What would a terraformed Mars look like from Earth?

To terraform Mars, broadly speaking would make it more like an analogy of Earth. As you say, it might be a little colder, but we would expect a terraformed Mars to have oceans, snow and maybe some imported plant life.

How would Mars look from Earth? A good starting point would be to see it the opposite way around. Here is how Earth looks from Mars.

Like Earth from a distance, a terraformed Mars would likely appear as a pale blue or white dot. If you want more snow or ice than water, it would surely lean towards white rather than a deep blue.


For colonization purposes, what is so good about Titan?

There is nowhere in the solar system except for parts of Earth where humans have any hope of surviving without heavy technological support. There is next to no free oxygen anywhere, nowhere has a surface whose temperature remains in a range we can tolerate (generously let's say 230-320K) and nowhere has liquid water on the surface. So, at a minimum, we will need a sealed habitat and some form of protective suit when we travel outside it.
Terraforming -- transforming the environment of some planet to be close enough to Earth's to allow us to live without protection is much discussed, usually for Mars or Venus. Even the most optimistic detailed proposals I have seen suggest it would take centuries or millenia to complete.
Even more speculative is the idea of transforming humans to be able to live on other planets relatively unchanged. Although popular with SF, at the moment I don't think we really have the understanding to say anything specific about this.
So for short-term colonisation, we are left to decide which planet would present the least difficult technological challenges for hi-tech base.  There are a few somewhat serious contenders:

Mars -- relatively close, has hydrogen, carbon and oxygen accessible on the surface (as ice and CO2), atmosphere protects against micro-meteorites and some radiation, surface temperatures not too far from habitable.
Venus (upper atmosphere) -- pressure and temperature are both reasonable; fairly close; limited hydrogen availability; no easy supply of metals; corrosive atmosphere; good micrometeorite and radiation protection; not much to do
Ceres -- very low gravity, no atmosphere, plentiful water ice, easy access to space, easy access to metals etc (from other asteroids), good for supporting space activity
Moon -- very close, no atmosphere, almost no hydrogen, enough of a gravity well to make landing and taking off non-trivial
Titan -- rather remote, plenty of oxygen, hydrogen, carbon and nitrogen, atmosphere protects against radiation, comfortable pressure -- you'd need a heated dry suit or equivalent, lots of interesting stuff in the neighbourhood, cold.
Europa -- plenty of oxygen and hydrogen, lots of radiation on the surface, deep in Jupiter's gravity well and no atmosphere for aerobraking, so not very appealing on the surface, but a "submarine" colony under the ice might be viable, the ice will provide radiation shielding and some warmth and you might be able to swim outside in just a dry suit and breathing gear.

You pays (a lot of) your money and you takes your choice.


When would Oumuamua have entered the solar system?

One problem with answering this question is that it's not precisely clear where the solar-system ends or when (if) Voyager I crossed that border.

If we set the estimate at 19 billion km (per article) or 127 AU. (about 4 times the distance between Neptune and the Sun), then an estimate is possible.

Some conversions first.  Voyager left earth in 1977 and the announcement was in 2013, so it's usually listed as 36 years (it might be 35 and some months) but that's not all that relevant.

18 km/s x 31.5 million seconds in a year (close enough) is 567 million km per year.   It should have taken Voyager about 33.5 years at that velocity to reach 19 billion km.   However, Voyager's velocity wasn't consistent.  As it flies from the sun it slows down and it had several gravity assists along the way which sped it up and changed it's direction.   Mathematical calculations get a little tricky, but here's a chart of Voyager's velocity.

(Note: that's Voyager 2's velocity not Voyager 1.  I'm having trouble finding Voyager 1's velocity chart).

Here's it's direction, more of a fish-hook shaped path than a straight line, though it's been mostly straight since Saturn which is over 90% of it's journey.

Oumuamua's velocity at infinity, as it approached our solar-system was 26.33 km/s.   It speeds up as it approaches the sun and it will slow down as it leaves.   When it passed the 19 billion km "border", you have to add the escape velocity from our sun at that distance, which is about 3.7 km/s, so it entered the solar-system traveling at (close enough) 31 km/s or 976 million km/year.  At that velocity it would take 19.5 years to reach the sun (granted it will miss the sun), and in theory another 19.5 years to leave the solar system.   But that estimate doesn't take into account that it speeds up as it gets close to the sun, so it takes less time.   17-18 years (ballpark) might be a good estimate.  That's 17 (or so) years traveling in, another 17 (or so) to travel back out and reach the made-up border.

Unlike Voyager, nearly all of Oumuamua's journey is nearly straight, except for a quick curve around the sun.  That makes it's journey close to "V" shaped when viewed from a distance.  (I only mention that in relation to Voyager's more curvy fish-hook flight path, which was by design to pass by several planets).

There's probably someone here who can do the math (I'd prefer not to), and they can work out a more precise answer of when it entered the fuzzy 19 billion km border, when it will reach it's perihelion or closest to the sun, and when it will pass outside that border again.


Does mean solar time and sidereal time sometimes indicate the same time?

The 3m56s mean difference between solar and sidereal days is due to the Earth's orbital motion around the Sun.
This adds up to 1 day per year; for every 365¼ solar days there are 366¼ sidereal days.

Sidereal time equals the right ascension of what's on the celestial meridian at that time.
The Sun is on the meridian at 12:00 apparent solar time.
These align at the September equinox, when the Sun is at RA 12h.
Mean solar time is ~7.5 minutes behind apparent solar time on that date and aligns with sidereal time ~1.9 days earlier.


What are the Mathematical Algorithms/Formulas behind hindu Sunrise and Sunset Calculations?

In the linked algorithm,  the "zenith" is the position of the sun at sunrise.  This is the number you need to change.
If you want to calculate the time when the top of the sun reaches the horizon you use the "official" zenith of 90⁰50' This is intended to account for refraction and should estimate the time when the sun first appears over the horizon.
Since the sun in 30' across, if you want the time when the centre of the sun is on the horizon, just subtract 15'  ie use a zenith of 90⁰35' (90.583 degrees)
It isn't possible to be more accurate as the amount of refraction varies with the weather. But this should give values that are correct to the minute.


The detection of complex organic molecules, often abbreviated as COM in astrochemical literature, has only taken up pace in the last couple of decades since substantial advances were only made possible with the advent of new, high-end observational facilities (Green Bank Telescope (GBT), Very Large Array (VLA), Atacama Large Millimeter/Sub-Millimeter Array (ALMA), etc.) in the radio-regime, where specific fingerprints of rotational, vibrational or ro-vibrational transitions are possible to observe.
An regularly updated list of detected molecular species in space can be found at the Cologne Database for Molecular Spectroscopy (CDMS), maintained by the Universität zu Köln. In the table below, you can immediately see that only four organic molecules have so far been detected that are made out of more than 12 individual atoms. However your molecule of interest, Tetrahydrocannabinol (THC), is constituted by 53 (!) individual atoms as one can infer from its chemical forumlae: $C_{21}H_{30}O_{2}$
Its going to be almost if not certainly impossible to ever detect THC in an extraterrestrial environment through remote observations (i.e. sending rovers or other space exploration mission for in-situ explorations is another story altogether). Furthermore, THC would need to be in detectable quantities in an exoplanets atmosphere or within a molecular cloud in interestellar space. A scenario which is missing any arguments or evidence about a process or mechanism that could provide such a high density. The existence of detectable amounts of THC in interstellar space is hindered by the harsh enviroments, e.g. X-ray and UV-fields from nearby stars that would not allow any high concentrations of very large molecules. Furthermore, I believe THC is a biogenic molecule, i.e. not being created except for biochemical processes in terrestrial lifeforms or - of course - as synthetical variant in chemical laboratories. I am not aware of any non-biogenic process that could provide for high THC concentrations in a planets atmosphere. That would make it a dream candidate as a biomarker, i.e. a molecule whose presence alone would strongly hint at biological/xenological activity.  Anyhow as a consequence, the feasibility of the detection from THC fails on numerous levels, thus making the detection via telescopes virtually impossible, from at least what we know so far about how, where in what typical quantities THC is formed.


And the whole barycenter thing too.

You're throwing out the definition of orbit at this point, and moving into a different realm. If you throw out barycenter, you have to leave me with my frames of reference.

A star can definitely orbit a planet, if one such as this did in fact exist. You are assuming it is simply a planet-star system, which makes things easier. If you're on the planet, the star is orbiting you, if you're on the star (ouch), the planet is orbiting you.

Once they are of comparable size, the barycenter thing has to be taken into account, you can't just throw out this crucial part of the physics, even in a thought experiment. If you want that part out, the planet must be much bigger than the star.


Do meteor showers start slowly each night?

The number you are able to detect increases as it gets darker.

The number occurring is apt to be greater when the zenith of the sky is closest to pointing in the direction the earth takes orbiting the sun
(like a windshield hitting more bugs than back window when a car is moving forward)

The last factor I can think of at the moment is the density of comet tail material earth happens to be passing through (in its orbit around the sun)
Which could unfortunately be not when your local the zenith is 'forward in our orbit' and when it is dark.


Are galaxies expanding away from us at more than the speed of light?

You've been confused by a subtle error that seems always to be made when trying to express the speed of light limitation of Relativity in words.

What's fundamental is something called the Lorentz Symmetry AKA Lorentz Invariance which is the basis of Relativity. This symmetry is a local symmetry and it's a strong symmetry of of the physical universe. Basically, what it says is that the laws of physics are independent of velocity. A consequence of this symmetry is that two objects can't pass each other at a speed greater than that of light. (The reason why is not especially difficult, but best explained using mathematics rather than words.)

General Relativity is not a local theory -- it's a theory of the structure of spacetime -- but its local structure is that of Special Relativity, and hence it has Lorentz symmetry.  One of the surprising (utterly surprising -- Einstein himself didn't believe it for many years) consequences of General Relativity is that spacetime isn't stable, but it must either expand or contract.  This expansion is not the galaxies flying further and further into the void like the bits of an exploded hand grenade, but is spacetime itself expanding everywhere (there is no center of expansion) and the galaxies everywhere more-or-less sorta sitting still in the expanding spacetime and being carried along with it.

If you pick two points which are far enough apart, they can be receding from each other at a speed greater than that of light -- or, alternatively and just as correctly -- more than 300,000 km of new space can be created between them each second.  Because nothing is moving past each other at FTL speeds, local Lorentz symmetry is preserved and Einstein is happy and his equations are satisfied and all's well with the universe.

