Could Ganymede's aurora be seen with the naked eye?

The aurora depicted in the artist's conception is of ultraviolet light.  It would not be seen with the naked eye.

"NASA's Hubble Space Telescope observed a pair of auroral belts encircling the Jovian moon Ganymede. The belts were observed in ultraviolet light by the Space Telescope Imaging Spectrograph and are colored blue in this illustration. They are overlaid on a visible-light image of Ganymede taken by NASA's Galileo orbiter. The locations of the glowing aurorae are determined by the moon's magnetic field, and therefore provide a probe of the moon's interior, where the magnetic field is generated. The amount of rocking of the magnetic field, caused by its interaction with Jupiter's own immense magnetosphere, provides evidence that the moon has a subsurface ocean of saline water."


Why temperature of dark side of moon is not 3 Kelvin

The "dark" side of the Moon is only truly dark during full Moon. Everywhere on the Moon there is day and night as well. The dark side of the Moon is called like that because we do not see it from Earth, since the Moon shows us always the same side due to tidal locking - not because it is always dark there.
In other words: During (lunar)  day, the Sun heats up the surface of the Moon and it gets pretty warm, even on the (not so) "dark" side of Moon. This explains the $100 {\rm K}$ you are citing.


When will all eight planets in our solar system align?

This is a low accuracy - yet simple - answer

It allows you to calculate only radial alignment configuration of the planets.

If you would like an approximation, let's say, you approximate the position of the planets as hands in a clock, you could work the math out by something like this.

Assume $\theta_i$ is the initial angle for planet $i$ at time $t_0$ - measured from an arbitrary but fixed position, and $l_i$ is the length of the year - in days - for planet $i$.

Then it resumes to solving this system of equations:

$$
x \equiv \theta_i \left( \ mod\ l_i\right)
$$

From here you would then simply apply the Chinese Remainder Theorem.

Finding the minimum x, will give you the angle that the planet that at $t_0$ had angle $\theta_i = 0$ would have travelled until an alignment configuration was reached. Asuming you choose Earth as the mentioned planet, then divide that angle by a complete revolution ($360^{o}$) and you will get the number of years for that configuration to be reached - from the $t_0$ configuration.

The different $\theta_i$ in degrees for all the planets on 01 Jan 2014 - you can use this as your $t_0$:

\begin{align}
Mercury &\quad 285.55
\\Venus &\quad 94.13
\\Earth &\quad 100.46
\\Mars &\quad 155.60
\\Jupiter &\quad 104.92
\\Saturn &\quad 226.71
\\Uranus &\quad 11.93
\\Neptune &\quad 334.90
\end{align}

Source

The different $l_i$ in days for all the planets:

\begin{align}
Mercury &\quad 88
\\Venus &\quad 224.7
\\Earth &\quad 365.26
\\Mars &\quad 687
\\Jupiter &\quad 4332.6
\\Saturn &\quad 10759.2
\\Uranus &\quad 30685.4
\\Neptune &\quad 60189
\end{align}

Finally under an integer values aproximation and using this online solver for the system of equations the answer is $x = 4.0384877779832565 \times 10^{26}$ which divided by $360^{o}$ gives you roughly $$1.1218 \times 10^{24} \quad\text{years}$$

Edit 1

Just found this site you may like to play around with. It's an interactive flash application with the accurate position of the planets.

I also know that the all the information can be obtained from this NASA page and that is as accurate you can get, but it is just incomprehensible for me now. I will try to revise it later when I find time.

Also this book by Jean Meeus called Astronomical Algorithms covers all the fundamental euqations and formulas - it has nothing to do with programming algorithms though.

Edit 2

Seeing that you are a programmer, it could be worth for you checking out the NASA site I mentioned above, the data for all the planets can even be accessed via $\tt{telnet}$.
Or this Sourceforge site where they have implementations for many of the equations described in the book also mentioned above.


If they knew exactly where to point Mr. Hubble, then yes, it should be easily visible, though at that distance, it would still be blurry, not a clean image.

The estimated apparent magnitude of Planet Nine varies based on which website you believe, Google provides ranges from 20 to 25 with Wikipedia saying >22.  Higher numbers mean less visible.

Hubble can see things up to apparent magnitudes of about 30, so, yes, it should be easily visible by Hubble or (perhaps) even more visible by an infra-red telescope or radio telescope that knows where to look.  But in any case, with current technology all images would still be blurry.   Hubble's pictures of Pluto are blurry and Pluto's much closer.


I've checked several sources. Where this term appears in dictionaries, it is translated with a definition, and not a gloss (usually if a direct translation exists in English only a gloss is given) For example https://en.wiktionary.org/wiki/mezzocielo intersection of the meridian of a given place with the celestial equator.

I conclude that English speaking astronomers have not had the benefit of single word translation of mezzocielo.


In theory, is there anywhere in the universe where velocity=0?

Yes.

The velocities you list (X, Y, …) are all velocities with respect to some reference frame. But all reference frames are arbitrary, and you can always define a reference frame where the velocity of some object is exactly zero, as long as it's not accelerating.

For instance, Earth's velocity in the Sun's reference frame is X, but in its own reference frame, it's zero.

This is not just semantics; it is probably one of the most profound concepts in physics that there is no such thing as an "absolute frame of reference". Before Einstein, people thought that time and space are absolute, but Einstein taught us that everything is relative.

In cosmology, there is a special reference frame in which all matter on average is zero. The coordinates of this frame are called comoving coordinates, and in many cases it makes sense to regard this as the most natural reference frame. But it is important to remember that it is still just a choice, and that any other reference frame is equally legit, although a bad choice of reference may make certain calculations harder or impossible to solve. In comoving coordinates, the cosmic microwave background (CMB) is isotropic, i.e. it looks the same no matter which direction you look. But since Earth is traveling with a non-zero velocity in comoving coordinates (at an average of 369 km s$^{-1}$), the CMB is slightly blueshifted in the direction in which we're traveling, and slightly redshifted in the other direction.

What would time look like? Time runs slower, the faster you move. In any non-accelerating reference frame (an inertial system), time runs "as it should", but an observer in this frame who measures how time runs in a frame that has a non-zero velocity with respect to his/her frame, will see it run slower.


How long will star formation endure before the eventual heat death of the observable universe?

This is actually very well researched. Wikipedia has quite a good summary, stelliferous era ends after about $10^{14}$ years, assuming an expanding universe.


How to deal with shifting CCD bias (zero exposure) levels?

Looks like you've got some flaky electronics attached to your CCD if the bias level is changing by several units in consecutive exposures. However, all is not lost if your CCD has an overscan region.

This is normally several columns of "pixels" that appear appended to each row of your data, but which do not correspond to physical pixels on the CCD. They are actually dummy readouts to assess the actual bias level in each particular row.

The overscan region should be clearly visible in any real exposure of a light source, since the overscan region will be un-illuminated.

You can use the overscan in two ways. The conventional method is to overscan-correct your images before doing a bias subtraction. You should also do this to your bias frames before combining them into your "master bias" frame.

So, take your overscan region and either (i) calculate a median (or clipped mean) within it and subtract this value from every pixel in the image before use (including the bias frames); or (ii) fit something more complicated as a function of row number and then subtract this from your actual data on a row-by-row basis.

Method (ii) is required if you notice that there is any structure or drift/trend in the overscan values as a function of row number.

This link gives plenty of details, some pictures and a recipe.


Can you see city lights on the Moon from Earth?

It is possible but unlikely. Here is a really good 'Science 2.0' article about the possibility (http://www.science20.com/robert_inventor/could_you_see_moon_city_lights_or_a_greenhouse_from_earth_just_for_fun-157480). Essentially, you likely wouldn't see the light on Moon settlements because there would need to have many thousands of bright lights and windows- both of these would be impractical on a moon base (windows let in radiation and are expensive, and bright lights waste valuable power). Furthermore, in order to reach a magnitude of brightness to constitute as visible, there would have to be a very large amount settlements concentrated together.


How to determine atomic number density of an element in a star based on equivalent width measurements

The measurement of a chemical abundance is not a question of using a simple equation.

The simplest it gets is using a "curve of growth", which relates equivalent width to abundance and assumes you already know the temperature of the star and its surface gravity.

For the Li 6708A line, the relevant tables, that can be interpolated, are found in Soderblom et al. (1993). http://adsabs.harvard.edu/abs/1993AJ....106.1059SS . These are LTE curves of growth based on Atlas atmospheric models.

As an aside, the equivalent width of this line in the Sun is 3 mA. The formula you quote looks like some sort of linear approximation, but this resonance line rapidly enters the saturated (non-linear) portion of the curve of growth for equivalent widths of 50 mA or more. As the equation is couched in terms of ratios, then the units you use are entirely up to you.


Could an astronomical process create elements in the islands of stability?

To this point, this is purely theoretical.  We have no solid evidence of islands of stability.  They may or may not exist to begin with.  It's hard to speculate on the methods of creating something we don't even know exists.

That said, currently, the heaviest elements found in nature appear to be synthesized in the mergers of neutron stars.  It might be possible that a similar process could create even heavier elements.  It's possible that some of the discovered transuranic elements we've discovered might be formed in such mergers, however their half-lives are so short that they break down before they're detectable.  If there are stable transuranic elements, then they might be formed in this manner, but we've never seen any evidence of them.  Me might not know what to look for, but so far we've detected no suggestions of such elements.

Another option is the accretion disks of black holes.  It is believed that nucleosynthesis occurs in accretion disks, and one could speculate that there may be conditions in accretion disks that could make it possible.

But without any evidence of the islands of instability, this is entirely conjecture.


Intention to answer

"Is there still a possibility that Makemake is not ellipsoidal but asteroid-shaped?"

Probably not under reasonable assumptions.
From Ortiz et. al. (2012) "Albedo and atmospheric constraints of dwarf planet Makemake from a stellar occultation" https://www.nature.com/articles/nature11597

Our preferred solution that fits the occultation chords corresponds to a body with projected axes of 1,430 ± 9 km (1σ) and 1,502 ± 45 km, implying a V-band geometric albedo pV = 0.77 ± 0.03. This albedo is larger than that of Pluto, but smaller than that of Eris. The disappearances and reappearances of the star were abrupt, showing that Makemake has no global Pluto-like atmosphere at an upper limit of 4–12 nanobar (1σ) for the surface pressure, although a localized atmosphere is possible. A density of 1.7 ± 0.3 g cm−3 is inferred from the data.

