Are the viewable nebulae only in our galaxy?

1) The nebulae are spread out over the universe in many different galaxies. Nebulae called extra-galactic nebulae are nebulae outside of our galaxy. They aren't common, but we can see them.

2) Most of the ones we can see, such as the Helix nebula or the Orion nebula, are in our galaxy, but the ones outside of our galaxy are just more harder to see - there is a same dispersion of nebulae in our galaxy than in other galaxies.

3) Credit to @AtmosphericPrisonEscape (thanks!) - many nebulae outside of our galaxy are too dim for us to see. I assume that since the Tarantula Nebula is so massive and prominent, it is able to be visible from such a far distance.

For instance, NGC 2070, or the Tarantula nebula, located in the LMC. (Link: Here). The LMC is rather close to us, and the nebula is around 160,000 light years away.

I really can't find other major nebulae that are outside of our galaxy, but there are a lot of research papers on them, so they are visible - just not as prominent as the ones in our galaxy.

Some research papers with extra galactic nebulae include: Here, Here, Here, and there are many more online.

I don't understand the 2nd part of your question with the coordinates, but I hope this at least helps!


From a quick skim of the paper, it appears they using data from a previous paper by Smartt et al. (2009). That paper compiled a sample of observed supernovae (SNe) for which there are good images of the parent galaxies taken before the actual SNe. In some cases, Smartt et al. found plausible progenitor stars for the SNe; those are the "detections". Cases where they couldn't find a plausible progenitor on pre-SN images are the "non-detections".

(The authors of the paper you are talking about appear to be suggesting that Smartt et al. underestimated the masses of progenitor stars because they failed to account for extra local dust extinction from the more massive red supergiant stars.)


It doesn't work like that. An observer at the light source (and indeed any observers anywhere else) will always see light travelling (in vacuum) at the speed of light locally.

There is also a major problem with your thought experiment. It is not possible for you to have a stationary light source within the event horizon of a black hole. It, and everything else in its vicinity, must be moving inwards. This is as inexorable and unavoidable as is the passage of time for an observer outside the event horizon.

In my opinion, the best "visual" way of thinking about the situation inside the event horizon is to imagine your photons of light like salmon trying to swimming upstream, whilst you are on a boat flowing with the stream and releasing the salmon into the water. You will always see the salmon swimming at some speed with respect to your boat. Unfortunately if the stream flows fast enough then the salmon will make no progress and you will both be swept over a waterfall (the singularity) a little further downstream.

Likewise, your common-sense fails with the situation of firing light towards a black hole. Light is always measured to have a speed of $c$ locally. It is following through with the consequences of this principle that leads to all the weird behaviour that black holes exhibit.


What star system/Galaxy did LIGO observe the gravity wave?

The location of GW150914 was rather poorly constrained by the 6.9 ms time difference between the event when it was observed by the two separated LIGO detectors at Livingston and Hanford.

Further details are here
https://www.ligo.caltech.edu/image/ligo20160211b and this site provides a visualisation of where the waves came from.

The location OF GW150914 can only be constrained to be somewhere within 600 square degrees with 90% confidence. There are millions of possible host galaxies within this region.


Is the radius used in the formula for the escape velocity the average radius of the celestial object or the radius at the starting location?

Just to provide an official answer, the radius $r$ in that equation is formally the distance between the center of masses between the two objects. That does come with some assumptions of course, but for most standard physics problems that concept is perfectly fine. As you stated, the equation does not assume anything about air resistance, and to extend that, it does not assume anything about continued propulsion. If you're at some radius $r$ from the center of the Earth and you have some velocity $v>v_e$, then just coasting along in your trajectory will allow you to eventually escape the Earth's gravitational pull (ignoring the obvious case where you intersect the Earth with your trajectory).


This is not an answer, but it is too long to put in the comment.

There are formulas for cooling timescales of different radiative processes. Cannot recall right now about sources for reading.

One thing about nebulae. Some of them might have compact objects such as pulsars or black holes feeding energy from the centers. In those cases, you have to consider the energy input from the central sources and radiative cooling.

Another thing about X-ray emission processes. There are thermal X-ray emission (i.e., bremmstrahlung) and nonthermal ones (i.e., synchrotron and inverse Compton; IC). Each process dominates in different environment. Regarding to SN nebulae, nonthermal processes are the dominant ones, the electrons involving in the processes are relativistic and thermal process requires dense electron medium (for collision) to maintain the thermal equilibrium. IC normally dominates at early ages of nebulae due to the compactness of the system, while synchrotron self-absorption is also strong at these ages. At older ages, synchrotron dominates, while the IC can interact with the background photons (i.e., cosmic background radiation) and results in X/gamma-ray photons as well.

Last about IR emission. Since IR emission is mostly from thermal process while X-rays are from nonthermal processes, X-rays tend to last longer.


What is the maximum transmission distance of the radio signal in the outer space which could still be understood?

It cannot be said correctly, since we humans have hardly traveled to the moon and sent space probes to explore other planets in our solar system. So, theoretically anything might be possible. I'm trying to be a bit practical here. The only man made object that has gone really far is Voyager 1, which is at a distance of 18.7 billion kilometers (125.3 AU) from the sun. Although launched in 1977, it is the only live transmitter and receiver which is that far.

The radio communication system of Voyager 1 was designed to be used up to and beyond the limits of the Solar System. The communication system includes a 3*.7 meters (12 ft) diameter parabolic dish high-gain antenna* to send and receive radio waves via the three Deep Space Network stations on the Earth. Voyager 1 normally transmits data to Earth over Deep Space Network Channel 18, using a frequency of either 2296.481481 MHz or 8420.432097 MHz, while signals from Earth to Voyager are broadcast at 2114.676697 MHz.
As of 2013, signals from Voyager 1 take over 17 hours to reach Earth.

I agree that there are powerful transmitters in the world than what is present in the Voyager 1, but, most of them still remain untested. So, we can be exact with the measurements.


Properties of low-mass stellar remnants vs the Earth

Stellar remnants are completely different from planets.

The Earth was never a star and fusion has never occurred in the Earth's core at any time in its history. When a small to medium sized star dies, and the outer layers are lost, what remains is a white dwarf. It is still much more massive than the Earth, and is very hot. It is crushed by its own gravity, causing the matter to become "degenerate".

On Earth, atoms are held together by chemical bonds. Lots of interesting patterns can be formed by the atoms, producing minerals, rocks, seas and life. Nothing like this can happen in degenerate matter.

In degenerate matter, the atoms are pushed together by gravity. Degenerate matter is unlike regular matter. It is much much more dense, and chemical bonds are not a significant force between atoms.

White dwarfs are formed of carbon, oxygen, hydrogen and helium (with the lighter elements on the surface). Even after a white dwarf has cooled, it would still be degenerate, and quite unlike the Earth, in composition, properties and in density.


Please Guide me to buy my first Telescope

It's a broad question, but I'll take a stab trying to provide the essentials for a good start. Also see this post which contains important additional information:

Best telescope for the viewing of Nebulae, Stars and Planets

There are many, many factors involved in choosing a telescope. You seem settled on a dobsonian. That's not the only possible choice, but a dob is a good scope to start with.

You are correct that aperture (diameter of the primary lens or mirror) is the primary driver of telescope performance. Bigger aperture = better performance. However, performance is not the only thing you should worry about. Think of a car - do you only care about speed when buying a new car?

Any size scope will show you something. I've a 50 mm (2") achromat refractor that does double duty as finderscope. I can push it up to 100x magnification, and it will show lots of craters and mountains on the Moon, will show Jupiter as a disk with two equatorial belts, will show the rings of Saturn, will show the M13 globular cluster (barely) and the M31 Andromeda galaxy (again, barely). Not bad for a little scope.

A small 150 mm ... 250 mm (6" ... 10") dob will show you more. When seeing is good, you see several belts on Jupiter, and even individual curls on the big belts. Saturn's rings start to show the divisions, such as Cassini. These days Mars is at opposition - such a scope will show you the polar ice cap and the major land features. M13 fills the eyepiece with a dust of stars. The Orion Nebula looks big and clear even in the city. The Ring Nebula is visible, etc. Under a very dark sky, you can see (barely) the 3C 273 quasar, at 2.4 billion light years.

A mid-size dob (12" ... 18") will show you even more. More details on the planets, when seeing permits. More faint fuzzies (galaxies, nebulae, star clusters) especially out and far away from the city.

A large dob (20" and over) is an amazing instrument. On those very rare occasions when seeing cooperates, Jupiter at 1000x in such a scope is unforgettable. Also, under a very dark sky out in the desert, point it anywhere and the sky is full of faint fuzzies.

So, which size should you choose? It depends on price, maintenance, and ease of use (size, etc). Let's talk about maintenance.

As a dob owner, there are two things you must never forget: collimation, and cooling.

Collimation

It means "keeping all the optics aligned and centered". The mirrors will move around a bit, and in time will lose collimation. As a dob owner, you have to collimate the scope once in a while (and every time before you observe high-resolution targets such as planets). Once you learn how to do it, it takes about 2 minutes, it's super-quick and easy.

This is why collimation is important:

http://legault.perso.sfr.fr/collim.html

This is an introduction to dobsonian collimation:

http://www.cloudynights.com/documents/primer.pdf

Here are some simpler collimation techniques, which you can use until you acquire all the tools:

http://www.etnaastros.com/documents/No_Tools_Collimation.pdf

http://www.skyandtelescope.com/astronomy-resources/how-to-collimate-your-newtonian-reflector/

Also see the manual that comes with the scope.

Cooling

If you keep the scope in the house or garage, and then take it out for observation, the primary mirror will be much hotter than the night air. As a result, air convection will appear on the mirror and distort the image - just like the shimmering you see in the summer along the hot walls of a house under sunlight. The mirror is too "hot" for the ambient.

This is what happens:

http://www.garyseronik.com/?q=node/55

http://www.garyseronik.com/?q=node/69

To combat convection, take the scope outside at least 1 hour prior to observation, and let it cool off. Very small dobs (up to 6" or so) will probably work fine just with passive cooling. Small-medium and above (10" or above) basically require a fan on the primary mirror (forced cooling).

Always collimate and cool off the scope before you do an observation requiring high resolution (when observing planets, the Moon, double stars). If you don't need a lot of high resolution (like when observing galaxies, nebulae, star clusters) then collimation and cooling are not so important.

