(1) is very doable. The others don't seem doable, at least not easily. If you practice astrophotography for a while you'll have a better understanding of the difficulties involved.

It's best (but not mandatory) if you begin doing plain visual astronomy, with something like an 8" dobsonian, for a year or two, to get yourself familiarized with how it all works. Once that's accomplished, the transition to astrophotography should be easier.

For AP (astrophotography) you'll need: a mount, a camera, a telescope. The mount is by far the most important. The camera is very important. Oddly enough, the telescope is the least important.

The accuracy, overall quality, and load carrying capacity of the mount will determine how well the whole system works. Just get the best mount you can afford.

Entry level mount examples: Celestron CGEM, Sky-Watcher EQ6.

Midrange mounts: Losmandy

High-end mounts: Software Bisque Paramount, Astro-Physics GTO

Cameras: for solar system objects like planets or the Moon, you'll need cameras with smaller sensors that can take lots of frames quickly (planetary cameras). For galaxies and nebulae (DSO - deep space objects) you'll need cameras that perform well with long exposure (DSO cameras); the best DSO cameras have cooler elements installed on the sensor.

There are OSC (one-shot color) cameras that don't require filters. There are monochrome cameras that require a filter box with R, G, B filters; you need to take separate exposures through each filter. OSC cameras are easier to use; monochrome cameras tend to provide higher performance.

Planetary camera examples: ZWO ASI 224 MC, ZWO ASI 178 MM

DSO camera examples: ZWO ASI 1600 MM, ZWO ASI 071 MC-Cool

Telescope:

For DSO it is very highly recommended to start with a relatively small refractor with short focal length. Something like the Orion ED80. Add a focal reducer / field flattener too. The long exposures require very high precision tracking from the mount, and a long focal length only creates problems. Keep it short until you learn how to make everything work. You could use a larger instrument like an SCT, but make sure your mount can carry that weight, and make sure you've fine tuned the process so the very long focal length of the SCT doesn't amplify the tracking errors.

For planetary imaging, the bigger the aperture the better. Use an SCT such as a Celestron C8 or bigger (up to C14 if the mount can carry it).

Capture software:

For DSO: Nebulosity (beginner), Sequence Generator Pro (advanced)

For planetary: FireCapture

Image processing software:

For DSO: Deep Sky Stacker (beginner), PixInsight (advanced)

For planetary: AutoStakkert for stacking. Subsequent processing could be done with Registax, or Astra Image.

Guiding:

For DSO, you'll do very long exposures. The mount alone is not precise enough. You need to guide it. This is done by using a smaller telescope (guidescope) installed on the main telescope, using a small camera (guide camera).

The finderscope that's normally used for visual astronomy could be used as a guidescope.

If you have a DSO camera and a planetary camera, you could use the planetary camera as a guide camera for DSO photography.

You'll have to connect the camera to a software that can control the mount and issue corrections to it. The most popular app is PHD2 a.k.a. Open PHD.

When doing DSO imaging, you'll spend most time during your first year basically learning all the subtle aspects of guided exposures.

Learning:

'The Deep-sky Imaging Primer' by Charles Bracken. A bit old but still very relevant.


How far away would a supernucleus comet be visible?

Even 100 meter NEOs could cause significant damage, and there are far
  more of them. LSST will be able to detect objects as faint as 24.5 in
  magnitude in a 30s visit, enabling it to detect 140m NEOs as far away
  as the Main Belt asteroids.
  [...] During its survey of the sky, LSST can find 90% of the PHAs over
  140 meters in diameter.

Earth is 1 AU from the Sun, so that would be 1.2 to 2.2 AU away from us:

  The Asteroid Belt is located in an area of space between the orbits of
  Mars and Jupiter. That places it between 2.2 and 3.2 astronomical
  units (AU) from the Sun. The belt is about 1 AU thick. The average
  distance between objects in the Asteroid Belt is quite large. If you
  could stand on an asteroid and look around, the next one would be too
  far away to see very well.

Timewise, it would be about 8 years away:

  The bottom right panel shows an example of a very different hazardous object - the 3 km large comet C/1996 B2 Hyakutake, which passed within 0.10 AU from Earth
  in 1996. The apparent magnitude vs. distance diagram indicates that the current
  generation of wide-angle imaging surveys, such as SDSS with V < 22,
  could have provided an early warning about 3 years before the closest
  approach (if they had sufficient cadence and sky coverage). With LSST,
  the warning time would be ∼8 years, with over 500 observations over
  that period.

Fortunately, your comet is 60 km, making the comet tail easy to spot:

  While the solid nucleus of comets is generally less than 50 km across,
  the coma may be larger than the Sun, and ion tails have been observed
  to extend 3.8 astronomical units (570 Gm; 350×106 mi).

This table of impact risks shows that the most likely (5.8e-02) impact is between 2095 and 2113.


The name of a particular low extinction region

The Lockman hole?

http://en.wikipedia.org/wiki/Lockman_Hole

It has a column density of neutral hydrogen of $6\times10^{19}$ cm$^{-2}$, which will correspond to very little dust.

Two others I've heard of towards the Galactic bulge are known as the Sgr I and Sgr II windows.

Finally, I offer "Stanek's window" towards the bulge, which has been referred to in a number of the Chandra Galactic plane X-ray surveys but appears to have no refereed publication associated with its discovery. I found this on arXiv.  http://arxiv.org/abs/astro-ph/9802307  It is at $l=0.25$, $b=-2.15$ and has $E(B-V) \sim 0.6$.


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

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


Explanation for the mathematics behind Venus' retrograde motion

The really simple explanation is that Venus' orbital period is 225 days. Earth's is 365 days.

That ratio is 13:8 (well, more accurately 12.9777777777..:8) so you can see that your complete cycle is 13 orbits of Venus to 8 of Earth, but within that cycle our view of Venus travelling round the sun gives the pretty pattern as shown in your image.

This YouTube video is quite a nice visualisation - should make it easy to understand.


Does the accelerating expansion of spacetime mean that the pace of time is changing?

To talk about 'the rate of time', we essentially need at least two different time coordinates. For example, this happens in special-relativistic time dilation, which is equivalent to $\mathrm{d}t'/\mathrm{d}t$ across two different inertial frames. Fortunately, we can do something similar here.

  Space expands everywhere, also here. And time is inseparable from space. Does this mean that time also "expands" as in changing its pace? ... In a similar way that the expansion of space is compared relative to, well, to itself I suppose.

A spatially isotropic and homogeneous universe has the metric in the form
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\mathrm{d}\Sigma^2\text{,}$$
where $a(t)$ is the scale factor and $\mathrm{d}\Sigma^2$ is the metric of an isotropic and homogeneous Riemannian manifold: the 'open' hyperbolic $3$-plane, the flat Euclidean $3$-space, or the 'closed' $3$-sphere (or real projective $3$-space, but that's usually not considered because it's non-orientable). If the scale factor is ever zero in the past, the cosmological time for this is conventionally chosen to be $t = 0$.

The cosmological time measures the proper time of an observer at rest relative to the bulk of the matter in the universe, so in some sense it's the most intuitive choice of a time coordinate, but like all coordinates, it's not sacred. We can, for example, define a conformal time coordinate $\eta$ such that $\mathrm{d}\eta = \mathrm{d}t/a$, in which the metric takes the form
$$\mathrm{d}s^2 = a^2(\eta)\left[-\mathrm{d}\eta^2 + \mathrm{d}\Sigma^2\right]\text{,}$$
and so all of the dimensions of spacetime are affected by cosmic expansion in the same way. Therefore, I think conformal time satisfies the requirements in your question, although it is not measured by any local clock.

  Is the changing rate of time also astronomically observable?

The scale factor is astronomically observable, and $\mathrm{d}\eta/\mathrm{d}t = 1/a$, so yes.

  How did time behave during the radical inflation shortly after Big Bang?

The conformal time essentially uses the particle horizon as a measure of time, i.e. the furthest distance from which an ideal lightlike signal could have travelled since $t = 0$ in order to reach the observer by the present time. During inflation, the particle horizon rapidly expanded.


How did they measure the distance to the center of our galaxy to 0.3% accuracy?

The orbit of star S2 is completely determined by the astrometric observations. i.e. One has the orbital period, the angular scale of the orbit and the inclination of the orbital plane. With this information alone, the distance and central mass are degenerate. If you know the distance you get the mass and vice versa.

However, here the mass is not needed. Full knowledge of the orbit means that you can predict the radial (line of sight) velocity of the star at any point in its orbit. The conversion from the angular scale of the observations to an absolute velocity, depends (linearly) on the distance to the source.

Hence they fold a set of radial velocity measurements into their analysis and get an estimate of the distance.

The distance is that to Sgr A*. The paper discusses that this is compatible with other measurements of where the dynamical centre of the Galaxy is.

The Galaxy is a proper noun (like the Sun), referring to our Milky Way, and hence is capitalised. When referring to the Galaxy it is conventional (though not strictly necessary I think) to use the word Galactic to indicate a property of the Galaxy (as opposed to some other galaxy).


What's that ring that only now appears in black hole illustrations?

This is due to photons actually going in circles around the black hole multiple times.

Far away from the BH, the image is only slightly distorted. These are photons that deviated only a little bit.

As you get closer, you're seeing photons that deviated more and more. At some point, you'll see photons that came from behind you, went around the BH, and hit your eye. Eventually, close enough to the BH, you'll see photons that did a full 360 degree turn. The BH acts like a fisheye camera.

Looking closer and closer to the disk, you get photons that did 1.5 turns, 2 turns, 3 turns and so on.

What you see is concentric ring images, each one of them containing the full picture of everything around the BH but compressed in a ring. There's essentially an infinity of concentric images, but you only see clearly the first few of them.

https://imgur.com/VZnkFxP

It should be noted that the black disk in the middle is not the event horizon. It's simply a region in space where no light comes from that could reach your eye. The actual event horizon is about 2x smaller. You could think of it as the image of the event horizon "magnified" by the spacetime distortion, but the analogy is not perfect.

A black hole is, in many ways, like a very strange "lens".


