for sure, but I assume that it includes all of the planets, and that everything else is negligible at the scale of the thickness of the line.


It appears to be an older version of the Bushnell Voyager Sky Tour 800x70mm Refractor Telescope

Full manual here


Stellar mass limits for Neutron Star and Black Holes

A succinct summary of supernova types is given in the following image based on Heger et al. (2003):

Image courtesy of Wikipedia user Fulvio 314 under the Creative Commons Attribution-Share Alike 3.0 Unported license. The graph is based on the graph in Fig. 1 of the linked paper.

The pair instability realm is upwards of ~100 solar masses, though it is metallicity-dependent (Question 3). As Figure 1 (below) shows, neutron stars form in the mass range of >9 solar masses - again, this is metallicity-dependent (Question 1a). Starting at around 25 solar masses, black holes will form (Question 1b).

It is thought to be possible for a black hole to form without a supernova (see the section of the graph marked "direct black hole"). This has not been confirmed observationally, although there are some possibilities. I've written about this in more detail here.


dull rock/potato-shaped object seen with naked eye, moving in straight line during night. What could it be and is size and speed estimation possible?

Rock-shaped doesn't mean very much, rocks come in all shapes and sizes. It couldn't possibly have been an asteroid. It must have been quite large for you to see it at night, but it is surprising you could see the colour of it at night, when colour vision fades. Large, fast, elongated in the direction of motion - it sounds as though it could have been a dirigible, aided by a tail wind. Dirigibles are airships vaguely similar to a zeppelin. The ones with no rigid skeleton are called blimps. They have come back into fashion for certain tasks in the last 20 years or so; there are some in Europe and in USA and probably other countries too, Russia for example. Whether there are any in Greece, I couldn't say, but one thing I can say is that it is far more likely to have been an airship than an asteroid. One possible reason for its rapid disappearance is that it could have disappeared into or behind a cloud.


This shows that the file contains 3 Header-Data Units (HDUs; you can read more about this here). The headers contain meta-data. The primary HDU (0) has some 'cards' (basically keyword/value pairs most of the time, or comments) but no data. The next HDU (1, called 'EVENTS' in your file) is a binary table which contains 19 columns. Finally, the last HDU (2, also called GTI) contains another table with 2 columns. You can access the data in one of the columns of the EVENTS table using for example:

from astropy.io import fits
hdulist = fits.open('myfile.fits')
hdulist['EVENTS'].data['colname']

where colname is the name of one of the columns. You can find out the column names by looking at:

hdulist['EVENTS'].data.names


Why can't we observe the Oort cloud with a telescope?

The angular resolution of the telescope really has no direct bearing on our ability to detect Oort cloud objects beyond how that angular resolution affects the depth to which one can detect the light from faint objects. Any telescope can detect stars, even though their actual discs are way beyond the angular resolution of the telescope.

The detection of Oort cloud objects is simply a question of detecting the (unresolved) reflected light in exactly the same way that one detects a faint (unresolved) star. Confirmation of the Oort cloud nature of the object would then come by observing at intervals over a year or so and obtaining a very large ($>2$ arcseconds) parallax.

The question amounts to how deep do you need to go? We can do this in two ways (i) a back of the envelope calculation assuming the object reflects light from the Sun with some albedo. (ii) Scale the brightness of comets when they are distant from the Sun.

(i) The luminosity of the Sun is $L=3.83\times10^{26}\ W$. Let the distance to the Oort cloud be $D$ and the radius of the (assumed spherical) Oort object be $R$.
The light from the Sun incident on the object is $\pi R^2 L/4\pi D^2$.
If we now assume that a fraction $f$ of this is reflected uniformly into a $2\pi$ solid angle. This latter point is an approximation, the light will not be reflected isotropically, but it will represent some average over any viewing angle.

To a good approximation, as $D \gg 1$ au, we can assume that the distance from the Oort object to the Earth is also $D$. Hence the flux of light received at the Earth is
$$F_{E} = f \frac{\pi R^2 L}{4\pi D^2}\frac{1}{2\pi D^2} = f \frac{R^2 L}{8\pi D^4}$$

Putting some numbers in, let $R=10$ km and let $D= 10,000$ au. Cometary material has a very low albedo, but let's be generous and assume $f=0.1$.
$$ F_E = 3\times10^{-29}\left(\frac{f}{0.1}\right) \left(\frac{R}{10\ km}\right)^2 \left(\frac{D}{10^4 au}\right)^{-4}\ Wm^{-2}$$

To convert this to a magnitude, assume the reflected light has the same spectrum as sunlight. The Sun has an apparent visual magnitude of -26.74, corresponding to a flux at the Earth of $1.4\times10^{3}\ Wm^{-2}$. Converting the flux ratio to a magnitude difference, we find that the apparent magnitude of our fiducial Oort object is 52.4.

(ii) Halley's comet is similar (10 km radius, low albedo) to the fiducial Oort object considered above. Halley's comet was observed by the VLT in 2003 with a magnitude of 28.2 and at a distance of 28 au from the Sun. We can now just scale this magnitude, but it scales as distance to the power of four, because the light must be received and then we see it reflected.
Thus at 10,000 au, Halley would have a magnitude of $28.2 - 2.5 \log (28/10^{4})= 53.7$, in reasonable agreement with my other estimate. (Incidentally my crude formula in (i) above suggests a $f=0.1$, $R=10\ km$ comet at 28 au would have a magnitude of 26.9. Given that Halley probably has a smaller $f$ this is excellent consistency.)

