How close to circular is the Earth's equator

My answer won't be complete, from lack of time and resources, but I still wanted to share some interesting aspects here that could be helpful.

The difficulty in aswering this question revolves around the complex and irregular shapes involved here. Also, finding a "best-fit" ellipse for comparison is not as easy as it seems, because it depends on what and how you want to model. The actual shape of the equator is rather complex and irregular. Generally speaking, it is pretty much circular, but indeed topography and the geoid complicate matters. The Earth's movement in the solar system is not a perfect ellipse either, because of the gravitational interactions with other celestial bodies. Let's review the different irregularities involved here.

The Earth is usually modeled as an ellipsoid of revolution (oblate spheroid), and the equator as a circle. A good example of that is the WGS 1984 geodetic reference system used by the Global Satellite Navigation System. Of course, the equator is not a perfect circle, it has irregularities mainly because of topography, and even sea level itself is a little irregular too. We can approximate sea level with a geoid, for example, here is a map of the EGM2008, a geoid used with WGS 1984 to transform ellipsoidal heights to geoid heights:

Basically, this map shows the height of the geoid (the idealized sea level without the effects of tides and currents) with respect to the WGS84 reference ellipsoid of revolution (semi-major axis 6,378,137 m, semi-minor axis 6,356,752.314 m). The differences are mostly less than 100 meters, and are caused by the irregular distribution of mass inside the Earth itself.

Now, some studies show that the Earth's shape could be slightly better modeled by a triaxial ellipsoid, and one could try to model the equator as an ellipse, and the Earth as a triaxial ellipsoid, however, even with a best-fit triaxial, we would still need geoid corrections for the irregular mean sea level, let alone topography, and geodetic computations would be more complex on a triaxial. Other funny models and names have come up over time, like a pear-shaped (because of a slight bulge in mid southern latitudes) model as a best-fit shape. But if you look at the map above, good luck visually finding the pear shape in there, or other mathematically modelizable aspects of these bumps. We are talking about very subtle differences here, that do not necessarily need to be taken into account for most purposes when describing the general shape of the Earth.

So depending on how you consider the shape of the equator (i.e. by topography and ocean floor, or sea level) you will arrive at a shape that is mostly circular with irregular bumps along the way. There is no authoritative agreement that I know of about an eccentricity of the equator. For instance, This study proposes a flattening of about 70 meters for the equator's ellipse, This article on Encyclopaedia Britannica proposes 80 meters.

For the Earth's orbit, for the sake of this comparison, we can use an best-fit ellipse of 149.598 million km by 149.577 million km. Of course, that is only a idealized ellipse, the real movement of the Earth in the Solar System is more complex.

Finally, say we scale down and superimpose the Earth's orbit's ellipse on the equator to compare. The eccentricity of the Earth's orbit is 0.0167, the semi-major axis is 149.598 million km and semi-minor axis is 149.577 million km. Scaled down by a factor of 23,455 to the equator's size, this corresponds to a difference of about 900 meters in both orbit axes. So I think we can agree that a best-fit ellipse of sea level along the equator is more circular than the Earth's orbit.

However, topography-wise, there are bumps up to over 4,000 meters in the Andes, and the sea floor reaches 5000 meters below sea level in several places. So the "topographic surface" equator would, for one thing, definitely appear more bumpy than the Earth's orbit. Unfortunately, I haven't found a example or study showing what a best-fit ellipse of the equator (including topography and bathymetry) could look like, mainly because we tend to approximate sea level, not topography itself, but with more time, tools and data, it might be possible to work out the answer with a "topographic best-fit" ellipse.


What forces expelled these huge clouds, then blocked further progress, yet allowed it to maintain its threads?

This was a coronal mass ejection.

Those 1973 astronomers weren't looking at the picture correctly. They didn't have the tools at that time to look at the picture correctly. Coronal mass ejections (the term used now) were only discovered a couple of years prior to that picture taken from Skylab, via the Orbiting Solar Observatory 7 satellite. Those early 1970s astronomers called such events "coronal transients". OSO-7 took 45 minutes to transmit a single 256x256 image back to Earth. A proper look at the picture took a few more years to develop. Those coronal mass ejections (the term used now) eject massive amounts of solar material into space.

When looked at via a single still image, it does indeed appear that material is looping up only to fall back to the Sun.

Image credit: SOHO (ESA & NASA)

A better view of a coronal mass ejection became available once those satellites had the ability to send more than an image per hour back to Earth. The image below portrays a coronal mass ejection that occurred in 2001.

Image credit: SOHO (ESA & NASA)


Could a gas giant orbit a star with a 30 year period at a distance of just 1 AU?

Gas giants can be found very close to their suns. The closest ones are then called Hot Jupiters.
But not with a orbital period of 30 years. Everything that orbits a star with one solar mass at 1 AU will also have an orbital period of 1 year, according to Keplers 3rd law:
$$
T^{2}=\frac{4 \pi^{2}}{G M} a^{3}.
$$


Comet Neowise c2020 f3 will return in ~6800 years, how can we be sure Earth will be in the right place to see it then?

The time of the orbit is not well known but the inclination can't change as much (small changes in velocity due to outgassing can significantly change the orbital period, but can't change the inclination much. So, while we have a good idea of it's orbital track close to perihelion, we just don't know when it will be there.
We don't know where the Earth will be in its orbit when it returns, but a comet is typically visible for several months. It is possible that the Earth will be particularly badly placed, but even then it will be visible
The Earth doesn't need to be in an exact position for the comet to be visible. It may not be a naked eye object next time, but it should still be visible, no matter where Earth is on its orbit.


If the Milky Way galaxy is as big as planet earth, how big is Earth?

Well, they are different shapes, but we can get a rough answer by doing a ratio calculation based on their radii.

Let $g$ be the radius of the galaxy, about $5\times 10^{20}\ \mathrm m$.
Let $e$ be the radius of the Earth, about $6371\ \mathrm{km}$.
Let $s$ be the radius of the scaled-down Earth.

Then,
$$\frac e g = \frac s e$$
So $$s = \frac{e^2} g = \frac {(6371\ \mathrm{km})^2} {5\times 10^{20}\ \mathrm m}$$
which is approximately 80 nanometres. So it's about 1000 times the size of an atom.


This would be more appropiate at Physics.s.c but here is the answer nonetheless.

Your object may be perfectly rigid or not.

If the object is perfectly rigid, you would not be able to accelerate it to the point at which the extreme points of the spires reach the speed of light, because when they are approaching it, their mass will increment towards infinity and the energy you will need to continue accelerating will increment to infinity as well.

So you will simply not be able to accelerate the whole thing to a rotation speed where the extreme points will reach the speed of light.

If the object is not rigid, well, it will simply bend, plus the previous explanation.


Why does the moon not leave its orbit around the earth due to the gravitational force from the sun?

The speed of the Earth (and Moon) around the Sun is $\approx 29.8 \frac{km}{s}$.
The speed of the Moon around the Earth is $\approx 1.05 \frac{km}{s}$.

The result is that the Moon has a nearly perfect circular orbit around the Sun, the effect of the Earth is only a perturbation in it:

In fact, the "waves" are much smaller (400.000km Moon orbital radius to  150.000.000km Earth orbital radius)

This is mainly because $R_{Earth-Moon} \ll R_{Earth-Sun}$, and $M_{Moon} \ll M_{Earth} \ll M_{Sun}$. In general, 3-body gravitationally bound systems tend to be unstable, i.e. one of the components tends to fly away with time. But this system seems stable, because the ratio of the masses are big.

The same can't be said from the whole Solar System, (it is a notoriously unsolved problem) although also this seems stable in human time scales.


Not all planets are exoplanets, but all exoplanets are planets.

When we differentiate the two, we use "planets" to refer to the eight planets revolving around our Sun. The Greek prefix "exo" (outside) should indicate that exoplanets orbit other stars.

In reality, you can call exoplanets "planets", and you'll often hear "Astronomers have found x planets around star y". The word "exoplanet" is used interchangeably with planets from other planetary systems.


Can an observer on Earth only see half of the sky?

At any given instant of time, in any place on Earth, if the sky is clear and the horizon is low and flat, you see half of the celestial sphere - at that very instant.

But as the Earth keeps turning, you may end up seeing more, depending on where you are.

If you're at the North or South Pole, you see exactly half of the sky no matter how long you wait. That's all you'll ever see from there.

If you move closer to the Equator, you'll end up seeing more than half, if you're willing to wait.

At the Equator, you see pretty much all the sky, if you wait as the Earth keeps spinning around, revealing all the sky to you eventually.


Will 1036 Ganymed's orbit be changed by its close encounter with Mars in 2176?

Ganymed's Mars approach in 2176 is only moderately close: 4.4 times Mars's Hill sphere radius or 11 times the average Earth-Moon distance.
JPL HORIZONS predicts these heliocentric orbital elements (J2000) before and after the encounter:

Date
e
a
i
Ω
ω

2176-12-06
0.525078
2.65922 au
27.7778°
212.085°
136.031°

2176-12-26
0.525093
2.65922 au
27.7772°
212.085°
136.025°

The change is slight.
With respect to Potentially Hazardous Asteroid criteria,
Ganymed is large enough (H = 9.4 < 22)
but can't get close enough to Earth (MOID = 0.34 au > 0.05 au).