If you keep your dob collimated and cooled, it will provide great views of the planets. There is this myth out there that dobs "are not great for planets", and refractors are somehow magically "so much better for planets". This is not true. The myth exists only because of the large number of people out there who keep their dobs in terrible shape; those poor scopes underperform by a very large margin. Don't be one of those folks.

Let me just draw a line in the sand for you: An 8" f/6 dob is a good start overall. Why is that?

At 8", it is big enough that it will be seeing-limited most of the time. In other words, it is so big, atmospheric turbulence will limit the resolution more often than the size of the scope.

It is big enough to show you some faint fuzzies even from the city. All the Messier objects should be visible no matter where you are.

It is not too big to fit on the back seat of a sedan car, if you want to drive to a place with a darker sky. It is not too big to move around in the backyard easily.

The mirror is not so big and thick as to mandate active cooling (fan on the mirror); you can pretty much get by with passive cooling only - take it out at least 1 hour before observation. However, a fan may improve things a bit anyway (you need to experiment with that, but it's not a priority at this size).

At f/6 it's not too hard on the eyepieces, and it's not too hard to collimate. A shorter scope (f/5 and below) produces a steeper light cone; you need higher quality (and price) eyepieces to use it, and collimation requires much better precision.

Finally, the price is right - a few hundred dollars, which is affordable for quite a few people. Take care of it and it will last you a lifetime.

Now, if you can afford it and think you can deal with the steeper learning curve, could you buy a bigger instrument? Sure. 16", 18" - the wallet is the limit. Such big dobs are made differently - they are called "truss dobs". They don't have a big tube where everything is contained. Instead, the tube is replaced by a truss made of a bunch of poles - looks like the arm of a crane machine.

When you're done observing, you disassemble it. It all fits in a sedan car. It's not too heavy for transport, because you handle each part separately.

However, such an instrument is usually f/5 or less. Collimation is much more difficult. You need a coma corrector (like a Paracorr) to fix coma at the edge, and that's additional money. You need better eyepieces. Cooling doesn't work unless you use active cooling (fan). And it's just more expensive overall.

Can you start on a large dob like that? It's not impossible, but it requires more money and a much more significant commitment. This is not a toy that you can discard after two week-ends.


Why not use only one support vane for the secondary mirror, to avoid multiple diffraction spikes?

I can't speak for the JWST, but I do work at a company that designs telescopes for spacecraft and I can tell you that a lot of the structure of the telescope is to protect the instruments during the initial ascent to space where the launch vibrations and g-loading will be the worst. Plus mirrors for telescopes that big have really precise alignment tolerances and if the secondary mirror isn't very rigidly attached to the spacecraft even simple microstrains from heat loads or spacecraft pointing can move the mirror out of alignment. Even the fact that the telescope is going from a 1g environment to a 0g environment can move the mirror because it was aligned on Earth when the beam was slightly bent and then it goes back to being straight in microgravity and the mirror is misaligned! It stinks that there is stuff in the way of some beautiful science that you're trying to take a picture of, but it's the only way to make sure the telescope survives up there!


What is so special about the Ultima Thule passed by New Horizon spacecraft?

Ultima Thule is the first small Kuiper Belt Object (KBO) we have visited. For the first time, we can see a KBO as more than a tiny point of light in a telescope.
it seems to be flat, rather than roughly round. This is really unusual (we've never seen that shape before in planets, moons, asteroids etc.).


There are a couple different ways this could go, depending on where in the optical train (of other powered items such as mirrors and lenses) this particular mirror is.

In a simple case, such as when you view a "funhouse mirror," the mirror causes pure distortion, and not any defocus or aberrations.  This is easily corrected either with a mirror of opposite shape, or by remapping the image based on the distortion parameters.  Remember: distortion does NOT blur anything.

If the mirror is in an intermediat focus plane, then you essentially have created the heart of an adaptive optic system -- albeit your mirror is not "helping" .  In this case, the mirror's shape is causing the optical wavefront (phase) to get degraded. The phase delay, which is a function of optical path length, varies across the mirror.  In this case, you will need some sort of phase sensing device, such as a Shack-Hartmann sensor, to identify the errors and possibly drive correction systems.

I suspect your overall question is more closely related to the first of these cases, so a simple remapping will suffice.  The typical approach taken by us optickers is to image a large picture consisting of dots on a grid.  We expect the image dots to be on a rectangular grid, so when they aren't we know how much to "move" each pixel in the camera to get the correct image.


There isn't a lot of engineering here. Just use a grating dense enough to give you a comfortable angle. Less dense grating (fewer lines per mm) means a more shallow angle and narrow spectrum, and that can be harder to use. Super-dense grating might be expensive and fragile. Stay somewhere in the middle - you'll have to do a few trial runs on your own, see what works for you.

I made this spectrograph for my kids, works well enough:

http://sci-toys.com/scitoys/scitoys/light/spectrograph/spectrograph.html

I used the cheap grating that they provide on their own store, 1000 lines / mm, not bad:

http://www.scitoyscatalog.com/product/DIFFRACTION.html


The Milky-Way does not orbit the Andromeda galaxy, they both move under the influence of all the members of the local group. Even if one were orbiting the other the orbit need not be near circular but could be a very eccentric (elongated) ellipse.

The projected merger is because the tangential component of Andromeda's velocity with respect to the Milky-Way is small compared to its radial component That is Andromeda galaxy appears to be moving almost directly towards the Milky-Way (which is what it says in the Wikipedia page you link to, but it is not too difficult to find primary sources using Google, here is an arXiv paper reporting a proper motion study of Andromeda and reporting such).


Metallicity: What are the metallicity values of various galaxies?

metallicity is a complicated issue, and giving just one number (usually Z, but also 12+log(O/H), ...) is very rough. But, here are the standard values for the Magellanic clouds:

LMC: Z~1/2

SMC: Z~1/5

with Z=1 being the solar value. Practically, the values will vary across any given galaxy (the Milky Way also has a metallicity gradient).

Reference:

Madden et al. 2014, PUBLICATIONS OF THE ASTRONOMICAL SOCIETY OF THE PACIFIC, 125:600–635, 2013 June
http://adsabs.harvard.edu/abs/2013PASP..125..600M

The article is also available on astroPH


What's the difference between the Julian and the Julian Ephemeris Day?

Is it the only difference?

By definition, yes. It's important to keep in mind that the difference between TT and UT (UT1) is not constant.


Calculate object apparent diameter by distance and radius

If I understand correctly, you are trying to create a realistic looking view of the stars rendered (in the game) on a dome surrounding the game play area.  So you don't need to fly through the stars or interact with them in a 3D environment, is that right?
If so, it isn't the star size you want to reproduce - it is the brightness.  We can't see the width of a star from Earth and until very recently no telescope could, either.  It's the brightness of stars which makes them seem "bigger" or "smaller" than other stars.
The good news is this makes the calculations much easier for your game.  You want to represent the apparent magnitude as seen from Earth.
Planetarium programs, such as Stellarium or The Sky, achieve nice results with a combination of making the star dot circle radius larger/smaller as well as brighter/dimmer.  The planetarium programs which look the best also change the brightness level of the pixels across the radius of the star circle (the radial profile) which is called the point spread function in astronomy.
The easiest way to get a realistic star is with a simple bell curve shape (radial profile).  Make the width (in pixels) and height (brightness of pixels) of your bell curve variables and play around with them until you get the look you are happy with for your brightest star... those are the trickiest ones to render in a realistic way.  The faint stars are much easier.
When you have found how big the radius can be (and still look nice), assign it to Sirius' (the brightest star in our night sky) apparent magnitude  of ~ -1.5 and go down from there.  In a very dark location, our eyes can see stars down to a magnitude of 5 or 6.  Positive magnitudes are fainter; a magnitude 0 star is 100x brighter than a magnitude 5 star.
In a nutshell, you want to find the best bell curve width (diameter in pixels of your drawn star dot) and height (brightness of pixels at the center of your drawn star dot) to represent a star with an apparent magnitude of -1.5 (Sirius) and the best bell curve width and height to represent stars around magnitude 5 or 6.  You can, of course, include even dimmer stars but then the difference between the bright stars and faint stars becomes less obvious.  You'll have to find a balance you think looks nice.  The more stars you draw, the less of a difference there is between the brightest and faintest. If you increase the diameter in pixels of the bright ones to compensate, at some point they begin to look very fake.
Since you already have the YBC, you already have the stars positions and magnitudes (the YBC magnitude is the apparent magnitude and is listed as "V Mag").  Hope this helps, good luck with your game!
Edit: I was looking at the YBC earlier and had that stuck in my head, but the HYG should also have an entry for "V Mag".


During an eclipse, how big is the shadow of the moon on the earth?

But how big is this shadow? How many kilometers is its diameter?

That's a photo of the umbra and penumbra on the surface of the Earth taken from Space. It's a little distorted because it's not directly under the ISS but far off near the terminator.

It's hard to pin down the size of the penumbra because it's fuzzy and fades near the edges, but if you could see the very edges then it would be twice† the diameter of the moon or very roughly 6900 kilometers total.

†Actually this is only true by coincidence because the angular diameter of the Sun happens to be the same as that of the Moon. @Flaffo's answer does a good job of explaining how to calculate the diameter of the umbra, and that math could probably be extended to calculate the diameter of the penumbra as well.

The umbra has a well defined diameter but the size varies a lot due to variation in the distance from the Moon to the Earth since it's orbit is not circular. Sometimes the Moon is so far away that it can't fill the Sun and there is no umbra at all, that's called an annular eclipse.

Wikipedia says:

  Typically, the umbra is 100–160 km wide, while the penumbral diameter is in excess of 6400 km.

Source: Geometry of a Total Solar Eclipse

You can see an example of a very detailed simulation of just the umbra moving across the Earth's surface in the NASA Goddard video Tracing the 2017 Solar Eclipse The precise 3D shape of the Moon generates the shadow and it then moves over the contour of the Earth's topography. If you think the shape is weird I agree! See answers to The Moon's shadow could not possibly look like this — could it? (also see What are the “Moon L, B, C” angles shown in this solar eclipse simulation?)

Here's a screenshot:


Why can't we travel faster than light? And is the theory of relativity absolutely flawless?

No scientist really considers a theory flawless, but as theories put together by mankind go, it is one of the best ones put out there.  One of the things that makes for a great theory is not only being able to describe what's already happened, but to make predictions about new stuff, and general relativity did that with predictions of things like gravitational waves that took 80 years for technology to get good enough to see.

General relativity also passes an aesthetics test, where some people feel that the most fundamental laws are beautifully simple and elegant, and though as far as we know, there is no scientific reason that says the Universe is simple and elegant, most of the best theories turn out to be that way, including general relativity.