Concerning the irregular shape of Haumea there are some uncertainties, it might not be a homogenous body but one with a ring and it has satellites.
From Ortiz etl a. (2017) "The size, shape, density and ring of the dwarf planet Haumea from a stellar occultation" https://www.nature.com/articles/nature24051

Secondary events observed around the main body of Haumea are consistent with the presence of a ring with an opacity of 0.5, width of 70 kilometres and radius of about 2,287 kilometres. The ring is coplanar with both Haumea’s equator and the orbit of its satellite Hi’iaka. [...] The occultation by the main body provides an instantaneous elliptical projected shape with axes of about 1,704 kilometres and 1,138 kilometres.

The paper's also on Arxiv it seems.


Would a black hole passing next to a star create a deadly focal point due to gravitational lensing?

tl;dr: seen at a distance, the lensed object will appear as an annulus or "ring" around the lensing object, and while that will be brighter than if there were only empty space, sadly it won't be death-ray bright!

Let's first think about what makes a familliar lens a lens. Near the center the thickness doesn't vary much, but as you move farther from the center the thickness changes faster and faster.
If we measured the slope or angle of the surface, we'd see that the angle increased roughly linearly with distance from the center.
In the thin lens approximation the angle that the lens bends the light $\Delta \theta$ is proportional to the distance from the center of the lens $r$ where the light hits it. The bending power increases linearly with distance.
$$\Delta \theta \approx \frac{r}{f}$$
where $f$ is the focal length of the lens.
How do concentrated points of mass bend light?  Wikipedia's Gravitational lens give us
$$\Delta \theta \approx \frac{4 G M}{r c^2}$$
and that's a problem because now $r$ is on the bottom!
Single concentrated objects like black holes do not act like the familliar lenses we use to focus. They do have some ability to concentrate somewhat compared to empty space, but nowhere near as much as real lenses do. From a source that's a certain distance away, there will be only one angle that gets bent parallel into a "beam", angles slightly larger or smaller passing slightly farther or closer from it will be bent much less or much more, either diverging or converging from the axis afterward.
Thus seen at a distance, the lensed object will appear as an annulus or "ring" around the lensing object, and while that will be brighter than if there were only empty space, sadly it won't be death-ray bright!
In the case of your star/BH pair, exactly how bright it will be along that axis depends on the details, but one can get a good estimate by ray tracing, either with a few lines of Python or pencil and paper for estimate purposes.
If the star were close then it is an extended object, a wide disk, and those can't be focused to points even by lenses (we can't concentrate blue sky with a magnifying glass) so there wouldn't be much of a death effect. If it were far, you could concentrate it better (because it's more point-like) but it would be a lot dimmer to begin with because it would be farther away.

A Horseshoe Einstein Ring from Hubble  The bright thing in the middle is the lensing mass, the ring is the lensed object behind it. It's not a "beam" but just a distorted view, but it is brighter than if the lens weren't there.


"Supernova" is the explosion or the resulting celestial body? Is it incorrect to call the explosion "supernova"?

A supernova is the explosion, visible on Earth as a apparently new star that appears to brighten over a few days, and then fade. There have been no visible supernovae in our galaxy for several hundred years, and we are well overdue another one, though they are quite often visible in other galaxies.

After the supernova there will be a remnant, a small nebula of hot gas that expands, eventually taking the form of a ring. In the centre of the remnant there may be neutron star or a black hole. A spinning neutron star may be a pulsar.

Supernovae that occur when a old, giant star explodes at the end of its life can produce a neutron star or black hole. But some supernovae form when a white dwarf accumulates matter from another star onto its surface, and then explodes, in which case the star is completely destroyed.

The term "nova" was previously used for all stellar explosions. However the term "nova" is nowadays limited to smaller explosions, caused by the fusion of hydrogen on white dwarf (but no sufficient build up of hydrogen to cause the white dwarf to be destroyed)


In what units to quote the thermal Blackbody temperature

I'm not entirely sure what you mean, but the (Planck's) formula for blackbody radiation is given by

$$S_{\lambda} = \frac{8 \pi h c}{\lambda^5} \frac{1}{e^{hc/\lambda kT} - 1}$$

where $h$ is in $\mathrm{J \cdot s}$, $c$ in $\mathrm{m/s}$, $\lambda$ in $\mathrm{m}$, $k$ in $\mathrm{J/K}$ and $T$ in $\mathrm{K}$.
So, the temperature is just in kelvin, not in energy.
This formula, with these units, gives $S$ in $\mathrm{W/m^2/m}$, which describes the amount of energy per temperature and wavelength. There is no need to convert a given temperature to energy.


What would happen to a gas planet if its core mass goes beyond the Chandrasekhar limit?

Gas giants like Jupiter consists mainly of Hydrogen and some Helium. If you gradually add mass to these planets then core temperatures will rise gradually and a stage will come where they will ignite like normal stars. Unlike white dwarfs where accretion of mass leads to type 1A supernova.


Question on Equatorial and Galactic coordinate systems

Yes, because the Milky Way galaxy and the Earth's equator are in different planes, currently about 63 degrees apart. To avoid confusion, rename the galactic Cartesian coordinates to (u,v,w). The transformation from (u,v,w) to (x,y,z) can be expressed as multiplication by a rotation matrix.


What would happen if a Sun-like star were to consume a Jupiter-like planet?

There are scenarios where the effects of planet accretion on a sun-like star would be very significant indeed, at least in the short term. Whilst the amount of mass accreted by the star would be a tiny perturbation, the amount of accreted energy and/or angular momentum may not be.

Scenario 1: The scenario where a hot Jupiter just drops into the star from a distant radius would certainly have drastic short-term effects. But short-term here means compared with the lifetime of the star.

The kinetic energy of say Jupiter falling to the star's surface from a distant (more than a few solar radii) would be of order $GM_{\odot}M_\mathrm{Jup}/R_{\odot} \sim 4\times 10^{38}$ joules.

For comparison, the solar luminosity is $3.83 \times 10^{26}\ \mathrm{J/s}$.

The addition of this much energy (if it is allowed to thermalise) would potentially affect the luminosity of the Sun for timescales of tens of thousands of years. The exact effects will depend on where the energy is deposited. Compared with the binding energy of the star, the additional energy is negligible, but if the energy is dissipated in the convection zone then the kinetic energy would do work and lift the convective envelope. In other words, the star would both increase in luminosity and in radius. If the effects were just limited to the convective envelope, then this has a mass of around $0.02 M_{\odot}$ and so could be "lifted" by $\sim 4\times 10^{38} R_{\odot}^2/GM_{\odot}M_{\rm conv} \sim 0.05 R_{\odot}$.

So in this scenario, the star would both expand and become more luminous. The relevant timescale is the Kelvin-Helmholtz timescale of the convective envelope, which is of order $GM_{\odot}M_{\rm conv}/R_{\odot} L_{\odot} \sim $few $10^5$ years.

If the planet somehow survived and punched its way to the centre of the star, then much less energy would be deposited in the convection zone and the effects would be lessened.

On longer timescales the star would settle back down to the main sequence, with a radius and luminosity only slightly bigger than it was before, commensurate with its tiny (0.1%) increase in mass.

This all assumes that the planet can remain intact as it falls. It certainly wouldn't "evaporate" in this direct infall scenario, but would it get tidally shredded before it can disappear below the surface? The Roche limit is of order $R_{\odot} (\rho_{\odot}/\rho_{\rm Jup})^{1/3}$. But the average densities of the star and (gas giant) planet are almost identical. So it seems likely that the planet would be starting to be tidally ripped apart, but as it is travelling towards the star at a few hundred km/s at this point, tidal breakup could not be achieved before it had disappeared below the surface.

So my conclusion is that dropping a Jupiter into a Sun-like star in this scenario would be like dropping a depth charge, with a lag of order $10^{5}$ years before the full effects became apparent.

Scenario 2: A hot Jupiter arrives at the Roche limit (just above the stellar surface) having lost a large amount of angular momentum. In this case the effects may be experienced on human timescales.

In this case what will happen is the planet will be (quickly) shredded by the tidal field, possibly leaving a substantial core. At an orbital radius of $2 R_{\odot}$, the orbital period will be about 8 hours, the orbital speed about  $300\ \mathrm{km/s}$ and the orbital angular momentum about $10^{42}\ \mathrm{kg\ m^2\ s^{-1}}$. Assuming total destruction, much of the material will form an accretion disc around the star, since it must lose some of its angular momentum before it can be accreted.

How much of the star's light is blocked is uncertain. It mainly depends on how the material is distributed in the disk, especially the disk scale height. This in turn depends on the balance of the heating and cooling mechanisms and hence the temperature of the disk.

Some sort of minimal estimate could be to assume the disk is planar and spread evenly between the solar surface and $2R_{\odot}$ and that it gets close to the solar photospheric temperature at $\sim 5000\ \mathrm K$. In which case the disk area is $3 \pi R_{\odot}^2$, with an "areal density" of $\sigma \sim M_{\rm Jup}/3\pi R_{\odot}^2$.

In hydrostatic equilibrium, the scale height will be $\sim kT/g m_\mathrm H$, where $g$ is the gravitational field and $m_\mathrm H$ the mass of a hydrogen atom. The gravity (of a plane) will be $g \sim 4\pi G \sigma$. Putting in $T \sim 5000\ \mathrm K$, we get a scale height of $\sim 0.1 R_{\odot}$.

How long the accretion disk would remain, I am unsure how to calculate. It depends on the assumed viscosity and temperature structure and how much mass is lost through evaporation/winds. The accreted material will have radiated away a large fraction of its gravitational potential energy, so the energetic effects will be much less severe than Scenario 1. However, the star will accrete $\sim 10^{42}\ \mathrm{kg\ m^2\ s^{-1}}$ of angular momentum, which is comparable to its current angular momentum of the Sun.

The accretion of an exoplanet in this way is therefore sufficient to increase the angular momentum of a star like the Sun by a significant amount. In the long term this will have a drastic effect on the magnetic activity of a Sun-like star – increasing it by a factor of a few to an order of magnitude.


Is Earth unique in its fairly clear atmosphere?

Our atmosphere is only transparent to visible light, In most other wavelengths, some or all of the light is absorbed

Image from Wikipedia, adapted from image by NASA

Our eyes have evolved to take advantage of the transparency at these wavelengths. If we had evolved in an atmosphere with a very different mix of gases. One in which visible light was absorbed, we would have evolved eyes that see different wavelengths.