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

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

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

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

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

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

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


Of course you will get multiple peaks in the periodogram. The Fourier series representing a non-sinusoidal signal will contain frequencies at multiples of the fundamental frequency. Similarly, you can have a periodic signal with double or treble the period which will look identical, but where the phenomenon causing the signal repeats either two or three times during each cycle.
The solution is of course to plot the data folded on the proposed period, as you have done.
In the examples you have shown, clearly the period of 3.23d is the only valid period for a claimed exoplanet, since the transit only takes place once per cycle and there is no sign of any other features or excessive scatter in the folded light curve at that period.
For the 9.71d period to be valid you would either (i) have to have three planets, equi-spaced around the same orbital circle, each with a 9.71d period, and of identical size so that the transits are the same depth. That is not a stable situation. Or (ii) you are unfortunate enough to have one planet with a  3.23d period and another with a 9.71d period that transit at exactly the same time. But it is not possible to do that without having unequal transit depths and causing scatter in the transit shape when folded at 3.23d.
I cannot make out what your point is about Kepler 20f. This would produce a single feature in a light curve folded on a period of 19.48d, but the amplitude of that signal is small (and I can't see it). Both Kepler 20c and 20d produce transits that will be $\sim 10$ times as deep, with periods of 10.8d and 77d respectively.


Elliptical craters can be studied three ways:  Laboratory simulations, computer simulations, and observations.
Laboratory simulations have not really been done in detail for this problem in decades (I don't know why, I'm not an experimentalist).  The most recent big modeling study that I know about is Collins et al. (2011) and Eibeshausen et al. (2013).  The former tried to determine what fraction of craters should be elliptical on different solar system bodies, while the latter looked more at that impact angle dependence as a function of other model parameters like pressure.  I suggest going to the Eibenhausen et al. link which is open-access and Fig. 5 shows ellipticity versus impact angle (yes, there is a continuum to answer your question) for a few different pressure regimes.
Observationally, this is rarely studied when building impact crater databases.  Primarily, the reason is that it's of less interest and it requires more effort.  Usually, people will just try to find craters that look elliptical and then, effectively, eyeball the axes and record the major and minor axes lengths.  The problem with this approach is that one must visually recognize the crater is elliptical and then one must subjectively choose those axes.  This then could be compared with a "background" crater population if one has also measured craters that are more circular.  An example of this work would be by Bottke et al. (2000).
A better way to do this (in my opinion) is how I have attempted to study this question.  It begins by tracing the crater rim and then fitting an ellipse to that trace as well as a circle, and doing this for all craters.  I did this with my Mars crater database (Robbins & Hynek, 2012), though the ellipse analysis is in that Collins et al. (2011) paper.  In doing this, there are some problems, though.  One is that any bias in how the rims are traced is going to bias the ellipse results, which are much more sensitive to inaccurate rim traces than circle fits because there are more free parameters.  Another problem is exactly what technique is used to fit an ellipse.  There are lots of different algorithms out there, and many that are the most popular are severely biased in various ways.
I tried to quantify both of those effects in my lunar crater database (Robbins, 2019) by doing a lot of different analyses and playing with a lot of different ellipse fit algorithms.  The supplemental material that discusses most of that is not behind a paywall.  After all those things are factored in, the conclusion is that more craters are "elliptical" than are predicted by simulations, but here again there's a problem: What is "elliptical?"  Is it when the major axis is 10% larger than the minor?  20%?  50%?  100%?  This gets to that problem I mentioned with the Bottke et al. (and similar) studies – when is something visually elliptical enough to count it as "elliptical?"
That said, there is also a problem of studying elliptical craters in the question of: Did the craters form that way?  Differential erosion could cause craters to erode more in one direction and thus become more elliptical as they degrade, especially if they are on slopes.  Another problem is that secondary craters – craters that form from the ejecta of larger, primary craters – form at lower velocities / energy and so are almost always more elongated, so removing those from a primary crater population before studying their ellipticity would also be desired.  This gets to be a problem on the Moon where we might have lots of unrecognized, multi-tens-of-kilometers secondaries from basins that affect what would otherwise be interpreted as primary elliptical craters.
I suppose if I had to summarize all of this, it would be that elliptical craters are, indeed, studied in and of themselves, but there are relatively few of those studies, in part because there is not necessarily too much new science there (is it worth spending a year on a computer to slightly refine percentages? or $millions in a vertical gun lab to refine things slightly?), and in part because the observational study of them is quite difficult due to the time-consuming nature of gathering the data and then understanding the biases in those data before any analysis can be done.


That's mainly because of the gravity, and of size. Small objects, e.g. a stone, create only a very weak gravitation. The stone is much harder than necessary to withstand this force caused by its own gravity.

Earth is much larger. The gravity at the surface causes the weight we feel. A huge mountain on Earth would be flattened under its own weight, because the matter it's made of isn't hard enough to withstand this weight.
By hills filling up valleys the planet as a whole becomes spherical.

A fast-rotating planet gets a slightly flattened ellipsoid shape.


SDSS - getting separate u, g, r, i, z frames from SWARP tool and astropy

What they did was the following: for each individual filter, they assembled overlapping frames into a combined, single-filter image. E.g., they combined several g-band images into a single g-band image, combined several r-band images into a single r-band image, etc. (So they ran SWarp five times, once for each filter.) They then put each combined single-filter image into a separate layer to create a five-layer "datacube"; this is their "64x64x5 pixel datacube".
What you are telling SWarp to do is something different: combine five individual-filter images into a single "$u+g+r+i+z$" image. (People do sometimes do this sort of thing, though with SDSS images they would normally just use the g, r, and i images, since the u and z images are so low-S/N that they mainly just add noise to the result. And in any case this isn't what the paper you're interested in was doing.)


Serious alternate form of the Drake Equation, or graffiti?

Given the font, it's xkcd 384, The Drake Equation.
The two $X$'s are the other $f$ terms from the original Drake equation, while the $B_S$ is "amount of bullshit you're willing to buy from Frank Drake"


If the solar system is nearly flat, then why don't all the planets appear to lie on the same axis when viewed from earth?

The first image is from a July article by TV meteorologist Brent Watts.
Approximating that view in Stellarium:

The yellow arc is the ecliptic, the plane of the Earth's orbit around the Sun.
It looks curved due to the map projection.
If the view is centered on it, the ecliptic looks straight:

Unlike the Sun, which appears on the ecliptic at all times, the planets can appear a few degrees north or south of it because their orbital inclinations are all slightly different.
The more another planet's orbit is inclined relative to ours, and the closer it approaches us, the more its ecliptic latitude varies.


Understanding the theory of an expanding Universe

Hubble discovered that he could measure the approximate distance to some galaxies by observing a type of variable star. The absolute brightness of these stars is known, so by measuring their apparent brightness Hubble could find how far away a galaxy was.

It was also discovered that most galaxies were "red-shifted", which was interpreted as meaning that the other galaxies were moving away from the Milkyway. Hubble plotted a scatter graph with distance on one axis and speed on the other, and noted that the more distant galaxies were moving faster, and the relationship was approximately a straight line. This meant that all galaxies were moving away from all other galaxies.

This expansion could be predicted by general relativity, which allowed for  the space between galaxies to expand. It also suggests that the galaxies were closer together in the past, suggesting that the early universe was hot and dense.

The Milkyway does not have a special position in this expansion, all galaxies are moving away from each other. There is no implication that the Milkyway is central to the universe.


Difference between "planetary ring" and "circumstellar disk"?

From Physics SE:

  A circumstellar disk is a set of objects that orbit about a star-like object and that collectively look disk-like in some way.

From Wikipedia:

  A ring system is a disc or ring orbiting an astronomical object that is composed of solid material

So, a circumstellar disk may have gas, dust, asteroids, etc., but it must orbit a star-like object. A ring system must have something solid--dust or moonlets--but it may orbit any astronomical object.

Other than that, you specifically asked about a planetary ring. That would mean a ring system that specifically had a planemo as the central body.

In other words, a disk around a brown dwarf could be described as a circumstellar disk (a brown dwarf is a star-like object) or a ring system (as long as the disk contained solid material). It could not be called a planetary ring, because a brown dwarf is not a planet.


How many stars are visible to naked eyes around Equator area?

The numbers you're quoting are extreme, more like a theoretical limit, and meaningless in practice.

The 44k number would be true if you assume a limiting magnitude of 8. There aren't many people alive today who have seen at least one object at mag +8. This is not so much due to vision problems (although that may be important for some people), as to the fact that a mag +8 sky is an incredibly rare occurrence today due to light pollution.

If you live on a continent with a low population density (such as North America, Australia, or the remote parts of Asia), then yes, maybe you have a chance of catching a mag +8 sky - if you drive far out into the desert, many hours away from any city or town. But if you live in Europe or the like, forget it.

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

http://www.jshine.net/astronomy/dark_sky/

Furthermore, the 44k number is obtained by assuming the mag 8 conditions, and then counting all known objects in the sky whose brightness exceeds that limit. It is very unlikely that any one person would see them all in one lifetime. Seeing even a single star at mag 8 can be a difficult undertaking for many people, even in ideal conditions.

It is very hard to quote a number, since it is so much dependent on light pollution, transparency, visual acuity of the observer (nearsighted people can't focus well enough to see very faint stars), the observer's age (acuity gets worse with age), the observer's experience (visual astronomers with years of experience will be able to see much fainter stars because they know how to look), etc.

If you're on a farm, away from cities, in a place with reasonably low light pollution, and your eyes are good, and you've been sitting in perfect pitch black darkness for at least 30 minutes prior - when you look up you can reasonably expect to see a few thousand objects, mostly stars. Keep looking, and after a while you will distinguish one or two thousand more stars, very faint, that you could not see at first sight. Practice this steadily for a few years, and you'll add maybe another thousand; but you won't be able to see those all at once - only one at a time.

Now travel to the Cerro Tololo site in Chile, up in the mountaineous desert, zero light pollution, excellent transparency, and you'll multiply all those numbers by a factor of 2x ... 5x.

As you can see, the numbers are very flexible because there are so many factors involved. You can't just slap a 44k label on it and call it a day; that doesn't make any sense in reality. Astronomers know that the pure magnitude number doesn't mean much by itself, because it is just one factor among many.

In a place with very high light pollution (like where I live, in the middle of a large, dense, sprawling urban area in California), you'd be lucky if you can see a hundred stars at night.