Note that everything I've said is wrong, but I hope it's not misleadingly wrong.
You really need to do the math to see what's actually happening.

But the upshot is that we can't fly to the Moon (or Mars or Alpha Centauri or wherever) and come back at a speed faster than light, but many distant galaxies -- which we will never see or hear from again -- have just passed c moving away from us, and more do so every year.


How much clearer are stars in earths orbit?

It varies.

The best astronomical sites have a visual band extinction of 0.1 mag, which means that only $\sim 10$ per cent of light is absorbed/scattered in the atmosphere.

In dusty, smoggy or polluted sites, this can easily reach one magnitude of extinction, which means that 60 per cent of the light is scattered.

These numbers are per airmass - which means looking straight up at zenith. At lower altitudes, the baseline extinction figure would be multiplied by $\sec z$ where $z$ is the angle from the zenith.

This amount of absorption doesn't sound like a huge difference, but what it means is that a large number of extra stars become visible, because roughly-speaking, you get a factor of two increase in the number of stars for every 0.5 magnitudes deeper you can see, and I think that is what is most striking about going to a clear, dark astronomical site.

Thus there wouldn't be much of an improvement going into space over the best astronomical sites (in the visible band), at least not due to the decrease in extinction. You do also escape light pollution, which is a different effect. With light pollution it is the fact that the background sky becomes too bright for your eye to pick up the fainter stars. [As Carl Witthoft notes, it is a different ball game at other wavelengths because the atmosphere is significantly more opaque to UV, X-ray and IR wavelengths.]

Another thing to consider is what you are breathing! Lack of oxygen can hinder visual acuity. http://navyaviation.tpub.com/14020/css/Effects-Of-Hypoxia-141.htm


If we "turned universe upside-down" and changed all matter to anti-matter instantly

It would look very similar, but tiny differences are suspected, see e.g. this LHC experiment.

Here a list of more antimatter experiments at CERN.

See also CP violation on Wikipedia hinting towards tiny differences between matter and antimatter.

Immediate consequence: The CP violation parameter of the CKM matrix in this Wikipedia article would be different, because of different quark generation probabilities.

If you also reverse time, you'll get to the CPT theorem (Charge, Parity, Time). Matter and antimatter are thought to be symmetric in this sense.

Or to state it with simpler words: Antimatter behaves like matter travelling backward in time.


Do comets have a limited number of orbits?

Everything in the universe has a limited number of orbits.   But the way to make sense of your question is to ask how long comets last.  Comets lose ice when they fly past the sun, so they get smaller each time.   How much smaller depends on how close they get to the sun and their composition, or how easily they "melt" or break apart.   (Ice doesn't exactly melt in space, it sublimates when it heats up)

Sungrazers have a very limited number of orbits.   Halley's also is expected to have a limited lifespan of about 100 more orbits before it stops being a comet and basically becomes a rocky body.   Halley's passes closer to the sun than Venus at its perihelion, so it loses a lot of ice, each pass, which helps make it's tail impressive. (when Earth and Halley's are properly aligned that is).

Your question also depends on your definition of a comet.   Will Halley's still be a comet after it loses it's ice and it's tail? (sometimes called a Manx comet).  That's a matter of interpretation.   It will get smaller, but it will likely remain in orbit for some time after it loses it's ice.

Orbiting objects that cross planetary orbits are also, on average, less stable, but I suspect that's another subject.   Halley's might be destined to crash into the Earth, or the Moon at some point.  It's orbit nearly intersects the Earth's orbit in two places.  But don't worry.  That's not going to happen anytime soon, if it happens at all.


Sounds like Julian Calendar slippage.

The Gregorian Calendar, the one we use now, was created to fix a problem with the Julian Calendar: The fact that the Solar Year wasn't exactly 365.25 days. As a result, compared to the calendar year, the date of the Vernal Equinox (and more importantly at the time, Easter) was slipping forward.

To remedy this, ten calendar days in 1582 (or up to thirteen calendar days in 1918, depending on how long a particular country waited to adopt the Gregorian Calendar) were omitted, and the Gregorian Calendar added a modification to the Julian calendar's rule of One leap year every 4 years: No leap year on century-years that aren't divisible by 400.

If your code automatically switches from Gregorian to Julian Calendar at the appropriate year, the adoption of the Julian Calendar probably winds up at about the expected days when the Julian calendar was adopted mid-first-century BCE.

But every 400 years further back, your calendar is slipping about three extra days.

And at 4000 BCE, that's thirty days.


What is the light source in the black hole images?

The radiation being detected in this case is mostly synchroton radiation, caused by energetic electrons spiralling around magnetic field line, rather than thermal radiation, but it comes from the same place (the disk). The actual evidence for light bending is, I believe the dark area in the middle, which is several times larger than the actual event horizon and is caused because the light that would have been coming to us from that direction has been bent away.


How can clouds form in Jupiter's atmosphere of Hydrogen and Helium?

First, it's a great question.  Mostly the answer is straight forward, so I can answer it, but it's still a great question.

and I'll add a similar, but slightly more detailed picture to the one you posted.

Source

You're right that there is a clear difference between Earth's surface where liquid water can exist, evaporate, make clouds, rain and repeat.   Earth's water cycle in theory, could go on indefinitely so long as Earth's atmosphere and the solar input are maintained (and lost hydrogen is replaced), but it's a circular system that only needs solar input.

Jupiter is different because over time, the heavier gases in Jupiter probably will sink deeper towards the center and Jupiter's cloud forming gases should decrease given enough time.   Some of Jupiter's "rain" probably does fall too deep in its swirling mix of gases, and leaves Jupiter's cloud cycle permanently, similar to water seeping underground and leaving Earth's water-cloud cycle.    So, in 100 billion or a trillion years or so, Jupiter could lose its clouds and cloud forming gases in its upper atmosphere for the reasons you suspect.

The reason this hasn't happened yet is simply mixing.  While density of gas tends towards layers of increasing density, the internal heat inside Jupiter also wants to even out, so there's enormous convection going on pretty much all the way through the planet.   This keeps some heavier gases in Jupiter's upper atmosphere.   Jupiter's much too turbulent to have just hydrogen and helium in its upper atmosphere.

So, once we begin with the observation that Jupiter's upper atmosphere is (about) 90% hydrogen, 9% helium, 1% other gases and the mixing maintains the 1% of other, after that it's just cloud physics.

Clouds look like puffy collections of water vapor (tiny droplets of ice or water, as water vapor is actually transparent).  They look like objects with shapes, but that's not entirely accurate.   If you are up close to a cloud (flying in a plane for example), the clear edges disappear.  A cloud isn't an object so much, it's a visible phase change.

Atmosphere on Earth is about 78% nitrogen, 21% oxygen, 0.9% argon and (usually not listed because it's so variable), about 0.4% water vapor on average, as high as 1% with high temperature and high humidity and close to 0% in cold temperatures or dry-deserts.   When you take warm surface air that's 0.6-0.8% water vapor, and that air rises (as hot air does), it's the phase change that creates the clouds.   The cloud forms in the hot rising air as it cools.  There's some electrostatic attraction, but mostly it's just a block of similar air undergoing cooling and the cloud looks like it has solid edges, but it doesn't.

The same exact thing happens on Jupiter, different gases phase change at different temperatures/pressures, but the process is the same.  And, just as on Earth, once the droplets or "icelets" form, they are denser and they begin to fall, but falling droplets are very small so they fall very slowly and for the most part, they are falling through rising atmosphere.  Also, as they are a phase change, new cloud is being formed and old cloud is being disbursed or returned to gas all the time, kind of like sea ice.   The clouds have the appearance of semi-permanence, but clouds are dynamic.

If my explanation doesn't work for you, here's an explanation on clouds and how they really aren't bound together even though they look that way.

But that's the gist of it, mixing keeps the upper atmosphere of Jupiter from being pure hydrogen and helium (or pure hydrogen), and after that, the cloud formation is pretty much the same as on Earth, just without a surface.   Some of the heavier gases probably do get lost from the cycle, but the loss is slow enough that Jupiter still has some heavy clouds forming gases in its upper atmosphere and will for probably billions of years to come.

The greater density variation between H/He and other gases likely plays a role in how the clouds behave, as the density variation is greater, but the wind speeds are also higher on Jupiter.   All that's really needed is mixing.  After that, with gases that can become liquid or ice under temperature/pressure variations, the phase changes create the clouds.

It's also possible that Jupiter's cloud forming gases, from time to time, are replenished by asteroid and comet impacts.   Shoemaker-Levy 9 was about 5 km in diameter and a fair percentage of that was probably ammonia and water ice.   That's a lot of cloud forming gas added to Jupiter's upper atmosphere.   Jupiter's faint ring system, which might have been much larger millions of years ago, but since rained onto Jupiter, and eruptions from Io might also play roles in keeping Jupiter's upper atmosphere rich enough in cloud making elements like water and ammonia.


Here's a NASA movie showing Libration: http://svs.gsfc.nasa.gov/cgi-bin/details.cgi?aid=4000

As given in the details here, the position angle is the angle between the Moon's north and celestial north. This changes because of the inclination of ecliptic to equator + inclination of moon's orbit to ecliptic and plus, there might be some contribution from the inclination of moon's rotation axis to its orbital plane. Note that all these changes are apparent. The pole need NOT physically shift at all for these effects.

Having said that, whether the pole does shift, requires one to correct for all those effects and the remaining effects would be cause by some precession or nutation of the lunar poles, which is likely to be negligible given that it is tidally locked to the earth.


Radial Velocity Graph and Rossiter Mclaughlin Effect

What's happening is that (i) the star is spinning and (ii) it is being eclipsed (by an exoplanet perhaps?)
As the exoplanet (or whatever dark object eclipses the star, the same explanation would apply to a brown dwarf or very low-mass companion) traverses across the face of the star it will eclipse parts of the stellar surface that gave their own specific line of sight velocity.
For example if we are seeing the system edge-on to the stellar equator and the planet orbits in prograde fashion with zero inclination, then it will first eclipse the side of the star that is spinning towards us. This takes a blue-shifted chunk out of the stellar spectrum, leading to a slight redshift (compared to what might have been expected). As the transit continues, the exoplanet moves across the centre of the star and then to a position blocking the redshifted side of the star, leading to a slight blueshift compared to what would be expected.
The image below (by AMitchell and shared under CC Attribution-Share Alike 4.0 International shows this situation.