The observation of Halley by the VLT represents the pinnacle of what is possible with today's telescopes. Even the Hubble deep ultra deep field only reached visual magnitudes of about 29. Thus a big Oort cloud object remains more than 20 magnitudes below this detection threshold!

The most feasible way of detecting Oort objects is when they occult background stars. The possibilities for this are discussed by Ofek & Naker 2010 in the context of the photometric precision provided by Kepler. The rate of occultations (which are of course single events and unrepeatable) was calculated to be between zero and 100 in the whole Kepler mission, dependent on the size and distance distribution of the Oort objects. As far as I am aware, nothing has come of this (yet).


Are there equal number of planets, stars, galaxies etc in observable universe spinning in both directions?

Angular momentum of the entire universe remains constant. Assuming that the big bang was the only  thing present initially, it wouldn't  posses any angular momentum. Now, if there are unequal number of planets, galaxies, stars,etc. in the observable universe spinning in both directions, there'd be some net angular  momentum which would violate the conservation of angular momentum. Hence, in my opinion, there are equal number of celestial bodies spinning in both directions.


Advice on a thread size and fitting for a Celestron binocular tripod mount

The standard thread for these is 1/4 - 20 UNC.  The 20 means 20 threads per inch - you could try measuring that. The same size fastener should be what's on the Opticron, so it should "just work".  It's possible that there is something in the hole in the binocular that's blocking the thread.

However (and if you are not the first owner),  someone might have tried to use the wrong fastener and damaged the thread.  Threads can be recut with the right tool.


How fast do the poles of Jupiter and Saturn precess?

You’ll find the answers you want in the Report of the IAU Working Group on Cartographic Coordinates and Rotational Elements 2015, published by Archinal et al in 2018. I found a copy at https://astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/WGCCRE2015reprint.pdf. A new version might be available this year, as the committee meets every three years… and take about three years to publish its report!
What you’re looking for are on pages 22 and 23 and are the mathematical formulas for the direction of the North poles of Jupiter and Saturn. These start by $ \alpha_0 $ and $ \delta_0 $ and are rather long for Jupiter, quite short for Saturn.
As for how long it would take their pole to fully precess by 360°, it’s exactly 1 million years for Saturn… as for Jupiter, because the formula is more complex, it’s more difficult to say, but I’d guess around 5.5 million years from the first term should be a good ballpark.
Clear skies!


There are three kinds of different hyperbolic in this question and the previous answer by AtmosphericPrisonEscape: the behavior of Lorentz transformations, the spatial hypersurfaces in cosmology (which seem to be flat), and the hyperbolic appearance of spacetime when tracing world lines in a universe with accelerated expansion.

TL;DR: Spacetime is expanding exponentially and if you choose an unconventional hypersurface, that hypersurface is hyperbolic.

Lorentz transformations: Lorentz covariance is the mathematical expression of Einstein's principle of the constancy of the speed of light. Changing the frame of reference of an observer is done by a Lorentz transformation which preserves the light cone and is closely related to the concept of causality.  The answer to this question explains this for the Minkowski space, where a Lorentz transform is a hyperbolic rotation.

In this flat (Minkowskian) spacetime:

$$ ds^2 = c\, dt^2 - dx^2 $$

you can use a Lorentz transform to switch to different coordinates $dt$ and $dx$ while the line element $ds$ is invariant, analogous to the hyperbola

$$ 1 = a^2 - b^2. $$

This kind of hyperbolic is not the point of the question.

Cosmology: If we assume large scale homogeneity and isotropy (Copernican principle), we can choose isochronous coordinates.  There you have a synchronous time $t$ and spatial hypersurfaces for given $t$ ("time slices").  Freely falling observers are comoving, i.e. have fixed spatial coordinates $\mathbf x$.  Their world lines are perpendicular to the spatial hypersurfaces and their proper time matches the synchronous time $\tau = t$.  Galaxy clusters are such freely falling observers.

If you take these assumptions and allow time dependence of spatial coordinates on time, you end up with the FLRW metric:

$$ ds^2 = c\, dt^2 - a(t)^2 d\Sigma^2$$

where $d\Sigma^2$ describes the spatial hypersurface.  Now additionally to this being a hyperbolic equation, the hypersurfaces $d\Sigma^2$ can have curvature, especially constant positive (closed sphere), zero (Euclidean), or constant negative (open hyperboloid) curvature.

A flat universe is, for example, given by

$$ d\Sigma^2 = dr^2 + r^2 d\Omega,\\
   \text{where}\quad d\Omega = d\theta^2 + \sin(\theta)^2\, d\phi^2, $$

which is the metric of Euclidean 3-space in spherical coordinates.  Here the only relative acceleration between freely falling observers comes from the scale factor $a(t)$.

A hyperbolic spacetime could be

$$ d\Sigma^2 = dr^2 + \sinh(r)^2 d\Omega $$

which is a hyperbolic plane in spherical coordinates.  In such a spacetime, freely falling observers do not only measure a relative acceleration due to the scale factor, but an additional contribution due to the curvature of space.

Expansion: In the FLRW metric you see the scale factor $a(t)$.  The Friedmann equations describe the evolution of the scale factor.  It does not influence the qualitative shape of the spatial hypersurfaces, but in an expanding space their curvature becomes more and more flat (just like the surface of a sphere locally becomes more and more flat as the sphere expands).

