What reference frame do age-of-the-universe calculations assume? I'm thinking, in particular, about general relativity. When we speak, for example, of neutrino decoupling, what do we mean when we say this happened in the first second after the Big Bang? Do we mean one second from the point of view of an imaginary observer subject to the same relativistic effects on time? How much can we even say about the affect of relativity in this context?

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

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

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

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

  [...]

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

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


Jupiter-Venus conjunction is vertically aligned. I'm at 49° N. How is this possible? The very beautiful Jupiter Venus conjunction of ~ 2019-01-20 (practically something out of "2001 A Space Odyssey") is seen as "vertically aligned" in the dawn sky, Venus "above" Jupiter.

See this picture by Jeff Majewski at EarthSky.com to see what I mean (not sure I can copy it in here directly)

This should mean my "vertical" is aligned with the Ecliptic. I'm in northern latitudes (49° N), how is that possible?

And is this alignment geometrically linked with the "Super Wolf Blood Moon" which was very visible at the same time?

Update

The problem seems hard to explain so here is a little free-hand drawing.

The observer sits on the equator and is just being moved into sunlight by Earth's rotation (the image shows Winter Time for the Northern Hemisphere). The observer's "UP" vector lies in the Ecliptic Plane. This being so, and Jupiter and Venus also being in the Ecliptic Plane, the observer can see Jupiter and Venus aligned "vertically" (same azimuth), separated by a couple of degrees in elevation. As the day progresses, that vertical alignment will evidently cease as the line connecting the planets rotates relative to the observer.

This works on the Equator, but what if the observer is at higher latitudes? I can't get my head around the idea that the planets can be vertically aligned while still being on the Ecliptic Plane.

A combination of several movements can lead two objects in the sky to appear vertically aligned. The planets are constantly moving relative to each other, they are orbiting on slightly different planes, the Earth is orbiting and rotating, so such configurations between planets can occur frequently.

An observer does not need to be in the tropics to see two planets being vertically aligned at some point. The planets orbit close to the ecliptic, but their orbital planes are slightly inclined relative to ours. Jupiter's orbit is about 1° from the ecliptic, and Venus is about 3°. This means that both planets can move above or below the ecliptic by a few degrees. When they pass each other as seen from the Earth in a conjunction, they are rarely on top of each other, most of the time, they are separated by a few degrees, and they are not exactly on the ecliptic line. (Such occurences of planets occulting or transiting each other do happen, but are rather uncommon: the next time Venus will transit in front of Jupiter will be on 22 November 2065)

I'll add a few pictures to help visualizing the positions, movements and alignments involved here.

Here is a view of the sky on 20 January at 49° North latitude:

As you can see, Jupiter and Venus appear pretty much vertically aligned in the morning sky. The yellow line is the ecliptic. Note that the ecliptic does not pass through the planets here. Jupiter is pretty close, but Venus is a few degrees north of it. As both planets pass each other in the next days, at some point, their alignment will be perpendicular to the ecliptic (called the ecliptic conjunction), as we can see from this January 22 simulated image:

But they won't appear vertical anymore from your vantage point. (In fact, they would apprear vertical again later during that day, because of Earth's rotation, but the Sun will be up, so it wouldn't be observable directly)

A few days later, on January 26, they will appear horizontally aligned!

(Images generated in Stellarium)


Why isn't the day backwards between leap years? So every four years, we add an extra day. A leap year. This is because a year is actually $365_{1/4}$ days long. But what happens between leap years? After two years, it would get dark at 8:00 AM instead of 8:00 PM. This is obviously not the case. What's going on here?

Leap seconds keep the clock in sync with solar time of day. Leap years keep the calendar in sync with the seasons. The two (leap seconds and leap years) have nothing to do with one another. We have leap seconds because a day is now a tiny bit longer than 86400 seconds. We have leap years because a year (tropical year) is about 365.24219 days long.

If we didn't have leap years but did have leap seconds, solar noon and noon by the clock will continue to occur at more or less the same time (ignoring cyclical changes such as the equation of time and daylight saving time). The calendar would become out of sync, becoming almost a month off from the seasons in just a century. Adding an extra day every four years doesn't quite do the trick; this is why 2100, for example, will not be a leap year.


Can a star emit heat but no visible light? Can a star emit enough heat to allow human life on its planets, however emit no visible light? If so, what problems would this pose?