So if you acknowledge that relativity handles all the many many tests that have been thrown at it, the answer to your first question has to do with the Lorentz factor, which goes as $ \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $, so you see that as your velocity $v$ gets closer and closer to the speed of light $c$, the bottom of the fraction gets closer to zero and the Lorentz factor blows up.  Since the kinetic energy of something moving really fast involves the Lorentz factor, that means as you approach the speed of light, our energy blows up, and to actually travel at the speed of light, you'd need an infinite amount of energy.


Can a planet in our system eclipse the sun as seen from another one?

In our solar system, it is possible for one planet to partially eclipse the sun, but it is not possible for any planet to cause a full solar eclipse as seen from another planet.
The sun is too big and the planets are too small and too far apart.  Transits occur, and can occur for any pair of planets, but they are very rare.  As seen from earth, Mercury only transits a dozen or so times per century, and before 2004, the last time Venus had transited the sun was 1882. http://www.eclipsewise.com/transit/transit.html Orbital periods increase for the planets further away from the sun, so transits across the sun by the outer planets (while possible since no planets are in orbital resonance since Pluto got kicked out) are increasingly rare. Michael Seifert dug up these treasures, documenting transits of the outer planets: Jupiter, Saturn, Uranus.  As expected, the transit of Uranus from Neptune is the rarest, occurring next in October, 38172 CE.
Proof that no planets could fully eclipse the sun from another planet:
To rule out the possibility that any eclipses are possible, We only have to verify orbits pairwise in order of increasing distance from the sun. Since, if the $nth$ planet can't fully eclipse the sun from the $n+1_{th}$ planet than it certainly won't be able to do so from any other planet.  For example, if Jupiter can't eclipse the sun from Saturn, it certainly won't be able to from Neptune.
As seen from a planet, if the apparent size of the sun in the sky is smaller than the apparent size of the next inner planet, then a full eclipse is possible.  The largest an inner planet can appear from an outer planet is when the inner planet is at aphelion and the outer planet is at perihelion and the pass is directly overhead of the observer.
The formula for angular radius of a sphere is $\delta = \arcsin (d/2D)$ where d is the diameter of the sphere and D is the distance between the observer and the center of the sphere.
I ran these calculations using matlab and data from NASA. https://nssdc.gsfc.nasa.gov/planetary/factsheet/
For my input data, the rows are planets, and the columns are diameter (km), perihelion (e6km), and aphelion (e6km).
Here are my results.  Since the values in the second column are smaller than the first for each row, no inter-planetary solar eclipse is possible. As an example of how to read this table, an observer on Venus will see an angular radius of 6.48e-3 radians for the sun in the sky, but an angular radius of only 6.47e-5 radians for the next inner planet, which is Mercury.

Here is my matlab code for posterity:
sunDiameter = 1.3927e6;
planetVals = [4879,12104,12756,6792,142984,120536,51118,49528;...
46.0,107.5,147.1,206.6,740.5,1352.6,2741.3,4444.5;...
69.8,108.9,152.1,249.2,816.6,1514.5,3003.6,4545.7];
results = zeros(2,7);
for i=1:7
  innerPlanetDiam = planetVals(1,i);
  innerPlanetApogee = planetVals(3,i)*1e6;
  outerPlanetPerigee = planetVals(2,i+1)*1e6;
  outerPlanetDiam = planetVals(1,i+1);
  sunApparentSize =
    asin(sunDiameter/(2*(outerPlanetPerigee-outerPlanetDiam/2)));
  innerPlanetApparentSize =
    asin(innerPlanetDiam/(2*(outerPlanetPerigee-outerPlanetDiam/2-innerPlanetApogee)));
  results(1,i) = sunApparentSize;
  results(2,i) = innerPlanetApparentSize;
end
results'


That sounds very much like a 22° halo.

It's not an astrophysical phenomenon; it results from the refraction of light by ice crystals in Earth's atmosphere.

If this is what it is, it should be a fuzzy but regular circle centered on the Moon, with a radius of about 22 degrees. If you hold your fist out at arm's length with the thumb extended, the angular distance between the base of your fist and the tip of your thumb is typically close to 15 degrees; you can use that for visual comparison. Remember that 22° is the radius of the halo; the diameter is 44°, close to three time the width of your fist+thumb.


Disk-integrated vs disk-resolved, what is the difference?

The way disk-integrated and disk-resolved is used in your cited paper is nothing special, just indicating optical resolving power in relation to the image of the object of interest:
With a disk-integrated signal they mean a single value as measured over the whole disk (of an image a planet makes - or whatever is imaged as a disk).
When they talk about disk-resolved data, they mean that they have a large enough image of the planet they image so that they can start to discover regional inhomogeneity of the signal, thus make out different regions with different properties.
So the distinction is just that of a spatial resolution: either you can resolve the object of interest and see surface (or cloud) structures (disk-resolved). Or your signal is too faint, your optics too small, and you just get spatially unresolved data, thus disk-integrated over the image of the disk, the whole planet (thus you cannot distinguish anything based on location on the planet - you just measure overall albedo, spectral properties etc)


What kind of telescope do I need to see most of the Jupiter's moons?

Unless you're really rich, unfortunately you won't be able to see all of them.

Jupiter's fifth largest Moon, Amalthea, has an apparent magnitude of $m$ = 14.1. Comparing this to the magnitude of Europa, the dimmest of the Galilean moons, which is 5.3, tells us that Amalthea is roughly 3000 times less bright. Your telescope thus needs to have an area 3000 (or radius ~55) times larger for Amalthea to have the same apparent brightness.

In general, without a camera on your telescope, the dimmest object you can see depends on your vision, but on average, humans are able to see objects of magnitude 6. That means that in order to detect an object of magnitude $m_\mathrm{obj}$, you need a light-collecting area which is larger than you pupil by a factor of
$$f = 10^{(m_\mathrm{obj}-6)/2.5}.$$
Thus, in order to be able to just barely detect Amalthea, you need a telescope which is larger than you pupil (radius = 6 mm) by a factor of ~1738, i.e. has a radius of 25 cm.

It quickly becomes really difficult to see them; for instance, Elara, which is the eighth largest moon, has an apparent magnitude of 16.3 and thus requires a 70 cm telescope. And remember, this is just on the threshold of what you can see.

Of course, if you mount a camera to your telescope and take images with long exposure times, you can get away with smaller telescopes. In fact all moons smaller than Amalthea were discovered this way, some of them by cameras on the Voyager space probes.


Was the "green star" event in NGC 3314 ever figured out or named?

tl;dr No additional data of interest, but I explain how I searched, and I can explain the green color.
It seems that there have not been any additional observations of that object, which perhaps isn’t too surprising given how faint it is - 21st magnitude is possible only with big telescopes.
Simbad lists that object under the name of “EQ J103712-274051” but doesn’t list any other names, nor any references besides the IAU Circular already mentioned.  That’s a purely coordinate-based designation, with the “EQ” meaning it is based on equatorial coordinates.  Simbad says that that designation is internal to Simbad for objects of unknown nature and is to be avoided in the literature.
Looking at ADS, the only citation for the IAU Circular is a 2019 paper by Bonanos et al., which is a catalog of HST variable sources.  That seems promising on its face, but digging into the paper it turns out that the sources they flag are different ones, and the citation is in a footnote that says, “This object is unrelated to the transient in NGC 3314 reported in the IAU Circular 7388 by Keel & Frattare (2000).”
I also looked for further observations in Vizier- that can be a good way to find observations of a given source, even if it wasn’t called out specifically in a publication. While that search yields lots of objects within 5”, especially in HST observations, none of them seems to quite match up in coordinates and magnitudes with what Keel & Frattare report in the IAU Circular.
Regarding the green color that drew their attention to the object in the first place, here’s my interpretation of what they are saying there.  First, note that HST filters are (usually) named with “F” for filter (as opposed to “G” for grism), a number that gives the filter effective wavelength in nm, and then “N”, “M”, or “W” for “narrow”, “medium”, or “wide”, indicating the width of the filter bandpass.  So the F450W filter is similar to Johnson B, for example, and the F555W and F675W filters are green and red, respectively.
From 1999, when the object wasn’t bright, they had blue and red images.  From 2000, after it brightened, they had blue, green, and red images.  (The 1999 “red” data, F814W, are actually near-infrared, but I’m guessing they used those in the red channel for the Heritage image.)  So in combining to make an RGB (color) image, they had two each of red and blue images (one before and one after brightening), but only one green image (after brightening), which they would have needed to weight doubly in the composite to get a reasonable color balance. That’s why that object ended up looking green - they weighted the “bright” green data twice as much as the “bright” red and blue data.
EDIT:
I just realized that it would be possible to test the idea above
by downloading the individual filter images from the HST data and
looking at them directly.  Fortunately, the Hubble Legacy
Archive makes this pretty easy to
do, and even provides pre-combined color images for datasets where
multiple filters were taken.  So we can directly download images made
from data only before and only after the outburst of that object and
see what they look like.
This
search
gives all WFPC/2 images of NGC 3314, and we can see that there are
some from 1999 and some from 2000.  Those are the pre- and
post-outburst data, respectively, that were combined in the original
press release images shown above.   If you look in the column that says "Level", the Level 4 data products are color composite images.  The two options there are composites of the 1999 data and of the 2000 data.  Let's look.
Here are the 1999 (pre-outburst) images, which are only in red and
blue filters.  Presumably the color composite fills the green channel
with some combination of the red and blue:

And here are the 2000 (post-outburst) data, combining images in red,
green, and blue filters:

Below I've zoomed in and taken a screenshot, trying to get the images
centered as close to the same as possible.  I've positioned them so
the object in question is just to the left of the red crosshairs.  Top is 1999, bottom is 2000.
1999:

2000:

You can see that the newly-brightened object is basically white, not
green.  The individual, isolated red, green, or blue pixels are noise.  In any image, you will have some pixels that are extra-bright, either from noise in the detector, or from a cosmic ray hit that was not removed entirely.  Since they are random noise, they aren't in the same place in every image.  So if you make a color composite image, then those randomly-bright pixels will show up as the color of whatever filter they appeared in, since they are bright only in that image and dark in the images taken in the other filters.


Relationship between temperature of nebula and size of star

This is such a complex issue that I'm not going to try to be comprehensive. One major thing you are missing is that stars tend not to form in isolation, especially massive stars, but in groups and clusters.