There is a notable "window" at about $10\mu m$ in the diagram above. And you might wonder if any animal has evolved to see in this window. However, our own bodies emit thermal radiation at about $10\mu m$, eyes wouldn't work as they would be swamped by their own glow. However this window is used by thermal imaging devices.


Which planet is also the name of an element?

The answer is Mercury.

Plutonium was made after Pluto was discovered.


How to calculate the pixel size and beam size of an image?

If you know the angular size of your image (e.g., 10", 1' etc), then pixel size  (in "/pixel units) is just image size / number of pixels.
For beam size, depends a lot on the wavelength-range of your observations (for instance sub-mm/radio observations are conceptually very different from UV/optical), but to have an idea of the beam/psf size, you can fit a 2D gaussian/moffat profile to the emission of a point-like object (i.e., a star) in the field of your image.

Note that most of the time these info are contained in the header.


Usually the separation exists, but is negligible compared to the size of the pixels.  Hence, the loss due to photons "falling through the cracks" is usually very small, and can be well accounted for using either Monte Carlo-based methods or experimentally using a focused source of known brightness. For example, attached is the user manual for the WIYN One-Degree Imager (ODI), a CCD detector that I have done some work on. There is a section on detector geometry that will probably help.

http://www.wiyn.org/ODI/ODIUserManual.pdf


Meaning of critical density in radiative transfer

Your misunderstanding is in the second paragraph. If $n<n_c$ then there are insufficient collisions to excite the molecule into the upper level. Thus in equilibrium there are very few excited molecules and since the luminosity is proportional to the number of molecules in the excited level then little radiation is seen.

The critical density is defined as that (temperature-dependent) density at which the rate of collisional excitation to the upper level of a molecular (or atomic, I suppose) transition, equals the rate of radiative de-excitation.

The rate of de-excitation is given by $n_2\ A_{21}$, where $n_2$ is the number density of the molecular species in the excited state and $A_{21}$ is the relevant spontaneous emission coefficient.

The rate of excitation is given by $\gamma_{12} n_1 n$,
where $\gamma_{12}$ is the collision probability (which depends on the speed of molecules in the gas and hence temperature), $n_1$ is the number density of the molecule of interest in the lower energy state and $n$ is the number density of all molecules in the gas.

There is also a rate of collisional de-excitation given by $\gamma_{21}\ n_2 n$. However there must be a relationship between $\gamma_{12}$ and $\gamma_{21}$ because for a gas in thermodynamic equilibrium, we know that
$$n_2/n_1 = \exp(-E_{21}/k_BT),$$ and so if collisions are in detailed balance then
$$\gamma_{21}\ n_1 \exp(-E_{21}/k_BT) n = \gamma_{12} n_1 n$$
$$ \gamma_{21} = \gamma_{12} \exp(E_{21}/k_BT)$$

OK, so now consider a gas where there are radiative losses from state 2 and allow it to come into equilibrium such that
$$\frac{dn_2}{dt} = 0$$
$$ -A_{21}n_2 - \gamma_{21}n_2 n + \gamma_{12} n_1 n = 0$$
$$ \frac{n_2}{n_1} = \exp(-E_{21}/k_BT)\ \frac{1}{1 + A_{21}/\gamma_{21}n}$$

If now we say that $A_{21} \ll \gamma_{21}n$, then radiative de-excitation is (relatively) negligible compared with collisional de-excitation then the $n_2/n_1$ ratio approaches the level expected by the Boltzmann ratio.

If on the other hand $A_{21} \gg \gamma_{21}n$, then the ratio of $n_2/n_1$ becomes much lower and radiation removes excited molecules faster than collisions create them.

Now we ask how much radiation we are going to see from this gas.

(1) If $A_{21} \ll \gamma_{21}n$, which we can restate as $n \gg n_c$, the "critical density".

The luminosity from the molecules will be
$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT)$$
If $k_BT < E_{21}$ this means that $n_1 \gg n_2$ and the luminosity per molecule in the gas has a fixed value that is independent of density $n$. There is just a fixed amount of radiative losses per molecule of the species in question.

(2) If $A_{21} \gg \gamma_{21}n$, which we can restate as $n \ll n_c$, The luminosity from the molecules will be
$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT) \frac{\gamma_{21} n}{A_{21}}$$
and so this is smaller than for the first case by a factor of $n/n_c$. So now there is less radiation per molecule of the interesting species because fewer of those molecules are in an excited state.

NB This is a different thing to the "quenching" density for forbidden lines that I refer to in my comment.


How are exoplanets with 10,000 day periods discovered with the radial velocity method?

You don't need to collect data for one cycle in order to determine the period.

From Newtonian's physics, we know precisely how a trajectory of one particle orbitting around another one would be. Since the instantaneous velocity (v_inst) at any point on the trajectory is tangential to the point on the trajectory, we can decompose the v_inst into the radial velocity which we can match with the observation from, e.g., redshift. With assumption of constant v_inst and trajectory, we can determine the period from just a couple of observations (depending on how accurate the redshift can be measured, and the number of variables in the equation).


is there a way to determine a distant galaxy’s speed relative to the Hubble Flow by measuring time dilation effects between there and this galaxy?

No. Time dilation is expected and observed as a result of universal expansion (e.g. in the light curves of type Ia supernovae and the duration of gamma ray bursts, Blondin et al. 2008; Zhang et al. 2013). There is no way of distinguishing between a Doppler shift and cosmological redshift for an individual source.


How many satellites orbit the Earth? At what rate do they orbit the earth?

There are over 1000 operational satellites, and probably as many again defunct or non-operational satellites in orbit, plus a great deal more small pieces of space junk.

About half of these are in a low earth orbit (LEO). Satellites in LEO have a period of between 90 minutes and 130 minutes. But you would not expect a satellite to follow the same path in the sky in two orbits, as the earth will have moved between orbits of the satellite. So it is very unlikely that you will see the same satellite pass in front of the moon twice.

Satellites in LEO are moving fast! They will cross the moon's disc in less than a second, and transits are rare, as each is a path only 7km wide. at any one time, a transit is visible from less than 100th of 1% of the Earth's surface. To obtain the awesome photo that you have linked, the astrophotographer would have had to use some orbital prediction software to calculate the exact time of a solar transit by the ISS and travel to a carefully calculated position (and prayed for clear skies). You could be observing the moon all year and never see a transit.

Most of the remaining satellites are in Geostationary orbits, They orbit at a great distance over the equator. As their orbital period is exactly one day, they appear to hang in space above a point on the Earth. Such satellites are too distant to be seen as "dots".


Why is the Sun so bright, but you can feel it far away?

The answer to your first question has to do with luminosity. It's a measure of power, the energy given off by an object in a certain amount of time, which you can think of as brightness. The more luminous the object, the brighter it appears.

We can treat the Sun as an idealized object called a black body, which emits thermal radiation according to something called the Stefan-Boltzmann law. Assuming the Sun is spherical, with a radius $R$ and temperature $T$, then its luminosity $L$ is
$$L=4\pi\sigma R^2T^4$$
where $\sigma$ is the Stefan-Boltzmann constant. The important thing here is that $L\propto R^2$. If the Sun was the size of one of your furnaces, it would not be very bright; if the furnace was spherical and could be treated as a black body, the two would have the same luminosity.

The Sun, however, has a radius on the order of about $7\times10^8$ meters. I don't know how big the furnace is, but I'm assuming it can't be larger than ten meters, to an order of magnitude. First, that's a size difference of about $10^7$. Second, we square the radius when determining luminosity, meaning that there's a difference of a factor of $10^{14}$. To make this a bit more meaningful, the Sun is then essentially one hundred million million times more luminous than the furnace, even though the temperatures are the same.

I feel like your second question may be a duplicate of What is the temperature of outerspace? (see also the questions linked therein), and Run like hell has already tried to address it. barrycarter's comment, though, is spot-on. The flux from the source, $F$, drops off via the inverse-square law - that is, at a distance $r$ from the source,
$$F\propto\frac{1}{r^2}$$
Remember how we just established the power (pun intended) of squaring a large quantity? Interstellar distances are enormous; the nearest star system, Alpha Centauri, is roughly $10^{16}$ meters away, or ten thousand million million meters. Now square that distance. That's a tiny factor - even when multiplied by the luminosity of all three stars, which combined is still on the same order of magnitude as the Sun.


The simple answer is (as with so much in astronomy):
We Don't Know

Parallel universes may or may not exist. There is no definitive way to prove that these universes do or don't exist.

A parallel universe is a separate existence to ours. The Theories that suggest that there may be parallel universes are classified as theories of multiverses. There are many theories of multiverse, all of which propose different ideas about what could exist beyond the limits of our universe. There are also theories that suggest that the multiverse doesn't exist, although the theories with most support are by far the multiverse theories.

For a nice reference in book form, see Steinhardt and Turok's "Endless Universe: Beyond the Big Bang". Also, see Max Tegmark's work on multiverses levels I-IV (Max Tegmark -> See his Scientific American article entitled Parallel Universes).


Why does each and every planet of our solar system have an unique axial tilt angle?

In the early stages of the formation of the solar system, planetesimals start condensing and everything rotates with angular momentum inherited from the collapsing cloud of gas and dust, so the planetesimals all have their orbit and spin axes closely aligned with that of the proto-Sun. And while a planetesimal continues to grow by attracting nearby material its spin orientation isn't affected very much.

But eventually the planetesimals get large enough that they start attracting one another, and collisions occur. These collisions can disturb the spin axis of a planetesimal, especially if the collision is off-centre, and if the two bodies have roughly similar momentum.

Big planetesimals grow faster than small ones, and are less likely to have their spin axis disturbed. So proto-Jupiter manages to capture most of the matter that doesn't end up in the proto-Sun, and suffers little disruption to its spin axis. The smaller planets aren't so lucky.

That's the rough picture, and it misses a lot of details. Some of the details are still unclear, and more work is needed to clarify them.

The biggest mystery is Uranus. We don't know why Uranus is rolling around on its side, but we presume that was caused by some interaction fairly late in its history because its major moons orbit in its equatorial plane, pretty much. One theory is that it suffered collateral damage in whatever event upset Saturn's spin axis. Note that Saturn's rings and large moons orbit in Saturn's equatorial plane.

Mercury has a short orbital period and hence a high orbital speed. It's tidally locked to the Sun, so that it spins 3 times on its axis in 2 revolutions around the Sun; that tidal locking also locks it's spin axis fairly closely to its revolution axis.

