The NOAA formula sheet
expresses some quantities in radians, some in degrees, and some in minutes of time.
It does say which is which, but it's easy to miss. A few minor code changes can help expose bugs:

Store angles in radians. Trigonometric expressions free of Math.toRadians calls are easier to verify against a reference document.
Put that document's URL in a comment so it's easy to find.
If you need to store something in other units,
make it obvious
with a naming convention such as long_deg for longitude in degrees.

The other quirks are the $\cos(\phi)$ and $\cos(180 - \theta)$
on the left hand sides of those equations.
$\arccos$ has a restricted range of $[0, \pi]$,
which is OK for zenith angle but leads to 180-degree errors in azimuth
unless you code east/west cases or derive new formulas where you can use Math.atan2.

The solar hour angle should equal 0 at noon and increase by 1 degree every 4 minutes.
Sunset hour angle should be about 90 degrees near an equinox, smaller in winter, larger in summer.

The sunrise/sunset formulas produce times in minutes past midnight UTC, so you'll need to apply a time zone offset to get local times.


Timekeeping on other celestial bodies than Earth

The HORIZONS web interface allows for you to change the observer location to other bodies in the solar system.  Select "Observer location [change] and read the instructions carefully.  I was able to generate an ephemeris showing the Sun's position for the Viking lander location by entering in "Viking 1@499".  Mars is designated by "@499" and it allows you to enter in any Mars surface coords (Mars lat/long) before the @499 to generate the ephemeris.  There is a search function to find the code for nearly anything in the solar system; it will always be @ something.
To determine when the Sun rises and sets on Mars, you'll also need to change the columns returned to include "Apparent AZ/EL" (azimuth and elevation over the horizon).  Also, you'll want to have it run for the accuracy you're looking for, such as every 5 mins or every 1 min, etc. Here is a screenshot using 10 mins for the step time.


Impact of inclined orbit satellite on a planet

Yes, yes and yes.

Your main misconception is that satellites follow a sinusoidal path. That is just a distortion of the map.

The path of the satellite is in a single plane. When that plane meets the Earth produces a great circle (which is effectively straight, there are no straight lines on the curved surface of the Earth, great circles are the closest that you get to straight)

When you project that great circle onto a map you may get a sinusoidal path, it depends on the projection. Your statement "The ground track of a satellite follows a sinusoidal path" is not correct, that is just an effect of the transformation from a globe to a plane, not "real".

The path of the fragments is, compared to the size of the Earth is very short, it is on the same great circle that the satellite was orbiting in (except as the fragments get closer to the ground they start to move with the atmosphere). If projected they would be on the same track as the satellite. But a great circle is the closest thing to straight on the surface of the Earth.  The rotation of the Earth is much slower than the orbital speed. By the time the fragments have slowed to terminal velocity then they fall roughly in a parabolic curve to Earth, the same as if you threw a ball off a building.


The synodic period of the moon is $29.53$ days, a little shorter than a calendar month, which is on average about $30.4$ days. This is slightly longer than its orbital period, but corresponds to the periodic visual appearance of the moon as viewed from Earth. I mention this to make it clear that we should be forgiving of a little imprecision.

Conventionally, the moon's appearance is divided into four phases: first quarter, full, last quarter, new. That means that on average, each phase lasts about $7.4$ days. Since calendars count days in integer amounts, a $7$-day period seems to be a natural choice.

The social importance of the seven-day period in Western cultural probably has much more to do with its religious significance in Abrahamic religions than astronomy per se (although certainly not unique to it). But its ultimate origin probably does lie in the natural division of the moon's appearance into four phases, which correspond to an apparent geocentric celestial latitude difference between the Moon and Sun of $0^\circ$, $90^\circ$, $180^\circ$, and $270^\circ$.

That, the explicit answer to your question is

1 week = 7 days = one lunar phase.


Naming convention for multiple star systems

The purpose of the notation is to indicate how the system is physically constructed, to indicate which stars orbit which - especially useful in hirachically organized stellar systems. So both ways are correct as their meaning is understood.
The notation I encounter most often is the first one you indicate: indicating with capital letters the systems which orbit the common pericenter of the overall system. With non-capital letters the components of the upper level are distinguished. And keeping it more readable, numbers are used for the 3rd level if further distinction is needed for a 3rd level.
However this nomenclature is not unique - and there is so far no general consensus. See the Washington Multiplicity Catalog for a take on getting some order at a given time and the IAU working group on stellar names.
For a given system the designation might be more traditional and depend on order of discovery and naming by discoverer than following a fixed scheme. See also the discussion of future nomenclature on stellar system components in the corresponding wiki article.


Is there any practical use for astronomy?

This question begs the question, does everything need a practical use? The answer is a resounding no. What's the practical use of the Louvre, or of your local neighborhood public park where you enjoy weekend barbecues?

There are some things that are very worthwhile that have little or no economical gain. Your local neighborhood public park in fact has negative economic gain. Admission is free, but maintenance is not. Think of how much money your city would make if they sold it to a condominium developer, and how much money it would save by not having to pay to have the park maintained.

Despite having no obvious economic gain, some things are nonetheless worth quite a bit. Many of the sciences fall in this category. For example, what is the practical use of archeology? (There are some, but that's not the point.)

Astronomy, like archeology, the Louvre, and your local public park, doesn't need a practical economical purpose. The purpose of the science is good enough.

That said, there are practical applications of astronomy. The key application has been and still is navigation. Knowing the location of a ship at sea or the orientation of a vehicle in space requires astronomy.

A less direct but still very important application of astronomy is in how it informs physics. Kepler was an astronomer, not a physicist. (Those two disciplines were very, very distinct in Kepler's day). Yet Kepler's work informed Newton on how to describe gravitation. More recently, astronomy has informed physics that its standard model was not quite correct. The observed neutrino flux from the Sun (see http://en.wikipedia.org/wiki/Solar_neutrino_problem) was a third of what physics at the time said it should be. This resulted in a change to the standard model. Neutrinos have a small but non-zero mass, and they oscillate from one form to another.

Astronomy continues to inform physics to this day. Physicists (and astronomers) remain clueless with regard to what constitutes dark matter and dark energy. But whatever they are, they certainly do exist.


At noon on 1 Jan 2000 (UT) the sun's RA & dec were (J2000):

18hrs 45min 9.36s

-23deg, 2 m, 8.2s

This is per Guide9 software.


Natural units of star clusters angular momentum and energy

Using unit analysis, I've found that a natural unit of angular momentum should be
\begin{equation}\tag{1}
L_0 = \frac{G M_{\odot}}{c} \approx 8.816 \times 10^{41} \, \text{kg} \, \text{m}^2/\text{s}.
\end{equation}

In the case of a bounded system of two equal masses, circularly moving around their center of mass, the total angular momentum (relative to the center of mass) is
\begin{equation}\tag{2}
L_{\text{bound}} = 2 M \, v \, r = \frac{2M \, v^2 \, r}{v}.
\end{equation}
Now, using Newton's equation and the gravitationnal force, we have
\begin{equation}
M \, v^2 = \frac{G M^2}{4 \, r}.
\end{equation}
Thus, equation (2) becomes :
\begin{equation}\tag{3}
L_{\text{bound}} = \frac{G M^2}{2 \, v}.
\end{equation}
The classical theory cannot be trusted when $v \sim c$, and since $v < c$ (forgetting the 2 factor) :
\begin{equation}\tag{4}
L_{\text{bound}} \approx \frac{G M^2}{v} > \frac{G M^2}{c} \equiv L_0.
\end{equation}
So $L_0$ appears to be a good unit.  When $L \sim L_0$, we then need relativity.  We're in a classical regime when $L \gg L_0$.

The system isn't bounded when $L \gg L_{\text{bound}}$, which implies $L \ggg L_0$.

I don't know if this is a good way to introduce the $L_0$ unit defined by (1).  Any opinion/suggestions/comments on this ?

Do astronomers use $G M^2 / c$ as natural unit of angular momentum, when they study star clusters and globulars ?


How to mathematically describe a planet's eccentric orbit?

The formula is Kepler's equation, but to understand it you need to know three values:

$M$ is the "Mean Anomaly". It increases linearly from 0 to 360 over the period of one orbit, measured from periapse to periapse. So if a planet has a period of 100 days, then the mean anomaly at day 0 is 0, at day 50 it 180degrees, at day 25 it is 90 degrees. This is only the position of the body that is in a circular orbit. It is the first stage of calculating the position of the body in an elliptical orbit.

$E$ is the "Eccentric Anomaly". This is also not the actual location of the planet. It is defined by $\cos\,E = x/a$ where $x$ is the x-coordinate of the of the planet, measured from the centre of the ellipse not the focus, and $a$ is the semi-major axis of the ellipse

$T$ is the "True anomaly", the angle between the actual position of the planet and the line through the major axis of the ellipse, measure at the focus of the ellipse (ie at the sun).

These are related by the Kepler equation. $$M = E- e\sin E,$$where $e$ is the eccentricity of the ellipse. You can calculate M directly, then solve this equation (numerically) to find E. Providing the eccentricty isn't very close to one, Newton's method is effective at finding a solution to this equation.

Then the true anomaly is given as
$$
T   = 2 \, \mathop{\mathrm{arg}}\left(\sqrt{1-e} \, \cos\frac{E}{2} , \sqrt{1+e}\sin\frac{E}{2}\right)
$$
(where $\mathrm{arg}(x,y)$ is available as atan2(y,x) in many programming languages)

It is possible to use the Eccentric anomaly to calculate the coordinates of the planet directly using $\cos\,E=x/a$and $\sin\,E=y/b$ where a and b are the semi-major and -minor axes of the ellipse.

The word "anomaly" doesn't mean "false result", but "angular deviation (from periapse)" and has been in use since Kepler's time.


Caveat: I'm not a cosmologist, so this answer may not reflect the forefront of scientific knowledge/accuracy, but I have some knowledge so thought I'd share it and hopefully someone can correct me if I'm wrong.

The Big Bang theory states that everything is moving away from everything else, so it must have started closer together. We know from the Cosmic Microwave Background that everything was very dense and hot and small.