How to model and subtract nebular emission lines from stellar spectrum

In general, you can't.  If obtaining spectra in regions where there is expected to be a spatially varying background then you either need to do long-slit spectroscopy so that you have a good measurement of the ISM contribution either side of your source, or you do integral field spectroscopy with the same idea.

The problem is that the line strengths for the emission lines you see from the ISM are integrated along the line of sight and are very dependent on the physical conditions of the ISM. This means that you cannot really assume any particular ratio of line strengths to assist in any subtraction.

In some circumstances you may be fortunate that there is a relative velocity between your stellar source and the background - in which case, separation of the nebular and stellar components may be possible. Or it may be that the stellar features are broadened by rotation (if you have the resolving power), whereas the nebular contribution is usually very narrow/unresolved.

It is difficult to provide more details without a lot more information about the observations, the target and the background.


What is the word for space that is in the solar system, but outside the heliopause?

There doesn't seem to be such a word. Interstellar space within the Solar System is still just interstellar space.

There doesn't seem to be a demand for the word you're looking for either. We distinguish space by its contents; the space within the heliosphere is called the interplanetary medium (it contains solar plasma, dust, etc.), while the interstellar medium comprises sparse gaseous molecules from the Local Fluff and (if you go far enough) the rest of the Local Bubble.

Whereas there are about 10 oxygen molecules per cubic centimeter in the interplanetary medium, there is only 1 oxygen molecule per cubic centimeter in the interstellar medium. Seeing that there is no clear boundary for the Milky Way (nor is there any single definition for a boundary for galaxies), the intergalactic medium just vaguely describes the medium within galaxy filaments; here, the space is so rarified that there is only 1 oxygen molecule per cubic meter.


Could dark energy (the mysterious accelerating expansion of the universe) be explained by "negative gravity"?

But it already is "negative gravity". In general relativity, the stress-energy tensor $T_{\mu\nu}$ describes the energy, momentum, and stress of matter in spacetime. Through the Einstein field equation, it is connected with Ricci curvature $R_{\mu\nu}$, which consists of half of the twenty independent degrees of freedom of spacetime curvature.

The intuitive geometrical meaning of Ricci curvature is as follows: suppose you have a small ball of test particles, initially all at the same velocity with respect to one another, and allow them to fall freely under gravity. Then the Ricci curvature (contracted with their intitial velocity) measures the initial acceleration of the volume of the ball of test particles divided by the initial volume. Thus, this is a good way to tell if something is locally "attractive" or "repulsive", based on whether a ball of test particles around it starts shrinking or expanding.

Through the Einstein field equation, stress-energy is connected to Ricci curvature. As a result, the volume of the small ball of test particles shrinks proportionally to $\rho+3p$, where $\rho$ is the energy density and $p$ is the average of the principal stresses--for a perfect fluid, the pressure. Naturally, if $\rho+3p < 0$, this is "negative gravity" in the sense that the test ball will start expanding instead. This quantity is also what's seen in the cosmological Friedmann equations. Since the Friedmann-Robertson-Walker family of solutions assume that the universe is homogeneous and isotropic on the large scale, it's sensible that treat the large-scale the same as the above local condition.

Dark energy has $\rho+3p<0$, and in particular (for cosmological constant), $\rho = -p > 0$.

To recap: (1) in general relativity, gravity depends on the stress-energy tensor, which has more than just the mass-energy density, and numerically pressure is more important because there are three spatial dimensions while only one time dimension, (2) we can characterize locally "negative gravity" as $\rho+3p < 0$, or, more generally, as any the violation of the strong energy condition, because this characterizes whether a small ball of test particles is attracted or repulsed by the local stress-energy content.

  Does antimatter have negative gravity?

Not in general relativity. Energy is the gravitational charge, and antimatter has positive energy; thus it should gravitate in the same way.

  Gravity is the flow of space into matter, and it flows out of antimatter.

If you're thinking of some analogy with electric charge, with electric field lines flowing out of positive charge and into negative charge, antimatter has positive mass, which is required by quantum field theory.

  ... I just searched for this online to see if anyone else has this theory, and yes, someone has something very similar here: ...

Some particles, such as photons, are their own anti-particle. This has immediate implication for Villata's claim that particles and antiparticles behave oppositely in a (normal) matter-produced gravitational field: it predicts that gravitational lensing cannot happen, since if light is bent one way by a gravitational field, Villata's stance predicts it's also bent in the the reverse direction.

Your stance is different from Villata's, but the corresponding statement for your theory is that light cannot produce a gravitational field. That's a lot harder to specifically test for, but it's clear that it's massively inconsistent with general relativity.


What would harnessing of gravitational waves look like?

The mass of fundamental particles, like the proton or electron is very very small in comparison to the charge. This means that the motion of electrons in atoms can produce significant amounts of high frequency electromagnetic radiation.

Quantum effects stop electrons from falling into the protons they orbit.

To produce and control gravitational radiation in significant amounts you would need something very small, but very massive: a microscopic pair of black holes.

Such things would be rather hard to manage. In short it is beyond any conceivable current or future technology.


What will the PRECISE time of the Summer Solstice 2020 be?

TL;DR:
In a couple of weeks it might be possible to say precisely (within a second, perhaps a fraction of a second) when the summer solstice did occur. But until then, sub-minute estimates should be treated as fraudulent.

The reason you are seeing different times for when the summer solstice will occur is that different websites use different models of the orbits of the bodies that comprise the solar system and different models of the Earth's orientation in space.
It's very important to remember that "all models are wrong, but some are useful".
There are several issues with finding the exact time of when solstices occur. One is that the solstices are extrema of the Sun's declination. Without a model of derivatives, finding extrema (e.g., the solstices) mathematically is a much harder problem than is finding zero crossings (e.g., the equinoxes).
More importantly, a very precise prediction would require multiple very precise models. While many models of the orbits of the bodies in the solar system are very, very good, none are good enough to get the timing of the solstices down to the millisecond. Even more importantly, the tilt of the Earth's orientation in space with respect to its orbit about the Sun needs to be modeled. The best model, the IAU 2006/2000A precession-nutation model has over a thousand terms (1365 terms, to be precise).
And even then, "all models are wrong." The people at the International Earth Rotation and Reference Systems Service (IERS) know this very well. The IERS is the organization that is responsible for modeling time as measured by the Earth's rotation and for modeling how the Earth is oriented in space. There are some aspects of the Earth's orientation in space that are not quite predictable. The US Naval Observatory, one of the key members of the IERS, publishes daily updates of observed deviations from model predictions. It take a couple of weeks to fully digest measurements made by astronomers worldwide.


How exactly does inflation convert random gravity fluctuations into coherent gravitational waves?

When inflation occurs the stretching of the gravitational field is dependent on the perturbations within that field originally. These perturbations are caused by quantum fluctuations. This leads to fluctuations in the magnitude of the inflation itself at a given point in space, causing gravitational waves due to the gravitational potential between neighboring points. The waves are fed by the inflationary energy, but are seeded by these quantum fluctuations.

This site gives a good explanation, as well as this particular quote:

  Gravitational-wave perturbations are different. They are not modulated
  by some unknown potential; they are produced by inflation, and we
  observe them directly. In straightforward models of inflation, the
  amplitude of the gravitational waves is directly proportional to the
  inflationary energy scale.


Could dark matter particles be unstable?

Yes, as mentioned elsewhere, it is possibly possible. Dark matter particles may be intrinsically unstable (though having long lifetimes, which are at least significantly longer than Hubble time).

Check for more info here:
http://arxiv.org/abs/1307.6434


Brightest star visible from Italy and Argentina

Sirius is the brightest star in the sky (excluding the Sun, of course), with a visual apparent magnitude of −1.46, and is visible to everyone in both Argentina and Italy, depending on the season.

In fact, Sirius is visible from anywhere in the Southern Hemisphere, including all of Argentina (which, I might add, is an enormous country stretching from just above the Tropic of Capricorn, all the way south to the Drake Passage separating Tierra del Fuego from Antarctica).

Sirius doesn't rise above the horizon for locations in the Northern Hemisphere north of 73°, but this is well beyond the northernmost point of mainland Europe at around 71°N, so Sirius is visible in all of continental Europe. As the northernmost location in Italy is around 47°N, Sirius at its zenith appears well above the horizon to anyone in Italy.


The color you can see in the sky depends on two things, first the color (if you want to be more specific, the emission spectra of the light of the star) of the light shining, this is the Sun in our case (but may be another star in order to answer your question about outside the solar system), and second depends on the chemical components of the atmosphere of the planet in question, because this atmosphere is the reason for what is called the absorption lines.

Absorption light is something like that, if you have white light and a gas with green absorption lines, then after crossing the gas the light will turn the opposite of green, I don't know what color that is.

In conclusion if our atmosphere changes (for some unknown reason) here on planet Earth, then the color of the sky is going to change.


Solar System formation, considering its and the universe's age

Early in the life of the universe it is thought that star formation in "metal-free" gas probably favoured larger stars. These Population III stars had short lives and very quickly enriched the interstellar and intergalactic medium with nucleosynthesis products.

In fact the main enrichment of the interstellar medium (ISM) in our own Galaxy is thought to have taken place some billions of years before the Sun was born. Stars that are being formed now have basically the same composition as the Sun. The plot below shows the iron abundance (on a logarithmic scale) as a function of deduced age for stars in the Gaia-ESO spectroscopic survey (from Bergemann et al. 2014). As you can see, there has not been too much change in iron abundance over the last 10 billion years.

I think the key to this is that most of the heavy elements in the gas in our Galaxy come from stars more massive than the Sun, and which have much shorter lifetimes as a result. [Stellar lifetime is roughly proportional to $M^{-2.5}$]. So several "generations" of these have lived and died (and must have done so).

To add some more detail. Almost all the elements heavier than Helium (known as "metals") are produced inside stars. To get back into the ISM you have to get the "processed" material out of the centres of the stars. This occurs in basically three ways. (i) Supernovae - the explosions associated with the final collapse of very massive stars ($>10 M_{\odot}$). The supernova progenitors have short lives $<50$ million years, so many generations of such stars can have lived and died before the Sun was formed, and probably of order a few hundred million to a billion had done so. (ii) Giant star winds. These occur towards the ends of the lives of stars with mass of $<10M_{\odot}$. Some fraction of processed material gets dredged out of the centre and expelled into the ISM. The lifetimes of these stars is 50 million to $>10$ billion years. Most of the enrichment is caused by stars about in the middle of this range. These stars go on to fade as white dwarfs. (iii) Type Ia supernovae. Explosions triggered by mass transfer on to a white dwarf (probably in binary systems) causing a detonation and the scattering of metals into the ISM. Progenitors have lifetimes similar to the giant stars in (ii).