Or, in a place with zero light pollution, shine a flashlight into someone's face, and you've temporarily blinded them. You've reduced the number of stars they could see by an order of magnitude for the next half hour (night vision gradually recovers, and it takes 30 minutes to fully recover, according to US military manuals and visual astronomers practice).

Now, Equator versus poles.

The spin axis of the Earth has a fixed direction in space. The north - south axis always points at the same places in the sky, near Polaris in the north, and near the Southern Cross in the south.

For this reason, at the North Pole, you only ever see half the sky - the northern half, centered on Polaris. At the South Pole, you only see the other half. You'll never see Polaris from Australia.

At the Equator, you pretty much see the whole sky - not all at once, of course, but as the Earth rotates, it gradually exposes all parts of the sky at different times. However, the parts near Polaris and near the Southern Cross are always low in the sky and a bit hard to see.

So, bottom line is this:

In any place on Earth, in any given instant, you see exactly half of the sky. At the poles, that half is all you ever see. At the Equator, if you're patient, you'll eventually see the whole sky in time, as the Earth rotates and gradually exposes different parts of the sky at different times.


Is the earth bombarded equally in all directions by neutrinos?

There are only two types of neutrino source that are "bright" enough to be reliably detected. The sun and nearby supernovae.

The source of solar neutrinos is nuclear fusion, which is also the source of most of the star's energy. Neutrinos also spread out in all directions, so their intensity follows an inverse square law. So the amount of neutrinos is proportional to the brightness of the star. With current detectors no star is bright enough to be observed except for the sun. Other stars do produce neutrinos, and stellar neutrinos come from everywhere (probably more from the milky way) but there are not enough of them to be detected.

Supernovae in the milky way and neighbouring galaxies produce ridiculous amounts of neutrinos, and a spike in neutrinos was observed from SN1987A the closest recent supernova.

Since the sun is the brightest source of neutrinos you might think that the Earth would block out neutrinos during night. However, neutrinos pass right through the Earth almost without noticing. The Earth is transparent to neutrinos. So we detect as many neutrinos at night as during the day.

One thing is certain, there is absolutely no effect on the Earth's rotation or anything else from neutrinos, they just pass right through.


Is the location that published calendars use to calculate the seasons one specific location on the earth?

Astronomical events such as the solstices and equinoxes occur at a specific instance in time on a specific day, often given in Universal Time.

For example, this winter's solstice occurs on 21 Dec 2014 at 23:03 Universal Time. Since most of Europe is an hour ahead, this may occur on 22 Dec 2014 at 00:03 (and subsequent hours) given in local time based on the time zone. So British and American calendars will show the solstice occurring on 21 Dec, and in countries east of Greenwich on 22 Dec.


How Are Radioactive Decay Rates Influenced by Neutrinos - On Earth and Other Dense Planets

Resurrection alert.

I was Googling the subject and found this paper refuting an earlier paper claiming to have observed a neutrino-based effect on beta decay in radon.

Edit: another study here finding there to be no correlation.


Finding luminosity of a star with magnitude

The ugriz values are just measures of the stellar brightness taken in 5 bands from the near ultraviolet u through to the near infrared z.

You need the distance to the star to convert the apparent SDSS magnitudes to absolute magnitudes.

Then you need to find a relationship (or relationships) between bolometric correction in a particular band and the colour of the star (by colour I mean something like g-i). Possibly instead you might find relationships between bolometric correction and temperature and will need to first estimate the temperatures of the stars from their colours.

Once you have absolute magnitudes and a bolometric correction, then the absolute bolometric magnitude follows.

In principle you could try this with a number of different colours and see if the results agree. Alternatively, you would fit a library of stellar models to the spectral energy distribution defined by the 5 magnitude values.

A table of bolometric corrections for the SDSS filters was calculated by Girardi et al. (2004) and can be found here. (Other tables/formulations are available, there is no universally accepted version).

Some bolometric correction tables can be found in Schmidt et al. (2014) for very cool objects, which are more problematic.


Are there sufficient observational data to measure non-Newtonian perihelion advances of any Asteroid and Comet orbits?

I am by no means an authority on this area but prompted by user /u/called2voyage I will refer the 1994 Astronomical Journal paper by Shahid-Saless (Colorado) and Yeomans (JPL) Relativistic Effects on the Motion of Asteroids and Comets.

To paraphrase their abstract: They study the predicted effects arising from relativistic perturbations on the motions of asteroids and comets and show that for a number of such objects, inclusion of relativistic contributions in the equations of motion give rise to significant improvements in the orbital solutions.  They go on to argue that ignoring such corrections to the equations of motion gives incorrect solutions.  They point out how the use of masses derived from relativistic ephemerides together with purely Newtonian equations of motion results in an inconsistent hybridized non-Newtonian, non-relativistic model.

With respect to the posted question:-
(a) are the existing observational data on any Asteroids or Comets sufficient (in terms of accuracy and extent of observations) to determine/estimate rates of non-Newtonian perihelion advance?

...the authors short-list 15 asteroids with the largest predicted relativistic perihelion precession rates (from a long-list of 156 commonly-studied objects).  For example Icarus has the greatest rate (0.101 arcseconds/year) and had (in 1993) been observed over 43 years thereby producing a predicted cumulative relativistic precession over this period of 4.34 arcseconds. Observations referred to are a mix of optical and radar (no spacecraft telemetry) and are those used to produce the ephemerides model from which orbital elements are extracted.

The authors note that, assuming a (then) current observational accuracy of a few tenths of a second of arc, the relativistic contribution to precession should be detectable for Icarus and other asteroids near the top of the (ranked by precession-rate) list. As with the planets the majority of asteroid perihelion precession will be caused by the perturbing influences of other planets, but the authors do not present calculated values for these contributions).

(b) are any such determinations publically available?

...the authors selected six asteroids whose motions are significantly affected by general relativistic effects. They then calculated a set of orbital elements for each orbit using JPL's development ephemeris DE200 (with Earth and Moon perturbations treated separately).  They did this in two ways, firstly with Newtonian equations of motion alone and secondly with the addition of non-Newtonian General Relativisitic equations of motion.

The authors indicate that the residuals (differences between observed and predicted positions at various times) are improved (i.e. reduced) by application of non-Newtonian corrections to the equations of motion. For example the RMS residual for Icarus is improved by 30%. Thus it is technically possible to calculate differences in the orbital behavior (such as perihelion advance) predicted by (i) the pure-Newtonian model and  (ii) the Newtonian plus non-Newtonian model.

However the paper does not actually present explicit accounts of total observed perihelion precession or a breakdown into Newtonian and non-Newtonian components.  In any case, due to complexity, such a breakdown could only apply over the period of observations which is limited (maximum 61 years in 1993) compared to the timescale (a few hundreds of years) over which (predominantly Newtonian) planetary-induced perturbations might be expected to average out.

However it would be possible (assuming there are no major perturbing encounters with planets or the asteroid belt) to use the provided asteroid orbital element data in a numerical orbit simulator program to roughly model the perihelion precession rate of an asteroid orbiting the Sun in the presence of the other planets using Newtonian equations of motion (i) without and (ii) with relativistic modifications calculated from $\frac{F}{m} = \frac{GM}{r^2} \left(1+\frac{3V_t^2}{c^2}\right) = \frac{GM}{r^2} \left(1+\frac{3GM.SLR}{c^2.r^2}\right)$ (see Walter-Goldstein-Schwartzchild Formulation ).  That way the amount of asteroid precession resulting from each of the two different sources (Newtonian, non-Newtonian) could be modelled.

Note: As /u/UhOh/ commented, an alternative, possibly superior, equation for the extra, relativistic acceleration is given in the Shahid-Sales & Yeomans paper cited above as Eqtn 3.11, on page 1886.


What object in the universe is most opaque to neutrinos?

... So what object will absorb the highest fraction of neutrinos passing through it, or at least be a good candidate? Feel free to assume a certain energy range of neutrinos. Exclude black holes because they just absorb everything and that's not as interesting.

Neutrinos have amongst the smallest mass and travel at nearly the speed of light, this property, along with their weak interaction allows them to travel through all but the densest objects.

You've asked for an answer that excludes trapping by gravity, ridiculously long objects should also be excluded. That leaves reasonable sized (existing) objects of extreme density.

If the progenitor star is in the mass range 8–25 M$_\odot$ the core slowly contracts to a neutron star, and it is therefore called proto-neutron star (PNS). Once it contracts and becomes particularly dense it becomes opaque to neutrinos. It also need to release a huge amount of angular momentum, through neutrino emission, to balance the Equations of State (EoS). In the first tenths of seconds after core-bounce, the PNS is turbulent and wobbly, but during the next tens of seconds, it undergoes a more quiet, “quasi-stationary” evolution (the Kelvin-Helmholtz phase), which can be described as a sequence of equilibrium configurations.

This phase is characterized by an initial increase of the PNS temperature as the neutrino degeneracy energy is transferred to the matter and the PNS envelope rapidly contracts, and then by a general deleptonization and cooling. After tens of seconds, the temperature becomes lower and the neutrinos mean free path is greater than the stellar radius. The PNS becomes transparent to neutrinos, and a “mature” neutron star is born.