You may have heard that the universe started in a singularity. This means one point at which all matter was in the same place. In this state, all information in matter is lost. No particle can be labelled as having a position or size or spin, because every particle is in the same state as every other particle. This means we lose all possibility of having any information about what happened before this. There are many theories of what could have been before this, but no matter how simple/beautiful/mathematically rigorous the theory, it is physically impossible to know anything about it so they are impossible to test. In some definitions this makes these theories not actually science, as the scientific method involves making testable predictions.


Does tidal heating imply orbit degradation?

There is a wonderful post at Physics Stackexchange:

  Gravitational coupling between the Moon and the tidal bulge nearest the Moon acts as a torque on the Earth's rotation, draining angular momentum and rotational kinetic energy from the Earth's spin. In turn, angular momentum is added to the Moon's orbit, accelerating it, which lifts the Moon into a higher orbit with a longer period. As a result, the distance between the Earth and Moon is increasing, and the Earth's spin slowing down.

You can find more reading material and information in the original thread.


How can I contribute to the scientific community using my telescope?

Amateurs do useful scientific work by being many-handed and widely spread.

Many Supernovae are discovered by amateurs. You need a set-up that can image one galaxy after another and look for any "new stars" that appear in them. As you are looking for the appearance of a Mag-14 star you need a big mirror; 90mm might not be enough. However look to Backyard observatory supernova search for more information. With the right equipment, you might find 1 SN every 5000 galaxies imaged.

You can also contribute by observing asteroid occultations. These occur when an asteroid passes in front of a star, blocking out its light. By timing the occultation you can get an idea of the size of the asteroid in one cross-section. If multiple people observe the same occultation from different locations you can get a picture of the shape of the asteroid. See the Asteroid occultation site which links to FAQs and a page for submitting reports.


Gravity doesn't affect the speed of light. It affects the space-time geometry and hence the paths of light. However, this can have a similar effect.

Light emitted at source $S$ to pass a massive object $M$ that is very close on the otherwise (if M weren't there) straight path to an observer $O$ has to "go around" $M$, which takes longer than following the straight path in the absence of $M$. The light that reaches $O$ from $S$ is not emitted from $S$ in the "straight" (in the absence of $M$) direction to $O$, but slightly off that direction, such that the "bending" of its path by the gravity of $M$ "deflects" it onto $O$.

Of course, light is never bent, but always follows a straight path. What is bent is the space-time as compared to Euclidean space-time in the absence of distorting masses (see: geodesic). This distortion in the fabric of space time is called gravitational lensing.


Is there a mechanism that makes small moons more rounded than comets?

I can only speculate but here goes nothing.

67P/Churyumov–Gerasimenko apparently came from the Kuiper Belt. There are a lot of KBO (Kuiper Belt Objects) in that region, so it is not unreasonable to expect some of these objects to have crashed into other objects of about the same mass, resulting in irregular shapes.

Small moons (or other moons in the Solar System) have been isolated from other objects of the same size. This allows gravity work work undisturbed for a longer period of time than those in the Kuiper Belt, molding the object into more or less a spherical shape.

That is my best stab at the question, though I can't imagine some other cause that would result in such a drastic irregular shape in something as large as a comet (relatively speaking).

I just found something interesting. So the smallest moon in the solar system is Aegaeon. It happens to be a couple of orders of magnitude smaller than the comet mentioned above. In addition, this moon appears to have an oblate spheroidal shape to it. It may just be the case that we have only seen spherical moons, but there are many more moons out in the cosmos that have odd shapes to them.


The "Friedmann model" is a model of the Universe governed by the Friedmann equations, which describes how the Universe expands or contracts. These equations are a solution to Einstein's field equations, and with two very important assumptions they form the basis for our understanding of the evolution and structure of our Universe. These assumptions, together called "the cosmological principle", are that the Universe is homogeneous, and that it's isotropic. This is a layman's explanation, so I won't write the equations here, but farther down I've added the equation and go a bit more in detail what it means.

The cosmological principle

Homogeneity

That the Universe is homogeneous means that its "the same" everywhere. Obviously, it's not, really. For instance, under your feet is a dense, rocky planet, whereas above your head there's thin air. We live in a galaxy full of stars and molecular clouds and what not, while 100,000 lightyears from the Milky Way, there is virtually nothing. But on very large scales, say above half a billion lightyears, the Universe actually looks the same all over.

Isotropy

That it's isotropic, means that it looks the same in all directions. Again, obviously it doesn't on small scales, but on large scales, it does. If it didn't, it would mean that we occupied a special place in the Universe, and we don't think we do.

So, neither of these assumptions have to be true, but observations tell us that apparently, they are. See this image, where every dot is a galaxy (Maddox et al. 1990):

You might think that a homogeneous Universe would also be isotropic and/or vice versa, but that's not the case.

Three possible solutions

It turns out that, for these assumptions, there are three possible solutions to the Friedmann equation. We call the three possible universes flat, positively curved (or "closed"), and negatively curved (or "open"). Which of these possible universes we live in, turns out to depend upon the average density in the Universe, so by measuring this, we can determine the "geometry" of our own Universe. And it seems that it's "flat".

A flat universe

The reason it's called flat is that the geometry is like that of a flat 2D table, only in 3D. That is, a triangle has 180º, parallel lines never meet, etc. And it's infinitely large. Intuitively, we'd think that this is the way the Universe is, and definitely on small scales (say within our own Galaxy), it's an adequate approximation. In the 2D analogy, the 2D surface of Earth seems flat locally, and for all practical purposes the parking lot outside is flat. But if you draw a triangle from Congo→Indonesia→the North Pole→Congo, you'd measure it's sum of angles to be roughly 270º. That's because the geometry of the surface of Earth is not flat, but is "closed".

A closed universe

If the Universe were "closed", it's geometry would, in the 2D analogy, correspond to that of the surface of a ball, i.e. a triangle has more than 180º, lines that start out being parallel will at some point meet, etc. But like the surface area of a ball is finite (but doesn't have a border), so is the Universe. So if you take you spaceship and fly straight away from Earth, you would end up back here (assuming the Universe doesn't collapse before you get back, or expand too fast for you).

An open universe

If it were "open", it's geometry would, in the 2D analogy, correspond to that of the surface of a saddle, i.e. a triangle has less than 180º, lines that start out being parallel will diverge, etc. And it's infinitely large.

This picture from here visualizes the 2D analogies.

In 3D, only a flat geometry can be pictured, and this doesn't look "flat" by any means; it is simply your good old 3D ("Euclidian") space that you know from your everyday senses.

Expansion of the Universe

The Friedmann equation, together with the densities of the constituents of the Universe (radiation, normal matter, dark matter, and dark energy) tell us how the Universe expands. So, again by measuring these densities, we can prediect the evolution of the Universe. And it seems that the Universe not only expands, but actually expands faster and faster.

Beyond the layman's explanation

Here, I'll expand a bit on how to understand the equation:

The first Friedmann equation is intuitively most understandable, I think, when written like this:
$$\frac{H^2}{H_0^2} =
\frac{\Omega_\mathrm{r,0}}{a^4} +
\frac{\Omega_\mathrm{M,0}}{a^3} +
\frac{\Omega_k}{a^2} +
      \Omega_\mathrm{\Lambda}.$$
This equation tells us the connection between the expansion rate of the Universe (the left hand side), and the density of its components and its size (the right hand side). Below, I'll go through the components of the equation.

The Hubble parameter
In the equation, $H$ is the Hubble parameter describing how fast a galaxy at a given distance recedes (or approaches, for a collapsing universe), at a given time in the history of the Universe. $H_0$ is its value today, and is measured to be roughly $70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$. This means that a galaxy at a distance of, say, 10 Mpc ($\simeq33$ lightyears), moves away from us at a current speed of 700 km/s. A galaxy 20 Mpc away recedes at 1400 km/s, and so on (and galaxies farther away than roughly 4.3 Gpc recede faster than the speed of light, but that's no problem and we can still see them).

Size
The size of the Universe is unknown, and quite possibly infinite (lest it wouldn't be homogeneous, but to be fair, we only know that the observable Universe is homogeneous). Thus, we cannot talk about its absolute size. But we can talk about how much a certain volume of space expands in a given time. We us the parameter $a$, called the expansion factor. Defining $a$ to be 1 today, that means that at the time when the Universe was so small that all distances between the galaxies was, say, half of today's values, $a$ was equal to 0.5 (this happens to be 8 billion years ago).

Density parameters
Whether the geometry of the Universe as described above is flat, closed, or open, depends on whether the total density $\rho_\mathrm{tot}$ is exactly equal to, above, or below a certain critical threshold $\rho_\mathrm{cr} \sim 10^{-29}\,\mathrm{g}\,\mathrm{cm}^{-3}$. It is customary to parametrize the density of the $i$'th component as $\Omega\equiv \rho_i / \rho_\mathrm{cr}$.

Matter
The term "matter" includes "normal" matter (gas, stars, planets, bicycles, etc.), and the mysterious dark matter. As the universe expands, the volume grows as $a^3$. That means that the density falls as $\Omega_\mathrm{M}=\Omega_\mathrm{M,0}/a^3$.

Radiation
Photons redshift as space expands, and this redshift goes as $1/a$. This is in addition to having their number density decrease, so the total energy density of radiation decreases faster than matter, namely as $1/a^4$. Today, the energy density of radiation is dominated by the CMB, and can be neglected, but in the early times, they would dominate.

Curvature
If space is not flat, its curvature contributes to $\Omega_\mathrm{tot}$. The reason is that the curvature affects the volume in which we measure densities (thanks to John Davis for this explanation). This scales as $1/a^2$.

Dark energy
Finally, there's the magical dark energy, of which even less is known than the dark matter. If it exists, it's thought to be a property of space itself, i.e. its energy density grows proportionally to the volume of the Universe, and thus there's no $a$-dependency.

Interpretation

From the equation, it is readily seen that if we're able to measure all the Omegas, then we know how fast the Universe has been expanding at all times. That means that we can integrate backwards in time and calculate when $a$ was 0, i.e. we can calculate the age of the Universe. Also, from the $a$-dependencies we can see when the Universe went from being radiation-dominated to being  matter-dominated. We can also see that not only is it now dominated by dark energy (since $\Omega_\mathrm{\Lambda}\simeq0.7$, but $\Omega_\mathrm{M}\simeq0.3$), but due to the $a$ factor, it will only get "worse". That is, all other densities will keep decreasing, but $\rho_\Lambda$ stays the same, and since dark energy has a repulsive effect rather than attracting, the expansion of the Universe accelerates.