If you look carefully at the colour scheme of the plot above, you will notice that ancient stars have a higher ratio of magnesium to iron. This supports the ideas in the paragraph above. Magnesium is an alpha-capture element, produced mainly in short-lived massive stars and gets into the interstellar medium through type II supernovae. Conversely, although iron is also produced in massive stars, it is primarily produced and dispersed in type Ia supernovae that have much longer lived progenitors. Thus Mg enrichment occurs more rapidly than Fe enrichment in the early Milky Way.

The numbers of white dwarfs we see in the Galaxy and the deduced rate of supernova explosions that are occurring (e.g. from counting the number of neutron star remnants) also matches up with what is required to produce the metallicity of the ISM and the Sun. But there are plenty of variables and unknowns and not everything is solved. For instance it looks quite likely that star formation needed to be higher in the past; and some parts of the Galaxy (e.g. the bulge) have a higher metallicity, thought to be due to a period of very vigorous high mass star formation 10 billion years ago.


Calculate Day of the Year for a given date

This would really be better off in Stackoverflow, or another page.  Presumably, you assumed some of the calculations rely on concepts in Astronomy, which as far as I can tell having gone through them, they do not with the exception of the concept of leap years; which I think is general enough that it doesn't really mandate being in Astronomy.  Regardless:

The equation is a little abstract, so probably the easiest way to understand it would be through understanding each bit of data. Let's start with N2, I'll come back to N1:

N2 = floor((month + 9) / 12)

N2 will equal 0 if month is less than 3, and will equal 1 if it is greater.  This formula is to determine whether February has passed.

N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))

N3 will be equal to 2 if the year is not a leap year, and will be equal to 1 if it is a leap year. The math here is just to determine whether the current year happens to be a leap year.

N = N1 - (N2 * N3) + day - 30

Now we take N1 (which I haven't covered yet) and we subtract it by the product of N2 and N3, which will equal 0 if we haven't reached March, or if we have it will be 1 on a leap year, and 2 on a non-leap year. We then add the number of the current day of the month, and subtract 30.  So what does N1 do?

N1 = floor(275 * month / 9)

Well, lets look at the values returned compared to days in a non-leap year:

Month     N1         Day       Diff
1.        30         31          -1
2.        61         59          +2
3.        91         90          +1
4.        122        120         +2
5.        152        151         +1
6.        183        181         +2
7.        213        212         +1
8.        244        243         +1
9.        275        273         +2
10.       305        304         +1
11.       336        334         +2
12.       366        365         +1

A bit messy, right?  Now, remember that you're subtracting 30 from the total at the end to get N, and we're adding in the current date. This means that although we multiply our current month to get N1, we're actually using this to calculate the dates from the months prior to our current month! Thus if we take the value of N1, subtract it by 30, and compare it to the preceding month, the chart will come out like this:

Month     N1         Day       Diff
1.        31         31           0
2.        61         59          +2
3.        92         90          +2
4.        122        120         +2
5.        153        151         +2
6.        183        181         +2
7.        214        212         +2
8.        245        243         +2
9.        275        273         +2
10.       306        304         +2
11.       336        334         +2
12.       ---        365         ---

From this, you can see that the value of N1 will equal 2 greater than the actual date for any day in which it is March or later.  This is perfect, as N2 is already a formula determining this for catching leap days. Note, these would all equal +1 on a leap year, as in another day would have been added in February. Thus coming back to the final calculation:

N = N1 - (N2 * N3) + day - 30

We take N1 and subtract in by the value of (N2 * N3), which will be 2 on a non-leap year if February has passed, 1 on a leap year, or 0 if we haven't passed February.  We add in the days in the current month, and we subtract 30 to get the offset in N1 values.

This should consistently give you the day of the year.  As for special significance, they're being used only for mathematical properties. Dividing a number between 1 and 12 by 9, just so happens to equal less than 1 for numbers less than 3, and greater than or equal to 1 and less than 2 all the way up to 12, which is why it is we have started March. The 275 * (month / 9) formula I'm not sure where they came up with. AFAIK, they have no special meaning to time or the Sun-Earth system - rather, they were chosen because they calculate well in this particular usage.


How to use an app to help find deep sky objects in general, either attached to a telescope as a finder scope or held on hand?

You can do this. It's possible to mount a phone on a telescope and use it as an electronic finder with an app like Sky Safari or Stellarium Mobile.
However:

The accuracy of it very much depends on the accuracy and sensitivity of the phone's sensors, and this varies quite a lot between phones
If the telescope tube contains ferrous metal then it'll throw off the magnetic sensor
The light of the screen adversely affects your eye's night adaptation

These things may or may not be issues depending on each individual situation.
I've tried it a couple of times but never been happy with the results. The two phones I tested weren't accurate enough.
Some people do use their phones as finders, but they tend to do it with digital setting circles (DSCs) such as the Nexus 2. This has a pair of 'encoders' which you attach to your telescope, and these are used to track the movements of the telescope's axes, and update a mobile application accordingly.

Personally, I use a Telrad non-magnifying sight to get myself in the general vicinity, then move over to a magnifying finder scope, and finally to the eyepiece.
I star-hop using a combination of three, depending on my target. A lot of the time I can go straight from the Telrad to the eyepiece. Other times I'll star-hop with the finder telescope.

If your friend is short on available observing time, then it can help to plan sessions beforehand: a list of targets and directions to them from the nearest brightest recognisable star.
A program like Kstars can be used to assist with this: it has a star-hopping planning tool as well as a "what's up tonight?" tool.


Has J been used by itself as a spectral type?

As far as I can tell, the J-type was introduced by Bouigue (1954), and have subsequently often called "J-type stars" or "J-stars", e.g. Abia & Isern (2000). This likely led to the J-type designation being put as "in use", though the spectral types themselves appear to always have been used in combination with "C".


Why does the path of the moon in an eclipse curve (relative to the sun)

You have misunderstood the website. When the eclipse begins, in the morning, the sun will be in south-east, at a bearing of 135 degrees.

At maximum eclipse (at about midday) the sun will be almost due south, at a bearing of 178 degrees.

When the eclipse ends, in the afternoon, the sun will be in the south-west, at about 217 degrees bearing.

The sun and the moon will move in the sky, due to the rotation of the Earth, this says nothing about the straightness of the path of the moon relative to the sun.


How did they make a video of the center of the galaxy, and what is it exactly that's flashing there?

Question: How are these images obtained?

Later in the video the narrator says they took the images using ESO's VLT.

  03:40 [Narrator] 14. Making these measurements pushed the power of ESO’s Very Large Telescope to the limits.

(Source: ESO transscript)

Over the whole observation period multiple telescopes and imaging instruments were used. Early observations were accomplished using the NTT. Since around the year 2002 the VLT observed Sagittarius A* with the NACO and GRAVITY instruments too.

This graph puts the rotation period of Sagittarius A* in relation to the time and the observing instruments:

(Source)

Question: What process is it that is believed to be causing that flashing?

ESO provides older footage over a minutes-timescale showing a flare of Sagittarius A* in May 2003: Flashes of light from disappearing matter

  ESO Press Video eso0330 shows the detection of a powerful flare from the centre of the Milky Way galaxy. These and other adaptive optics (AO) images (with resolution 0.040 arcsec in the near-infrared H-band at wavelength 1.65 µm) of the central region of the Milky Way were obtained with the NACO imager on the 8.2-m VLT YEPUN telescope at the ESO Paranal Observatory on May 9, 2003. [...] The position of the 15-year orbiting star S2 (cf. ESO Press Release eso0226) is marked by a cross and the astrometric location of the black hole is indicated by a circle.

The cause of the flickering is probably the same for the video sequence you showed.

A October 2018 ESO publication states:

  ESO’s GRAVITY instrument on the Very Large Telescope (VLT) Interferometer has been used by scientists from a consortium of European institutions, including ESO [1], to observe flares of infrared radiation coming from the accretion disc around Sagittarius A*, [...]. The observed flares provide long-awaited confirmation that the object in the centre of our galaxy is, as has long been assumed, a supermassive black hole. The flares originate from material orbiting very close to the black hole’s event horizon — making these the most detailed observations yet of material orbiting this close to a black hole.

(Source)

ESO also has another video showing the therory with more detail: skip to about 1:00.

Flares associated with Sgr A* are also described in this answer.

Artist's illustration of heated gas orbiting the black hole:

(Screengrab from 1:25 of that video)

See also

Phys.org (2019): Black hole at the center of our galaxy appears to be getting hungrier
Do et al. 2019 Unprecedented Near-infrared Brightness and Variability of Sgr A*
Chen et al. 2019 Consistency of the Infrared Variability of SGR A* over 22 yr

Figure 1 from Do et al. 2019

  Figure 1. Top row: a series of K' images taken on 2019 May 13 centered on Sgr A* showing the large variations in brightness throughout the night. The first image from the left is the brightest measurement ever made of Sgr A* in the near-infrared. Also labeled are nearby stars S0-2 (K' = 14 mag) and S0-17 (K' = 16 mag) for comparison. Bottom panel: K' (black) and H-band light curves of Sgr A* from 2019 May 13. On this night, we alternated between H and K' observations. The H-band magnitudes are offset using H − K' = 2.45 mag. There appear to be no significant color changes during the large change in brightness. Red circles show the location of the four images in the panels above.


The convective overturn time is the typical timescale for a convective cell to rise in a gas. Imagine a "lava lamp" - it's the time for one of the blobs to rise from its lowest to highest point.

I am most familiar with its use in stars, where convective energy transport is modelled using a mixing length. This posits that the typical height travelled by an adiabatic convection cell is some multiple $\alpha$ of the pressure scale height $H_p$ (the typical length on which the internal pressure changes significantly).