Venus is also a bit of a mystery. It was formerly believed to be tidally locked, with its year equal to its day, but actually its rotational period is about 243 (Earth) days, about 20 days longer than its year, and it's spinning backwards! We don't know why, but suspect it may be due to a strong coupling between it's surface and very dense atmosphere. Also note that Venus experienced planet-wide severe tectonic activity not so long ago (on an astronomical timescale), so its surface is relatively new, and a lot of the planet's angular momentum may have been transferred to its atmosphere.


How did DSCOVR capture the Moon's shadow on Earth during the August 21 2017 solar eclipse without the Moon itself also being visible?

DSCOVR is in a large Lissajous orbit around the Sun-Earth L1 point, about 1,500,000 kilometers from the Earth. At this point it is far from the line drawn between the Sun and Earth; if the Sun were behind you and you were looking at the Earth-Moon-DSCOVR system, you would say that the spacecraft is down and to the right by several hundred thousand kilometers.
I've used the Python package Skyfield plus a location for DSCOVR from JPL's Horizons to plot their positions. Earth is large and blue, the Moon is green (cheese) and smaller, and the very tiny black spot is DSCOVR. In each case I've annotated DSCOVR's position.
I've done some rotations so that my X-axis passes through the centers of the Sun and the Earth during the eclipse.
If I get a chance I'll make a 3D plot of DSCOVR's orbit. The script I used can be found in https://pastebin.com/AXsJM8J1


Why is magnitude of Beta Lyrae given as 30.0 in Henry Draper catalog?

30.0 mean "variable".
If you click on the "Ptm" column header, you get a pop-up window including "Note (2) : codes used for the magnitudes:"; below that is a list of codes, including "30.0 = variable (var. in published catalog)"


What are the aberrations of an SCT? And how can they be eliminated?

The primary aberration present in a Schmidt–Cassegrain telescope is spherical, due to the primary mirror being spherical in shape; that causes light at the edges to have a different path length than light at the center, and thus a different focal point.

The image above, from Wikipedia's "Schmidt corrector plate" webpage, shows an exaggerated corrector; they are actually much flatter.

The secondary aberration is coma, though this is present to a much lesser extent. Coma is a spherical aberration from rays that come in off-axis. Coma shows up as little off-axis comet-shaped blobs that point inwards towards the center of the field and get bigger as you look towards the edge of the field of view.

Persons wanting a lengthy technical and mathematical explanation will enjoy reading sections: "10.2.2. Full-aperture Schmidt corrector: Schmidt camera" and "10.2.2.4. Schmidt-Cassegrain telescope (SCT)" of the https://www.telescope-optics.net/ website.

Recent improvements in the design are offered using aplanatic flat field Schmidt-Cassegrain optics. Here's a screenshot from Celestron's video showing the difference between a standard versus aplanatic Schmidt telescope design:

The competing system from Meade is called Advanced Coma-Free (ACF™) optical design. Claiming also to offer coma-free pinpoint star images and flatter image fields while reducing astigmatism and eliminating diffraction spikes.

The website forum "Starry Nights - Cats & Casses" compares Celestron and Meade.


How do you convert an observed magnitude to a flux?

The reference magnitude and flux really depends on what filter (i.e. range of wavelengths) the observation was made at (e.g., see the UBV system, or a more complete set of photometric passbands here). If you know the filter, and it is in one of the common ultraviolet, optical, or near infrared filters, then this website and this website provide useful reference fluxes. Using the common AB magnitude system then a source with a measured magnitude of zero in any band by definition has a reference flux of 3631 Jy (where 1 Jansky = 10-26 W Hz-1 m-2).


Is the time lapse considered when estimating the age of the universe?

As relativistic effects will cause clocks to run slower, a frame of reference must be chosen when considering the time of the "big bang". There is a natural and convenient choice of reference frame, based on the cosmic microwave background. The cosmic background appears to be extremely red-shifted light, indicating it is receding from us very fast, due to the expansion of the universe. If we choose a frame in which the CMB is receding equally fast in all directions, we have a convenient frame of reference. It is called the Comoving frame.

Now that we have a frame of reference, we can talk about time and distance in a way that all observers that share this frame can agree. In the comoving frame, the "big bang" occurred about 13.8 billion years ago.

To directly answer the question: The time measured is the time in the comoving frame and relativistic time dialations (time-lapse) are considered.


Converting parsecs to light years in the dumbest way possible

You asked for a dumb, tedious way of doing it. So here it is, in all its glory
One parsec is defined as $$\dfrac{1}{\tan(1'')} \text{ AU}=\dfrac{648000}{\pi} \text{ AU}$$
As $1 \text{ AU} = 149597870700 \text{ m}$, one parsec is equal to $$\dfrac{648000}{\pi} \cdot 149\,597\,870\,700 = \dfrac{96\,939\,420\,213\,600\,000}{\pi} \text{ m}$$
One light year can be calculated by multiplying $c$ by the number of seconds in a year: $$299792458 \cdot 86400 \cdot 365.25 = 9\,460\,730\,472\,580\,800 \text{ m}$$
Dividing the former by the latter, we get $\dfrac{1 \text{ pc}}{1 \text{ ly}} = 3.261564$


Why does matter stay collapsed in the core, following a supernova explosion?

In order to "blow something up" you need to release more energy than its binding energy and have a way of trapping that energy so it can't escape in another way.

At the centre of a core collapse supernovae is a 10 km radius, $1.4 M_{\odot}$ ball of (almost) neutrons.
Its gravitational binding energy is $\sim GM^2/R = 5\times 10^{46}$ J.

This is almost exactly how much energy is released by the collapse of the core from a much larger size (i.e. the energy of the supernova is gravitational to begin with) and since some of that energy goes into dissociating iron nuclei and making neutrons (both endothermic processes) and most of the rest escapes in the form of neutrinos, then there can't be enough energy to unbind the core. Only a tiny fraction (1%) of this energy is transferred to the envelope of the original star, which since it has a much larger radius (by at least 5 orders of magnitude), is enough to overcome its gravitational binding energy and blast it into space.

The case of a type Ia supernova (an exploding white dwarf) is quite different. Here the energy source is not gravitational collapse, but from a thermonuclear detonation of all the carbon and oxygen that make up the white dwarf, to form iron peak elements. This exothermic process rapidly releases enough energy to unbind the original star (e.g. see here) and it is completely destroyed.


Caves traced in nine planets of solar system

A key requirement for caves is solid substance: while there may be rock cores inside the giant planets, the rest is liquid or gas. Metallic hydrogen may be solid or liquid in Jupiter, but presumably the pressure is high enough to prevent cave formation. So no for Jupiter, Saturn, Uranus and Neptune.
A second requirement is some factor digging cavities. On Earth water, chemical erosion and lava are factors.
The terrestrial planets have solid substance and not too extreme gravity. There are suspected "cave skylights" seen on Mars. Lava tubes are expected on Venus. Mercury volcanism was very different, but lava tubes may have existed. There are also weird hollows.
That leaves Pluto. Pluto has had cryovolcanism, so it is not implausible that "lava tubes" have been dug by molten water and ammonia.
As a bonus: smaller asteroids are often "rubble piles" of rocks loosely held together. It is not hard to imagine that there are empty spaces between big rocks that could count as caves. Same for comets, where evaporation of ice may contribute to making cave-like structures.


What are those long, smooth, longitudinally-opposing areas from pole to pole on Mercury?

During the Messenger mission, there were (among many others) two global images of Mercury taken.
One is the so called "Low Incidence Angle Mosaic" which is created from combining images taken at local noon when the Sun is high in the sky and the amount of shadows is reduced. This image is useful to see different types of surfaces.
The other one is the "Morphology Mosaic", taken at a higher incidence angle at local morning / afternoon. This shows enhanced shadows and therefore provides a better view of craters, but there are some areas were not photographed at the right time.
The Mercury 3D model is filled with images of the Morphology Mosaic.
You can see the same smooth areas present in the mosaic if you view it in higher resolution at the link below.
https://messenger.jhuapl.edu/Explore/Images.html#global-mosaics
And, comparing the two mosaics, the smooth looking area is a copy of the low incidence angle image that has been used to replace missing images. So, the only difference between the areas is the incidence angle of the light, making many of the craters almost disappear. The resolution of the image of both areas is the same, but shadows are not.
This is also described in the image caption provided by NASA:

Any remaining gaps in the global mosaic were filled by using images
obtained in support of the high-incidence imaging campaign.

https://photojournal.jpl.nasa.gov/catalog/PIA17230


What is the typical resolving power of the largest optical telescopes?

The angular resolving power of a telescope is approximately $1.22 \lambda/D \times 648000/\pi$, where this is the Airy disc diameter in arcseconds, $\lambda$ is the working wavelength in metres and $D$ the diameter of the telescope primary mirror in metres. For an 8-m telescope working at 500 nm, this gives an angular resolution of 0.05 arcseconds.

There is then a separate question about whether this theoretical limit is ever achieved because of detector limitations or degradation by blurring due to atmospheric turbulence (seeing).

In practice, at visible wavelengths, seeing always trumps the intrinsic angular resolution and has a typical value of 0.5-1 arcsecond at the world's best astronomical sites. There is no set number for this. It varies with atmospheric conditions from night-to-night or even hour-to-hour. Most observatories publish statistics on this so that you can see what the mean (or median) seeing is. For instance here are some statistics for the Paranal (VLT) site, where the median is about 0.8 arcsec.

Special techniques can mitigate the seeing degradation. You may have heard of adaptive optics - you can analyse the incoming wavefront and send a correcting signal to a tip-tilt or fully deformable mirror that removes the distortion, based either on an adjacent star or a laser guide star. However, these techniques are most effective in the near infrared and are not usually attempted at visible wavelengths. See http://en.wikipedia.org/wiki/Adaptive_optics

Another method is Lucky Imaging - which can be used by amateur astronomers too. Take many pictures with short exposures. If there is a bright point source in the image somewhere, you can use this to select the best images and to shift and stack the images to produce a cleaner final product. The shorter the exposures the better this works. Very close-to the theoretical maximum resolution can be obtained in this way - but it needs a special high speed, low noise CCD and you end up throwing away more than 90% of the images.


Why both sunrise and sunset use the position of the sun's upper limb?

When were these definitions formulated?

Multiple religions have rather strict requirements based on sunrise / sunset, for example, when one must start or can stop fasting, or when exactly to sacrifice a lamb (or a human in some religions). The definition has very deep roots.

Is this done purely because sunrise and sunset are the main examples of such events and it is easier for the naked eye to tell when sun is completely below horizon?

Yes.