Observationally, it is found (in several independent ways) that all the $\Omega$s add up to one, i.e. that the total energy density of the Universe happens to be exactly equal to the critical density. This is pretty amazing. This figure, taken from here, shows the contribution from the different components now (top), and at the time of the CMB emission (380,000 yr after the Big Bang; bottom):


How will planets behave in the night sky as seen by Mars colonists?

This should be a relatively simple problem to solve - which is the Academician's way of saying "do it as a homework problem" :-) .

First thing: find some celestial mechanics calculator and determine when each outer and inner planet is visible in Mars' night sky.  That gives you timings.

Next thing: compare the distance from each outer planet to Mars vs. Earth. Maybe start with closest and farthest approach as examples. That'll tell you both the angular extent of the planet as well as the relative brightness compared with Terran observations.  Don't forget to add a fudge factor for the atmosphere.

In fact, probably the most obvious difference will be that stars no longer twinkle. (yeah I'm ignoring during serious dust storms).  You could use a wicked large primary and have no need of adaptive optic corrections.

BTW there are photos of Earth as seen by the Mars rovers (or maybe just the surveying satellites, I forget) so you can get an idea of the color/features available.


Could the Earth be used to cast an arago/poisson spot on something

One constraint is the recommendation $F=\frac{d^2}{b\lambda}\geq 1$, in this case with $d=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, and $b$ the distance between circular obstacle and observer.
The distance between Earth and the observer should hence be

$b\leq \frac{d^2}{\lambda}=\frac{(12700\cdot10^3\mbox{ m})^2}{600\cdot 10^{-9}\mbox{ m}}=481.67\cdot 10^{18}\mbox{ m}$

Another constraint is the surface roughness of the circular object:
$\Delta r < \sqrt{r^2 + \lambda\frac{gb}{g+b}}-r$, with $r=6350\mbox{ km}$ the radius of the circular obstacle (here Earth), $g$ the distance between the point light source and the circular obstacle, and $b$ the distance between the circular obstacle and the screen.

To simplify calculations, say $g\gg b$. Then approximately
$\Delta r < \sqrt{r^2 + \lambda\frac{gb}{g}}-r = \sqrt{r^2 + \lambda b}-r$.

After adding $r$ and squaring you get
$(\Delta r +r)^2<r^2+\lambda b$.
This simplifies to
$(\Delta r)^2+2r\Delta r<\lambda b$.
Assume $\Delta r\ll r$, and neglect the second order $(\Delta r)^2$ to get
$2r\Delta r<\lambda b$. Divide by $\lambda$ to get an approximate constraint for $b$ as

$b>\frac{2r\Delta r}{\lambda}$.
With $2r=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, we get

$b>\frac{12700\cdot 10^3\mbox{ m}\cdot\Delta r}{600\cdot 10^{-9}\mbox{ m}}
= 21.1667\cdot 10^{12}\Delta r$.

The two constraints allow for reasonable values of $\Delta r$.
Assume a surface roughness of Earth of e.g. $\Delta r = 1\mbox{ km}$.
Then a valid range of distances of observers would be between
$0.00224$ and $50912$ lightyears of $9.4607\cdot 10^{15}\mbox{ m}$ from Earth.

In astronomcal units of $149597870700\mbox{ m}$ the closest distance of an observer would be $141.49 \mbox{ au}$ from Earth.

Due to Earth's oblateness, however, you would get a point spread function significantly different from a dot for this "short" distance from Earth. It might be possible to correct this by an appropriate telescope optics.

The effect of gravitational lensing is
$\theta=\frac{4GM}{rc^2}=2.969\cdot 10^{-27}\frac{\mbox{m}}{\mbox{kg}}\frac{M}{r}$, after applying the constant of gravitation $G$ and the speed of light $c$. With the mass $M=5.97237\cdot 10^{24}\mbox{ kg}$ and a radius of $r=6350000\mbox{ m}$ of Earth, we get an angle of
$\theta=2.793\cdot 10^{-9}$ by gravitational lensing at the surface of Earth.

This would focus parallel rays of light to a point near a distance of
$b=\frac{r}{\tan \theta}=130.27\cdot 10^{15}\mbox{ m}$, or $13.77$ lightyears, hence well beyond the minimum distance where an Arago spot could form. But, of course, the innermost peak of the point spread function would be closer to a circular disc at this larger distance with relevant gravitational lensing.


Are any Pluto-sized objects remaining to be discovered in the Kuiper Belt?

This is a part answer to your question, as it is difficult to answer without speculating, so here are some facts/observations related to your question.

Asides from Pluto/Charon, Eris, Triton (could be a captured Kuiper Belt object), Makemake and the football shaped Haumea, most of the Kuiper Belt Objects (KBOs) are according to the article "Kuiper Belt Objects: Facts about the Kuiper Belt & KBOs" (Redd, 2012):

  thousands of bodies more than 62 miles (100 km) in diameter travel around the sun within this belt, along with trillions of smaller objects, many of which are short-period comets

and is believed to have a total mass of only a tenth of the Earth, according to the article "Forming the Kuiper Belt by
the Outward Transport of Objects
During Neptune's Migration" (Levison and Morbidelli).

Here is a list Of the many Transneptunian Objects that have been documented, detailing their absolute magnitudes.

In regards to one of your main questions - according to Redd (2012), the challenge in their detection is

  Because of their small size and distant location, Kuiper Belt Objects are a challenge to spot from Earth. Infrared measurements from NASA's space-based telescope, Spitzer, have helped to nail down sizes for the largest objects.

I would add, their irregular elliptical orbits* and extreme (compared to the major planets) inclination to ecliptic make it that much more difficult to detect them.  Additionally, according to "The Edge of the Solar System" website, further difficulties include low surface reflectivity.

An example of a possible KBO with an extremely elliptical orbit is Sedna, which is believed to take over 10,000 years to orbit the sun; is smaller than Pluto, but was observed at about 90AU (3 times further than Pluto).

So, there could be many small Pluto-sized 'dark' worlds in highly elliptical irregular orbits in the Kuiper Belt and beyond.  However, beyond those listed, we have not seen that many and the total mass theorised does not support the idea of too many in existence, but that does not mean that they are not out there.


Is the wind's intensity on Mars similar to Earth?

Credit to this question for inspiration, though my calculation methods differ.
The dynamic pressure equation is $q=0.5\rho v^2$ where $q$ is the pressure, $\rho$ is the atmospheric density, and $v$ is the wind speed. If we want to know what wind speeds give us equivalent pressures on Earth and Mars, we simply generate dynamic pressure equations for each of them: $q=0.5\rho_e v_e^2$ and $q=0.5\rho_m v_m^2$, set them equal $q=0.5\rho_e v_e^2=0.5\rho_m v_m^2$, and solve for $v_e$ to get $$v_e=\sqrt{\frac{\rho_m}{\rho_e}}v_m$$ where $\rho_m=0.020 \space kg/m^3$ is the atmospheric density for Mars, $\rho_e=1.225 \space kg/m^3$ is the atmospheric density on Earth, $v_m$ is the wind speed on Mars, and $v_e$ is the equivalent wind speed on Earth.
With a velocity ratio of about 7.826 we can plug in a few values for wind speed in kilometers per hour for Mars to get:
v_mars    v_earth equivalent
   10           1.28
   50           6.39
  100          12.8
  200          25.6
  400          51.1

These could be kph, or in fact, any units of velocity. screeenshot
and here's what hat looks like in a plot:

So the 400 kph gust on Mars only has equivalent pressure of a 51 kph gust here on Earth


What is the ratio of cosmic microwave background radiation to normal radiation?

Photons from the CMB outnumber all other photons by more than 200:1.  This is contained in Pela's excellent answer to: Does the number density of photons  nγ≈108m−3  refer to CMB photons only?.  Pela uses this graph from Hill et al. 2018 to derive this ratio.  Note that the CMB forms the largest contribution to the electromagnetic spectrum.


Which kinds of astronomical observations most need to avoid the Moon being up?

Imaging and spectroscopy of very faint objects, especially towards the bluer end of the spectrum. The effects of trying to subtract the background sky brightness also become more demanding if the object is extended in size or if the angular extent of sources is broadened by atmospheric turbulence (aka "seeing").

The reason that the effects are worse at the blue end of the spectrum is that the presence of the Moon has a greater effect (because scattering is stronger) on the dark sky surface brightness, increasing it by the equivalent of about 4 magnitudes (per square arcsec). Thus the sky brightness is a factor of about 40 higher (at a good site, with little dust and at 90 degrees to the Moon) than when the Moon is well below the horizon.

The contrast is much lower in the red and near infrared. There is less scattered light and in the infrared, the sky background is dominated by airglow, and unaffected by the Moon.

Broadly speaking, observations of bright stars ($V<15$) or observations at near infrared wavelengths can be done at any time, including "bright time" when the full moon is up. Deep imaging, especially of extended objects, and spectroscopy of faint objects ($V>19$), especially at bluer wavelengths requires "dark time", with no moonlight. An exception might be observations of emission line objects (e.g. planetary nebulae), where all the flux is in very narrow wavelength intervals. In between, there is "grey time", when the Moon is up, but less than half illuminated.

For completeness - radio, mid-infrared and mm-wave observations are unaffected (unless the Moon is in the way!)


How do tadpole, horseshoe co-orbital states arise and how are they stable?

If the body is in front of the planet (relative to the planet's orbital motion) and a little further from the sun, it will orbit the sun slightly slower than the planet.

As it is slower, the planet will slowly catch it up. (It takes many "years" for the planet to get close to the body.)

As the planet catches up with the body, the gravitational effect of the planet will tend to pull back on the body, this causes the body to lose energy and begin to fall slighly towards the sun.

The slight fall towards the sun cause the body to move to a lower orbit (relative to the sun) and so speed up. It appears paradoxical, but it the fact that the gravitational drag of the planet on the body actually causes it to speed up.

The body speeds up and so begins to move away from the planet.