The basic unit in star formation is the Jeans mass.
$$ M_J \propto T^{3/2} \rho^{-1/2}$$
where $T$ is the temperature, $\rho$ is the density and $M_J$ is the minimum mass for a cloud to gravitationally collapse.

So from this point of view, a higher temperature cloud must be more massive in order to collapse in the first place.

But as the cloud collapses, the density goes up and the Jeans mass decreases - and the cloud fragments into smaller sub-units and ultimately into a cluster of stars.

Exactly how this pans out in practice is basically the whole research field of star formation. It depends not only on relatively simple things like the temperature of the gas, its density, the equation of state; but also on much more complicated ideas like magnetic field support, competitive accretion between stars, feedback from newly-formed stars and protostellar outflows and the rate at which turbulence is generated and then dissipated.

There almost certainly is no simple relationship as you propose. In fact the outcome of star formation seems to produce a mass function (number of star per unit mass) that is only weakly dependent on the initial conditions, but may differ a bit in low density and high density environments.


I can personally attest that the Starlink satellites are visible with the naked eye even after they have spread out, however they are quite dim - with apparent magnitudes near 5.5. Here is a brightness comparison drawn from a widefield capture where two Starlink satellites are can be seen (albeit barely) juxtaposed with another satellite with apparent magnitude near 3.

The region in which this was taken is Class 1 on the Bortle Scale and the entire 60-satellite train was visible, though only 4-5 were visible at any given time. In brighter regions the satellites would not be as readily visible. Furthermore, since they have such low orbits, they were only visible for ~105 degrees in the sky before they passed into Earth's shadow.


Creating a basic "fixed" solar system to host a 3D space travel simulation

The easiest approach is to do a numerical integration of the equation of motion of your system. For that you need to remember that the gravitational force of $a$ on $b$ is $$\vec{F}_{a\rightarrow b} = -G\frac{M_a M_b}{r_{a\rightarrow b}^{3}} \vec{{r_{a\rightarrow b}}}$$
where $\vec{{r_{a\rightarrow b}}} = \vec{x_b} - \vec{x_a}$ and that
$$ M\vec{a} = \sum \vec{F}$$
where $M$ is the mass of your object, $\sum\vec{F}$ is the sum of the forces applied to the object and $\vec{a}$ is the acceleration.

To do the simulation, you then have to follow this numerical scheme

set some initial conditions to your problem: initial position $\vec{x_p}$ of the planet and $\vec{x_\star}$ of the star and initial velocities $\vec{v_p}$ and $\vec{v_\star}$.
compute the force $\vec{F}$ of the star on the planet and the accelerations of the planet and the star:
$$ -M_\star \vec{a_\star} = M_p \vec{a_p} = \vec{F_{\star\rightarrow p}}$$
update the speed of the planet and the star using the respective acceleration:
$$ \vec{v}(t+\Delta t) = \vec{v}(t) + \Delta t\times \vec{a} $$
update the positions of the planet and the star using their respective velocity:
$$ \vec{x}(t+\Delta t) = \vec{x}(t) + \Delta t\times \vec{v} $$
recompute the distance between the planet and the star $\vec{r_{{\star\rightarrow p}}} = \vec{x_p} - \vec{x_\star}$ and go back to 2.


Are there any known neutron star - supergiant binaries?

One possible formation pathway for TŻOs involves a massive star undergoing mass transfer to a neutron star companion. A common envelope forms, and subsequent friction leads to a merger. High-mass x-ray binaries (HMXBs) are therefore good candidates for progenitor systems. Hutilukejiang et al. 2018 simulated TŻO formation through a number of channels, and found that 3 of 114 known HMXBs in the Milky Way involve a massive supergiant companion capable of filling its Roche lobe and engulfing the neutron star, potentially leading to a TŻO. The three systems listed are  SMC X-1, Cen X-3 and LMC X-4.
Progenitors of TŻOs through other channels may be harder (if not impossible) to find. Random collisions between neutron stars and supergiants in globular clusters are effectively unpredictable on an individual basis. Pulsar kicks from a supernova in a supergiant-supergiant system can also lead to TŻOs but are also difficult - again, if not impossible - to identify, given that kicks are not well understood. I would therefore assume that common envelope evolution is the best bet for identifying progenitors.


The "diffuse ionized gas" (DIG) is another term for the phase of the interstellar medium (ISM) usually called the warm ionized medium (WIM). With a temperature of the order $10^4\,\mathrm{K}$, but extenting to lower and higher temperatures, it is hot enough to keep hydrogen ionized, and various metals exist as low-ionization species, such as S II, N II, and O II, and even (weak) O III (e.g. Hill et al. 2012; Zang et al. 2017; Weilbacher et al. 2018).

Rough pressure equilibrium with the other phases of the ISM results in characteristic densities of $0.1\,\mathrm{cm}^{-1}$, but typical ranges are an order of magnitude to both sides.

A review of the DIG/WNM can be found in Haffner et al. (2009).


Several pulsars have double-peaked profiles: you get two peaks per rotation, with one peak stronger than the other. An example of this is the Crab Pulsar. This suggests that at least some pulsars have two beams. Remember that the gravitational bending of light in the vicinity of the neutron star makes it possible to see a larger fraction of the surface, which increases the chance of both poles being visible.

The single-peaked light curves likely represent the case where the second pole is not visible. Of course, if you wanted to play Devil's advocate, you could argue that there might be single-beam pulsars lurking in that sample... you'd probably need to come up with some kind of physical mechanism by which such an object could exist, given the physical theories that have so far been used to explain pulsars predict two beams.


Is it true that we can estimate the mass of a star more accurately if it has a companion star?

In general, masses can only be precisely estimated for stars in binary systems (asteroseismology and some other indirect techniques can yield estimates too).

There are two broad types of binary where this is possible - astrometric binaries, where the motion of both stars can be seen on the sky and where the distance to the binary is known (e.g. Sirius AB); or eclipsing binaries, where the stars eclipse each other and their motions are inferred from their line of sight velocities, measured by spectroscopy (e.g. YY Gem).

You are correct, that the relative motions of the stars only yields their mass ratio. What is needed to get the individual masses is something that fixes the absolute sizes  of the orbits and the orbital period, which can then be used with Kepler's third law to give the total mass of both stars and hence combined with the mass ratio to give the masses of the individual components.

In astrometric binaries you can measure the angular size of the orbits and calculate an absolute size using the known distance. In close (unresolved) binaries, that size is inferred from the velocities and orbital periods, but also requires knowledge of the inclination of the orbital plane. This is known quite accurately (is close to 90 degrees) only for eclipsing binaries.

I could present the Maths if you really like, but these are quite standard results.


Is liquid water on Mars such a big leap compared with ice?

We have not found life, bacterial or otherwise, anywhere other than Earth ... yet. I think you mean the difference between liquid water and water ice. This difference relates to the hospitality to life. Life of the type we are used to needs liquid water, even if this is just trapped moisture, unless the life is temporarily in suspended animation. If there was no liquid water on Mars, there would seem to be no possibility of ongoing life there.

It is important to distinguish hospitable conditions from evidence for life. While having a hospitable environment is necessary for life, it is far from proof of life. The big hurdle is starting life, or bringing it from somewhere else.

There are a variety of hospitable environments in the solar system and on exoplanets, such that if certain life from Earth (such as extremophile bacteria) were somehow placed there they could live. But that does not mean there is in fact life there.

Another important point, Mars was way more hospitable early in it's history, with bodies of liquid water. Which is important because even though starting life on Mars under current conditions is unimaginable, early Mars was rather Earthlike and could possibly start, or receive, life.


No, you can't use them to identify your star. Indeed, you have too many stars in the universe. Even in our galaxy you can find two stars which have pretty much the same spectrum. You can classify the star though.

There is another problem with redshift and blueshift. It change the spectrum of the star depending its movement.


The main method by which heat is transferred in a white dwarf is thermal conduction. Degenerate electrons have very long mean free paths and can conduct heat excellently.

White dwarfs lose energy by emitting photons from their outer, non-degenerate surface layers. They may also emit neutrinos from their interiors when they are really young and hot.

Gamma rays and X-rays are of course emitted by nuclei. This works just the same as atomic emission - charged particles (protons) change energy levels within the nucleus.

Most white dwarfs are not hot, they are similar to the Sun or cooler. Hot ones are easier to see, but only young white dwarfs are hot. But yes, your scheme is approximately (in that white dwarfs are not blackbodies) correct.


Why do linear velocity redshifts correspond to linear pixel shifts when the spectra are binned in constant log wavelength?

All this means is that you need to bin your spectra in equal intervals of log wavelength for each pixel to be a constant interval in velocity.

First consider the case whereeachpixel is worth a constant interval in linear wavelength. Here we have
$$ \frac{\Delta \lambda}{\lambda} = \frac{\Delta v}{c},$$
and the $\Delta v$ represented by each pixel depends on $\lambda$, which changes across the spectrum.

But now if you bin in equal increments of
$$\Delta \log \lambda = \frac{\Delta \lambda}{\lambda} = \frac{\Delta v}{c}$$
and each pixel has the same fixed velocity interval, $\Delta v = c \Delta \log\lambda$, independent of $\lambda$.

This log wavelength binning is a prerequisite for cross-correlation procedures that yield velocities.


What safety concerns should an amateur astronomer take?

The only thing you need to protect your eyes from is the Sun. A lunar eclipse, no matter what phase it is in, is not dangerous to look at. A partial or an annular solar eclipse are dangerous because the Sun is still visible. A total solar eclipse is perfectly safe to look at—however, as the phenomenon starts and ends with a partial eclipse (as the Moon slowly blocks the view, then slowly moves away), those parts of the (otherwise) total solar eclipse are dangerous to look at directly.
DO NOT LOOK DIRECTLY at the Sun with the naked eye, even though some people have done it for VERY SHORT times (seconds, at most) without permanent damage. Again, DON’T DO THIS! The problem is the eye doesn’t have pain receptors, so you wouldn’t know when you had been looking at the Sun for long enough to be hurting your eyes. This can range from a small “blind spot” in your field of view, that goes away after a few days or weeks, to complete, permanent blindness.
There are safe ways to view the Sun through an optical instrument (which, because it collects more light than the eye, would be even MORE dangerous to look through without protection!), and some of them are really affordable financially. The best, however, is a solar filter that goes in front of the objective, NOT at the eyepiece (as the temperature there is so high, it could melt and make you permanently blind INSTANTLY)! Such filters come in many different qualities and many different prices, but are generally rather affordable as well. They are made by applying a special metallic coating to an optical-grade piece of glass or even plastic (Mylar).
DO NOT LOOK AT THE SUN THROUGH AN EXPOSED PHOTOGRAPHIC EMULSION (yes, those still exist!) and DO NOT LOOK AT THE SUN THROUGH A POLARIZED FILTER (unless you’re also using a safe filter as described above).