Your plot shows this happening the other way around: a blueshift followed by a redshift. This can be explained if the exoplanet orbits in retrograde fashion, so it obscures redshifted, followed by blue-shifted stellar light.


Why does the Solar Wind consist of charged particles?

You are right about the kinetic energy (ie the fast motion) of the particles being the reason, but wrong when you say

  They must be of almost the same energy if so there will be no relative kinetic energy,

The individual particles are moving fast, but there is huge variation between them in just how fast and in what direction. In other words, the solar wind is very hot. The extent of disorderly motion in a gas is measured by its temperature.  The wikipedia article you link gives temperatures of 100 000 to 800 000 Kelvins. So if an electron and proton did happen to collide gently enough to form an atom of hydrogen, another particle would probably smash into them before very long and knock the electron off again.

You might think they are all moving away from the Sun, but you have to allow for the effects of magnetic fields. Charged particles moving in a magnetic field curve around in all sorts of ways, and, additionally, moving charged particles create a magnetic field which affects other particles. So  the movement is extremely turbulent.


Available energy for potential life on Titan

I can give a rough answer to your question and some brief thoughts on the paper, and I invite correction from anyone smarter than me.   Your article talks about energy per mole, but energy of production on a daily basis would be a product of solar energy.

To produce hydrogen in Titan's atmosphere requires UV light, from this article, UV light of 1600 Angstroms (160 NM) (Source)

Quote:

  Methane is a carbon atom surrounded by four hydrogen atoms, and it can
  be broken apart by ultraviolet light at wavelengths of about 1600
  Angstroms. The fragments, or radicals, that are produced from this
  process are very chemically reactive. They’re things like CH, CH2, and
  in some cases CH3.

Only a small percentage of the sun's light is in that range.   8%-10% of the sun's light is in the UV range, but maybe only 1%-2% in the upper UV3 range of 160 NM.  (Source)

The same Wiki article, using Saturn as a guide, Saturn receives between 13.4 and 16.7 Watts per square meter (Source), so figure an estimate of 15 watts per square meter total solar energy, 1%-2% of that strong enough to pull a hydrogen atom off Methane, so 0.15 - 0.3 watts per square meter.   That might not sound like much, but on a moon the size of titan, that's a lot of square meters so the low wattage isn't a big deal.   That's a peak estimate though as some of the hydrogen would recombine with the CH3 it split from, some energy would likely be converted to heat, and perhaps the biggest problem, mentioned in the article above, a share of the hydrogen, perhaps the lion's share would simply be lost off the planet.

  In the atmospheres of the giant planets, the hydrogen stays around
  because of the high gravity of those planets. Their atmospheres are
  primarily hydrogen anyway, and after the fragmentation occurs, the
  products sink into the deeper atmosphere, and methane is
  reconstituted. It’s a complete chemical cycle.

  On Titan, as far as we understand, that does not occur. Because the
  gravity is low, hydrogen should escape. The Voyager ultraviolet
  spectrometer saw a corona of hydrogen around Titan, which is a good
  indication that hydrogen is escaping. Now, if hydrogen is going away,
  the products that can be made from the methane are going to have a
  higher carbon to hydrogen ratio than methane itself, and you’re not
  going to be able to remake methane. So as far as we understand the
  photochemistry, Titan should be destroying methane and making more
  carbon-rich products, the simplest of which are acetylene – C2H2 – and
  ethane – C2H6. These are made directly from methane

This is my problem with the article.  I don't see any way the hydrogen finds it's way down to the planet's surface to be used in the way the article suggests, life in the lakes of Titan.    It's an interesting idea and the people who wrote it are probably PHD educated and I'm not, but I suspect the majority of hydrogen produced high in Titan's atmosphere would remain high in it's atmosphere and over time, lost to the solar wind.

It does raise the question, because Titan should lose it's Methane over time, why does it still have so much Methane - a very good question.

Lets look at the timing.   Huygens landed on Titan January 14th, 2005.

Your article was written  14 January 2005; revised 18 April 2005

and here's an article from a few months later, November 2005.

http://www.spaceref.com/news/viewpr.html?pid=18410

  The origin of methane in Titan's atmosphere is a mystery because it
  gets broken down by sunlight and particle radiation from space in the
  upper atmosphere. If surface lakes and pools were the only source, all
  of Titan's methane would be lost by this mechanism in less than a
  hundred million years, a short time for a moon that's been around
  since the formation of the solar system 4.5 billion years ago.�
  Components of the methane molecules react with each other and
  atmospheric nitrogen. As they descend, they form larger and heavier
  molecules that comprise the orange haze that blankets the moon.
  Because Titan is very cold (292 degrees below zero F, or minus 180
  degrees Celsius) these heavy compounds condense and rain out on the
  surface.

  "We have determined that Titan's methane is not of biological origin,
  so it must be replenished by geologic processes on Titan, perhaps
  venting from a supply in the interior that could have been trapped
  there as the moon formed,"

So, I'd chalk this one up to timing.   January-April 2005, life on Titan was a viable theory, which has since been de-theoried, or, whatever the correct term is.


Can we see evidence of sunspots on other stars?

Strictly speaking, no. Since there is one and only one Sun in the entire universe, there is one and only one star in the universe that has sunspots. That's rather pedantic. Going beyond that, the answer is almost certainly an emphatic yes. The generic term is starspots rather than sunspots, and they have been observed on other stars. For a very nice review, see the open source, peer reviewed article on starspots by K.G. Strassmeier.

The key issue with observing starspots on other stars is that even the largest sunspot observed on our Sun would be far too small / far too dim to resolve with even the largest of telescopes. But some stars apparently do have starspots much larger and much cooler than the largest / coolest sunspot on our Sun. Other than being larger and cooler, these very large starspots on other stars exhibit behaviors analogous to the sunspots observed on our Sun.

Starspots the size of the sunspots observed on our Sun almost certainly exist, but they currently can't be detected given the limitations of current technology.


To answer your second question first: Yes, antimatter does exist in the same space as matter. In fact, the universe creates antimatter (and an equal amount of matter) every day as a matter of course in events like lightning strikes and supernovae, and even in certain nuclear decays. Humans create it in particle accelerators for research and for commercial/medical applications such as Positron Emission Tomography. The thing is, when we create antimatter, we also create an equal amount of matter.

In the hot flash of energy after the Big Bang, particle-antiparticle pairs were popping into existence and annihilating each other constantly. There were almost exactly equal amounts. For some reason, though, for every 100 trillion (10^11) particles of antimatter, there were 100 trillion and one particles of matter. In the ensuing few minutes, all the antimatter and all but that tiny fraction of matter annihilated each other and turned back into energy. Everything we can see today, all the galaxies, stars, and planets, are made up of that tiny amount of matter that was left over. Particle physicists still aren't sure why there was this tiny imbalance in the amount of matter and antimatter, because all interactions we've seen so far produce equal amounts of both. This is one question particle colliders like the Large Hadron Collider are attempting to answer.


What reference frame do age-of-the-universe calculations assume?

Do we mean one second from the point of view of an imaginary observer subject to the same relativistic effects on time?

That's basically correct. Typically, cosmologists use comoving coordinates:

  Although general relativity allows one to formulate the laws of physics using arbitrary coordinates, some coordinate choices are more natural or easier to work with. Comoving coordinates are an example of such a natural coordinate choice. They assign constant spatial coordinate values to observers who perceive the universe as isotropic. Such observers are called "comoving" observers because they move along with the Hubble flow.

  A comoving observer is the only observer that will perceive the universe, including the cosmic microwave background radiation, to be isotropic. Non-comoving observers will see regions of the sky systematically blue-shifted or red-shifted. Thus isotropy, particularly isotropy of the cosmic microwave background radiation, defines a special local frame of reference called the comoving frame.

  [...]

  The comoving time coordinate is the elapsed time since the Big Bang according to a clock of a comoving observer and is a measure of cosmological time.

So we use the frame in which the cosmic microwave background and cosmic neutrino background are isotropic. In practice, we use the isotropy of the CMB because the cosmic neutrino background is so redshifted that the neutrinos have very low energy, so they're impossible to detect with current technology.


How far apart are stars in Terzan 5 cluster?

You have to assume a characteristic mass for the stars in the cluster. Usually, to get an approximate I go with stars of mass $1 M_\odot$.

Then, given the density of $10^6\ M_\odot.\mathrm{pc}^{-3}$ you have a density of stars of $\rho_\star = 10^6\ \mathrm{pc}^{-3}$. In a box of $1\ \mathrm{pc}^{3}$, you have $10^6$ stars, so the characteristic distance between stars is simply given by the density of stars to the minus one third:
$$ d_\mathrm{carac} = \rho_\star^{-1/3} \approx 10^{-2}\ \mathrm{pc} \approx 2000\ \mathrm{AU}.$$

Addendum: if you want to do a more correct calculation, you should take into account the different masses of stars to estimate the correct number of stars, using for example a Chabrier initial mass function and try to get the mass function at the age of the cluster. Because of the fact that stars of about $\geq 1 M_\odot$ reach their red giant branch and lose much of their material in less than the age of the cluster, it will decrease the mean star mass of the cluster to something between $0.1$ and $0.8 M_\odot$ ($0.3 M_\odot$ according to Rob Jeffries). However it won't change the order of magnitude of the mean distance between stars.


How wary should I be of old cosmology books?

It's a sliding scale. Generally the older the book, the more the chance of something being wrong. But it depends what books you're talking about. Are they text books? Or are they popscience books? Some books, even newish books, have a habit of presenting speculation as established fact. There's been a lot of trouble about inflation and the multiverse recently. Peter Woit talks about it on his Not Even Wrong blog. He says “the political campaign for the multiverse continues today with a piece by Amanda Gefter at Nautilus. It’s a full-throated salvo from the Linde-Guth side of the multiverse propaganda war they are now waging”.

  I know, from a recent book, that the inflaton theory is probably newer than both books

Inflation started in circa 1980. Guth coined the phrase and generally gets the credit, but it arguably goes back to Erast Gliner in 1965 and his paper algebraic properties of the energy-momentum tensor and vacuum like states of matter. It didn’t get much attention.

  and that the Higgs Boson has been discovered

When inflation was first mooted Alan Guth talked about the Higgs field and GUTs, saying magnetic monopoles were "topologically stable knots in the Higgs field expectation value”. But that was old inflation, which was replaced by new inflation, which was replaced by eternal inflation. As far as I know the Higgs field is not thought to be linked to the inflaton field.