When the body is makes an angle of more than 60 degrees from the planet, the gravitational pull of the planet now has the opposite effect. It combines with the pull of the sun to cause the body to accelerate, and (apparently paradoxically) this acceleration causes the body to move to a more distant orbit, and slow down.

There are two things key to understanding this. First remember that the diagrams you show are drawn with a fixed planet. I.e. the "camera" is rotating at the same rate as the planet takes to orbit the sun, so the planet appears fixed. If you don't rotates the camera, you would see that the body actually has a nearly elliptical orbit, relative to the sun. Secondly you need to grasp the idea that, in orbit, drag doesn't slow you down. Drag causes you to fall towards the sun and speed up.

How did bodies get into such orbits? Basically by luck. They happened to have orbits that were close to Earth's orbit, and by a lucky effect of the gravity of the other planets they were dislodged from their regular orbit of the Sun to a tadpole orbit. Once there, they can remain in their orbit for tens fo thousands of years. Even so, only one object is known to exist in a tadpole orbit relative to Earth.


Will the Magdalena Ridge Optical Interferometer be able to image extended objects like the surface of the Moon?

According to this site The 10 telescopes will be optically linked together in order to make images of astronomical objects with unprecedented detail. The interferometer will have a resolution 100 times greater than the Hubble Space Telescope and will be able to make accurate images of complex astronomical objects many times faster than other existing interferometric arrays.

And: The Interferometer will take delivery of the second telescope enclosure in February 2020 and the second telescope in August 2020. They expect to fully incorporate the second telescope by the end of 2020, which will allow the instrument to produce “fringes,” using the proprietary fringe-tracker called ICoNN.

There is also this headline:

  How America's Spooks Seek to Spy on Distant Satellites The
  intelligence community has plans for a telescope network that can see
  not just a blob in orbit but details such as a satellite’s solar
  panels.

from here.

That's one reason why the U.S. Air Force, which wants to monitor its own orbital assets and presumably those of others, is funding MROI. "They want to know: Did the boom break or did some part of the photovoltaic panels collapse?" says Michelle Creech-Eakman, an astronomer at the New Mexico Institute of Mining and Technology in Socorro and project scientist on MROI. But if the facility succeeds, its biggest impact could be on the field of astronomy, by drawing new attention to the promise of optical interferometry, a powerful but challenging strategy for extracting exquisitely sharp images from relatively small, cheap telescopes.

Radio astronomers have had it easier. The long radio wavelengths mean data from separated dishes can be recorded, digitized, time-stamped by an atomic clock, and combined later for analysis. But optical interferometry is far trickier: The short wavelengths of visible light, running at terahertz frequencies, cannot yet be digitized by any electrical system. So the light must be merged in real time, with nanometer precision.

From here.

Also Reference 3.

http://www.mro.nmt.edu/about-mro/interferometer-mroi/


At what speed does something have to travel away from us for it to red shift enough that it becomes invisible to the human eye?

The following deals only with redshift caused by motion (Doppler effect).

Wavelength (wl) shift for an object moving away from a stationary observer is calculated by following formula: shift = wl x V/C
with V speed of the moving object and C speed of light
(source: https://en.wikipedia.org/wiki/Doppler_effect)

Let's say we want a Sun-like star to become invisible to the human eye. Our Sun's emission starts around 250 nm and human vision ends around 700 nm (source: https://en.wikipedia.org/wiki/Sunlight#/media/File:Solar_spectrum_en.svg).

So we want a minimum shift of 700 - 250 = 450 nm for wl 250 nm.
Formula yields V/C = 450 / 250 = 1,8
which
(1) is impossible because nothing moves faster than light
(2) makes the classical formula irrelevant

The relativistic formula (for objects moving at speeds > C/10) is:
shift / wl = SQRT( (1 + V/C) / (1 - V/C) ) -1
with SQRT the square root
(source: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect)

Rounding (shift / wl) up to 2, the relativistic formula yields V/C = 8/10

The speed V would have to be even higher for a hotter (white to bluish) star with UV emission starting below 250 nm.

So my answer to the question is:
very close to the speed of light (80% in above calculation)
and therefore not a real-life scenario (see below).

"Among the nearby stars, the largest radial velocities with respect to the Sun are +308 km/s" (source: https://en.wikipedia.org/wiki/Doppler_effect), that is 1000 times smaller than C = 3e5 km/s. Rounding down to 300 km/s, it would produce a maximum shift of 700 x 300 / 3e5 = 0,7 nm, several hundred times smaller than required for invisibility.

Besides, the star would still appear on an infrared image.


Effects of a rogue planet passing between the Earth and the Moon?

It doesn't look good.

Tidal effects. The passing planet will certainly cause disruptive tides as it passes. The exact height of the tides would be hard to calculate as it depends on how the tides flow around the land masses, but the effects would be much greater than lunar tides. The tidal effect could also have consequences for the earthquakes.

Disruption to the orbit of the moon. The moon's orbit will be significantly altered. It could cause the moon to enter an elliptical orbit (with consequences for tides).  There would not be enough disruption to cause the moon to leave Earth orbit immediately

Disruption to Earth's orbit.  The orbit of Earth would also be significantly changed, it could enter a significantly elliptical orbit, this would change the amount insolation over the year and greatly change the Earth's climate

If the disruption were significant enough, it could bring the Earth's orbit uncomfortably close to that of Venus, this could result in further disruption to the moon's orbit

Fortunately space is big, and even if there are a lot of rogue planets out there, it would be a very rare event for one to pass so close to star


Is it possible to observe strong gravitational lensing with amateur telescopes?

The cosmic horseshoe is beyond amateur instruments. It is a magnitude 20 object. In a large (2.5 m) professional telescope it looks like:

This image taken from the SDSS III data.  It is small (10'', half the size of Mars at opposition) but that is not insurmountable for amateur equipment. But is it very dim.  If an object like this was visible in moderate telescopes, it would have been found long ago.  It is really only the sky surveys, and computerised searches of the survey data that allows for such objects to be found.
Processsing the SDSS data can bring out more detail, as can be found on the discovery paper https://arxiv.org/pdf/0706.2326.pdf
While this isn't visible, another example of gravitational lensing can be: the Einstein cross is a gravitationally lensed quasar, and it is at the limits of amateur equipment.  It is even smaller, but as it is composed of four point-sources it can be seen. Skyhound recommends at least 18 in telescope, but practically 24'' to see the four components, exceptional dark skies, fully adapted eyes and perfect conditions (especially perfect seeing: this object is small). Your reward for this is "four smudges of light embedded in the galaxy" The SDSS view gives an indication of how hard this object is.


Yes, one can superpose these two noise sources: add a Gaussian readout noise (with a spatially constant sigma) to a Poisson noise (with spatially variable number count k).

In practice, as the number counts are high for optical CCD images, one often approximates the Poisson noise by a spatially-variable Gaussian noise, using the square root of the photon number counts as sigma. If your images have some significant sky background level (e.g., hundreds of photons), this is fine even for faint sources that add only few photons.  A Poisson distribution gets very close to a Gaussian already for k > 20. Note that despite this common approximation by a Gaussian, such a simulated noise contribution is still often called "Poisson noise", to stress that it originates from the spatially variable signal.

One warning: using the square root of the signal as sigma of a Gaussian to approximate Poisson noise only holds for a signal in units of photons (or almost equivalently, in units of electrons). If an image is given in ADU (analog to digital unit), the gain is required to convert the pixel values into electrons. The same holds when using a true Poisson distribution: one has to be careful that the counts have not been rescaled.


Relation between angular resolution and aperture?

The angular resolution only tells you your ability to distinguish ("resolve") two separate points of light. For example, your ability to distinguish the two distinct headlamps on a car from being a single light in the distance.

The angular magnification has nothing to do with the angular resolution: it is determined by the focal lengths of your primary and objective, and the eyepiece if applicable.


Your intuition is correct.  It does not really make any sense for the universe to have a border.  Would it be an unbreakable wall? What would be on the other side?  Models with a finite universe would not have a border.  The universe would be curved into a closed loop like the surface of a sphere.  In other words, if you go far enough in any one direction you come back to the starting point.


What is an asteroid called if it is in deep space?

We have only discovered one such object so far: ʻOumuamua. (Although a second, 2I/Borisov, was discovered in 2019) The general practice has been to call it an "interstellar asteroid" or "interstellar comet", even though it is not in orbit.  You could just call it an "interstellar object".


Present distances between planet. How can I find them?


I think we can broadly distinguish two classes of effects here:

Flares directly affecting the planet
The by-products of the flare affecting the planet

I can loosely tell you what I know (although my knowledge here is far from being complete)

Direct effects include:

Magnetic field geometries: Far-away from their host star (>0.05AU) planets usually have isolated dipoles. The solar wind then must penetrate the magnetosphere of a planet to erode its atmosphere. This is however much easier for the wind if due to small distance both star and planet-magnetospheres connect. What this will lead to, is quasi-free flow of a significant portion of the stars total ejected mass towards the planet and huge energy input into the planet's upper atmosphere.

Indirect physics can be:

An active young star can have frequent outbreaks of CMEs, those are accompanied by strong UV / X-ray emission and this will affect the planet:

Increased photochemistry: This can generate a very active photochemistry in the upper atmospheres. Doubly ionized molecules like $O_2^{++}$ or $N_2^{++}$ can then be produced more efficiently and eject their constituents into space upon dissociation due to high dissociation energies
UV-X-heating: Additional heat through thermalized high-energy photons can add to the already happening 'natural' atmospheric erosion.

We don't know however, to which extent those effects play a role. I'd add tomorrow some references if you're interested.

There could be signs however of such activity in young stellar systems, as verious combinations or extensions of the above have been proposed to explain the radius anomaly in super-close extrasolar giant planets.


Converting ångström spectral dimension to galaxy speed (km/s)

If you want bins that have an equal step in km/s then you will have to rebin in steps of equal log wavelength. On this scale the rest wavelength of H alpha is at zero km/s and then each bin will represent a increment in velocity with respect to that.

See How do I apply a velocity shift to a wavelength array with uniform logarithmic spacing?


What is the standard reference point for measuring speed?

As Einstein realized and like you correctly state, you indeed can't measure the speed of an object by itself since it has to be measured relative to something else.