The convective overturn (or turnover) time is then $\alpha H_p/v$, where $v$ is the velocity at which convective cells rise.


Can we know where (South, East, West...) the moon will be at a specific time?

A planetarium programme will tell you exactly where to look, and can be invaluable when looking for the location of planets, but the Moon is very easy to find in the sky, even after several glasses of Flanders Red Ale.

So what you really want to know is a) is the eclipse visible from Belgium and b) roughly where in the sky will the moon be.

This map shows that the Eclipse will be fully visible from Belgium. So that's good.

The the Moon will be full (a lunar eclipse always happens at full moon) so the moon will rise in the East as the sun sets, and travel into the West. At Midnight (12 AM, but many people make that mistake) the Moon will be in the South, and at 4AM it will be South West, and quite low in the sky. But even eclipsed the moon hard to miss!

Don't go out too early. Nothing much will be visible until the "Umbral eclipse" starts at 1:07 UTC (=3:07 am in Belgium)

I wish you clear skies.


The "local bubble" is a region of the galaxy where the density of interstellar gas is lower than average. It has about 50 thousand atoms per cubic metre, compared with 500 thousand averaged over the whole of the milky way.  It was probably formed by one or more supernova explosions (10-20 million years ago) physically pushing gas away.

There are many such bubbles all over the galaxy, carved out by supernova and other energetic objects. See https://en.wikipedia.org/wiki/Superbubble


It depends what lines you are measuring and in what kind of star.

When you measure the RV from a spectral line, you are measuring an intensity-weighted average RV over the region where the line is formed.

For a star like the Sun, the photospheric lines are all formed within a layer no thicker than about a 1000 km and the differences in RV with depth in the atmosphere are small (though not negligible if you are interested in finding low mass exoplanets using the RVs). On the other hand, if you were to start measuring RVs from lines originating in the chromosphere or Corona then you could find different RVs caused by mass motions of the emitting gas.

Similarly, hot stars are surrounded by fast moving winds with a temperature structure. Different lines can originate at different heights from gas with different velocity. This can lead to different RVs from different lines.

A better way to measure RV is by cross-correlation with a synthetic or observed template of a similar star with a known RV.


The Hubble tuning fork classifies galaxies based on whether they are elliptical or have spiral arms, and whether there is or not a bar.

A historical aside

Edwin Hubble thought that galaxies started out as ellipses and evolved into spiral galaxies. This is why ellipticals are also called early-type galaxies, and spirals are spirals are called late-type galaxies.

As we studied galaxies more, we figured out that Hubble was wrong on that point, because elliptical galaxies typically have older stars and spiral galaxies still have new stars forming in them.

Elliptical galaxies

In elliptical galaxies, the velocity dispersion of the stars is high: the stars have random orbits around the center of mass of the galaxy. This is seen to be evidence for the fact that elliptical galaxies are the result of collisions and mergers between many galaxies.

When two non-elliptical galaxies merge, their planes are rarely aligned. So when many galaxies merge together, like what happened in the early universe, you end up with stars moving in many different directions. This is how a galaxy can end up being elliptical.

Spiral galaxies

The spiral arms of a galaxy might come to be through two different mechanisms. It is possible that some spiral arms are caused by the first mechanism, and others by the second.

First mechanism: The arms of a galaxy are density waves.

Picture a traffic jam: all the cars are slowed down on a portion of the highway, but the cars that are in the jam aren't always the same ones. Each car reaches the traffic jam, slows down, gets to the end of the jam and speeds back up. This is what the stars are doing in the spiral arms: a star reaches the arm, slows down because of the higher density, gets to the end of the arm, resumes its normal speed. As a result, the arms spin slower than the individual stars.

Second mechanism: The arms appear as star formation propagates through the galaxy.

The leading edge of the arm is where star formation occurs. The trailing edge is where stars die (so are no longer visible). In this case, there is not more matter in the arms, it simply is more visible.

As stars form in one region, the largest ones cause shock waves to spread through the interstellar medium, igniting the formation of new stars. This causes the star-forming region to propagate through the galaxy. Combined with the differential rotation of the galaxy, this makes spiral arms appear.

Other shapes

You mention ring galaxies in your question. These are possibly due to one galaxy having gone right through another.

Many galaxies have bizarre shapes (e.g. the Antenna Galaxies) that are caused by interactions between galaxies.


Do massive OB stars accrete mass the same way as the black holes they turn into do?

Many, if not all, stars will, in their birth phase, have an accretion disk, or circumstellar disk, around them, formed of the material from which the star forms. This disk dissipates in a few million years, both due to the material accreting onto the star, due to material being blown away by radiation pressure, possibly assisted by dust grains, and due to depletion of the material to rocks and planets. Friction in the disk causes it to heat up and be observable in the infrared. After this phase, the star usually won't have an accretion disk. That is, until they die.

As you suggest, a star that later on goes supernova can accrete matter from its "reverse shock", i.e. matter that is falling back due to the shockwave hitting the surrounding interstellar medium. More often, though (I think), will the accretion disk be made of material from a companion star: If one star of a binary system turns into a black hole, and then the other star starts to evolve into a red giant, threspassing the Roche limit, it will transfer matter to the black hole.

As you say, the black hole's mass isn't stronger than the original star's mass, and thus the gravity at a given distance is also not stronger (in fact it's smaller, since it has blown out most of its mass in the explosion). However, due to its much smaller size, the accreting matter can come much closer before being sucked in. Coming much closer means being in a much steeper gravitational potential and reaching much higher speeds, i.e. being heated to much higher temperatures. Hence, black hole accretion disks can be detected in X-rays


How can astronomers pinpoint the location of the source of a neutrino?

You correctly state that neutrinos do not interact too often. The physical parameter describing that is the effective cross-section. So what you observe in a detector is not the neutrino itself, but secondary particles, e.g. muons. Colloquially put, you may regard anything with high mass (density) in between the neutrino source and your instrument (to detect mainly the secondary particles) as detector for the neutrinos. This might be a large chunk of ice (like for the icecube experiment), a mountain range (as for the Gran Sasso lab), or even a complete planet, if the incident neutrinos originate from a star located on the other side of Earth at the moment of observation. The latter is possible since neutrinos e.g. from the Sun may pass through the whole planet without interacting with its matter (at night-time for the scientists working for the detector).
Let me try to explain with a drawing I took from a 2018 blog entry of the Antarctic Muon And Neutrino Detector Array (AMANDA). The red dots are on the ice surface, the vertical lines are the boreholes into the Antarctic ice shield and the little black dots are the PMTs, the Photomultiplier Tubes which detect the Cherenkov light generated by some fast muons, in the picture on the top of the blue cone. The readings of the PMTs (detecting the Cherenkov light) are displayed as color-code, I assume the color scale somehow relates to the time passed since the trigger event.

So how do we now know where the neutrino came from? We can use particle physics to derive the secondary particle's velocity (vector) and from that where the neutrino actually came from. This is not trivial and requires quite a bit of computer power, but it is possible.


If the universe keeps expanding, can we travel to the edge of the universe?

When we talk about the universe, we are really talking about one of two things:

The observable universe, which is everything we can possibly see.
The Universe, which is everything that has ever existed, currently exists, and will exist.

The observable universe has its own center, usually the Earth. It is a spherical region of everything that we can see, essentially anything whose light has reached us. We usually refer to this when we say things like "there are $10^{86}$ atoms in the universe."

In reality, everyone has their own observable universe, and it can change depending on where you are. An exoplanet far away has its own observable universe, and can receive light from different places. Essentially, you are the center of your own observable universe. As more light reaches us, our observable universe will grow (or expand, if you will). If this is what you are referring to, then your answer is here.

If you are referring to the latter, then there is a totally different answer and "expansion" refers to something completely different. The Universe (notice the capital "U") is all of space and time and its contents. Anything that has existed, will exist, and currently exists is part of it.

The Universe is thought to be infinitely large, so it can't have a center. The center of something is the point equidistance from the edges, but if something spans infinitely long, it would just keep going. It wouldn't have an edge, and thus it wouldn't have a center. You couldn't find the point equidistant from the edges if it just spans infinitely.

The Universe is not like a ball. Rather, you can think of it like a flat grid, and its "expansion" just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. That's what we mean by expansion — that objects are moving away from each other, since more space is being created between them.

Here's an easy analogy: imagine you are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other.

That's essentially happening everywhere: space is expanding between everything, so we are drifting away from other galaxies. The Universe is infinite, and we can constantly drift apart from other objects because space is being created in between us. Here's a GIF I made that might help you get it:

You can see how the galaxies drift apart as the space between them increases. And this happens everywhere in the Universe. The Universe is infinite, but more and more space is being created between matter.

Fun fact: These objects can actually drift away from each other faster than the speed of light. That is, light from them eventually won't make it to us, since they'll be drifting away too quickly.

Now, this doesn't actually go against Einstein's theory that the speed of light is the fastest thing in the Universe. Einstein said that nothing can travel through space faster than light — but here, space itself is actually being created between the objects. Distances are increasing because space itself is dilating, and thus we can drift apart from other objects faster than light.


Assuming you are calling cEci and have the units and sign conventions correct (longitudes are notoriously inconsistent as to which way is positive) then you should be able to calculate it almost as you have as:

distance = sqrt(x*x + y*y + z*z)
RA = atan2(y/x)
# Normalize right ascension into the 0..2*PI range
if ( RA < 0. )
  RA += PI + PI
Dec = asin(z/distance)
# Convert radians to hms and or degrees to taste

This code was taken from Bill Gray's sat_code, specifically observe.cpp. Other bits of that code handle the conversion from a latitude, longitude, height (geodetic coordinates) to a X,Y,Z vector from the Earth center to the observing site (geocentric co-ordinates) and then the rotation by the Local Mean Sidereal Time (site longitude + GMST, where GMST is the Greenwich Mean Sidereal Time) and then the 2 vectors (Geocenter->Site and Geocenter->Sattelite) are added. However it looks cEci already has this as I see references to the flattening factor and LMST.

One thing to note with TLEs is they are not a very good representation of the satellite motion/orbit over long periods of times. You should make sure your TLEs are up-to-date or use something like SGP4 to apply the perturbations and update the TLEs.


