A quick (elementary) check on arsecond conversion to sky distance in parsecs

The pedagogical answer is this.  You're confusing the
width
of something, with it's
parallax!
Here's the fixed background

Here's something that happens to have width ...

Here's something that has parallax ...

That's the deal!
Note that width can just be measured, using one printout of the photo and a wooden ruler.
But parallax means two photos taken 180 days apart, and you measure how much the thing has moved.
(The mind-boggling GAIA does not actually use wooden rulers!)

One huge confusion with "parsecs" is this:
Parallax means the distance an object moves in two photos take six months apart. That technique can be used with objects from 1 to about 100 parsecs away from us.
However, astronomers (for convenience) still use the measure "parsecs" even when the distance is far more than 100 parsecs.  And, they even use the measure "parsecs" for things like "size", which just have no connection at all to parallax photo measurements.
If you hear that something is "5000 parsecs away", it is inconceivable the distance was determined using parallax photo measurements!!  And, if you hear that something is "1000 parsecs wide", that just has no connection at all to parallax photo measurements (it's a width - nothing to do with the distance from Earth!!)
In the question at hand here, the "width" of an object was quoted in parsecs. The OP was originally assuming that, since parsecs were mentioned in relation to the width, you could use something related to the technique of parallax photo measurement, in relation to the width.  In fact, parallax photo measurement just has no connection, whatsoever to width. You literally might as well use smell or taste - heh!  The confusion is that parsecs happen to be also used (very confusingly) to measure things like width: but remembering that the technique of parsecs (parallax photo measurement) has absolutely no connection at all to width, and no connection to anything more than 100 ps from us.


We do not have direct data on the age of any exoplanets, but we can estimate the age of their host stars, around which they presumably formed at the same time. These methods are uncertain and may have systematic biases but gives at least some information. If we use the Wikipedia page and plot the ages (again, both old data and now extra selection biases!) the resulting distribution looks like this:

The mean age is 4.9161 Gyr and the median age is 4.5260 Gyr, comparable to the age of the Earth. The skew distribution means Earth is younger than the mean, but fairly typical.

Another way of estimating whether Earth is young or old is to compare to star formation rate estimates across the history of the universe. This is of course also presently uncertain and may be different in different galaxies (elliptic galaxies have low star formation, so their planets would tend to be older than the ones in spiral and irregular galaxies). Overall, it appears to have peaked 3.5 Gyr after the big bang and has declined since. That would perhaps make the average planet that ever formed much older ($\approx 10$ Gyr old) than Earth.

However, many of these have also been "lost" since they formed around bright stars that died long ago and are not measured in the exoplanet surveys. There is also the issue of whether early low-metallicity stars form many planets: it might be that planet formation got going much later (Lineweaver has a -- somewhat old -- paper arguing that Earth formed after 75% of the other planets). Again, this is a very active research field and future answers are likely to be better.


Stellar nurseries in the "Shedding New Light on the Whirlpool Galaxy" video

Do stars tend to leave "these stellar nurseries" after a while, and it's only the short lifetime of the most massive stars that keeps them from leaving before going supernova?

Yes. The lifetimes of very massive stars (10-50 Myr) are comparable with the dispersion timescale for young clusters and associations. So massive stars tend to die near where they were born and they will be surrounded by their (more numerous) lower mass siblings. Lower mass stars have long lives and end up dispersing and making up the general population of a galaxy.

  What impact does the heating of surrounding gas to X-ray temperatures have on the formation and evolution of other stars within a given "nursery"?

It is unclear. The injection of energy and heating of gas ought to make it less susceptible to collapse and fragmentation. On the other hand that doesn't matter isf most of the gas that was going to turn into stars has already turned into stars. On the other hand, it is possible that the shock waves and outflows from supernovae actually compress surrounding structures and trigger new star formation.

  Are there really such things as spatially well defined "nurseries"? Are they distinct and roughly countable, or is this a subjective popular term? Can one point to something and say "this little blob here at x, y, z in the spiral arm of the Whirlpool Galaxy, it's a classic nursery"?

To some extent yes. There are some massive clusters of young stars that can be observed in other galaxies and can be certainly identified as the birthplaces of tens of thousands or even millions of stars. On the other hand, in our own galaxy it appears that although most stars are probably born in clusters and associations, there is no such thing as a "typical" cluster and that star formation takes place at a range of scales and at a range of densities. i.e. Much of the star formation taking place in other galaxies may take place away from some of the obvious blobs of star formation.


What happens if one day aliens want to talk to us. How can we talk to them?

Some basics:

Whether we listen for or send signals into space, something to keep in mind is that space is full of EM radiation of all wavelengths, from very long wavelength background radiation, to a range of microwave, infrared, visible and UV light from stars to gamma-rays from accretion disks and collapsing stars.   Any signals we send and perhaps any signals we could receive would be like raindrops in an ocean.   There's an obvious range to listen called the hydrogen line, which is the emptiest region of EM radiation in space.

Even so, across light years, signals would still be faint, so the first signal that's sent needs to be an easily recognizable signal and math is very good for that.   Pi in binary or a series of repeating prime numbers wouldn't happen naturally, so signals like that are a great way of saying "We are here".

Once the first signals of prime numbers or other math are exchanged, and we see what direction the signal comes from and we know where to listen and where to send, then the communication can begin, presumably using more traditional language.


Projecting an image of the sun / eclipse

I've done this using two lenses.  In my case they were a 300 mm and a 50 mm.  I built a jig to hold the lenses in-line with each other, to allow for small adjustments in the placement, and for a tripod mount.

Put the long lens, in my case the 300 mm, in front normally.  That will form a focused image behind it.  Put the short lens backwards behind that.  This lens is used in reverse.  It takes the small image created by the long lens and projects it out what is normally its front.

The ratio of the front lens focal length to the back lens focal length is how much angular "gain" you get from this setup.  In my case, the cone of project of the sun was 6 times wider coming out of the 50 mm than going into the 300 mm.  This means you don't have to get as far away from the overall setup to get a decent size picture.

I'm not where this jig is now.  I'll try to remember to take a picture of it and post it here within the next day or so.

One thing you'll find is that the earth rotates annoyingly fast when you are looking at a magnified picture of the sun.  Use a telescope mount if you can.

Another issue is blocking out as much ambient light as you can.  The projected image of the sun is reasonably bright, but so is daylight.  Some sort of baffle over the top and the front except for a hole for the front lens helps a lot.  People can then view the projected image from the sides.


Might someone see changes to the corona during a solar eclipse?

Yes. There is a chance of one or more solar flares or CME's occurring during the eclipse and these would be noticeable during totality. There are actually two NASA jets equipped with telescopes and high-speed cameras to film the Corona and any such event. The plane will track the eclipse throughout its duration.


How to find inverse steradian from $\text{arcmin}^{-2}$ for density of galaxy

There are 60 arc minutes in a degree and $180/ \pi$ degrees in a radian.

So 1 radian is 57.2957795 degrees or 3437.746771 arc minutes.

As long as you are talking about units, you can square any of these angular units to obtain solid angle units.

So 1 steradian is 3282.80635 square degrees, or 11818102.860 square arc minutes.

Let's try your theory:

1/354543085.801 steradians is 9.259259E-06 square degrees, or 0.0333333 square arc minutes.

The answer is that it is 1/30th of a square arc minute, so YES it is 30 inverse square arc minutes!

Then 48 per square arcminute is:

$$48 \times 60^2 \times \frac{180^2}{\pi^2} \approx 567268937.282 \ \ \text{sr}^{-1}.$$


Does a magnet outside a black hole affect one inside it?

The tidal forces emerging from the black hole would at first break the magnet flung towards it into two pieces. Those two pieces would themselves undergo deformation and splitting, as the process of spaghettification will continue beyond the molecular, atomic and even sub-atomic levels, reducing everything that crosses the Event Horizon to a Singularity.

Note:

You won't actually see the magnet thrown towards the black hole crossing the event horizon, but that's a different matter altogether - please see this if it might interest you.

Now, magnetic attraction is essentially just an Electromagnetic Force, the range (not to be misunderstood as strength) of which is infinite - see reference.

In the context of the scenario stated in the question, the fate of attraction between a pair of magnets, one of which is thrown into a black hole, could be visualised by the following analogy: imagine holding a magnet in your hand and the other attached to a train which is accelerating downhill into a gorge. The two possible results are:

If the attraction is strong enough to pull the magnets together before the train falls into the gorge, then they both crash into the abyss together (assuming that the magnet attached to the train would easily be able to pull the one that's being held by hand). In case of a black hole, this would mean that both the objects (magnets) would be crushed to singularity. However, we might safely say that they both were "attached" before crossing the Event Horizon.
Or, the acceleration of the train reduces the attractive force infinitesimally, with the magnet in hand barely having any effect on the other magnet that crashed (and vice-versa). In a classical sense, this means that the force of attraction would still exist, even if it's reduced infinitesimally, to almost zero. However, with regards to a black hole, Event Horizon ensures that anything inside the horizon can no longer communicate with or influence anything outside the horizon (and vice-versa). This means, if the magnet somehow crosses the event horizon as a single piece, before losing its existence to singularity, its magnetic attraction would still never be able to cross the boundary defined by the horizon, even though the range of influence of the electromagnetic field is infinity. Ultimately, this would only last for a very brief moment and the magnet would be unable to prevent being pulverised to singularity, either by itself or by any attempts from any other entities outside of the Event Horizon.

Hope this helps.


Can we hear something on Venus, Mars and Titan?

It depends on the composition of the atmosphere. In a helium (or other light gas) atmosphere, your voice would have a higher pitch. If the gasses are heavier, your voice drops.
The density is also a factor: on Mars, sounds can't be as loud as on Earth. Sound is transmitted as pressure differences, and the largest pressure difference you can get is that between the atmospheric pressure and a vacuum.
The speed of sound also changes with the composition and pressure of the atmosphere. This would change the way you perceive the distance and direction of an object that makes a sound: we normally use the tiny differences in time between a sound reaching our left ear and our right ear. A different speed of sound screws up those time differences.