As a logic result, if you ask the question "How fast is X moving?", you will have to specify that you want the speed with respect to another object because motion cannot be measured without a reference point.

Some examples:

If you ask how fast Earth is moving with respect to its own axis, the center of the Earth will be your reference point.
If you want to know how fast the Earth orbits the Sun, the Sun will be your reference point.
If you want to know how fast the Moon's distance from Earth increases, Earth will be your reference point.
If you want to know the speed of our solar system in the milky way galaxy, the center of the Milky Way will be your reference point.

"standard" reference point

If no reference point is given, it could be assumed that the "standard reference point" is the location of the observer. In the realms of astronomy, that would be the location of the astronomer.

It should be noted that the astronomer could well be on an space mission outside Earth's atmosphere, or he/she could be using a telescope in Earth's orbit. Therefore, "Earth" can not generally be assumed to be a standard reference point for astronomy because, depending on the precision of the measurements and the location of (for example) the telescope, assuming "Earth" may result in inexact measurements.

EDIT

As @TidalWave correctly commented, there's also the International Celestial Reference Frame (ICRS) which can help you find reference points, calculate distances, etc. according to a celestial reference system, which has been adopted by the International Astronomical Union (IAU) for high-precision positional astronomy. The origin of ICRS resides at the barycenter of the solar system.

Wrapping it up:

If no reference point is defined and if we can assume that the astronomer works according to the rules and definitions of the International Astronomical Union (which would be the regular, if not ideal case), the International Celestial Reference Frame provides you with a "standard reference point" at it's center (which is the barycenter of the solar system).

In rare cases where IAU compliance can not be assumed (something that "might" happen in amateur realms), it has to be assumed that the reference point is the point of observation.


How much time does it take for a meteor (shooting star) to touch the ground after its combustion?

I think the best model we could use for this is nuclear test fallout, whose dispersion was monitored quite extensively.  It's very, very dependent on weather patterns at altitude.

For example, the Castle Bravo test mushroom cloud reached at altitude of about 40km and hence hit the Jet Stream.  In addition to this the wind speeds at lower altitudes were higher than ground speeds and this was not checked for, so the combination of factors spread the fallout downwind about 280 miles (about 430 km) in about 16 hrs.  This was one reason for the irradiation of many people.

What these tests and other nuclear incidents show us is that material ejected into the high atmosphere can remain there a long time, even circling the globe like traces from Fukushima.

In theory trace particles from a meteor could remain in the upper atmosphere for days, months, even years.  Most would be expected to fall within days, however.


How massive does a planet need to be to create gravitational lensing?

You could google "Maccone focal" for the most ardent proponent of this project (and I love it too, the most furthest meaningful mission possible within a life time, and what a view!) But others may have variations of his idea. One should be a bit wary about enthusiasts.

Here's a link to some article where Maccone is cited to explain that planets need 6-17 greater distance to become gravitaional lenses than the Sun needs. Neptune surprisingly happens to be next best after Jupiter: http://www.centauri-dreams.org/?p=15290

Paper (might not be available for free):
Maccone, “A New Belt Beyond Kuiper’s: A Belt of Focal Spheres Between 550 and 17,000 AU for SETI and Science.”

Here are slides from one of his presentations where he explains his calculations more generally, but I think only for the Sun:
http://www.spaceroutes.com/astrocon/AstroconVTalks/Maccone-AstroconV.pdf

And then there's micro gravity lenses! A very different thing in terms of our observation. I now think my answer above misunderstood what you were looking for.


I think you just need to keep winding the focusser knob, you are way past focus in the first picture but you can see the image get smaller as you approach focus. The last image looks like you may be there or thereabouts, I think the colored diffraction patterns are caused by the camera optics.


Software with simple representation of all stars

Guide9 from projectpluto.com will do this.  You choose Chart mode (Display, Background, Chart mode).  Right clicking on a star or any other object will give you the information you want.
It runs on all versions of Windows.  It is not freeware, and is a bit dated in some aspects.


V471 Tauri's circumbinary brown dwarf non-observation; Applegate, or over-restrictive assumptions?

For all of these direct imaging results, the critical parameter is the contrast as a function of separation. This lets you know how much fainter an object you can see around the much brighter primary object whose light has been suppressed by the coronograph (the black circle in the center of the star).

From the change in eclipse timings (Figure 1 in their paper) you can predict the expected mass and separation of the third object (proposed brown dwarf) using their equation 1 since you know the mass of the binary. From evolutionary models of brown dwarfs, you can make predictions of how bright the proposed brown dwarf should be for its predicted mass.

You can then look for an object at the right separation in the SPHERE images. This is shown in their Figure 3 (right hand figure of the 2) where they show how bright an object they could see as a function of separation from the primary star; everything above the solid curve is something they should see. The vertical dotted lines show the boundary on the separation of the brown dwarf, predicted from the eclipse timings. The diamond symbol is the predicted brightness of the brown dwarf from the evolutionary models given its mass.

Since this is well above the contrast curve (by a factor of about 15x) and nothing is seen in the images at that separation, this strongly suggests that the proposed brown dwarf does not exist. The only "get-outs" are if the SPHERE team have measured their contrast curve incorrectly (unlikely at the level needed) or our models of how bright brown dwarfs should be are also wrong by about a factor of 15x and the brown dwarf is much, much fainter than predicted.

The Applegate mechanism is a bit weird. The idea is that as the star goes through magnetic activity cycles (as the Sun and a lot of other stars do) and the magnetic field strength grows and shrinks, this causes the shape of the star to change, bulging more or less at the star's equator as the cycle progresses. This changes the amount of angular momentum in the star, which has to couple into the binary orbit as the angular momentum has to be conserved, causing the binary orbit to shrink or expand. This would then explain the change in timings of the eclipses without the need for the third body brown dwarf tugging on binary. Unfortunately the Applegate mechanism, although it may work in V471 Tau, can't explain the variations in a bunch of other binary systems (magnetic field strength not strong enough) so it can't be the only solution for eclipse time variations.

The latest paper I could find that references the Vaccaero et al. 2015 paper, is Vanderbosch et al. 2017. They argue that because of the timing of the eclipses change but the spin period of the white dwarf secondary doesn't change, a third body such as a brown dwarf can't explain the changes as it should change the timing of both "clocks".


By looking at the link provided by James K, I was able to write a Python 3 script that calculates the duration of the longest night for a predefined location (Boston, MA) for 160 years. The script uses the PyEphem module. The longest night becomes shorter by ~72 msec per year, and its length also oscillates with the period of ~19 years.

Here's the script that I used:

import ephem
import matplotlib.pyplot as plt
import matplotlib
import numpy as np

location = ephem.Observer()
location.lat, location.lon = '42.3601', '-71.0589'

years = range(1900, 2060)
longest_nights = []
for year in years:
    location.date = ephem.next_winter_solstice('{}-12-12'.format(year))
    sunset = location.next_setting(ephem.Sun())
    sunrise = location.next_rising(ephem.Sun())
    if sunrise < sunset: # Look at the night before!
        location.date -= 1
        sunset = location.next_setting(ephem.Sun())
    longest_nights.append(sunrise - sunset)

longest_nights = np.array(longest_nights)
longest_nights = (longest_nights - longest_nights.mean()) * 24 * 3600 # In seconds

# Find the slope
xp = np.poly1d(np.polyfit(years, longest_nights, 1))
np_years = np.array(years).reshape(len(years), 1)

# Plot
matplotlib.style.use("ggplot")
plt.plot(years, longest_nights, "-o")
plt.ylabel("Deviation from the mean, sec")
plt.xlabel("Year")
plt.title("Duration of the longest night in a year in Boston")
plt.plot(years, xp(years), "-")
plt.arrow(2017, plt.ylim()[0], 0, plt.ylim()[1] - plt.ylim()[0], color='k')
#plt.savefig("solstice.png")
plt.show()


In a finite universe, what happens when light reaches the boundary?

One important concept that has not been mentioned yet is the "cosmological principle."  This is the key simplifying idea for our cosmological models, it says that the universe is the same everywhere on the largest scales at a given age, so alien astronomers 50 billion light years from us that also conclude the universe is 13.8 billion years old will be observing pretty much all the same things we are.  This principle is not proven by data, but the data is consistent with it, most notably the Hubble law and the homogeneity of the cosmic microwave background.  It allows us to understand the past of our own part of the universe by seeing the past of distant parts (a constraint we are stuck with given the speed of light).

So that's why a universe model that is finite in size must curve back on itself-- nothing else would satisfy the cosmological principle.  This doesn't mean the principle is true, it means we are not going to part with it unless we have to.


In what ways can force be applied to a black hole

Aside from gravity, electromagnetic forces can change the momentum of a charged black hole.

When something falls into a BH momentum is conserved according to every observer, although they may have very different ideas of how that happens. A distant observer will see the new black hole as having some momentum and so moving in some direction relative to the trajectory of the old black hole. It's not really appropriate to describe it as pushing, though. From the object's perspective, nothing special happens as it passes the event horizon. From a distant observers perspective, the gravitational field of the moving object merges with that of the black hole, and the result is the field of a slightly bigger and slightly moving black hole.


Missing Terms in Weinberg's treatment of perturbations on Newtonian Cosmology

The answer turns out to be embarrassingly simple, and I suspect nobody came up with it because of my poor communication in the question.

The $a$ is the occurring in the Fourier transform is the scale factor which has a time dependence. So if we look at the term $\frac{\partial}{\partial t} \rho$ we will get,

$$
\frac{\partial}{\partial t} \rho = \frac{\partial}{\partial t} \int  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) \ \mathrm{d}^3 \vec{q}
$$

$$
 = \int \frac{\partial}{\partial t}\left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q}
$$

$$
 = \int \left(-\frac{\dot{a}}{a} \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q}
$$

$$
 = \int \left(-H\frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q}
$$

The left hand term in the integrand matches the Fourier transform of $H (\vec{X} \cdot \nabla) \rho$. Which is why the terms cancel.


What is an Ideal apparent field of view for viewing planets through a zoom eyepiece?

