How could time only have started with the Big Bang?

This is both a physical and philosophical question because it depends what you mean by "time".

From a physical perspective, time doesn't really exist. We know that things change, and we measure a rate of change in relation to other things that also change (like clocks) but how we define time as an absolute concept is not something we have a firm answer to. It may not exist at all. To understand the details fo this you need to learn a whole lot of quantum mechanics, string theory and theoretical physics in general, which will take about a decade, two degrees and a doctorate. Alternatively, I've been reading a great book called "The Order of Time" by Carlo Rovelli. He is a theoretical physicist who has worked for decades to understand this stuff and then written all about it with hardly any maths and some drawings of Smurfs. I would highly recommend it.

Also, from a more philosophical perspective, if the Big Bang started with a singularity in which time and space and all matter were compressed into one single point with no dimensions, then that point contained no information about what was before it. If we can never know what was before the singularity (or even if there was anything before), then there is very little point in discussing it, so we can choose to set that as the zero point of time and call it the beginning because anything that happened before that is irrelevant. Like starting a stopwatch to time a race when the cars set off - there was time before, but that's not important for the race so we call the start of the race a time of zero.


What is a parsec and how is it measured?

A parsec is approximately 3.26 times a light year or 206,265 Astronomical Units (AU).

According to "Cosmic Reference Guide" (Caltech), the word 'parsec' stands for "parallax of one arc second", and is (according to the website):

  the length of the long leg of a right triangle, whose short leg is one astronomical unit when the angle between the Sun and the Earth, as seen from an object in space (a star for example), is one arcsecond

Illustrated (from the link above):


Why do our clocks not go out of sync with the day over a year?

You are correct, and in fact the clocks are offset. But most people never notice this because of the way days are defined.

If you measure the "day" length by the stars (instead of the sun) you will see it's not 24 hours at all. It's around 23 hours 56 minutes. (Check this by going out at night, picking a bright star, and noting the exact time it passes due south. Then observe it the following night and note it passes 4 minutes earlier.)  By the way this shortened "day" is the sidereal day.

Each day, the Earth shifts along in its orbit (by about a degree, as you correctly observed in the question). So if you go out and point at the Sun at Noon every day, you are pointing in a slightly different direction. (I'll also neglect the Solar Analemma at this point too as it's just too confusing.)

Really, the "day" of 24 hours is not related to the true rotation period of Earth at all; loosely speaking it's just the average time between sunrises, or sunsets, or noons.


Inflation cosmology: slow-roll inflation versus tunneling between two vacua

I...think you've got it backwards? Slow-roll inflation is alive and well, unlike models which involve tunneling between two vacua. The model of inflation first proposed by Alan Guth back in 1980 was a tunneling model. But it had a serious problem: it didn't reheat. Tunneling from a false vacuum to a true one (or a lower-energy false one) wouldn't release any radiation. And we need that radiation, because we are that radiation, cooled down and condensed after nearly 14 billion years of expansion. The solution was slow-roll inflation: as you mentioned in your question, rolling down the potential well releases a great deal of energy, which in turn leads to the wide variety of matter and radiation in the universe today. Slow-roll inflation also does a nice job of predicting the tiny density fluctuations in the early universe that seeded the formation of all the structures in the universe. Wikipedia has more on all of this.

A small disclaimer: there's a lot of inflationary models out there in the literature (there's a hilarious partial list in footnote 3 of this excellent paper by Yadav and Wandelt), and for all I know, one or more of them involve vacuum decay. But the dominant paradigm for inflation today is definitely slow-roll.


How would we detect an Earth doppelganger planet?

Using current technology (and by that I mean experiments and telescopes that are available now) we have not detected an "Earth-like" planet and we would probably be unable to detect life on Earth even if observed from a distance of a few light years. Therefore there is currently no prospect of detecting life on an "Earth doppelganger". I elaborate below:

No planets like the Earth have yet been detected around another star. That is to say, none that have a similar mass, radius and orbit at 1 au (or close to it) from a solar-type star. With current technology, it is just out of reach. Therefore any directed search for life on an Earth-like planet wouldn't actually know where to start. If you can't detect the planet at all then there is absolutely no chance of looking at its atmospheric composition to look for biomarkers (e.g. oxygen along with a reducing gas like methane, or chlorofluorocarbons from an industrial civilisation - Lin et al. 2014). The only exoplanets for which atmospheric compositions have been (crudely and tentatively) measured are "hot Jupiters". - giant exoplanets orbiting very close to their parent stars.
A "blind" search could look for radio signatures and of course this is what SETI has been doing. If we are talking about detecting "Earth", then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio "chatter" and accidental signals generated by our civilisation.
The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from Cullers et al. (2000): "Typical signals, as opposed to our strongest signals fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star". Quoting from Tarter (2001): "At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...". The equivocation in these statements is due to the fact that we do emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches.

Hence my assertion that with current methods and telescopes there is not much chance of success. But of course technology advances and in the next 10-20 years there may be better opportunities.

The first step in a directed search would be to find planets like Earth. The first major opportunity will be with the TESS spacecraft, launching in 2017, capable of detecting earth-sized planets around the brightest 500,000 stars. However, it's 2-year mission would limit the ability to detect an Earth-analogue. The best bet for finding other Earths will come later (2024 perhaps) with the launch of Plato, a six-year mission that again, studies the brightest stars. However, there is then a big leap forward required to perform studies of the atmospheres of these planets. Direct imaging and spectroscopy would probably require space-borne nulling interferometers; indirect observations of phase-effects and transmission spectroscopy through an exoplanet atmosphere does not require great angular resolution, just massive precision and collecting area. Spectroscopy of something the size of Earth around a normal star will probably require a bigger successor to the James Webb Space Telescope (JWST - launch  2018), or even more collecting area than will be provided by the E-ELT in the next decade. For example Snellen (2013) argues it would take 80-400 transits-worth of exposure time (i.e. 80-400 years!) to detect the biomarker signal of an Earth-analogue with the E-ELT!

It has been suggested that new radio telescope projects and technology like the Square Kilometre Array may be capable of serendipitously detecting radio "chatter" out to distances of 50 pc ($\sim 150$ light years) - see Loeb & Zaldarriaga (2007). This array, due to begin full operation some time after 2025 could also monitor a multitude of directions at once for beamed signals. A good overview of what might be possible in the near future is given by Tarter et al. (2009).


Is Optical VLBI theoretically feasible? If not why not?

This question amounts to: is optical interferometry possible when detecting photons? The answer to this is yes. Many experiments have been done using interferometers where one photon at a time is passed through the apparatus and still reveals an interference pattern when detected on the other side.
The problems with optical interferometry are manifold: optical interferometry at the moment relies on bringing the signals together in hardware; there is an immediate problem with applying the radio interferometry technique of "taping" the data and doing the correlation offline, which is simply that the bit rate of the higher frequency optical light is far too high to do this at the moment. I have to admit ignorance as to whether this will be possible in the forseeable future. There is an obvious issue in recording the time of arrival of photons more accurately than the uncertainty principle allows for even the fastest optical transitions of $\sim 10^{-8}$ s, which is well below the resolution needed to sample frequencies of $\sim 10^{15}$ Hz. (Edit: the fastest detectors don't use transitions in atoms and can achieve timing of $10^{-10}$ s; still too slow by orders of magnitude.)
However, even if you could "tape" the data, there are many other problems to be overcome. These difficulties mostly relate to the coherence length and coherence time of phase changes introduced in the atmosphere above each telescope and the inability to use phase-reference sources as a result; and the inability to use phase-closure techniques unless the sources are very bright. These difficulties are explained in more detail in https://astronomy.stackexchange.com/a/35215/2531
To summarise, telescopes in different places observe the same object through different atmosphere. This introduces phase errors. These can be calibrated out in radio observations by using a nearby reference source. This is because the phase errors persist over a reasonably long time and are similar over a reasonably large ("isoplanatic") patch of the sky. For optical observations, the isoplanatic patch may only be a few arcseconds across, so there is rarely a bright reference source, and the atmosphere changes in any case on timescales of tens of ms.
Optical interferometers get round this by using phase closure techniques. The phase error can be eliminated by comparing the phase differences from at least three stations. But again, you have to do this on very short timescales and this means optical interferometry is limited to bright objects in order to get enough signal.
Where is the cut-off? Well certainly beyond the near-infrared, because those interferometers also use similar techniques (delay lines, phase closure) in their techniques.


Why doesn't Titan have a carbon dioxide atmosphere?

The chemistry of Titan's atmosphere is complex, with reactions occurring between carbon dioxide, oxygen, carbon monoxide, hydroxl, and other compounds. This means that carbon dioxide production and destruction takes place through a variety of reactions (Samuelson et al. 1983), some spurred by ultraviolet light from the Sun (and hence photodissociation). In other words, comparing Titan to Earth and Mars ignores a variety of rich chemical processes; right away, the comparison shouldn't hold.

All that said, the primary source of carbon dioxide loss on Titan is condensation, with photolysis playing a secondary role (Horst et al. 2008). With a surface temperature of 94 Kelvin, Titan is much colder than either Earth or Mars - well below the freezing point of carbon dioxide. Even if you were to add a large influx of carbon dioxide to Titan, much of it would very quickly (from an astronomical point of view) condense into solid form and fall to the surface. Indeed, much of Titan's surface contains significant amounts of dry ice.


Is there a way to factor age into the mass-luminosity relationship for stars?

What you are looking seems to be the vertical path of a star in the Hertzsprung-Russel diagram (HRD).

The only problem is that stellar evolution is pretty complicated. See here a few simulated trajectories for different masses and chemical compositions from  Bertelli et al. 2008

Especially before and after the main sequence (i.e., regular boring hydrogen burning) the luminosity evolution becomes extremely variable.

From the simulations above one can also derive respective age-luminosity relations (taken from  Danchi 2013):

Again, you can see that the relation is very complicate for young or old stars and is fairly constant during the main sequence. The problem is, that it changes too little during most of the main sequence. So for most stars even with good mass and metallic measurement you won't be able to accurately estimate age just from luminosity. In addition to that, there are still uncertainties in our models.