A flat $d\Sigma$ and a constant $a$ gives you no acceleration between freely falling observers whatsoever.  An $a$ with constant change manifests as all observers flying apart, but no relative acceleration.  Finally, an $a$ with constant acceleration makes everything fly apart ever faster (currently observed).  The Friedmann equations tell us, that the behaviour of $a$ changed during the lifetime of the universe as different effects where dominating (radiation, matter, cosmological constant).

Currently we observe an accelerating universe, so $a$ is exponentially increasing (just at the right end of the illustration).  If you look at the world line of two Galaxies, their physical distance increases exponentially.  Therefore the circumference of a chunk of space increases exponentially, so the hypersurface spanned by a line of freely falling observers is actually hyperbolic (white grid in the illustration).  However, I struggle to put that into an equation that clearly shows the hyperbolicity, so don't trust that too much, yet.


What are the minimum element requirements for a star?

Yes, a star can exist that's entirely hydrogen. Hydrogen is the fuel that makes stars happen unless they're very large. A star made entirely of hydrogen, so long as it was massive enough, would be very similar to the stars we see. The "metalicity" which refers to non hydrogen-helium elements has some effect on the star's rate of fusion, density and spectroscopy but it doesn't have a major effect.

When the universe was very young and almost entirely hydrogen and helium in terms of regular matter or elements, star formation was difficult because those light gasses were difficult to clump together.   The first stars were probably very large as a result, but once heavier elements began to be released in early supernovas, then star formation became easier because there were elements that could clump together and help smaller stars form.   So the other elements play a key role in star formation, but a much smaller role in the life of the star.

Very large stars fuse heavier elements, but as I understand it, fusion of heavier elements is more explosive. Hydrogen is wonderful in the way that it's very unlikely to fuse, so even large stars take millions of years to fuse their hydrogen because something like 99.9999% of the time, when two hydrogens get close enough to fuse, they just return to being two hydrogens, and only 1 in a hundred million billion billion times (1 in 10^26), credit to @PM2Ring.  It's only very rarely that the proton-proton collusion undergoes the fusion process and becomes a deuterium.

This very slow process doesn't happen with the heavier elements, so a star made entirely out of elements heavier than hydrogen and helium would begin it's life as a kind of white dwarf star, that is, the mass of a star but the size of a planet, and technically, it wouldn't be a star at all because it wouldn't be undergoing fusion and when the fusion process finally starts, it would be more like a type 1A supernova, happening very quickly and explosively.

The CNO process which creates some of the heavier elements in larger stars doesn't happen without hydrogen either, or, more accurately, free protons.

Basically, you can make stars out of pure hydrogen. Other elements would be made inside stars, but you can't make stars out of any other elements, unless you consider variations of a type-1a supernova a "star".  I personally don't.


If Saturn's rings cannot coalesce into a moon because of tidal forces, then how are shepherd moons able to exist?

There's two parts to this question, the first part is relatively easy.  The 2nd part, more tricky.

  How are shepherd moons such as Pan and Daphnis able to exist without
  being disintegrated by Saturn's tidal forces despite literally
  orbiting within the rings themselves?

I'm going to focus on Pan because it's larger and closer.  It's unusual, ravioli shape suggests, at least to me, an unconventional origin.  It's unusually symetrical for an object that size and probably formed while rotating, but that has nothing to do with your question.

Pan is small enough that it's able to withstand the tidal forces inside the Roche limit.   Phobos, around Mars is in a somewhat similar situation, though not as deep inside Mars Roche limit and outside the solid Roche limit, the principal is the same.  The integrity of the moon is strong enough to resist the tidal forces.  The relatively small size of the object is a big help in that regards.  Smaller objects experience smaller tides.  That's basically why oceans have visible tides and lakes have tiny tides, often not visible at all.

Pan doesn't look like a normal 30 km object in space

Source.

Source

This is what a 30 km object should look like:

An interesting property of being on a small moon inside the Roche limit is that you'd float off the surface on either the planet side or the far side of the moon.   Stand on the right spot on Pan and you'd float away.   Pan doesn't break apart because of the integrity of it's ice, but it, in effect, has negative gravity over parts of it's surface - the tidal bulge parts.  That's true with Phobos as well.

But the forces are small.  You'd drift off the moon very slowly, unless you were to find something wet and stick yourself to the moon - the wet would freeze, creating a grip that would hold onto you.

It's thought that Pan accrued it's ice-ring around the moon by holding onto ice particles it collected from Saturn's rings.   Ice is sticky and while the Moon does clear out a band in Saturn's rings, any wayward ice doesn't need much velocity to reach the moon and upon making contact, some of the ice sticks.    As a result, Pan kind of, almost magically, slowly accrues ice in a region where a larger moon would break apart.

The more tricky part of the question is how Pan got where it is, because it couldn't have formed inside the Roche limit.  The origin of Pan is a mystery.


What range of exit pupils work for observing the full moon?

Light pollution does not matter for the Moon. Even transparency doesn't matter that much. What does matter is seeing, a.k.a. air turbulence.

It is very rare that an exit pupil smaller than 0.5 mm is useful for anything - perhaps for some tight double stars, but that's about it. So take that as a hard lower limit.

In terms of a "soft" limit, it depends. If seeing is excellent, if the optics are very good, if the instrument is in perfect collimation and at perfect thermal equilibrium, then you can push e.p. below 1 mm, especially for a small aperture like the range you mention.

If seeing is less than perfect, if optics are so-so, if the instrument is miscollimated or too hot, then the minimum usable e.p. goes up. Each one of these factors influences the observation, so it's hard to pin a number.

Bottom line is, use what works for you.

  claims that 1-2mm works best for the moon. Is that correct?