Wouldn't this offset the symmetry of rotations (since the "day" is always longer than the "night" by the duration it takes the background object to move fully below the foreground horizon)?

Daytime is not longer than nighttime in the middle of winter, in places that do have winter. The definition of sunrise and sunset does make daytime be a bit longer than nighttime on the equinoxes.
Suppose sunrise was instead defined as the time at which the entire Sun first appears above the horizon and sunset was instead defined as the time at which the first bit of the Sun drops below the horizon. Even with these extreme definitions, daytime would still last longer than nighttime on the equinoxes thanks to atmospheric refraction.


How spectrographs that measure radial velocities manage to translate variations in the stars' spectrum lines into the "speed" of the star?

There are many ways to measure a star's motion, radial velocity (wobble) can be measured using doppler spectroscopy. The first exoplanet discovered by this method was 51 Pegasi b by Michel Mayor and Didier Queloz, who discovered the planet on December 1995.

A drawback of this method is that it can only detect the movement of a star towards or away from the Earth. If the orbital plane of the planet is "face on" when observed from the Earth the wobble of the star will be perpendicular to an observer's line of vision, and no spectrum shift will be detected.

In most cases a distant planet's orbital plane is neither "edge-on" nor "face-on" when observed from the Earth. Most likely it is tilted at some angle to the line of sight, which is usually unknown. This means that a spectrograph would not detect the full movement of the star, but only that component of its wobble that moves it towards the Earth or away from it.

The ESPRESSO instrument uses two 90x90 mm CCD detectors, one red and one blue sensitive. The detectors view the light after it is reflected off of an echelle grating, which is optimized for use at high incidence angles and therefore in high diffraction orders. Higher diffraction orders allow for increased dispersion (spacing) of spectral features at the detector, enabling increased differentiation of these features.

The equations are relatively simple. The observed Doppler velocity is $K=V_{\mathrm {star} }\sin(i)$, where $i$ is the inclination of the planet's orbit to the line perpendicular to the line-of-sight.

Reference:

"A Jupiter-mass companion to a solar-type star", Nature volume 378, pages 355–359 (1995), by Michel Mayor and Didier Queloz

An improvement to Mayor and Queloz's equations is offered in:

"The Rossiter-McLaughlin effect and analytic radial velocity curves for transiting extrasolar planetary systems" (Mar 25 2005), by Yasuhiro Ohta, Atsushi Taruya, and Yasushi Suto

  "7. Conclusions and discussion
  We have discussed a methodology to estimate the stellar spin angular velocity and its direction angle with respect to the planetary orbit for transiting extrasolar planetary systems
  using the RM effect previously known in eclipsing binary stars (Rossiter 1924; McLaughlin 1924; Kopal 1990). In particular we have derived analytic expressions of the radial velocity
  anomaly, $\Delta_{vs}$, which are sufficiently accurate for extrasolar planetary systems. If the stellar limb darkening is neglected, the expression is exact. We have extended the result to the case with limb darkening and obtained approximate but accurate analytic formulae. For a typical value of $\gamma = R_p / R_s \sim 0.1$, the formulae reduce to a simple form (eqs. [40], [43], [44], [45], [48], and [49]):

  $$\Delta v_s = \Omega s X_p \sin I_s \frac{\gamma^2 \{1-\varepsilon(1-W_2) \} }{ 1-\gamma^2-\varepsilon \{ \frac{1}{3}-\gamma^2 \} } \tag{56}$$

  during the complete transit phase and (the following is scrollable):

  $$\small { \Delta v_s = \Omega_s X_p \sin I_s \frac { (1- \varepsilon) \{ - z_0 \zeta + \gamma^2 \cos^{-1} (\zeta / \gamma) \} + \frac { \varepsilon }{ 1 + \eta_p } W_4 }{ \pi (1- \frac {1}{3} \varepsilon ) - (1 - \varepsilon) \{ \sin^{-1} z_0 - (1 - \eta_p) z_0 + \gamma^2 \cos^{-1} (\zeta  / \gamma) \} } \tag {57} }$$

  during the egress/ingress phases, where

  $$\begin{align}
W_2 & = \frac { (R^2_s - X^2_p - Z^2_p)^{1/2} }{ R_s }, \tag {58} \\
W_4 & = \frac {\pi}{2} \, \gamma^{3/2} (2 - \gamma)^{1/2} \, (\gamma - \zeta) \; x_c \frac {g(x_c ; \eta_p , \gamma) }{ g(1- \gamma ; - \gamma , \gamma) }, \tag {59}
\end{align}
$$

  where $g(x; a, b)$ is defined in equation (A17). The definition and the meaning of the variables in the above expressions are summarized in Table 1.

  The numerical accuracy of the above formulae was checked using a specific example of the transiting extrasolar planetary system, HD 209458, and we found that they are accurate
  within a few percent. Our analytic formulae for the radial velocity anomaly are useful in several ways. One can estimate the planetary parameters much more efficiently and easily,
  since one does not have to resort to computationally demanding numerical modeling. Furthermore, the resulting uncertainties of the fitted parameters and their correlations are easily evaluated.

  ...

  Table 1. List of notation
  Variables    Definition     Meaning

  Orbital Parameters
  $m_p$        Sec.2        Planet mass
  $m_s$        Sec.2        Stellar mass
  $a$          Fig.1       Semimajor axis
  $e$          Fig.1        Eccentricity of planetary orbit
  $\varpi$         Fig.1       Negative longitude of the line of sight
  $i$            Fig.2        Inclination between normal direction of orbital plane and y-axis
  $r_p$         Eq.[1]       Distance between star and planet (see Fig.1)
  $f$          Eq.[2]        True anomaly (see Fig.1)
  $E$         Eq.[2]       Eccentric anomaly
  $n$          Eq.[3]       Mean motion
  $M$        Eq.[4]       Mean anomaly

  Internal Parameters of Star and Planet
  $Is$        Fig.2         Inclination between stellar spin axis and y-axis
  $λ$          Fig.3         Angle between $z$-axis and normal vector $\hat{n}_p$ on $(x, z)$-plane
  $Ω_s$         Eq.[12]       Annular velocity of star (see Fig.2)
  $R_s$         Sec.4      Stellar radius
  $R_p$         Sec.4      Planet radius
  $\varepsilon$         Eq.[38]       Limb darkening parameter
  $V$           Sec.6      Stellar surface velocity, $R_sΩ_s$

  Mathematical Notation
  $X_p$         Sec.4      Position of the planet
  $γ$         Eq.[25]       Ratio of planet radius to stellar radius, $Rp/Rs$
  $η_p$          Eq.[28]       See Fig.6
  $x_0$          Eq.[33]       See Fig.6


'spacetime' is real, however the word 'fabric' is made so that anyone can 'visualise' the meaning of space-time
You cant touch or feel the fabric, but you can perceive it through the bending of light over it. When someone says that the spacetime is curved around a planet, the following statement holds true:

the time is slower near the surface of the planet relative to far away from the planet

the space is 'curved' towards the planet, so if you send a space ship to the planet, the ship will curve towards the planet, even of it is travelling on a straight path. So the straight path is not so very straight to an external observer even if it is straight for a observer going through the path

The word 'fabric' is used so that anyone can easily 'visualise' the spacetime
Our Universe has a large fabric of spacetime, with several heavy objects bending and curving it here and there.
The word 'spacetime' has also a non-gravity related meaning. Einstein stated that as the speed of an object increases in space, the time experienced by it decreases
that means that if an object travels at the speed of light through space, it will not travel through time at all. And if an is stationary in space, it will travell the fastest through time


What does Gauss' divergence theorem say about compression of a body under self-gravitation?

Gauss's divergence theorem applied to the gravitational field $\vec{g}$ is that
$$ \oint \vec{g} \cdot d\vec{A} = \int \nabla \cdot \vec{g}\ dV,$$
where the left hand side is the flux of gravitational field into/out of a closed surface and the right hand side is the integral of the divergence of that field over the volume enclose by the surface.

The fundamental definition of how mass produces gravitational field is that
$$ \nabla \cdot \vec{g} = -4\pi G \rho,$$
where $\rho$ is the mass density.

If you are at a point outside the mass then the right hand side of the divergence theorem  becomes a constant.
$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV = -4\pi GM.$$
Therefore the left hand side is also constant when calculated over any surface that encloses all the mass.

If you allow the mass to contract in a spherically symmetric fashion, then as the gravitational field on the left hand side is then always in the radial direction then $\vec{g}$ must stay the same at any position in space outside the mass distribution.


To what extent do the spin of smaller celestial bodies reflect the spin of the larger system’s of which they are a part?
What about the rotations at other scales of the larger systems we are in, such as our local star cluster, our local galactic cluster, and so forth? Thank you.

There's an amusing explanation here: Changing the rotational rate of a natural body, it's about this much (as far as most people are concerned):

Simple and incorrect answer: In Newtonian mechanics the gravitational field of a body depends only on its mass, not on it's rotation. If you go a bit further and assume that the spinning object has perfectly uniform density then it matters not whether it spins nor the direction.
More precise answer: The Lense–Thirring effect is very small—about one part in a few trillion. To detect it, it is necessary to examine a very massive object, or build an instrument that is very sensitive. The non-static stationary distributions of mass–energy causes frame dragging resulting in mass–energy currents and what is known as gravitomagnetism.

"This approximate reformulation of gravitation as described by general relativity in the weak field limit makes an apparent field appear in a frame of reference from that of a freely moving inertial body. This apparent field may be described by two components that act respectively like the electric and magnetic fields of electromagnetism, and by analogy these are called the gravitoelectric and gravitomagnetic fields, since these arise in the same way around a mass that a moving electric charge is the source of electric and magnetic fields. The main consequence of the gravitomagnetic field, or velocity-dependent acceleration, is that a moving object near a massive rotating object will experience acceleration not predicted by a purely Newtonian (gravitoelectric) gravity field. More subtle predictions, such as induced rotation of a falling object and precession of a spinning object are among the last basic predictions of general relativity to be directly tested.".

The equations for calculating the effect are straightforward but perhaps more than you wanted to know. Modelling this complex behaviour as a curved spacetime problem has yet to be done and is believed to be very difficult.
If you're trying to point a camera at a distant object it's a big deal:


Congratulations, you have seen a fireball. An especially bright meteor. It was bright and fast. This is a rock, a very small asteroid, hitting our atmosphere and "buring up". It is also possibly a satellite or space junk re-entering. Re-entering satellites look move slower than asteroids. You can see fireball videos on youtube, as well as re-entries, to compare.