Another thing one could do is trying to improve the mass-luminosity-relation by including age. I think this is also what the title of your question implies. The problem here it simply that we generally don't know the age of a star. But if we do, as Rob Jeffries explained in his answer, it is in principle possible.

One think that is done to get age from luminosity is to go all in and calculate so called isochrones in the HRD. These are lines with stars of same age but varying masses and metallicities and can be derived from simulations. If one then measures the luminosity, temperature and metallic one can look on which isochrone the star falls in the HRD (and  therefore what age it has). This however is still pretty inaccurate, especially on the main sequence, and is mostly done with whole star cluster, where statistics makes things easier.

This is however not my field of expertise so I would be happy it an actual expert could chime in. :)


Where can I find the image of MY Camelopardalis?

The work on the recent paper was done using spectra and light curves, not via direct imaging.  There may not be a good image to view.  I checked the Hubble Legacy Archive and could not find a Hubble image of the object.


Your problem is that Jupiter is completely out of focus.  You need to turn the focus knob so that the bright ring gets smaller and smaller, and then you will find that the bit in the middle is bright, and is Jupiter.  Try turning the focus knob one way, and if the central dark circle gets bigger, then turn the other way, slowly.

What your pictures are showing is a magnified image of the central reflector that is located on the front corrector plate of your scope - that's the dark central circle - surrounded by the unfocussed light of Jupiter.

You could start by practising during the day, focus on distant objects (trees etc) and without the camera. Then try at night, again without a camera. Once you can use the telescope to see, try taking a photograph, again you can practise during the day. The moon might then be a good target before trying to get images of planets.

Good luck


Could an asteroid knock the moon out of its orbit?

Basically no and not by a long shot.   The asteroid's measurements are pretty vague, but lets give it a nice clean 600 meters in diameter and the same density of the moon.   The moon is 3,474 km in diameter or 3,474,000 meters.   So the moon is nearly 5,790 times the diameter and that gives is 194 billion times the mass as that asteroid.    Conservation of momentum, if the asteroid transfers all of it's 78,000 mph into the moon, with no energy going into angular momentum, 78,000 / 194,000,000,000 = .0000004 mph.   That's roughly the speed your fingernails grow.

To knock the moon out of orbit, it's velocity would need to be increased by oh, very rough ballpark, maybe 400-500 mph.   The additional velocity needed for the moon to reach escape velocity from the Earth is .414 x 2288 or about 947 mph, but given that it has an elongated orbit and that it would escape the Earth's sphere of influence well below escape velocity, maybe as little 400 extra MPH could knock the moon into a far enough orbit that it could escape the Earth.

Figuring 78,000 MPH, and you want to move the moon some 400 or more MPH with ah it just right, you'd need an object about a 5th or 6th the diameter of the moon or 600 KM in diameter give or take, assuming it was straight on impact as a glancing blow would lose a share of energy into angular momentum and assuming it hits the moon away from the Earth, not towards as towards would require even more energy.

This space pebble doesn't even come close.   You'd need something the size of a small dwarf planet to have a chance at knocking the moon out of the Earth's orbit and that speed.

300-600 meters is huge, if it hits the Earth.  That's probably a few square miles of earth practically leveled to the ground and broken windows and knocked down trees a good deal further than that.   If it hits the moon it would leave a nice crater, maybe a miles or two across but that's about it.

  It seems like you're only talking about the moon being kicked out into
  space. Would the same amount of energy be needed to knock the moon out
  of its orbit in towards Earth?

The mathematics of crashing is a bit more complicated.   Escape Velocity vs Orbital Velocity is easy, just straight multiplication.   More details here, and because the Moon wouldn't need to reach full escaple velocity, I cut the number in half - and it might be a bit less than that.

Moon's orbital Speed (2,288 MPH) x (Square root of 2 minus 1, or 0.414).

Crashing the Moon into the Earth - a bit harder.  There's probably a simple enough formula, but I'm not sure what it is, but I believe that would take even more energy.

It's discussed here without math and here, with math.  But it would take a huge amount of energy, probably 3 or 4 times more energy to push the Moon into the Earth than to push it out of orbit if I was to guess.


Third rotational angle in equatorial coordinates

This angle is dependent on the specific positioning of the eyepiece on your telescope. For models on which the tube can be rotated or custom-positioned, this means there are even more possibility—technically, an infinity.
However, no matter which telescope you use, the angle of rotation of the field of view will not change as objects move across the sky due to the Earth’s rotation if you are using an equatorial mount. Of course, using an altazimuth mount would make this angle change with time.
Usually, angles increase eastwards at the eyepiece. See for example https://en.wikipedia.org/wiki/Position_angle
Clear skies!


Why isn't the moon dark in a solar eclipse?

Light refracts in the atmosphere, causing the light to scatter in many directions. This light scattering is what causes our day to be blue, our sunsets to be their red-orange-violetish colors and allows us to see the light of the Sun shortly before sunrise and after sunset. As this only begins happening in any significant amount after the light enters Earth's atmosphere, there should not be any notable difference in the color or intensity of light between you and the moon and any other part of the sky at a relative angle from the Sun.

On the other hand, the foreground does not have direct light on the photographer's side (because it is nearly between the sun and the photographer) and there is a very limited amount of atmosphere to scatter light. As a result, less light will be coming to the camera from that direction.


Is the Apophis asteroid a concern for planet Earth?

Apophis is of no real concern to us as far as we can tell.  That said, it doesn't mean another asteroid doesn't have our name on it...

According to the most reliable data we have regarding this particular asteroid it only has a 1 chance in about 250,000 of hitting the earth.  University of Hawaii states:

  “Our new orbit solution shows that Apophis will miss Earth’s surface in 2036 by a scant 20,270 miles, give or take 125 miles,” Tholen said. “That's slightly closer to Earth than most of our communications and weather satellites.” He credits the large telescopes and superb atmospheric conditions on Mauna Kea for being able to make these determinations.

Considering how big space is, this is pretty much a hit in cosmic terms.  Keep in mind that this information has been readily available since 2009 at the NASA website as well.


At the Big Bang, when everything was close together, why did it not "collide", violating Planck length or Pauli Exclusion Principle?

The Pauli Exclusion Principle forbids two indistinguishable fermions occupying the same quantum state. It does not prevent them getting arbitrarily close together so long as they have very different momentum states.
The big bang model relies on classical General Relativity. When we go back to scales where quantisation of space might become important (i.e. Planck length scales, corresponding to the first $\sim 10^{-43}$ s of the universe) then GR needs to be replaced by some quantum theory of gravity (which we don't have).


Why does time get slow near a black hole?

The simplified answer is: Due to gravitational time dilation.

That's a consequence of general relativity. One fundamental idea of general relativity is the equivalence of inertial and gravitational mass, meaning forces induced by acceleration aren't distinguishable from gravity.
Persuing this principle together with the constant speed of light lead to Einstein's field equations. One of the solutions of these equations is the Schwarzschild solution for non-rotating, electrically neutral black holes.
This solution can be used to derive gravitational time dilation, as part of spacetime warp near mass/energy.
(Introductory video to general relativity)

A black hole provides a way to go faster into the future of a region of higher gravitational potential away from the black hole.
But the body near the black hole needs to withstand the high gravity. Circular orbits around a black hole are only possible down to 3/2 times, stable only down to 3 times, the Schwarzschild radius.


Where can I find data for fast radio bursts like FRB 180916?

There are parameters from the data listed here There is a site with raw data for FRB 121102, another repeating FRB, but the data set is 400 TB. The first site has links and may take you to the site where the data is stored. There is a Wikipedia page with your specific FRB listed among others. The sources listed for 180916 may take you to the data site.


Is it possible to get the distance to a star in the IPHAS DR2 catalog?

Only by knowing what kind of a star it is, or by inferring that information from the available colours. From there you use a photometric or spectroscopic parallax calibration. ie. There is a relationship between the absolute magnitude and colours of a main sequence star. Of course there is no guarantee you are looking at a main sequence star.

You can get more idea by cross correlating with the 2MASS JHK catalogue, as the J-H vs H-K colours of giants are usually distinctive.

More specific help needs a more specific question.


Position of Neutron stars in H R diagram

The HR diagram is an observational diagram. Whilst neutron stars could be placed in the HR diagram in the same way as white dwarf stars are, it turns out to be impractical to do so because the photospheric luminosity and photospheric temperature of neutron stars is next to impossible to determine. The reason for this is two-fold: (i) Neutron stars start off very hot (interior temperatures of $\sim 10^{10}$K and photospheric temperatures of $\sim 10^{7}$K, but they cool very rapidly. Within 10,000 years after the originating supernova they will have cooled below a million degrees, then photon cooling takes over from neutrino losses and they cool to a few thousand degrees within 10 million years (e.g. http://adsabs.harvard.edu/abs/2004ARA%26A..42..169Y ). There are many uncertainties and unknowns in these processes. (ii) the photospheric emission is usually dwarfed by emission from the magnetosphere or luminosity due to accretion from a companion or the interstellar medium.

One can theoretically work out where neutron stars should be by assuming that the emission is like that from a blackbody and that the radius $R \sim 10$ km.

In that case neutron stars lie on a locus defined by
$$ \frac{L}{L_{\odot}} = 1.9\times 10^{-9} \left(\frac{T}{10^{4}K}\right)^4 $$
So, contrary to what you you say in your question, most neutron stars could be cool and very, very faint and spend the majority of their (cooling) lives at the bottom-left or even bottom-right of the HR diagram. There is actually a huge uncertainty over where neutron stars would appear on this locus. Their very low heat capacities means that any "reheating processes" could very effectively raise their temperatures. Such processes include Ohmic dissipation of the magnetic field, some kind of thermalisation of their rotational energy or accretion from the interstellar medium. For the latter, luminosities of $10^{20}-10^{21}$ W may be possible, implying effective temperatures of tens of thousands of Kelvin.