Earth's perihelion timings and distances are based on the distance between the center of the Earth and the center of the Sun. They are effectively the times and distances of the closest approach between the two bodies, not the barycenters. As we can see on Fred Espenak's astropixel page about perihelion and aphelion, the time intervals between two successive perihelions tend to vary between 363 and 368 days.

The data on astropixels seems perfectly valid. With JPL Horizons, for example, the next perihelion is calculated to be on 2020 Jan 5 at 07:47:42, at a distance of 0.98324356482 AU (147,091,144 km) which matches astropixel's values.

I analyzed the perihelions of the 21st century to see how much the actual timing can differ from an idealized "average" perihelion moment (based on the anomalistic year length). Perihelions this century remain within 1.4 days of that average. The differences from the average are mainly due to the gravitational effect of the Moon.

Using all perihelion times in the 21st century, and the average perihelion interval (anomalistic year, 365.2596 days) I constructed the following graph showing these differences:


Calculate eccentricity from apside altitudes?

The major axis is just the sum of the distance at perigee plus the distance at apogee. This is 400 km + 1500 km + the diameter (not radius) of the Earth, and the semi-major axis is half of that.

However, the semi-minor axis is harder to determine. Therefore, why not use an alternative method to calculate the eccentricity?

$$e = \frac{r_a-r_p}{r_a+r_p}$$

Here, the $r$'s are the distance to the center of the Earth, not the apside altitudes. Can you take it from here?


How does dark matter interact with black holes?

It interacts gravitationally.That's all there is to it.

There is a big difference though to the way that normal and dark matter interact with black holes - dark matter is far less likely to be captured.

Given two lumps of matter, one normal one dark, with the same mass and angular momentum with respect to the black hole; only the normal matter is capable of shedding its angular momentum (normally an accretion disc is involved), which allows its orbit to shrink enough to be captured (within three Schwarzschild radii). Dark matter is dissipationless; if it has too much angular momentum it won't get captured.


Yes, very much so.
The lines or circle of constant longitude always form a great arc which intersects with the poles; thus it always has the same length. A circle of constant latitude varies in circumference: the largest is at the equator while it has intermediate length at intermediate latitudes and no length anymore when you reach the pole.

See e.g. that image from wikipedia and follow a circle in longitude, thus from North to South. And compare the different sizes of different circles for different lattitudes (thus the circles which are parallel to the equator)
Mind though that the nautical mile is no longer defined via the fraction of the polar circumference of Earth. It is defined as exactly 1852m in length, and thus can in principle be determined as accurately as one can measure time (1852 / 299 792 458 seconds).
In earlier times it was a MUCH easier task to determine a difference in latitude as it directly relates to the (culmination) height of celestial objects above the horizon. It was VERY DIFFICULT to impossible to determine difference in longitude unless you had a very accurate clock and an almanach of the rise and set times of celestial objects.


Looking at this site, there is a table, which has some information.

  Rotation period: 12.4043 ± 0.0007 * hours

It also rotates around a fixed axis

  recent observations suggest that the axis is tilted about 40 degrees

There are actually only two objects known to rotate chaotically in the solar system - one is Saturn's moon Hyperion, and the other is a small asteroid called 4179 Toutatis (source: this link from Guinness World Records)


What leads to increase in opacity in kappa mechanism?

It's a bit subtle, the key thing the partial ionization does is to keep the temperature from changing much.  What you really want is an increase in density, not temperature.  The reason the kappa mechanism is important is that it allows heat to be added to the gas when it is compressed, and removed when the gas is expanded, that's what allows for the pulsation to have energy added to it.  Heat is added when the opacity is high, and removed when low, so you need the opacity to go up when the gas is compressed, and go down when the gas is expanded.

Now, normally opacity goes down when gas is compressed, because the rise in temperature causes ionization, which actually hurts metal opacity by reducing the number of bound electrons (metal electrons have higher opacity in general when bound, due to resonances, than when free, though the hydrogen and helium we had in the Big Bang are an exception that works the other way-- in a star, it is usually metal opacity that is important and variable).  But increases in density also tend to produce more bound metal electrons, so there's a competition there that temperature usually wins and stars are pulsationally stable.  But in partial ionization zones, temperature changes are reduced by the phase changes eating up the excess energy, and that gives the density increase the edge that allows opacity to rise when the gas is compressed.


What is the orientation of the M87 black hole image relative to the jet?

The jet is projected onto the image roughly in the E-W direction with the (main) jet coming towards us to the right of the black hole.

The brightening in the ring is perpendicular to this. It is probably caused by the spin of the black hole (aligned with the jet), which drags emitting material around and Doppler boosts the emission seen to the south.

This picture is from the fifth event horizon telescope paper published today. The jet is E-W (actually, I believe it is at PA=288 degrees, so rotated 18 degrees anticlockwise) and the black hole spin axis (indicated with the black arrows) is projected at the same PA and into the screen (the two left hand plots).


Is it possible to know if you are travelling close to the speed of light if you have nothing to compare your speed with?

Velocity is relative. From your perspective, your velocity is zero. Without windows (or sensors) on your spaceship, you cannot know the difference. To measure your velocity, you have to define a reference point, eg a star, the average velocity of surounding stars or the galaxy center. If you have such a huge velocity relative to them, from you perspective, they are this fast and not you. So you can observe the time and space dilation of these objects, while your time appears to be normal. You ship is an inertial reference frame.

As for throwing the baseball: From your perspective, you have the speed zero, and the baseball eg. the speed 0.5c. A "fixed" observer, which observes you at 0.99c will measure the balls velocity at 0.997c. With those relative effects, speeds don't add up, only kinetic energy does. http://curious.astro.cornell.edu/about-us/139-physics/the-theory-of-relativity/special-relativity/1016-why-can-t-relative-velocities-add-up-to-more-than-the-speed-of-light-intermediate The increased mass is only a relativistic observation, resulting from the kinetic energy (relative to the observer).


What's the distinction between "planetary formation mechanisms" and "gravitational collapse"?

"Gravitational collapse" is not the generally used name of any planet formation model it simply refers to the collapse of a cloud of gas or collection of objects under the influence of their mutual gravity. "Planetary formation mechanisms" simply refers to mechanisms which form planets, all of which involve the mutual gravitational attraction of the planetary components at some phase.

In the context of the answer you refer to, there "gravitational collapse" refers to where a small object (like a brown dwarf) forms through the same direct collapse and fragmentation process that forms stars.

Planets on the other hand are thought to form in the disc that is inevitably created around a protostar during the star formation process. This might be through something called gas disc instability, or the more likely model at present is something known as the core accretion model.


How can we infer the mass of SMBH in galaxies that are not active anymore?

I can think of two methods.

Both rely on the dynamics of material surrounding the SMBH, which is affected up to a distance of the order of the "sphere of influence". This is the region where the BH dominates the dynamics as compared to the enclosed mass of the galaxy. The sphere of influence is:

$$ R_{\mathrm{infl}} \equiv \frac{G M_{\mathrm{SMBH}}}{\sigma_{\mathrm{bulge}}^2} >> R_{\mathrm{horizon}} \approx R_{\mathrm{Schwartzschild}} \equiv \frac{G M_{\mathrm{SMBH}}}{c^2} $$

While typically $\sigma_{\mathrm{bulge}} \approx 250 km s^{-1}$, it is well known that $c \approx 300000 km s^{-1}$. This means that the influence of a BH can be felt much further away than its event horizon, which is where the accretion takes place. In fact the ratio between the two distances is about 1000000.

Exploiting this fact, astronomers have been using two methods to probe extragalactic*, quiescent SMBH:

The first method is to observe CO lines (radio astronomy) to trace gas circling the BH. The gas does not need to be near the event horizon, which is much smaller than the sphere of influence. In fact CO observations rely on the gas being relatively dense but cold. Essentially the speed at which the CO gas will rotate is a (quadrature) sum of the declining component due to the stellar mass, plus the Keplerian component due to SMBH mass.
The second method is completely analogous, but relies on measuring the unresolved kinematics of the stars surrounding the SMBH. This can be done in various bands, but most authors use visible line absorptions to measure the velocity and velocity dispersion (and other moments) of the stars in the regions surrounding the BH. If the kinematics cannot be explained without including a point mass in the middle, then you are done.

See this work, for a comparison of the two methods.

*(extragalactic means outside of our own galaxy)


A contrast stretch reveals stars down to magnitude 4 or 5.
The stars you asked about are Deneb (center) and Vega (bottom).
The constellation in the center and below is Cygnus; we also see Cepheus at upper right and part of Draco at lower right.

The stars appear as blobs 10-12 arcminutes wide due to several factors:

Rotation around the north celestial pole, to the right of the upper right corner of the image. In your 30 second exposure, Deneb and Vega drifted 5-6 arcminutes in the 5 o'clock direction.
Diffraction due to finite aperture. The width of the ring pattern is inversely related to the width of the aperture. These rings are consistent with a 2-3 mm aperture.
Optical aberrations such as coma, increasing toward the edge of the image.
Vibration as noted in another answer.


What is the detector size of LANDSAT 8's detector array?

The answer can be deduced from the main landsat 8 page, where you find the following picture of  QWIP detector arrays, and QWIP is short for Quantum well infrared photodetector, I added a gray arrow pointing to one of the 3 arrays.

I cannot give you the exact dimensions in mm, but it is clear from the SMD chip elements (SMD = surface mounted device) - one of them I highlighted with a gray ellipse, that each sensor is probably coin-sized, if not smaller. Or in numbers: I guess a side length $1.5\pm1 {\rm cm}$ seems reasonable, it depends a bit whether the sensor has the same size as the silver square or whether there is a border around.
References

Some introduction to SMD resistors


How much more atmosphere is the sun passing through at sunset as opposed to at its peak in the sky?