The creation of a proto neutron star is explained in "Neutrino Emission from Supernovae" (Feb 28 2017), by H.-Th. Janka. It has this simple illustration on page 4:

  Fig. 2 Evolution of a massive star from the onset of iron-core collapse to a neutron star. The progenitor has developed a typical onion-shell structure with layers of increasingly heavier elements surrounding the iron core at the center (upper left corner). Like a white dwarf star, this iron core (enlarged on the lower left side) is stabilized mostly by the fermion pressure of nearly degenerate electrons. It becomes gravitationally unstable when the rising temperatures begin to allow for partial photo-disintegration of iron-group nuclei to $\alpha$-particles and nucleons. The contraction accelerates to a dynamical collapse by electron captures on bound and free protons, releasing electron neutrinos ($v_e$), which initially escape freely. Only fractions of a second later, the catastrophic in-fall is stopped because nuclear-matter density is reached and a proto-neutron star begins to form. This gives rise to a strong shock wave which travels outward and disrupts the star in a supernova explosion (lower right). The nascent neutron star is initially very extended (enlarged in the upper right corner), and contracts to a more compact object while accreting more matter (visualized by the mass-accretion rate $\dot{M}$) within the first second of its evolution. This phase as well as the subsequent cooling and neutronization of the compact remnant are driven by the emission of neutrinos and antineutrinos of all flavors (indicated by the symbol $v$), which diffuse out from the dense and hot super-nuclear core over tens of seconds. (Figure adapted from Burrows, 1990b)

Notice that the image in the lower right blocks $v$ (neutrinos).

Text on page 2:

  "... [Lots of interesting information] ... [shortest possible quote] ... With neutral-current scatterings of neutrinos off nuclei and free nucleons being possible, it was recognized that the electron neutrinos, $ν_e$, produced by electron captures can escape freely only at the beginning of stellar core collapse (which starts out at a density around 10$^{10}$ g cm$^{−3}$), but get trapped to be carried inward with the in-falling stellar plasma when the density exceeds a few times 10$^{11}$ g cm$^{−3}$. At this time the implosion has accelerated so much that the remaining collapse time scale becomes shorter than the outward diffusion time scale of the neutrinos, which increases when scatterings become more and more frequent with growing density. Shortly afterwards, typically
  around 10$^{12}$ g cm$^{−3}$, the electron neutrinos equilibrate with the stellar plasma and fill up their phase space to form a degenerate Fermi gas. During the remaining collapse until nuclear saturation density (about 2.7×10$^{14}$ g cm$^{−3}$) is reached, and the
  incompressibility of the nucleonic matter due to the repulsive part of the nuclear force enables the formation of a neutron star, the entropy and the lepton number (electrons plus electron neutrinos) of the infalling gas (stellar plasma plus trapped neutrinos) remain essentially constant. Since the change of the entropy by electron
  captures and $ν_e$ escape until trapping is modest, it became clear that the collapse of a stellar core proceeds nearly adiabatically (for a review, see Bethe, 1990). The proto-neutron star, i.e., the hot, mass-accreting, still proton- and lepton-rich predecessor object of the final neutron star, with its super-nuclear densities and extreme temperatures of up to several 10$^{11}$ K (corresponding to several 10 MeV) is highly opaque to all kinds of (active) neutrinos and antineutrinos. Neutrinos, once generated in this extreme environment, are frequently re-absorbed, re-emitted, and scattered before they can reach semi-transparent layers near the “surface” of the proto-neutron star, which is marked by an essentially exponential decline of the density over several orders of magnitude. Before they finally decouple from the stellar medium closely above this region and escape, neutrinos have experienced billions of interactions on average. The period of time over which the nascent neutron star is able to release neutrinos with high luminosities until its gravitational binding energy (Eq. 1) is radiated away therefore lasts many seconds (Burrows and Lattimer, 1986; Burrows, 1990a).".

In the study "Observing Supernova Neutrino Light Curves with Super-Kamiokande: Expected Event Number over 10 s" (Aug 22 2019) by Yudai Suwa, Kohsuke Sumiyoshi, Ken'ichirō Nakazato, Yasufumi Takahira, Yusuke Koshio, Masamitsu Mori, and Roger A. Wendell they investigated neutrino properties observable by Super-Kamiokande up until 20 s after the bounce using the database of Nakazato et al. (2013). It includes this text and accompanying graphic:

Page 4:

  "While the neutrino-radiation hydrodynamics (RHD) simulations account for the neutrino emission before shock revival, the neutrino light curves from the PNS cooling simulations are reasonable for times after the shock revival. On the basis of these considerations, the neutrino light curves of the early and late phases are interpolated by an exponential function assuming shock revival at either $t_{revive}$ = 100, 200, or 300 ms after bounce. In Figure 1, a typical neutrino light curve obtained by this procedure is displayed."

Page 6:

  Figure 1. Neutrino luminosities (top panels) and average energies (bottom panels) as a function of time after bounce for the
  13M⊙, Z = 0.02, trevive = 300 ms model.


I am assuming that the question you want answered is how to calculate the elevation of an orbit above a reference plane given the orbital inclination with this plane. If so, please update your question to reflect this, heeding the advice given in the comments.

Kepler's first law tells us that planets move in elliptical orbits, which we can define as follows,

$$r = \frac{p}{1+\epsilon \, \cos\theta}$$

where $p$ is the semilatus rectum, $r$ is the orbital distance from the central body, $\epsilon$ is the orbital eccentricity, and $\theta$ is the angle to the orbiting bodies current position from its closest approach.

Using $r$ along with the orbital inclination with respect to a given plane, we can calculate the height above this plane at any point in the orbit using simple trigonometry.

The image above shows the relevant variables for a given orbit when the body is both above and below the reference plane. $r$ is again the orbital distance, $i$ is the orbital inclination, and $d$ is the height above the reference plane asked for in the question. $d$ is then given by

$$d = r \, \sin \, i$$


What projection would I use to map the northern night sky onto a geodesic dome?

The mathematically ideal scheme would be a separate gnomonic projection
(see Wikipedia or
IMO)
for each tile of the dome.
Many planetarium programs implement this projection as an option.
Fortunately, each tile boundary is a straight line on both adjoining maps.
Unfortunately, continuity across tile boundaries requires some care in choosing each tile's scale and center of projection.


Create Position-Velocity diagram from a velocity field

Such diagrams are usually called rotation curves, and show the velocity of stars/gas in a disk galaxy viewed edge-on along the line of sight (LOS) as a function of distance from the center.

That is, the $x$ axis gives the distance of the stars from the center ($x=0$) of the galaxy, and the $y$ axis gives the measured velocities at a given distance. Along a given LOS you'll find not a single velocity, but a (Gaussian-like) distribution of velocities, both because there is a certain scatter in velocities, but also because you only measure the component of the velocities projected onto your LOS, while both speed and direction change along the LOS, as seen here:

If they all had the same velocity along the LOS, the plot you show would be a thin line; instead it is smeared out.

Due to mass being concentrated in the center, there's a steep increase in velocity at low $x$. If there were only the mass you can see, the velocity would decrease at larger $x$ due to the larger distance from the central mass. The reason it stay more or less constant is thought to be the extra, invisible mass known as dark matter.

The reason the left side of the plot is similar to the right, but mirrored not only across the $y$ axis but also the $x$ axis, is that on this side, the stars move toward you, while on the right side they move away from you. The whole galaxy evidently moves away from you at $500\,\mathrm{km}\,\mathrm{s}^{-1}$, since this is the velocity at $x=0$.

So, to recap: How do you generate such a plot? You measure the distribution of velocities along many lines of sight in the galactic plane, from one side ($-R_\mathrm{gal}$) to the other ($+R_\mathrm{gal}$). For instance, at $x=14\,\mathrm{kpc}$, you'd measure $V=675\pm12\,\mathrm{km}\,\mathrm{s}^{-1}$:

In practice, images like this is made by using a grism which disperses the light according to its wavelength, together with a slit placed along the galactic plane, such that you block the light from the rest of the field. This is called slit spectroscopy, in contrast to "normal" spectroscopy where you block all light except for a little hole placed on some object (e.g. a star), or integral field spectroscopy, where you get the whole spectrum behind every pixel in an image.


How would a small nearby black hole be detected and confirmed as such?

I would say that the only way this can be done is making ever more careful observations of the TNOs that are claimed to be affected by this putative black hole.

Ultimately such observations should narrow down the location of the object. Even then, confirming it's existence will be very difficult. It will have a small lensing/magnification effect on background stars, but it would have to get very close (in terms of angular separation) to a background star for this to be observable. It's unlikely that the position could be pinpointed accurately enough to predict when such a microlensing event would take place.

Perhaps a better bet would be to send a probe, like New Horizons, to its likely location, where the gravitational effects could be studied in situ.


Try https://www.coursera.org/courses?query=astronomy

Online courses from several universities. Some are on a schedule with online lectures, etc. Others you can just do yourself at any time.

Also there are extensive libraries on YouTube, for instance from this guy Michel van Biezen.
http://www.youtube.com/playlist?list=PLX2gX-ftPVXWddG6sGq6H-3cN0IQzfQmG

That link goes to the first of his playlists about astronomy. His astronomy section has about 20 parts!


At what distance could a supernova damage the Earth's ozone layer?

I came across your question and figured I would reply to it because doing a project on the subject. Explanations here are draw from the website astronimate.com

Would a supernova of Betelgeuse only 440 light years from Earth have any dangerous effects on the Earth's ozone layer (damaging it dangerously) and on orbital spaceflight?

No. A star would need to explode within 30 to 50 light-years to jeopardize Earth. Fortunately, no stars even remotely close to such distances are prime to explode any time soon. In fact, the most likely bomb candidate, IK Pegasi, is a safe 150 light-years away. And, at such distances, we would see the spectacular celestial object, but avoid virtually all danger!
However, were a star 30 light-years away to explode, Earth would be in major danger. Some, or all, of our ozone layer, would disintegrate, leaving us vulnerable to lethal radiation from our Sun.
Not to mention, phytoplankton and other aquatic food chain staples would completely parish. Ultimately, our ocean life and primary food chain would die off, creating massive evolutionary problems.
Plus, essential gases in our atmosphere, like nitrogen and oxygen would likely be ionized by bombarding radiation. Eventually, this would cause obvious dangers to most living organisms. Alas, Earth seems to be quite safe from supernovae for several million, or billions of years. Yet, hopefully, we will experience a supernova from a safe distance. After all, this would be one of the most notable sights in recent human history!