The minimum temperature of an object classed as a "star" is something like 2700 K. Such an object, although emitting the bulk of its radiation in the infrared, would still emit something like a few per cent of its energy at visible wavelengths.
Note that the visible and near infrared spectra of cool stars are nothing like blackbodies, so if the calculation is important then you would need to look carefully at empirical or model spectra.
Lower temperature objects do exist - these are called brown dwarfs. Some have (a subset called Y dwarfs) been found that are only a few hundred kelvins, so a planet could orbit not far above their surface and receive a kW per square metre of infrared light.
A plausible scenario might be a brown dwarf of spectral type T7 (a T-dwarf), with a temperature of about 1000 K (e.g. Burgasser et al. 2002). Such an object has a luminosity of only $\sim 10^{-5} L_\odot$, so a planet would have to orbit at about 0.003 au (about 0.5 million km) to receive the same bolometric flux as does the Earth from the Sun. Such objects cannot be described as "invisible" at visible wavelengths, but a typical $R-J$ (red minus near infrared colour) is about 9 or 10 - indicating that there is thousands of times more flux in the near infrared than there is in the red part of the spectrum. An example of the spectrum of such an object is shown below (GL229b is a T7 brown dwarf). Note how the spectrum is nothing like a blackbody.

Edit: There is likely a limit to how close a planet can get to its parent body, depending on a comparison of its density with the density of "star" it is orbiting. Too close and it will break up due to tidal forces, but for an Earth-like planet, whch would be less dense than a brown dwarf or of similar density to a very low-mass star, then that limit is either about a stellar radius above it or very close to the surface respectively.
Edit 2: NB cool white dwarf stars might be considered, but the oldest, coldest ones in the Galaxy are still around 3500 K and their spectrum encroaches significantly into the visible band. It is possible though that the most massive white dwarfs ($>1 M_\odot$) in old binary systems could be cooler than 3000 K, though no definitive detection has been made (e.g. Kaplan et al. 2014).


How do satellites impede current telescopes? As we send up more and more satellites, it stands to reason that our view of the universe becomes more and more obscured. I would certainly expect for example that a large telescope such as the LSST would be able to see the International Space Station, and since the ISS is not transparent, it wouldn't be able to observe the patch of the sky beyond it.
How do current telescopes currently deal with satellites obstructing their view? Do they simply not observe where the satellite is? If we send up more and more satellites, does it mean that telescopes will have to focus on narrower and narrower fields of view to avoid the swarm?

Satellites, even in geostationary orbits, move with respect to the background stars and make "trails" on telescopic images that are tracked at the sidereal rate.
Removal of these can be as straightforward as taking the median of a set of exposures.
What I mean by this, is that in a sequence of exposures, the satellite trail will pollute different pixels in each exposure. Therefore if you take a sequence of exposures, tracked to keep stellar positions identical, then take the pixel-by-pixel median of the sequence of frames, the presence of the satellite will be removed. i.e. Each x,y pixel in your stacked image is formed by taking the median of the values of that pixel (scaled by the exposure time if necessary) in each image. It is essentially the same procedure by which cosmic ray strikes are removed.


Observation of Gravitional waves by LIGO from binary black hole merger I read the paper "Observation of Gravitational Waves from a Binary Black Hole Merger"
https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102

I tried to understand all the graphs and data observed in the paper.
I am curious how these data and graphs conclude that the gravitational waves observed were because of binary black holes merger? Also what calculations and equations led to the same?
Also how will the calculations and equations look if neutron stars were there instead of black hole in the same merger?

(graphs and data can be seen in the observation section in the paper)

Neutron stars are less massive but larger than black holes. The frequency of the gravitational waves is due to the orbital period of the objects as they spiral together. A neutron star merger would have a different orbital speed, so the frequency would be different and lower. It would rise at a different rate. The signal from neutron stars is also much weaker, so couldn't be detected from the same range.

Prior to the detection of the signal, numerical simulations of merging black holes of various sizes were done, so the theoretical shape of the waveform expected from the merger of black holes (and neutron stars) was known. It was possible from this theoretical work to estimate the mass of the two black holes taking part in the merger by matching the observed signal to a theoretical one. You can see the theoretical signal for black holes with 36 and 29 in the diagrams labelled as "numerical relativity"


How big can a crater get? What is the biggest possible crater from an asteroid impact? Not asking for the biggest existing crater in the solar system. I should be able to find plenty of lists online. What I'm curious about is how big a crater can form before the impact results in both bodies reforming destroying near all evidence of the crater itself.

I don't mind erosion or normal geological processes  destroying the craters. That is just a matter of time.

Eg when the Mars size object hit the Earth. Both were substantially altered and incorporated in each other. Earth experienced a near complete reformation (in the spherical sense). Also resulted in the moon's formation. Obviously that sort of impact doesn't leave an 'obvious' crater.

I assume the makeup (mineral, size, speed, angle of impact etc) of both bodies will change the answer, so maybe a just a short list of sizes for the various xommon combinations?

Eg what would be the biggest crater possible on Earth before the Earth experienced severe...Um, reshaping?

Theoretically, if you're talking crater to planet size, I (think) a crater that covers half or a bit more than half of the planet is possible if you have a dense planet (Say, Mercury) and a lighter, more icy object (say, Titan).   Mercury would break Titan apart like a hammer hitting glass (er, maybe), but Titan, which is slightly larger than Mercury but weights only 41% as much, with a lower velocity impact, might not make the entire surface moulton and could leave an impact crater over 50% of the surface with debris scattered unevenly leaving something like an impact crater.   That's, granted, pretty speculative, but I think that type of impact, a rocky world and an icy one of similar size might provide a collision that could create an impact crater a bit larger than 50% of the planet's surface.   Such an impact would be interesting to observe, but . . . I probably won't see one in my life time, unless they get some very good telescopes out there and get lucky.