The amount of atmosphere that light from outside Earth has to pass through before it reaches the ground is given by the air mass, $X$. It is normalized to unity for a source directly overhead (at "zenith"), and increases as the source "sets".

If you don't look too close to the horizon, you can use the formula $X = \sec z$, where $\sec \equiv 1/\cos$, and $z$ is the angle from zenith. For instance, if the Sun is $30^\circ$ above the horizon, so that $z=60^\circ$, then $X = 2$, i.e. its light passes through twice the amount of atmosphere compared to when its at zenith.

The formula above makes (at least) three approximations, however. Firstly, it assumes a plane-parallel atmosphere (an infinite, flat Earth), and so $X\rightarrow\infty$ as the Sun sets. Secondly, it assumes that the atmosphere is homogeneous, whereas in fact its density decreases with distance from Earth. This not only causes a non-uniform absorption$^\dagger$ along the ray, but also causes refraction which increases the path length for high $z$. For astronomers, these approximations don't matter much, since they're quite accurate out to $z\sim70^\circ \text{–}\,80^\circ$ and we usually don't observe at greater angles anyway, precisely because of the large air mass. But several more exact formulas based on various models of the atmosphere have been created to fit observed air masses. These models converge roughly to $X \simeq 38$ for $z=90^\circ$. In other words, the amount of atmosphere that light goes through for a source at the horizon, compared to a source at zenith, is roughly 38. However, note that just $10^\circ$ above the horizon, $X$ is already $<6$, so it's the last few degrees that really matter.

The exact value depends on many factors, including temperature (and its change with height), position on Earth, and the altitude of the observer.

$^\dagger$Most photons are actually not absorbed, but rather scattered out of the line of sight. Effectively, the result is the same; photons are removed from the ray. The scattered photons then become part of the background (and since blue photons scatter more than red, the sky is blue, and the Sun becomes redder as it sets).


Where can I find permissibly licensed database of stars and constellations?

I have an idea how to do this, but it requires a few steps:

Yes, the IAU set down which connections make up the classical constellations, but I didn't see an easy way there to get their data.
Wikipedia has a list of stars by constellation. The stars there per list are many more than you need, BUT: the primary connectors of the constellation will always be made up by the brightest stars in a constellation, which will be the first entries in each list.
To access each list's first entries, I would recommend to modify the simple script from this question's accepted answer, which accesses a website and then datamines it for the interesting stuff.
Now not every constellation has the same amount of stars, but probably mining the 20 brightest ones each should be enough (brightness designators are the greek alphabet in order, $\alpha$ for the brightest, $\beta$ for second and so on.).
Then adding the connections: I don't have a good idea for that one. Possibly you have to bite the bullet there, grab a bottle of wine in the evening, and mark the connections by hand. 88 times 5-10 pair connections. Depending on the efficiency of how you mark the connectors in your database, you could be done in 1-2 hours. Still much faster than without any preselection. Should no other answer cover this step in a more efficient way, at least you could now get back to the IAU website and get the official connectors.


How much of the Universe is invisible to us, and how does it affect our theories?

It well might be the case that universe beyond our observable universe is filled with pink flying elephants, but this just seems highly unlikely.

Our current models are usually based on the fact that universe is isotropic, ie there’s nothing special about our observable universe vs everything else there is. Obviously there are smaller and larger deviations, but  on larger scales these are thought to cancel out. This means the simplest position here is to assume all of universe looks more or less the same.

While we can’t currently / no longer observe what is going on on the other side of horizon, there is some evidence supporting above view. Firstly, cosmic microwave background measurements seems to support isotropic universe. Secondly, inflation models seem to provide explanation for isotropic
universe via early causal connection beyond our current observable universe, and as our early universe model is able to explain many other things to great detail, it gives some confidence to this claim as well.

Since early universe models do support the view universe outside our observable universe is pretty similar to what we can see now, this helps us to rely on observable universe for overall model building.

Also, there does not seem to be any large mass concentrations outside the observable universe, since these would be seen as bias for movement of galaxies and galaxy groups, which also supports that the rest of universe is at least very close to what we can observe directly. This gives further confidence that our current models are probably not affected much by things beyond our currently observable universe.

However, since everything outside our observable universe is now causally not connected to us, anything that might be going on over there does and will not affect us - thus it is not necessary nor valid to include any of that in our models either.

To summarize my answer: no, there’s probably no effect on universe beyond the observable horizon that would need to affect our models.


How long can you be trapped orbiting around a black hole for?

You're right to mention time dilation, because the point of view is important. The standard metaphor is that of an astronaut (A) falling towards the event horizon, wearing a watch, while another astronaut (B) watches from a distance. Astronaut A will simply fall through the horizon and hit the singularity, but B will never actually see A cross the event horizon, instead they will see A's watch run slower and slower.

In terms of survival time, bigger is better. Getting too close to a really small black hole could rip you apart with tidal forces without even having to cross the event horizon (just as moons and planets exert tidal forces on each other) - but with a supermassive one you'd not even notice you'd crossed it.

For a relatively "standard" black hole of about 30 solar masses the infalling astronaut will have around 0.0001s between crossing the EH and hitting the singularity (from their point of view....see above, B will never see them cross the EH).


Difference between J2000, FK5 and ICRS coordinate systems? Which one does the Yale Bright Star Catalog use?

FK5 is based on optical observations. ICRS is based on radio data. Radio observations (ironically) are much higher accuracy than optical through the application of long baseline interferometry.
Other than extremely accurate astrometry (fractions of an arcsec), they are essentially identical.


I think I get it - you are trying to calculate a luminosity based on some assumption about how much light is reflected/scattered from the Sun towards the Earth - hence the relevance of the flux from the Sun at the comet.
I think it's very hard to adopt this approach. There is uncertainty about what is the appropriate geometry to use for the comet and how the reflection/scattering should be treated. It is doubtful that assuming a comet diameter of 5 km will be appropriate because I would have thought most of the light we see is scattered from a coma around the comet, not from the surface of the comet itself.


North Pole Right Ascension/Declination to axial tilt conversion

This is wrong, but may help someone find the right answer:

You can convert right ascension and declination to a 3 dimensional
unit vector in the J2000.0 ICRF reference frame using the standard
formula for converting spherical coordinates to rectangular
coordinates and using a radius of 1:

{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}

The ra and dec of the north ecliptic pole is ra,dec of
{270, 66.5607083333} per Wikipedia, so the unit vector representing
the north ecliptic pole is:

{0, -0.3977771648286046, 0.9174820582119942}

As it turns out, there is conflicting data for Ceres, perhaps because
of the semi-recent DAWN flyby.

Instead, I'll use Vesta, where wikipedia explicitly states (https://en.wikipedia.org/wiki/4_Vesta#Rotation):

north pole pointing in the direction of right ascension 20 h 32 min,
declination +48 [... which] gives an axial tilt of 29 [degrees]

Converting to degrees, Vesta's north pole's ra,dec is
{308,48}. Using the formula above to find the unit vector, we get:

{0.411957936296447, -0.527282113378167, 0.743144825477394}

If we take the dot product of two vectors and divide by the product of
their lengths, we get the cosine of the angle between them. In this
case, both vectors have length 1, and the dot product is 0.891563
whose arc-cosine is right around 27 degrees.

So, if Vesta's orbit were the same plane as Earth's orbit, the axial
tilt would be 27 degrees.

However, since Vesta's orbit is inclined 7.14043 degrees to the
ecliptic, this answer is incorrect.

I don't think you can find the correct answer without using Vesta's
inclination and the longitude of its ascending node.

Both of these values are known, but I can't figure out how to use them
to get the correct asnwer.


Has this image of the Draco Constellation been altered

The image of Draco has definitely been altered.  Most of the stars in Draco are quite faint, even the ones making the dragon shape, whereas in the image they are all shown as bigger/brighter than the stars in Ursa Minor (the quadrilateral in the middle with 3 stars trailing down to the lower left.  The last of the 3 is polaris, which is in reality brighter than all the stars in Draco.

The same is true for Kocab.
The image of Orion is typical of probably thousands taken by amateur astronomers.  If you want to find others, try a search engine.  There may be some specific sites that collect all or most of the constellations, but I'm not aware of them.


Exoplanet dip in transit light curve when the planet passes behind the star

Just before the planet goes behind the star, we see the light directly from the star as well as the light reflected from the planet's surface.  When the planet is behind the star, we no longer see the reflected light.
Note that the top of the line also curves to reflect how much of the planet's surface is illuminated from the star.  The most illumination is when it is almost "full" just before it passes behind the star.  The least illumination is when it is "new" just before it passes in front of the star.
Edit (to address Fraxinus's comment below): Here is an example from Zhang et. al: Phase Curves of WASP-33b and HD 149026b and a New Correlation between Phase Curve Offset and Irradiation Temperature.

The above graphic shows processed data from the Spitzer space telescope.  From the paper:

Observations were timed to begin before a secondary eclipse and end
after the following secondary eclipse,

These light curves help aid in estimates of planetary albedo, day/night temperature, and atmospheric composition.


Was the radiant of the Perseid meteor shower ever near Vega thousands of years ago?

Okay, this turned out to be quite a long answer describing a few arguments I've com up with. The short answer is: maybe, maybe not, but I think it's possible. There are arguments to be made for either side of this discussion, and I don't think you you can ever be absolutely sure unless you find an expert on both astronomy and Chinese folk tales (which I don't claim to be), or invent a time machine to go back 2600 years and check for yourself.

The most obvious argument to be made in favor of the hypothesis that these magpies are the Perseid meteor shower, is that folk tales referencing stars are very likely to be based on astronomical occurrences during that period (such as a meteor shower, perhaps).

The (in my mind) most obvious argument against your hypothesis is what you suggest yourself: Perseus is quite far away from Vega and Altair. However, Perseus is the radiant of the shower, which means that all perseids will appear to come out of Perseus. Because of this, it is perfectly possible for a lot of perseids to appear between Altair and Vega, parallel to the Milky Way. It is, however, impossible for them to go from Vega to Altair, or vice versa, since they wouldn't come out of Perseus anymore.