This event was also witnessed from Puducherry, the details seem to match up quite well. The time is slightly out, but the location is spot on. You can add your report at http://fireballs.imo.net/members/imo/report_intro

No, its not Florence, which will not be entering our atmosphere, and so will not appear to be burning, it will be a single point of light. It will be moving much slower, and won't be visible with the naked eye.


The magnitude of the force of gravity between two bodies is proportional to the product of their masses:
$$F=G\frac{m_1m_2}{r^2}$$
This doesn't change depending on which body you're applying the force to, i.e. if you interchange the masses. The magnitude is the same.

What does change is the direction of the force. Force is a vector quantity, denoted as $\vec{F}$ or $\mathbf{F}$. If we write the equation for gravity using proper vector notation, we have
$$\mathbf{F}=G\frac{m_1m_2}{|\mathbf{r}_1-\mathbf{r_2}|^2}\frac{\mathbf{r_1}-\mathbf{r_2}}{|\mathbf{r_1}-\mathbf{r_2}|}$$
Here, the positions of the objects are represented by vectors, $\mathbf{r}_1$ and $\mathbf{r_2}$. Additionally, $|\mathbf{x}|$ denotes the norm of a vector $\mathbf{x}$ - its magnitude.

Now, if you interchange the masses, the direction of the force changes, although $|\mathbf{r_1}-\mathbf{r_2}|=|\mathbf{r_2}-\mathbf{r_1}|$, because this refers to the magnitude of the vectors. So the force applied on one object is the opposite of the force applied on the other object. This is Newton's third law.

The acceleration is more interesting. The force on object $1$ due to gravity is
$$F_1=m_1g_1$$
Here,
$$g_1=\frac{Gm_2}{r^2}$$
where $m_2$ is the other mass. This should tell you that $g_1\neq g_2$, except when $m_1=m_2$.

I'm not an expert in general relativity, but I do know that it describes how spacetime curves due to the presence of one body. The solution to the Einstein Field Equations, the metric, is different for different bodies, because one piece of it, the stress-energy tensor, is different for objects of different mass/energy/etc.


How does the sun move over the year in the sky of the Moon?

That's right. Unlike the Earth, which has a substantial inclination of 23.4 degrees to the ecliptic, the Moon is inclined by only 1.5 degrees. This means that there are no substantial seasons. The moon also spins much more slowly than the Earth. Each "day" lasts for one Earth month. It also means that at the poles there are craters that are completely shaded from the sun. In these craters, there may be useful amounts of water ice.

Now as noted by the Wikipedia site, this is due to two motions being linked. The moon's orbit is inclined by 5.1 degrees to the ecliptic, and the moon's equator is inclined by 6.7 degrees to its orbit, however the direction of the pole moves to keep these two angles always in opposite directions to each other (There can be no pole star on the moon) The result is that the inclination of the moon to the ecliptic is fixed at 1.5 degrees.


What direct or indirect observations of dust can one make by eye or with binoculars?

If you can see the Milky Way, then in Summer through early Autumn you should be able to see the "dark rift" which runs through the middle of the milky way from Saggitarius to Deneb in Cygnus. This is a dust band in the Milky Way.

If you live far enough South, you can see the Coalsack, a quite dense dust cloud near the Southern Cross, and also clearly visible without optics.


How might WFIRST help us better understand dark energy?

The WFIRST's study of dark energy is pretty straight forward.   It's no different than the previous ways of studying dark energy.   It's not studied directly, cause we can't see it or re-create it, it's studied by the effect it has on the relative motion of distant galaxies.

WFIRST will get much more accurate readings on the relative velocity of galaxies both relative to each other in their neighborhood and relative to us, this will get a significantly more accurate estimate for dark energy than was previously possible and questions can be answered such as, is it speeding up.

NASA's WFIRST Web page discusses this in more detail

  The High Latitude Spectroscopic Survey will measure accurate distances
  and positions of a very large number of galaxies. By measuring the
  changes in the distribution of galaxies, the evolution of dark energy
  over time can be determined. The High Latitude Survey will measure the
  growth of large structure of the universe, testing theory of
  Einstein's General Relativity.

  Type Ia Supernovae (SNe) Survey uses type Ia SNe as "standard candles"
  to measure absolute distances. Patches of the sky are monitored to
  discover new supernovae and measure their light curves and spectra.
  Measuring the distance to and redshift of the SNe provides another
  means of measuring the evolution of dark energy over time, providing a
  cross-check with the high latitude surveys.

  High Latitude Imaging Survey will measure the shapes and distances of
  a very large number of galaxies and galaxy clusters. The shapes of
  very distant galaxies are distorted by the bending of light as it
  passes more nearby mass concentrations. These distortions are measured
  and used to infer the three-dimensional mass distribution in the
  Universe. This survey will determine both the evolution of dark energy
  over time as well as provide another independent measurement of the
  growth of large structure of the universe.

Dr. Yun Wang's comments appear to be written for a Science-TV show and/or youtube video, and video like those are designed for a general audience so I wouldn't analyze that particular sentences meaning too closely.

As for Erik Verlinde's . . . ideas.   I'm not a fan of his proposals, but the answer is mostly the same.   WFIRST will get more accurate readings and better readings should help get more specific answers.


If you're dead broke (like most people) I would suggest you take look at a small refractor telescope, you could pick up a cheap-o Celestron Powerseeker 60/70 for something like 40 bucks. The useability of such a small telescope is very limited, but you'll be able to get a nice clear image of the rings of Saturn, four of Jupiter's moons, or nice views of the moon, maybe even a few Messier objects if you're patient enough.
-but-
If you're willing to drop a bit more coin, then I would suggest nothing smaller than an 8" reflector. For less than 200 bucks you can find something that will really impress you.

I used to take my crappy little telescope and drive up into the hills where light pollution was at a minimum, where I would just sit for hours trying to find Messier objects. Loads of fun!

As far as joining an amateur astronomy society, I would suggest you check with your local Community College's astronomy professors. At least here in California, one of the requirements of the astronomy classes is that there be opportunities to participate in actual observing. This means that there will more than likely be an astronomy group having "star parties" and other observing events, which will give you some good connections to competent groups.


Question about number density of electrons

The 2 in your equation is the average number of mass units per electron. e.g. totally ionised carbon has 6 electrons and 12 mass units. In your formula, the mass of a proton is used as a mass unit (close enough in this case).

In your second equation $n_e^{-1/3}$ is an approximation for the inter-particle separation. This is an approximation and there is a small numerical factor that would need to be introduced in an accurate treatment.


The answer to this is that stars are effectively point sources, but planets have disks.  Borrowing the link to Sky and Telescope's explanation from @N. Steinle's comment.

The way I envision it is to think of the light from a celestial object as a fiber optic cable, continually being snipped at with scissors (which stand in for atmospheric turbulence changing the path of light traveling through it).  The cable from a star is a single glass fiber with a very bright beam shining down it.  The cable from a planet is a thick bundle of fibers with a much dimmer beam shining down each fiber in the bundle (compare the surface brightness of Mars with the surface brightness of the Sun)-- the total amount of light coming through the cables is comparable.  The scissors can cut through the single-fiber cable completely, but they can only take out a few strands of the thick bundle.  So atmospheric turbulence can cause the light from a star to be drastically interrupted, while the change for a planet is much less.  Therefore, there is visible twinkling from stars, but not from planets.


Why are there no stars on New Horizons images of Pluto

The answer by @mso is very good. Another contributing factor is the exposure time of the camera on New Horizons.

The amount of light reflecting from Pluto is greater than that from the background stars. As a result the exposure time is set to achieve a good image for the desired target and detail required.

Over the integration time the amount of light from background stars is not enough to register onto the focal plane and subsequent image.

In order to get good pictures of stars one needs to integrate over long period of time while maintaining accurate pointing.


It seems very unlikely because while a supernova which sheds enough mass int he explosion can scatter matter from the planets of its system into interstellar space, it also tends to irradiate that matter and to heat it to a very high temperature, so at least as far as a supernova's planets being the source of life, it's difficult to see how that might work.


It's at Jupiter's distance and beyond that ices were able to form out of the disk of material surrounding the early sun.  Go much further in and there is too much energy from the sun for them to stay as solids (and will sublimate into gases); this is why asteroids are principally rocks and metals.  So at this distance more of the materials of the planetary disk can form the base planetesimals.

At this point, your question is answered largely by a geometric consideration (or two) and one of Kepler's laws.

First, the geometric consideration.  A circle of radius $r$ has area $\pi r^2$.  The bigger the radius, the more area.  The material for Jupiter (or any other planet) came from an annulus: stuff outside one circle, but inside a slightly bigger circle.  This annulus had a lot more area for the out planets than the inner planets, and so could contain a lot more mass.

Of course, that could makes us think that Jupiter shouldn't be the largest of the gas giants: it's the closest of them all to the sun, after all.   The density of the disk needn't have been approximately constant throughout these regions, though.  Quite possibly the density was such that Jupiter's region had more mass than the areas for the other planets.  As HDE's answer (posted as I was finishing this) points out, these ices probably also helped stop materials from passing into the inner solar system, maintaining a higher density than you might otherwise expect in the inner solar system, as well as causing materials to sort of "dam up" right around Jupiter's orbit.

Now for the Kepler's law.  The further you get from the sun, the slower your orbital period.  Picking the correct units, we have $P^2=a^3$, where $P$ is the period measured in years, and $a$ is the semi-major axis of the orbit measured in AU.  The further out you go, the slower you go around the sun; indeed, it's not just that it takes you a longer total time, but your actual velocity goes down.  We can also see this as a Newton's law of gravity consequence.  At Jupiter's furthest point from the Sun, the escape velocity is a little more than 18 km/s.  At Saturn's maximum distance, the escape velocity drops to around 13.25 km/s.  So things can go roughly 35% faster within Jupiter's orbit than they can closer to Saturn's, and they have less distance to travel to make a complete orbit.

What this means is that it takes longer for planetesimals to get close enough to each other to accrete together the further out you go, and there is a longer mean time between collisions.

Now, eventually, the Sun "turned on" and started blasting space with it's solar wind (before that, the heat came primarily from thermal radiation from the gravitational contraction of the sun).  This ended up clearing out most of the unaccreted particles out of the solar system, stopping planetary growth (and removed portions of existing atmospheres; a very young Earth probably had a lot of H and He in its atmosphere, until the sun hit it with enough energy to knock any of it not locked up in rocks away).