A neutron star at the same temperature as Sirius would have an absolute visual magnitude that was about 22 magnitudes fainter, $M_V \sim 23$. Another way of visualising this is that the neutron star cooling sequence is roughly parallel with the white dwarf cooling sequence but about 13 magnitudes fainter.

You never see this locus shown on an HR diagram because it is usually way off the bottom of the plot.


Why do neutron stars collide instead of just revolving around each other like planets revolve around the Sun?

Neutron stars are one of the possible end products of the evolution of stars greater than around 8 solar masses.

If you start out with a close binary pair of these fairly massive stars -- not common, but not rare, either -- the more massive star will evolve to a red giant and tides (or even friction) in the extended envelope will pull them closer together.  The red giant will eventually supernova and may produce a neutron star. Later the originally less massive star will do the same, sometimes resulting in a very close binary pair of neutron stars. (Or sometimes a neutron star/white dwarf pair.)

The close binary will emit significant orbital energy in gravitational waves, causing the neutron stars to spiral towards each other and, eventually to coalesce.

The reason they don't just sit there orbiting stably like planets do is because they and very massive and very close together, and consequently orbit each other very rapidly.  Since the intensity of gravitational wave emission depends on the radiating body's mass and its acceleration, close neutron star pairs can radiate away their orbital energy in "mere" hundreds of millions of years.  The exact same effect happens for planets, but because of the lower mass and lower acceleration, the time to radiate significant energy is far longer than the current age of the universe.


Although you might think that Y-dwarfs (brown dwarfs with temperatures $<500$ K) might become cool enough to support liquid water. In fact that does not happen. As discussed by Burrows et al. (2003) and Morley et al. (2014), in Y-dwarf atmospheres, when the temperature becomes low enough for water to condense out of the gas phase ($<375$K), it goes straight to water ice particles.

Thus, although water vapour will condense  in clouds in a cold brown dwarf, it will be in the form of ice particles, and there is not expected to be liquid water in a brown dwarf atmosphere.


When was the concept of rogue planets first theorized?

I don't know the first use of the time rogue planet or the first use of such planets called something else.
But if science fiction is anything to go by, rogue planets are as thick as fleas in interstellar space and they are all heading for Earth's solar system.
The Wanderer (1965) by Friz Leiber involves a planet entering the solar system and causing devastation on Earth.
The Fifth Planet  (1963) by Sir Fred Hoyle and Geoffrey Hoyle involves another star passing close to ours, and an expedition to a planet of that star.
Battle of the Worlds (1961) is a movie in which a rogue planet called "The Outsider" enters our solar system heading toward Earth.
Back in the 1960s I read a juvenile science fiction novel in which a planet with human like inhabitants entered the solar system.  I don't remember the title or author, but it is in the future when Earth has space travel, and strange spaceships are seen. The boy protagonist becomes friends with a boy from the alien planet.  I remember a scene where an Earth spaceship was struck by an energy beam hear its needle nose, and incandescent vapor sprayed out from the ship.  I think that the aliens looked like humans with orange skin.  This book could be the Winston science fiction The Mysterious Planet, 1953, by Kennith Wright (Lester Del Rey).
In The Secret of the Ninth Planet, 1959, by Donald A. Wollheim, it is discovered that there are Plutonians and they are up to no good.  It is discovered that Pluto was lost from its original star system and wound up in our solar system.
In The Mysterious Planet, 1953, by Kenneth Wright (lester Del Ray), a new planet, Planet X, is entering the solar system and there are mysterious attacks by space pirates on Earth spaceships. This is a Winston juvenile science fiction novel, and possibly the one I mention two paragraphs above.
When Worlds Collide  (1951) is a movie based on the novel, Two rogue planets named Bellus and Zyra are entering the solar system and Bellus will collide with Earth.
The Man From Planet X (1951) has an alien land in a small village in Scotland. Apparently he is from a rogue planet passing through our solar system.
In Chapter One, "The Planet of Peril", of the movie serial Flash Gordon (1936) The planet Mongo is headed on a collision course with Earth.

The planet Mongo is on a collision course with Earth. Dr. Alexis Zarkov takes off in a rocket ship to Mongo with Flash Gordon and Dale Arden as his assistants. They find that the planet is ruled by the cruel Emperor Ming, who lusts after Dale and sends Flash to fight in the arena. Ming's daughter, Princess Aura, tries to spare Flash's life.

https://en.wikipedia.org/wiki/Flash_Gordon_(serial)
As I remember, Ming says he controls the course of Mongo.  When Ming says he plans to destroy Earth Dr. Zarkov asks him why destroy Earth, why not conquer it, and Ming agrees.  So presumably Ming changes the course of Mongo and the planets don't collide.
The Flash Gordon Sunday newspaper comic strip ran from 1934 to 2003, and a weekday Flash Gordon comic strip ran from 1941-44, and 1951-1993.

The comic strip follows the adventures of Flash Gordon, a handsome polo player and Yale University graduate, and his companions Dale Arden and Dr. Hans Zarkov. The story begins with Earth threatened by a collision with the planet Mongo. Dr. Zarkov invents a rocket ship to fly into space in an attempt to stop the disaster. Half mad, he kidnaps Flash and Dale and they travel to the planet. Landing on the planet, and halting the collision, they come into conflict with Ming the Merciless, Mongo's evil ruler.[1][3][9]

https://en.wikipedia.org/wiki/Flash_Gordon#Comic_strip_characters_and_story
When Worlds Collide  (1933) is a novel by Philip Wylie and Edwin Balmer, in which planets Bronson Alpha and Bronson Beta enter the solar system, with Bronson Alpha on a collision course with Earth.
In "Wandl the invader" Astounding Stories (1932, book  version 1961), by Ray Cummings, the tiny planet Wandl is sent from another star system to invade our solar system.
A comment mentioned rogue planets in works of Olaf Stapledon.  Possibly one or more might be in Last and First Men (1930).
In "A Brand New World" Argosy All-Story Weekly (1928, book version 1964) by Ray Cummings, a new planet with humanoid inhabitants wanders into our solar system. If I remember correctly, the new planet is named Xeneprene.
In "The Star" by H.G. Wells (1864), a star enters the solar system on a collision course with the Sun and causes great damage when it passes close to Earth.

This story is often credited with having created a science fiction subgenre depicting the impact event of a planet or star colliding, or near-colliding with Earth—such as the 1933 novel When Worlds Collide by Philip Wylie and Edwin Balmer (made into a film in 1951), Fritz Leiber's The Wanderer (1965), and Lucifer's Hammer by Larry Niven and Jerry Pournelle (1977).
However, it was preceded by two stories in 1894: Omega: The Last Days of the World by Camille Flammarion (the astronomer of the Flammarion Catalog) and Olga Romanoff or, The Syren of the Skies by George Griffith. In 1895, Griffith used an comet disaster again in The Outlaws of the Air.

https://en.wikipedia.org/wiki/The_Star_(Wells_short_story)
I note that the comets that strike Earth in the Flammarion and Griffith stories are not necessarily described as from beyond the solar system.  By the 1890s it was probably realized that the vast majority of comets are solar system objects in orbit around the Sun.
I believe that Edgar Allen Poe wrote a sort story in which humans became extinct or Earth was destroyed, possibly by collision with another astronomical body.  This may be "The Conversation of Eiros and Charmion" 1839, in which two spirits of dead persons discuss the past destruction of the Earth by a comet. I don't know if the comet was supposed to be a solar system comet or a rogue comet.
And those are all the examples I could remember or dig up.
In real life interstellar objects should often pass through the solar system, most of them tiny asteroids and comets ejected from their star systems.  A number of candidates for those objects have been suggested, and there are two confirmed cases: Oumuamua, an extra solar asteroid in 2017, and 21/Borisov, an interstellar comet in 2019.
https://en.wikipedia.org/wiki/Interstellar_object#Confirmed_objects


Sky photography through compositing images in software?

What you are talking about is making a 'mosaic'. This technique is commonly used in astronomy to stitch together many images that are off-set from each other of some (apparently) large astronomical object (e.g., a star cluster or a nearby galaxy). In addition to covering a large field of view, making a mosaic generally also improves signal to noise where the images overlap.

One way to realign the images is to uses cross-correlation to find the optimal shift for consecutive images in order to align them. This can be done in common data analysis languages such as python (free), matlab or IDL (not free).

I have not used black-box software to do this, but here is a list of programs that might help: http://astronomyonline.org/AstronomySoftware.asp#CCD_Control_and_Imaging_Software:_

Hopefully some astrophotographers here can recommend some software.


Well, you are not mistaken that travelling at low speeds you would never get there.
However, there are moves afoot to send (uncrewed) probes to the nearest stars at significant fractions of the speed of light and to send back pictures to the Earth - (e.g. the Breakthrough Starshot project). In which case, information might be returned to Earth on timescales of 100 years.


The stereographic projection
(Wikipedia,
Wolfram MathWorld)
is commonly used for all-sky maps
(e.g. Skymaps.com,
Your Sky).
As distance from the center of the map increases,
constellations grow larger but keep their shapes.


Where are Moon's Apogee and Perigee? Do they rotate too?

Yes they do rotate!
Apsidal precession is the rotation of the line of apses (line connecting apoapsis and periapsis and passing through the Earth). Lunar precession takes this line about nine years to rotate once around the Earth, referenced to the celestial sphere (the stars).

Source


Can we see individual stars in other galaxies?

Yes, Edwin Hubble did that for the first time in 1919. Before that time, it was thought that the galaxies we can observe were just nearby gas nebulae located inside our Milky Way. But Hubble was able to resolve the nearby galaxies like the Andromeda nebula into individual stars. By measuring the brightness of so-called Cepheid variable stars, he was able to calculate the distance to the Andromeda galaxy. Here one uses the fact that the total power radiated by the star is related to the period of the brightness oscillations, so by observing such stars you can deduce the distance to these stars and hence the distance to the galaxy. But later it was found that there were two different types of Cepheid stars and the wrong relation had been used; the distances were actually about twice as large.


Why is there "modulus" word in the "distance modulus" term?

According to Oxford Dictionary of English, the word "modulus" is the diminutive of the Latin modus, meaning measure (modus, in turn, comes from Proto-Indo-European mod-os, Nocentini & Parenti 2010.