  Has anything else changed since then?

Yes, like pela said, before 1998 people thought the expanding universe was slowing down. There are other changes too. Some people used to say the universe was once the size of a grapefruit. Now since WMAP some people say the observable universe was once the size of a grapefruit, and the universe is infinite. For myself I cannot see how that fits with big bang cosmology, which I think is largely correct.

  Which topics of the books should I understand are considered outdated?

It's hard to say. Dark matter, dark energy, cosmological constant, flatness of the universe, size of the universe, age of the universe. If you could name the books maybe I could look at the chapters.


Is it true that the sun goes down to the west?

Earth rotates. That gives us the feeling of Sunrise and Sunset. Actually there are no such things. See Below animated image. Earth completes one rotation in a day (24hrs). (Actually 1 rotation of earth is called a "Day").

This image also depicts the cause of Solstice and Equinox, earth's tilted axis.


It looks likely to be one of the bolts for adjusting the polar alignment of the scope. There are usually two opposed hand operated bolts for fine adjustment of the left/right pointing of the mount on the tripod (so it points north/south - you're expected to position the tripod so its roughly correct to start with), and another two opposed bolts (might only be one in some designs) for tilting the top part of the mount  up/down so that the rotation axis in the top part (parallel to the telescope tube in the photo) (which pivots around the silver bolt left and level with the top of the red circle) points at the celestial pole (how much tilt this needs varies with your latitude).

If that bit is missing - (it looks most likely to be the one you screw in to tilt the top part of the mount upwards more) - you won't be able to set the scope up properly.

If you check the instructions on how to set the mount up, you should be able to tell which bit it is.

(The reason why it's important, is that if you set the mount up properly, then when something starts to drift out of view due to the earth's rotation, then you can keep it in view by only adjusting one of the slow motion controls).


The drift of the Moon's equatorial rotation velocity

Since the Moon is tidally locked to the Earth, it rotates with the same orbital period as it orbits the Earth.  Thus, tracking the Moon's rotational velocity is essentially the same thing as measuring the changes in its orbit - if you know one, then you know the other.
The recession of the Moon has been measured using laser ranging (see links in the answer to this question) to be about 3.8 cm / year.  From that rate of change of orbital semimajor axis, you could use Kepler's third law to work out a rate of change of the Moon's orbital period, and thus of its rotational period.


I suspect that The Sky Live got their comet IDs mixed up.
The orbital elements show perihelia in Aug 2015 and Jan 2022,
aphelion 5.69 AU, period 6.45 years, consistent with other sources e.g.
Minor Planet Center.
However, the distance chart
shows a perihelion in mid 2018, aphelion about 11 AU, period about 16 years,
clearly not the same object.
Report the bug and they'll probably fix it.


What is the coordinate acceleration for pure radial motion?

As Rob Jeffries suggested in a comment, formulating a solution for arbitrary dr/dt requires another parameter, either the energy per unit mass, E/m, or the radius of the shell where the object was dropped.

A shell observer at $r_0$, sees an object passing by at proper radial velocity $v_0$, with corresponding energy/momentum as measured in the local Minkowski frame (using units with c = G = 1)
$$m^2=E_0^2-p_0^2=E_0^2-\gamma _0^2 m^2 v_0^2$$
$$\frac{E_0^2}{m^2}=\gamma _0^2 v_0^2+1=\frac{v_0^2}{1-v_0^2}+1=\frac{1}{1-v_0^2}=\gamma _0^2$$
where the subscripts indicate local shell coordinates. In Schwarzschild coordinates,  energy per unit mass is constant and defined as
$$\frac{E}{m}={ \left(1-\frac{2 M}{r}\right)}\frac{dt}{d\tau}$$
From equations 11 and 12 in section 6.3 of the Taylor, Wheeler, Bertschinger book (TWB) linked in the original question
$$\frac{E_0}{m}=\left(1-\frac{2M}{r_0}\right)^{-1/2}\frac{dt}{d\tau}$$
$$\frac{E}{m}=\sqrt{1-\frac{2M}{r_0}}\frac{E_0}{m}=\gamma _0 \sqrt{1-\frac{2 M}{r_0}}=\sqrt{\frac{1-\frac{2M}{r_0}}{1-v_0^2}}$$
The equation for direct radial inward motion is obtained from TWB section 8.4, equation 17, by setting the angular momentum to zero and taking the negative root .
$$ \left(\frac{dr}{d\tau}\right)^2=(E/m)^2-\left(1-\frac{2M}{r}\right)$$
$$\frac{dr}{dt}=\frac{dr}{d\tau}\frac{d\tau}{dt}=
\frac{1-\frac{2 M}{r}}{E/m}\frac{dr}{d\tau}$$
$$\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\sqrt{1-\frac{1-\frac{2M}{r}}{(E/m)^2}}$$
Differentiating, and a bit of algebra yields the following result for the acceleration.
$$\frac{d^2 r}{dt^2}=-\frac{2M}{r^2}\left(1-\frac{2 M}{r}\right)
\left(1-\frac{3}{2}\frac{1-\frac{2 M}{r}}{(E/m)^2}\right)$$
These can be rewritten in terms of the Schwarzschild radius $r_s=2M$ and proper velocity, $v_0$, at $r_0$.
$$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-\frac{1-\frac{r_s}{r}}{(E/m)^2}}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}}$$

$$\frac{d^2 r}{dt^2}=-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right)
\left(1-\frac{3}{2}\frac{1-\frac{r_s}{r}}{(E/m)^2}\right)
=-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right)
\left(1-\frac{3}{2}(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}\right)$$

For known $dr/dt$ at a radius $r$, $E/m$ can be determined by inverting the previous expression for $dr/dt$. Then substitute $E/m$ in the equation for acceleration above.

$$(E/m)^2=\frac{(1-r_s/r)^3}{(1-r_s/r)^2-\left(\frac{dr}{dt}\right)^2}$$

The alternate approach, dropping an object from rest at radius $r_0$, uses equations 37 and 38 from section 6.7 of TWB, resulting in

$$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)
\sqrt{\frac{\frac{r_s}{r}-\frac{r_s}{r_0}}{1-\frac{r_s}{r_0}}}$$

$$\frac{d^2 r}{dt^2}=\frac{r_s\left(1-\frac{r_s}{r}\right)}
{2 r^2\left(1-\frac{r_s}{r_0}\right)}
\left(1+\frac{2r_s}{r_0}-\frac{3r_s}{r}\right)$$

The shell radius $r_0$ necessary to obtain a desired velocity at shell $r_1$ can be found using
$$\frac{E_0}{E_1}=\sqrt{\frac{1-\frac{r_s}{r_1}}{1-\frac{r_s}{r_0}}}
=\frac{\gamma_0}{\gamma_1}$$
where $\gamma_0=1$ for an object dropped from rest at $r_0$.

As an example, a plot is shown below using $v_1=1/2$ at $r_1=2r_s$. For these values $\gamma_1=2/\sqrt{3}$, $E/m=\sqrt{2/3}$ and $r_0=3r_s$. For reference, the green and red curves are Schwarzchild coordinate velocity and acceleration for an object dropped from rest at infinity, $E/m=1$. The dashed blue and brown curves are for an object dropped from rest at $r_0$, and the solid blue and brown curves for an object launched from $r_1$ with a Schwarzschild velocity of
$$\left(\frac{dr}{dt}\right)_{r_1}
=-\left(1-\frac{r_s}{r_1}\right)v_1
=\frac{1}{4}$$
Note that the dashed curves are overlaid by the solid ones from $r_1$ to $r_s$, since the initial conditions were chosen to result in the same velocity at $r_1$.


What is the official term for a relationship between a planet and its moons?

I don't believe there's an official term but there are a few familiar/common terms.

Parent Body/Satellite is one.

Central Object/Orbiting object or Central Object/Satellite.

Star/Planet or Planet/Moon imply one orbits the other.   For a non orbiting Moon or Planet "Rogue" is used before their name.

A grey area comes into play when the two objects can be considered binary as opposed to one orbiting the other.   Some people define a binary orbit as one where the barycenter is outside the body of the more massive object, but I don't like that definition because by that definition, Jupiter and the Sun are a binary system and the Sun weighs 1100 Jupiters.   The Sun also has planets that don't meet that criteria, more distant than Jupiter.

Another problem with the Barycenter outside so it's binary argument is that when a star dies and goes white dwarf or even denser, then pretty much every planet now orbits a barycenter outside the star, but when it was a main sequence star, this wasn't the case.   The orbits didn't change, just the size of the central object.  That shouldn't change the definition of the orbital relationship.   I think the Barycenter argument, while convenient, is highly flawed for that reason.

Pluto/Charon could be called a binary dwarf planet, but because Charon was likely created from a collision on the Body that's now Pluto, I think that's a poor definition and I think Charon should be called a moon or collision moon.   That said, because four tiny moons orbit both Pluto and Charon, calling them a binary-system has some merit too.  There's no neat and tidy way to make a precise dividing line, besides the barycenter, which is neat and tidy, but as I said above, inaccurate.

Asteroids are sometimes found orbiting one another and I rarely hear that defined as a central asteroid with a smaller asteroid orbiting it, it's more often called a binary asteroid.  I've never heard it called a baby asteroid orbiting it's parent, but again, definitions fall into a grey area if one asteroid is quite a bit larger than the other.

This XKCD comic addressed circumbinary orbits and "the other kind", which doesn't have a name.   Not everything needs a name, and in this case, I don't there is an official one.  "Central body" "Central object" or "Parent" work just fine as familiar and common terms.


Are there small voids of 10 - 20 megaparsecs also within, and not only between superclusters?

My only goal is basically to find out whether I would be wrong or not if I state that small voids of 10 - 20 megaparsecs, really actually exist within, and not only between superclusters.

The largest structure is called a filament.

We are in the supercluster Laniakea, in the smaller supercluster Virgo, in the Milky Way galaxy, in the Solar system, on Earth.

It looks like this:

$$\text{Click to zoom in - 23,624 × 2,953 pixels}$$
$$\text{Laniakea}$$

Here we are:

Wikipedia says:

Distribution: cosmic voids and sheets

Research has tried to understand the way superclusters are arranged in space. Maps are used to display the positions of 1.6 million galaxies. Three-dimensional maps are used to further understand the positions of these superclusters. To map them three-dimensionally, the position of the galaxy in the sky as well as the galaxy's redshift are used for calculation. The galaxy's redshift is used with Hubble's law to determine its position in three-dimensional space.