2 mm seems too big for a small 4...6" aperture when observing the Moon. It might be okay if you want to capture the whole disk, but it seems large if you're interested in small details.

Again, it's all relative.

I have a 6" reflector with a self-made primary mirror of excellent quality. I always take care to collimate the scope, and to keep it at thermal equilibrium. I live in California, so seeing is often good. It's not uncommon for me to use less than 1 mm exit pupil when observing the Moon. I use 0.84 mm as a starting point, sometimes pushing it down as low as 0.6 mm.

If I use an e.p. of 1.1 mm for the Moon that means seeing is pretty bad - but in those cases I'd rather do something else, instead of being frustrated by bad seeing. But at that e.p. in a wide eyepiece (82 degrees apparent field) you see the whole Moon at once, which is always nice when you do a public demo (sidewalk astronomy).

For much larger apertures, the ideal exit pupils tend to be somewhat larger even when observing the Moon.

To predict seeing, go on the Clear Sky Chart site, find a location near your home, and look at the third row in the chart (labeled Seeing); when it's dark blue, seeing is good.

For the largest practical e.p., don't worry about it. The Moon is so bright, losing light is not something you should pay any attention to at all.


How to project an arbitrary elliptical orbit onto the night sky plane?

The (x, y, z) position of an object on its orbit with reference to a reference plane are given by:
u = ω + ν
x = r (cos Ω cos u − sin Ω sin u cos i)
y = r (sin Ω cos u + cos Ω sin u cos i)
z = r sin i sin u
(MEEUS, Jean. Astronomical Algorithms, Second Edition, p. 233)
This should help you find what you need.


How much dimmer is a partial solar eclipse that goes from -26.71 to -25.89?

There are two ways to think about this change: firstly in terms of the amount of light energy.  The rule for magnitudes is that "five magnitudes is a factor of 100" So to find the change in luminosity between two magnitudes you can calculate
$$\Delta L=100^{(M_1-M_2)/5}$$
With your values I get 0.47, or about half as bright at full eclipse.
The second way is to note that your eyes (and brain) actually do a good job of compensating for huge variation in brightness.  Unless you are looking towards the sun (with your safety glasses of course) you probably won't notice the difference in illumination. In the 1999 eclipse, it was only at 90%+ coverage that there was any apparent change in illumination. The burghers of Vlissingen may have been blissfully unaware.


Details on the telescope(s) on the Chinese Space Station 天和

The Chinese Space Station (CSS) has many instruments. It is planned to conduct a spectroscopic survey of hundreds of millions of galaxies. The Chinese Space Station Optical Survey (CSS-OS) is a planned full sky survey with applications to, e.g., cosmology and photometry. It will "probe weak and strong gravitational lensing, galaxy clustering, individual galaxies and galaxy clusters, active galactic nuclei..." This effort has been in the works for over a decade, e.g., here. There are also many laboratory experiments planned.
The CSS is the culmination of the Tiangong program, which began in 1992, and this is why it's known as the Tiangong Space Station. The word "tiangong" means "heavenly palace."
Xuntian is a planned telescope that will co-orbit the CSS for periodic docking. It is expected to have a much larger field of view than the Hubble Space Telescope, and may rival the James Webb Telescope.
Keep in mind that the efficacy of Chinese cooperation with international science programs is controversial.


How much iron does a star fuse before going supernova?

Wikipedia says the iron-nickel core is inert, and doesn't mention any endothermic iron fusion reactions, just the core collapse, which converts the core to neutrons (and a not insignificant proportion of protons & electrons), and releases a lot of neutrinos. The nickel is radioactive, decaying to cobalt & iron.

The silicon burning phase leading to nickel only lasts about 5 days. The inert core collapses when it exceeds the Chandrasekhar limit of about 1.2 M$\odot$.


How can I observe the Orionid Meteor Shower?

While this shower is now largely over, the way to best observe meteors is the same for all showers in principle (including coming showers like the Leonids around November 17 and Geminids around December 14).

The name of most showers reference the name of the constellation their radiant (point in the sky the meteors seem to be coming from) lies in. In the case of the Orionids, this is the constellation of Orion, so look in the general direction of (and around) Orion.

As you suggest, meteors appear unexpectedly and move too fast to be caught with a telescope.  The best way to observe them is to lie down comfortably and use the naked eye.  The key word is "comfortably": find a comfortable reclining garden chair or lilo and lie down.  That's one half of the comfort; the other is to stay warm.  Even when watching the Perseids in summer you're lying still under an open sky, and for the fall/winter showers mentioned here you'll really want to bundle up and perhaps take a thermos with a hot drink.

The comfort is important - there may be several minutes between two meteors and hence lying down is a good way to relax while watching, which will allow you to stay out longer and see more.  It is definitely more relaxed than standing up with your head in your neck.

One other tip for "comfort": many people will enjoy this more in the company of other enthousiasts, and if you're one of them you should consider going out with a few other people.  It will allow you to discuss the meteors just sighted to pass the time between events, or even to help each other to see one (if they're not too fast).


How could lithium burning take place in a quasi-star?

Lithium, along with Hydrogen and Helium, was one of the 3 elements created in the Big Bang. Thus, it should exist to some part in any star that hasn't burnt all of it out, and as mentioned, it's not an easy thing to do.

Population III stars are expected to contain Lithium, and Beryllium as well. The amount, however, is not particularly high.


Why did Mercury not appear to transit through the middle of the sun?

Mercury's orbit is inclined 7 degrees relative to Earth's orbit.