You suggest that this could be explained by the precession of the equinox, but a quick lookup on Stellarium (which is a great (and free) tool if you're into astronomy) tells me that even 2600 years ago Perseus was quite far away from Vega and Altair (actually, that's quite obvious, since the precession of the equinox causes stars to move relative to the extension of the north pole, but not relative to each other, as far as I know).

Another argument to be made against this hypothesis is that the perseids are by far not spectacular enough to be seen as a bridge over the Milky Way, even when they all appear in the required spot. During the yearly peak of the perseids, you are expected to see about 1-2 meteors per minute, which can be noticeable at best. This argument is then countered by the fact that the ZHR (how much meteor you see per hour) is not the same every year, but can peak on years when Earth passes through a fresh dust cloud left behind by the comet that causes the shower (which, in this case, is the comet Swift-Tuttle, if you're interested). Therefore, it is perfectly possible that the perseid meteor shower was extremely active 2600 years ago, causing the creation of this story. Such peaks are somewhat rare, but not too rare for this hypothesis to actually make sense (in fact, we will be seeing such a peak in 2028).

The fact that the dates of the festival and the peak of the perseids are a few days apart and not exactly the same is not quite an argument against your hypothesis, since it is perfectly possible for the date of the festival to have moved slightly due to differences between the (old) Chinese and Western calendars. I'm not an expert on calendars, however, so I can't say that for sure.

Given all these uncertainties, your hypothesis (along with what you mentioned yourself in your own comments) seems plausible, apart from one thing: we don't know whether Swift-Tuttle was already active 2600 years ago. If it wasn't, the perseid meteor shower obviously did not exist (note that comets don't exist forever: some gravitational variation causes them to exit the Oort cloud and enter the solar system). We don't know when exactly Swift-Tuttle started being a comet.

That is only a weak argument, however, and when it comes to astromical hypotheses, the most convincing argument is usually a principle which is referred to as Ockham's razor: if given multiple explanations, choose the one that requires the least amount of assumptions (or the most plausible ones).

Personally, since the date of creation (2600 years ago) is very vague (the uncertainty on this must be at least a hundred years), I think it is perfectly possible to assume that the perseids peaked somewhere during that period, causing the creation of this story.


What wavelength to best detect the "9th planet"?

The possible planet 9 is thought to be about 10 Earth masses and is unlikely to be a gas giant (it may be the core of an "interrupted" gas giant). As such, it will not be generating significant luminosity itself and would be rocky, or more likely, icy in character. It would thus only be seen by reflected light.

The considerations for what wavelength to search in balance the sensitivity of the instruments at hand with the likely spectrum of the object. This in turn depends on the solar spectrum and the wavelength dependence of the reflectivity (albedo).

For most icy objects, including Pluto and Trans-Neptunian objects, the reflectance increases to the red and near-infrared, whilst the solar spectrum peaks at shorter wavelengths. This suggests that searches are best carried out with wide field optical instruments in the R or r' bands at around 600 nm.

A further factor in finding a candidate is that you are going to have to cover a large area. This is only feasible at optical and NIR wavelengths unless the object was bright enough in the mid-IR to show up in WISE (which I'm sure is being thoroughly checked). A press release I saw said SUBARU is being used for the search. I would bet they are using the half degree field of Suprime-Cam at optical wavelengths and not pursuing COMICS mid-IR imaging with it 42x32 arcsecond field!

Confirming a candidate should be easy, given the enormous parallax and proper motion expected.


The simple answer is that light from the distant parts of the Milky Way Galaxy have already been traveling for that amount of time. Because light does not travel instantaneously through space, we are seeing the galaxy not as it is now, but as it was 100,000 (or so) years ago.

Incidentally, this is also how we are able to see back to earlier moments of the universe's history. In terms of distance, the further we look, the longer light would have had to leave in order to reach us today.

The universe did have a beginning (at least in the sense of having an earliest time). Because the universe is not infinitely old, there is actually a limit to how far we can see back into the universe's history (see: Cosmological horizon). We are able to make maps of the universe, like this one from SDSS:

by running what's known as a survey. These are large projects which maps out the positions (and potentially other properties, depending on what their science goals are) of galaxies in the universe. How you should look at this picture is the following: 1) You're at the center of the image, 2) Each point in the picture represents an individual galaxy, 3) Distance from the observer, the redshift of a galaxy, increases with increasing distance from the center, and 4) as you travel around the circle in angle, you're sweeping through the right ascension coordinate of the celestial coordinate system (think longitude). You're seeing a pizza slice instead of a sphere because this is a particular slice in declination (lattitude). As you look to higher redshifts, you are looking further back into time, and earlier into the universe's history. This is how astronomers/cosmologists can say something statistically about models of structure formation of the universe (how objects like galaxies, and clusters of galaxies came to be).

The problem with looking at galaxies at very high redshifts is that they become
very faint. At a certain point, you need more powerful telescopes to continue looking further back in time. Though, we can get around this somewhat by looking at the radiation component of the universe at microwave wavelengths (most surveys are in the optical region of the electro-magnetic spectrum). This radiation is called the cosmic microwave background radiation (CMB), and can also be mapped out (see: COBE,WMAP, and most recently Planck satellites).

This is also a map of sorts. This map of low energy photons is a snapshot of the universe approximately ~380,000 (redshift of z=1100; meaning the universe was smaller by about a factor of 1101) years after the big bang, when the universe had very little of what we would recognize as structure. The job of the cosmologist is to connect the picture we see from the CMB to the picture we see from our surveys of galaxies through the use of physics.


In fact time always "moves" at the same speed to any given observer. It appears to move at different speeds to an external observer, but you cannot experience slowed time as such - except in the sense that you might observe external clocks ran faster than the watch you had your wrist.

An attempt to explain: The supposition - very strongly supported by the experimental evidence (the latest discovery of gravity waves being the jewel in the crown) is that we live in a four dimensional spacetime. An individual always experiences time as "moving" at a constant speed, but as has been remarked by others, externally gravity appears to add curvature to the "straight line" we are moving along on the time dimension and so we "gain distance" in this direction more slowly when seen externally.


How does the Eddington limit make the gravitational field of stars asymmetric?

Hot, massive stars tend to be (a) asymmetric, due to their rapid rotation (at least on the main sequence); and (b) very luminous, and thus nearer to the Eddington limit, but the two aren’t causally connected to each other.  So perhaps that answer was using correlation in place of causation.
The rapid rotation is due to the fact that hot stars lack the dynamo-driven magnetic fields that let lower-mass stars shed angular momentum and spin down over time.  The high luminosity is just a natural consequence of maintaining the higher temperature and pressure needed to support the higher mass, i.e. to maintain hydrostatic equilibrium.
It is closer to being true the other way around: because of rapid rotation, the effective gravitational field at the star’s equator is lower, and thus that part of the star will be closer to the Eddington limit at a given luminosity than a non-rotating star would be.


Why can we see stars but not astronaut on the moon

An astronaut on the moon could only be seen by reflecting the Sun's light towards Earth. Stars on the other hand emit their own light.

To first order, the amount of flux incident upon the Moon from the Sun is the same as that at the Earth - about 1.4 kW/m$^{2}$.

Let us assume that an astronaut is perfectly reflective and that the relevant reflective area that we can see from the Earth is 1 m$^2$. NB: If the astronaut is not lit up by the Sun, then there is obviously no way that they can be seen.

Treating the astronaut on the Moon as an isotropic point source emitter of reflected light, we have a light source of power 1.4 kW at a distance of 400,000 km. The flux at the Earth is therefore $7\times10^{-16}$ W m$^{-2}$.

How does that compare with starlight? Well, the total luminosity of the Sun is $3.8\times10^{26}$ W. It has an absolute magnitude of 4.8. This means that if we put the Sun at a distance of about 20 pc, it would be about as faint as the faintest naked eye star in the sky. The flux received at the Earth from such a star would be $8\times 10^{-11}$ W m$^{-2}$ and thus 100,000 times brighter than the astronaut.

No need to worry about the resolution of the eye, since both the star and the astronaut (at the distance of the moon) are unresolved points.

Also no need to go into the problems of contrast against the moon's bright surface (which you would need to consider if an astronaut's reflective area was 100,000 times bigger), the reflected light from the astronaut is just too faint to be seen at that distance.


Is time taken into account by astronomers when creating and using maps of the universe?

For the first part of your question, @userLTK has already provided a very good explanation, and I believe that by introducing redshift, the second part of your question is also partially answered. I will try to expand a little on it, hoping that I have understood your question correctly. To start off, I'm going to go ahead and quote my own thesis (not yet published);

  [W]hen moving to celestial distances we quickly realise that time measurement ambiguity also arises. To combat this we tend to not describe distances to objects directly, but relate these properties to an object’s relative line-of-sight velocity with respect to us.

This is why, when NASA earlier this year announced the Hubble Space Telescope having observed the most distant galaxy yet, the distance to the galaxy was never mentioned. The distance is not a direct observable, and has to be calculated based on certain assumptions. One direct observable we do have, is the line-of-sight velocity, which we express in terms of redshift. By describing objects in terms of redshift, we simultaneously have a "distance" measurement to the objects, as well as knowing how fast they move away from us.

When we make observations and try to learn more about things like the energy content of the universe (radiation, regular matter, dark matter, dark energy), the redshifts of the objects we observe are crucial. We consider various models of the universe, and see how they compare with what we observe on different redshifts. I very quickly found back to a paper written by my supervisor some years back; figure 7 in there shows you that kind of comparison.

Personally, I am working on something which I hope will lead to us being able to see whether dark energy evolves over time. In doing that, knowing the redshifts of the objects are crucial to my work, and I constantly need to keep in mind the implications of the distant objects' movement away from us. It is simply a factor I need to multiply or divide by in my calculations, to make sure I am comparing the correct things with each other.


How to convert the observed magnitude of a Quasar to it's bolometric luminosity?