Theoretically if you count surface melt, it's arguably possible to have an impact that would melt 75% or more of the surface but leave 25% or less intact, but I don't consider surface melt the same thing as an impact crater, so I wouldn't count that, but that's optional.

As far as largest known craters, if we measure crater diameter against planet or moon radius (I'm going to ignore depth cause depth is easier on lower gravity bodies), but just diameter to radius.

Hershal on Mimas (the death star crater)  (Props to Andy)

Diameter of Hershal, 139 km.    Radius of Mimas, 198 km.   Ratio, 70%.

But there's an even bigger one.

The South Pole–Aitken basin on the moon.

Article here

It went undetected until relatively recently because it's large enough that it no longer looks like a crater, but it is, and it affected the Moon's density where one half of the moon has a measurably thicker crust than the other half. (The crater half, I presume, the crust is thinner).   This makes orbits around the Moon weird cause the moon is gravitationally quite lumpy.  (Proximity to the Earth also makes orbits around the Moon weird, but that's another story.

This ancient crater has a diameter of about 2,500 km compared to the radius of the moon of 1,737 km a ratio of about 144%, over twice the relative diameter and 4 times the relative size relative to the crater on Mimas.

and Mars might have an even bigger one.   See here.  It's size isn't listed, but Mars (like the Moon) has a "split personality" where one side has a measurably thinner crust than the other side and one possible, perhaps likely cause of that is a giant impact.   This "Crater" if we want to call it a crater covers about 1/2 the planet.   This is different than the theorized Earth-Theia impact, which liquefied the surface and left no visible crater, but it may have liquefied half the surface and not left a very visible impact.     One problem with measuring craters as they get this large is they look less like craters.


Is it possible that we see our solar system as a star? A few days ago I just read here that our galaxies (one of which we are part of) move or expand faster than light,
And read here that light we see from stars is extremely old, maybe more than several years.

If galaxies' expansion is faster than the light of stars, then is it possible that we see our own Sun as a star somewhere because light may reach us later than expansion?

I think the original answer has confused you somewhat - it is spacetime that is expanding faster than light - when viewed from sufficiently far away - we certainly are not moving away from the Sun at anything even remotely close to the speed of light (in fact the expansion of spacetime is not a motion at all in the true sense, but a change in the underlying metric).


Kepler's equation and eccentric anomaly From the wiki article about eccentric anomaly follows:

$$\cos E = \frac{x}{a}$$
$$\sin E = \frac{y}{b}$$

where E - eccentric anomaly, a - semi-major axis, b - semi-minor axis, P = P(x,y) a point on elleptic orbit. So if you know an E, you can find a position on an orbit:

$$x = a \cos E$$
$$y = b \sin E$$

However, from article on Kepler's eqution follows:

$$x = a (\cos E - e)$$
$$y = b \sin E$$

Please, help to understand why these formaluae differ? I expect them to be the same.

Both formulae are correct. The discrepancy is because the formula from the eccentric anomaly article uses the centre of the ellipse as the origin, but the formula from the Kepler's equation article uses a focus of the ellipse (i.e, the central gravitating body, eg the Sun) as the origin. Note that $c = ae$ is the distance from the ellipse centre to a focus


Future redshifting and effect on the 'pitch' of CMB radiation After discovering this question exploring the sound of a blackbody, I started wondering about the sound of the Cosmic Microwave Background radiation from the Big Bang, specifically what the current pitch might be at its peak frequency and how that might evolve over time due to redshifting. Here's what I've been able to deduce so far:

$$
pitch_{CMB} = 160.4\,GHz \approx D_{33} = 157.7\,GHz
\\
pitch_{CMB}\,lowered\,by\,30\,octaves = {160.4\, GHz \over 2^{30}} = 149.38\, Hz \approx D_3 = 146.83\,Hz
$$

Building on this result:

How much time will need to elapse before the CMB redshift will cause this pitch to drop by an octave (or even by a semitone)?
How much variability would this pitch have depending on the dipole anisotropy of the CMB from our perspective as well as due to more generalized anisotropies in the CMB?
What pitch would the peak value of this blackbody have had at earlier points in the history of the Universe?

I've tried to apply some of the information in this question and elsewhere, but would greatly appreciate hearing from someone with more expertise in the subject area if this unorthodox comparison has piqued your interest like it has mine.

The wavelength of the CMB increases linearly with the scale factor $a$, which is defined as 1 today, so the "sound" of the CMB has dropped an octave when the Universe has doubled in size (in all three directions), i.e. when $a = 2$, which will happen when it is roughly 25 billion years old (see e.g. Fig 1 of Davis & Lineweaver 2004).