This drawing is exaggerated. When Mercury is "between" us and the Sun it might be below, above, or passing through the Sun's disc from our point of view.

Another post from Universe Today explains this also.


Cosmological redshift vs doppler redshift

After considering @benrg's comments, I realize that my first answer contained too strong statements about the relation between the two redshifts. I try here to moderate my answer, but you might want to accept their answer instead.

It is common to think of the two redshifts as having nothing to do with each other. Doppler shifts arise when the observer and/or the emitter moves through space, whereas the cosmological redshift can be derived considering stationary emitters and stationary observers in an expanding space.
Because the cosmological redshift doesn't involve movement through space, it is often considered completely different from the Doppler. However, it is also possible to derive the cosmological redshift by considering it as infinitely many infinitesimally small Doppler shift (e.g. Lewis 2016). I admit that I'm not well enough versed in general relativity to be certain about my statements, but just because an infinitesimally small patch of spacetime is flat doesn't necessarily mean that infinitely many such patches add up to be flat. However, as @benrg says, in GR there is only one redshift.
Different or the same?
The reason I think it makes sense to view the Doppler shift and the cosmological redshift as two separate mechanisms is the following:
In principle you could have a universe (non-capitalized, since it's not our universe, the Universe) that were static when a distant galaxy emitted a photon, then at some point expanded quickly by a factor of 2, and then again is static. In this hypothetical case, the observer would still measure the photon to have been redshifted by a factor of 2 (i.e. $z=1$).
That this is true can bee seen from considering the mathematical derivation of the cosmological redshift (see e.g. here) which involves an integral, the result of which only depends on the initial and the final state, not on the expansion history.
In contrast, if you and your friend stand still with respect to each other while your friend shines her flashlight at you, then run away from each other with a relative velocity of $0.6c$, then stand still before you receive the light (i.e. analogously to the hypothetical universe above), then you would measure no redshift; you wouldn't measure the special relativistic Doppler shift of $z+1 = \sqrt{\frac{1+v/c}{1-v/c}} = 2$ that you would if you were receding from each other while either emitting or observing.
In the real Universe, galaxies move through space (i.e. they change their comoving coordinate $\chi$), and space expands (i.e. the scale factor $a$ evolves). If the physical distance to a galaxy is
$$
d = a \chi,
$$
then the change in this distance gives their total velocity wrt. us, and is obtained through differentiation:

$$
\begin{array}{rcl}
v_\mathrm{tot} & =      & \dot{a}\chi + a\dot{\chi} \\
               & \equiv & v_\mathrm{rec} + v_\mathrm{pec},
\end{array}
$$
where dots denote differentiation with respect to time, and the two terms have been identified as the cosmological recession velocity, and the peculiar, "normal" velocity. Each of these terms give rise to a redshift, but through two very different mechanisms. Only the latter term is called a Doppler shift.


Where can I look up the 3D positions of the closest stars?

You can determine the 3D position for a star, $p$, by using the distance from Earth, $R$, as well as the right ascension, $\Omega$, and declination, $\delta$, within the following formula.

$p = R \begin{bmatrix} \cos \Omega \cos \delta \\ \sin \Omega \cos \delta \\ \sin \delta \\ \end{bmatrix}$

Note this will yield positions within the Earth-centered inertial reference frame.


How much mass do the volcanoes of Io erupt past escape velocity?

The velocity required to escape the gravitational attraction of a massive body is given by the following equation:

$$ v_{\mathrm{escape}} = \sqrt{\frac{2GM}{R}}$$

where $G$ is the gravitational constant ($G = 6.67 \times 10^{-11} \; \mathrm{Nm^{2} {kg}^{-2}}$), $M$ is the mass of the body from which you are escaping, and $R$ is its radius.

Inputting the values for Io's mass and mean radius, $M = 0.015 \, M_{\oplus}$ and $R = 0.286 \, R_{\oplus}$,1 we get an escape velocity of

$$ v_{\mathrm{escape}} = 2560 \; \mathrm{m/s}$$

However, the explosive ejecta are ejected from the top of volcanoes, so we should strictly add this to our radius. The tallest volcano on Io is approximately $2.5 \;\mathrm{km}$ above the surface; including this we get a marginally lower velocity

$$ v_{\mathrm{escape}} = 2559 \; \mathrm{m/s} $$

which is higher than the $\sim 1000 \; \mathrm{m/s}$ maximum velocity of ejecta calculated in McEwen & Soderblom (1983). Therefore, no mass is ejected from the surface of Io through volcanic eruptions.

For comparison, the earth's escape velocity is much higher, $11.2 \; \mathrm{km/s}$.  As discussed in the linked paper, the most extreme ejecta can reach heights of $500 \; \mathrm{km}$ before falling back to the surface.

1where $M_{\oplus} = 5.972 \times 10^{24} \; \mathrm{kg}$ and $R_{\oplus} = 6371 \;\mathrm{km}$


There are multiple solutions to general relativity which allow for multiple different types of black holes. The "normal" black hole you see most people talk about, with a zero-volume, point singularity, is known as the Schwarzschild black hole. If the black hole is spinning, the Schwarzschild solution no longer applies and you're talking about a different type of black hole. This new type is referred to as a Kerr black hole (other types include the Reissner-Nordström and the Kerr-Newman).

In a Kerr black hole, the singularity at the center is still zero-volume, but it is no longer a point. Instead it is a disk of zero height, often referred to as a ring singularity or "Ringularity". The angular momentum of the spinning black hole is then the angular momentum around the axis of rotation passing through this ring.