The bolometric magnitude of a quasar can be determined by the same method as a star (using total energy radiated)......but for the fact that quasars do not radiate isotropically and this distribution is dependant on wavelength
A further complication comes as many photons are absorbed and re-emitted , etc
Some standard methods of calculation can be found here......
Quasar bolometric corrections


Account for radiation energy in the statement "total energy of universe is zero"

The total energy of the universe consists of the mass energy of all the matter (both normal and dark), the mass-energy of the radiation and of the dark energy plus the gravitational potential energy of the universe (which is negative).  The hypotheses is that this sums to zero.  As far as I know, there is no experimental evidence for this.

But there is good evidence for the size of the various mass-energy terms.  The best current estimates are that the mass-energy of the universe is:

73% -- Dark energy
23% -- Dark matter
3.6% -- Interstellar gas
0.4% -- Stars (includes black holes)
v. small -- Relativistic neutrinos
v. small -- Radiation (CMB and stellar radiation combined)

This balance changes.  As the universe expands, the density of matter (dark and normal) varies as the inverse 3rd power of the size of the universe. (Matter stays the same and the volume expands as the cube of the radius.)

But the density of dark energy stays constant, since it is a property of empty space.

The density of radiation is proportional to the inverse 4th power of the radius, because not only does the increased volume dilute the radiation like it dilutes the matter, but also the radiation gets red-shifted, contributing an extra power of the radius.

(Incidentally, that means that the fraction of the total mass-energy of the universe in radiation decreases with time, and is estimated presently to be 1000:1 matter:radiation.)

So radiation seems simply to be too small to matter.


Seeing a galaxy (quasar) greater than 46.6 billion light years away

The edge of the observable universe is actually 46.6 billion light years away, despite the Big Bang being only 13.8 Billion years ago. This is because the light which we are now receiving as the furthest visible stuff had to travel through ever expanding space in between, being redshifted down into what we call the Cosmic Microwave Background Radiation (CMBR). There is a little bit further than that which we are technically receiving, but it has been redshifted infinitely.

To see anything further away than 46.6 Bly, it would have had to existed literally before time itself, or travelled faster than the speed of light. Two highly improbable things


Why is Gaia operating around Earth orbit? Why not send it to Neptune's orbit?

Well, you thought about the spatial aspect of a parallax measurement, but not about the temporal one.

Gaia's intention is to measure 3D positions as well as 3D velocities. For the distance, you need accurate parallactic measurement, which come in with your orbital period.
For a typical Gaia-star with several measurement per year, you'll get 5 values of the parallax after 5 years of time, which you then average. If you'd send Gaia towards Neptune (besides the fact that no one has ever sent an orbiter, to say nothing of a L2 mission that far out) that has a period of 168 years, then after 5 years you'd get... 5/168 th of one paralactic measurement.

It simply couldn't achieve its science goals if put around the L2 behind Neptune. Also no one on this planet has any experience in putting something into a outer system L2 point. This is different than putting it into Earth's L2, because reaching the L2 around one of the giants has vast and very precise $\Delta v$ requirements. This would be a massive technological leap, and things don't work that way in space. Small, incremental technological steps are required in an anyways unfriendly environment, to make sure everything works properly and no millions of dollars have been wasted.
Compare that to Gaia's predecessor, the Hipparcos satellite, which was parked in geostationary orbit.

Now you could still say, why not use Jupiter hypothetically anyways. Well, the orbital period there is still 11 years, and Jupiter's L2 still suffers from the intense radiation environment that is provided by Jupiter's magnetosphere. This would lead to rapid degradation of the CCDs used for scanning across the sky.


The Sun's radiation spectrum is close to thermal at $T_\odot = 5778\,\mathrm{K}$. That means that that power output, given by the Stefan–Boltzmann law, is around $L = \sigma A T^4$, where $\sigma = 5.6704\times 10^{-8} \,{\mathrm{W}}/{\mathrm{m}^2\mathrm{K}^4}$ is the Stefan–Boltzmann constant and $A$ is its surface area:
$$L_\odot = \sigma(4\pi R_\odot^2)T_\odot^4 = 3.85\times10^{26}\,\mathrm{W}\text{.}$$
However, at the Earth's orbital radius $r = 1\,\mathrm{AU}$, this radiation is spread across a surface area of $4\pi r^2$, so the power incident on an Earth-sized sphere would be just
$$P = L_\odot\frac{\pi R_\oplus^2}{4\pi r^2}\text{,}$$
so approximating the Earth as a spherical blackbody, the equilibrium temperature is determined by its own thermal radiation balanced by this incident radiation:
$$\sigma(4\pi R_\oplus^2)T^4 = P\text{,}$$
which gives a surface temperature of
$$T =  T_\odot\sqrt{\frac{R_\odot}{2 r}} = 279\,\mathrm{K}\text{.}$$
Therefore, no special mechanisms are necessary to prevent the Earth from frying.

In reality, the Earth is not a blackbody because of its albedo and atmosphere, but its actual global temperature is not too far off from this estimate. The average surface temperature is close to $288\,\mathrm{K}$. Earth's atmosphere actually raises the average temperature, not lowers it.


What record do we have of the length of supernovas?

I would expect that the further away a supernova is, the "longer" we observe it will be active because of an increased red shift.  Objects in the universe that are further away are generally going away from us faster than objects that are close, hence the increased red shift.  This increase in time due to redshift is very similar to the Doppler effect, where a siren is high pitched (and faster) when approaching you (e.g. blue shift with light), and low pitched (and slower) when going away from you (e.g. red shift with light).  The faster the siren is moving away, the more the effect (both lower pitch and longer time) will be increased, likewise with light.

Here are some sources that may help with what you are looking for (although I haven't been able to find a list exactly like you are looking for):

http://www.wolframalpha.com/input/?i=supernova

http://en.wikipedia.org/wiki/Supernova#Light_curves

http://en.wikipedia.org/wiki/Cosmic_distance_ladder#Type_Ia_light_curves

http://curious.astro.cornell.edu/question.php?number=49


If our galaxy is mostly empty space then why does it look so solid?

You are mostly empty space.  Every atom in your body is very tiny compared to the relatively vast spaces between it and its nearest neighbors.  And the same goes for every "solid" object you ever saw.
But when you look at yourself in a mirror, the number of atoms you see is so incredibly vast that you don't see the spaces between them but instead think that yhour body is a solid object.
And probably the wavelengths of visible light are too long to fit between atoms and so get bounced back from the surface of your body.
That is not the case with the galaxy, since the spaces between stars in the galaxy are wide enough for all wavelengths of light to pass through instead of being reflected.
But if you look at our galaxy from a distance of about one million light years, the average separation between stars of about 6 light years will be a very small angle.  I believe that at one million light years distance, one arc second will be about 4.848 light years wide.

The maximum angular resolution of the human eye is 28 arc seconds or 0.47 arc minutes,[19] this gives an angular resolution of 0.008 degrees, and at a distance of 1 km corresponds to 136 mm. This is equal to 0.94 arc minutes per line pair (one white and one black line), or 0.016 degrees. For a pixel pair (one white and one black pixel) this gives a pixel density of 128 pixels per degree (PPD).

https://en.wikipedia.org/wiki/Visual_acuity#Physiology[1]
So at a distance of one million light years the smallest detail a human could see would be 28 arc seconds, or 135.7 light years wide.  So such a smallest possible thing you could see would be a square about 136 by 136 light years wide, and possibly containing about 350 stars in a single layer and all the vast interstellar spaces between them.
Furthermore, galaxies are not a single layer deep.  Depending on the size of a galaxy, the distribution of stars within it, and the angle from which it is viewed, a viewer might see  tens, hundreds, or thousands of layers of stars one behind the other.  The stars are not arranged in any sort of regular lattice pattern, but orbit around the center of the galaxy in separate orbits and their relative positions slowly change, thus making their relative positions random.
Thus the empty spaces between stars in the outermost layer of a galaxy will show the light of stars farther behnd them.
So when you look at galaxy from a distance of one million light years, the smallest part of the galaxy you can see will have the light of hundreds or probably thousands of stars blended together with the black space between them to make a pale grayish light.
The farthest objects which humans can see with the unaided eye are the Andomeda Galaxy M31 about 2,540,000 light years from Earth, and the Triangulum Galaxy M33 about 3,200,000 light years from Earth.  Since they are several times a million light years away, the smallest part of them which can be seen with the unaided eye will combine the light of even more thousands of stars together.
Astronomers have photographed galaxies tens of millions, hundreds of millions, billions, and tens of billions of light years distant.  With astronomical telescopes they can detect some individual stars in galaxies tens of millions of light years away.
But any photograph which shows an entire galaxy will show the light of millions or billions of stars blurred together into a blob of light.


What would happen to the Galilean moons and Titan if Jupiter and Saturn disappeared?

They might escape from the solar system, if the angles are right. If not, they'll probably wind up in elliptical orbits around the Sun. We'll use a simplifying assumption that the orbits are circular to make the calculations easier; All objects mentioned have orbital eccentricity of less than 0.05.

The mean orbital velocity of Jupiter is 13.1 km/s, which makes Solar escape velocity at Jupiter's mean distance from the sun 18.5 km/s, a difference of 5.4 km/s.

The mean orbital velocities of the Galilean Moons of Jupiter are:

Io: 17.3 km/s
Europa: 14.3 km/s
Ganymede: 10.9km/s
Callisto: 8.2 km/s.

Which means, if the geometry is right, with the direction of travel of the moon in the same direction of the direction of travel of Jupiter at the time of the disappearance, the resulting vectors could add up to any of the Galilean moons exceeding solar escape velocity if Jupiter suddenly vanishes, and exiting the solar system.

Doing some trigonometry to find the limiting angle:

Io: Reaches Solar escape velocity  if the moon was traveling within 106° of Jupiter's direction.
Europa: Escapes if traveling within 95°
Ganymede: Escapes if direction within 79°
Callisto: Escapes if direction within 62°

There's also a tiny, tiny chance of Europa or Io being in the right positions and having the right direction to hit the Sun if their directions at the time of disappearance were exactly right, but the chance of that is extremely unlikely.