The magnitude of the dipole of the CMB is $3.4\,\mathrm{mK}$ (Kogut et al. 1993), i.e. $\sim10^{-3}$ of the peak value. The primordial fluctuations are even lower, roughly a factor $10^{-5}$ of the peak value. The dipole is partly due to our motion round the Milky Way, and so will fluctuate somewhat with the period of a Galactic year, which is 225 million years.

The figure shows how the peak frequency evolved in the past. Its current value is 160.4 Ghz. To calculate how a human would perceive the sound at a given time, I suppose you'll have to multiply the spectrum with the frequency-dependent transmission function of the human ear, but that's physiology which I know nothing about.

The figure was produced by calculating the frequency at a given time as
$$
\nu(t) = \frac{160.4\,\mathrm{GHz}}{a},
$$
where $a$ runs from $1/1100$ to $4$,
and calculating the age by integrating the Friedmann equation, assuming a Planck 2015 cosmology.


Moment of inertia factor of Venus and Pluto Wikipedia's Moment of inertia factor begins:

In planetary sciences, the moment of inertia factor or normalized polar moment of inertia is a dimensionless quantity that characterizes the radial distribution of mass inside a planet or satellite. Since a moment of inertia must have dimensions of mass times length squared, the moment of inertia factor is the coefficient that multiplies these.

The article lists values for 18 solar-system bodies including the Ceres and many moons.
But Venus and Pluto are conspicuously absent from this table.
What are the RI factor values for Venus and Pluto? Is there any source of reasonable calculation done for these two?
C/MR2
According to this note on the page Venus has a range of 0.327–0.342. Pluto is not mentioned anywhere.

Measuring the moment of inertia of a planetary body is not straightforward and simple, but instead it requires a lot of data about how the body rotates, models for how mass is distributed across the body, and the gravity field of the body (these are related things, i.e. the gravity field can tell you how mass is distributed).  We have so many satellites in orbit of Earth and so much ground research that we know this well enough to measure the moment of inertia of our own planet.  The same is reasonably true for the Moon, and more recently for Mars.
The MESSENGER spacecraft had a highly elliptical orbit around Mercury for several years and so was also able to provide data to reasonably estimate Mercury's moment of inertia, too, but to far less precision:  Based on the Wikipedia article from the question, the Moon's moment of inertia has a relative uncertainty of ±0.2%, while Mercury's is ±4% (for completeness, Mars is ±0.1%).  Mars' is better known, despite having had fewer orbiting spacecraft, because it has two moons that help.
Reading through the paper on Mars' moment of inertia gives some idea of everything you have to do to really get at this value:  Love number, second-degree gravity field, correct things for atmospheric drag (including tides within the atmosphere), account for orbital changes in the moons, map the precession rate, measure the solar torque since Mars is oblate, look at all of Mars' orbital parameters (eccentricity, obliquity, mean motion, spin rate, precession) ... it's a lot.  Once they had all that, with their associated uncertainties, they could calculate the moment of inertia.  But, they suggested this could then be modified based on mass asymmetries across Mars, such as the giant Tharsis volcanic region providing a load on that side of the planet, which could tweak one of the terms that goes into the calculation and change the moment of inertia from 0.3644 to 0.3652.
I didn't provide links to what all the terms mean in the above paragraph (someone can edit if they wish), the real point is that we need to know a lot to actually measure the moment of inertia of a planetary body, and we simply don't have those for most bodies in the solar system.  The Wikipedia article that's linked in the question does list values for 17 (not 18) bodies, noting that Venus is unknown (hence 17 bodies), but then the other 13 are listed as "not measured" but rather calculated through other means.
For example, several list the Darwin-Radau equation method, which relates the body's shape and spin to moment of inertia.  Since we can relatively easily see those, we can guess at the moment of inertia via those means, but it makes a lot of assumptions about the interior structure.  Since the interior structure of Venus is likely more complicated than something like Io, whomever wrote the Wikipedia article chose not to list them in the table, though two references are given with values of 0.327–0.342.
The internal mass distribution of Pluto is also quite complicated given what we know about Sputnik basin and the glacier within it.  We also don't even have a full shape model for Pluto or Charon because of the lighting geometry during the New Horizons flyby in 2015, where south of about 45° was in permanent night for this part of Pluto's orbit (same with Charon).  Without even the shape, you can't use the Darwin-Radau equation to estimate the moment of inertia, though based on limb profiles no flattening was detected so I suppose some estimates could be made.  But, like the Venus estimates, how accurate those would be is debatable might not be at all useful.
Keep in mind that the moment of inertia for a perfectly uniform, solid sphere is 0.4, so a planetary body is going to be less than that (unless it's perfectly uniform and solid).  Values less than that indicate the mass is more concentrated toward the center, but even with Earth's large metallic core, the value is still 0.3307, which is not hugely different from 0.4 when you're trying to precisely measure all of these different values that go into the calculation.  So, if you calculate something like 0.35±0.05 for Pluto (to make up numbers) based on your limited data, is that really useful?  Maybe.  Or, maybe not.