Refinement to sun/wristwatch estimate of N/S direction

The method is generally reliable between 9 am and 3 pm.
The other consideration is when one is using the method in either of the tropics: Cancer or Capricorn.
It is generally accepted that in the northern hemisphere, the sun is located in the southern part of the sky and the opposite in the southern hemisphere. This applies to latitudes greater than the latitudes of the Tropics. Between the equator and the Tropics, depending on the time of year, the position of the sun in the sky alternates between north and south.


Your question is quite complex, since dark matter and dark energy are two different things, that only have in common that they are called "dark" and are hypothetical quantities. They are hypothetical, because we only defined them as mathematical quantities to make physical laws, we invented, comply with the observations.

what is the origin of dark matter? In 1930 Fritz Zwicky observed the coma cluster. The coma cluster is a cluster of galaxies that move amongst each other based on their gravitational pull to one another. He found to his surprise that the galaxies in this cluster were moving faster than the gravitational pull based on their visible mass would predict. Since the visible mass of the galaxies couldn't explain the movement of the galaxies, he suggested there was another, yet unseen, matter in that cluster that contributed to the gravitational pull. He called it "Dunkle Materie" (he was Swiss) - Dark Matter. Later studies on the rotation of galaxies also showed that the rotational speed of the galaxies couldn't be explained by the visible matter in those galaxies, so another dark matter needed to be present to explain for the anomaly.

after adding the hypothetical dark matter in the laws of nature many things could be explained on the theoretical level. The forming of galaxies clusters in the current observable universe and the fluctuations of the cosmic microwave background could be explained with dark matter. The theory here brings the existing physical laws in accordance with the current observation. The problem is that dark matter has not yet been proven to exist in experiments. It has not been observed and many theories exist on what that dark matter could be.

Dark Energy has a different history, leading to a similar thing. When Einstein laid out his theory of general relativity he included in his field equations a cosmological constant, lambda, to be able to come to a static universe based on his equations. He assumed the constant to be zero and therefore come to a static universe. A negative constant would mean the universe would shrink and a positive constant would mean the universe would expand.

Unfortunately, Hubble showed that the universe was not static, but expanding and in the 1990's measurements showed this expansion was accelerating. Now the cosmological constant was not zero and an explanation needed to be found why the universe was expanding ever faster. For this expansion an energy is needed that counters the gravitational pull of the (dark) matter in the universe and increases with each expansion of this universe. Later observations from a supernova in 1998 and the Boomerang and Maxima experiments on the cosmic microwave background gave strong evidence of the existence of dark energy. This is, when we believe the standard model of cosmology is right.

The problem with dark energy is that in the most cases the laws of cosmology are correct and explain the universe. This means that dark energy must be real, or the cosmological laws must be altered. Currently, the search for dark energy is going on. The definition of dark energy in the models explains the expansion in of the universe and also the fast expansion during the early years of the universe. However, dark energy is a hypothetical quantity and has not been observed yet.

I hope this gave you a brief history how dark matter and dark energy have come into astrophysics. Don't mix it up with the dark force. That's not even hypothetical and only exists in certain movies.


If a black hole has the same mass as me, then why dont i suck everything at my center of mass?

Because you are somewhat larger than your Schwarzchild radius.
In order to turn into a black hole and start experiencing exciting things like Hawking radiation, you'd need to be compressed down into a ball about $10^{-25}$ meters in diameter, about one ten-billionth the size of a single proton.  At that size, you'd have a hard time sucking anything in: your gravity falls off quickly with distance, and even most subatomic particles will just bend their paths slightly as they pass you by.


What are these "nonstandard format" lines in Distant Artificial Satellites Observation circulars?

Those are data from "roving observers", essentially any ground-based observation that isn't from an observatory blessed with a three character observatory code. The observatory code is columns 78 to 80 of the entries. Note that the observatory code for those roving observer entries is always 247".

The longitude, latitude, and elevation of the known observatories are known. This isn't the case with those "roving observers", which is why they take two lines. The first line of a "roving observer" entry contains a "V" in column 15 and "247" as the observatory code. The second line of a "roving observer" entry contains a "v" in column 15. The bulk of the second line contains the longitude, latitude, and altitude of the observer.


What are the so called clouds of dust and gas made of?

The clouds of gas and dust that form stars are usually what are called Molecular Clouds and Giant Molecular Clouds (GMCs).

The "Molecular" means that most of the atoms are combined into molecules rather than being free atoms.  So the hydrogen is mostly H2 and not H.  This is really important, since a cloud comprised of molecules can radiate away the heat of collapse much easier than a cloud of atoms can. If that heat isn't radiated away, the cloud won't collapse.

As far as the contents go, clouds int he solar neighborhood today have a composition very similar to that of the outer layers of the Sun: around 90% (71%) hydrogen, 9% (27%) helium .08% (1%) oxygen, .04% (0.5%) carbon and lesser amounts of things like nitrogen, silicon, magnesium, neon, iron and sulfur.  Note that these are atom% followed by (mass%).  See this for a table.

Astronomers would call it 90% (71%) hydrogen, 9% (27%) helium and less than 1% (2%) "metals".

Much earlier in the universe's history the percentage of hydrogen was higher and the percentage of everything else was lower.   Since the Sun is a relatively new star, the gas it condensed from was "polluted" by the gas emitted by generations of older stars which enriched it with helium and "heavy metals".


Is there still a red shift when moving perpendicular to the direction of incidence?