Take another photograph of the same field that is less exposed. (Doesn't matter if you do this by shortening the exposure time, decreasing the ISO, etc.) This will give you an image with far fewer stars so you can easily pattern match your images.


What are the metrics of observation conditions?

I'm not an observer, but there are a couple of quantities I know:

seeing: is a measure of atmospheric turbulence and airmass. Its units are arcseconds and is the measured size of a point source, typically a star, as it appears in the image. In short: atmospheric turbulences shift slightly the position of the star, so if you observed it for long enought, the star will appear blurred. Good places for telescopes have seeing of few arcsecond; Excellent ones reach about 0.6 arcsec. This link can give some more info
fog, cloud, rain, snow: you don't even bother opening the telescope domes as you won't be able to observe much
humidity: I think that part enter in the seeing. If you do just imaging what you get is some extra blurring of the image. If you do spectroscopy you will see a lot of atmospheric lines (due atoms and molecules, e.g. water, in the atmosphere) that you want to remove from the measured spectrum.


What does "additive fringing from the (multiplicative) twilights" mean?

The fringing pattern is caused by thin-film interference within the CCD. The signal received in a pixel will be proportional to the light falling on it, multiplied by a sensitivity, but then some extra signal is added or subtracted which depends on how much of the incoming light is at particular wavelengths that are affected by the interference (i.e. the whole broadband signal does not contribute, only the flux at certain wavelengths).

The fringing pattern is therefore an additive modulation of the sky background, which in general, for broadband filters (e.g. the i filter is 100 nm wide), has both a broadband component (scattered sunlight, moonlight) and a component at discrete wavelengths (airglow lines). Only the latter make a significant contribution to the fringing.

The fringing does not affect stellar photometry measurements, other than by introducing structure in the background that can be difficult to deal with when performing background subtraction. However, it does affect the images cosmetically and disrupts photometric measurements of extended objects.

It is not possible to remove perfectly because the amplitude (and also the shape) of the fringing pattern depend exactly on the spectral make-up of light illuminating the pixels and this will vary depending on (e.g. time of night, whether the Moon is up).

A first-order correction is to make a "fringe frame" by median stacking all the moonless night time exposures, after scaling them by the median sky signal (or the exposure time). A scaled version of this can then be subtracted from every processed image frame to remove the background component that is affected by fringing. It is hence referred to as an additive correction.

Twilight flat fields are used for an entirely different purpose. At twilight the sky background is close-to-uniform and has a broadband continuum spectrum that is dominated by scattered sunlight. It is therefore very difficult to see any fringing in it with broadband filters.

That is good, because the purpose of the twilight flats is to calibrate out the broad band sensitivity of each pixel by dividing by the flat field (hence referred to as a multiplicative correction, and you would normally multiply by a "balance frame" constructed by dividing a smoothed version of the median twilight flat by the unsmoothed version). The twilight flats can also be used to calibrate any large spatial scale sensitivity variations caused for example by vignetting. Twilight flats that do show evidence of strong fringing in broadband filters should not be used to remove pixel to pixel sensitivity variations. In those cases one usually resorts to "dome flats".


Does the CMB pattern evolve in a human lifetime?

In principle yes, in practice no.

As seen in the temperature power spectrum below, the Planck satellite detects power (i.e. "a signal") even on the smallest probed scales, which is a few arcminutes.

Fig. 1 from Planck Collaboration et al. (2016) with my own approximate corresponding angles annotated in red.

The smallest scale shown in the figure is $\ell = 2500$, corresponding to an angle of roughly $4'$. At the redshift of the CMB ($z=1100$), this angle subtends 57 million lightyears (Mlyr) today. When the CMB was emitted, such a region was only 52 thousands lightyears (klyr) across).

In other words, we can see that at least down to regions that were roughly 50 klyr across, the CMB pattern is anisotropic.

As time goes, the CMB we see will come from an increasingly larger distance. Hence, when this distance has increased by roughly 50 klyr, the CMB pattern should have evolved. This will take a bit more than 50 thousand years$^\dagger$(kyr), i.e. much longer than a human lifetime.

To be able to see changes within 100 years, we would need an angular resolution of 0.5 arcseconds. But angular resolution is not enough — the power seems to fall off as a power law on these scales, so a higher sensitivity would also be needed.

$^\dagger$A bit more, because space expands somewhat during the time that light travels that distance, so in total roughly 55 kyr.


$\xi$ and $\eta$ are just arbitrary names of coordinates, used because the actual position on the sky doesn't matter in this context, i.e. only the relative position in the plot is important. An analogue would be to label the coordinates of your living room $x$ and $y$, instead of using Earth's longitude and latitude.
If you really want to transform, you can use the information given by the authors that an "over-dense feature" is located at $(\xi,\eta) = (0.25,0.06)\,\mathrm{deg}$, which they say corresponds to $(\mathrm{RA},\mathrm{dec}) = (2^\mathrm{h}21^\mathrm{m}2.2^\mathrm{s}, -34^\circ 27'17.6'')$.
This transformation requires that positive $\xi$ and $\eta$ values point toward east and north, respectively. This is not explicitly written in the reference you give, but is seen to be the case in the paper they quote (de Boer et al. 2012). Thanks to @planetmaker for digging this up!


An ADS search on refereed astronomical papers including the words "Boötes Void" in the title returns only two papers in this millennium, in 2001 and 2002, and even these do not report any new data, but use data from the beginning of the 1990's. I can't find any newer references for the number density of galaxies in the Boötes Void in particular, but a typical value is roughly one-tenth of the average number density in the Universe.

Theoretical approach

I previously answered the similar question of the number of galaxies in an overdensity. This answer is based on an anlytical fit to (observationally calibrated) simulations of galaxy halo formation, giving the so-called halo mass function, i.e. the number of galaxy halos per halo mass. The total number $N_\mathrm{gal}$ of galaxies in a volume $V$ can be expressed as
$$
N_\mathrm{gal} = n_\mathrm{gal} \times V \times \delta,
$$
where $n_\mathrm{gal}$ is the average number density of galaxies in the Universe, and $\delta$ is the relative density in the volume. As I wrote in the other answer, $n_\mathrm{gal}$ is a number that impels you to define a lower threshold of galaxy size. The reason is that the lower you go, the more there are, and there is no formal threshold for how larger a clump of stars you need before you call it a galaxy. But as in the other answer, for the sake of this discussion we may use Small Magellanic Cloud-sized galaxies as our lower minimum. In that case, with $\delta\sim0.1$ and $V\sim236,\!000\,\mathrm{Mpc}^3$, the total number becomes
$$
N_\mathrm{gal,Böotes} = 0.17\,\mathrm{Mpc}^{-3} \, \times \, 236,\!000\,\mathrm{Mpc}^3 \, \times \, 0.1 \, \simeq \, 4000\,\mathrm{galaxies}.
$$

Units of galaxy number density

The way I write this result also answers your question 3: Galaxy number densities are almost always written in $\mathrm{Mpc}^{-3}$. In theoretical/numerical work, you'll often see the factor $h^3$ in front of the unit. This is simply the Hubble constant divided by 100 (i.e. $h=0.7$ for $H_0 = 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$), allowing people to compare results more easily without knowing the exact value of $H_0$.

Observational approach

The observations of the Boötes Void are old, and seem to have been carried out on 1m-class telescopes. Hence, they won't be able to observe the smallest galaxies. In addition to telescope specifications, weather, etc., the exact detection limit (in terms of a limiting magnitude $m_\mathrm{lim}$) depends on how long they integrate (i.e. expose). Without reading the old papers in detail, I can't say what this is, but a typical value for such surveys would be, very approximately, $m_\mathrm{lim} \sim 20$ (if someone has a more realistic value than this, please edit). That is, objects fainter (i.e. larger values due to the backward astronomical system) than $m=20$ would be missed.

The distance to the Boötes Void implies a distance modulus of $\mu\sim37$, so the minimum absolute magnitude is
$$
M_\mathrm{lim} = m_\mathrm{lim} - \mu \simeq -17,
$$
which is somewhere between the Small and the Large Magellanic Cloud.

The figure below (from Wyder et al. 2005) shows the local-Universe luminosity function for UV-selected galaxies.
That is, it shows the number density of galaxies at a given magnitude. For instance, it shows (with the green dashed line) that the number density (here called $\Phi$) of galaxies with magnitudes around $M=-17$ is roughly $10^{-2.5}\sim0.003\,\mathrm{Mpc}^{-3}\,\mathrm{mag}^{-1}$.

Integrating over magnitudes from the $M=-17$ doesn't change the 0.003 much, since the density of brighter galaxies quckly declines. I get 0.004, i.e. the number density of galaxies at least as bright as $M=-17$ is $0.004\,\mathrm{Mpc}^{-3}$, smaller than the theoretical result above by 1.5 orders of magnitude. Multiplying this by the volume $V$ and the relative density $\delta$ yields $N_\mathrm{gal,Böotes} \simeq 100$ galaxies, not far from the 60 you quote.

To conclude, the number 60 seems in rough accordance with what is expected observationally, but theoretically, we'd expect there to be many more galaxies (although they're very small).

Location of the galaxies

They detected galaxies seem to be lying in a "tube" extending across the void. In general, galaxies and the underlying dark matter mass field tend not to be evenly distributed, but to form knots, sheets and filaments, separated by voids. My guess is that this "tube" is such a filament. Outside this filament, the void is voidier, but not completely void. There will still be galaxies, although few and small.


What will succeed the Arecibo Observatory?