Position of Neutron stars in H R diagram Why is that Neutron Stars are never depicted in an H R diagram? They can be placed in the bottom right corner but you will never find any diagram in literature showing neutron stars.

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.


How long do other planets in our solar system have total solar eclipses? Jupiter, Saturn, Uranus and Neptune — can all have total solar eclipses so how long do they last?
https://www.livescience.com/amp/60037-do-other-planets-have-solar-eclipses.html

TL;DR: Last for couple of seconds except for Jupiter where it lasts for minutes.

Mars

Phobos and Deimos are too small to totally eclipse sun but there are annular eclipses. See this video which Curiosity Rover took on Mars September 13, 2012 where Phobos is eclipsing Mars.

Jupiter

Only four moons are large enough to create totality: Io, Ganymede, Europa, and Callisto. But three can pass between Jupiter and the Sun at the same time. A triple eclipse happens on the gas giant once or twice every 10 years. Last for some minutes. For more information, see OP's answer.

Saturn

Seven of Saturn’s 62 moons (Janus, Mimas, Enceladus, Tethys, Rhea, Dione and Titan ) are big enough to create a total solar eclipse. But because of Saturn axial tilt of 26.7 degrees, solar eclipses on Saturn are much more rare than solar eclipses on Jupiter (only occurs once every 15 years). This article discussed about eclipses by Pandora and Epimetheus (the former one lasted for split second and the latter one lasted for 15 seconds)

Uranus

Twenty-seven moons circle Uranus, but more than half of them are too small or too far away to completely cover the Sun.  Also, eclipses can occur only when the ring plane-crossing of Uranus (equinox) occurs, approximately every 42 years, with the last crossing being in 2007/2008.

Neptune

Neptune’s six inner moons and Triton can create a total solar eclipse. Because of the planet’s distance from the Sun, Neptune only receives 1/900th of the light Earth does. At 0.018 degrees in apparent diameter, the Sun looks like a dot. That’s why Neptune’s solar eclipses start and end in a matter of seconds. Also, eclipses of the Sun from Neptune are even rare due to the planet's long orbital period and large axial tilt of 28 degrees. In addition, the largest moon, Triton, has an orbital inclination of about 25 degrees to Neptune's equator and has a retrograde orbit. So, eclipse occurs for only a couple of seconds.

Pluto

Pluto also experiences total solar eclipses but every 120 years, Pluto and its largest moon, Charon, take turns eclipsing each other once a plutonian day: that’s approximately 6.4 Earth days. But because Pluto only shows one side of its face to Charon, solar eclipses only happen on that half of the planet.
Source: https://astronomy.com/news/2017/06/total-eclipses-planets


Determine sun azimuth and elevation for shading a window throughout the year I was wondering how I could get the sun's azimuth and elevation values to shade a particular window with a roller shutter.

At first I thought azimuth is good enough; of course, it wasn't, as I found out in practice.
While there are calculators to find a and e for a specific location at a certain date, I lack the maths to figure out when the sun will hit a western window.

The window faces 270 deg, I know how far the eaves stick out, and the top and bottom edges of the window (and of course location).

I get the idea the maths will be complicated... however, (it won't matter much) as I want to program this into a controller to operate the shutter.

Any hints appreciated.

You can use a package such as the pyephem package for python. It can calculate the position of any astronomical object.

However for the sun since it moves along the ecliptic, roughly at a uniform rate, you can approximate the position by doing coordinate transformations. The sun's longitude moves by 365.24/360 degrees a day from 0 at the vernal equinox and has an ecliptic latitude of 90 degrees. This can be converted into equatorial coordinates (right ascension and declension), and then, given the hour, the alt and az positions can be found. The formulae can all be found on https://en.wikipedia.org/wiki/Celestial_coordinate_system


Is it difficult to see DSO in your eyepiece? I want to know if bringing the image of a DSO in eyepiece generally represents a problem for beginners in astronomy.

It's a pretty broad topic. It depends a bit on the type of DSO. Speaking in general, most DSOs lack brightness, and so to observe them you need two things:

lack of light pollution
lots of aperture

Light pollution

Observing from the city, DSOs are challenging. The further away you are from city lights, the better your views. Deserts, national parks, sparsely populated areas are all good. Driving at least 1 hour away from the city usually provides very noticeable improvements. Light pollution maps help in finding places with a dark sky.

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

http://www.lightpollutionmap.info/

Aperture

The bigger the aperture, the better you can see DSOs. If you're a DSO hunter, you must be focused on increasing aperture before you do anything else. What helps here is having an instrument with a good aperture / cost ratio, such as a dobsonian reflector.

There's essentially no limit here - more is always better.

It should be noted that a larger aperture will always perform better than a smaller one, even when light pollution is very heavy - but of course it's best if both aperture and dark sky cooperate.