If Jupiter was placed in Mars' orbit, how would Jupiter appear from the naked eye?

Since the question is actually asking for a size of Jupiter as seen from Mars, there is a very simple equation that will give you a good idea.

$$\frac{\alpha}{360^{\circ}}=\frac{\ell}{2\pi r} $$
$\alpha$ is the angle that Jupiter would subtend in the sky, or the angular size, and $\ell$ is the actual diameter, or linear size, of Jupiter. $2\pi r$ is the circumference of Jupiter's orbit about Mars. Now I know what you're thinking, that this doesn't make any sense because of the vast differences of Jupiter's orbit when it is close compared to when it is at opposition. It'll work out...

The following are all approximations.

Mars orbits the Sun at $\sim1.53AU$ and Jupiter at $\sim5.2AU$. Because $1AU \approx 150Mkm$, The difference in their orbits are roughly $550.5Mkm$.

$$\alpha=\frac{140000km}{2\pi(550.5\times10^6km)}\times360^{\circ}=0.015^{\circ} $$

That is just the average angular size, at the average closest distance, but given that the only real tool to measure angular size with the naked eye (that isn't a sextant) is your hand, and your smallest digit (the pinky) at arms length marks out $1^{\circ}$ you'll never be able to tell the difference in the variation of size due to the change in distance.

Now answering your other questions directly:

Jupiter would look larger, but not enough to actually tell a difference in size.
As shown above, Jupiter would be much smaller than the moon appears to us from Earth, $0.5^{\circ}$.
No disk, as this is very much a point-like size as far as human eyes are concerned.


Why are explosions always used to represent the Big Bang?

No, it's not the best reproduction of explaining the big bang. Generally when you think logically, it was an explosion of matter, however it was not a fireworks show.

To best explain it simply, it is conceived as a singularity that rapidly expanded. An important thing to understand is that within the beginning of the expansion, the matter was quark–gluon plasma and opaque, meaning there was no 'BANG'. Intensity of the heat was too great to allow quarks to form hadrons. Hadrons comprise of baryons such as protons and neutrons, at the earliest stage of inflation protons had not formed, being a representation of the 'big bang' they aren't implicitly going to introduce these factors. A general representation if the bang factor itself, if the bang is presented as an inflating balloon filled with gas, it's not going to attract. To be a true analogue, the balloon would simply appear into existence (between 10^-6 sec and 1 sec), already expanded and have pockets of matter now cool enough to begin producing light, within the next 10 seconds electrons begin to form attracting photons and neutrons, the first atomic bonds (hydrogen) would continue for the next 380,000 years. Much of the earlier matter has been annihilated by antimatter and we begin to see matter in the form it is today.

In a nutshell, it's hard to present this theory when something just 'becomes', so I believe they use the explosion as a way of demonstrating the nature of forces rather than an analogue of mere inflation.


The short answer is no. Take Mercury for example in this comparison of Earth Mercury core. Mercury is thought to have a liquid outer core and solid inner core. The gas giants like Jupiter are thought to have a relatively tiny rocky core but the convective motion in the metallic hydrogen is what gives them the strong magnetic fields.

See also: Is Mercury's core liquid? for more on Mercury's core.


During night on the Moon is there Earth light and Earth phases?

From the perspective of the Moon, the Earth exhibits phases just as does the Moon when viewed from the perspective of the Earth. There are however some differences. One difference is that the phase of the Earth at a fixed local lunar time will be pretty much the same from one day to the next. This is a result of the fact that the Moon is tidally locked. Another difference is that earthlight is considerably brighter than moonlight. This is a consequence of the Earth's significantly larger albedo and significantly larger diameter compared to the Moon. A third difference is that except for locations close to ±90° lunar longitude, the Earth doesn't rise and set.

  I assume the Earth isn't as reflective as the Moon since the Moon is much lighter in colour, however the Earth is also fairly larger, does this make any difference in the amount of light reflected?

You have this backwards. The Moon is rather dark. It's reflectivity is about that of old asphalt. The Earth is significantly more reflective than is the Moon.

  Finally, if the Earth does appear in phases, does that mean that the two week long nights on the Moon are sometimes pitch black (new Earth) and sometimes less than pitch black (full Earth)?

No. This is the first difference between Moon phases and Earth phases. Earth phase on the Moon is a function of lunar longitude and time of lunar day. You won't see a full Earth one night and a new Earth some other night.


Does the recent news of "ten times more galaxies" imply that there is correspondingly less dark matter?

All Conselice et al. (2016) appear to suggest is that when you look at something like the Hubble deep field, there are many faint (and presumably low mass) galaxies that are not seen. This has absolutely no effect on the need for dark matter.

The main results are: (i) as you look back in time, the overall (co-moving) density of galaxies (more massive than a million times that of the Sun) increases. (ii) But the density of more massive galaxies actually decreases. This is consistent with hierarchical merger picture where small galaxies merge to become larger galaxies.
This really doesn't have any influence on the need for dark matter.

First, the presence of dark matter is inferred from many different observations. Some of these (e.g. galaxy rotation curves) are not influenced at all if there are lots of extra galaxies.

Second, the "missing" galaxies are at high redshift, not (or not all) in the present day universe, so they cannot significantly affect a calculation of how much normal matter there is in the universe today. Presumably, many of these small galaxies then merge to become larger galaxies and the total mass is conserved.

Third, just because there are lots of them does not mean they contain much mass anyway. The "mass function" (number density as a function of mass) of galaxies goes roughly as $\phi(M) \propto M^{-1}$ at low masses. This means the mass contained in any interval is
 $$M_{\text{tot}}\propto \int^{M_2}_{M_1} M\phi \ \mathrm{d}M \ = M_2 - M_1$$
So although, low mass galaxies may be ten times more frequent, they are ten times less massive and so don't change the total mass very much. I will need to read the paper more carefully to see if the authors are suggesting that low mass galaxies are much more common in the early universe than was already supposed.

Fourth, primordial nucleosynthesis calculations tell us that only 4 per cent (as a fraction of the critical density) of the energy density of the universe is in the form of baryonic mass. Observations of gravitational lensing, cluster dynamics and the cosmic microwave background tell us that the mass density is actually around 30 per cent of the critical density. Thus most of the dark matter is non-baryonic and cannot be in the form of missing faint galaxies, or any other form of normal baryonic matter.


Possible search terms to use:

Vintage Russian CCCP Spotting Scope 3PT-460 20x
Komz Russian Military Telescope - 3PT-460 20x
ZRT 460 Russian Scope

Images of item:

Prices I found range from \$50 to less than \$100.


First we need to decide what you mean by "year"

The first time period you mention "365.256" is the Orbital period of the Earth. It is equal 365 days, 6 hours, 8 minutes, 38 seconds.

The second period is the tropical year, the time between two equinoxes. They are not equal because this period depends on the tilt of the Earth. As the tilt changes (called precession) the tropical year is about 20 minutes shorter than the orbital period of the Earh. Since this is the time between seasons, this is the length of the year that is more important practically, and is the basis of most calendars.

These periods do change over time. This is due to solar system dynamics. The Tropical year can change because the rate of precession is not constant.

The Orbital period also changes due to perturbations from other planets (mostly Venus and Jupiter) The Earth can gain some orbital Energy from other planets, causing it to move slightly further from the sun and increase the length of the year. It could also lose some energy, causing it to approach the sun and shorten the length of the year.

This rate of change is not constant. And over the long term the Earth sometimes gains energy and sometimes loses. Currently, the Earth is gaining energy (mostly from Jupiter) and the distance to the sun is increasing by 0.0005% each year

However if you look at the average changes over a period of 6000 years, there is almost no change in the length of the orbital period. In fact, on average over a period from 3000BCE to 3000CE the Earth has actually lost energy and the distance to the sun has dropped by 0.000003% per year. (Figures from https://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf)

Your value of 5.5 seconds relates to the tropical year and so includes second-order precessional effects.

While these can be usefully modeled by a linear formula over 1000 years, this is just an approximation and over the longer term the actual orbit of the Earth is pretty stable.

Over the very long term (billions of years) other factors, such as loss of mass by the sun need to be considered, but that is not the cause of the short term variation in year length.


Sirius is a binary system, composed of a main sequence star (Sirius A) and a white dwarf (Sirius B). The two orbit each other with a period of about 50 years, and the eccentricity given on Wikipedia is the eccentricity of the stars' orbits around one another. The inclination, according to Bond et al. 2017 - the paper Wikipedia references for both numbers - is described as the inclination of the "relative visual orbit", leading me to believe that the inclination is defined based on a plane relative to the point of view of an observer on Earth.

Something it's important to realize is that stars don't follow Keplerian orbits, because of the non-point-like distribution of matter in the galaxy; the motion of a star around a galaxy is therefore somewhat more complicated than the motion of a star in a binary system.


why do we look at dark energy as an energy?

Since relativistic physics almost always takes units of $c = 1$, there is no difference between units of energy density and mass density. Conceptually, however, the gravitational charge is energy, not mass.

In general relativity, a perfect fluid has no viscuous shear stresses and no heat flux, and can be described by the stress-energy tensor $T^{\mu\nu} = (\rho + p)u^\mu u^\nu + pg^{\mu\nu}$, where $\rho$ is density, $p$ is pressure, and $u^\mu$ is the fluid four-velocity. In the case of vanishing pressure, the result is called a dust, and is appropriate for modeling cold dark matter, galaxies on the cosmological scale, etc. On the other hand, a radiation fluid has $\rho = 3p$, and is appropriate for modeling incoherent electromagnetic radiation, or a fluid made of strongly relativistic particles (e.g., hot dark matter), and so forth.

Thus, in the specific context of cosmology, the matter-dominated regime has $p\sim 0$ and radiation-dominated regime has $p\sim\rho/3$. Note that in the latter case, the fluid particles have zero (or at least negligible) mass. In-between, we can talk about a fluid of particles that are not strongly relativistic, and hence have some mass.