For comparison, Saturn has a mean velocity of 9.5 km/s, (Solar Escape Velocity at that distance is about 13.4 km/s) and Titan has a mean velocity relative to Saturn of about 5.5 km/s.

Titan could escape under the right conditions (Within 56° of Saturn's direction of travel), but doesn't have enough Saturn-relative velocity to have a chance of crashing into the Sun without help.


It would appear to have been developed by Carl Pulfrich working for Zeiss in 1904.

Alternatively Max Wolf in 1900 again working with the Zeiss company.

Looks like the idea was Wolf's and the realisation Pulfrich's. From the second link we have:

  Wolf was a codeveloper of the stereo comparator together with Carl
  Pulfrich from the Zeiss company. The  stereo comparator consists of a
  pair of microscopes arranged so that one can see simultaneously two
  photographic plates of the same region taken at diff erent times. Wolf
  seems to have experimented with such techniques as early as 1892, but
  without success. When Pulfrich approached him to adapt the technique
  from geodesy to astronomy, Wolf was delighted. A steady exchange of
  letters followed. Wolf and Pulfrich then worked together to analyze
  the rapidly growing accumulation of photographic plates. Tragically,
  Pulfrich lost one eye in 1906, preventing him from using the
  stereographic tool from then on.


Gain in electronics can have different meanings:

Gain is a measure of the ability of a [...] [device] to increase the power or amplitude of a signal from the input to the output port. [...] The term gain alone is ambiguous, and can refer to the ratio of output to input voltage (voltage gain), current (current gain) or electric power (power gain).

For optics, I found the light-gathering power the most appropriate measure, which relates the apparture of the telescope with the one of the human eye:

The light-gathering power of an optical telescope, also referred to as light grasp or aperture gain, is the ability of a telescope to collect a lot more light than the human eye. Its light-gathering power is probably its most important feature. [...]
The gathering power $P$ compared against a human eye is the squared result of the division of the aperture $D$ over the observer's pupil diameter $D_{p}$ [...] An example gathering power of [a telescope with] [...] $254\,{\rm mm}$ [aparture] compared to an adult pupil diameter being $7 \, {\rm mm}$ is given by:
$$P = \left(\frac {D}{D_{p}}\right)^2 = \left(\frac {254}{7}\right)^2 \approx 1316.7$$

Edit: After the question changed slightly, I guess I should add a bit on "how to calculate gain from an image": Gain requires always two measurements, e.g. the image from your telescope and a base of comparison, e.g. the human eye, that's why I found the aperature-based-gain definition appealing.
If you have a datafile with intensities from your telescope - basically an image in grayscales depicting the spatial light intensity distribution - you would have to compile an image of the same area of the sky which serves you as a base of comparison.


Astronomy as a profession for moderately good software engineer

A profession in astronomy can entail various things.  Unlike what other people are saying, you don't have to have a PhD in astronomy to do work in this field, to be an astronomer.  There are other options, but you will have to work hard.

Study as much physics and mathematics as possible.  Astronomy is essentially a branch of physics.  If you're still in school, then take classes in these areas.  If you're not in school anymore, then there are plenty of online resources to continue your education.  This will allow you to understand what people are talking about when they discuss technical things.
Decide what exactly you're interested in.  There are lots of different areas of study within astronomy.  Decide what appeals to you the most.  Do you like planets?  Galaxies?  Cosmology?  All of these different focuses require more specialized knowledge.
Your skills are already marketable.  Being an astronomer can mean lots of different things.  There are some astronomers who hardly ever look through telescopes, and spend most of their time reducing data.  A large part of astronomy is computer programming.  As a software engineer, you may already have the skills you need to work with astronomy research.
Keep an open mind.  There are lots of jobs in astronomy research where you might not be called an "astronomer".  For example, as a software engineer, you could find work with NASA supporting space research.  Your job title might not be "Astronomer", but you'll still be in the field.
Be prepared for hard work.  Being an astronomer can be difficult, not just because of the subject matter, but also because of the effort you have to put in to get a job.  It can be relatively difficult to find an employer looking for people with your particular skillset.  My recommendation would be to keep an eye on the NASA Jobs website to see when they're hiring.  You can also search for internships on NASA's OSSI site.
Try some amateur astronomy. There are lots of ways you can get involved in astronomy without having it as a job.  A quick Google search for "Amateur astronomy projects" yields many different results.
Join a local astronomy group. If you live in a reasonably sized city, there should be an astronomy group near you.  Find them and join them.
Don't give up! Having astronomy as a profession can be difficult, but it is also rewarding.  If you truly want to do this, you'll need to work at it.  Just don't give up!


Declination and Ascension - the Sun and Andromeda

If you mean solar time midnight than the answer is simple (as commented by @rgettman). The Sun should then be at the opposite side of the sky at $\alpha = 13^{\textrm{h}}$, where $\alpha$ is the right ascension. There is a problem, however, when you use local time (using a timezone). If, for instance, you mean midnight GMT in Liverpool, which is at 3° west then solar midnight will happen at 12 minutes past midnight GMT and the Sun would have $\alpha \approx 13^{\textrm{h}}12^\prime$.


How can the life time of a multiple star system, such as for example the trinary system PSR J0337+1715 be derived?

Such a three-body system is in principle expected to show a chaotic behavior sooner or later. Nope. Hierchical multiple systems (such as this), where the semi-major axes differ by a factor ten or larger may well be stable for ever (never become chaotic), in particular if the eccentricities are low and if the most massive object is in a tight binary.

An unstable three-particle system will eventually result in (typically) the two most massive objects in a tight binary and the third particle ejected (unbound). The time scale for this to happen is of the order of several (10-100) dynamical times and is indeed a highly chaotic process.

The concept of the Lyapunov time scale is not too useful here. One issue is that as soon as one object is ejected (unbound), the system is no longer bounded, when the concept of Lyapunov becomes problematic. Another issue is that the Lyapunov time is defined in the limit of infinite time and does not necessary reflect the behaviour of the system over any finite time.

Finally, to answer your question. I think there is no stringent way. What one can do is to integrate numerically many realisations of the system, each equally in agreement with the data (and their uncertainties). Then one can see whether there are any stable configurations and how frequent they occur. Given that the system didn't form yesterday, it appears likely that it is indeed stable.


Determining distance from semi-major axis and eccentricity

Supposing that the mass of the object is negligible compared with the mass of the Earth, you can derive the orbital period $T$ from the 3rd Keplero's law:

$\frac{T^2}{a^3} = \frac{4\pi^2}{G(m_E + m_b)} \approx \frac{4\pi^2}{Gm_E},$

where $a$ is the semi-major. With $T$, for each time istant you also know the mean anomaly $M$, given by (suppose $t = 0$ at perigee):

$M(t) = \frac{2\pi}{T}t$.

Solving numerically the Keplero's equation for the eccentric anomaly $E$ (where $e$ is the eccentricity)

$M = E - e\sin E$

and then use the following equation to derive the true anonaly $\nu$, which is the angle between the direction of periapsis and the current position of the body, as seen from the Earth:

$ \cos \nu = \frac{\cos E - e}{1 - e\cos E}$
and
$ \sin \nu = \frac{\sqrt{1-e^2}\sin E}{1 - e\cos E}$.

The distance from the Earth is just given by the orbit equation

$ r = \frac{a(1-e^2)}{1 + e\cos \nu}$.

If I'm not wrong with calculation, it should be:

$ T = 9364 s = 2.6 h.$

$ M(90min) = 207.60°$

$ E(90min) = 203.11°$

$ \nu(90min) = 198.95°$

$ r = 11'366 km$

To get the covered distance you should calculate the line integral of the orbit equation.


$\nu_0$ is the frequency you would expect for the absorption/emission line in the absence of any broadening. i.e. It is the centre of the line profile.

$\delta \nu$ is just $\nu - \nu_0$. (i.e. it is the separation in frequency from the centre of the line profile and is equivalent to $f -f_0$ in the second expression).

$\Delta \nu_{th}$ is a measure of the broadening due to (I guess) thermal motions in whatever gas/plasma is emitting/absorbing the radiation. It is similar to the Gaussian "sigma" term in the expression
$$ f(x) = \exp(-x^2/2\sigma^2)$$

If you have a gas that follows Maxwell-Boltzmann statistics, then $\Delta \nu_{th}$ will be related to the speeds of atoms/ions/molecules in the gas and hence to the temperature of the gas.
$$ \Delta \nu_{th} \simeq \sqrt{\frac{2k_BT}{m}}\times \frac{\nu_0}{c} \,$$
where $m$ is the mass of the particle emitting the light.


Are stars NML Cygni, UY Scuti, VY Canis Majoris and VV Cephei near the ends of their lives?

NML Cygni and VY Canis Majoris
Zhang et al. (2012) considered several values for $T_{\text{eff}}$ (effective temperature) and $L$ (luminosity) and found that NML Cygni matches up with the evolutionary track of a star of mass $\sim25M_\odot$, near the HR-diagram location of a similar star, VY Canis Majoris (denoted as VY CMa). Using these data points, the model estimates that NML Cygni is approximately 8 million years old and in the post-main sequence phase of its life.
However, NML Cygni is thought to be related to the Cygnus OB2 association, which has an age of 2-3 million years. This is strange because all the stars in the association should be nearly the same age  (see Knodlseder (2008)). That age may be questionable, however, as evidence for older stars nearby has been presented (see Wright et al. (2010)).
The exact location of VY Canis Majoris on the evolutionary tracks is not certain. Massey et al. (2006) created models that placed it on a lower-mass track to help it avoid the "forbidden zone" beyond the Hayashi limit.
VV Cephei
Based on standard parameters, stellar wind measurements and Geneva grids, Bennett (2010) suggested that the red supergiant component of VV Cephei could be near the end of its life, assuming a mass of $\sim20$-$25M_\odot$.
UY Scuti
Wikipedia claims

UY Scuti should fuse lithium, carbon, oxygen, neon, and silicon in its core within the next million years. After this, its core will begin to produce iron, disrupting the balance of gravity and radiation in its core and resulting in a core collapse supernova.