It was discovered from those maps that superclusters of galaxies are not spread uniformly across the universe but they seem to lie along filaments. Maps reveal huge voids where there are extremely few galaxies. Some dim galaxies or hydrogen clouds can be found in some voids, but most galaxies are found in sheets between the voids. The voids themselves are often spherical but the superclusters are not. They can range from being 100 million to 400 million light-years [that's 30.660139381 to 122.64055752 megaparsec] in diameter. The pattern of sheets and voids contains information about how galaxy clusters formed in the early universe.

There is a sponge analogy used often that compares a sponge to the pattern of clusters of galaxies in the universe – the holes are the voids and the other parts are the locations of the superclusters.".

So, yes - there are small and large voids.

See also this Wikipedia page:

Galaxy distribution

The number density of galaxies in the LS falls off with the square of the distance from its center near the Virgo Cluster, suggesting that this cluster is not randomly located. Overall, the vast majority of the luminous galaxies (less than absolute magnitude −13) are concentrated in a small number of clouds (groups of galaxy clusters). Ninety-eight percent can be found in the following 11 clouds (given in decreasing order of number of luminous galaxies): Canes Venatici, Virgo Cluster, Virgo II (southern extension), Leo II, Virgo III, Crater (NGC 3672), Leo I, Leo Minor (NGC 2841), Draco (NGC 5907), Antlia (NGC 2997) and NGC 5643. Of the luminous galaxies located in the disk, one third are in the Virgo Cluster, while the remainder are found in the Canes Venatici Cloud and Virgo II Cloud, plus the somewhat insignificant NGC 5643 Group. The luminous galaxies in the halo are also concentrated in a small number of clouds (94% in 7 clouds). This distribution indicates that "most of the volume of the supergalactic plane is a great void." A helpful analogy that matches the observed distribution is that of soap bubbles. Flattish clusters and superclusters are found at the intersection of bubbles, which are large, roughly spherical (on the order of 20–60 Mpc in diameter) voids in space. Long filamentary structures seem to predominate. An example of this is the Hydra-Centaurus Supercluster, the nearest supercluster to the LS, which starts at a distance of roughly 30 Mpc and extends to 60 Mpc.


When we talk about the "shape" of the universe. It is its intrinsic shape, not the embedding of the universe within a larger space.
It does make sense to talk about the intrinsic shape.  Properties like "curvature" can be measured from inside a shape, you don't need suppose that the universe is "in" something to talk about its shape.
Now the space of the universe is three dimensional, this creates visualisation problems.  So I'm going to first talk about an imaginary 2d universe:
A 2d universe could be flat, with a curved boundary (like a disc)  Or it could be flat with no boundary (an flat infinte plane).  Or a 2d universe could be intrinsically curved.  If it has positive curvature it could be like the surface of a ball (positive curvature, finite with no boundary) Or shaped like a bowl (positive curvature)  Alternatively it could have negative curvature: it could be shaped like a "Pringles crisp"  These have negative curvature (this can happen with or without a boundary). Finally there could be regions with different curvature: positive in some parts, negative in others.
The same possibilities exist for our universe:  It could be flat (with or without boundary)  It could be positively curved (This is intrinsic curvature and your brain can't visualise this for a 3d shape)  Or it could be negatively curved (again, don't try to visualise this for a 3d shape, you can't).  In the case of positive curvature, it could be finite and unbounded, but for negative curvature, it must either have an edge, or be infinite.
So mathematically these are the possibilities. What does the science say?
We can measure the curvature on large scales.  Our measurements are not perfect, there is some room for error.  We measure the large scale mass/energy density of the universe, since curvature is caused by gravity, and gravity is caused by mass and energy. If the mass/energy density is > 1 then space would be positively curved.  We actually find that the energy density is 1.00±0.02  That is the universe is either flat, or very nearly flat.
No there are profound problems with supposing that the universe has a boundary. But there are also problems with supposing it to be infinite.  Scientifically, no edge has ever been detected. So when we build models of the universe, we will generally suppose to have no edge.
Thus our "best guess" is that the universe is uncurved and infinite.


No complete answer from me, but some obvious examples of "hair" on a neutron star are:

The temperature - both internal and surface temperature can be measured (in principle). These are mainly a function of the age of the neutron star.

The magnetic field - young neutron stars have very strong magnetic fields that are probably an inherited property from the composition and rotation of the progenitor. The field decays with time.

The surface composition may vary from neutron star to neutron star. In principle this can be measured with X-ray spectroscopy. The composition may be connected with the progenitor, but could also reflect accreted material.

Neutron stars have a radius! The radius may or may not be a unique function of mass. It could depend on the rotation, composition and age.

I think though, the gist of your question is could one determine the progenitor of a neutron star from observations of the neutron star? At present the answer to that is no. Even the relationship between neutron star mass and progenitor mass is not understood; it is also unclear why some neutron stars have very high magnetic fields and so on. But ultimately that would be an aim.


This is a spectral energy distribution (SED). Since astronomical objects don't emit light a single frequency an SED tells you how much emission you're getting across a range of frequencies.

"log" indicates it's on a logarithmic scale
$\nu$ represents frequency in Hz (e.g. $log~\nu=12$ is $10^{12}$ Hz)
Jy is a unit of specific flux density $f_{\nu}$ is the specific flux
MRK 1014 is the name of the object
Each of the points are observations there is a link on the page for "photometric data points" to see where everything comes from. All the positions have a point, which is measurement of the specific flux density at that given frequency. The vertical lines are the error bar for the flux measurement and the horizontal is the range of frequencies a measurement covers (e.g. a filter might have a central frequency but it filters light over a range of frequencies).
There are a few upper limits which are indicated by vertical lines with arrows pointing down.
LINE indicates that this is an estimate from a spectrum and it is the measurement of the flux of that line, not from a imaging filter.


How to calculate position of an unknown star knowing positions of some other stars from an image?

First, let assume your image geometry is homogeneous, and has no peculiar distorsion in either direction.

Second, let assume you have the resolution of your image: the number of arcseconds / pixel.

Now, take one 'red-cross' star, call it A. It will be the origin of the triangle we will draw. Name your 'yellow-cross' star B. Now, take a new point, called 'C', that is at the same pixel-y coordinate of A, and the same pixel-x coordinates of B.

Drawing lines between A, B and C gives you a right triangle. You cannot simply apply flat geometry equations, since you are on the celestial sphere.

Hence, one must compute the 'Bearing' angle between A, B and C. See for instance here: (A bearing is an angle, measured clockwise from the north direction).

Here is a little piece of code you should be able to read:

double adjacent = sqrt(pow(B.x-C.x, 2) + pow(B.y-C.y, 2)); // dBC
double opposite = sqrt(pow(A.x-C.x, 2) + pow(A.y-C.y, 2)); // dAC

double theta = atan2(opposite, adjacent) * ONE_RAD_IN_DEGREES;

theta is the bearing angle, here expressed in degrees, thanks to the conversion constant ONE_RAD_IN_DEGREES. atan2 is the Arc-Tangent function that takes care of which quadrant you are in (it computes arctangent(opposite / adjacent), correcting for the quadrant, see the wikipedia article for instance).

Now, depending on whether you have East to the left or not (astro images have East to the left usually), you need to correct your angle. Again this little piece of code:

BOOL eastLeft = <true or false>

if (B.x < A.x && B.y > A.y) {
    theta = (eastLeft) ? theta : 360.0 - theta;
}
else if (B.x < A.x && B.y < A.y) {
    theta = (eastLeft) ? 180.0 - theta : theta + 180.;
}
else if (B.x > A.x && B.y < A.y) {
    theta = (eastLeft) ? theta + 180. : 180.0 - theta;
}
else if (B.x > A.x && B.y > A.y) {
    theta = (eastLeft) ? 360.0 - theta : theta;
}

Now, we have the correct theta value. Now, compute the distance (below, in degrees) between A and B, and call it delta.

Assuming the R.A. and Declination of A are called lambda1 and phi1, you can compute the R.A. and Declination of C, lambda2 and phi2, using the formulae given in here, under the section "Destination point given distance and bearing from start point".

In my code:

double phi1 = declination_A * ONE_DEG_IN_RADIANS;
double lambda1 = rightAscension_A * ONE_HOUR_IN_RADIANS;

double delta = degrees * ONE_DEG_IN_RADIANS;
double theta = bearing * ONE_DEG_IN_RADIANS;

double phi2 = asin(sin(phi1)*cos(delta) + cos(phi1)*sin(delta)*cos(theta));
double lambda2 = lambda1 + atan2(sin(theta) * sin(delta) * cos(phi1), cos(delta) - sin(phi1) * sin(phi2));

with the usual meaning of trigonometric functions (sin is sine, asin is arcsine, etc).


Effect of the obliquity of the ecliptic / tilt of the Earth on the equation of time

As the solar time
Wikipedia page illustrates, we can split a solar day into a constant 23h56m to complete a rotation relative to the stars
(a sidereal day)
and a variable ~4m to compensate for one day's orbital motion around the Sun.
The Sun's apparent motion around the equator determines how much more than 360$^\circ$ the Earth must rotate from one solar noon to the next.

But the Earth orbits the Sun in the ecliptic plane, which is oblique to the equator.
If that orbit were circular, the Sun's
ecliptic longitude
would change at a constant rate (see Note below).
One degree along the equator spans exactly 4m00s of
right ascension;
one degree along the ecliptic, measured around the equator, may span ~8% more or less than that.

At an equinox, the ecliptic crosses the equator at a 23.4$^\circ$ angle,
and one degree of ecliptic longitude spans 3m40s of right ascension.
The Sun appears to move slower than average around the equator,
so the apparent solar day is slightly shorter.

At a solstice, the ecliptic is tangent to a $\pm$23.4$^\circ$ declination parallel,
and one degree of ecliptic longitude spans 4m22s of right ascension.
The Sun appears to move faster than average around the equator,
so the apparent solar day is slightly longer.

Stellarium images around the March equinox and the June solstice,
showing 30 degrees of ecliptic longitude and 2 hours of right ascension for comparison.
Equatorial grid is blue, ecliptic grid is orange, ecliptic is yellow.

The equation of time is the cumulative sum of differences between mean and apparent solar day length.
Like a mathematical function and its derivative, zero crossings of one should roughly correspond to maxima or minima of the other.
Differences of a few seconds per day, several weeks in a row, add up to a few minutes.