Olber's paradox describes an infinite, static, eternal, and homogenous universe. If there were a cloud of opaque gas situated in some region of the sky, light coming from all conceivable directions via the paradox would bombard the cloud until either it begins emitting light itself via the radiation it absorbs or its particles scatter to allow light to pass through.


Current best-known Drake Equation values?

Have the many exoplanets we've discovered in the last several decades improved our estimates of , and ?

Absolutely! The quantity $f_p n_e$ is a quantity of much interest in investigations of extrasolar planets, which effectively equates to the fraction of stars that have a habitable planet in their star's habitable zone. Plenty of studies have examined this specifically for Earth-sized extrasolar planets and place the value of $f_p n_e$ between ~$\frac{1}{6}$ (i.e. one in six stars have an earth-sized planet in a relatively close orbit to the host star) and ~$\frac{2}{5}$ [1][2]. A 2013 comprehensive study by Petigura is a great paper to get familiar with the specifics and assumptions of how the Kepler surveys are conducted and the quantities of interest calculated [3].

What is our current best estimate for ⋆ for the Milky Way?

Unfortunately the answer is: it varies. Roughly, we think that the Milky Way puts out about 3 solar masses worth of new star every year [4]. However, these don't have to be in one star. Since we think most stars are probably ~1 M$_\odot$, this works out to ~3 stars/year. However, some estimates go less, and some estimates (based on chemical decay of elements like aluminum) go as high as 7 stars/year [5].


How can we detect water on Mars-like exoplanets?

It would be difficult to detect the water-by-weight in the dust of another planet to the precision that has been done for the Martian soil.

However, having said that, there are a couple of ways to detect the presence of water, and that would be, according to "60 Billion Alien Planets Could Support Life, Study Suggests" (Gannon, 2013), to detect the presence of clouds.

To confirm that the clouds are water clouds, not something the iron-vapour etc, according to web article "Hunting for Water on ExoPlanets" (bolding mine),

  Using ESO’s Very Large Telescope (VLT), a team of astronomers have been able to detect the telltale spectral fingerprint of water molecules in the atmosphere of a planet in orbit around another star. The discovery endorses a new technique that will let astronomers efficiently search for water on hundreds of worlds without the need for space-based telescopes.


I think the most common explanation for this kind of structure is ram pressure from the galaxy's motion through the surrounding intracluster medium (ICM). In the rest frame of the galaxy and jet, the ICM is blowing past with a velocity of several hundred km/sec, and this is enough to divert the flow of the jet. In this picture, the jet isn't moving back toward the galaxy; this is a projection effect (the inner jet is thought to be pointed toward us, only 30 degrees or so away from our line-of-sight to the galaxy).

A second factor may be that the point where the jet seems to bend is where its smooth, relativistic flow breaks up due to the Kelvin-Helmholtz instability. The initial small kink visible at the right-hand edge of the HST-WFPC2 image (upper right) is probably a sign of the instability starting to disrupt the jet. Once the smooth flow of the jet breaks up, it is likely to slow down, making it more easily diverted by the ram pressure of the ICM.


What is the force of a CME on objects in space?

This isn't a complete answer, but it's a partial one:   http://pwg.gsfc.nasa.gov/istp/outreach/sunearthmiscons.html

  Despite the electromagnetic havoc they play with satellite electronics
  and with Earth's magnetic field, the solar wind and CMEs barely exert
  a pressure or force that you could measure. In fact, they couldn't
  ruffle the hair on your head. The solar wind has fewer particles per
  cubic centimeter than the best vacuums scientists have ever created on
  Earth. Our own air is billions of times denser than the solar wind,
  such that a cubic centimeter of air has as many particles as a cube of
  solar wind measuring 10 kilometers on each side.

So, for your question on changing orbits - not very much.   Granted everything moves a little under any pressure.

Now, the lighter the object is to it's surface area, the more a CME can move it, that's why a solar sail, very light weight and large area, might work.    CMEs can also move gases, like the tail of a comet can be pushed visibly away from the sun (I read that, but misplaced the article), or lighter, high atmospheric gas can be blown away from smaller planets.

Also, no 2 CMEs are exactly alike.   A smaller and/or a younger star will emit much stronger CMEs, so around a red dwarf for example, the effect might be more noticeable for tiny asteroids and the like.


Picking two random points in space and drawing a line between them, what is the chance the line intersects anything?

Indeed this is correct.  Photons can travel distances greater than the radius of the observable universe without interacting with matter.  This has been true ever since the period know as "recombination" when electrons became bound to hydrogen and helium nuclei.  Prior to that time there were lots of free electrons that could interact with light.  After that time the electrons became bound to protons, and the electrically neutral gas was effectively transparent.  We can still see the photons emitted at that time, although now they have been cooled by the expansion of the universe and appear in the microwave part of the radio spectrum (This is called the cosmic microwave background)
Now the matter in the universe is concentrated into galaxies within galaxies it is consentrated into stars,  so the mean free path of photons in the universe is still much larger than the scale of the universe.
The region that is very distant is also very far back in time, and before recombination we can't "see" because the universe was "foggy" because of the free electrons.  Most of what we know about the universe before this time is inferred from observations of how matter behaves at high energies,


Let's apply @uhoh's answer's equation to a few bodies.  Numbers come from the obvious Wikipedia pages unless otherwise stated.  G = 6.6741e-11, in kg-m-s.

Verona Rupes on Miranda.
f = 20 km / 236 km, ρ = 1200 kg/m^3.  So t = 711 s.

Enceladus has rifts 1 km deep.
f = 1 km / 252 km, ρ = 1600.  So t = 133 s.

Those are the two bodies with smallest mean radius, of all solar system bodies in
hydrostatic equilibrium.

Let's try a much smaller eccentric body, just to check.
The nucleus of the comet Wild 2 has cliffs "hundreds of meters tall" (Ulivi's "Robotic Exploration of the Solar System," part 3, p. 228).
f = 200 m / 6400 m, ρ = 600.  So t = 610 s.

So Miranda's Verona Rupes beats even a comet.  Compared to Miranda, any other round body is much bigger (f is much smaller) and significantly denser (ρ is bigger), so to beat this it would need truly brobdingnagian cliffs, which flybys would have seen by now at least with radar if not with visible light.

I could grind through rest of that table of bodies in h.eq., but it's a safe bet that Verona Rupes wins by a long shot, Enceladus is a distant second, and everything else is a giant hardly differentiated peloton.


Some trajectories need less energy than others, so to save on fuel, scientists often choose the lengthy, fuel efficient journey rather than a shorter one. In the case of the Apollo astronauts, it was important to get them to the moon in the shortest possible time, so a short trajectory was chosen.


Clarification on calculating IR Dust Extinction $A_K$/$A_V$

The extinction $A$ of a medium (e.g. dust) is defined as the number of magnitudes an object is increased (because higher magnitude means less light), as the light from that object travels through the medium. It is usually (but not always) a function of wavelength $\lambda$, and is thus written $A_\lambda$ (less often $A(\lambda)$).
The wavelength can be a single wavelength such as the Lyman $\alpha$ line at $\lambda=1216\,\mathrm{Å}$, in which case we may write that exact wavelength as a subscript, $A_{1216}$. Alternatively it can be the total, average extinction in a wavelength band such as the $K$ band, i.e. the near-infrared wavelength region roughly between $\lambda = 2\,\mu\mathrm{m}$ and $\lambda = 2.5\,\mu\mathrm{m}$ (depending on the manufacturer of the filter).
In this case we write $A_K$.
I can't seem to find the paper you're referring to, but you ask "How can I get that?", and the answer is that you measure it. To measure an extinction, you must know how much light you'd expect in the case that some object would not have been obscured (it doesn't necessarily have to be at the particular wavelength you're interested in; it could also just be at a set of other wavelengths, and then you fit an extinction law to these points, under certain assumptions of the properties of the medium). Or you could simply not measure anything, and the assume that the extinction is "similar" to some "standard" extinction law, such as the Cardelli (1989) law, or the Calzetti (2000) law.
Relative extinction
Because extinction depends not only on the dust properties, but also one the distance traveled, it makes sense to measured it relative to a fixed wavelength, e.g. the optical band $V$ centered on $\lambda \sim 5500\,\mathrm{Å}$, thus factoring out distance. Depending on available data, sometimes the $B$ band at $\lambda \sim 4400\,\mathrm{Å}$ is used as reference, but since $A_B$ is usually larger than $A_V$, the flux is larger in $V$ and hence more precise. From the definition of the color excess
$$
E(B-V) \equiv A_B - A_V
$$
and the absolute-to-selective extinction
$$
R_V \equiv \frac{A_V}{E(B-V)},
$$
the relation between the two ratios is
$$
\frac{A_\lambda}{A_B} = \left(\!\frac{R_V}{1+R_V}\!\right) \frac{A_\lambda}{A_V}
$$
You could call the function $A_\lambda$ the extinction law, but I think it's more common to use the term for the quantity
$$
k(\lambda) = \frac{A_\lambda}{E(B-V)}.
$$


Check out the on-line SIMBAD astronomical data base. You can even make separate lists based on criteria (like: list all of the stars between 10 and 20 parsecs (have to convert parallax to light years/ parsecs = 1/parallax (in arcsec/ 1 parsec = 3.262 LY) with class between G0 and G9 from 0 to +30 deg declination). It's pretty cool and shows numerous listings of distance values, brightness, metallicity, etc. complete with date of study and source. I just looked again and masses aren't listed (sorry). Masses aren't directly measured anyway (except in the case of binaries where we know the distance). Masses are essentially determined from the Hertzprung -Russell (mass/ luminosity) diagram and can be refined if we can determine age by looking at spectra. Back to Sol-Station or Wikipedia or I should say their references.