As far as I know, that claim is made only through rough estimates of evolutionary tracks, not precise calculations. We do know that UY Scuti is certainly in the later period of its hydrogen-burning phase, if it is indeed at that point in its life. Arroyo-Torres et al. (2013) believe observations place it near evolutionary tracks of stars originating with masses of $\sim25$-$40M_{\odot}$. Therefore, it may have already lost a substantial fraction of its original mass - not surprising for a red hypergiant.


Neptune and Uranus have estimated core temperatures of 7273 K and 5255 K respectively. Uranus and Neptune however likely do not have liquid water oceans. The combination of pressure and temperature on both planet results in unsuitable conditions for liquid water oceans. In fact for liquid water oceans to form on Neptune, it's predicted it would have to cool down.

Paper on this at: Cornell University

(Irrelevant, but fun) Diamonds oceans might exist. Uranus and Neptune are made up of 10% carbon some of which is methane. Methane when exposed to the conditions available in Uranus and Neptune can form diamond. Diamond can melt and form liquid diamond under again, the conditions on Neptune and Uranus. Therefore there could potentially be diamond oceans on Neptune and Uranus.

Paper on this at; Nature Materials

Two plausible theories considered for Neptune source of heat is that as you said, the methane concentration in the atmosphere is high enough to retain the formation heat after the formation of the planet. The other one is that the tidal forces generated by Triton's retrograde degrading elliptical orbit, but the paper on it concludes that it doesn't explain the difference between Neptune and Uranus.

Another thing to consider is that Uranus might be cooler than it should be rather than Neptune being warmer. This could be a result of an Earth sized
indirect collision earlier in its lifetime. The collision causes this decrease in emitted temperature because it deposited a significant amount of material around the core which made heat transfer from the core to the outer parts of the planet much slower. Note that if this collision were to occur, it is what would've caused the unusual tilt of 97.7°.

Paper on this collision at: The American Astronomical Society

To the second question, the methane in the atmosphere of Neptune, as covered before, can produce a significant greenhouse effect.

  Neptune emits 2.7 times more energy than it receives from the Sun. This access energy powers the atmosphere to produce the storms that are not seen on its twin planet Uranus. The source of internal energy cannot be due solely to leftover energy from formation (i.e. Jupiter) since Neptune is smaller and would have radiated away the energy long ago. Nor is it due to an unusual chemical change, such as the helium rain for Saturn. Rather, it appears that Neptune is more efficient at trapping leftover formation heat due to the fact that methane is highly abundant in Neptune's atmosphere, and methane is an excellent insulator of heat (i.e. the greenhouse effect). Neptune has a sub-zero type greenhouse effect that is trapping formation heat that should have been radiated billions of years ago like Uranus.

This extract is from:
University of Oregon

In regards to why Uranus atmospheric methane doesn't appear to have as much of an effect as on Neptune, it could be that either Neptune simply has more methane in it's atmosphere or that a combination of factors, some of which mentioned above causes the difference. Beyond that I do not know.

Paper on Tritons Tidal Heating:
University of Texas


How is the field of view of a radio telescope determined?

Single "pixel" dish
If you have a single point feed at the focal plane (one single-mode waveguide or one $\lambda/2$-sized antenna) then the antenna's reception pattern (or radiation pattern) will be similar to it's diffraction pattern. For a circular aperture and ignoring perturbations due to the small secondary or its supports this pattern will be the Airy disk.
The half-angle of the first minimum is given by the familliar equation used to express the resolution of any telescope, optical or radio:
$$\theta = 1.22 \frac{\lambda}{d}$$
where $\lambda$ is the wavelength and $d$ is the diameter of the telescope's aperture.
So if you have a 20 meter dish and you are receiving a signal at 3 GHz, your wavelength is 0.1 meters and the half-width of the angular acceptance is 0.006 radians or 0.3 degrees.
Arrays of telescopes
If you have an array of radio telescopes and you are using interferometry to generate a synthetic aperture (like the VLA or ALMA or the Event Horizons Telescope (EHT)) then your field of view can be much smaller but it is defined in the software running in the correlator computer.
Dish with a focal plane array
Some radio telescope dishes are equipped with an array of receivers, analogous to a CCD array of pixels on an optical telescope, though much fewer pixels.
In this case the half-width of the field of view (in radians) is the half-width of the focal plane array divided by the effective focal length there, or more precisely the arctangent of that ratio.
For more on that see:

What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE?
How do ASKAP's focal plane phased array feeds interact with the entire array phasing?

Cropped from CSIRO ScienceImage 2161 Close up of a radio astronomy telescope with several more in the background


The various moons are on orbits that are mostly a lot more than 1km apart from other moons. The orbits are stable, so the moons just keep going round Saturn on the same orbit. The moons are far enough apart that they have negligible gravitational influence on other moons. So the moons never get close to each other and don't collide. There are moons, like Daphnis and Atlas that can come with 1000km of each other, but they are very small. No moons come within 1km of each other.

There are some exceptions to this: Janus and Epimetheus orbit at nearly the same distance from Saturn. If they stayed like this they would collide within each other within about 4 years. However as one moon starts to catch up with the other, it is pulled forwards, causing it to move out, and slow down. Letting the other moon move inwards and faster. The two moons essentially swap orbits. Although there is only about 50km between the orbits, the two moons never approach closer than 10000km, so are not at any risk of collision.

Moon collisions can take place, one theory on the origin of the rings is they they were formed from a moon that broke up, either from a collision or by Saturn's gravity (It's not the most popular or recent theory). Collisions have never been observed, nor are any forecast to occur for as long as we can predict.

Of the 62 moons, remember that some of them are extremely small: a few tens of metres across, and there is no clear division between "small moon" and "large clump of ring material"


What was the absolute limit to the possible sizes of the first stars formed from "primordial material with no metals"?

First off, let us try to clarify a few terms:

As usual in astrophysics, metal-free star means atomic number $Z \leq 3$, i.e. it only consists of the primordial elements hydrogen, helium, and lithium.
Primordial star literally means original star and refers to the first star(s) (generation) formed after the big bang. It is IMHO equivalent to metal-free, and a primordial metal-free star would be a pleonasm.
Star can be most generally be defined as massive, self-luminious gas sphere (see Lexikon der Astronomie, p. 412), which leads to the question what each of these four terms mean exactly. Massive would probably mean at least about  $0.09 M_\odot$ as 0.09 solar masses is the weight of the smallest star observed, AB Doradus, which is undergoing nuclear fusion.
A quasi-star is also worth a definition for the current question. The first paragraph of that Wiki-page summarizes it pretty neatly:

A quasi-star (also called black hole star) is a hypothetical type of extremely massive and luminous star that may have existed early in the history of the Universe. Unlike modern stars, which are powered by nuclear fusion in their cores, a quasi-star's energy would come from material falling into a black hole at its core.

ProfRob already pointed out that a key issue is the definition of star.

Or is something that starts nuclear fusion, but is always collapsing, a star?

Related is the question about the stability of a star, which is studied by Isabelle Baraffe et al. in arXiv:astro-ph/0009410 :

The stability of metal-free very massive stars ($ Z= 0; M = 120 \ldots 500 M_\odot$) is analyzed and compared with metal-enriched stars. Such zero-metal stars are unstable to nuclear-powered radial pulsations on the main sequence, but the growth time scale for these instabilities is much longer than for their metal-rich counterparts.

The metal-free stars analyzed in that manuscript would still be smaller than the quasi-stars, which require at least $1000 M_\odot$. Again a Wikipedia-quote:

Quasi stars would have had a short maximum lifespan, approximately 7 million years, during which the core black hole would have grown to about $10^3 \ldots 10^5 M_\odot$ for modern stars.

To summarize: The maximal size limit varies, depending on what we exactly regard as primordial metal-free star, and also how long such an object has to exist/ be stable. I would assume that the upper limit of primordial metal-free stars is probably larger than the upper limit of stability of $80 \ldots 100 M_\odot$ which holds for modern stars (see p. 458 in Lexikon der Astronomie).
References

Helmut Zimmermann, Alfred Weigert: Lexikon der Astronomie. Edition 8. Heidelberg/ Berlin 1999. ISBN 3-8274-0575-0. (in German)


Are red giants stationary or do they move?

Everything moves.  Entire galaxies move.  Our galaxy is moving towards Andromeda at over 100 km/s and our entire local group is moving towards the great attractor even faster than that.

When a less massive object orbits a more massive one, both move, but the less massive one moves more.  Often the calculation is simplified to make the central body unmoving, but that's a simplification and not accurate.

Our Milkyway has a center of mass with a giant black hole in that region.   If anything in our galaxy didn't move, it would slowly fall into the center.   Stars don't move like that, they move around each other while orbiting within the Milky-Way at high speeds.

Movement, I should add, is defined from a certain point of view, so it's always possible to pick one thing and say "this isn't moving" and that would be correct from that thing's point of view, but that object still sits in spacetime (relatively speaking), and space time defines a force, which (shockingly) moves things, so even if you say "Lets say this object isn't moving", the unevenness of space time will move it.

So, everything moves.   That's a universal law.


The matter that the black hole is ingesting begins with some initial conditions outside of the black hole's event horizon, and due to numerous processes this surrounding matter loses energy and angular momentum and so falls into the black hole. The physics of this infall process can be quite complicated - especially if an accretion disk forms - but essentially, viscous friction between the particles that compose the infalling matter will cause it to lose momentum, among other possible processes.

There is a theoretical limit of how much matter a black hole (or any astrophysical object) may accrete: the Eddington luminosity limit. Basically, as the black hole swallows the matter it illuminates electromagnetic radiation due to the intense collisions the infalling matter is experiencing near the event horizon. Eddington found that once the luminosity of this emitted radiation reaches a certain limit, the accretor (in this case a black hole) will actually begin to expel the surrounding matter rather than continue swallowing it, due the fact that once the limit is reached the surrounding matter will be blown away by the radiation. See here and here for deeper explanations of this.