Other factors that may help:

Filters

These are the most over-utilized and over-hyped accessories, especially for beginners (most people are too much focused on filters and not enough focused on things that really matter). That being said, some nebulae do look better in some filters.

Dave Knisely is one of the foremost experts in the amateur astronomy community in terms of DSOs and filters. Read anything he writes on the subject, e.g. this long article:

http://www.prairieastronomyclub.org/resources/by-dave-knisely/filter-performance-comparisons-for-some-common-nebulae/

In a nutshell: If you can only get one DSO filter, get a UHC. If you can get two, also get an OIII filter. Of course, dark sky and large aperture should always come first. Don't bother with filters if you're observing from the city using a small aperture.

Filters are useful for nebulae. They don't help with galaxies, star clusters, individual stars, etc.

Books

Find a book called Turn Left At Orion. It's a fantastic introduction for beginners to DSOs. It will help you find the easier targets, and essentially open up the sky for you. Later you'll be able to figure things out on your own.

http://www.amazon.com/Turn-Left-Orion-Hundreds-Telescope/dp/0521153972/

It should be noted that the entire Messier catalog can be observed in a relatively small telescope even in the city - and all Messier objects are DSOs. I've seen the Dumbbell Nebula (M27) and the Ring Nebula (M57) in 6" of aperture in a very light-polluted place (Silicon Valley). So, no, it's not that hard to observe DSOs.

Of course, if the sky is dark and/or the aperture large, all DSOs look better. This is especially true of globular clusters, such as M13. They already look pretty spectacular in any aperture, even in a small scope - but the view is mind boggling in a very large dob.


What are the masses of the two stars (given the information provided)? If anyone would at least be willing to point me in the direction of finding the answer, or help me solve it I'd appreciate it. I don't know how to go about this, and need to know how to solve this on my own (plus I need the answer for a homework assignment, but just an answer isn't gonna help).

Question:

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 10 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of $~8.0\cdot 10^4~ \mbox{m}/\mbox{s}.$
What are the masses of the two stars? Assume that each of the two stars traces a circular orbit around their center of mass.

You need to use the generalized (a la Newton) form of Kepler's third law; see here.


How fast does Venus move as seen from the earth? So last night I was about to sleep with the window in front of me. there was a 3 inch gap between the curtains that let me see a small portion of the sky. I could see something magnificently bright in the sky that couldn't be anything but Venus. So while I was looking at it for about 1-2 minutes it gradually moved out of my sight. That seemed pretty fast. I mean I kinda estimated that with such a speed it could move from east to west in less than 2 hours.
So is that normal?

The Earth rotates on its axis once every 24 hours. (Well, actually a little less, by about 17 minutes, but close enough.) That's $360^\circ$ in 24 hours, or about $15^\circ$ per hour. $15^\circ$ is 30 times the width of the Sun or Moon. The Sun moves exactly that fast (on average), the stars a smidgen faster, and the Moon a tad slower. (The Moon lags behind the stars by approximately its own width every hour.)

Since Venus is never more than about $45^\circ$ from the Sun, it always sets at most 3 hours after the Sun does, and rises at most 3 hours before.


Terminology: Is there a name for the points on the surfaces of tidally locked parent/satellite bodies that always face each other? This is purely a question about terminology, one that has eluded my googling efforts.

When a satellite and its parent body are tidally locked to each other, there is (in an ideal case) a single point on the surface of the parent body and on the surface of the satellite body that remain the closest two points on the two bodes at all times.

Is there a specific name for those two points? If so, what is it? I think "antipodal points" is descriptive, but I wonder if there is a more specific term.

Thank you!

For tidally locked binary stars, the two points in question are known as the substellar points.

For a tidally locked exoplanet, the point closest to the star would also be known as the substellar point. If the star was also tidally locked to the planet, then there would be a subplanetary point.

For a moon locked to a planet, the point on the moon would also be the subplanetary point. If the planet was mutually locked, it would have a sublunar point (not sure whether this latter only applies to the Earth's moon).


Calculation of right ascension and declination I am confused about this problem:

If I see an object from Mount Teide (longitude is 16"30'E and latitude is 28"18'N) that passes the meridian (azimuth=0) at 5h (am) UTC, and I also know that the elevation of the star is 43"40', the stellar time at Greenwich at 0h UTC is 22h20min.

How can I calculate the right ascension and declination?

The answers to this question are very clear and much better than I could have written.

I'm not sure what you mean by "the stellar time at Greenwich at 0h UTC is 22h20min", but I'm assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your body can be easily calculated. All we need to do is figure out what RA is passing your meridian at the time in question. This is probably a good time to point out that the longitude of Mt Teide is 16° 30'W not E.