The earliest/latest sunrise and latest/earliest sunset of the year do occur several days on either side of the summer/winter solstice.
However, on any given day, sunrise, noon, and sunset are all behind or ahead of mean solar time by approximately the same amount.

Note: the Earth's orbit is not quite circular.
Following Kepler's second law,
the Sun's ecliptic longitude changes ~3% slower at aphelion and faster at perihelion,
making the apparent solar day slightly shorter or longer respectively.
This Wikipedia entry
discusses it in more detail.


Yes, there are many. One of the first, if not the first was the radio lensed quasar B1938+666, in which a near-IR Einstein ring was reported by King et al. (1998).

The answer to your real question though is not as straightforward as that. Although lensing should be wavelength independent (according to GR), it is not independent of the size and position of the source. The radio waves and optical emission arise from different source sizes and locations. In addition, if you are interested in testing whether the surface brightness magnification is the same, well that relies on knowing the intrinsic source properties; and the radio to optical brightness of quasars varies wildly.


What is the term for astronomical objects outside the solar system that are smaller than dwarf planets?

Just add the prefix "exo-". At least exocomet and exomoon seem to be established.


Is there any official star gazing location list?

You may be out of luck. If you live pretty much anywhere in the North East, there are very few locations where you can see a truly dark sky. You might be able to get somewhere where you can see the faint glow of the milky way band, but to see a really dark sky, you pretty much have to be out west in the middle of no where.

One resource to help you out is to look up the Bortle Scale. This is a metric used to describe how dark the sky is in a given area and thus what you're likely able to see with the unaided eye. There are numerous sites which attempt to predict or inform what your Bortle scale number is at your specific location.

In general though, you can get a feel of the light pollution in a particular area by using light pollution maps like this one or this one. Attached below is an image from the first link showing the general light pollution level across the US.


How does a neutron star emitting magnetic-dipole radiation lose angular momentum?

Radiation carries a momentum per unit volume given by
$$ \vec{p} = \epsilon_0 \vec{E} \times \vec{B}$$

It therefore carries an angular momentum of
$$ \vec{L} = \vec{r}\times \vec{p} = \epsilon_0 \vec{r} \times (\vec{E}\times \vec{B})$$

For plane-polarised transverse electromagnetic waves propagating in the $\hat{r}$ direction, then $\vec{L}=0$. However, if the axis of a magnetic dipole is inclined to the rotation axis then the radiation is not a single pure transverse wave; the dipole moment is the sum of two oscillating magnetic dipole moments that are perpendicular and oscillate $\pi/2$ out of phase. These produce two transverse waves, with arbitrary amplitudes, perpendicular polarisations and a $\pi/2$ phase shift between them - a.k.a. elliptically polarised light. Elliptically polarised light carries angular momentum.


How do we know background radiation does not come from undiscovered galaxies?

Assuming you mean the Cosmic Microwave Background ...

The CMB is entirely in a very narrow range of the microwave part of the spectrum. Galaxy emissions are all over the spectrum.

The CMB comes uniformly from all over the sky. It's like looking at painting that is all one color. Galaxy light looks like pinpoints on a dark background. (Nearby galaxies are not pinpoints, but those at cosmological distances look that way.)

The CMB matches the exact wavelength we expect for light that is heavily redshifted from the moment of last scattering. That moment was hundreds of millions of years before galaxies ever formed. Light from oldest galaxies are at a redshifted wavelength from 500 million years after the last scattering.

If the CMB was due to early starlight reflecting off dark objects, you would see a very different spectrum. It would contain many absorption lines as the dark objects absorbed some wavelengths better than others. Also the reflected starlight would be across a wide range of frequencies, unlike the CMB which has a very narrow range.


tl;dr The answer is opinion-based, but choosing an NFW profile probably won't offend anyone.
Pseudo-isothermal profile
There are many more or less physically motivated functional forms for the density profile of a dark matter halo.
The pseudo-isothermal sphere is a less unphysical profile than the simple, isothermal profile (which is unphysical in the center and has a diverging integrated mass for $r\rightarrow\infty$). It gives good fit to the rotation curves in the optical disk of galaxies, but not to the light distribution.
NFW profile
Arguably, the most popular profile is the NFW profile (Navarro, Frenk, & White 1996;1997) — with a total of ~13,000 citations, these two papers are among the most cited papers in astrophysics ever. Its popularity is probably partly due to its simplicity. It isn't really physically motivated, but simply turned out to give good fits to numerical simulations.
Einasto profile
However, subsequent N-body simulations with higher resolution
(
Navarro et al. 2004;
Hayashi & White 2008;
Gao et al. 2008;
Springel et al. 2008
)
showed small, but systematic, deviations from the NFW profile which are more accurately modeled with an Einasto (1965) profile:
$$
\rho(r)=\rho_{-2} \exp \left[\frac{-2}{\alpha}\left\{\left(\frac{r}{r_{-2}}\right)^{\alpha}-1\right\}\right],
$$
where $r_{-2}$ is the radius at which the slope of the profile is $-2$, $\rho_{-2} = \rho(r_{-2})$, and $\alpha$ is a fitting parameter with best-fit values that typically lie in the range $[0.12,0.25]$, increasing with halo mass.
However, since there isn't much physics here either, and this form is more complicated, the NFW profile seems to remain the profile of choice for most people.
Other profiles
Many profiles can be written as double power-laws
$$
\rho(r)=\rho_{0}\left(\frac{r}{r_{0}}\right)^{-\gamma}\left[1+\left(\frac{r}{r_{0}}\right)^{\alpha}\right]^{(\gamma-\beta) / \alpha},
$$
where the parameters $\alpha$, $\beta$, and $\gamma$ can take a variety of values (Tab. 5.1 from Mo et al. 2010):
$$
\begin{array}{lll}
\hline
\hline (\alpha, \beta, \gamma)  & \text{Name} & \text{Reference} \\
\hline(1,3,1) & \text{NFW profile} & \text{Navarro et al. (1997)} \\
(1,4, \gamma) & \text{Dehnen profile} & \text{Dehnen (1993)} \\
(1,4,1) & \text{Hernquist profile} & \text{Hernquist (1990)} \\
(1,4,2) & \text{Jaffe profile} & \text{Jaffe (1983)} \\
(2,2,0) & \text{Modified isothermal sphere} & \text{Sackett & Sparke (1990)} \\
(2,3,0) & \text{Modified Hubble profile} & \text{Binney & Tremaine (1987) }\\
(2,4,0) & \text{Perfect sphere} & \text{de Zeeuw (1985)} \\
(2,5,0) & \text{Plummer sphere} & \text{Plummer (1911)} \\
\hline \hline
\end{array}
$$
At small and large radii $\rho$ is proportional to $r-\gamma$ and $r-\beta$, respectively, while $\alpha$ gives the sharpness of the break.
Some profiles are better at describing the inner structure, some are better farther out, some are better when you include gas, etc. What you choose depends on your particular problem, and your mood.


Dark matter has different properties than ordinary matter. It cannot collide, and thus it alone cannot coalesce to form a star. Furthermore, dark matter and dark energy are unassociated properties. There is not currently any reason to assume dark matter can be converted to dark energy.

That said, there is a strictly hypothetical entity referred to as a dark star, which does involve dark matter. While composed mostly of ordinary matter, this entity is hypothesized to include a high concentration of dark matter, generating heat and preventing fusion of the ordinary matter. If dark stars existed, it is likely they existed in the early universe and did not survive to the present. They are modeled to be large clouds--4 to 2000 astronomical units in diameter--of hydrogen and helium, with a surface temperature low enough to make them invisible to the naked eye.


What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE?

I am not a professional astronomer, so take this answer with a grain of salt, but just from visiting a few facilities, I know some.
For single dish applications, the 100m telescope in Effelsberg, Germany, uses a 7-beam receiver -- fewer than the Parkes array you mention, and I dont't know of any single dish setup with a larger number of beams.
Regarding arrays, the Westerbork Synthesis Array in the Netherlands uses the APERTIF arrays in most of its individual 25m dishes. With 121 elements per array, this seems to be on the higher-end side of granularity.

update:
From astron.nl/dailyimage for 31-01-2017 First image with Apertif: a new life for the Westerbork radio telescope

The first images made with the upgraded telescope that demonstrate this new 'wide-angle' capability is shown here. The first image shows the dwarf galaxy Leo T. The image is colour-coded and shows the gas (in blue) in this galaxy together with many distant radio galaxies in the background shown in orange. For comparison, the field of view of the previous Westerbork system and the size of the full moon are also indicated.
To make this new capability possible, ASTRON developed and built the hardware in-house. 121 small receivers are used in each telescope whose signals are combined electronically to produce the large field of view.
The upgraded telescope will also be used to search for and study new variable sources in the radio sky. With the new Apertif receivers, observations of large parts of the sky can be done much faster, and projects that used to be impossible as they would take tens of years can now be done in a much shorter time. Westerbork is therefore poised to make many new discoveries in the radio sky.


Rule of Exception in All Nearest Objects

I have not heard of any idea that the nearest object is exceptional in any significant way, and in the case of the nearest pulsar, nearest star, nearest (large) galaxy etc. the properties are usually typical of the underlying population, though perhaps not the observed population because of biases and selection effects.

Take Proxima Centauri as an example. It is unlike any star that can be observed with the naked eye. However, it is a very good example of a "modal" star - much less massive than the Sun, about half the age of the Galaxy and a member of a multiple system. The reason for the discrepancy is of course that stars have a wide range of luminosities and distances; in a "flux-limited" survey, we always pick out the brightest sources, which tend to be the most luminous sources at large distances, where our survey volume is largest.

So, it is true that nearby objects are different to the observed population.

The same may apply whenever some unusual chacteristic makes a type of object more likely to be detected. There is certainly a phenomenon in astronomy where the prototype of a class of objects (usually the first discovered) turns out to be atypical of the class as a whole.

As an example we could consider RS CVn binaries, named after the object RS CVn, which is a close binary system containing an evolved (subgiant) component, and exhibiting extreme levels of magnetic activity (star spots, coronal X-ray emission).

It was initially believed that the presence of an evolved component was a defining feature of the class. However, over the years it has become apparent that there are many more examples of close, magnetically active binaries with main sequence components. The prototype is atypical in having an evolved component - the explanation is that, although short-lived and rarer, evolved magnetically active stars are much more luminous and easier to find/study.