There's no simple answer. In the immediate future, different radio telescopes around the world will pick up the slack in various ways; how that happens will depend on the needs of individual observers and collaborations. Unless someone was to build an identical observatory at the same latitude as Arecibo, with the same frequency range, receiver options and field of view . . . we'll have to spread out the observing. Given that we didn't really expect to lose Arecibo for more than about a year$^{\dagger}$, there are a lot of things up in the air. Insert "probably" and "maybe" into this answer wherever you please.
I do pulsar timing, so I'll talk about what a future without Arecibo might look like from our perspective. My collaboration's two main instruments were Arecibo and the 100-meter Green Bank Telescope (GBT), with roughly half of the observing time at Arecibo and half at the GBT. There aren't firm plans in place for the exact strategy we'll take going forward, but we'll have to decrease our observing cadence at the GBT (i.e. how often we observe certain pulsars) so we can use some of the time to observe pulsars we'd normally observe using Arecibo. So the GBT can take some of the load, but as we only have a finite amount of observing time, that impacts our other observations. We'll likely have to decrease the cadence of observations and the number of pulsars we monitor regularly.
Pulsar timing arrays will also have options with other existing instruments in the immediate future. A key one is the Canadian Hydrogen Intensity Mapping Experiment (CHIME) in Canada. The folks at CHIME will actually be able to observe all of our northern hemisphere pulsars (CHIME Pulsar Collaboration, 2017), which is excellent. In the short-term, CHIME will slightly lessen the burden of losing Arecibo. We've also done a little bit of timing with the Very Large Array, although I don't think that'll be a major player in the future.
That brings us to telescopes currently in the planning stages or under construction. The DSA-2000 (Hallinan et al. 2019) will hopefully see first light in the late 2020s and would likely become a major part of our ongoing observations. The Square Kilometer Array will also be extremely important for radio astronomy as a whole - I don't know whether we in particular will be using it a lot. The SKA should also reach completion in the next decade.
These are just the current potential options from my corner of the astronomical world - and they're quite tentative. Arecibo was our most-used screwdriver in the toolbox, so we've certainly been hit. We'll hopefully be able to somewhat supplement the GBT with CHIME and ideally DSA-2000 a decade from now - so in that sense, those will effectively succeed Arecibo for our observations. In reality, all that means is that we won't lose quite as much of our observing capacity - but it doesn't come close to making up for the loss. Other collaborations and astronomers will presumably increase their usage at the observatories that are best equipped for their observations, depending on frontend/backend requirements and field of view.
It's not going to be fun. We'll be set back. But science will go on. We'll figure it out.

It's worth updating this answer now that the white paper for the proposed Next Generation Arecibo Telescope (NGAT) has been released. The Arecibo Observatory has proposed a phased array of small dishes mounted on a circular platform ~300 meters in diameter (or potentially multiple smaller platforms) which would be angled in various directions to provide reasonable sky coverage. One variant involved 1,112 9-meter dishes, as smaller dishes provide a higher packing efficiency and therefore maximize collecting area. Regardless of the precise configuration, the sinkhole that housed the former 305-meter dish would ideally be the new array's home.
The aim is for frequency coverage from 200 MHz to 30 GHz via four broadband receivers and a frequency-dependent field of view ranging from 6 degrees at 300 MHz to 3.5 arcminutes at 30 GHz, a significant field of view increase over its predecessor by a factor of 500. The dishes would also be equipped with transmitters for radar observations.
Whether the NGAT will be built is a question for which nobody has any answers just yet. It would cost roughly $450 million, which could take money away from other projects unless it received special Congressional funding or support from private sources. Additionally, the setup has not been tested on such a large scale, so there are some technical and engineering concerns (the group did consider but reject single-dish and classical array designs). All that said, the design was put together in an impressively short amount of time, and even if the NGAT isn't build, the scientific considerations that went into it may inspire future designs for replacement telescopes at Arecibo.

$^{\dagger}$ This isn't totally true - a scenario where we lose Arecibo had already been studied in detail. But . . . I think a lot of folks felt that it was more likely that it would be out of commission for a bit but could be salvaged in some way, and hopefully brought back online before too long. This was not the Thursday we were expecting.


How do we know the universe's expansion is speeding up?

As a fellow layman, I'll give this my best shot.

  The further away a galaxy is from us, the faster it is moving away
  from us. But the galaxies we see exist in the distant past because it
  takes a long time for their light to reach us. The closer galaxies
  are, the more recent their light is, and the slower they move away
  from us.

All true.   Think of a single point explosion, at any given time, the object twice as far should be moving twice as fast, cause they both started in the same place at the same time, so twice as fast moves twice s far.   There are, complications - with the explosion there's air resistance, with the universe well, I'll get to that.

  Doesn’t this seem to indicate that the expansion of the universe is
  slowing down, because the more recent light from galaxies, appears to
  be moving away slower? Does this make sense?

OK, it sounds like you have an incorrect assumption there.   Expansion means the far away is moving away faster and the closer is moving away slower, that's true with accelerating or decelerating expansion.

What they look for when studying expansion is measuring the precise speed and comparing that to distance.   In steady expansion, you expect to see the galaxy 8 billion light years away to be apparently moving away twice as fast as the galaxy 4 billion light years away.  (I say apparently cause we can only observe the relative velocity from 8 billion years in the past).

So, whether the galaxy 8 billion light years is traveling slightly more or slightly less than twice as fast away from us as the galaxy 4 billion light years away - that's what tells us about acceleration vs deceleration.

Imagine a car driving 60 MPH, and the driver puts the ever so slightest pressure on one of the pedals, but you don't know which one, and similarly the car driving 30 MPH does the same, but we can only see how fast the 60 MPH car was driving 6 minutes ago and we see how fast the car driving 30 was driving 3 minutes ago - that gives the 30 MPH car more time to apply the acceleration or deceleration.    It's that time differential that gives us the information.

So, like the cars above, the galaxy 4 billion light years away has been in the accelerating expansion of space longer than the one 8 billion years ago and that's what tells us that the acceleration of expansion is happening, the nearer galaxies are moving a bit faster away from us than they would in a steady expansion or a gravitationally slowed expansion.  (edited to make a bit more clear)

  The newer light is moving slower. The older light moving faster. So
  from this it seems that the universe was expanding faster in the past?

light doesn't move slower or faster.  it red-shifts when objects are moving away from each other.  The faster they move away the greater the red-shift.  That's one way relative velocity can be measured.


The DS9 Reference Manual, p42, 'Numeric Formats'. http://ds9.si.edu/doc/ref/ref.pdf#page48 has a bunch of information.

Based on that, I expected that "tlhm.1" would achieve what you proposed.

However, it seems that SAOImage DS9 is very keen on dropping leading hours it sees as unnecessary, even if you have specified them in a format specifier.

I haven't been able to figure out a format specifier combination that retains the hours for every gridline while also showing the minutes or seconds.

I have searched elsewhere for advice on this and haven't found it either. In fact, I haven't been able to find any comment on the hours being suppressed, nothing that explains why it does this, nor even anyone else remarking or complaining about it.

I suggest sending an email to the DS9 Help Desk address found in the menu under Help > Help Desk.


Do star systems stripped from galaxies have different expected behavior?

The local environment is very important to star formation because it requires material as well as something to perturb the nebula to start collapsing and forming star(s). Therefore, both the type of galaxy as well as the location in the galaxy are very important to creating stars.

However, once a star moves from protostar onto the main sequence, external influences don't much affect its development. This is how we can identify stellar streams because the stars come from a similar population when they are created, but some tidal forces cause them to spread out around the galaxy.


No, the light doesn't just disappear, it adds to the black hole's mass in accordance with $E=mc^2$. So if a star produced light from $m$ kilograms of mass and all that light fell into a black hole, the black hole's mass would increase by $m$ kilograms.


How are solar systems found again after being discovered?

I think you're overestimating how quickly stars appear to move as viewed from Earth.  Once a telescope like Kepler identifies a solar system around a star, we know where that star is.  Stars have coordinates associated with them that astronomers can use to know where to point their telescopes to see it.  For example, look at the star Betelgeuse.  Look in the info box on the right, the right ascension and declination are the coordinates to find it.  On the timescale of one Earth year, or even a few hundred Earth years, the stars are mostly fixed, and that coordinate to find them still works.  Sure they move slightly, but not enough to lose track of it, or even enough to dramatically change how the constellations appear to us.  Even if one does somehow move dramatically, it would take decades to centuries, and astronomers would be tracking it and updating the coordinate information.

Think about it this way.  The constellations used in the sky today are named after many ancient Greek and Roman mythological characters.  This isn't something we decided recently, they were named hundreds or thousands of years ago and they're still recognizable in the same form they were then.  If stars moved across our sky as quickly as you appear to think, those constellations would all have to be renamed.  You need timescales of tens of thousands of years to make the sky unrecognizable, and even then it would have more to do with the precession of the equinoxes.


Is the measurement of distance and position of remote celestial bodies accurate?

Large masses can bend light, but space is largely empty.  The light from distant stars and galaxies rarely passes close enough to another star or galaxy to have deviated. On the few occasions when it does, it is special and notable.
For example, the Einstein cross looks like four quasars in a (very small) square, with a galaxy in front of it.  In fact it is four images of one quasar, the light having been bent by the gravity of the foreground galaxy.  In this case the image of the quasar is split up and moved by a few thousanths of a degree, because there is a very exact alignment of a quasar and a galaxy.
Such examples are rare.  For nearly everything else the light has travelled in a straight line through flat and empty space. The light we see pinpoints the location that the object was when the light was emitted.
An exception to this is that our own sun (and to a lesser extent the other planets) create local distortions. In very high accuracy measurements this can be taken into account. But the distortions are very small and as we know the location of the sun, they can be fully taken into account.
Although the light travels in straight lines, it can be very hard to measure the distance to stars and galaxies.  Often there is considerable uncertainty on the distance of astronomical objects. But this is not a result of gravity, it is just because measuring distance is hard.


Why solar eclipse paths are symmetrical?

The orbit of the moon is inclined by 5.14° to the ecliptic.

As you may know, the ecliptic is the apparent path of the Sun across the sky, so it is also the plane of Earth's orbit around the Sun. The plane of the Moon's orbit is inclined by 5.14° to the plane of the Earth's orbit. The intersection of the two plans is a line bisecting both planes.