We already know that RA 22H 20m is passing over Greenwich at the time of interest (or, at least that is what I am assuming as your explanation is not at all clear.), as we are to the West of Greenwich, then we know that any RA on our meridian is earlier and we can calculate how much earlier simply by dividing our longitude by 15 (as the earth spins at a nice steady 15°/hour):-

diff in RA = Long/15
           = 16° 30'/15
           = 1H 06m

Then:-

RA = RA at Greenwich - diff in RA
   = 22H 20m - 1H 06m
   = 21H 14m

Now for Declination:-

The first stage is to take a piece of ruled or graph paper and near the middle mark a short, horizontal line. This is your horizon, and is marked 'H'. Now draw upwards 9 lines and mark the top ,'Z' and downwards 9 lines and mark the bottom 'N'. You should end up with something that looks like this:-

Which is, basically a picture of your meridian as you sit and gaze south at it. 'Z' is your zenith, the point above you in the sky and 'N' is the nadir, it's opposite out on the celestial sphere through the earth beneath your feet.

Now we need to add a little more detail. Imagine that the celestial sphere is actually a glass sphere surrounding the earth with the stars and stuff stuck to it that we can project anything we want on to (we'll ignore porn and coke adverts for now), where would the celestial equator show and where would it go on our picture of the meridian? Well, it would be a distance above the horizon equal to our latitude (lets round it off to 30°), so we now have:-

Now we put in the body we are looking at. We know it's elevation, just over 43°, we can just call it x as you didn't specify a body. We can also label some of the distances we know, to end up with:-

We already knew that Lat was H -> Q, that's how we place Q, we know that elevation, or altitude is between the horizon and the body and that declination is measured from the celestial equator. We want to find dec and the diagram shows us that:-

dec = Alt - Lat
    = 43° 40' - 28° 18'
    = 15° 18'N

As 'x' is to the north of 'Q', then Declination is north.

So now we know that we are looking at a body with RA 21H 14m and Dec N 15° 18'.

This is much more complex to explain here than it is to do, or to teach in person. With a bit of practice you'll be doing it in your head. I hope this has helped a bit.

I should point out that I have ignored observation errors, such as height of eye, which may be significant from the place you specify and parallax but I wanted to keep it as simple as possible.


What was the "optical illusion" that led to erroneous metal concentrations in stellar atmospheres in the galactic center? Articles summarizing a recent result about certain heavy element concentrations in stars near the galactic center all say that earlier reports about high levels in stellar atmospheres were the product of an "optical illusion". Examples include

UPI.com Study explains optical illusion at the center of the Milky Way
Science Daily Mystery at the center of the Milky Way solved
Phys.org High levels of scandium near the galaxy's giant black hole were illusory, astronomers find

At least Phys.org goes on to explain with real "science words":

  According to the new study, the lower temperatures of the giant stars helped to create the optical illusion that appeared in the measurements of spectral lines. Specifically, it means that the electrons in the elements behave differently at different temperatures, which in turn can be misleading when measuring the spectral lines of elements in different stars. The conclusion is the result of a close collaboration between astronomers and atomic physicists. (emphasis added)

Question: What was the "optical illusion" that led to erroneous metal concentrations in stellar atmospheres in the galactic center? Beyond just "electrons in the elements behave differently at different temperatures" (because how electrons behave in atoms at different temperatures should be at least fairly well understood by astronomers) what specifically is the source of the misunderstanding? Is it atomic physics, incorrect assumption just a mistake, something else?

The mention of "close collaboration between astronomers and atomic physicists" suggests that the answer will turn out to be fairly interesting!

The journal paper is Thorsbro et al. (2018).

The facts are somewhat mundane. The atmospheres of cool M-giants are not well understood in detail. The infrared lines of neutral Scandium that had previously been used to claim massive overabundances in stars near the Galactic centre are problematic.

The paper by Thorsbro et al. shows that the same "anomalies" are present in other M giants nowhere near the Galactic centre. Instead they argue that the details of line formation, especially the formation low excitation lines formed high in the atmosphere, like the IR transitions of neutral Scandium, are particularly vulnerable to poorly modelled NLTE (non local thermodynamic equilibrium) effects. If analysed with atmospheric codes that assume LTE, then you get the "illusion" of a much higher abundance.


Crab Nebula Time Period I've learned that the time period for a rotating neutron star, i.e., a pulsar is very accurate and can be used as an interstellar clock. But according to Walter Lewin's 8.01 Lec 19 on classical mechanics, he says that the rotational time period of neutron star in the Crab Nebula loses up to 364 ns everyday in the form of neutron star emissions (this comes from rotational kinetic energy of the neutron star). So, how are pulsars used as precise clocks?

All pulsars are observed to slow down gradually,albeit with a wide range of rates.The following diagram,known as the $\dot{P}-P$ diagram for pulsars illustrates this.The crab pulsar yes indeed would be a bad choice as a clock due to it's high $\dot{P}$.

But there are pulsars to the bottom left of the diagram which have periods of the order of milliseconds(hence called millisecond) pulsars and their $\dot{P}$ is also very low.These are the pulsars which can be used as atomic clocks,for example:MSP J0437− 4715.

  We can see that the both the period and the period derivative of this
  particular pulsar are very low providing great precision to the period
  of the system.So it can be used as a precise clock.