As a final word on this, there is a set of nearby/nearest objects, including the Sun and Moon, that are somewhat unusual. The Sun is more massive than 90+% of stars in the Galaxy; no other planets in our solar system have a moon that is so (relatively) large. An explanation for these things may lie with the anthropic principle, which argues that we would not observe any circumstance that would preclude the development of intelligent life on Earth or, to turn that around, we are more likely to find ourselves in a situation that is favourable for the development of intelligent life on Earth.


Do lower mass stars consume more hydrogen than higher mass stars?

Stars less than about 0.3 solar masses are fully convective, so hydrogen throughout the entire star can be transported to the core and consumed.  At higher masses, the core is radiative, so only hydrogen that starts in the core is consumed.


How much of a molecular cloud can end up as "starstuff"?

This quantity is referred to as star formation efficiency (SFE) by astronomers who study star formation and galaxy evolution.   Estimates can vary but typically are around a few percent.  In Sec. 4.1 of this paper Inoue et al. review  some estimates from the literature.  Those numbers are for the central regions of spiral galaxies, and the rate may be lower in the outer regions where the density is lower.
Note that even if the rate is a few percent in a given cloud for a given episode of star formation, the rest of that gas mass will be dispersed into the interstellar medium and eventually find its way back into another cloud to form stars later - star formation is ongoing in spiral galaxies.
Note that you will often see SFE expressed as star formation rate (solar masses per year converted to stars) divided by total gas mass.  That has units of (mass/yr) / mass, which reduces to units of 1/yr.  You can think of that as the fraction of gas that forms into stars per year, or alternatively, the inverse of that number gives you a timescale in years to deplete the gas entirely.


Why the blank wedges in this very early 21 cm map of the Milky Way? (Oort et al. 1958)

TLDR: these wedges are bits where things are moving around the centre of the galaxy at about the same speed as us, so we can't understand what is there.

As it states on page 4 of the paper you have linked,

  the great gap between 315 and 340, where, except at small R, the differential rotation is too small to separate the various arms

The method used to deduce this structure is described on page 2:

  Observations in the decimetre continuum can only give the integrated radiation over the line of sight. The 21-cm line gives discrimination in distance. But although the 21-cm observations give discrimination in distance they cannot by themselves provide actual distances. The distance distribution in a given direction can only be inferred from radial velocities. For this we have to suppose that in each part of the Galactic System the average motion of the gas coincides with the circular velocity at the corresponding distance from the centre.

The map is given by combining several observations at different wavelengths - the 'decimetre' gives the total amount of stuff along the line of sight (because it isn't absorbed much along the way), the 21-cm line gives a measure of distance (because it is absorbed so predictably wakened by the amount of stuff it has passed through), and then to we have to assume a model of Keplerian rotation as a guess at how things are moving around the centre of the galaxy. Combining these three things allows you to map out the material even though you are seeing different bits stacked together along your line of sight.

This doesn't work in the blank wedges because the speed of rotation is too similar to ours, so we can't use the rotation velocity to disentangle distances.


How to determine the ellipticity of galaxies in SDSS

Regarding your original question about the "correct value for ellipticity", keep in mind that a galaxy has no objective ellipticity. The inserted image nicely shows that the light distribution of a real galaxy is way more complicated than a smooth profile with isophotes of identical ellipticity. In other words, the ellipticity will in fact be defined through the measurement process.

Flux-weighted second moments (which use the pixel brightness as weighting function) or adaptive moments (which use an elliptical Gaussian weighting function) are two ways of doing this. And there might be nothing wrong with the observed difference in ellipticity measurements, as the galaxy light profile is very different from a Gaussian!

The flux-weighted measurement is more sensitive to the ellipticity at inner radii, while the adaptive moments give more weight to the disk (assuming that the algorithms converge well).


Do Mercury and Venus exhibit retrograde motion?

As noted by PM2Ring, the notion of "retrograde motion" predates the "Copernican revolution" So you should think about the motions of the planets from a Ptolemaic view:

  The planets have two major motions (in addition to the diurnal motion around the Earth) motion with a period of one year, and a second motion with a period that is specific to each planet. For Mars, Jupiter and Saturn, the main orbit has a long period, and the epicycle has a period of one year.  For Mercury and Venus, the main orbit has a period of a year (keeping them aligned with the sun) and the epicycle has a period of less than a year.

When the right ascension of the planet is decreasing, then we say it's motion is retrograde. Retrograde periods occur during inferior conjunctions of the planet with the sun, and so the planet cannot be easily observed while it is retrograde.


Are gas giants supported by thermal pressure?

I am not sure what you mean by "thermal" pressure. Jupiter is supported by pressure, just like all objects that are in (approximate) hydrostatic equilibrium.

That pressure is provided by your everyday, temperature-dependent Maxwell-Boltzmann ideal gas pressure in the outer parts, but the free electrons in the interior become degenerate and so in these regions I suppose you would more accurately describe the pressure as being due to (partial) electron degeneracy pressure.

At the very centre there may be a liquid or even a solid core. In the metallic hydrogen model then this would still be degenerate electrons contributing the pressure. For a solid, rocky core, well solids are rather incompressible.

You can only describe Jupiter as being in approximate hydrostatic equlibrium. It is losing energy from its surface; this energy is supplied from gravitational potential energy and Jupiter is shrinking at a rate calculated to be about 2 cm per year. As it does so, the interior gas becomes denser and more and more degenerate and the pressure will become more independent of temperature. As a result, the rate of contraction will slow down and Jupiter will tend towards the radius of a "cold" body of that composition.


What are the 4 candidates for Planet Nine reported in 2017?

A frequently cited study about Planet Nine is arXiv:1601.05438: Evidence for a Distant Giant Planet in the Solar System, but I can confirm that after 2016 there are not too many studies on the subject, at least I could not find many. There are two interesting studies I would like to share, but none of them spoke of four candidates.

Spitzer Follow-up of Extremely Cold Brown Dwarfs Discovered by the Backyard Worlds: Planet 9 Citizen Science Project. The abstracts reads

We present Spitzer follow-up imaging of 95 candidate extremely cold brown dwarfs discovered by the Backyard Worlds: Planet 9 citizen science project, which uses visually perceived motion in multiepoch Wide-field Infrared Survey Explorer (WISE) images to identify previously unrecognized substellar neighbors to the Sun. [...]

Even more interesting is What If Planet 9 Is a Primordial Black Hole? by Jakub Scholtz and James Unwin from 2020:

We highlight that the anomalous orbits of trans-Neptunian objects (TNOs) and an excess in microlensing events in the 5-year Optical Gravitational Lensing Experiment data set can be simultaneously explained by a new population of astrophysical bodies with mass several times that of the Earth (M⊕). We take these objects to be primordial black holes (PBHs) and point out the orbits of TNOs would be altered if one of these PBHs was captured by the Solar System, inline with the Planet 9 hypothesis. Capture of a free floating planet is a leading explanation for the origin of Planet 9, and we show that the probability of capturing a PBH instead is comparable. The observational constraints on a PBH in the outer Solar System significantly differ from the case of a new ninth planet. This scenario could be confirmed through annihilation signals from the dark matter microhalo around the PBH.

I hope that this is helpful for you.


Why is the magnetic axis of Uranus and Neptune off center?

Disclaimer: This answer is inspired by Magnetic fields of Uranus and Neptune: Metallic fluid hydrogen by Nellis [2017], who explains the center offset of the ice-giants' magnetic fields by contrasting them with Earth's magnetic field.
Earth's axisymmetric magnetic field
The most researched magnetic field in our Solar System is, of course, Earth. Nellis 2017 explains how coupling between the rotational motion and convective dynamo motions in the Earth's outer core cause the magnetic axis to drive towards alignment to the spin axis:

Because rotational motion (RM) of Earth is strongly coupled into
convective dynamo (CD) motions of its fluid-Fe outer core, planetary
RM stabilizes convective motions that generate a dipolar magnetic
field. If a convective fluctuation occurs which tends to destabilize a
given dipolar axis, then strong RM-CD coupling either drives
convective motions that essentially restore the initial orientation or
CD fluctuations that drive the initial magnetic axes out of
orientational equilibrium are so strong that RM-CD coupling eventually
drives the dipolar axis into an alignment anti-parallel to its initial
one.

Uranus/Neptune non-axisymmetric magnetic fields
Voyager 2 measured both gravitational moments as well as the magnetic fields of both Uranus and Neptune.  These measurements helped develop models of planetary composition as well as models of magnetic field generators. For Uranus/Neptune, the fields are generated near the surface at radii as large as $\approx0.9R_U/R_N$, while the Earth's magnetic field is generated much deeper at $\approx0.5R_E$.
The magnetic generation near the surface, as well as the H-HE envelope (again according to Nellis) means that:

local convective dynamo motions of fluids that produce the magnetic
fields are essentially decoupled from global rotational motions [...].
The dynamos of U/N would then be relatively free to wander as local
convective fluctuations dictate.  Thus, tilt angles and center-offsets
of their fields would vary slowly over the age of the Solar System.

Summary
The Earth's composition and convective liquid outer-core currents cause a coupling between the rotational motion of the Earth and the convective dynamo motion.  This drives the magnetic axis to align with the spin axis, which passes through the center-of-mass.  No similar strong coupling exists for Neptune and Uranus, which means the magnetic axis on each of these planets is free to wander both in orientation and offset from the spin axis and center of mass due to local surface convection currents.


It is possible to combine a reflector and a refractor telescope together?

...is it possible to connect a reflector telescope and a refractor telescopes (like monocular for bird watching) together...

Yes, a properly focused telescope (for someone who's not near-sighted) will have infinity conjugate focal points; parallel rays in make parallel rays out.
If the telescope is set up for 50x magnification (f=600 mm objective, f=12 mm eyepiece) then a 1 degree field of view (FOV) in the sky will be a 50 degree apparent FOV at the output of the first eyepiece.
Alas, a second 8 x bird-watching monocular may only have a 7 degree input FOV so we will indeed get 50 x 8 = 400x magnification, but of a 0.14 degree field.
That's just what you would get looking through a single telescope set up for 400x (say f=1200 mm objective, 2x Barlow and f=6 mm eyepiece).
In other words, you could do this, but...

...to build a better telescope?