Hence, the distance modulus is a "measure of distance", just like the modulus of a vector is a way of measuring its size, and Young's modulus is a way of measuring the stiffness of a material.


What's the oldest possible age of the universe? (LCDM parameter range)

Q: What would happen if you took today's six-parameter LCDM model to the 95% confidence level? Understanding that the number of permutations is high ...

NASA's Graphical Parameter Comparisons webpage has this to say:

Sources in the literature for values of these selected parameters determined between the years 1996 and 2016 are shown in the Data References Table. It would be a formidable task to include all results from all experiments and every paper. Therefore, results presented here are representative, and there are multiple caveats associated with those choices. For example, we do not guarantee that the datapoints are statistically independent of each other, although the choice of data sources is motivated to some extent by a wish to include independent determinations. There also has been no attempt to estimate what fraction of the parameter uncertainties are due to cosmic variance, although this should clearly be a consideration.
We discuss the individual parameters in pages that follow, and include an accompanying graphical history for each. Vertical gray lines shown in the graphical histories are weighted averages of the WMAP and Planck data points rather than the entire sample. This is done as a visual aid, rather than a statement that this is the current best value: the WMAP and Planck data values serve as a common thread in each of the plots. In a few cases, liberties were taken in the quoting of asymmetric error bars from published results as symmetric error bars on the plots. For illustrative purposes, the differences are negligible. Quoted uncertainties are 68% confidence limits.
Data references Table - Contributed by the NASA LAMBDA Archive Team.

The published data isn't accurate enough to reach the confidence level you require.
Prof. Wright of UCLA has a webpage about the age of the universe using actual age measurements, not estimates from cosmological models, but a more authoritative source would be NASA's Lambda Archive Team which has a page on $t_0$.
There are many ways to tweak the values and make the calculation, the computed value is somewhat dependent of the exact cosmological model assumptions. Within the standard ΛCDM model framework, CMB measurements, combined with BAO and SNIa measurements of the expansion rate, are presently in agreement and tightly constrained. The gray vertical line, representing the weighted average of WMAP and Planck data points, is positioned at $t_0$ = 13.796 Gyr.
This graphic from their website sums it up:


How far away are objects whose light will never reach us, because of the expansion of the universe?