Planets are very small objects, so they'll fit into whatever AFoV the eyepiece provides.
One highly-rated planetary eyepiece, the Vixen HR, has an AFoV of only 42°.
The main problem with low fields of view and high magnifications, however, is keeping the planet in the centre of the field of view.
The hardest type of mount to do this with is a manual altitude/azimuth mount, because you need to move in two axes at the same time.
In your case, with a manual German equatorial mount, you only need to worry about tracking with the right ascension (RA) axis, and your mount may well have a slow-motion control to help you out with that.

So, when looking to buy a zoom eyepiece for looking at planets, don't worry about the FoV it has. Almost all zoom eyepieces have quite small fields of view when compared with wide-field fixed-length eyepieces. I believe there's only one zoom - a Leica - which has a comparable FoV.

(And if tracking planets at high magnification manually still isn't for you, then consider a wide-angle eyepiece (82° or greater) to give yourself more "drift time" before you have to nudge the telescope)


Solar Wind and Asteroid orbital behavior

Sort of, via radiation pressure and heating.
The Sun emits photons, which carry with them energy and momentum. Any flux of photons applies pressure to an object it hits; this is the basic principle behind solar sails. In general, the larger the object, the more force it feels, since pressure is force per unit area. There are two main ways light from the Sun can significantly affect minor planets, such as asteroids:

The Yarkovsky effect: For a rotating body, there is a delay between when a portion of it receives heat and when the heat is re-radiated. The spinning object then has an offset between the angles of absorption and emission, which means that the radiation applies a net force. This can cause slow but steady changes; over 12 years, the asteroid 6489 Golevka deviated from its predicted orbital path for 15 kilometers.
The Yarkovsky–O'Keefe–Radzievskii–Paddack effect: The YORP effect occurs when irregularities in the surface of an object causes incoming light to be scattered in different directions. This changes the rotation direction and rate of the body. This has been observed in asteroids 54509 YORP (a rotational change of 250 degrees over four years) and 1862 Apollo.


Why does any planet look like a (not so) big white ball through my telescope?

It seems that you haven't align properly your finder (or) laser. You can align at day with a chimney, but at night you must start with the moon and after that you are sure that you are pointing correctly move to a well known feature (like Pleiades cluster in Taurus) and try to find it with the scope (after that you can move to a star (maybe Mizar in Ursa Mayoris or Albireo in Cygnus if you are in the northern hemisphere) after that you must be sure that the finder and laser are pointing correctly.

Other question is that (probably) you are using a very high magnification with that telescope. I have myself a RTF newtonian 114 / 500 and I usually works with 20 mm eyepice (that is 25x), you are using about 216x that is almost the maximum theorically magnification for that aperture. Maybe you should use the 6 mm eyepiece without the barlow (that is 108x more or less) this should be enough but with cheap optics that could be a very high magnification also. I think that you could try with a bigger eyepiece... maybe 15mm plössl (30x and about 1.5º of size)


Why do we believe that the super massive black holes at the centers of two merging galaxies would themselves merge?

The SMBHs reside in the bottom of the galactic potentials, which are dominated by the galaxies' dark matter halos. But although dark matter dominates gravity, collisions between gas and dust particles in the interstellar medium causes enough friction that the baryonic component of the galaxies is decelerated. This will cause the other components of the galaxies to decelerate as well, through gravitational attraction.

Moreover, despite dark matter (and, in practice, stars and black holes since they're so small) being collisionless, there are several ways of "relaxing", i.e. to evolve towards an equilibrium. In the context of galaxy merging, the most important mechanism (I think) is "violent relaxation", where the rapid change of the gravitational potential causes particles to relax, e.g. more massive particles tend to transfer more energy to their lighter neighbors and so become more tightly bound, sinking towards the center of the gravitational potential.

Although SMBHs are…, well, supermassive, the potential will (usually) be dominated by dark matter, gas, and stars, so the new gravitational potential will also cause the SMBHs to seek towards the bottom in the same fashion, and eventually merge.


That the rotation period of the bulk mass of a planet is estimated through something with the magnetic field is true.

But let me maybe elaborate a bit in-depth on that.

No planetary magnetic field is a dipole. Earth has strong magnetic anomalies over the Atlantic, there may be ionic disturbances by moons like Io, or your magnetic field is just really crazy, like Uranus and Neptunes.

(A simple, low-order model of Earth's magnetic field, from GFZ Potsdam)

But no matter how it looks, it will usually be constant on lines of latitude and longitude. This in particular signifies that the field and its distortions are created somehow by the bulk mass of the planet, else the field would move relative to a surface map, in the case of Earth.
So all this talk just to establish the following: We can take the shapes of the magnetic field as if they were surface shapes, and by following them determine a rate of rotation.

Now the story doesn't end there. While this sounds nice in principle, to do this in practice, we would need to send a probe and do in-situ measurments of this changing magnetic landscape.

But what can be done from afar, and from Earth is observe ionospheric patterns. Because the ionosphere of a planet is well, ionized gas, at super-low pressures (roughly on the order of microbar to nanobar) it is frozen-in with the magnetic field that makes the particles of the plasma gyrate. This means, the ionospheric plasma follows the magnetic landscape, now translating into a plasma landscape. This plasma landscape is then usually observable from Earth by the radio signals they emit. It is also possible to see this in the ultraviolet, if there is strong plasma activity through auroral bombardment of a planet's upper atmosphere.

For your second question, that's a simple no.
First of all there is no 'uppermost' layer of atmosphere, as the atmosphere is continously decreasing in density with increasing distance from the planetary body in question.
What you're possibly thinking about are the upper cloud layers, as they visibly end at some height.
But wind speeds are much more determined by pressure gradients created by the central star heating the planet, and the atmospheres ability to cool.
There is one class of planet, however where a relation between the wind speeds at roughly 0.1 bar pressure and magnetic dynamo might exist. Those are Hot Jupiters which host atmospheres that might be sufficiently ionized for its winds to be braked down by the planetary magnetic field.
Then there would exist a limiting speed, given a certain planetary field. However this is still a field of active research, but it would give you 'a' relation between the two.


what would the mass and density of Sirius A be if the mass of Sirius B is roughly the same as the Earth?

The masses of the two objects are completely determined by a careful measurement of their relative positions on the sky over decades, combined with an accurately known distance.
There is no degeneracy in this solution. For a given orbital period and separation of the two objects, the masses are entirely determined. If Sirius B were not a massive white dwarf then the orbit would be different.  Specifically, if you made the total system mass smaller, but with the same orbital period, then the size of the orbit would be smaller. If you tried to compensate by adding mass to Sirius A, then it would almost coincide with the centre of mass of the system and would show almost no orbital motion.
The density is then found by combining these masses with radii estimated from the measured luminosities and temperatures of the components and is therefore also fixed (to about $\pm 10$% because of the observational uncertainties in the mass and radius determinations).


How to install Distro Astro operating system?

Distro Astra is a Linux distribution for Astronomy.

Unless you are very familiar with Linux distributions be vary careful about installing it you main computer. If you have an old computer not being used I would suggest you first installed Distro Astra on that computer.

If it is installed as the only operating system for the computer it will wipe whatever was on the disk prior to installing Distro Astra and you will lose everything that was on that disk.

Being an operating system, once installed, it will make permanent changes to the hard disk drive. It may be possible to have a dual boot system that gives the user the option to use the computer via Distro Astra, or the existing (most likely) Microsoft based operating system originally installed, when the computer is booted.

If you do not have an old computer keep your existing hard disk drive in storage but install a new hard drive for Distro Astra. If the installation of Distro Astra becomes totally messed up you will still have the original hard drive to reinsert and your computer will be as it was prior to dabbling with Distro Astra.

If you have no knowledge of Linux operating systems, find a friend to help you or be prepared to learn, make mistakes and sometimes be very frustrated.

Installing Linux on older computers that utilize a BIOS is easier than installing on newer computer that use a UEFI.


Where do the heavier elements come from?

The processes at work are the r-process and the s-process.

The r-process occurs in supernovae. Under the extreme conditions leading up to it, heavy elements are initially blocked from undergoing beta decay to other elements. If a heavy atom captures a nearby neutron, it can go to a heavier isotope or element. This happens multiple times to an atom, producing heavy elements.

The s-process is slower, and occurs under less dramatic conditions in asymptotic giant branch (AGB) stars - a certain type of red giant. It also uses neutron capture and the subsequent decay of that neutron into a proton. Over thousands of years, atoms climb to become heavier elements. The s-process, though, requires that there already be some heavy elements in existence, unlike the r-process.

For more information, Hyperphysics, this lecture and this German site (in English) are very useful.


Has Hubble ever been used to try to image a near Earth asteroid?

[rewritten to address the revised question]

Maybe, depending on how fussy you want to be about "resolved".

This is a study from 1995, using observations of asteroid 4179 Toutatis made in 1992 with HST. They reported marginal resolution of the asteroid, as suggested by this figure comparing a deconvolved image of a star (observed with the same filter and imager location) and a similarly deconvolved image of the asteroid itself (each pixel corresponds to about 450 m at the distance of the asteroid):

The appearance of the asteroid is pretty clearly not a point source, but it's also fair to say it's only partly resolved, and mostly just in one direction.

(My admittedly vague impression is that this is one of the best, if not the best, case of HST "resolving" a near-Earth asteroid.)

Most observations of near-Earth objects with HST are, I think, aimed at getting optical information on compositions not possible from other wavelengths, and sometimes refining estimates of rotation rates, as was done this year (using data from 2012) for the asteroid Bennu, currently being visited by OSIRIS-REx.

In practice, you get much better spatial resolution using radar (including line-of-sight distance variations due to the structure of the asteroids from time-of-return measurements, allowing you to construct 3D models of them), so there's not much point in trying to resolve them with HST.


Question about the Figure-eight periodic planar three-body motion

There may be some three-body periodic solutions that orbit around a common point, but in general they get a little crazy-looking and may not always contain an immediately obvious center of mass from casual observation. But of course it will always exist.

If you watch closely, you'll see that whenever one body reaches the intersection point, all three are on a straight line with the other two equidistant (symmetric). If you look where this orbit was originally described mathematically, they probably defined the initial conditions with the green one in the middle - so that the y coordinates of all there were zero and the line at that point is horizontal.

At those moments (all six permutations are present) the center of mass is at the intersection point since the masses are equal.