Yes, there is a transverse relativistic Doppler shift. You can think of it as being caused by time dilation. https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

There can be a redshift or a blueshift depending on when, where and who does the measurement.

e.g. a receiver with a source going around it in a circular orbit. The receiver sees a lower frequency (redshift), by a factor $\gamma = (1 -v^2/c^2)^{-0.5}$. On the other hand, a receiver orbiting the source would receive a blueshifted signal by the same factor. When $v \ll c$ then
$$\gamma \simeq 1 + \frac{v^2}{2c^2}\ ,$$
so whilst the standard (longitudinal) Doppler shift is of order $v/c$, the transverse Dopper shift is of order $v^2/c^2$.

NB: This scenario is chosen so that the relative motion of the source and receiver are perpendicular to the line between them. All other scenarios are complicated by the usual Doppler shift you would see because there is a component of the velocity along the line joining the source and receiver.


Galactic extinction as a function of distance

If your sources are within 100pc then the best thing to do is assume the extinction is zero, unless you are looking for absolute magnitudes that are a lot more precise than say +/- 0.1 mag. I think you would be very unfortunate if the extinction was any more than about ~0.1 mag in the V-band (and less than that at R and I).

Beyond that you are really struggling because the structure of the ISM is very non-uniform, so you would have to go with the kinds of maps that are produced (for example) using IPHAS - but these are only in the galactic plane. But even if you could find some general map covering all positions it would be unlikely to give you an accurate answer out to 100pc, because there isn't enough extinction to give an effect that is big enough to be measured with any precision (i.e. the uncertainty might be bigger than the value).

If you have spectral types for your objects then you could always compare the V-I colour with what you expect for a star of that spectral type. This gives you the reddening $E(V-I)$ and then (roughly speaking) $A_V \simeq 2E(V-I)$.


A "Celestron Powerseeker 114AZ" Google turns up old reviews at Amazon.com and .ca. Most are somewhat critical of the mount.
In all honesty that is to be expected at this level. Many amateur astronomers are unhappy with the amount of wiggle in mounts worth many thousands of dollars. A little tweaking might help just don't use much more than finger pressure on the tools.
The telescope is supposed to be OK optically. Putting the optical tube on a dobsonian mount might be a good father and son astronomy project.
For what it is worth that 114 telescope is probably ten times better than my first telescope. I was as proud of that thing as any kid could be. As adults we tend to forget those sorts of memories.


In 2016, the summer solstice will coincide with a full moon. How often does this happen?

Florin Andrei what evidence can you supply to charge the oldest scientific body on this planet (The Royal Society at Greenwich) to be a profit-driven media entity.

Bishop, if you are asking when the 2 specific points in time match, it is true that they have a 'astronomical' infrequency to look at. But if you look at what the common terms for the definition of a solstice and full moon occurring at the same time, then it is not too infrequent. Matching a 24 hour period (solstice day) against a 72 hour period (a full moon) then they occur with a much more reasonable frequency. There are lots of lunar cycle calculators around. The current cycle we are in has the same middle day of the full moon  period In 2004, 2007, 2015, 2026, 2034, 2042 and 2045. further dates can be found using this http://aa.usno.navy.mil.
Just chose any date, calculate the drift on your 400 year cycle then calculate how often that drift occurs at the same time as the 3 day phase of the full moon drift. There will be large periods when the match occurs rarely (The center point of each phase occurring furthest apart). As the centre points close on each other the Frequency will increase. The recurring pattern of all astronomic movements.


How did Johannes Hevelius' long telescope work? Why all the round holes?

In general,  the less the support structure weighs, the easier it is to move the whole thing around.  Home-user telescopes usually have a solid body because it's by far the cheapest and simplest way to get a small rigid structure.

Large observatory 'scopes often have a minimal framework supporting the secondary (the far element in a reflector);  here the primary refractor is supported with a similar structure.

Most certainly the support blocks do not contain lenses.  For one thing, just imagine trying to align and focus a 12 or 15-element system!  For another, producing the lenses for telescopes of that size was incredibly tedious back then (and now, really).

There are a couple great chapters in "The Age Of Wonder" by Richard Holmes about a couple of the early astronomers & scope builders in Great Britain for those who are interested.


There are a couple of components to your question, so rather than directly answering, I need to discuss a few things in a more narrative form.
Mercury was first spacecraft-imaged in the 1970s, with only about 45% coverage.  From those images, some topography could be generated through stereogrammetry (two images taken with different viewing geometries that use parallax to reconstruct topography) and photoclinometry ("shape-from-chading").
The MESSENGER spacecraft orbited for several years, 2011–2015.  Its orbit was designed, in part, for global stereogrammetric coverage and it had a laser altimeter that could operate at high- and mid-northern latitudes.  As data were collected, early results were published.
The Wikipedia page you link to for Mercury's highest point, the Caloris Montes, was based on earlier results.  It goes at least as far back as 2015, I didn't check further.  Caloris Montes is named because it is an important feature on Mercury due to its association to the giant Caloris impact basin.  It's not named because it's a high point, specifically.  "Mons" is a singular mountain, while "montes" is used for a mountain range.
The article you link to is based on the full MESSENGER mission dataset, which was used to construct a new global mosaic and digital terrain model (available here).  From the new, better, global model, there is a higher point in the southern hemisphere than the Caloris Montes.  To me, it looks like a random point between some large craters, not anything especially important.  Therefore, it doesn't have a name.  Planetary nomenclature is used only when features are considered important and so need a name for easy referencing.  So, instead of saying, "That really big basin on Mercury," I can say "Caloris" and people will know what I'm talking about.  Similarly, as I noted above, the Caloris Montes are important as a mountain range, but a specific highest point among them is not.
As for a YouTube video saying the highest point is "Mt. Hermes," that sounds dubious to me with respect to anything official (for one, mountains on other bodies are "montes" after the Latin, if they have a formal name).  A search shows no one else is using this name.
Regarding "sea level," it's what Wikipedia said: You use a reference ellipsoid, usually.  The reference ellipsoid is made usually by fitting many different limb profiles to an ellipsoid, and/or via a laser altimeter.  The best fit is the ellipsoid.  Any deviations from that would be topography.  Mercury has plenty-enough data to make a reference ellipsoid, indeed you linked to one in the comments above.  Once these are published in the scientific literature, assuming they are generally accepted, the International Astronomical Union makes that a standard and publishes it in various reports.  They don't often update their standards, though I think the last one was in 2015.  So, it would not reflect the latest you found.
Practically all resolved solar system bodies have a reference ellipsoid that's used and published.  How good it is, though, is another question.  Asteroids, given their highly irregular shapes, for example, that were imaged by flyby missions would generally have a much more poorly constrained ellipsoid than, say, Earth's moon that has been observed through telescopes for over 400 years, orbited by numerous spacecraft, has retroreflectors on it, and has >8 billion laser altimeter points from the Lunar Reconnaissance Orbiter Laser Altimeter.