There is a nice picture of this at http://www.cnyo.org/tag/orbital-plane/. [I'll insert it here after checking the copyright]

So, it is only when the Earth is very near to the intersection line that an eclipse can occur, because it is only then that the Sun, Moon and Earth can be in a strait line. This occurs for slightly more than a month (called an eclipse season), twice a year.

Another angle comes into the picture too. The axis of the Earth is tilted, so the equator is at 23.4° to the plane of the ecliptic.

So, the pattern you have observed comes about because of that geometry, where mirror image eclipse paths occur about six months apart.

The linked pages explain a lot more of the complexities.


What do the calendar rounds in Tomás Filsingers "Map of the Universe 2.000" mean?

The star map is centered at the north celestial pole.
A thin red circle marks the celestial equator.
The Roman numerals around the edge of the star map are hours of right ascension, measured along the equator and subdivided into minutes.
Sidereal time runs slightly faster than solar time and corresponds to the right ascension of celestial objects currently crossing the overhead meridian.
The hour labeled XIV here is more commonly designated 0h.
The months in the blue outer ring show when the Sun appears along those parts of the ecliptic, marked on the star map with a white dotted circle.
Numbers at intervals along that circle indicate the approximate day of the month when the Sun passes those points.
Meteor showers are marked when the Earth encounters those meteoroid streams, e.g. the Orionids in late October.
The orange triangles pointing outward for each day may be mostly decorative.
As you inferred, the thin circles on either side of the month ring are for the planets in order of their distance from the Sun.
I would expect the number of dots punctuating each of these lines to match that planet's synodic period in days, indicating the planet's average rate of apparent motion along the ecliptic.
Depending on the Earth's and the planet's relative positions in solar orbit, this motion may at times be faster, slower, or retrograde.
The gray ring in 12 segments shows the signs of the tropical zodiac, 30° blocks of longitude measured along the ecliptic.
They are named after constellations which were in the same parts of the sky ~2000 years ago; precession since then has changed that alignment.
Some sign boundaries correspond to the equinoxes and solstices.
The Sun passes those points about 70% of the way through those calendar months.
Since the ecliptic is oblique to the equator, ecliptic longitude is slightly condensed around 0h and 12h right ascension and slightly expanded around 6h and 18h.
This chart may instead show ecliptic and equatorial coordinates aligned all around for aesthetic reasons.
The days of the week are named after the Sun, Moon, and bright planets in Latin and related languages.
Other than dividing the circle into 7 equal parts, their placement on this chart may be arbitrary.
The rainbow-colored ring shows the spectrum of the Sun, with thin dark lines where the electrons of gas atoms and ions absorb photons of particular wavelengths.
Astronomers use this information to identify elements in the atmospheres of stars and planets.
They also use displacements of spectral lines to measure radial velocity.
The dashed yellow circle in the outer part of the star map, tangent to the ecliptic at 18h right ascension, is the circle of -23.5° declination.
The corresponding circle of geographic latitude is called the Tropic of Capricorn.
South (outward) of -35° declination, the star map shows a few highlights in addition to Crux and α and β Centauri.
These include the Large and Small Magellanic Clouds and the globular cluster ω Centauri.
Crux may be plotted a few degrees east (clockwise) of its actual position for aesthetic reasons.


What is the accuracy of an artificial satellite at a certain magnitude?

What you are referring to is the precision with which the astrometry (position) of stars will be measured with the Gaia satellite.

The sentences you refer to in Perryman et al. (2014)  are "Gaia will achieve accuracies of some 10μas (microarcsec) in positions and annual
proper motions for bright stars (V ∼ 10), degrading to around 25μas at V = 15, and to around 0.3 mas (300μas) at V = 20 (Lindegren et al. 2008)."

What this means is that after the 5-year Gaia mission is complete, the angular position (or coordinates) of 10th magnitude stars on the sky will be determined with a relative precision of 10 millionths of an arcsecond.  This will also be the accuracy with which the annual tangential motion (the so-called proper motion) of stars can be measured. i.e. 10th magnitude stars will have an uncertainty in their proper motion of $\pm 10 \mu$as/year.


What is the lifetime of an x-ray binary?

I am not an expert on the topic but probably you have to specify in your search which mass regime are you looking for (low or high mass x-ray binaries). Accretion time must be important and depends on the mass. High-mass stars are young (very short life) < $10^7$ years. This reference is quite old but probably helps a bit:

The Average X-Ray Lifetime of Massive X-Ray Binaries
Observations of accretion disks in X-ray binaries
X-Ray Binaries


Why do small mirror imperfections matter with modern computers

correct for the distortion

An imperfect mirror does not produce a distorted image - it produces a blurry image. With light-field sensors and phase imaging, one could possibly correct for the blur, but it is much more challenging problem than normal lens distortion correction.
Distortion refers to a systematic change in how shapes are projected in an image. It results from a lens or mirror with good, accurate geometry that just does not produce a rectilinear projection.
Random imperfections in a mirror do not cause distortion. Every point in the surface of a mirror contributes to every pixel in the result image. If a single part of the mirror is at slightly wrong angle, it does not cause a distortion in one point of the image. Instead, it projects the same image at a slightly different alignment on the same sensor. (1)
In the case of a starfield, this would cause ghost images of very dim stars to appear next to the real stars. Repeat this for a thousand imperfections, and the result is just blurry dots. Deconvolution is a process that can be used to remove blurriness, but noise and other uncertainties limit its effectiveness.
(1) This may be a bit unintuitive if you think about funhouse mirrors where the image is distorted. Those work differently because they act the part of a planar mirror, where indeed each part of image is reflected by a single part of a mirror. But planar mirrors cannot form an image by themselves, instead the lens in your eye is the critical component of the image accuracy.


Could stellar occultations be used to survey Kuiper belt objects?

Yes this is possible and has been tested by the TAOS (Transneptunian Automated Occultation Survey) group among others. That group is in the process of building the TAOS-II project which will use several telescopes and fast readout CMOS cameras (since the occultations only last a few seconds) to search for occultations caused by unknown Kuiper Belt Objects. Multiple telescopes are very useful to build confidence that you have a real event and not atmospheric fluctuations or other instrumental effect. A detailed workthrough and calculations of how many objects you would see, what is needed to detect them and resolve the Fresnel diffraction and other issues is given in a section by Roques et al. in the Solar System Beyond Neptune book and which is available here.

In general for surveys such as occultations by unknown KBOs, it is etendue (product of telescope collecting area and amount of sky covered in a single exposure) that matters (but it is often more complicated as survey depth doesn't just scale with telescope area) in that higher etendue value equals "better" survey. So for the camera(s) in a survey, it is sensor area that matters, which is a product of pixel size and no. of pixels (and number of sensors and size of gaps between if you are tiling them). The telescope area/diameter tends to be set by how bright your targets of your survey are and practicalities such as cost and manufacturability tend to influence the f/ratio and therefore focal length and pixel scale at your detector. Rarity of your targets and how many you want to find, drive how much area you need to cover. (The telescope area also influences this since in general you get more targets per area if you use a bigger telescope and go fainter). So you end up trading all of these things off to get your best survey for your particular science and type of targets, normally constrained by available or potential budget.

In the case of trying to time-resolve occultations by KBOs, your exposure time is set by the event timescale which depends on the velocity of the KBO across the line of sight and the diameter of the KBO. For typical KBO velocities of 20 km/s, this means that a Pluto (approx 2000km diameter) occultation lasts $2000\,km/20\,  kms^{-1}=100\,s$. If you wanted to find small 20 km KBOs similar to 2014 MU69/Ultima Thule, then the events are $20\,km/20\,  kms^{-1}=1\,s$ long (both of these calculations assume the occultation goes over the largest part of the KBO and isn't grazing).

Given that you will want several points during the occultation, both to convince yourself it's real and to measure diameters etc, you will need at least 5 points across the event. For the 20km KBOs/1s events, this means at least 5 frames per second, ideally more. This means that you need to get enough signal-to-noise on your target stars in a 0.2 or 0.1s exposure.

As discussed in the reply comments to the question author, scientific CMOS sensors such as the Andor Marana (a commercially available closeish match to the custom detectors used by TAOS-II with ~$31k pricetag; Jan 2019) will be better at both the 'signal' part (due to higher quantum efficiency) and the 'noise' parts (larger, less noisy pixels, cooling and temperature stabilization) than a consumer DSLR sensor. Looking at a study of DSLR dark current by Photonics.com, the dark current is ~50 counts/pixel/second (the Photonics.com figure is for 30s exposures and assuming 1500DN) rising to ~100 counts/pixel/second (averaging at ~3000DN for 3 of the pixels) after 100 frames as the sensor heats. For the Marana, the figure is 0.26 counts/pixel/second (taking datasheet value of 0.2 electrons/pix/s and assuming a gain of 1.3 to match the full well capacity of 85k electrons to the 16 bit ADC (65536 levels)), an almost 400x lower value. (The ADC data range is another area of difference; the Marana can do 16bit digitization whereas DSLRs are typically 12 or 14bit, giving only 4096 or 16,384 levels to record all the signal levels in the image)

The other use of occultations is to monitor predicted occultations by known asteroids and Kuiper Belt Objects. This can be used to get a better idea on the size and shape of the object (as you note for Ultima Thule/2014 MU69), search for additional satellites or rings around the object (as has been done for the KBO Haumea (Nat Geo article) and the Centaur Chariklo (Space.com article)) and look for and monitor changes in the atmosphere (as has been done for Pluto since the first occultation in 1988).


The resource you cite, in addition to this full list of crater's names are the official, definitive lists of named craters. Note that you have the coordinates of each crater in the complete list; I guess you could reuse these informations for your project. As for the atlas, it is specified in the "introduction" of the atlas that the list is up-to-date with the approved nomenclature of lunar craters.