'P' is the penumbra, the zone where the Earth blocks some, but not all, of the Sun's light from reaching the Moon. 'U' is the umbra, the zone where the Earth blocks all of the light from the Sun (see this primer). As the Moon moves across the two zones then you will get various times as the Moon touches those zones. So P1 is where the Moon just starts to touch the penumbral shadow, U1 is where the Moon just touches the umbral shadow cone, U2 is where all the Moon is inside the umbral shadow and the total eclipse begins. Similarly U4 and P4 are where the Moon last touches the umbral and penumbral shadows. Note that the times in the chart are in Universal Time so you need to account for your timezones' difference from UTC.


a 1.7X extender

By 1.7X extender I think you mean a Teleconverter; a projection lens that re-images your main lens's focal plane back on to the sensor, but magnified.

a Tameron 150 - 600mm lens (very heavy),

This says Tamron SP 150-600mm f/5-6.3."
So 600 mm  x 1.7 = 1020 mm focal length, and I'm guessing f6.3 at 600 mm so 95 mm aperture.
This combination is a f=1020 mm f/10.7 telescope!
If you magnify the planets they will get dim at f/10.7 so you'll have to expose longer, meaning atmospheric seeing will cause some trouble.
From here I see that

Canon 7D Mark II comes with a 22.4 x 15 mm CMOS sensor, which has a diagonal of 26.96 mm (1.06") and a surface area of 336.00 mm²... The sensor has a surface area of 336 mm². There are approx. 20,200,000 photosites (pixels) on this area. Pixel pitch, which is a measure of the distance between pixels, is 4.08 µm.

So with f = 1020 mm and 4.08 E-03 mm pixels, each pixel will be about 0.83 arc seconds.
You are ready to go, and you should probably skip the teleconverter at first and just gain some experience; setting up, avoiding dew, pointing at the planets, watching them drift by as the Earth rotates, then re-pointing over and over.
Then taking photos without vibrating the whole thing! DSLR's usually (always?) have mirrors that flip, you may want to delay the exposure until it stops vibrating.
Then taking many photos and learning to stack them, or sort them for lucky imaging.
Photography SE has lots of astrophotography questions and answers I recommend you go there and read a bunch before asking further so as not to ask a duplicate question.
-https://photo.stackexchange.com/questions/tagged/astrophotography
Have fun!


VizieR is an online source for all sorts of astronomical data published in scientific papers. As you mentioned, The HIPPARCOS catalogue contains visual magnitude data.

Open the query page for the main HIPPARCOS catalogue
Select the fields you want (defaults are ok for you)
Hit submit to see the results
You can limit the number of results and format under Preferences on the left

This table gives you the measured visual magnitude, i.e. the Apparent Magnitude ($m_V$, V column). To convert that into Absolute Magnitude ($M_V$) you need to know the distance to the star. This can be calculated using the Parallax field (Plx column).

Here's the formula for you:

$$M_V = m_V + 5 * log_{10}( Plx / 100 )$$

You can easily dump the data into Excel or something, put a formula into an extra column and calculate the Absolute Magnitude.


Yes, light gravitates. The gravitational charge is energy. Well, gravity is a spin-2 force, so you really have momentum and stress as well, but they are analogous to a generalization of electric current.

In general, anything that contributes to the stress-energy tensor will have some gravitational effect, and light does that, having both an energy density and putting a pressure in the direction of propagation.

  But while that light is travelling through space, its energy is "unavailable" to the rest of the universe.

Not quite. It still gravitates. However, the radiation-dominated era was before about 50k years after the Big Bang, but it is long past. Today the gravitational effect of radiation is cosmologically negligible. We live in a transition between matter-dominated and dark-energy-dominated eras.

  Given that all stars on the sky is sending photons that reaches every square centimeter of the earth surface, the amount of pressure should sum up to be quite large.

The light pressure on any surface is proportional to the light energy density incident on it. Thus we can check this line of reasoning directly by observing that the sky is dark at night.

Why it is dark at night is probably deserves its own question (cf. also Olbers' paradox), but it is pretty clear that it is in fact quite small. To be fair, we should check more than the visible range, but even so the sky is pretty dark. Thus on average, light pressure is very small.

We have the privilege of being close to a star, but even during the day, the light pressure due to the Sun is on the order of micropascals.

  ... there should be energy amounting to the gravity of about 1.7*10^22 suns distributed throughout the universe.

And this is a tiny amount. As you just said, this is the equivalent of about 0.034% of the total mass of stars in the universe, which is in turn constitute but a fraction of the matter in the universe. So why you are surprised that its effect is negligible? It's literally thousands of times less than the uncertainty in the measurements of the amount of matter in the universe.


How is the gravitational effect of galaxies outside of the visible universe on galaxies within the visible universe currently modeled?

We don't know…

We don't know how much Universe there is outside our observable Universe.

The observable Universe seems to have a "flat" geometry (in the 3D sense, not in 2D). If it really is globally flat, then it just goes on and on forever, i.e. there is infinite mass. However, just as Earth looks flat on small scales, it might very well be that the Universe has another geometry on scales larger than we can observe.

A recent paper (Di Valentino et al. 2019) claimed evidence for a "positively curved" geometry, implying that the Universe is only (very roughly and with high uncertainties) three times bigger than what we see (see this answer and comments for the calculation). However, the authors sort of cherry-picked one out several of the data sets of Planck Collaboration (2018), and the evidence is not convincing.

…and we don't care

However, no matter if the Universe is finite or infinite, we do not take into account the mass outside the observable Universe in any calculations. This mass does not affect us in any way. Gravity travels at the speed of light, and hence hasn't had time to reach us yet. More importantly, due to the inverse square-law, gravity on any object is by far dominated by nearby sources.

Well, we care a little

There is a sense in which you can say that the gravity is taken into account, though: So-called "cosmological simulations" aim to simulate the formation and evolution of the large-scale structure and the galaxies of the entire Universe. Of course, we cannot simulate the entire Universe, so instead we simulate a large chunk of it; large enough that another similarly-sized chunk would not be statistically different from it (preferably several hundred million light-years on each side).

A galaxy close to the edge of the computational box, say its left side, doesn't only feel the gravity of the galaxies on the right. This would make the simulated universe "implode". Rather, we use a computational trick called "periodic boundary conditions" which means that a particle that happens to travel through the left side of the box enters immediately through the right side. And a galaxy close to the left side feels the gravity of matter lying close to the right side.

Snapshot from the Virgo simulations (Jenkins et al. 1998), where I've tiled the box in several copies, emphasizing the periodic boundaries.


Why you use log to measure metallicity in galaxies?

The distribution of metallicities appear to be more evenly spread out in logspace than in linear space. The reason for this can be ascribed to there being no preferred scale for the abundance of a given element; rather they span several orders of magnitude. The same can be said for instance about the distribution of dust grain sizes, the distribution of the mass of dark matter halos, and the distribution of the area of lakes on Earth.

So, if you measure the number of oxygen atoms (or ions) in ten stars, and divide by the number of hydrogen atoms, you may get values like
$$\mathrm{O}/\mathrm{H} = \{0.03,3.5,25,0.003,0.9,0.4,0.09,0.01,8,0.02\}\times10^{-4}.$$
Plotting this in a linear and a logarithmic plot, you see that on a log scale, the values are easier to tell apart:

(Another reason, as zibadawa timmy writes in his/her answer, is that taking the log you don't need to write all those $10^x$ factors).

Now why add 12? This factor corresponds to measuring the number of a given atom per $10^{12}$ hydrogen atoms. I've been asking colleagues from students to professor level, and the best answer we can come up with, is that the least abundant elements in the Sun (uranium, rhenium, thorium, …) have abundances of the order of one per $10^{12}$ hydrogen atoms, so adding 12 to $\log(X/\mathrm{H})$ makes all values positive. This angst of negative values, I don't understand, though. Usually, we're okay with minuses. And measuring metallicities in stars or gas of sub-solar metallicity yield negative values anyway. Moreover, the scale was in use even when we had only lower thresholds for uranium and bismuth of $\log(\mathrm{U}/\mathrm{H})+12<0.8$ and $\log(\mathrm{Bi}/\mathrm{H})+12<0.6$, respectively (Grevesse 1969) and didn't know if they were actually smaller.


What is the danger for Ceres of collision with other asteroids?

Sadly, a well referenced answer to this question likely lies behind academic publisher paywalls.
However, Wikipedia claims:

  Collisions between main-belt bodies with a mean radius of 10 km are expected to occur about once every 10 million years.[63]

-Source for citation 63 has disappeared from the internet.

However, the Veritas asteroid cluster, and the Karin cluster reportedly formed by the collisional disruption of 140 km diameter and 25 km diameter diameter objects only 8.3 and 5.8 million years ago. These numbers obviously cast doubt on the Wikipedia claim.

This piece Small asteroid collisional characteristics from 2002 states:

  Orbits of nearly 88,000 astreroids from the main belt are needed to find statistically one encounter within 1000 km distance during 50 years.

Naturally the more abundant smaller asteroids collide more frequently. In fact Hubble seems to have caught the aftermath of such an event back in 2010.

To answer the question. It seems Ceres will remain intact for some millions of years to come. But it does get hit fairly regularly, like the moon. If Ceres regularly took impacts large enough to preclude habitation, we'd expect to see co-orbiting debris clouds like the Veritas cluster, or in the 2010 Hubble images. We do not see such debris clouds near Ceres.


The Schwarzschild metric can be written as
$$ c^2 d\tau^2 = \left( 1 - \frac{r_s}{r}\right)\ dt^2 - \left(1 - \frac{r_s}{r}\right)^{-1}\ dr^2 - ...,$$
where $r$ is the radial coordinate, $t$ is the coordinate time, $\tau$ is the proper time (that measured on an observer's own clock) and $r_s = 2GM/c^2$ is the Schwarzschild radius. I have left out the angular terms on the right hand side which contribute a further negative term independent of whether $r$ is greater or less than $r_s$.
For an observer with mass, $d\tau>0$; for a massless particle $d\tau=0$ (e.g. a photon).
When $r<r_s$ the first term on the RHS is negative, while the second term becomes positive. In order for the LHS to be $\geq 0$, then
$$ \left(\frac{r_s}{r}-1 \right)^{-1}\ dr^2 \geq \left(\frac{r_s}{r}-1\right)\ dt^2 + ...$$
$$ \left| \frac{dr}{dt}\right| \geq \left(\frac{r_s}{r} -1\right)$$
What this means is that $dr/dt$ can never be zero, which means the direction of radial travel can never reverse. ie. Anything that enters a black hole (i.e. for which $r<r_s$) and has $dr/dt<0$, can never have $dr/dt >0$.
There is a slightly more satisfactory "proof" using Eddington-Finkelstein coordinates, that shows that all future light cones point inwards and that $dr<0$ when $r<r_s$.


Does Mars contain more iron than the Earth?