So Jupiter was probably in a bit of a goldilocks situation.  The average density of the region it formed in was probably higher than where the other giants formed in, it was at the perfect spot for lots of materials to start accreting early, and the accretion process would have been faster.  So Jupiter is growing faster, and this gives it a competitive advantage: the bigger the growing planetesimals get the further their influence extends and the faster they can pull in more materials, and subsequently interfere with the growth of other planets (or planetesimals).  Somewhere around a mass of 10-15 earth masses, the giants can start pulling in large quantities of the hydrogen and helium gasses.  And, again, Jupiter likely hit this mass well before the other giants, and had more material to pull from, so it became much larger than the others could before the solar wind stopped the process.


Are Barlow Lenses Good For Deep Sky Observing?

The Failures of High Magnification

Higher magnification doesn't help you observe deep sky objects better. Deep sky objects unlike stars are extended objects. They subtend a finite solid angle on you. This ensures that the surface brightness(brightness per unit solid angle) of extended objects remains constant. Hence, a higher magnification would not make it brighter for you to see.

It is worse for another reason. After a certain magnification, the apparent angular size of the extended object becomes comparable to the field of view, or sometimes even greater. Your eye won't be able to distinguish the object against the background since most of the background is the object itself.

Good Uses of Barlow

Some of the uses that you can put your Barlow:

Observe the Moon, Solar System objecs
Some of the clusters which seem like point objects really "explode" when using a Barlow. The best example is Omega Centauri.

Experiences

I have a 8" Dobsonian Newtonian Reflector. The viewfinder is 20x80. I really get a better picture of deep sky objects like Ring Nebula, which have a small angular size, when using Barlow. But for cases like Lagoon and Trifid Nebula; they are visible to me from my viewfinder but not through the eyepiece(even in the lowest magnification).

Conclusion

Barlow really helps if you have a deep sky object with small angular size. For larger objects, a binocular may do better, especially for city conditions.


No, for several reasons.

there are no radial velocities in the required range observed, for example in the Milky Way galaxy.
the reservoir of stars moving out from the centre would quickly be drained, so you would need to magically generate them at the centre -- this is much worse than postulating dark matter.
you'd expect huge amounts of stars and gas that have moved out to large radii.

The only serious alternative to dark matter is some modification to Einstein's/Newton's theory of gravity. There are several candidates, but none is really convincing (though arguably, dark matter is not too convincing either).


Are there any reliable optical measurements of the radius of a neutron star?

Yes, the neutron star RX J1856.5-3754 has an observed* radius of 17 km. After accounting for general relativity, its actual radius is calculated as 14 km. It is not the only neutron star whose radius is known.

* It's worth mentioning how astronomers arrived at this radius. Technically, fitting the X-ray data to a blackbody arrives at a calculation of a 5 km radius, which seems unusually small, so for a time it was believed that this star may be something more exotic, like a quark star. However, after atmospheric modeling based on temperature, a more accurate estimate of the emission radius was derived to be 17 km. Once gravitational redshift was accounted for, the value of 14 km was yielded. See Magnetic Hydrogen Atmosphere Models and the Neutron Star RX J1856.5−3754 - Wynn C. G. Ho, et al.


Is it possible to see stars that are gravitationally lensed by the sun from the ground?

The answer is no. But there's an interestign alternative.

The Eddington's experiment was not about seen a gravitationally lensed star but to see a small displacement of its position (just below 2 arcseconds) due to the solar gravitational field. Lensing involves considerably magnifying the source and that can only be done if the star is directly behind the Sun (or very close to that direction).

Only close to the Observer-Sun line the star would be considerably magnified and the flux increased. The result would be an Einstein ring, an hyperdistorted version of the image of the star built with the light-collecting area as large as the Sun (at least).

For this to happen you need to locate yourself at the solar gravitational focus. Theoretically this is not a single point, in fact it is an entire line extendign from the Sun outwards (we call it the gravitational focal line). The thing is that if you are too close to the Sun the Einstein ring is "located inside the solar disk". So you need to get very far away to be able to see the focused light. The point at which the Einstein ring grows larger than the apparent diameter of the Sun is located $550$ astronomical units away from the Sun. This is the minimal distance you would be able to exploit the gravitational lensing effect of the Sun.

A distance of $550\; AU$ is insanely huge. The farthest probe we have sent, the Voyager 2, is currently located $147\; AU$ from Earth, so you would need almost 4 times that to reach the point where this could become feasible. Also you would probably prefer to position yourself a bit farther since the Einstein ring would still be located inside the Solar Corona and a lot of information could be washed out by the intense light (and in the case of a radio telescope you would see an opaque wall). There are a lot of important things that could be done at the focal line of the Sun, you would have a crazy huge magnification (just as if the object was as close to you as the Sun from that same distance! which equates to tenths of millions magnifications in many cases), and the flux would be equivalent to the one obtained by a $1.000.000 \; km$ lens or even more.

It has been said that exploring the solar focal line will be the most important milestone between the exploration of the outer Solar System and the firsts attempts to explore the closest stellar systems. The main problem is that you need to position yourself with incredible alignment accuracy to get the Einstein ring of an object behind the Sun. A probe like that would be usefull to observe just a single object during the entirety of the mission (maybe an exoplanet wich might be crucial for our understanding of habitability), since the delta-v needed to change position and get to another part of the Solar System $550 \; AU$ from the Sun would be insane and it would be more economically feasible to lunch a another probe to the next direction just to observe the other target. A good place to place a telescope like this would be an possible moon around the hypothetical Planet 9 (which is considered to reach $800\; AU$ at perihelion), the problem been again that you would only observe breifly objects that happen to fall at the opposite position relative to the Sun.

A mission to the focal line has been under study for years now. You can read about the details in a presentation that was given by one of the main proponets, astrophysicist Claudio Maccone. There's also a nice talk about the topic on Keck's YouTube channel.


Binocular highlights from Tasmania next few weeks?

My favorite resource for this sort of thing is the Skymaps.com evening sky map.
The southern hemisphere edition is drawn for 35°S latitude.
The November issue doesn't show the Milky Way, which hugs the horizon at that time.
However, both the Large and Small Magellanic Clouds are well placed in the evening, and the back page lists several other objects "easily seen with binoculars."
For example:

47 Tucanae, a globular cluster visually near the SMC
NGC 2516, an open cluster in Carina
M42, the famous nebula in Orion


What explains the distribution of new impact craters on Mars?

I was on the targeting team for one of the cameras that discovered most of these new impact craters. The reason for the distribution is simply because it's easiest to find "new" impact craters in the dusty regions of Mars. Often what we see is the dark-toned blast zone created by the impact in lower-resolution data, which has a large areal footprint. Then, we target those dark-toned splotches with the higher resolution cameras---which have MUCH smaller footprints---to confirm whether or not the splotches are craters. See the attached image as an example. The dark splotches really stand out in dusty regions, but in the non-dusty areas there's nothing obvious to pick out to say "hey, there's a new crater" as a human without tediously comparing images.


See What do a galaxy's colors mean? Are they its true colors? This article should help answer your question. There's really nothing I can add to the article. It lists all the reasons I could think of.


Sorting out Julian Day, Julian Date, Julian Day number, Julian Day Calendar, and Julian Day Table

The Julian Date (JD) is the real number of days since 12:00 UT, January 1, 4713 BCE (proleptic Julian calendar).
The Julian Day Number (JDN) is floor(JD), the nearest integer ≤ JD.
The Modified Julian Date (MJD) is JD - 2400000.5, or the real number of days since 1858-11-17 0:00 UT.
The IAU officially approves either JD or MJD as long it's clear which quantity is given.
A number of 366 or less in a bureaucratic or informatics context is a day number or day of the year.
In his MJD explainer, the late USNO Time Service director Gernot Winkler strongly opposed calling this a "Julian" date:

The MJD (and even more so the JD) has to be well distinguished from this day of the year (DOY).
This is also often but erroneously called Julian Date, when in fact it is a Gregorian Date expressed as number of days in the year.
This is a grossly misleading practice that was introduced by some who were simply ignorant and too careless to learn the proper terminology.
It creates a confusion which should not be taken lightly.
Moreover, a continuation of the use of expressions "Julian" or "J" day in the sense of a Gregorian Date will make matters even worse.
It will inevitably lead to dangerous mistakes, increased confusion, and it will eventually destroy whatever standard practices exist.

Expressing the example date of 2019-06-08 13:26 UT in various forms:

Julian day number: 2458643
Julian date: 2458643.06
Modified Julian date: 58642.56
Day of year: 159
Ordinal date: 2019-159


What documented comet has/had the highest specific orbital energy?

That is a very good timing for this question.

There is a comet with the name C/2017 U1 (PANSTARRS), that was just discovered a few days ago (on 18th October) which has an orbital eccentricity of 1.18, which is by far highest eccentricity recorded so far.

The eccentricity describes how strongly "elliptical" an orbit is and refers to the variable e in the formula of Kepler's orbit. 1.18 makes the orbit a hyperbolic one, meaning it is not an actual orbit, but has higher velocity than the required escape velocity and is basically just passing by in our solar system. And it's the first recorded comet so far that has an hyperbolic orbit.

This comet is the first candidate for an interstellar comet ever in recorded history. It is already pretty certain that it originated from interstellar space, but not official yet.

Quote from minorplanetscenter.net:

  Further observations of this object are very much desired.  Unless there
  are serious problems with much of the astrometry listed below, strongly
  hyperbolic orbits are the only viable solutions.  Although it is probably
  not too sensible to compute meaningful original and future barycentric orbits,
  given the very short arc of observations, the orbit below has e ~ 1.2 for
  both values.  If further observations confirm the unusual nature of this
  orbit, this object may be the first clear case of an interstellar comet.

Sources:

 http://www.minorplanetcenter.net/mpec/K17/K17UI1.html
 https://en.wikipedia.org/wiki/C/2017_U1_(PANSTARRS)
 https://imgur.com/t/science_and_tech/vM4oh


Could this be a cosmic ray hit on my cameras sensor (CMOS, DSLR)?

It's a track about 50 pixels long, at 4.1 $\mu m$ per pixel that's 200 $\mu m$ in the plane of the sensor.

It's about 1 pixel wide, quite lumpy, and seems to change direction a bit. It really fits the description of an ionizing particle. There are multiple possible sources,

I'm pretty sure @JamesK has nailed it as an electron, but what else can we learn? There must be some way to get a very rough idea of how many $e^- h^+$ pairs correspond to a fully exposed pixel at "@ISO1600", then we could convert it to a dE/dx and see if it's minimum ionizing or if it's much higher.