Since there are no external forces, the center of mass doesn't change in between those moments, so the center of mass is always at the intersection point.


Is the universe macroscopically transparent to CMB? Is the fraction intercepted by stars and dust so tiny that it doesn't have a correction factor?

For the most part, the CMB photons travel directly to our telescopes from the surface of last scattering.  Some corrections need to be made to determine the blackbody nature of the spectrum, but they are not corrections for absorption of the photons.
The two main corrections are shown clearly by this COBE image sequence:

(from here).
First, the motion of us as observers, relative to the rest frame of the CMB, distorts its single-temperature blackbody shape, making it appear hotter in one direction and cooler in the opposite direction.
Once this is removed (by a sky-position-dependent frequency shift), the foreground emission of the Milky Way becomes apparent (middle panel), and that must be modeled and subtracted. The dust in our Galaxy is on average hotter than the CMB, so its contribution can be measured well by observing over a range of wavelengths, and of course it is most heavily concentrated toward the Galactic plane.
After that is subtracted, the remaining emission (bottom panel) is that of a blackbody to very high precision, with of course the remaining structure that gives us so much interesting information about the early universe.
Here is part of the key concluding paragraph from the Mather et al. 1992 paper on measuring the spectrum:

The FIRAS spectrum of the cosmic microwave background radiation  agrees with a blackbody spectrum to high accuracy.  The CMBR spectrum is the result of fitting a model including a dipole and a dust map derived from 240 micron data, excluding a region around the Galactic center. The final temperature is 2.726 +/- 0.010 K (95% CL systematic), where the error is dominated by our estimate of the thermometry errors.

So they are shifting the emission and subtracting the foreground emission, but not applying any overall correction for absorption of the emission.


Challenges when trying to build a telescope using two lenses

Diameter is important, but if you're trying to get any magnification out of that system, you need to look at the focal length.

If you have a very distant light source, like the Sun, focused through a convex lens, until the image of the source is as small and as clear as possible, then the focal length is the distance between the lens and the image you're creating. Like when you burn paper with the lens, focal length is the lens-paper distance.

For concave lenses focal length is a little more tricky to define, but they do have a focal length anyway.

If F is the focal length of the forward lens (the objective), and f is the focal length of the lens near your eye (the ocular, or the eyepiece), then the magnification of the instrument is:

M = F / f

So you need an objective with a long focal length, and an eyepiece with a short focal length.

Both the objective and the eyepiece could be convex actually. It's just that the image will be reversed. If the eyepiece is concave, the image will be straight-up, but the field of view of the telescope will be narrow (like looking through a peephole).

If you use both convex lenses, the distance between them need to be close to the sum of their focal lengths, F + f. Start there and adjust it slightly for best results.

If you use a concave lens for the eyepiece, then the distance between lenses needs to be the difference of their focal lengths, F - f.

Both lenses need to be perpendicular on their common axis, to minimize aberrations.

It helps if they are mounted in a tube, to reduce glare, but it's not absolutely necessary. They could simply be attached to a long rod or something - but you'll get a lot of glare into the eyepiece that way.

Do not necessarily try to get very high magnification. If the diameter of the objective is D, measured in mm, then the maximum usable magnification is 2D. So an objective 100 mm in diameter could never provide more than 200x magnification - I mean it could, but the image would be blurry. Even then, this is a theoretical limit, and in practice the image will get pretty bad long before you reach the limit, unless you use high quality lenses designed specifically for telescopes, etc. Also, at high magnification it's difficult to hold the scope steady, and the image is jumping all over the place.

Try and make a 5x ... 8x instrument first, see how that goes. A lot of binoculars are in this range.

I've built several refractor telescopes from lenses when I was in high school. It's totally doable. It just takes some patience to figure out how to solve all problems.


Why is Io's volcanism chemically so different from Earth's?

Two main reasons:

Location, location, location

The protoplanetary disk from which planets and satellites formed was not homogeneous.  Lighter elements and more readily dissociated molecules will be blasted away and/or apart if they are sufficiently close to the sun.  The sun emits a great deal of energy, and UV radiation in particular, which is great at breaking up molecules and providing an energetic kick to lighter atoms and molecules.

So objects like Mercury, which formed close to the Sun, will be composed of different material than will objects that form further away.  It's possible for objects to move from close to far, or far to close, though.  This is basically why the Earth has any meaningful amounts of water: the late heavy bombardment consisted of a relatively large number and volume of icy objects that formed further out having moved further in (or developing a large eccentricity that brought them inward) and ultimately colliding with objects such as the Earth.

To the best of my understanding, Io is believed to have formed in the general vicinity of Jupiter, so would have had access to a lot of materials that Mercury simply did not.  Sulfur and sulfur compounds in particular.

Why it has volcanism

Earth has volcanism, and a molten outer core, because of latent heat from its formation, radiocative materials in the mantle, and possibly other sources we don't completely understand.  That's a simplification, of course, as I'm not even touching upon plate tectonics and all that, but we just want to contrast it with an object like Io.

Io, on the other hand, has volcanism due to internal friction generated by tidal heating from its elliptical orbit around Jupiter, which is maintained as elliptical (instead of dissipating energy and circularizing) via an orbital resonance with other Galilean satellites.  The changes in the strength of Jupiter's gravitational pull on the planet variably stretches the planet, which stresses it and creates heat.

About that location...

So you might be thinking that if Io formed near Jupiter, which is beyond the (water) frost line, then why doesn't it contain large amounts of ice, or at least water?  Well, the location is important for this, too.

Io is the innermost of the Galilean moons.  Europa, the next closest Galilean moon (and one of the ones Io is in a resonance with), is more than 1.6 times as far from Jupiter as is Io.  This means it receives significantly more incident radiation from Jupiter than the other moons do (Europa would receive only approximately 37% of the radiative energy that Io does).  This radiation was sufficient in the early period of their lives to do what the sun's radiation does: blast water apart into hydrogen and oxygen (and then starts blasting the hydrogen away).  As such, while Io would have had access to a number of ices, including water, for its original formation, most of it would have been lost to this radiation.  Moons that are further out, or were captured by Jupiter much later in Jupiter's life, can retain their (water) ices, but Io was not so fortunate.

Thus you end up with a rocky, silicate-heavy world that's devoid of ices but has significantly more sulfur than an inner planet like Mercury.  The sulfur becomes pronounced in the volcanism for chemical reasons: it's volatile and pretty easily becomes gaseous.  And it doesn't need to have volcanism that behaves like volcanism on Earth because the source of its heat is fundamentally different.


Are all gas planets guaranteed to produce their own magnetic field?

It's not technically impossible, but it does seem incredibly unlikely that a gas planet would lack a magnetic field. This is because hydrogen (by far the most abundant element) attains metallic properties when put under extreme pressure. And a spinning metal/conductor generates magnetic fields via the dynamo effect. This is why earth has a magnetic field, except instead of metallic hydrogen spinning in our core, its our liquid iron outer core that does the trick.

Any collection of gas large enough to form a planet would certainly create the needed pressure to transform hydrogen in its metallic form. For a gas planet to have a magnetic field, it really only needs to meet 2 conditions:

The planet needs abundant free hydrogen (which is extremely
likely given that almost all matter is hydrogen).
The planet needs to be spinning (also extremely likely because a
planet retains the net angular momentum of all the stuff that fell
onto it while it was forming... the only way a planet avoids
spinning is by tidal locking, such as the moon always facing one
side towards the earth... and even then it's still spinning
slowly... it's just spinning once every time it completes an orbit.)

Ultimately, given the vast and possibly infinite size of the universe, and the great abundance of planets, I'd bet there is a planet somewhere that happens to be made of gasses other than hydrogen, or happens to not be rotating, and thus there would be a gas giant without a magnetosphere. But I'd be very surprised if we find one in my lifetime, or even in the the next 1000 years!


Telescope showing high chromatic aberration

Is this because I have not used achromatic lens?

An easy way to check for the effects of chromatic aberration will be to look at a narrow wavelength source. These days that's easy because the spread in wavelengths from single color LEDs is only 10 to 20 nm1 depending on the specific LED. If you can find something with red, green or yellow LEDs (not the white ones) can you check to see if those can be focused more clearly? All you need is one small LED to do this, maybe across a room. If your focal length is 1.1 meters and the LED is 10 meters away, you'll have to pull your eyepiece out by another 10 cm farther than you would for focusing at infinity.
1random example
Note also that chromatic aberration is stronger in the blue than in the red.
You can also estimate the effect with a bright white object with a sharp edge (like a white piece of paper) on a black or dark background. If the edges are blurry with strong color effects (blue and red in different places) then this may indeed be the problem.
If so, you might have to invest in an achromatic objective to get rid of this problem, or look only at things that are monochromatic (and in Astronomy there aren't many of those!)
In addition to making sure the eyepiece lens is centered on the objective lens' axis, you also have to make sure the eyepiece lens isn't tilted. Both of those will cause problems and the point-like LED spot will show this. It might look like a fuzzy comet coma aberration.

Can it be resolved if I used any coated eyepiece?

No, don't worry about that for now. Adding antireflection coatings to lenses reduces reflections which are a big problem when there are multiple lens elements with air gaps between them within the eyepiece or the objective. Those cause lens flare usually only a big problem if there are very bright, point-like objects in the field of view. But in your case each is a single element, so not an issue.

Does this smartphone photo show Mars just below the Sun?
Apparently two objects in Solar Eclipse image
Weird Blob near Polaris?

Is this issue is because the size of eyepiece is so small.

No, I think it's because of some combination of the eyepiece not being positioned properly and chromatic aberration, and a test on an LED might be one way to separate the two effects for diagnostic purposes. If the blurriness has a lot of color associated with it, especially around white objects on black backgrounds, this is your primary culprit I think.


What is a hard spectral state vs. a soft spectral state?

Active galaxies are known to change state as seen by a change in slope of their X-ray and gamma-ray spectra.  We say that a spectrum has become harder (or changed to its hard state) when the slope changes so that there are relatively more high energy photons, and it becomes softer when the ratio of low energy photons to high energy photons increases.  The physics of why they change state is not yet understood.
If the high energy spectrum is in a thermally dominated state, ie a Planck-like spectrum, then it is certainly soft.  If it is in a steep powerlaw dominated state with low  $\alpha < 1, I_\nu = f^{-\alpha}$, then it is still usually considered soft. But if $\alpha$ is higher, then it is hard.   But, it is also a relative term, so a system cycling between any two different values of $\alpha$ will be going between its hard state and its low state.