So the surrounding matter that is expelled (once the Eddington limit has been reached) did not ever reach the event horizon - it's expelled by radiation before it makes it that close to the black hole.

And here's a lovely image of a double burp! :)


Yes, gravitational waves can be redshifted. The answer of this question explains why: https://physics.stackexchange.com/questions/137292/can-gravitational-waves-be-red-shifted
The answer to the question above basicly states that doppler effect applies to any wave that propagates at light speed.

Actualy measuring the redshift is much more difficult, because neither redshift nor rest-frame mass can be extracted from the signal. Only the combination of the two can be obtained - M(1+z). This is called the redshifted mass. Deriving actual redshift from redshifted mass is quite complicated as this problem is relatively new and no efficient solution exists yet. This article here describes the derivation of the redshift in detail:
http://arxiv.org/pdf/1312.1862v2.pdf
From it you can see that it is indeed possible to derive redshift from redshifted mass, but the method is really imprecise as it has a massive 10 - 20% uncertanity.


Of course it is possible.

If the bigger moon (A) is far enough from a parent body so that the latter one cannot pull the smaller moon (a) away from it.
So the factors at play here are the mass of the planet, the mass of the moon (A) and the distance between them. These determine what is called sphere of influence (SOI).

So if the moon (a) orbits the moon (A) inside its SOI, it would be possible for (a) to exist there for a long time.
How long that time would be - it's a slightly more difficult question. As the tidal forces and various disturbances might eventually make the system unstable, and eventually (a) might be pulled away from (A) or even fall onto it etc.

I assume that by "unnatural" you mean "impossible"? Because usually when people talk about "non-natural" satellites, they mean "artificial", that is, "made by humans". So it has nothing to do with the type of the orbit. Natural satellites are formed, well, naturally, and artificial satellites are launched by us humans.


How do we know the current state of the universe by observing it from Earth?

Since light travels at a constant and finite speed, looking into the Universe means looking back in time.

No matter in which direction we look, on average we see the same: Locally, we see evolved galaxies, far away we see young galaxies, and even farther we see the so-called cosmic microwave background, which is light from so far back in time that galaxies hadn't yet formed.

That is, the Universe seems to be isotropic. Unless we occupy a special place in the Universe, this implies that the Universe is also homogeneous, i.e. it would look more or less the same if you were located in another part of it.

This is turn means that the current state of the Universe can be inferred by observering the local region. The most distant observed galaxies can then be concluded be in the same state (statistically speaking; of course they don't look exactly the same, but any given property, e.g. the distribution of their sizes, will be the same as in our neighborhood).


Probability of finding life-supporting exoplanets

The question can be rephrased into: How many exoplanets have been found that could harbour life? As of early May 2015 the number appears to be about 30, see the Habitable Exoplanet Catalog on the Planetary Habitability Laboratory's web site.

You should also note that exoplanet detection methods will bias this sample in favour of Super-Earths.


When the sun "enters" a constellation, doesn't it make it impossible to see that constellation?

It is true that, for an observer standing on Earth, the daytime sky blocks out our ability to see the stars (with the exception that stars are visible during a total eclipse).
It is possible to get pretty close.  In a different post, it was asked if the the center of the galaxy is only visible for 1/2 of the days of the year. That post is here:  Is it true that we see the center of the milky-way for only half of the year?  The center of the galaxy happens to be close to the ecliptic and located roughly between Scorpius and Sagittarius.  In that post I worked out that it is possible to see that spot in the sky (which happens to be near the ecliptic) roughly 10 months out of the year.
The point of the above post is that if we pick a specific location (for example... that spot in the sky between Scorpius and Sagittarius) you cannot see directly at that spot during the daytime.  But choose to observe merely a month earlier or a month later and you can see that spot either just prior to sunrise or just after sunset.  Since ancient observers would certainly have been able to observe these constellation side-by-side during months when they are visible at night, it would have been easy to derive (based on the first visible constellation after sunset or the last visible constellation prior to sunrise) where the Sun is located on the current day -- and to do this without needing to see the stars that beyond the Sun's current location.
Also consider that there are roughly 365 days per year and 360° in a circle.  Each day the Sun advances about 1° along the ecliptic.  The ancients would have understood this as well... and would have been able to pretty accurately plot the current location of the Sun without actually being able to directly observe those stars located in the sky beyond the Sun's current location.


When will the Milky Way "arrive" at the Great Attractor, and what all happen then?

I don't want to make any assumptions here regarding the Milky Way's presence in the Laniakea Supercluster simply because of how recent the discovery is. The findings could very well be accurate, but I don't want to base this answer off of them. Fortunately, I've found a few papers that get us around that little issue, as well as the University of Hawaii's site, that led me to them . Coincidentally, one paper bears the name of one of the researchers who recently announced the Laniakea findings, R. Brent Tully, who was also involved with the other.

The two papers (here and here) suggest that, no matter whether the Virgo Supercluster is or is not part of the Laniakea Supercluster, the Local Group (and by extension the Milky Way) is not being pulled toward the Great Attractor. Instead, it is headed for other galaxy superclusters beyond the Great Attractor, including the Shapely Supercluster.

The idea that the Milky Way is not moving directly towards the Great Attractor could render the question moot, except that it simply shifts the destination: What will happen when the Local Group reaches the further superclusters?

Well, we don't [yet] know. As the papers say, there are a number of sources "pulling" on the Local Group, and so we can't say for sure just where the Local Group will wind up. This means that we also can't set a definitive timetable for the arrival. Finally, modeling galaxies with precision is not easy. Some theorists have used models in general relativity that treat many galaxies as grains of dust, so getting more specific is tough! We won't know just what will happen during after the "arrival" until we get a lot more data. Even the composition of the Great Attractor itself is rather mysterious.

I'll end with that odd term, "arrival". As the article you referenced (which, unfortunately, misses a lot of the better points of the findings, such as actually describing Laniakea) says,

  The giant structures making up the universe often have unclear boundaries.

That's true on a smaller scale, too (see Where does the Milky Way end? for an interesting discussion). We do know which galaxies are in the Local Group, but not where the Group's gravitational influence ends and other galaxy groups begin to dominate. We could define "arriving", though, at the instant that the destination supercluster beings to attract one of the galaxies in the Local group that are further out, instead of the now-probably-merged Milky Way and Andromeda galaxies. At that point, the Local Group could be distorted by the gravity of the galaxies in the supercluster, or could simply merge into the supercluster without any problems. But like I said earlier, it's too early to do anything but speculate.


Check out the Wikipedia page on the missing baryon problem. It's how we compute the mass of ordinary matter. You can see there are methods of getting the mass of ordinary matter without adding them up. In other words, you can find ordinary matter even if they're invisible.

This is similar to how we find dark matter. Again, look at Wikipedia's article on dark matter, in particular the observational evidence section. Ignore the more technical methods if you're new to the field and just look at the first few:

With galaxy rotation curves, you know you need more matter to recover Kepler's laws. How much matter? You can tweak that number and therefore get an estimate for the dark matter needed. You can't prove that this dark matter isn't just invisible ordinary matter using this method, however.
The same goes for velocity dispersions.
Gravitational lensing can measure the total mass of an intervening cluster without knowing anything about the kinematics of the cluster.

If you're wondering how we know dark matter can't just be invisible ordinary matter, again look at the Wikipedia article, in particular this section.

Big bang nucleosynthesis constrains the amount of ordinary matter in the universe. If dark matter were just invisible ordinary matter, we also expect more helium, lithium, etc in the universe than we observe.
Gravitational microlensing constrains the amount of MACHOs (massive compact halo objects, such as black holes and invisible planets) we can have in a large mass range, covering most of the plausible candidates.
Data from the cosmic microwave background indicates a large fraction of matter does not react with photons.


How is the average distance between 2 objects orbiting around a third object calculated?

If we set the radius of Io's orbit to 1, then Europa's is about $a=$1.591. Since they are in a 2:1 orbital resonance we would expect that number to the 3/2 power to be exactly 2. It's close (2.007) but there's enough of a difference to make it interesting; since Jupiter rotates rapidly I'm guessing that it's related to this.
Okay based on @JamesK's idea of keeping one fixed I tried to get the analytical integral
$$<r_{12}> = \frac{1}{2 \pi}\int_0^{2 \pi}\sqrt{(a-\cos \theta)^2 + \sin^2 \theta} \ d\theta$$
from Wolfram Alpha but I got a

Standard computation time exceeded...

message which I've never seen before, (screenshot) so I did it in Python.
While @JamesK's answer is "the greater of the two" which here would be 1.591 or there would be 671,100 kilometers, because this is a 2D and not a 1D problem and Pythagoras has something to say, I get a different value.
('ratio: ', 1.5910384068278804)
('d.mean(): ', 1.7524934914237922)
('James_K: ', 1.5910384068278804)

The Python script below returns 1.752 or about 739,200 kilometers (solid line) versus the mean projected 1 dimensional distance (dashed line).

import numpy as np
import matplotlib.pyplot as plt

a = 671100. / 421800
print('ratio: ', a)

theta = np.linspace(0, 2*np.pi, 100001)[::-1] # don't double-count the endpoints
d = np.sqrt((a - np.cos(theta))**2 + np.sin(theta)**2)
print('d.mean(): ', d.mean())
theta_degs = (180/np.pi) * theta

plt.plot(theta_degs, d)
plt.xlabel('phase angle (deg)')
plt.ylabel("distance normalized to Io's SMA")
plt.plot(theta_degs, d.mean() * np.ones_like(theta_degs), '-k')
plt.plot(theta_degs, a * np.ones_like(theta_degs), '--k')
plt.ylim(0, None)
plt.show()

James_K = ((a-1.) + (a+1.)) / 2.
print('James_K: ', James_K)