On the other hand, the standard model of dark energy is a cosmological constant, which is equivalent to a perfect fluid $p = -\rho$. It does not fit into the above scale; it is not matter in the above sense, nor is radiation, nor anything in-between. Furthermore, it has the peculiar property of being exactly the same in every local inertial frame. This frame-independence enables us to think of the cosmological constant as an intrinsic energy density of spacetime, but it also makes it inappropriate to think of it as a collection of fluid particles that have some rest frame.

For your purposes, this important because $E = mc^2$ is only valid in a rest frame.


What's the estimated remaining life of Halley's Comet?

This paper estimates "another 2300 close perihelion passages". Multiplied with around 76 years per period, we get roughly 175,000 years. That's probably only a very rough estimate.

If we take the estimated mass of $2.2\cdot 10^{17}g$ and divide it naively by the estimated 1910 mass loss of $2.8\cdot 10^{14}g$, we get just 786 periods or roughly 60,000 years.

Hence, after about 200,000 years the remnant of Halley's comet will probably ressemble an asteroid, if any.


Same reason as the explosion at the Tunguska event years ago: Pressure, due to boiling water.

Meteors usually cme into our atmosphere at incredible speeds. At these speeds, the meteor is highly affected by the viscosity of air and is heated by compression (and a little bit of friction).

This starts boiling the ice present in the meteor, leading to a pressure buildup. If the conditions are right, it can just explode.

See also: http://www.nature.com/nature/journal/v361/n6407/abs/361040a0.html , https://physics.stackexchange.com/q/76045/7433


Excepting a Big Rip scenario, there is no eventual 'clash'.

Consider a Friedmann–Lemaître–Robertson–Walker universe:
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\right]\text{,}$$
where $a(t)$ is the scale factor and $k\in\{-1,0,+1\}$ corresponds to a spatially open, flat, or closed cases, respectively. In a local orthonormal frame, the nonzero Riemann curvature components are, up to symmetries:
$$\begin{eqnarray*}
R_{\hat{t}\hat{r}\hat{t}\hat{r}} = R_{\hat{t}\hat{\theta}\hat{t}\hat{\theta}} = R_{\hat{t}\hat{\phi}\hat{t}\hat{\phi}} =& -\frac{\ddot{a}}{a} &= \frac{4\pi G}{3}\left(\rho + 3p\right)\text{,} \\
R_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} = R_{\hat{r}\hat{\phi}\hat{r}\hat{\phi}} = R_{\hat{\theta}\hat{\phi}\hat{\theta}\hat{\phi}} =& \frac{k+\dot{a}^2}{a^2} &= \frac{8\pi G}{3}\rho\text{,}
\end{eqnarray*}$$
where overdot denotes differentiation with respect to coordinate time $t$ and the Friedmann equations were used to rewrite them in terms of density $\rho$ and pressure $p$. From the link, note in particular that $\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right)$.

If dark energy is described by a cosmological constant $\Lambda$, as it is in the standard ΛCDM model, then it contributes a constant density and pressure $\rho = -p = \Lambda/8\pi G$, and so no amount of cosmic expansion would change things. Locally, things look just the same as they ever did: for a universe dominated by dark energy, the curvature stays the same, so the gravitational tidal forces do too. Through those tiny tidal forces, the dark energy provides some immeasurably tiny perturbation on the behavior of objects, including atomic nuclei, forcing a slightly different equilibrium size than would be otherwise. But it is so small that that it has no relevance to anything on that scale, nor do those equilibrium sizes change in time. The cosmological constant holds true to its name.

On the other hand, if dark energy has an equation of state $p = w\rho$, then a flat expanding universe dominated by dark energy has
$$\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right) = -3\sqrt{\frac{8\pi G}{3}}\left(1+w\right)\rho^{3/2}\text{,}$$
and immediately one can see that there is something special about $w<-1$, leading to an accumulation of dark energy, while the cosmological constant's $w = -1$ leads to no change. This leads to a Big Rip more generally, as zibadawa timmy's answer explains.

  ? If the metric expands surely objects get further away from one another and that would include the stars inside galaxaies as well as the galaxies themselves?

Not at all. It wouldn't even make any sense: if you have an object like an atom or a star in gravitational freefall, by the equivalence principle only tidal forces across it are relevant. The tidal forces stretch the object until the internal forces balance them. But for a Λ-driven accelerated expansion, dark energy contribution to tidal forces is constant. Hence, an object already in equilibrium has no reason to further change its size, no matter how long the cosmic acceleration occurs. This also applies to galaxies, only that the internal forces are also gravitational and balance the dark energy contribution.

Looking at this in more detail, imagine a test particle with four-velocity $u$, and a nearby one with the same four-velocity, separated by some vector $n$ connecting their worldlines. If they're both in gravitational freefall, then their relative acceleration is given by the geodesic deviation equation, $\frac{D^2n^a}{d\tau^2} = -R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu n^\beta$. The gravitoelectric part of the Riemann tensor, $\mathbb{E}^\alpha{}_\beta = R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu$, represents the static tidal forces in a local inertial frame comoving with $u$, which will drive those particles apart (or together, depending). Hence, keeping those particles at the same distance would require a force between them, but it's not necessary for this force to change unless the tidal forces also change.

Galaxies don't change size through cosmological expansion. Stars don't either, nor do atoms. Not for Λ-driven expansion, at least. It would take an increase in tidal forces, such as those provided by a Big Rip, for them to do so.

A related way of looking at the issue is this: according to the Einstein field equation, the initial acceleration of a small ball of initially comoving test particles with some four-velocity $u$ (in the comoving inertial frame), which is given by the Ricci tensor, turns out to be:
$$\lim_{V\to 0}\left.\frac{\ddot{V}}{V}\right|_{t=0}\!\!\!\!= -\underbrace{R_{\mu\nu}u^\mu u^\nu}_{\mathbb{E}^\alpha{}_\alpha} = -4\pi G(\rho + 3p)\text{,}$$
where $\rho$ is density and $p$ is the average of the principal stresses in the local inertial frame. For a positive cosmological constant, $\rho = -p>0$, and correspondingly a ball of test particles will expand. That's cosmic expansion on a local scale; because of uniformity of the FLRW universe, the same proportionality works on the large scale as well, as we can think of distant galaxies as themselves test particles if their interactions are too minute to make a difference.

Again, we are led to the same conclusion: if the ball has internal forces that prevent expansion, then those forces don't need to change in time unless the dark energy density also changes, which doesn't happen for cosmological constant. If they're not 'clashing' now, then they won't need to in the future either.


60 second exposure photo with phone at eyepiece with an app produced giant ball of light, what next?

It's hard to tell, however the very likely solution of your problem lies in a combination of two or three effects:

The image is not focused. Did you focus on bright stars or the moon and not change the phone afterwards? This a common problem and (I personally) find it hard with a phone to get a sharp image through a telescope. On the telescop I have available, I need to hold my phone about 1cm away from the eye piece to get a sharp image. The focus WILL need to be different than for you looking through the telescope with your naked eye. An un-focused image will be just that: bright area where like arrives and black outside, showing vignetting.
the image is motion-blurred.
you really took too long an exposure with too high an ISO rating. Reduce either and see whether you get better results. At highest ISO rating my DSLR camera yields bright white images after a comparable time (I didn't try my phone but suspect similar sensitivity behaviour). How long exactly depends on how dark your local sky is.

From all these points the most important is: make sure you have your camera-telescope-combination focused. Test that on bright objects, thus a sharp moon or smallest-possible dot for a bright(!) star imaged. You can do the focusing in video mode and highest ISO setting. When that is ensured, then only start to image interesting objects.


If the earth stopped rotating, would we really fly inertially at 1000mph?

You would be sliding, because the friction between you and Earth would be too weak, or else you wouldn't survive the acceleration, if the rotation of Earth would be changed abruptly. That's like crashing into a wall with several hundred mph. This is with the exception of close to the poles.

Your weight wouldn't change much. On a smooth, plane, and horizontal surface you would slide (due to your inertia). Your velocity you would be sliding depends on your geographic latitude. At the equator it would be at the maximum. At the poles it would be zero. More precisely the velocity would be proportional to the cosine of the latitude (latitude 0 at the equator, and (+/-) 90 degrees, resp. $\pi/2$ at the poles).

Even more precisely, the speed you would be sliding would be $$2\cdot 465.1~ m/s \cdot cos ~\varphi,$$ with $\varphi$ the latitude.
Factor two, because superman accelerates to the opposite direction. Earth's equatorial rotation velocity is $465.1~m/s$.


What exactly is it that is being magnified 50 times in this gravitational lensing observation?

Reading through the full article, I see that the authors write

  we estimated the lensing amplification to be $\mu\sim52$.

This refers to the magnification factor. Essentially, it describes the solid angle of the image, as related to the solid angle of the source1:
$$\mu\equiv\frac{\theta}{\beta}\frac{d\theta}{d\beta}$$
Here's a diagram of the magnitudes of the angles:

Image courtesy of Wikipedia user Falcorian under the Creative Commons Attribution-Share Alike 3.0 Unported license.


Accurate way to programmatically generate a main sequence star?

This is a tricky question, not immediate.

If we talk about Main Sequence stars, yes, you can constrain your fictional star by two parameters. The crucial parameters in the diagram, as you said, are the mass and the radius. This is why the HR diagram well represents the stellar population with two axis. Of course, for your scope, we need to do some assumptions. For example, the metallicity, that for your purposes can be fixed as the Solar one. Or, that the initial mass is already in a condition to contract.

You can see it this way: the luminosity is connected with the radius (because it depends on the irradiating surface), while the temperature is connected with the mass (the more is the mass, the more is the pressure at the center of the star).

If you have $L \sim RT$ and $L \sim M$, then you also have $RT \sim M$.
The degeneracy from one parameter can be overcame if you have your reference star (the Sun), which allows you to trace isoradii lines on the HR diagram. See here for example:

Or, that is equivalent, you can express your first equation (Stefan-Boltzmann), in Solar units as well (as shown here).

Try it out just by your self: choose a mass, or observationally speaking, choose a luminosity. Then trace your line on the HR diagram, and find your star. That star has a defined temperature (spectral class), Mass and Radius as well.

If you need the exact equations, feel free to ask please.

Nice aim, anyway. Have fun!


Could a dwarf galaxy host a star at the center instead of a SMBH?