Mars definitely has far less Iron than Earth. Mars has 10.7% of Earth's mass. On the other hand, Iron comprises 32% of Earth since there is so much Iron in its inner core, outer core, and the mantle. That means if Mars was made entirely of Iron (which it is not), Earth would still have more than 3 times as much Iron.

You are right that Mars has more Iron at its surface than Earth so it looks very red. Earth has many other metals on its surface (not to mention water to make it look blue and plants to make it look green).


How can I accurately find & apply the white donut to the center of my mirror?

I make my own mirrors and telescopes.

The method recommended by Gary Seronik, that you linked above, is very good, and it's widely used. A few comments:

Don't be afraid to touch the mirror with clean paper. The vast majority of mirrors these days are over-coated with SiO(x) and are quite sturdy. Maybe wash the mirror before you do anything else, this way you'll remove potentially abrasive dust particles. This is how: https://www.youtube.com/watch?v=9Y8xFnXFVGQ - it's a good idea to wash the mirror anyway, like once a year or every other year.

When making the big paper template, cut the disk from a piece of paper without wrinkles. Draw a good circle, and cut it carefully. Make sure it's flat and it doesn't want to roll up. When centering it, go over the edge again and again, and adjust it so that it's perfectly centered. You center it by working out the edge alignment with the mirror. Take all the time you need.

You could also cut the template out of transparent plastic, larger than the mirror, and just draw a circle on it, the same size like the mirror. This method is recommended by Jim Fly (of Catseye) even for quite large mirrors.

Both versions (paper or plastic) work about the same IMO.

It may take a bit of trial and error to put the central dot on the mirror without smearing it. The marker tends to smear up underneath the paper. Practice on a random piece of glass or polished metal.

The reason why Bob Fies (of Alcoat) says it's not for lasers is that, the way he shows it, the paper dot is solid. Of course it wouldn't work with a bare laser - you need a donut shape, with a hollow center, to allow the laser to reflect (when using a bare laser, not with a tuBlug). Just use a little donut instead of a solid paper dot, for the final step - like Gary shows on his website.

If you never use a bare laser to adjust the primary, a solid dot is fine. But a donut gives you more options.

I've done business with Bob directly, he's very experienced. Like Gary Seronik and Jim Fly, he should be considered an expert in this field. But take his advice in context and look at all the details - he says what he says for a good reason.

The way to gauge the existing donut is also by using Gary's method. Make a paper disk with a hole in the middle, center it on the mirror, and put a little dot with a marker on the glass. Compare that dot with your existing donut.

Wipe the marker dot off with alcohol and a q-tip when you're done. You can't damage the mirror unless you press very hard. The center is shadowed by the secondary mirror anyway, so it doesn't matter.

Do everything very slowly and be as precise as humanly possible. The precision of the central donut is critical for the precision of your collimation. Since you only put the donut on once, it makes sense to do a very slow but very high quality job.

If the existing donut is off-center even a little bit, don't hesitate to remove it. If alcohol soaking doesn't remove it, try acetone. You may have to clean the center afterwards, but that's okay.

After everything is finished, clean the inside of the donut with alcohol and a q-tip. You need that surface relatively clean, for the laser to reflect back into your collimator when not using the tuBlug (just bare laser). It doesn't have to be squeaky clean, just remove the marker smears, more or less, and make it nice and tidy.

Congratulations on taking the right steps. Many, many people don't care about collimation, and it's so important for overall performance. Keep that scope collimated.

The tuBlug is an excellent tool for collimation, it makes the process a snap. Great choice. Catseye is another great collimator, but if you have the tuBlug and a Glatter laser (one of the most precise lasers out there, BTW) you probably don't need anything else, unless you have some extremely low f/ratio on your scope (not your case).

I use the tuBlug every time I do an observation. It's taking maybe 2 minutes to do the adjustments. Totally worth it.


What are the dimensions of LIGO Detector evacuated chambers?

The arms are $4\,\mathrm{km}\,\times\, 1.2\,\mathrm{m}$:

From the LIGO webpage:

  The 1.2 m diameter beam tubes were created in 19-20 m-long segments, rolled into a tube with a continuous spiral weld. While a mathematically perfect cylinder will not collapse under pressure, any small imperfection in a real tube would allow it to buckle (a crushed vacuum tube would be catastrophic). To prevent collapse, LIGO's tubes are supported with stiffener rings that provide a significant layer of resistance to buckling under the extreme pressure of the atmosphere. The tubes must withstand these stresses for at least 20 years.


Did the scientists really observed gold and platinum being created from kilonova?

No they didn't. The deduction that (very) heavy elements are being produced is somewhat indirect and involves creating a theoretical model of the expanding "fireball" and comparing that model against measurements of brightness vs time at a range of wavelengths and against spectra (mostly in the infrared) as the afterglow faded.

Normally, when we view a stellar spectrum, one can see individual absorption lines that correspond to particular, identifiable elements. In the afterglow of this explosion that is not possible. It is moving so fast that the absorption features that were seen are smeared out by the Doppler effect and the atomic physics required to predict the strength of absorption is too uncertain to make accurate identifications of particular elements.

What is likely though is that the heavy elements, known as lanthanides are providing the large opacity in the "photosphere" of the fireball and are responsible for the rapid fading of the visible signal, the rather more lengthy decline in the infrared and the appearance of extremely broad absorption features that develop in the infrared. The heating of the ejecta by the radioactive decay of very neutron-rich nuclei is also a vital component of explaining the development of the afterglow.

See for example Pian et al. (2017) and Chornock et al. (2017).


What is the difference between a dwarf spheroidal galaxy and a globular cluster?

There are many differences starting from size to the dark matter content... Please check the following link  it has given much information :
 http://www.answers.com/topic/dwarf-spheroidal-galaxy


Should the synodic period or the orbital period be used to determine the diameter of the Moon?

What you need, as Aristarchus, is a frame of reference in which the Earth's shadow is stationary. Otherwise you are trying to deal with a moving thing (the Moon) relative to another moving thing (the shadow cast by the light of the Sun as it orbits the Earth).
In a stationary-shadow frame of reference, the period you need is the time it takes the Moon to orbit from the middle of the Earth's shadow to the middle of the Earth's shadow. That is to say, from new moon to new moon: which is to say, the synodic month.


Can the Sun be used as a point source of light to achive better focus?

No, it won't work. After all, it is recommended to make a pinhole camera to watch a solar eclipse. What you suggested is similar to a pinhole camera. Why it won't work, is that the rays that pass through the pinhole are not parallel, but divergent. Light from a point source would have (nearly) parallel rays.


When and where are we exposed to most cosmic radiation?

You seem to have a good understanding of cosmic rays, but not all of them come from the sun, so there will still be some coming in at night. These will tend to be the most energetic ones emitted from supernovae and other violent cosmic events. You are most at risk on a long distance airliner, which flies at abut 33,000 feet, but the risk is small and nothing to get worried about. For airliner crew it is more serious, but for the vast majority the time spent at 30,000 feet doesn't seriously affect their health, otherwise they'd be getting danger money. If you want to avoid dangerous radiation, the thing to worry about is radon gas. This is emitted by the decay of uranium in granite rocks. Radon is an alpha emitter, and is dangerous when inhaled or ingested. You can find a map on the internet giving radon levels in your area (they are very variable). If you are in a high radon area, put a gas proof covering on your ground floor and make sure your house is well ventilated. In some houses, radioactivity can reach levels which would be unacceptable in a nuclear power station or research establishment. I don't know of any place on Earth which is especially safe from cosmic radiation except, perhaps, a coal mine or a nuclear bunker.


Why does the sky look like it has three suns in this video?

The ghost images in the video resemble the Sun and are separated by only 2 or 3 degrees (4-6 Sun widths), so we can rule out atmospheric phenomena.
Sun dogs or parhelia have shapes different from the Sun and appear 22 degrees away from it.

The effect is more likely due to looking through a double-paned window at an oblique angle.
The bright image on the right is the Sun viewed directly.
The image in the middle is a reflection by the inner glass surfaces facing each other, offset due to the gap width and the viewing angle.
The faint image on the left is a further reflection of the middle image.


Can one detect from earth that angular momenta of its axial rotation and its orbital rotation are "almost parallel"?

Certainly! Ancient astronomers, eg in Babylon ~3000 years ago were sufficiently familiar with the geometry of the celestial equator and the ecliptic to be able to predict eclipses, and discover the eclipse series now known as the Saros series. But of course, in those days astronomers (mostly) didn't conceive of them in terms of the motion of the Earth.

OTOH, there were a few ancient astronomers in ancient Greece (notably Aristarchus of Samos), and even in Babylon (Seleucus of Seleucia), who did believe in a heliocentric system.

The celestial pole is easy to discover, (especially if you're in the northern hemisphere), and it doesn't take a lot of observation to notice that the ecliptic differs from the celestial equator.

Although it's not possible to observe directly that the Sun's path is the ecliptic, it was known that all of the other celestial wanderers stay fairly close to the ecliptic, and that eclipses only occur when the Moon is very close to the ecliptic. And you can get a pretty accurate idea of the Sun's path by observing the rising & setting of stars and planets that are near the Sun's path. So astronomers across the ancient world recognised that the ecliptic is the Sun's path. And by the 2nd century BC, knowledge of these movements was sufficiently detailed that Hipparchus of Nicea was able to estimate the precession of the equinoxes.


How to calculate full (mechanical) energy on the hyperbolic orbit?

The total energy is the sum of the kinetic energy and the potential energy. It is constant, unless some interaction such as a collision between bodies occurs.

The kinetic energy is easy to understand. For a mass $m$ and velocity $v$ it is given by

$$E_k = ½mv^2$$

The potential energy formula is (technically) given as

$$U = -μm/r$$

where $μ$ is the Standard gravitational parameter. (I say technically because it can be hard to grasp the concept of negative energy. See the linked articles).

In the case of a hyperbolic orbit the velocity and distance from the sun must be measured; one cannot be derived from the other as is the case with a known elliptical orbit.

See also

Specific orbital energy
Gravitational energy
Gravitational constant


How much mass would have to be added to the Sun to significantly alter any of its characteristics?