The IAU (the International Astronomical Union) is the official organisation in charge of naming celestrial objects et al. You can rely on it.


How much difference did the supply of "fresh" r-process elements make to the Solar System.?

The discussion in that answer is slightly misleading. Here's the current picture, taking into account the Nature paper by Bartos & Marka referenced there.

Elements heavier than iron up through rubidium (atomic number 37) are produced by the r-process in supernovae.
Elements heaver than rubidium are produced by a combination of the s-process in the asymptotic giant branch stage of intermediate-mass stars (i.e., masses of 2-10 solar masses) and the r-process in neutron-star mergers.[1]
The argument in the Nature paper is that when the solar system formed, a significant fraction of actinide metals with short half-lives (e.g., isotopes of curium and plutonium, with half lives $< 100$ million years) came from one recent, nearby neutron-star merger. (These are the "fresh" r-process elements.)
Stable heavy elements produced by neutron-star mergers would have come from the cumulative contributions of hundreds (possibly a few thousand) NS mergers over the history of the Milky Way, mixed over hundreds of millions or billions of years into the Galaxy's interstellar gas (these are the "stale" r-process elements).
Unstable heavy elements with long half-lives -- e.g., U-238 (4.5 billion years) and Th-232 (14 billion years) -- would be mostly in the "stale" category, and so would not be significantly affected by the presence of said recent, nearby NS merger. In fact, the Nature paper estimates that their hypothesized nearby NS merger would account for only 0.3% of the solar system's initial Th-232. Even U-235, with a half-life of 700 million years, would probably only be moderately affected by that one NS merger.

In general, then, the presence or absence of metals from a single nearby NS merger would have made very little difference to the formation of the solar system or to the evolution of life on Earth.

[1] See this figure for an approximate breakdown by element -- note that the "dying low-mass stars" label in the figure refers to the AGB stage of what I call intermediate-mass stars, which are "low mass" only in comparison to stars with masses $> 10$ times that of the Sun.


Why do some electromagnetic waves continue travelling while others disappear?

Electromagnetic radiation will continue to travel until it is absorbed.  Some of your wifi signal is escaping to space where it may continue traveling for a very long time.
However, the strength of your wifi signal will degrade with distance according to the inverse square law.

So if you double the distance between your device and the wifi transmitter, your signal will be reduced by 3/4. When you move far enough away, eventually your signal will degrade enough that your device will no longer be able to decode the signal.  So the radiation doesn't "disappear" but instead, your device will just be able to pick up less of the energy.  Of course, if you separate the device and transmitter by a concrete wall, some of the signal will actually be "disappearing" as it is absorbed by the wall.
The CMB (Cosmic Microwave Background) on the other hand, was not emitted from a point source, and is not subject to the inverse square law.  Instead, as wikipedia points out, the CMB

is very nearly uniform in all directions, but the tiny residual
variations show a very specific pattern, the same as that expected of
a fairly uniformly distributed hot gas that has expanded to the
current size of the universe.

So one can imagine that wherever we are in the universe, we can point an antenna in any direction and get about the same amount of signal (excepting Doppler shift) from the CMB. Of course, if we point our antenna towards existing matter, we will expect some or all of the CMB to have been absorbed in that direction. In many cases, astronomers can tell a lot about the composition of gas in absorption based on the narrow frequencies of the CMB that was absorbed. Back in the early 2000s, I spent quite a bit of time looking for cosmic formaldehyde in absorption lines of the CMB near the center of our galaxy, as an indicator for early star formation.


Does the earth spiral around the sun's movement/motion path?

Does the earth spiral around the sun's path as it is shown in the
video (exact time is 19:49) and screenshots?

The Earth does spiral around the Sun's path, but not quite as it's shown in the video.

It isn't clear if the video accounts for the 60 degree tilt of the Solar System with respect to the galactic plane.

The sizes and distances are not to scale.  The Sun and the Earth are depicted far too large with respect to their distance from one another.

The relative motion is not to scale. The distance from one side of the spiral to the other side should be about 2 AU, or 3x10^8km.  But the Sun is moving at about 200km/s around the galaxy, or 6.3x10^9km per year.  So the distance between two consecutive spirals should be more than 20 times the distance from one side of the spiral to the other!  This can be visualized as an overextended slinky.

Is the spiraling movement caused by sun movement (on its path shown
above) where the sun is dragging the earth (and other planets)?

No, the Sun's and the Earth's mutual gravity cause the Earth to rotate around the Sun.  The Milky Way's gravity causes our Solar System to revolve around the galactic center.  The Sun and the Earth are on the same plane and moving through the galaxy together.  The Earth would still rotate around the center of the Milky Way if the Sun magically disappeared.

Does the earth go in front of the sun then back, then fourth and back
like it is shown on the video? I see both the sun and earth racing
with each overtaking the other over and over. Is this assumption true?
In this manner, would earth, at one point in time, be closer to the
destination where the solar system is heading (I think they call the
destination Vega) than the sun?

If you consider a plane tangent to the Sun's velocity vector, then the Earth will pass across that plane twice per year, appearing "in front of" and then "behind" the Sun.  This is simply a geometric result since the Earth and Sun share another plane. I wouldn't call any future position of the Solar System a "destination" without evidence the Solar System was sentient, but I am not a philosopher.

I am working on a poetry piece. If I mention that earth and sun are
locked in a never ending race with each taking the lead (earth moving
in front of the sun's path / winning) and earth actually dropping out
of the race (spiraling/moving out the race and falling behind), that
they are destined to never meet (collide), would I be correct?

We haven't ruled out the possibility that the Sun may someday envelope the Earth: What will "the Earth being consumed by the Sun" look like?. Also, the Earth isn't likely to leave its orbit before that time, but the Solar System is chaotic on long time scales. However, your poem would be consistent with current models, and we (humans) don't conclusively know how the Solar System will end.  So my two cents is that your poem doesn't contradict modern astronomy.  Besides, what's the use of a poetic license if you don't use it?


If our sun was already in the process of becoming a red giant, would the gradual rise in luminosity be noticeable to our eyes at some point in human existence?

I'd say no for a couple of reasons.

@PM2Ring's comment:

FWIW, in about 1.1 billion years it will be too hot for most lifeforms currently on Earth, long before the Sun starts moving off the main sequence and becoming a red giant. See Timeline of the far future.

Our eyes and vision processes accommodate (adjust for) changes in brightness, we walk indoors from a sunny day to a well lit room and the brightness of things has dropped by 99% or nearly a factor of 100 within a few seconds and yet though we are somewhat aware of the change of tens of percent per second we hardly notice it.

Our vision system is constantly adjusting its white point to accommodate changes in the color of available light.† A white sheet of paper looks white to us even as the color of ambient light changes because our vision is always "color balancing" (not exactly the right term but it's something like that).

†From @Luaan's comment

It's called "white balancing" - calibrating vision for ambient light to make white surfaces (hopefully) appear white. Sometimes it doesn't quite work properly, and you get fun things like The Dress (since different people's vision chooses different arbitrary ambient conditions, and end up disagreeing wildly about the colors involved)


The formation of absorption lines in solar spectrum

In the upper photosphere, the electrons join with the ions to neutralize the atom. When a photon strikes this atom (the atom absorbs the photon), it promotes a electron from a low energy shell to an higher energy shell. The electron then falls back to the lower energy shell emitting a photon in a different direction. When you break the sunlight into wavelengths (by a prism), you will see dark bands where the light was absorbed by the atoms and emitted in a different direction. These bands correspond to the energy absorbed and released when an electron changes energy shells. Here is the sun's light spread out to show the absorption bands.

Added: Here is a high resolution of the wavelengths from the sun.

Apr 20: You have the wrong interpretation of Figure 16.9. At level A, there is about the same number of photons at all wavelengths coming at the observer on Earth. As the photons pass through the Chromosphere, the photons with the wavelengths of the atom's electron transition are absorbed and redirected away from the observer. The photons with different wavelengths of the atom's electron transition eventually arrive at the observer. This is why there are dark lines in the spectrum. It is because most of the photons are absorbed and redirected in a different direction.

The absorption lines are formed at a higher region since the photons are removed from the spectrum.


Has the Milky Way collided with another galaxy?

According to our current understanding of physical cosmology (as contrasted with cosmologies of various cultural references) the Milky Way must necessarily have had collisions in a process called hierarchical clustering.
How we're certain of this is the following:

No matter what exactly happened at the big bang, we know the state of the universe at roughly the age of ~380.000 years, where the universe becomes transparent (more or less instantaneous, but let's omit that discussion). At this moment, matter emits light that teaches us the distribution of matter in the young universe.
This is the Cosmic Microwave Background Radiation.
In the CMB we see that matter is remarkably homogeneous, apart from very small fluctuations, that help us build the cosmological standard model. This means at the time of the CMB emission, the universe had nearly the same matter density everywhere. Which is the contrary of clustered matter blobs (galaxies) that we see today.

Fig. 1: CMB temperature fluctuations as seen by WMAP.
Those variations can be translated into density fluctuations, given a physical model. To our current understanding the form the seeds of galaxy growth.

Now we take those small fluctuations, add dark matter and evolve a model of the universe over time, in large-scale compuational facilities. Then we find that galaxies are a natural outcome of the small CMB fluctuations that become gravitationally unstable. Those unstable blobs of matter amplify the initial fluctuations, until whole pockets of matter collapse into dense, rotating objects.
Those objects would be dwarf galaxies, and those collide and merge and thus grow into larger, mature galaxies. This is the process of hierarchical clustering, I was referring to above. This process can be simplified to "small things collapse first", as opposed to "large things collapse first". In the latter case the Milky Way would not have had collisions before.

Summarizing
Yes, our galaxy had collisions before, and they were necessary to grow to it's current size. Scientists are searching for the relics of those collisions in a process called galactic archaeology (the link goes to a personal website, which I find neatly made).