Going diagonally, with a density of 2.3 g/cm${}^3$ that's about 1.3 mg/cm${}^2$ areal density per pixel. A [minimum ionizing particle] at 1.5 MeV / g cm${}^2$ would deposit only about 2000 eV which would make less than 1000 $e^- h^+$ pairs. That might correspond to a bright pixel. It's not out of the question that it's a minimum ionizing particle.


Question about Hubble parameter (Hubble constant) and measuring it

You can start by working out how fast a galaxy must be receding if its peculiar velocity (given as about 600 km/s) is less than or equal to 10% of its expansion velocity.

I'm not going to spell out the next step as you said you don't know where to start.


When quoted like this they would usually be the most likely value (peak of the probability distribution) or the median value (where half the distribution is above or below), and then the superscripts and subscripts would be increments that would contain $\sim 34$% of the probability distribution. In the M87 papers, the numbers are the 50th percentile (median) and lower (upper) limits corresponding to the 16th (84th) percentile.

The total range corresponds to about a 68% confidence interval. The reason for this is just that it corresponds to $\pm$ one sigma in a normal (Gaussian) distribution.

It is good practice to say what your error bars mean, but in the absence of that I would assume the above.


What is the magnification of my new telescope?

Magnification of your telescope depends on the ratio of the focal length of your primary optics and the focal length of your objective. We can represent this with this simple formula:
$$P=\frac{f_{objective}}{f_{eyepiece}}$$
where $P$ is magnification (power) and $f$ is focal length.
Your telescope has 900 mm focal length, so the magnification is $P_1=\frac{900\text{ } mm}{10\text{ }mm}= 90\text{x}$ (10 mm ocular) and $P_2=\frac{900\text{ }mm}{20\text{ }mm}= 45\text{x}$ (20 mm ocular).
A Barlow lens is used to increase the focal length by the given factor (thus 2 in this case), so directly impacts the magnification in the same way, increasing it to 180x and 90x - but gives you in return an equally smaller field-of-view and makes area-objects fainter as the same light is projected to a larger area. Note that there is a maximum useful magnification which is roughly twice the aperture (measured in mm), which is 120x in your case, so the Barlow lens will not be very helpful as it magnifies beyond reasonable due to inevitable diffraction on the clear aperture.
One usually compares them by their focal length and by their aperture - the ratio defines how bright the objects appear and thus how faint objects you can see.
With an aperture of 60 mm, it has an opening ratio of f/15 which is only a moderate one.
Such telescope is particularly useful for observations of the planets in our solar system as well as star clusters and the brighter deep-sky objects. Now in summer would be an excellent time to hunt for the globular clusters like the one in Hercules (M13). Also, you can download a useful program Stellarium to find more interesting features.


Where/how are metals distributed in our universe?

From the big bang nucleosynthesis (BBN) page is a handy chart of the sources of elements:

Since basically all metals come from some stellar process, the question of elemental distribution can vary greatly between different environments depending on when you're observing that particular galaxy, star/galaxy cluster, etc. Galaxy mergers, supernova, and stellar winds can be particularly good at moving metals around a galaxy or galaxy cluster.

Interestingly Wolfram has element data with Universe abundances but I'm not sure where they derive those.


The Newtonian gravitational acceleration of a test object (a small object with negligible mass) toward a point mass is given by

$$\vec g =\frac {GM}{||\vec r||^3} \vec r \tag{1}$$

where $\vec g$ is the acceleration vector, $G$ is the Newtonian gravitational constant, $M$ is the mass of the point mass, and $\vec r$ is the displacement vector directed from the test object toward the point mass.

Equation (1) cannot be used to compute the net gravitation acceleration of a test object toward a pair of point masses by pretending that the pair of point masses is equivalent to a single point mass located at the barycenter of the two point masses. Instead, what needs to be done is to apply equation (1) to the point masses individually and then form the vector sum of these individual accelerations. This in general will not point toward the barycenter.

There are two classes of points where the net gravitational acceleration vector to a pair of point masses is directed toward the barycenter of the point masses:

Points along the line connecting the point masses, less for two regions where the net gravitational acceleration vector points away from the point mass, and
Points away from this line and whose projection onto this line is exactly midway between the point masses.

For a test object located anywhere else, the net gravitational acceleration vector does not point toward the barycenter.


is there any mathematical formula, paper or anything that shows the probability of a meteor hitting a particular spot of moon's surface?

There is this source that claims 2800 kg of meteor material lands on the moon everyday.  They break down that amount to determine an average area per strike. There have been other studies like this one and this report.  I sure can't vouch for the accuracy of this first report but it's from a reputable (NASA) source.


What determines the surface material of the ALMA and Spitzer telescope?

They're all mirrors, they just operate in different wavelength ranges. A wire fence could be a mirror if the wavelength is big enough.

The surface is an easy problem to solve. Pick any material and do vacuum deposition with it on the substrate. It's the substrate, the base material that matters a lot - the thick part that keeps the shape.

A metal mirror is easier to make. Standard engineering principles apply. That's why this technology is used for ALMA.

But there's a catch - precision. This is dictated by wavelength. Imagine all the radiation captured by the mirror, focused in one point. If the mirror is perfect, it's all good. If the mirror has deviations from perfection, parts of the wavefront arrive too soon or too late. The wavefront begins to interfere with itself, reducing the amplitude.

Interference is completely destructive (the signal is completely lost) when the difference between the paths is exactly half the wavelength: λ/2. Then the peak wave (maximum) from one side falls into the trough wave (minimum) from the other side, and they cancel each other.

How do you get a λ/2 difference between two parts of the wavefront? If the surface error on the mirror is λ/4. One part of the wavefront begins to be reflected immediately, while the other keeps traveling λ/4 to get to the mirror, and then wastes λ/4 again on its way back up.

Mirrors with λ/4 errors are bad - since the error is never exactly half-and-half, parts of the wavefront are not cancelled so the mirror keeps working somewhat, but poorly. Around λ/8 the mirrors become decent. λ/16 are very good, and beyond λ/20 are essentially indistinguishable from perfection. This is for optical mirrors (visible spectrum).

This is why optical telescope mirrors are made of glass. It takes such a smooth, rigid material to reach the precision necessary for optical observations.

The Spitzer mirror is actually made of beryllium, cooled to 5.5 K. This is fine. Precision requirements are a bit more relaxed, compared to visible light. In fact, solid metal mirrors have been made for visible light telescopes even by amateurs - but in these conditions glass is preferred.

The ALMA would have to be manufactured to a precision better than 75 micron, in reality much better. That's doable with regular metal parts.

Other materials used for visible light telescopes:

Granite. It takes an excellent polish, just like glass, it's very rigid, and the coefficient of expansion is similar to glass. As long as it doesn't have voids, it's a decent material.

Marble - same as above.

Speculum metal - back in the days of Herschel they were casting mirrors out of this bronze-like alloy. Easy to work with but it tarnishes very quickly.

Special ceramics such as Zerodur.

Quartz - like glass, but better (low thermal expansion). Takes more work to grind and polish because it's harder than glass.

TLDR: It's mostly about precision and ease of manufacturing.

Once the mirror is made, it could be coated with anything you like, so it could reflect any wavelength up to hard UV. Like I said, the surface, the reflective layer, is the easy part.

Simply make a list of all materials that provide good reflectivity in the intended range of wavelengths. Most of these materials turn out to be metals, unsurprisingly. Now compare them based on such criteria as: durability, price, ease of coating. Pick one and drop it in the tungsten "boat" in the vacuum chamber. Put the mirror in the vacuum chamber and run some amperage through the tungsten boat until the metal layer on the mirror is thick enough so it's not transparent anymore - but not thicker than that (for precision). Done.

An alternative that is sometimes used in infrared and visible light is dielectric coating. It's a stack of fairly transparent layers, and there are multiple reflections from all the surfaces between layers. The thickness is such that the path difference between reflections is exactly one wavelength, 1 λ. That strengthens all reflections (constructive interference).

Dielectric mirrors have extremely high reflection coefficients, and are used for lasers, etc. Also tend to have preferred wavelengths where they work well (though broadband dielectric mirrors also exist). They're quite expensive and are not commonly used for telescopes.

Finally, after all that, telescope mirrors sometimes get a protective coating to shield the reflective metal layer from corrosion. A commonly used material is silicon oxide (SiOx, not SiO2 which is quartz and it's different). If the mirror operates in a vacuum or it can be recoated often then this is less important.

The vast majority of amateur telescopes nowadays use:

borosilicate glass (e.g. Schott Supremax) for the substrate; it's a low expansion glass, cheap enough. The original Pyrex glass (not made anymore) was almost the exact same thing.
aluminum for the reflective layer; sometimes silver is used, which is slightly more reflective initially, but decays more quickly
SiOx top coat for durability

Some mirrors have "enhanced coating" which is conceptually similar to dielectric mirrors, in addition to the metal layer.


What are the stages in the life of a universe?

Yes there are. They are mainly based on what dominates the energy density of the universe at the time and they are known as epochs.
Thus we have the inflationary epoch in the first tiny fraction ($\sim 10^{-32}$) of a second, when the energy density was dominated by an inflationary field.
Then we are in the electroweak epoch, when the weak nuclear and electromagnetic forces were united.
This is followed by the quark epoch when the electromagnetic and weak nuclear forces separate at about $10^{-12}$s, which is followed by the hadron epoch between about a millionth of a second and 1 second, when quarks were able to be confined into multi-quark hadrons.
Then the lepton epoch between about 1 second and a minute when most of the hadrons have annihilated with each other and the energy density is dominated by electrons, positrons and neutrinos, although this also coincides with the "epoch of primordial nucleosynthesis" when the first multi-nucleon nuclei were built.
From there until about 300,000 years, we have the radiation epoch, when photons dominated the energy density of the universe.
After that, the universe entered the matter-dominated epoch for about 5 billion years and now we are in the epoch where the universe is dominated by dark energy.
There are other schemes and other nomenclatures depending on what physics is being focused on. For example, people interested in the formation of the first stars will talk about the "epoch of reionization", which occurred some 100 million years after the big-bang. Those studying the cosmic microwave background talk about the "epoch of recombination", which coincides roughly with the transition from a radiation to a matter-dominated universe.


What's the name for [the other kind of planet] in a binary star system?

I don't know about names for the planets, specifically, but the orbits are called S-type and P-type:

S-type: The planet orbits around one star, and the host star has a binary companion (i.e., "the other kind" in the XKCD comic)
P-type: The planet orbits around both stars of the binary