How much mass would have to be added to the Sun to significantly alter
its characteristics

Asking how much would be significant is inexact.  The Sun is classified as a G2V main sequence star.   Though the chart lists one solar mass as G4V, so there's some variation in there.   The classifications seem to relate to temperature.    To go 1 step up (and using the sun or any G star), the increase of one ranking (decrease in 1 number after G) alternates between a 2% and 5% increase in mass.   I'm going to round that to 3%-4%, so for our sun, that's about 30-40 Jupiters (31.4-41.9 if you care about significant digits), but I'm going with 30-40 Jupiter masses added for each step up the classification ladder, ignoring the effects of impact, which would be considerable.
That increase of one in classification level viewed from Earth would be quite significant.   3% more mass corresponds to about the 4th power of that or 12.5% more luminosity.   Earth would be very different if the Sun was 12.5% more luminous.  I would think that would be enough to turn Earth at least into a steam-bath, perhaps a full blown runaway greenhouse effect due to the abundance of water and feedback mechanisms.   Adding mass would also change Earth's orbit, drawing it in slightly and shortening the year, so the increase from Earth would be a bit more than 12.5%.
The stellar mass math goes something like this.
Lifespan decreases by about the 3rd power of an increase in mass.
Energy output or luminosity increases by about the 4th power of the mass, around 1 solar mass.  This increase slows down to about the 3.5th power with more massive main sequence stars.
Surface temperature increases slightly with mass, perhaps 10-20 K per 1% increase in mass, or about a 1% increase in temperature for every 3% increase in mass would be about ballpark.
Radius increases with added mass too, so the heavier sun would be both brighter and larger.   The radius increases enough that the surface gravity decreases.  A 3% increase in mass corresponds to around a 2% increase in radius, so the size wouldn't appear very different on these scales.
If the Sun became just 0.5% to 1% more luminous  (about 1.5-3 Jupiters in mass) that would be in the ballpark to man made climate change so far.  Enough to melt surface ice over time and cause oceans to rise, so, significant depends a lot on point of view.
These numbers should be taken with a grain of salt because other factors like metallicity and age matter too.   Stars like our Sun also change in size and luminosity during their lifetime, growing both larger and more luminous over time.  There are more massive main sequence stars than our Sun that are less luminous because they're younger and the reverse as well, though off hand, I can't name any.
Errors fixed.  Thankyou for pointing out.
I want to add, since you mention XKCD in the question, if Randall Munroe was answering this question, he would take it to the extreme, answering what happens if you add million Jupiters, or a billion.   Fun times.  Without his clever diagrams, it wouldn't be the same, so I'll stop here.


Why are three year old eclipse glasses not recommended?

According to the American Astronomical Society (AAS), the warning that eclipse glasses should not be used if they are more than 3 years old is outdated (provided that they are compliant with the ISO 12312-2 standard).

A page on their web site says the following:

  Note: If your eclipse glasses or viewers are compliant with the ISO 12312-2 safety standard, you may look at the uneclipsed or partially eclipsed Sun through them for as long as you wish. Furthermore, if the filters aren't scratched, punctured, or torn, you may reuse them indefinitely. Some glasses/viewers are printed with warnings stating that you shouldn't look through them for more than 3 minutes at a time and that you should discard them if they are more than 3 years old. Such warnings are outdated and do not apply to eclipse viewers compliant with the ISO 12312-2 standard adopted in 2015.


How can I crop a fits file and get correct WCS header info without a new plate solve and WITHOUT USING PYTHON?

Skycat also stores the WCS information when selecting an image region from a fits file. Not sure how exactly this can be done in non-interactive mode but many features can in fact be "batched". Give it a try. Best luck!


Is the CMB the photons that were created at the birth of the atom?

The CMB consists of photons that were emitted very shortly before and during the recombination of electrons with hydrogen and helium nuclei. [Side note - why is it called the epoch of recombination when the atoms were never "combined" to begin with?]

Those photons were then able to travel, unabsorbed across the universe until they encountered our microwave telescopes 13.7 billion years later.

Although the photons are unabsorbed, the expansion of the universe causes a redshift as they travel. Their current wavelengths are approximately 1100 times larger than when they were emitted and the universe is almost completely transparent to photons of this wavelength - that is there is very little chance of a photon being absorbed even if it travels across the diameter of the observable universe.

There are scattering processes that can deviate the photons or even increase their energy (and decrease their wavelength) via inverse Compton scattering (the Sunyaev-Zel'dovich effect), but this does not affect the CMB spectrum a lot and is difficult to observe.

Our current ideas about the future of the universe are that it will continue to expand at an accelerating rate. The CMB photons will become even more redshifted - to radio wavelengths and beyond - and the universe will become even more transparent to them.

Related: Future redshifting and effect on the 'pitch' of CMB radiation


How bright is Andromeda's center, in magnitudes per square arcminute or something similar?

The surface brightness number depends on a few things: how large an area of the image is averaged, how close to the galactic nucleus it is, and in what wavelength band.
Some papers give a brightness profile along a line bisecting the galaxy image.

Sandage, Becklin, and Neugebauer 1969 report, for a 7.6 arcsec diameter circle centered on the nucleus and measured in the $V$ band, a total magnitude of 11.3 and an average surface brightness of 15.5 mag/arcsec$^2$.
The same aperture 30 arcsec east or west of the nucleus was 1.5 magnitude fainter.

Kent 1983 images the central bulge in the $r$ band and finds a surface brightness of 14 mag/arcsec$^2$ at the nucleus and 16 mag/arcsec$^2$ 10 arcsec away along the major axis of the galaxy image.

Walterbos and Kennicutt 1987
measure a 60 arcsecond circle and find a surface brightness,
1 arcminute away from the nucleus along the major axis,
of 18.5 mag/arcsec$^2$ in the $B$ band.


Why are there so few supernovae in M31 (Andromeda)?

The low supernova rate in M31 can be directly attributed to the fact that the galaxy's star formation rate is much lower than the Milky Way's.

Andromeda is currently in a relatively quiet phase in terms of star formation, currently experiencing rates of $\sim0.40M_{\odot}\;\mathrm{yr}^{-1}$. The Milky Way, on the other hand, has a star formation rate several times higher; as such, it contains more massive, short-lived stars which are supernova progenitors. This leads to a correspondingly higher supernova rate than Andromeda. Estimates of the supernova rate in M31 vary, but a value of 0.5 to 1.0 per century is reasonable. This in turn matches well with observations.


Is the lower mass limit of a neutron star the same as the upper mass limit of a white dwarf?

No, the two limits are not the same - there is some range of masses that both white dwarfs (WDs) and neutron stars (NSs) can have.

The Chandrasekhar mass limit suggests that WDs cannot be more massive than about $1.4\,M_\odot$.  However, this is true for non-rotating WDs.  Rapidly rotating WDs might be as massive as $2\,M_\odot$.  Accretion in a binary system could both spin up a WD and increase its mass.

NSs that have not undergone accretion are regularly observed with masses around $1.3\,M_\odot$, while NSs who have 'recycled pulsars' have an observed mean mass of $\sim 1.5\,M_\odot$ according to the same paper.


Are warm circumstellar debris disks still able to be detected at sub-millimeter wavelengths?

Such surveys are designed to detect an excess of emission over that expected from a stellar photosphere.

The emission from a circumstellar disk is often modelled as a blackbody at lower temperature than the stellar photosphere. A hotter blackbody with the same emitting area emits more flux than a cooler blackbody at all wavelengths.

Therefore, if you had a disk of fixed emitting area, it would be easier to detect a hotter disk at sub-mm wavelengths than a cooler disk.

However, if you have a lot of cold disk material and no warmer regions, then because the Wien tail of the blackbody distribution is so steep, it emits almost no radiation at near- and mid-infrared wavelengths, and sub-mm detection is the best way to detect it.


Orbiting supermassive black hole or galactic center of mass?

To a first approximation when we calculate how fast an object is orbiting around some mass distribution we can assume that the gravitational attraction it experiences is only that due to the mass interior to its orbit.

This approximation, known as the shell theorem should only be applied when the mass distribution is spherically symmetric, or most of the mass is concentrated well inside the orbit of the object.

The bulge of our Galaxy can be considered roughly spherically symmetric, and it appears that the mass distribution is strongly concentrated towards the centre.
Hence we can say that the centripetal force on an orbiting object is provided by the gravitational force due to the mass interior to its (again, I am assuming a circular orbit for the sake of argument, but this doesn't greatly matter) orbit is
$$ m\frac{v^{2}}{r} = Gm \frac{M(r)}{r^2},$$
where $m$ is the mass of the object (which cancels out) and $M(r)$ is the mass inside radius $r$. Thus the orbital speed $v$ is given by $\sqrt{GM(r)/r}$.

Thus by observing orbital speeds we can calculate roughly what mass exists interior to the orbit.

In the case of stars near the centre of our Galaxy, that turns out to be 4 million solar masses, packed into a space that is extremely small. The reason it is thought to be a black hole is not because of the mass, but because it is mass that is (i) densely packed and (ii) cannot be seen.

Even if it were a bunch of massive stars shrouded in dust, so that you could not see it in visible light, you can easily calculate that the heated dust should be kicking off huge amounts of infrared radiation; and that is not seen.

In the case of the Sun, well, the mass interior to the Earth's orbit is much larger than 4 million solar masses, it is many billions of solar masses. The contribution of the black hole at the centre is completely swamped by all the other mass in our Galaxy, so we are not directly influenced by the central black hole.

If you removed the central black hole, the stars that were near it would be orbiting way to fast for the new $M(r)$ that they experience. They would fly off and assume highly elliptical orbits with much larger semi-major axes, though I think they would still be part of the Galactic bulge.

Stars further out, including our Sun would have their orbits changed imperceptibly, because $M(r)$ will hardly have changed for them.


Force exerted by space debris on satellites

The force depends on the details of the collision. Depending on how hard the materials involved are, and the angles and everything, it might be a larger force for a shorter time, or a shorter force for a larger time.

What are relatively predictable is the impulse -- force times time, essentially (or the integral of force over time, if you want to be exact) and the energy released by the impact. For a small piece of debris these are determined by the mass and velocity of the debris with respect to the satellite.

Relative velocities are typically a few km/s, so, 1 mg piece of debris (a paint flake or something like that) will deliver about $10^{-3} kg m/s$ of impulse, and about $10--100J$ of energy.  The impulse will change the orbit of the satellite (typical mass perhaps 1 ton) by very little. Perturbations from tides and other effects will be bigger. The energy is more of a concern -- depending on what it hits something may get hot, or have a hole made in it.

A much larger piece of debris (say 1g -- a small screw) carries 1000 times the impulse and energy and may will wreck a satellite (although if it hits a solar panel or something it may just make a hole and go on through). If the satellite survived, then it the change to its orbit would be noticeable, but not a huge deal.