There are several different quantities of this sort that you can define, and the definitions are fairly confusing. Hopefully the following diagram will make things clearer.
                   Z                      <- future infinity
                  / \
                 /   \
                /     \
      D   C    B   A   B    C   D         <- now
      .   .   /   / \   \   .   .
      .   .  /   /   \   \  .   .
      .   . /   /     \   \ .   .
      .   ./   /       \   \.   .
      .   /   /         \   \   .
      .  /.  /           \  .\  .
      . / . /             \ . \ .
      ./  ./               \.  \.
~~~~~~d~~~c~~~~~~~~a~~~~~~~~c~~~d~~~~~~   <- last scattering (universe opaque below this)

The horizontal axis on this diagram is "comoving distance", with respect to which objects that move with the Hubble flow are at rest. The vertical axis is "conformal time", with respect to which light travels along diagonal lines of a constant slope (in combination with the comoving distance).
A is our current location; Z is the ultimate location in the far future of the same matter, supposing it doesn't deviate much from the Hubble flow.
Light emitted in the current cosmological era from closer than B will reach us at some point in the future. Light emitted in the current era from farther than B will never reach us (if we don't deviate too far from the Hubble flow). The distance from A to B is around 16 Gly (billion light years). This seems like the closest match to the quantity you were asking about in your question. I'm not sure that it has a name.
Light emitted at the last scattering time from c is just reaching us now. This is the cosmic microwave background radiation. The distance in the present era from us to the presumed current location of the matter than emitted that light (if it hasn't deviated too much from the Hubble flow) is the distance from A to C. That distance is around 46 Gly. This is what's normally called the radius of the observable universe. Note that the distance from a to c is around 1100 times smaller, or 42 Mly.
Light emitted at the last scattering time from closer than d will eventually reach us. Light emitted at the last scattering time from farther than d never will (if we don't deviate from the Hubble flow). The present-day distance from A to D, which is the extrapolated current location of that matter supposing it doesn't deviate from the Hubble flow, is around 62 Gly. Again, the distance from a to d is around 1100 times smaller. (And the distance at future infinity is ∞ times larger.)
So your statement that "if an object is currently 62 billion light years away, or more, the light it is emitting right now will never reach Earth" is incorrect, but it would be correct if you replaced 62 by 16, or if you replaced "is emitting right now" by "emitted at the last scattering time, supposing it didn't deviate too much from the Hubble flow". Also, you should probably replace "Earth" by "whatever matter is left at Earth's location in the distant future, where 'location' is defined by the Hubble flow".


Building a floating, ocean-going giant radio telescope?

I would be extremely concerned about the ability of such a telescope to make adequately precise measurements, given the motion of the water. The leading radio telescopes have their mirrors and receivers very exactly aligned. For example, Arecibo's Gregorian dome could be aligned with any location on the order of millimeters, while the Green Bank Telescope's mirrors are aligned to the tenth of a millimeter. I would certainly be concerned about replicating this performance in the ocean. It's fine for an aircraft carrier's surface to wobble by a couple orders of magnitude more than that; it's less so for a world-class radio telescope. I would expect to see mirror deformation and pointing errors due to the motion of the water.
There's a basket of additional logistical issues - constructing the telescope on water, transporting personnel to and from the structure, combating temperature variations, protecting electrical components from short-circuiting, staving off large waves, etc. All of these drive up construction and operating costs, and astronomers aren't exactly rolling in cash.
I can admittedly see two possible benefits from the plan. First, radio frequency interference (RFI) from artificial sources would presumably be nonexistent throughout much of the open issue (although Connor Garcia makes the excellent point that the increased amount of water vapor could be problematic!). RFI is the bane of the radio astronomer's existence, which is why radio quiet zones are so important. As a counter point, of course, there are available formal and natural radio quiet zones which could be taken advantage of. Second, you could change the telescope's latitude, which would give it a wider range of sources. On the other hand, it's not as if any significant parts of the radio sky can't be seen by radio telescopes, so it's unclear as to the benefits of this.
Overall, I'd say that the precision issues outweigh the potential benefits. If the National Science Foundation were to fund future replacements for Arecibo, there are plenty of better options, some of which are already being planned. If an aquatic radio telescope were ever to be built, I'd guess that it would be entirely privately funded by a billionaire with money to spare who would be willing to take significant risks. I wouldn't complain about something like that, but at the same time . . . I think a conventional radio telescope would be a much better choice.


Can rogue/nomad planets and other starless planetary bodies orbit each other in a stable system?

One of the main hypotheses for the formation of rogue planets is ejection from a preexisting planetary system. This is an unlikely scenario for the formation of a cluster of rogue planets, because it doesn't seem feasible for a whole bunch of bodies to be ejected together and remain in a stable system. If multiple planets were ejected from a planetary system, I would expect them to exit randomly and separately.

That said, a couple sub-brown dwarfs have been observed to host what may be protoplanetary disks (although that term may not be applicable in these cases). The important objects here are Cha 110913-773444 ($\sim8M_J$) and OTS 44 ($\sim11M_J$). These objects have not been classified yet; they could be considered rogue planets or sub-brown dwarfs (or even brown dwarfs, honestly). Assuming the normal processes active in protoplanetary disks take place - collisions, accretion and the like - then in the future, additional bodies could form.

OTS 44 is thought to have a disk of around 30 Earth masses (Joergens et al. 2013); depending on the efficiency of planet formation, we might see bodies similar in mass to terrestrial planets (note that this also depends on disk composition; much of it is likely to be hydrogen and helium, not silicates). Giant planets seem unlikely, as the disks mass is comparable to those of Uranus and Neptune, within a factor of two, and it is highly unlikely that the entire disk would coalesce into one planet.

If a rogue planet formed in situ, rather than being ejected from a system, we do have a different picture to consider. In this case, it seems more likely for binary rogue planets to form. At this point, of course, the label "sub-brown dwarf" is more appropriate than "rogue planet".


Rewriting the equation for the virial mass of a cluster

This is something that crops up a lot in Astrophysics and Astronomy, so here is the general approach.

Get your equation into SI units. Thus
$$M = \frac{5v^2R}{G} = 7.5\times10^{10} v^2 R,$$
with $M$ in kg, $v$ in m/s and $R$ in m.

Now if you wish to replace $v^2$ by $(v/1000)^2$ (which is velocity expressed in km/s), you would simultaneously multiply the numerical constant at the front by $1000^2$. Similarly, to replace $R$ by $R/3.09\times 10^{22}$ (which is $R$ expressed in Mpc), then you have to multiply the numerical constant by $3.09\times10^{22}$ to leave everything unchanged. This then leaves your result still expressed in kg. To express it in solar masses, simply divide by a solar mass of $2\times 10^{30}$ kg.

Thus
$$\frac{M}{M_{\odot}} = \frac{7.5\times 10^{10}\times 10^6 \times 3.09\times10^{22}}{2.0\times 10^{30}} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$
$$\frac{M}{M_{\odot}} =  1.16\times10^{9} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$

Finally, check your result makes sense. Galaxy clusters are of order a Mpc in size, have velocity dispersions of hundreds of km/s and consists of hundreds of galaxies with total mass of $\sim 10^{14} M_{\odot}$ or so. This seems to check out - usually if a mistake has been made you will be many, many orders of magnitude out.


Are space telescopes completely out of the earth's atmosphere?

A handful of space telescopes are located in Langrange point L2, 1.5 million km from Earth. This is much farther away than the Moon, and far outside Earth's atmosphere.

WMAP and Planck, which measure the cosmic microwave background (CMB), are located here because Earth is a hundred times brighter than the CMB in this wavelength region. Herschel observes in the infrared, which is heat radiation, so it was also put in L2 to avoid Earth's heat. Others can be found in the list provided in Keith's answer.

Another advantage of putting a telescope in L2 is that the Earth and Sun are always in the same direction, making it easier to observe most of the sky. The cost of putting it here is probably more than an order of magnitude higher, though.

Whether or not low Earth orbit telescopes like Hubble at 550 km are outside Earth's atmosphere is a bit subjective. The atmosphere doesn't stop at some particular point, instead gradually thinning. These space telescope are definitely far enough away that they are not heated up. Neither does the atmosphere affect their observations (by seeing). They are, however, slowly decelerated and thus need a boost once in a while not to fall down.


How was Io not torn apart by tidal forces during its formation?

No, it is not just a matter of migration. You need to take into account two facts.

One is that (as experience shows) Io's own gravity is enough to avoid it breaking by tidal forces. It has been like that through all its history: Io could not have been formed if it started aggregating today, but it was formed at the same time Europa and Ganymede did: they three were growing in parallel.

Other is that of the orbital resonances, which makes precisely that orbit with so simple integer number relations with those of Europa and Ganymede a stable one. Io could not have been formed in another place.


Each time Halley's comet passes us, we can make a pretty good estimate of its current orbit, and determine how close it will get to the massive bodies of the solar system like Jupiter, Saturn, Uranus, or Neptune on its next orbit. We can make good estimates of gravitational perturbation effects, and thereby know where to look for it.

In fact, astronomers have been doing exactly that for centuries with remarkable accuracy:

  Halley thus concluded that all three comets were, in fact, the same object returning every 76 years, a period that has since been amended to every 75–76 years. After a rough estimate of the perturbations the comet would sustain from the gravitational attraction of the planets, he predicted its return for 1758.

  Halley's prediction of the comet's return proved to be correct, although it was not seen until 25 December 1758, by Johann Georg Palitzsch, a German farmer and amateur astronomer. It did not pass through its perihelion until 13 March 1759, the attraction of Jupiter and Saturn having caused a retardation of 618 days. This effect was computed prior to its return (with a one-month error to 13 April) by a team of three French mathematicians, Alexis Clairaut, Joseph Lalande, and Nicole-Reine Lepaute.

However, Halley's position turns out to be hard to predict for time scales much beyond that of a single orbit; over the long term, the orbit is chaotically unstable and it's possible that it will collide with a planet, or be ejected from the solar system, within some hundreds of thousands of years.

You mention:

  Its not a very big object in space so it seems to me that its orbit could be changed very easily.

As it turns out, the mass of an object in space doesn't affect how the gravity of other bodies changes its trajectory. A feather, a baseball, a comet, and a small moon making a close flyby of Jupiter along the same trajectory, one after another, will all exit Jupiter's neighborhood in the same direction and at the same speed. The small moon will have more effect on Jupiter than the feather will, but aside from that there's no difference.


Early universe's first starlight (indirectly) detected for the first time, could it have been detectable with 1970's technology?

I could not find a freely accessible version of this article, so I am mostly guessing here.
From this article https://arxiv.org/pdf/1206.0267.pdf (especially sections 4 and 5), the main challenge seems to be distinguishing between the different components of the signal:

  (1) the 21 cm signal coming from the high redshift Universe,
  (2) galactic and extra-galactic foregrounds,
  (3) ionospheric influences,
  (4) telescope response effects,
  (5) Radio frequency interference and
  (6) thermal noise
  Distinguishing between all these components means understanding each of them equally well and probably requires recent knowledge (in cosmology, radio receptors, etc)

In any case, the idea of using the 21-cm line to probe the early universe seems quite old:
https://academic.oup.com/mnras/article/188/4/791/999622


Time of day of the K-Pg asteroid impact?

Given that the uncertainty in deltaT, the difference between the uniform dynamical time needed to predict the positions of the Sun and the Moon and civil (solar) time was already 4.5 hours by year -4000 (https://eclipse.gsfc.nasa.gov/SEcat5/uncertainty.html), then I would say by 65 million years ago we would have no idea of which way the Earth was facing at the time. Unless the dinosaurs left eclipse records that would let us calibrate the much faster spinning Earth back then, we would have no means to work out what time of day the impact happened.


How do one identify something they see in the sky?

There exist many astronomical catalogs, reaching from rather simple catalogs like the Bayer catalog for stars, or the Messier catalog for galaxies, nebulae, etc., to huge catalogs with millions of entries. A catalog with about a billion objects is scheduled for 2022.