I'm not sure it would be better in any possible way if it's for looking through.
However, if you have an old CMOS imager webcam from the 1990's with giant pixels and you want to blow up the primary focal plane of the first telescope to get some useful resolution, this might be a suboptimal solution. You could remove the eyepiece of the second bird-watching monocular and put the image sensor with too-large pixel size there and get better resolution than if you'd put it at the focal plane of the first telescope.
A better way to solve this unlikely problem might be to simply use positive or eyepiece projection with the first telescope's eyepiece moved outwards so that it's final focal point moves in from infinity to say roughly 8 times its focal length.
That would have the same effect but without adding the monocular.


The answer is yes time dilation does affect how much time an observer experiences since the big bang until the present (cosmological) time.

However there is a certain set of special observers called comoving observers, these are the observers to which the Universe appears isotropic to. For example we can tell the Earth is moving at about 350 km/s relative to a nearby comoving observer by measuring the anistropy in the cosmic microwave background (in fact it causes a relatively large anistropy, the pictures of the CMB you see from WMAP, etc are with this anistropy factored out).

One special property of a comoving observer in the Universe is that they maximize the age of the Universe, i.e. no other observer can experience more time since the big bang than a comoving observer (in case you ask there is no observer that minimizes the time since the big bang). When we talk about the age of the Universe we are talking about the age of the Universe from the pov of a comoving observer.


It doesn't exist.
It is actually rather difficult to see the shape of the milky way, because we are inside it. The more distant parts are obscured behind the nearer and very much of it is invisible. (We think that a supernova would happen every 50 years or so in a galaxy the size of the Milky Way, but none have been seen in nearly 400 years, so even massive bright events are hidden).
There are rough maps based on radio imaging of neutral hydrogen, but these are fuzzy and have artefacts, making part of the map invisible and part of it seems to point strongly towards the Sun.  The well-know images of the milky way on Wikipedia are as good an artist's impression as any other.


What is the completeness of an observation and how do I calculate it?

a) What is the definition of completeness?

Completeness is the number of objects in a data set that are detected over the number that exist.  In astronomy, completeness is often estimated for a particular apparent magnitude or flux density.  As an example, for sources that are as bright as the Sun (-27 magnitude), we have a completeness of 1.  That is, we’ve discovered all sources in the celestial sphere with the Sun’s magnitude.  As we get down to very dim objects like tiny pieces of space junk in Earth orbits, our completeness ratio drops dramatically.

b) How do you calculate the completeness of an observation?

On a ROC curve, completeness is the True Positive Rate or the y-axis if all the objects have been observed (thanks @ProfRob!). If only a fraction of the objects have been observed, completeness is that fraction times the true positive rate. Note that a classifier only yields better completeness at the cost of higher false positive rates.  Here is an excellent example of ROC curves for different algorithms which differentiate quasars from stars:

The left panel shows data used in color-based photometric
classification of stars and quasars. Stars are indicated by gray
points, while quasars are indicated by black points. The right panel
shows ROC curves for quasar identification based on u - g , g - r , r - i
, and i - z colors.

If you don’t know the true positive rate, you may be able to run a simulation in which you inject objects into your data set and determine what portion of these are found by your algorithm.  This will give you an estimate of your completeness.  Matlab has a specific function for ROC curve Monte Carlo simulation, for example.


Perhaps you are looking for Lagrange’s planetary equations? Given some perturbation $H_1$ to the Hamiltonian, they give six equations to calculate the change of the orbital elements over time; a thorough discussion is given in Goldstein’s textbook on classical mechanics.
The equation for $\Omega$ is $$\frac{d\Omega}{dt}=\frac{1}{\sqrt{GMm^2(1-e^2)\sin i}}\frac{\partial H_1}{\partial i}$$ where $i$, $e$ are orbital elements.


Is it possible that all dark matter is made of rogue planets (free-floating planet)?

First of all I'll start with a few ideas:

Baryonic Matter: Baryons are elementary particles made up of 3 quarks. This includes protons and neutrons, and the term baryonic matter refers to matter made of baryons, such as atoms. Examples of non-baryonic matter includes neutrinos, free electrons and other exotic matter.
Things like planets, stars, dust, etc. are all made of atoms, and so are classified as baryonic matter.

Now, how do we know that dark matter is present in the universe?

Astronomers measure the gravitational pull of galaxies and galaxy groups/clusters based on how objects behave when interacting with these objects. Some examples of this include tidal gas/dust stripping, the orbit of stars in a galaxy and gravitational lensing of distant light from a large cluster. Using this they determine the mass of the galaxy (or galaxy group).
We can also determine the mass of a galaxy or group by looking at it and adding up the mass of all the objects (like stars, dust, gas, black holes, and other baryonic matter). While these methods both give us approximations, it is clear that the gravitational mass of galaxies and groups exceeds the baryonic mass by a factor of 10-100.

When astrophysicists first found this phenomenon they had to come up with a plausible explanation, so they suggested that there is some new, invisible matter called dark matter. (Aside: some astrophysicists also came up with other explanations like modified gravity, but so far dark matter does the best job at explaining observations).

Okay, so now how do we know dark matter is not any sort of baryonic matter?

There are a few reasons astrophysicists know that it is extremely unlikely that dark matter is baryonic. First of all if all the stars in a galaxy shine on an object it heats up, this heat causes the release of radiation, called thermal radiation, and every (baryonic) object above zero kelvin (or -273.14 deg celcius) emits this radiation. However, dark matter does not emit any radiation at all (hence the name dark!)

If dark matter were baryonic it would also mean that it could become light emitting. If we got a clump of baryonic matter* and put it in space it would gravitationally contract, and would eventually form a star or black hole** - both of which we would be able to see.

So, because of these reasons the dark matter in galaxies and in galaxy groups/clusters cannot be baryonic, and so cannot be planets, dead stars, asteroids, etc. It would definetely not be planets as there is no way 10-100 times the mass of the stars in a galaxy would be planets, as the mechanism for making planets relies on supernovae, and the number of supernovae needed for the that many planets would be far too high to match our observations. I hope that this answered your question!

*provided the clump of baryonic matter was large, and the amount there is in galaxies definitely is!

** we don't observe black holes directly, but can see radiation from their accretion disks.


Formula to calculate the apparent magnitude of Earth from arbitrary distances

The absolute magnitude of the Earth is about $H = -3.99$. This means, from $1$ AU away from the Sun and the Earth, the Earth has an apparent magnitude of $-3.99$.
Voyager 1 is $152.26$ AU from the Earth, and $151.85$ AU from the Sun. Using the formula for apparent magnitude, $${\displaystyle m=H+5\log _{10}{\left({\frac {d_{BS}d_{BO}}{d_{0}^{2}}}\right)}-2.5\log _{10}{q(\alpha )}}$$
With $d_{BO} = 152.26, d_{BS} = 1, d_0=1,$ AU, and $q(\alpha) = 0.306$, we obtain that Earth's apparent magnitude from Voyager 1 would be about $\boxed{8.208}$.


Formulas to determine the illuminated phase and orientation of the Moon

I created the spreadsheet.  It's a bit complicated, to say the least.  Initially I was inputting the formulas by hand from the book I cited in the video, but the process was rather error-prone so I ditched those tables and simply scanned the pages electronically and used the numbers from the scans to construct the tables of the Fourier series used to compute the moon's position.  Some of the tables are therefore redundant and aren't used.  You may be able to use a much simpler method to compute the position of the moon though; for my purposes I needed arcsecond accuracy.  If all you're interested in is computing the approximate phase angle and apparent orientation you can get away with a much more simplified calculation for the moon's position that is accurate to within a few tenths of a degree.  In that case the main thing you need from my spreadsheet are the formulas for computing the phase angle and apparent orientation.  You'll need to calculate the approximate location of the sun as well in order to determine these things.  The spreadsheet does this as well, using a simplified method described by Jean Meeus in "Astronomical Formulae for Calculators."  He also gives a simplified method for computing the position of the moon.  Either "Astronomical Formulae for Calculators" or "Astronomical Algorithms" (both by Jean Meeus) would be excellent references for your project.  I can walk you through the formulas you need if you need help with that part.


2017 VR12
is expected to reach magnitude 11.8 on the night of March 6-7.
This is near the limit of a 4-inch scope; actual visibility depends on your observing site, your eyes, and the magnification.
You can get a more definite answer by trying to see stars of similar magnitude.

The other hard part is to acquire the target.
It will look like a star, so you need a finder chart showing stars at least as faint as the asteroid.
If you haven't observed one before, try finding a
named asteroid
around magnitude 10 first.
In this close approach, 2017 VR12 will move 0.9 degree per hour north-south,
fast enough that you need either quarter-hour marks plotted on the finder chart or a continuously updated display.

Stellarium
works pretty well for me; I downloaded some of the extra star catalogs, and I import the latest available orbital elements for the asteroid of interest.
If you wish to use your scope's go-to capability,
this guide
may help, but you still have to figure out which "star" in the eyepiece field is the asteroid.
This article
identifies a few brighter stars you can try to watch it pass.

I do recommend getting a ~10mm eyepiece in addition to the 25mm one you have.
It might help you see stars half a magnitude fainter, but the main benefit is higher magnification for planets.


If a hot Jupiter collides with the parent star, does it enhance or reduce the life of the star?

The short answer is pela's comment: more mass equals shorter life.

For a bit of a longer explanation...

The only way the life of the star could be prolonged is if the new material can be transported to the core.  This requires convection.  Most stars are not fully convective, having distinct layers of either convective or radiative zones. The net effect of the collision in this case is to simply add more mass to the star as a whole while adding no new fuel to the core (or a burning shell); ergo, shorter lifespan.

But some stars do (or may) have fully convective stages, such as the main sequences of very small (aka red dwarf) stars and very large stars with enough metals for a dominant CNO cycle.  So in such a situation the material of the colliding body would be transported to (and from) the core, giving more fuel to burn.  However, the star still has the same, increased mass, regardless of where the "new mass" currently is inside the star, resulting in higher core temperature. The net effect of which is that the increase to the fusion rate exceeds the proportional mass increase.  So, shorter lifespan.

The basic rule here being that core temperatures are dictated by the gravity well of the star, which is dictated by the mass (and a few other GR contributions, but usually predominantly mass).  The more mass, the deeper the well. So adding more mass just means the core (and areas near it) are in a deeper gravity well, meaning the temperatures go up.  Stellar fusion reaction rates are approximately non-linear polynomials in the temperature; some go as far as being proportional to $T^{40}$, the 40-th power of the temperature!  It's this non-linearity which results in the fundamental relation of the short answer: more mass equals shorter life.  Increasing the mass by a given proportion increases the fusion rates by a (much) larger proportion.