A galaxy's center of gravity is not determined by the most massive object, but by all objects in the galaxy. Even supermassive black holes (SMBHs) do not dominate the gravitational field except very, very close to the center. By far, most of the stars in a galaxy couldn't care less about the SMBH.

The region within which a BH dominates over that of the stars (the "sphere of influence"$^\dagger$) is given by (e.g. Peebles 1972)
$$
r = \frac{G M_\mathrm{BH}}{\sigma^2},
$$
where $G$ is the gravitational constant, $M_\mathrm{BH}$ is the mass of the black hole, and $\sigma$ is the velocity dispersion.

In the Milky Way (MW), there's a SMBH (Sagittarius A*) of roughly $M_\mathrm{BH} \simeq 4\times10^6\,M_\odot$. In that region, the stellar velocity dispersion is roughly $50$–$100\,\mathrm{km}\,\mathrm{s}^{-1}$ (e.g. Genzel et al. 2010). Plugging in those values, you'll find that Sgr A* dominates the kinematic out to roughly 3 pc, or 10 lightyears, which is nothing compared to MW's radius of $\sim10^5$ lightyears.

If you take the most massive conceivable stars (a hypothesized Pop III star of $M\sim10^3\,M_\odot$) in Willman 1, the smallest known dwarf galaxy — which has stellar velocity dispersion of the order $5$–$10\,\mathrm{km}\,\mathrm{s}^{-1}$ — you'll find that such a star will dominate the gravitational potential out to a distance of only $\lesssim0.1\,\mathrm{pc}$, again completely negligible compared to the galaxy's radius of $\sim25\,\mathrm{pc}$.

In other words, although it's possible for a dwarf galaxy to host a massive star, it will just be a star like all the others, and will in no way define the galaxy's gravitational center.

$^\dagger$Not to be confused with the event horizon which is even smaller.


What is the composition of an asteroid in percentages?

You might find the information you seek in this article by Nature on the compositional mass distribution of asteroids.

From Wikipedia Asteroid Spectral Types:

  The majority of bodies fall again into the three broad C, S, and X
  categories, with a few unusual bodies categorized into several smaller
  types:

C-group of carbonaceous objects including:[citation needed]
    B-type largely overlapping with the Tholen B and F types.
    C-type the most 'standard' of the non-B carbonaceous objects
    Cg Ch Cgh somewhat related to the Tholen G type
    Cb transition objects between plain C and B types.
S-group of silicaceous (stony) objects including:
    A-type
    Q-type
    R-type
    K-type a new category (181 Eucharis, 221 Eos)
    L-type a new category (83 Beatrix)
    S-type the most 'standard' of the S group
    Sa, Sq, Sr, Sk, and Sl transition objects between plain S and the other types in the group.
X-group of mostly metallic objects including:
    X-type the most 'standard' of the X group including objects classified by Tholen as M, E, or P-type.
    Xe, Xc, and Xk transition types between plain X and the appropriately lettered types.
T-type
D-type
Ld-type: a new type with more extreme spectral features than the L-type
O-type a small category (3628 Božněmcová)
V-type (4 Vesta)

You will have to go to the Wikipedia article to drill down for more detail on the probable minerals contained in these asteroid classes.  We really only have good evidence on those classes that we think we have meteor samples of.  The rest would be guesses.


Does this galaxy resemble the Milky Way, and is it relatable as the "template" for the Milky Way?

In the case of the image of the Milky Way Galaxy that is provided, no.
Wikimedia Commons lists the Milky Way Galaxy image as "Artist's conception of the Milky Way galaxy" and provides the following credit statement, labeling creation date as 25 June 2009:

Nick Risinger - Own work Adapted from the following NASA images:
236084main MilkyWay-full-annotated.jpg Messier51 sRGB.jpg

Given that about 70% of observed galaxies are spiral galaxies, it seems unlikely that 2MASX J10500165-0152029 in particular was specifically chosen to use as a model for formulating the general image of the Milky Way.


Implementation of formulae for Meeus for the Moon

Your understanding of the illumination is correct. k is the ratio of the illuminated length BC to the diameter AC. New Moon has an illumination of 0, Full Moon an illumination of 1, and the quarter phase an illumination of 0.5.

The position angle is measured counterclockwise from north (celestial north, along a line of right ascension) to the bright limb C. You need to calculate the parallactic angle, q, which is the angle from "up" to celestial north. From chapter 14 (in the 2009 version)
$$
\tan q = \frac{\sin H}{\tan \phi \; \cos \delta - \sin \delta \; \cos H}
$$
where $\phi$ is the geographical latitude, $\delta$ is the declination of the Moon, and H is the hour angle. The diagram below shows a positive value for both q and PA.


A colour index measures the ratio of brightnesses in two wavelength intervals, not the difference.

The reason an index is written as the difference between two numbers is due to the logarithmic astronomical magnitude system - the difference between two logarithms is the logarithm of the ratio.

The reason that a red colour is bigger than a blue colour is also because of the perverse way that objects with smaller magnitudes are brighter.

The calibration table you refer to appears to just show that stars of different spectral class and temperature exhibit different colours.

You do however need a calibration to relate real brightness ratios and real properties of stars to the colours. That is because each logarithmic magnitude scale has its own zeropoint (the flux associated with a zeroth magnitude star in that band). The baseline colours are defined using the star Vega as a standard, such that all the colours of an A0 main sequence star are approximately zero (as shown in that table).


Can a black hole with the density of water exist?

Applying a numerical density to a black hole isn't possible. The material inside the event horizon will fall to a "singularity" (or some other ultrahigh density state that we currently have no adequate theory to describe) on a relatively short timescale.

What you can do, is exactly what you have done, which is divide the gravitational mass of a black hole by the volume defined by a simple Euclidean estimate using the Schwarzschild radius$^{*}$.

If instead of substituting out the mass as you have done, you instead substitute out the Schwarzschild radius, then
$$ M = \left(\frac{3}{4\pi \rho}\right)^{1/2} \left(\frac{c^2}{2G}\right)^{3/2} = 1.4\times10^{8} \left(\frac{\rho}{\rho_w}\right)^{-1/2} \ M_{\odot},$$
where $\rho_w = 1000$ kg/m$^3$.

So, if you have a black hole of mass $1.4 \times 10^{8} M_{\odot}$, then it has the "density" of water if you calculate it like this.

There are lots of black holes with this mass or even higher that are situated in the centres of galaxies. The black hole at the centre of the Andromeda galaxy has a mass of about $10^{8} M_{\odot}$, so fits the bill perfectly.

Note that it is a vast simplification (and incorrect) to use $4\pi r_s^3/3$ as the volume inside the event horizon. There is in fact no uniquely defined volume; it depends on choice of coordinate system (see DiNunno & Matzner 2008).


Universe is expanding, why doesn't time expand with it?

The short answer is, "gravity".  The time of "co-moving observers" (observers moving with the expansion of the universe") is expanding, but our "proper time" is not, because we're within a local group of galaxies that are bound together gravitationally.  The principles involved are detailed in the Wikipedia article "Co-moving and proper distances", and its links.

However, in any reading about General Relativity--the currently-accepted theory of gravity-- it may be helpful to understand that "observers" are really any sort of hypothesized material or energetic object, however small:  I believe they were supposed to be hypothetical but possible beings, when use of that terminology began--back around the time when GR's predecessor, Special Relativity, was written--but are now involved in so many inaccessible and inhospitable situations that "they" are not necessarily considered to be sentient, or even alive.  It may also be useful to understand that, although gravity is realistically considered to be the curvature of spacetime, "time" itself is best viewed as comparable to the three well-known spatial dimensions, or at least as something comparably definite, like "duration":  The temporal aspect of spacetime actually includes most of its curvature, as described in a website about General Relativity by the University of Pittsburgh's Dr. Norton.

Regarding your secondary question, the delay in the arrival of particles (photons) from distant stars is actually affected by several factors, generally based on inflationary cosmology, that are in a well-known diagram shown in the first Wikipedia item I suggested, but visible in larger versions with explanatory text:  Searches by the names of its originators, "Lineweaver and Davis", will reach references on various levels of sophistication.


The destruction of the Universe by a bubble

I think I know what they mean. They're talking about a false vacuum scenario.
A vacuum state is a state of lowest energy. It's thought that the vacuum of our universe is in a lowest-energy state and is stable, and so nothing special will happen to spacetime. However, if our universe is actually a false vacuum, then it could merely be metastable, and some tiny perturbation could cause it to fall into a lower energy state - a true vacuum, or else a false vacuum of lower energy. The "bubble" that would appear would be a region of this lower-energy vacuum that would expand across the universe without stopping.
Scientists aren't sure if our universe is a false vacuum. Here's a chart of the masses of the Higgs boson and the top quark:

The latest measurements suggest that their masses lie in a metastable region of the graph (which could be bad), although it's towards the stable end of the region (which is good). This vacuum catastrophe could still happen, but the odds aren't in its favor. I also suggest reading some of the excellent papers in the references section of the Wikipedia article. They're quite comprehensive.

Better explanation:
The universe can be thought of as space containing a variety of quantum fields. A quantum field might be best thought of as something that has a value at every point in space. Put together a bunch of these fields and you can describe particles, particle interaction, and all the matter and energy in a given region of space - in fact, in the whole universe!
Picture a region of space with absolutely nothing in it. Nothing. (I'm ignoring vacuum energy, even though I really should discuss it) That's a true vacuum. Remove all the fields, particles, and other interesting stuff from our universe and that's what you'll get. Nothing.
The false vacuum scenario happens if there isn't actually "nothing" but "something". If our "vacuum" has some extra energy that it shouldn't (again, I'm not talking about normal vacuum energy). In other words, there's something where there should be nothing.
In this end-of-the-universe scenario, the universe goes from having this something to having nothing - from a false vacuum to a true vacuum. The region of space actually having nothing starts at a certain region and expands outward at the speed of light. That's what the scientists were talking about.

For anyone who wants a further explanation and/or could tell how many mistakes I made there: Yes, I'm aware that there were mistakes in that incomplete explanation. I didn't properly explain vacuum energy or an energy state, nor did I discuss stability or even properly touch on quantum fields. But quite frankly, I don't think this explanation needs that baggage. Does it add to the richness of the concept? Yes. But is it confusing? Also yes. I don't think working that in would be productive.