There exist also printed editions of some catalogs.

Objects in the sky have several properties, by which they can be described and mostly identified.

One of the most frequently used properties is the position in the sky. From the local geographic position and the time, the observed position of the object can be translated into a celestial coordinate system, e.g. the equatorial coordinate system. The position of an object is described in this system by declination and right ascension.

Two other frequently used properties are magnitude and color.

Also frequently used are patterns of stars, meaning their relative position to each other, beginning with star constellations.

The analysis of steller spectra allows a more detailed identification.

Longer term observations may uncover variability and proper motion as additional properties.

Here is a description of parameters (including properties) provided by the BSC5p catalog.


Is it possible to steer the sight of a Liquid Mirror Telescope using a plain mirror scheme? If yes, why hasn't it been done?

Designs where the curved primary mirror is fixed, and steering is achieved via a moving flat mirror (siderostat) in front of it (or a set of flats) have been done before. One example is the Pfund telescope. The main design constraint that led to the invention of the Pfund telescope was the need to have a focal point which is fixed in space and easily accessible.

Here's a variation of that design; the need for the fixed focal plane is driven by accessibility concerns here (the user cannot follow a moving focal plane):

Also see the related concept of heliostat - same idea, just applied to sunlight. A moving flat reflects sunlight in a fixed direction. Some solar power plants use heliostats.

So, in principle, fixed curved primary + moving flat is doable, and has been implemented a number of times. However, there are substantial issues with this design.

First off, the size and the quality of the flat mirror are not less than the size and quality of the parabolic primary. Let's say you build a liquid mirror telescope because it's easy. Then you have to make a flat mirror of the same diameter, with the same precision, and all the ease of construction is now lost. Flat mirrors are not easier to make than parabolic. In fact, as a mirror maker, I prefer to work on curved catopters - in some ways they are more predictable during manufacturing than large, high precision flats.

There are further issues down the road for the liquid mirror telescope. You would actually need 2 flats, both very large, both very precise. Both would have to be suspended in the air, above the liquid primary, and supported by complex mirror cells just like curved primaries in any telescope. Overall, it will be more difficult to make such a telescope, not less.

So why is the Pfund design used? Not for ease of manufacturing; as I said above, it's actually more difficult to make it than a regular scope. But in some situations you just need that fixed focal plane for some reason. In that case, you don't have much of a choice. Of course, it's not a design that scales well.

  I am assuming a large plain mirror is much cheaper than a parabolic
  mirror of the same size.

That was the incorrect assumption. Once that's taken care of, everything else should be clear.


How to understand the meaning of this International Occultation Timing Association (IOTA) lunar occultation of Mars map?

Unfortunately, there was a server problem in the days leading up to the exciting Mars occultation. (It didn't matter to me because it was, unfortunately, cloudy where I live.)

This description of the colored lines is from a temporary page provided by IOTA:

  The turquoise curves show where the disappearance or reappearance occurs at moonrise (left side) or moonset (right side); there is no northern limit as that misses the Earth to the north. The southern limit of the occultation, from which a partial occultation will be visible in a strip about 5 miles wide, crosses the n.e. Pacific Ocean and southern Mexico (white line, event at night), Central America (dark blue line that’s hardly visible, event during morning twilight), and northern S. America and the Atlantic Ocean (red dotted line, event during daylight).

If I am interpreting the image correctly, the limits are as follows (tl;dr 5: Entire occultation is visible)

No occultation is visible because the Moon has already set.
Mars disappears at moonset.
Mars disappears before moonset (so is visible), but reappearance occurs after moonset (so is not visible).
Mars disappears with the Moon above the horizon, but Mars reappears at moonset. (Entire occultation barely visible.)
Entire occultation is visible.
Mars disappears at moonrise.
Mars disappears before moonrise (so is not visible), but reappearance occurs after moonrise (so is visible).
Mars reappears at moonrise.
No occultation visible because the Moon has not risen.
Mars grazes southern limb of the Moon (occurs during daylight).
Mars grazes southern limb of the Moon (occurs during twilight).
Mars grazes southern limb of the Moon (occurs during night).

Sometimes the colors along the graze line (the southern limit in this case) represents whether the graze occurs on the dark limb of the Moon or the bright limb, but that does not appear to be the case here.


Assuming an open model of the universe and a long enough timeline galaxies which appear "normal" today would become more red shifted over the course of observation.  I put normal in quotes because there are no galaxies which are relatively static to ours.  Any galaxy which has movement will have some "shift" to them, and all galaxies are moving away from us.

It is possible that eventually the light from them could move into the infrared and even the microwave in extreme cases.


No. This is not true.
A star is much larger than a planet.  A star has nuclear reactions in its core, that prevent it from collapsing under gravity.
A planet is much smaller. It is small enough for "normal" pressure to be strong enough to stop gravity from making it any smaller, and there are no thermonuclear reactions at its core.
A star cannot form back to a planet. A planet cannot be "revived" back to a star.
(There is one class of bodies that may be star-like when young and planet-like when older: "Brown dwarfs".  These are objects with a mass of 15-75 times the mass of Jupiter. These can support deuterium fusion when young, but not hydrogen or even lithium fusion, when the deuterium is used up they don't have enough mass to collapse to a dense stellar remnant, and so become "planet-like".  Really they are objects sui generis, neither stars nor planets. Non of the solar system planets are brown dwarfs, and you can't revive a brown dwarf.)


As @Donald.McLean said in comments, the answer is yes, the Moon can and does occult the other planets in the Solar System. When something apparently big (like the Moon) passes in front of something apparently small (another planet) it's called an occultation. (I say apparently because from our perspective the Moon appears larger than the planets.)

The planets lie* in the ecliptic plane whereas the Moon's orbit is inclined at about 5° to the ecliptic plane, so the Moon crosses the ecliptic plane twice each orbit, so it's not as frequent as it would be if the Moon's orbit were in the ecliptic plane.

Occultations of the planets by the Moon can be found here. There is also software (Occult) that you can use to generate your own predictions.

* Although the planets are said to lie in the ecliptic plane, this is a generalisation - they actually have orbits that are within a few degrees of the ecliptic "plane".


Are planet-destroying quakes possible or observable?

A planet-ripping earthquake isn't possible on Earth.  For a magnitude 10.5 Richter scale earthquake, one would need twice the fault lines that exist on Earth according to the USGS.
The same website relates energy $E$ in joules to earthquake magnitude $M$ with Baths equation: $log E = 5.24 + 1.44M$.  A 10.5 magnitude earthquake would produce about 2.29e20 joules of energy.  The Earth's gravitational binding energy is about 2e32 joules. So if all possible earthquakes went off at the same time twice, it would only generate a trillionth of the energy necessary to rip the planet apart. Earthquakes occur just under the surface of the earth, and they release their energy over time. So, even if an earthquake could generate 2e32 joules of energy, it wouldn't be sufficiently directed to destroy the planet.
Earthquakes are the product of built up energy in tectonic plates by convecting magma.  This is a small effect compared to the power of gravity. The IAU requires a planet to have "sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape." That means gravity is a dominant force over the rigidity of the materials that compose the planet. But the seismic moment is governed by the modulus of rigidity of the materials. So localized earthquake forces and energies will be small compared to the gravitational forces from the planet itself.  So, we wouldn't expect planet ripping earthquakes on exo-planets either.
Notes:

If we consider nuclear fusion as a source of quakes, then supernovas might be called star-ripping quakes.

If we consider collisions as a source of quakes, then the asteroid 16 Psyche might be an example of a proto-planet destroyed by earthquakes.


Is Earth's orbital eccentricity enough to cause even minor seasons, without axial tilt?

Very cool question. I want to get into a little bit of detail here because otherwise there would be a one-paragraph answer, and I don't think that would cut it. So here goes.

The planets in the solar systems have orbits with pretty low eccentricities (see this for more eccentricity values). At the upper end is Mercury, with an eccentricity of 0.2056. At the lower end is Venus, at 0.00677. Earth is in between but moderately low, at 0.0167. The distance between perihelion and aphelion is 5 million kilometers - in an orbit with an average radius of about 150 million kilometers. Note, though, that eccentricities are always changing.

We would certainly have "seasons" if there was no axial tilt, but they most likely would not be dramatic. As Wikipedia says

  Because of the increased distance at aphelion, only 93.55% of the solar radiation from the Sun falls on a given area of land as does at perihelion.

But on this page, it says

  Orbital eccentricity can influence temperatures, but on Earth, this effect is small and is more than counteracted by other factors; research shows that the Earth as a whole is actually slightly warmer when farther from the sun. This is because the northern hemisphere has more land than the southern, and land warms more readily than sea.

So more land means that the planet can absorb heat better, which counteracts the change in distance.

But I think that's pretty boring. Don't you? So let's calculate how eccentric Earth's orbit would have to be for it to have seasons without axial tilt. From the chart here we can see that the highest average hemispherical temperature is 22 degrees Celsius, while the lowest is 8 degrees Celsius.

What about other planets?

There's a steady downward trend as we get further away from the Sun (Venus is an exception because of its runaway greenhouse effect). For Earth, the average surface temperature is 287 K (15 degrees C). So our maximum would be 295 K and our minimum would be about 281 K (Kelvin = Celsius + 273). Compare that to the other planets.

I made a graph, which unfortunately I cannot paste here, that shows that to have a surface temperature of 295 K, the planet would have to be - well, not much closer than we are. Relatively, that is. Roughly 0.975 AU from the star. To have a surface temperature of 281 K, it would have to be at 1.05 AU from the star. This neglects, by the way, the different levels of land/sea absorption.

So, actually, if the Earth's orbit had an eccentricity of about 0.35, it could have some moderate seasons.

I hope this helps.

Note: I used this eccentricity calculator for the sake of accuracy and time. I would still be obliged if anyone could check my overall calculations for this answer.


What is the difference between albedo, absolute magnitude or aparent magnitude?

Albedo is a measurement of reflected light. It varies between 0 and 1, with 0 being completely black (reflecting no light) and 1 being completely white (reflecting all light). The albedo of coal is about 0.04, which is actually very similar to the albedo of our moon. The Albedo of ice is about 0.7. Only planets and other small objects are described using Albedo because they cannot make their own light, and so its apparent brightness is due to reflecting sunlight.

Absolute magnitude is a measure of how much light a star is emitting overall. Absolute magnitude is related to a bunch of other properties of the star, mainly how massive it is and at what stage of its lifetime it's in.

Apparent magnitude can be used to describe both stars and planets. It is simply a measure of how bright the object appears to us. An illustrative example is Sirius and Canopus. Sirius is the brightest star in the sky, but it is bright because it is close, only being 8 light years away. Canopus is the second brightest star in the sky, and appears about half as bright as Sirius, but is about 300 light years away, which means the star itself must much brighter than Sirius. If we check the absolute magnitudes and do some maths (Sirius absolute magnitude = +1.42, Canopus = -5.71, the more negative the number the brighter the star, brightness difference = 2.5^(5.71+1.42)) we get a brightness difference of 680 times between Canopus and Sirius.


Is there a connection between black holes and dark matter/energy?

No, not really.

As Stan Liou hints at in his comment, massive objects such as black holes (BHs) were previously a candidate to dark matter (DM). But it can be shown that in order to explain the various problems that the existence of DM needs to solve, there would have to be so many, that their gravitational effect ("microlensing") on background sources would be much larger than what is observed. Today, most people think that DM is simply a particles with an extremely small interaction cross section, such that it virtually doesn't interact with other particles, except through gravitational attraction.