How is the Universe's Expansion Accelerating if the Hubble Constant is Decreasing?

The Hubble parameter is defined as the rate of change of the distance between two points in the universe, divided by the distance between those two points.
The Hubble parameter is getting smaller because the denominator is getting bigger more quickly than the numerator.

In the future, the cosmological constant, $\Lambda$ could result in an exponential expansion with time. A simple piece of maths shows you that the Hubble parameter is a Hubble constant only for an exponential expansion.

For example, suppose $H(t) = (da(t)/dt)/a(t)$, where $a(t)$ is the distance between two arbitrary points in the universe at epoch $t$. Now let's have an expansion rate $da/dt \propto t$ (ie accelerating). But in this case $a(t) \propto t^2$ and $H(t) \propto t ^{-1}$ (ie decreasing with time).

Some extra details:

The solution to the Friedmann equation in a flat universe is
$$H^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda}{3},$$
where $\rho$ is the matter density (including dark matter) and $\Lambda$ is the cosmological constant.

As the universe expands, $\rho$ of course decreases as $a(t)^{-3}$, but $\Lambda$ remains constant. Thus the first term on the RHS becomes unimportant if indeed $\Lambda$ is a cosmological constant.

Thus the Hubble "constant" actually decreases from its current value $H_0$ and asymptotically tends towards $ H = \sqrt{\Lambda/3}$ as time tends towards infinity.


Why do stars twinkle but the Sun doesn't (I'm asking this because the Sun is also a star)

Stars twinkle because they are effectively a point of light. This point of light can be distorted and magnified by movement of patches of varying density in the atmosphere. These act as lenses causing the twinkle effect.
If you get to space, the stars don't twinkle. Stars appear as a point only because they are so very very distant.

Planets twinkle less because they appear as little discs of light. If one point is being made brighter by the atmosphere perhaps another is being made less bright. By being a disc, the twinkling is smoothed out by the average amount of light making it to your eye.

The sun is a very large disc because it is closer than the other stars. The sun is so large in the sky that the distortion of the light is not very noticeable. However when the sun is setting and its light has to travel through a lot more air to reach the ground, you will notice a shimmering around the edge. This is due to the same kind of atmospheric distortion.


Is Earth's true anomaly roughly 1 degree currently?

I've found an almanac for Earth perihelion up to the year 2100. 2019 perihelion occurred at roughly 5:00 January 3, 2019. Thus the angle now of true anomaly is some extent past 0 degrees. And Wolfram Alpha is wrong by negative 180 degrees. By Wikipedia and other standard definitions of mean anomaly which have persisted since Kepler's time.


What is assembly bias in context of cosmology?

Early analytic calculations of the distribution of the masses of dark matter halos (in which galaxies reside) by Press & Schechter (1974), Bond et al. (1991), Mo & White (1996) and several others showed that the clustering of these halos depended only the redshift and the mass of the halos. However, later work showed that clustering depends on several other parameters, such as the assembly history of the halos (Sheth & Tormen 2004; Gao et al. 2005).

This fact is known as the assembly bias. I think the first to use the term was Croton et al. (2007). Later, the term was "expanded" to also refer to the dependence of the clustering on halo spin, concentration, and shape (ellipticity) of the halos.

You can find some more recent good discussions in Lazeyras et al. (2017) and Padilla et al. (2019).


Does the Giant Impact Hypothesis explain how the Moon circularized its orbit?

I want to point out that the Moon's orbit isn't circular now.   A 0.055 mean eccentricity isn't that circular.

But, onto your question.   I think you're making a bad assumption on the "must have started very elliptic.   Individual objects that are ejected from a planet need to follow their orbital path.   So any object flung from the earth would need likely either escape the Earth or fall back into it because you can't launch something, with a singular push, into a circular orbit.   All the individual bits of debris should have had highly eccentric orbits.

But the formation of the Moon was much more complicated than that.   First, it was enough matter to have it's own gravitational field, basically influencing itself, and the forming moon would be the combination of all those individual objects, so it wouldn't necessarily be eliptic at all.   The eccentricity of billions of objects could have (and likely did) cancel each other out to a large degree.

Because the impact was off-center, it set the Earth spinning and most of the debris moved in the same direction around the Earth.   But the coalescing mass of material that didn't fall back to the Earth could have been in a reasonably circular orbit at formation.

We should also define what number constitutes "highly" elliptic.   The Moon is thought to have started out 3-5 Earth Radii distant.   Any material that passed closer than 3 radii would have been inside the Roche limit and had a hard time coalescing into the Moon.    This article suggests that the debris would have had a hard time being ejected further than 5 Earth Radii, though I'm not sure how that conclusion was reached.  But you have a few things going on.  Debris colliding with other debris, and some (presumably quite a bit) of debris falling back to the Earth and, mentioned above, the debris having a gravitational effect on itself.   Ultimately momentum needs to be conserved and calculating the formation of the Moon requires a super-computer, but in general I think the initial orbit could have been relatively circular.

If the Moon formed close to the Earth, which most models suggest it did and since it couldn't pass inside 3 Earth Radii and stay solid, those numbers put a limit on how eccentric the initial orbit probably was.   If we use the 5/3 Earth Radii as an estimate, Ra=5, Rp=3, then the eccentricity is 0.25, and that's an estimate for the upper limit of the initial eccentricity after formation.  It may well have been quite a bit less.

But for the sake of argument, lets give the newly formed moon an eccentricity of .25 or, maybe a little higher with a more distant apogee, but we have to keep the perigee at 3 Earth radii or greater.   Your question still stands, how did the Moon's initial eccentricity become somewhat circular.

The answer is tidal circularization.   I can't find a good article on it, but for many 2 body systems, tides circularize orbits.   It's well known that the tides on Earth, caused by the Moon's gravity rotate ahead of the Moon and this creates a tug on the moon that pushes (pulls?) the Moon into a higher orbit, further away from the Earth.   This same secondary effect circularizes the Moon's orbit because the closer the Moon is too the Earth, the greater the push.  The math behind this gets very complicated and it's above my paygrade to put numbers behind it and the Wikipedia article I've linked is very lacking, so I invite anyone to provide more details if they can. But Tidal circularization is a commonly accepted theory and given the strength of the tides early on, it's likely any eccentricity in the Moon's orbit circularized relatively quickly, at least, astronomically speaking.   (Tens of millions of years is a long time to you and me, but not to a stable orbit).


How to find the distance between two quasars

Edit: I found you a very relevant paper!

http://arxiv.org/pdf/astro-ph/0007341v1.pdf seems to be solving exactly your problems for sources with arbitrary angular separation $\alpha$. Equations (12) and (14) will give you the comoving distance $\chi_2'$ and redshift $z_2'$ between them at the epoch when $Q_1$ received light from $Q_2$. You will have to solve these equations numerically to find the solutions, which will be more complicated depending on the cosmological model you use.

Have a read of these notes http://arxiv.org/abs/astro-ph/9905116. They are my go-to when trying to work out cosmological distances.


VSOP87D: What are the purpose of the initial columns in the VSOP87D data?

First, the data in most of these columns is not required for the computation of planetary positions using the theory if the 'second' form equation is used. The terms are, using the order of the example of Jupiter VSOP 87D's third line ...

4510 ...

"4" means the 'version code', i.e., this is the "D" version of the theory which allows computation of the heliocentric rectangular coordinates of the planet for the ecliptic of date. A "3" would be for the spherical heliocentric coordinates for the ecliptic of date.

"5" is the 'body' code, Jupiter being the FIFTH planet. Mercury would be planet '1', for example.

"1" is the 'index of coordinate', meaning in this example that this line is part of the 'longitude' coordinate calculation. Later lines show a '2' for the B (or latitude) coordinate or '3' for the R (radial or distance) coordinate)

"0" is the 'degree' or power of the time variable. Here, terms are multiplied by T to the ZERO power (or 1). Later terms are multiplied by successively higher powers of T.

3
... is simply the sequential number of this line in this series. Jupiter has 5 series (each multiplied by a higher power of T), so this number 'resets' for each series.

The next 12 terms, mostly zeros in this example, are the 'coefficients of mean longitude'. As near as I can tell, these coefficients are used to correct mean longitude quantities for secular perturbations. They are used only in the first form of the equation as shown below with the next two terms:

The next two terms are amplitudes S and K, used in the first form of the positional equation:

T**alpha * (S sin phi + K cos phi)

where phi = sum[i, 1 thru 12] [a(i) * lambda(i)] and lambda(i) values are given for each solar system body in the explanatory text.

The final 3 terms are Amplitude A, Amplitude B and Phase C for use in the second form of the equations to calculate position:

T**alpha * A * cos(B + CT)

... with T being time in Julian centuries, and alpha the 'power' of T.

My comment in the original question about the terms being 'off' by several orders of magnitude is moot ... for convenience either in calculating or printing, Meeus multiplied the terms by 10**8, then corrects for it later. This probably facilitated better computation on machines of the time it was written and also made for neater printing in the book.

The document provided by barrycarter, along with some other study of planetary motion allowed me to finally understand this, thanks!


Has anyone filmed a total lunar eclipse from the Moon's perspective?

What can you say but "Wow..."

http://www.space.com/32245-total-solar-eclipse-video-dscovr.html

"The moon's dark shadow travels across Earth's face in spectacular new views of Tuesday's (March 8) [2016] total solar eclipse captured by a satellite in deep space."

I'm sorry my answer is off by about 700,000 miles compared to what you asked :)

I'm fairly sure there did not happen to be, a time when NASA had cameras on the moon, when, there was a lunar eclipse.

Note that cameras today (such as this one on DSCOVR) are unbelievably better than cameras from Apollo era, so this would blow away any merely 700,000 miles closer imagery captured during the Apollo era.


You've put in the moon filter, right? The moon is awfully bright without it. In any case, take the scope out in the daytime, and try focusing on distant objects; the tree at the end of the block, or the radio tower on top of the hill at the edge of town. A scope is easier to learn how to use when you can see what you're doing.