Would a satellite that has a retrograde orbit and a shorter orbtial period than its planet's rotational period be tidally accelerated or decelerated?

In the first case, the planet's rotation will carry the tidal bulge ahead of the satellite,  this will tend to accelerate the satellite, causing its orbit to recede.  In the other cases, the planet's rotation will pull the bulge behind the satellite, causing the orbit to decay.  It doesn't matter if the period is greater or smaller in the case of a retrograde orbit. The bulge is behind the satellite, and so the orbit will decay.


No, this can never be stable over long periods of time.
First of all, drag from the air in from of the moon would cause it to quickly deorbit and plunge into the gas giant. No amount of wind will be able to keep the moon up in the air. A gas giant's orbital speed near the cloud tops will be over 30 km/s (See the orbital velocity of Metis, Jupiter's innermost moon). There will be astronomical amounts of drag if the object is large. If the object is smaller, then it will be irregular, causing the same amount of drag. Also, a gas giant's weather is never constant. There will never be a circle in the gas giant's atmosphere that has a constant wind speed. These fluctuations will probably cause the moon to deorbit over time, or, if it is high in the atmosphere, to be thrown out of the atmosphere into the vacuum of space. The latter is highly unlikely.


Well, it would be useful to define what a 'dead' galaxy is.  Probably the most simple method would be a galaxy that is no longer producing new stars.  We might also consider a galaxy that no longer produces significant light in the visual spectrum, or perhaps EMR across the entire spectrum.

Generally, there's unlikely to be a firm line between living and dead, and not nearly as dramatic as larger stars. More akin to watching a camp fire burn itself out.  Star formation is largely dependent available gases, but as more and more stars fuse those gases into heavier elements, there is less gas available for star formation. For your average sized galaxy, this will eventually result in running out of gas. Eventually the galaxy will dim and go dark, a process purported to begin at the center of the galaxy, where star formation is heaviest according to research based on Hubble images of giant galaxies. (Tacchella, et al.) The matter ought to (mostly) all still be there and still orbiting the (presumed) SMBH, but with no energy coming from fusion, it's going to be a dark, cold, and barren place. Sounds dead to me.

There are some complicating factors.  It's believed that encounters with nearby galaxies can affect available gases. The gravity from a larger galaxy could potentially strip the gases from a smaller one, a fatal blow for the smaller galaxy. Fortunately, it won't suffer much as the death will come (relatively) quickly. This process has been deemed 'strangulation' by a study published in Nature several years months days ago. (Ping, et al.) Note that as the study indicates, the methods of death are proposed solutions - not conclusive understanding of the exact processes that result in a galaxy's death.

S. Tacchella, C. M. Carollo, A. Renzini, N. M. Förster Schreiber, P. Lang, S. Wuyts, G. Cresci, A. Dekel, R. Genzel, S. J. Lilly, C. Mancini, S. Newman, M. Onodera, A. Shapley, L. Tacconi, J. Woo, and G. Zamorani. Evidence for Mature Bulges and an Inside-out Quenching Phase 3 Billion Years After the Big Bang
Science 17 April 2015: 348 (6232), 314-317. [DOI:10.1126/science.1261094]

Y. Peng, R. Maiolino & R. Cochrane. Strangulation as the primary mechanism for shutting down star formation in galaxies Nature 521, 192–195 14 May 2015 [DOI:10.1038/nature14439]

Andrea Cattaneo. Astrophysics: The slow death of red galaxies Nature 521, 164–165 14 May 2015 [DOI:10.1038/521164a]


The recent results on Hubble constant measurements

The Carnegie-Chicago Hubble Program. VIII. An Independent Determination of the

Hubble Constant Based on the Tip of the Red Giant Branch. I hope this helps.
https://arxiv.org/pdf/1907.05922.pdf


Why doesn't smashing larger and larger bodies incrementally into a rocky planet create a star?

Who can say for sure? I guess the physics in Universe Sandbox is not good enough.
What I would say though is that if you have a "protostar" that contains a higher proportion of heavy elements than a usual star, then the threshold for ignition of hydrogen will be lower than 75 Jupiter masses (e.g. How large can a ball of water be without fusion starting? ). If on the other hand it contains very little hydrogen (not enough to sustain a significant main sequence reaction rate), then it will be forced to rely on He, C or even O burning that require significantly higher temperatures and a much more massive object - probably in the region of 1000 Jupiter masses.
However, if you are telling us that some of your simulations collapse into a black hole at 90 Jupiter masses, that suggests that a more simple explanation is that the simulation physics is incorrect. I also think it would take a very peculiar composition indeed to avoid a nuclear burning phase prior to collapse to a black hole, even if you give the object several stellar masses of material.


Why do the gas giants in the Solar System have comparatively large orbits compared to the inner planets?

The current explanation for this is something called the frost line (which changes over time). At greater distances from the Sun, a body will receive less and less radiation, and so it will be colder than if it were closer to the Sun. Eventually, conditions become cold enough for volatiles to condense into grains. These volatiles make it possible for large-scale accretion to happen, and thus giant planets are formed. They may become gas giants (like Jupiter and Saturn) or ice giants (like Uranus and Neptune), depending on the composition of the protoplanetary disk at a given distance.

A graphic (from here) gives a better summary, although it's copyrighted.


Determining line ratios in planetary nebula

You need a model for the excitation of the line.  But since you say you have two forbidden lines, you can try assuming the density is well below the "critical density" (which is the density at which point the upper levels of the transitions get quenched by collisions instead of waiting as long as it takes for the "forbidden" radiative transition to occur).  Then if you are talking about two transitions from the same upper level, the line ratio is simply the ratio of the Einstein A values for the two transitions (note that the Einstein A value normally already includes the statistical weights of the lower levels).  If you are talking about two different upper levels, then you need a model for how those upper levels get populated, and the line ratios will be the ratios of the upper level populations times the Einstein A values.  It's not always easy to know how the upper levels get populated, so that's why it's hard to give you an answer to your question, it is very context dependent.  But the main simplification of using forbidden lines is that you don't have to worry about the photons getting reabsorbed before they get out.  (A detail is that line ratios are often given in terms of energy flux ratios, rather than photon ratios, so you'd need to multiply the Einstein A values by the photon energies if the transitions are at very different energies.)


Why aren't there blue elliptical galaxies or yellow spiral galaxies?

Galaxies consist of dark matter, stars, and gas. While gas is "collisional", i.e. it may interact hydrodynamically and cool, dark matter and, effectively, stars are collisionless. Hence, it is relative easy for an originally more or less spherical, gas-rich galaxy to collapse along the axis of rotation, while centrifugal forces prevents to collapse in the plane. In contrast, a spherical gas-poor galaxy will tend to keep its shape.
Spiral galaxies are blue
That is, a gas-rich galaxy will tend to settle down into a flat disk. Being gas-rich, such a galaxy will keep forming stars, and a spiral pattern will form.
Stars come in all sizes, but the most massive ones dominate the total luminosity (since $L\propto M^4$), and since massive stars shine with very energetic light, they are bluish/white.
Elliptical galaxies are yellow
Elliptical galaxies are the result of a merger between two or more galaxies. If the merging galaxies are very different in size, the smaller will just be "eaten" by the larger. But if the galaxies are similar-sized (and don't have too similar angular momenta), the event can be quite dramatic:
Tidal forces mess up the shape, gas is stripped from the galaxies, and collisions of gaseous clouds ignite massive starbursts. Such major mergers are quite efficient at suppressing angular momentum, allowing large amounts of gas to reach the galactic centers, feeding the central supermassive black hole leading to vigorous quasar activity (e.g. Hopkins et al. 2008).
Both the gas stripping, the radiation pressure from newborn, massive stars, the feedback from their fast and explosive deaths, and the active galactic nuclei, cause the merged galaxy to lose its gas. As the massive, blue stars die out, and with no gas to form new stars, only the smaller, yellowish/orange/reddish stars are left. The galaxy is said to be "quenched", ending up "red and dead".
Why not the opposite?
So the answer to your question is: The reason spiral galaxies are not red is that they keep forming new stars, the most luminous of which are blue.
And the reason elliptical galaxies are not blue is that they're unable to form new stars, so only the red ones are left.
Addendum thanks to Peter Erwin:
Spiral galaxies may eventually deplete most of their interstellar gas (partially through outflows) and stop forming stars. This cause them to gradually change color toward yellow. However, without the feedback from star formation, such a galaxy will also tend to lose their spiral structure, ending up instead as a lenticular (or "anaemic") galaxy (see e.g. Elmegreen et al. 2002).


Look at J2000 Solar System Barycentric (spatial reference) and Barycentric Dynamical Time (TDB).  Together they are good space and time coordinate system.  NASA/JPL has some good info and data on these.

In terms of time, TDB is rescaled so that from the Earth, it appears to be close to the same as TT (roughly ~UTC). The rescaling is due to the fact we are in a gravitational well, as well as moving at 30 km/s with respect to the solar system barycentre, due to relativity. The original unscaled is called Barycentric Coordinate Time (TCB) and differs by ~0.5seconds / year.


Can gravity cause a Big Crunch, if the cosmological principle holds?

You are thinking about gravity and how it affects the universe as a whole in the wrong way.

The dynamics of the universe are governed by the Friedmann equations, which can be solved to say how the scale factor of a homogeneous, isotropic universe behaves with time.

If there is enough gravitating matter/energy density in the universe then the scale factor may eventually decrease to zero in a finite time. The simplest case is a closed universe with no cosmological constant and a mass density greater than the critical density (which is not the universe we currently inhabit: our universe appears to be (a) flat and (b) expanding at an accelerating rate).

The interpretation of this is that the space between the galaxies is contracting. It is the reversal of what we currently observe in the expanding universe.

In the same way that you do not interpret the universal expansion of a homogeneous, isotropic universe as some sort of repulsion between galaxies; neither would the contraction be interpreted as some sort of attractive force between galaxies. As you rightly point out, in a strictly Newtonian sense, there would be no net gravitational force on any one galaxy and so its co-moving coordinates would be unchanged during the contraction.