You also ask if there's any spatial relationship, and in a way you could say there is. Most, if not all, galaxies sport a supermassive BH, and these tend to lie in the center of the gravitational potential wells created by the galaxies' DM. "Normal" BHs, however, just float around like stars.

BHs also tend to grow by accretion of matter onto them, and of course they can also eat DM, so to some extend BHs also consist of DM. But since the DM doesn't interact, a DM particle would need to fly directly toward the BH, lest it would simply be deflected and escape. Normal matter, on the other hand, is slowed down by friction and spirals toward the BH. Thus, BHs consist mostly of normal matter. It doesn't really matter, though, since after something has fallen into a BH, everything but its mass, charge, and angular momentum will be forgotten.

Dark energy is something completely different, better described as a negative pressure which is a property of space itself. I think. Or we think. Nobody really knows. But it presence doesn't correlate with that of BHs or DM. Ten years ago, there was a hypothesis that a collapsing star should be converted into dark energy rather than a BH, causing a so-called dark-energy star, but I don't think anybody believes this anymore.


Could a moon orbit an Earth size planet in such a way as to create a solar eclipse lasting for several weeks from start to finish?

Short answer: No, it could not.

The moon already has little enough mass relative to the earth so that its orbital period is not much shorter than a satellite with negligible mass. So, any object at about the distance of the moon would take around a month. If the object really had negligible mass it would be moving a little slower than the moon, so let's be generous and say 5 weeks. That means the object would have to be large enough to extend 1/5 of the way around the sky to block the sun for just one week. This is not possible. Such an object would be much larger than the earth.

If you imagine an object closer to the earth, it would be orbiting faster so it would have to be even larger.

Moving farther away is a little better. The orbital period is proportional to the radius raised to the 3/2 power, so moving the moon four times farther away would make its period eight times as long. But it would have to be four times larger, or 64 times the volume, just to eclipse the sun at all. And at that point it's approaching the size of the earth.

It may be possible to construct a synthetic object that orbits the sun at earth's L1 point (between the earth and the sun at just the right distance to stay in line with them) and permanently blocks the sun. You could tweak its solar orbit so that it only blocked the sun part of the time. It could be something like a mylar disk that rotates to avoid collapsing into a ball. But this becomes more of an engineering question than an astronomy one. No plausible natural object would have this configuration.


What does the 'amplification factor' of gravitational microlensing mean?

Gravitational lensing is just geometry, as is optical lensing. The amplification factor describes the increase of the area of the image (at constant surface brightness).

BTW, the amplification factor does depend on the mass $M$ of the lens, since $u=\theta/\theta_E$ and $\theta_E$ depends on $M$. Here $\theta$ and $\theta_E$ are the relative angular distance between source and lens and the radius (angle) of the Einstein ring for the lens, respectively.


Is it correct to assume that the relationship between an objects mass and size, is what decides if it’s a black hole or not?

Yes. A black hole is in some ways a very simple object. It has mass, charge, angular momentum.

These three values determine the size of the event horizon. If the Earth's mass were compressed to less than the size of a cherry it would be inside its event horizon and so would be a black hole.

The gravitational field of the Earth out at the moon would be (almost) unchanged. The moon would orbit an Earth-mass black hole in the same way that it orbits the Earth now.  The small differences are due to irregularities in the shape of the Earth, which wouldn't exist if the Earth were collapsed.

Above the Earth's surface the curvature of space would be (almost) the same if the Earth were to be collapsed. If a rocket were to use its thrusters to hold itself stationary 6400km above the black hole, the inhabitants would experience 1g of gravity, just as we do standing on the surface, 6400km from the centre of the Earth.

If you go far enough below the Earth's surface, gravity weakens, because there is mass above you.  If you collapsed the Earth, then gravity (and curvature of spacetime) would increase all the way to the black hole.


If the normalised filter response function is $R_{\nu}$ then the measured  flux is
$$ F = \int f_{\nu} R_{\nu}\ d\nu  $$
The integration is done over the frequency range of the filter.

If you measure a flux through a filter then the process cannot be inverted exactly. However the average flux density can be found by dividing the total flux by the effective frequency range
$$\langle f_{\nu} \rangle = F/ \int R_{\nu} d\nu $$


Why would a planet form a cometlike tail under conditions of weaker rather than stronger solar wind?

I think the first link that you sent has an interesting explanation that we can extrapolate from - here's the relevant section:

The new observations settle a debate about how the strength of the
solar wind affects the way in which ionospheric plasma is transported
from the dayside to the nightside of Venus.
Usually, this material flows along a thin channel in the ionosphere,
but scientists were unsure what happens under low solar wind
conditions. Does the flow of plasma particles increase as the channel
widens due to the reduced confining pressure, or does it decrease
because less force is available to push plasma through the channel?
“We now finally know that the first effect outweighs the second, and
that the ionosphere expands significantly during low solar wind
density conditions,” says Markus Fraenz, also of the Max Planck
Institute and co-author on the paper.

What I interpret is this: while solar winds do cause plasma to form a tail, or stream, a very strong solar wind will limit how voluminous that tail can be. Thus, a slightly weaker solar wind would have less of a limiting effect on the volume of the flow of plasma.
This pressure-reducing effect outweighs the effect of there being less force to actually move the plasma - as said:

“We now finally know that the first effect outweighs the second, and that the ionosphere expands significantly during low solar wind density conditions"

So, this comes down to which effect is stronger - is it the expansion of the tail due to less confining pressure, or the decreased force pushing the plasma into a channel/tail? Observationally, it seems to be the first.
Though, from here, I would also assume that there is some sort of lower limit - or equilibrium - at which the size of the tail/channel reaches its maximum. If the solar wind were far too weak, there would likely not be a tail at all. So, rather than "lower solar wind causes a larger tail", I think this phenomenon would be better phrased as "excess solar wind can decrease the size of a comet-like tail, specifically around planets". It's not that lower solar wind causes a larger tail, it's that there is a sort of mid-point at which the solar wind creates optimal pressure and conditions for the tail to reach its maximum size.


Relativistic effects in stellar dynamical systems

Stellar clusters around supermassive black holes are systems in which relativity likely plays a role. Currently, only bright stars can be seen in our own galactic center because there is a ton of neutral gas between us and the galactic center that obscures it. As a result, we only have a few "test particles" out of the many stars that actually orbit the black hole at close distances.

Nevertheless, measuring relativistic precession may be possible for a star with one of the closest-known pericenter distances to Sagittarius A* (the central black hole in our galaxy), S2, potentially within the next few years once enough data has been collected.

As to how relativistic effects can affect dynamics of the cluster, the precession induced by general relativity can suppress resonant interactions, including three-body resonances such as the Kozai. Depending on if these sorts of resonances are important compared to other relaxation processes, the relaxation time can increase significantly, resulting in the cluster evolving more-slowly over time. This can affect things such as the rate of mass segregation, tidal disruptions, and production of hypervelocity stars/S-stars.


What are the prospects for follow-up observations of phosphine on Venus?

This figure is from the paper "Phosphine as a bio signature gas in exoplanet atmospheres".  It shows the absorption cross section of Phosphine compared to other molecules. We can see that Phosphine has a distinct enough profile from the others molecules in the 7.8-11.5 microns range, with the exception of NH3.  Probing from 2-11.5 microns should then allow to distinguish Phosphine from other molecules. Simulations seem to indicate that current state of the art
nIR ground-based spectrometers could detect phosphine to a minimum of a few ppb, so expect a lot of follow-up observations from the ground in the coming months.


Creating an HR diagram from tabulated stellar data

With the Gaia Data Release 2, you can plot an HR diagram with a few million stars. From the Gaia website :

The data can be accessed in the search tab, where you can specify the columns you want to display (phot_bp_mean_mag and phot_rp_mean_mag). There is a button at the bottom to export the results to a file on your computer. You can then plot one of the columns as a function of the difference between the other two.


Get visible galactic longitudes in astropy

The following works for me; it uses the rootfinder from scipy to find the zero.

from astropy.coordinates import EarthLocation, AltAz, Galactic
from astropy.time import Time
from scipy.optimize import newton
import astropy.units as u

dwl = EarthLocation(lat=6.39688*u.deg, lon=52.6275*u.deg)
mytime = Time('2017-09-16T21:00:00')

def altitude_of_longitude(longitude):
    ''' Give the altitude (in deg) of a galactic longitude (in deg) '''
    return Galactic(l=longitude*u.deg, b=0*u.deg).transform_to(
                    AltAz(location=dwl, obstime=mytime)).alt.value

Now use the rootfinder from scipy, with two different initial guesses, hoping that they end up at the two different solutions where the galactic plane crosses the horizon. This gives:

newton(altitude_of_longitude, 0)
>>> 16.231753986033333
newton(altitude_of_longitude, 180)
>>> 196.23547201834657

The only remaining part is to check which part is visible:

altaz = Galactic(l=0*u.deg, b=0*u.deg).transform_to(AltAz(location=dwl, obstime=mytime))
altaz.alt
>>> −9°08′51.8623′′

Apparently the longitude at 0° is under (negative altitude), so the visible part is between 16° and 196° galactic longitude.


Who first observed another planet rotating?

Mercury's surface was virtually unknown until Mariner 10 flew by and Venus is covered in clouds. Of the terrestrial planets, that leaves Mars.

It was likely Christiaan Huygens in November 1659 who first realised Mars was rotating when he tried to map it. His estimate of roughly 24 hours was fairly close.

I believe Galileo's earlier observations on Jupiter were with a telescope not powerful enough to resolve smaller 'surface' features (like the Great Red Spot), which may have either proven that Jupiter rotates, or at least that something on the surface is moving across it.


Computing radial-velocities from Cross Correlation data

Why not fit the orders you are interested in separately and then use the standard error of the mean (possibly weighted by the signal-to-noise in each CCF) as the precision in the final, averaged RV.
In terms of what to fit, I don't see why a Gaussian is so bad? You probably need to limit the fit to the inner $\sim \pm 1 \sigma$ to avoid noise outside the peak pulling the fit one way or another. Other options are to just use the numerical centroid (but this may be affected by asymmetry) or you could use a sinc function to the central region (sometimes a better model of the peak of a CCF).
Edit: You haven't followed my advice, which was to limit the Gaussian fit to $\pm 1$ sigma from the peak (do it iteratively). At the moment, the "wings" of the Gaussian are just adding noise.
As a rule of thumb though, you are not going to do much better than $2.2\sigma$  divided by the signal-to-noise ratio. It looks like your $\sigma \sim 2$ km/s and your signal-to-noise ratio is about 30, so I don't see how you can have gotten a scatter as small as you have from data like this? And the graph below contains no points with the value $-10.76$. There is still something missing from your question.


What is the dimensions of the smallest object detectable by an optical fiber from a specific distance?

Assuming you are using single mode fiber, the entrance (mode diameter) is small, usually a few microns (extremes can range from 1 to say 30 microns, but 2-8 microns is common for visible SMF).

The only constraint: you have is to get enough light into that small spot. You can do that with a single lens or convex mirror, a microscope, or a telescope, or even by proximity (hold the fiber so very close to a tiny object that enough light reaches the core, works for certain nanoparticle or biological experiments).

Size of target object: So far I can not think of any constraint on the size of the object. Really the question is how bright the object is and how sensitive the instrument is at the other end. If you have a high-dispersion spectrometer, you probably need a lot more light than if you are doing broadband photometry by counting photons with a photomultiplier.

Object is moving: It doesn't matter if the object is moving or fixed, as long as your optical system can move to track it so that the focused image falls on the core of the fiber. That's simply a mechanical and tracking issue not related to the fiber.

The big advantage of using a fiber is now obvious, you can move your telescope, but keep your measurement instrument somewhere else and fixed, and the flexible fiber will guide the light to it.

Further background