Why is Proxima Centauri called Proxima Centauri?

So, how can we be that sure that Proxima is indeed the nearest one? Have astronomers indeed examined the complete sky in order to prove this?

The minimum absolute magnitude for a star is, according to this well-known diagram about +16. Proxima Centuari is at the very low end of this scale with an absolute magnitude of +15.60. Still, it has an apparent magnitude of +11.13. That means that if a star is closer to us, it needs to be +11.5 or brighter. Every object of that magnitude or brighter has been categorized, and a hypothetical close star would've stood out due to its large parallax. So it's safe to say there are no closer stars ... at this moment.

  you must be sure that in future, no closer star will be found.

Stars move (they rotate around the galaxy center), so in the (far) future, there might be a star closer to use than Proxima Centauri, just like sometimes Venus is the planet closest to Earth, and sometimes Mars. So it's possible that Proxima might not be the nearest star in the far future. This is not something we have to worry in our lifetimes, though :)


What telescopes have observed anomalies in the light curve of Tabby's star, KIC 8462852?

I do not know whether complimentary observations of the light curve are being done, but I will try to answer the rest of the question.

Is it easy to observe the light curve with ground-based telescopes? No, it's quite difficult. From the Kepler homepage:

  Since transits only last a fraction of a day, Kepler must monitor all target stars continuously. Their brightnesses must be measured at least once every few hours.

It goes on to explain how the observed field was chosen - no downtime due to it being too close to the sun, no obstruction by asteroids, etc.

A ground-based telescope has the obvious problem that it can't observe during the day, so you need several telescopes in different parts of the world, and combine their measurements. Since the distortion patterns, viewing conditions, instrument characteristics etc. are all different, combining the data is anything but trivial if you want a precision measurement.

In short, a change in the point spread function between telescopes might introduce both systematic and statistical errors in your brightness measurement. Kepler is only one instrument and optically highly stable.

Is KIC 8462852 bright enough for a small ground-based telescope? Certainly. Its magnitude is 11.7, while Kepler wants to observe stars brighter than 14th magnitude (link above). This magnitude calculator estimates a limiting magnitude of 17 for a 1m mirror.

It takes time to set up an observation. Large telescopes like the VLT accept requests for observation time twice a year. After your observing proposal is accepted, it will still take weeks or months until your observation is actually made. For smaller telescopes, it might be easier and quicker, but it will take time to set up a monitoring programme, even if telescope time is available.

However, there are ground-based, long-term monitoring programmes. COSMOGRAIL is one that comes to mind. From their homepage:

  Obtaining regular observing time on telescopes in good sites was (and is still) not easy

and they've been going for a number of years by now.

So, it would be technically challenging but possible. Politically, it might not be so easy: Every observatory wants to make the best use of the observing time they have, and most are oversubscribed (more time asked for than available). You need a fairly solid science case to block several telescopes around the world for several years, but the result you'd get if all goes well is merely confirmation of someone else's result.


Is it possible to use the stars to determine the passage of time?

In the night sky in 10,000 years, two things will have changed in relation to the stars. The first, the rotational axis of the Earth will have changed, shifting the celestial sphere. The second, the stars themselves will have moved a bit relative to each other due to proper motion. So, the night sky will be quite different in 10,000 years, but still recognizable, particularly the constellations which will have changed somewhat but will still be identifiable. I would posit that as long as the person or persons were at least casual star gazers or astronomy enthusiasts (not even necessarily professionals) they could estimate how much time has passed in the course of their slumber, I'd say with a margin of error of about +/-2000 years. Should they be trained astronomers that margin of error should fall.


What is the speed of light relative to (in space)?

To any observer between 'A' and 'B' where the light wave passes, it passes with the speed of light. Adding a velocity $v$ below the speed of light $c$ to the speed of light results in the speed of light, simplified $c+v=c$, where the '$+$' means relativistic addition of velocities, not the '$+$' of arithmetics.

Depending on the relative velocity of two observers at the (almost) same point they will observe different Doppler shifts of the light, more precisely of spectral properties of the light. By analysing the spectrum of the light the observers may be able to determine the Doppler shift, meaning relative velocity, to the object which caused the emission or absorption line in the spectrum.