The trick is to put a very large telescope between the object and the fiber, and to use a robot to place hundreds or thousands of fibers at the focal plane, each at the exact location of a galaxy or a star you would like to collect. This way you can line up the other ends of the fibers along a single slit of a spectrometer and study the spectra of thousands or even millions of objects over time without having to point the telescope to each object one-at-a-time.

Fibers have some big advantages:

Multiplicity and configurability (as mentioned above)
Flexibility (literally) it allows the telescope and the instrument to be mounted separately and not have to be mechanically aligned and stabilized.
Mechanical stabilization of source for instruments: even if the telescope drifts and the object moves slightly with respect to the entrance of the fiber, the physical location of the fiber exit can remain fixed. So tracking errors don't affect the distribution of intensity within the entrance slit (virtual or real) of the fiber and therefore don't cause instrumental spectral shifts.

above: From the astrophysicist Ángel R. López-Sánchez's enjoyable A 2dF night at the Anglo-Australian Telescope

above:: From AAO's The stellar cluster behind Sirius

above: GIF from the video (with exciting music) A 2dF night at the Anglo-Australian Telescope

And if you like that robot then consider the "magic fingers" at the focal plane of the Dark Energy Spectroscopic Instrument:

  DESI’s focal plane is designed to sit high atop the Mayall Telescope and to carry 5,000 robotic positioners, each holding a fiber-optic cable. Each of these fiber-toting robots is automatically positioned to fix on a preset sequence of individual galaxies and quasars so that the fibers can collect their light. The movements of these positioners must be carefully choreographed to avoid bumping into one another. The roundish focal plane, which measures nearly a meter in diameter, consists of 10 pie-shaped wedges that are fit snugly together. Each wedge holds 500 robotic positioners. The focal plane also contains sensors and light sources called field fiducials that help ensure the positioners are in proper alignment.

And also these robotic fibers in Australia's TAIPAN instrument at Siding Spring Observatory in North-Western NSW:


To find the angular diameter of a satellite you need to find
$$\arctan\left(\frac{\text{diameter of satellite}}{\text{distance to satellite}}\right)$$
As you have been using the diameter of the planet, your formulae are wrong.
You should also make use of the small-angle approximations $\arctan(x)\approx x$ for small x in radians. So to get the value in degrees you can do
$$\frac{180^\circ\times\text{diameter of satellite}}{\pi\times\text{distance to satellite}}$$
And you should round your values. You don't need 15 significant figures of accuracy!


Determination of High-Accuracy Distances of Terrestrial Planets from the Sun

The DE430 JPL ephemeris "memo" (https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf) has details of how it is constructed and what from. Mercury and Venus's positions are tracked by tracking the spacecraft orbiting them (to sub-km accuracy). Mars, Jupiter and Saturn are also measured from the orbiting spacecraft such as Galileo and Cassini. The outer planets plus Pluto positions are primarily done from astrometry so measuring the positions of the planets relative to the background stars, in the past through photographic plates or with CCDs in more modern times.

Positions of stars and planets are all measured in something called the International Celestial Reference System (ICRS) which is based on measuring very distant and fixed on the sky objects called quasars using a precise radio technique called VLBI.


What is the angle between the planes of Oumuamua's hyperbola and the Milky Way Galaxy?

The inclination of the orbit is 122.7 degrees (with respect to the ecliptic), whilst the Galactic plane is inclined at 60 degrees to the ecliptic. So there is 63 degrees between the two.

Such a high inclination by no means rules out that the object is orbiting within the Galactic disk. A study by Mamajek (2017) concluded that it's approach trajectory was entirely consistent with an object travelling at the median velocity of stars around us. It had insignificant vertical or radial motion with respect to the local standard of rest, but is travelling about 10 km/s too slowly for a circular orbit at the Sun's Galactocentric radius (whereas the Sun moves about 10 km/s too fast).

This appears to be discrepant with the inclination angles discussed above until you recall that the sun is moving up and out of the Galactic plane at around 7 km/s.

Have a look at the animation at https://www.cosmos.esa.int/web/gaia/iow_20180925 , it says a thousand words. It shows time running backwards from the time of encounter and shows how Oumuamua comes in from inside the Sun's Galactic orbit and from "above", with respect to the Galactic plane.


Why do some neutron stars become pulsars?

My understanding is that it would be quite usual for the majority of neutron stars to be pulsars shortly after they formed.

The requirements are that they should have strong magnetic fields and rapid rotation (faster than a rotation period of a few seconds). The latter seems an almost inevitable consequence of conservation of angular momentum and the former must surely be assisted (if not fully explained) by conservation of magnetic flux as the core of a massive star collapses (see the discussion in Reisenegger 2007).

But to be observed as a pulsar, the spin and magnetic field axes have to be oriented in a specific way, so that Doppler-boosted and beamed radiation sweeps over the Earth. This requirement means that we can observe only a fraction of the pulsars out there.

In fact there are a category of "radio quiet pulsars" - these are detected by their high energy radiation pulsations but not seen as radio pulsars. The idea here is that the beaming is less extreme for the high energy radiation, allowing it to be seen.

The pulsar phenomenon is also transient. Once the rotational kinetic energy of the pulsar has been extracted and it spins down beyond the "pulsar death line", or if magnetic field decay occurs rapidly, then the pulsar should switch off. Most of the billion (or so) neutron stars in our Galaxy are no longer pulsars. The slope of the death line is such (e.g. see Arons 1999) that even if the pulsar is born with a relatively weak field ($10^{6}$ T !) it should still be born with a rotation period short enough ($<1$ s) to produce the pulsar phenomenon initially.

You might find the discussion/simulations in this paper interesting.


Very large (for amateur) scopes in very dark places, with healthy eyes on a good night have credibly reported seeing color in nebula. The Orion is about as good as it gets,
lots of photons, my eyes are not particularly good, but have been out with others who claim noticing color in the Trapezium in a thirty-ish inch reflector


Looking for an equation that describes how star (x,y,z) positions are affected by expanding universe

The position of stars change very slowly for a couple of reasons, but not due to an expanding universe.

Galaxies distant from our own, are all moving away from our galaxy at a rate proportional to their distance. The equation is simple:
$$\mathrm{speed}=H_0 \times \mathrm{distance}$$

and $H_0$ is a constant of proportionality, with a value of about 70 km/s per kiloparsec. Since the galaxies are moving away, they don't change their position in the sky at all.

(The nearest galaxies are exceptions to this rule, they may be moving towards us)

Stars in our galaxy are also moving. They have their own proper motion, but the motions of the stars relative to our own are essentially random. You can look up the proper motion of stars on the SIMBAD database.

Stars also appear to wobble due to the annual rotation of the Earth about the Sun and our consequentially changing perspective. The amount of wobble is in inverse proportion to their distance (the nearest stars appear to wobble most)


What you're trying to do is change the pixel scale, but ultimately, you need to decrease the resolution of your image. I'm quite sure this can be done using a variety of image editing tools (even some free ones), but if you insist on doing this yourself, I can show you the general process. Let's say I have the image below, which for convenience is only 6x6 pixels in size.

I'll assume this image has a pixel scale of $p = 1\:\frac{''}{\text{pix}}$ and I want to convert it to an image with a pixel scale of $p'=1.5\:\frac{''}{\text{pix}}=1.5p$. This is effectively saying that I want the resolution of the image to decrease by a factor of $1.5$ and that each pixel in the $p'$ image will be comprised of $1.5^2$ pixels from the $p$ image. This is easily done by defining the new pixel grid over the old one and determining the new pixel "color" as the weighted mean of the old pixels which exist in your new, larger pixel.

For convenience, I'm going to define a coordinate system where the pixels are at position $(x,y)$ where $(0,0)$ is the top left pixel and $x$ and $y$ increase to the right and down respectively. This is a common coordinate system for images in programming.

Let's look at pixel $(0,0)$ in the $p'$ image. This is comprised of pixels $(0,0)$, $(1,0)$, $(0,1)$, and $(1,1)$ from the $p$ image. Specifically, the fractions of these pixels in our new pixel are $1$, $0.5$, $0.5$, and $0.25$ respectively. These fractions will become the weights.

What you need to do then is look at your four pixels in the $p$ image and determine their color/brightness/property-you're-trying-to-merge. For this simple case, I'll just use a number from zero (white) to one (black) that represents the magnitude of the pixel, but the same process applies no matter what your pixel information contains, be it an RGB value or whatever else. By looking at my four pixels in the $p$ image, I see the magnitudes are $0.05$, $0.05$, $0.25$, and $0.15$ respectively.

The new pixel's magnitude is then the weighted average of these.

$$p'_{00,\text{mag}} = \frac{1\times p_{00,\text{mag}} + 0.5\times p_{10,\text{mag}} + 0.5\times p_{01,\text{mag}} + 0.25\times p_{11,\text{mag}}}{1.5^2} = \frac{1\times 0.05 + 0.5\times 0.05 + 0.5\times 0.25 + 0.25\times 0.15}{1.5^2} = \boxed{0.1056}$$

The process is the same for each pixel in the $p'$ image. Hopefully you can see how this process would generalize for any new pixel scale. The hard part is determining where your new pixels will be and which pixels of your original image will comprise the new pixel. Once you know that, you simply average them together, weighted by the fraction of the original pixels that make up the new pixel. With a little bit of effort, this could all be fairly easily automated by a program and you'd only need to input the new pixel scale and the original image and it could output a new image at the new pixel scale.

One problem you might run into is that your new image can't be subdivided perfectly. E.g., if, in the above example, you wanted a new scale of $1.4$, you'd still wind up with a new $p'$ image of 4x4 pixels, but you'd only be able to cover 5.6x5.6 pixels of the original image and you'd necessarily have to cut off part of your image, thus losing some information. It'd be up to you to decide what you cut off.


Hydrogen and oxygen only react when there is sufficient energy. For instance, the autoignition temperature of hydrogen at 1 atmosphere is 536 °C. This is why you can do that experiment with mixed hydrogen and oxygen in a balloon, that only explodes when you touch the balloon with a lit taper.

Space is cold. Molecular clouds have temperatures in the tens of degrees Kelvin. In the rare (if not impossible) case that a molecular cloud of pure oxygen did exist, and it encountered a molecular cloud of hydrogen, there wouldn't be enough energy for an explosion.


A number of brown dwarfs have had 'surface maps' created using the light from those stars.
In 2013, observations of 2MASS J22282889–4310262, a brown dwarf 35 light years away, were published.  These were made using the Hubble and Spitzer space telescopes and were able to show changing light patterns and distinct layers of material at different altitudes in the atmosphere.  This provided evidence for textures but not necessarily banding in the atmosphere.  I  believe the Hubble has carried out other observations of brown dwarfs with the aim of determining their surface details.
In 2014 Luhman 16 A and B, a brown dwarf binary system, 6.5 light years from Earth, were imaged using the VLT CRIRES spectrograph.  Luhman 16B was found to have patches perpendicular to the plane of rotation.  Luhman 16A was found to be featureless at the frequencies observed.
Earlier this year, a paper was published describing how the VLT had been used again, this time employing polarimetry, to detect Jupiter-like banding on Luhman 16A.
The paper here describes modelling based on observations that indicate Jupiter-like cloud bands (page 13).
So there does seem to be mounting evidence that brown dwarfs have stripes.  The accuracy of the artists' impressions is something we'll probably find out in the future.

UPDATE - January 2021
A new study of Luhman 16 A and B by Apai, Nardiello and Bedin has been published.  They applied a 'novel photometric approach' to TESS data and found that we are viewing them at angles close to their equatorial planes and that they look 'strikingly similar to Jupiter'.  The paper at the Astrophysical Journal is here but is paywalled.

Last update, I promise!
This is an image from the Apai, Nardiello and Bedin paper that was reproduced at Centauri-Dreams.org.  The preprint of the paper is available for free download at Arvix.


Can you assume atmosphere height for the purpose of surface pressure calculation?

It all depends on what you are trying to do. For some purposes 6H might correspond to an adequate vacuum  for others not. The density is not low enough to ignore its effects on a space craft (we are talking of $\approx 48$ km in the case of the Earth). Indeed $48$km is less than the altitude record for an unmanned balloon, though greater than the altitude records for air breathing aircraft.


Horizontal to Equatorial Conversion - Code

HorizontalToEquatorial would be like EquatorialToHorizontal in reverse, something like this:

(x, y, z)_hor = rectangular form of (alt, az)

(x, y, z)_sid = (x, y, z)_hor rotated by geographic latitude

(ha, dec) = spherical form of (x, y, z)_sid

ra = local sidereal time - ha