References:

1.http://www.cv.nrao.edu/course/astr534/PulsarTiming.html

2.http://www.cv.nrao.edu/course/astr534/Pulsars.html

3.http://en.wikipedia.org/wiki/PSR_J0437-4715


Angular diameter of the Sun's reflection from the ocean, seen from Sun-Earth L1? I'm trying to understand how smooth the reflecting ocean surface would need to be to produce such a small bright spot as seen from the DSCOVR satellite at Sun-Earth L1. It appears to be only about 8E-05 rad, or about 0.3 arcminutes.

My question is primarily about the geometrical optics involved in the reflection of light from the Sun off of the Earth. Interpretation of the apparent size and comparing to the image is secondary.

At a distance of 151.3 million km from Earth, the 1.391 million km diameter Sun has an angular width of 0.009196 radians. The DSCOVR satellite is 1.586 million km from the Earth looking at the Sun's reflection from somewhere around the Sun-Earth L1 point. It's close to the Earth-Sun axis (the Sun-Earth-DSCOVR angle is about 7.6 degrees) so let's simplify and assume it sits on-axis.

The Earth's radius is 6378 km; as a convex mirror its focal length is $f=-r/2$ or -3189 km.

Question: What would be the angular size of the Sun seen from DSCOVR reflecting in a spherical mirror model for the Earth in this configuration?

What I'm looking at (for context):

This sequence of images taken by the EPIC camera on the DSCOVR satellite near the Sun-Earth L1 point is notable for showing the progress of the Moon's shadow across the Earth, but that's not the topic of my question.

The first image shows the back-reflection of the Sun from the Earth. It is likely to be from a particularly calm spot on the Ocean, but that's not for sure, there are some images that show reflections from horizontally oriented ice crystals as well. See Are there measurements or calculations that suggest atmospheric ice plates would be horizontal to within 0.1 degrees? and also How could the recently explained “glints” seen by DSCOVR appear so compact considering the finite size of the sun? for some interesting images and more on that.

But that doesn't matter for the purposes of my question.

Width of Earth 6378 km x 2 / 1586000 km = 8.0428E-03 rad corresponds to ~1500 pixels.

Bright spot is about 15 pixels wide, or about 8E-05 rad.

Could such a small angular width be a geometrical reflection of the Sun from a very smooth ocean?

Raw image at NASA seen here.

$d_o$ and $h_o$ are the distance and diameter of the Sun and $d_i$ and $h_i$  of the respective images. The Earth's radius is $r$ and $f=\frac{r}{2}$ is the focal length.

The mirror equation $$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$ in the approximation $d_o \gg f$ gives $d_i \approx f$. Then, using the magnification equation

$$h_i = \frac {d_i}{d_o}h_o \approx \frac {f}{d_o}h_o.$$

If $S$ is the distance to the satellite then the angular size is

$$\frac{h_i}{S} \approx \frac{h_o}{d_o}\frac{f}{S} = \frac{h_o}{d_o}\frac{r}{2S} \approx \frac{1.4\times10^6}{150\times10^6}\frac{6.4\times10^3}{2\times 1.6\times 10^6}\approx 1.8 \times 10^{-5}$$

where distances are in km. (We have also ignored that the image is $\frac{R}{2}$ closer to the satellite than is the Earth's center.)

On a more descriptive note, the reflection on perfectly smooth water would look like a very bright spot about 29 km in diameter. It is blurred out (due to sea waves) to about ten times that diameter on the image. It would be interesting to calculate the brightness (per area) of the (unblurred) spot compared to the brightness of the sunlit Earth's surface.


Density of stars on celestial sphere The pattern of stars on the surface of the celestial sphere seems perceptually to have an universally looking structure. If you pick a region of the sky at random, for instance by using the (excellent) set of tools from the stellarium suite you get for example this:

There are bigger structures, like the milky way or more clustered areas, but on a smaller scale (like that you will see in binoculars with a field of view of the size of the moon), you will very often get this similar pattern. This looks to me very much like a self-similar fractal and I am curious to know if anybody knows more about it.
Assuming a given density of stars in this patch of the universe, and  omitting inhomogeneities at first,  can we derive a mathematical expression for describing the density of stars in a given patch of the sky ?
As a bonus, is there a map estimating this density for different areas of the celestial sphere? Clearly, the milky way is more crowded and would perhaps have more structure...

If stars are uniformly distributed with density $\rho$ per cubic parsec, out to a distance $d$ parsecs, then that volume contains $4\pi \rho d^3/3$ stars spread over the whole sky.
The angular size of the whole sky is 41,252.96125 square degrees, so you would divide by that number to estimate the stellar density in stars per square degree.
This assumes that you can see all the stars. In practice you cannot, because of dust extinction and because stars become fainter the further away they are. You can account for these in any calculation but the details depend on the luminosity function of the stars, the distribution of dust and the instrumental properties of the telescope/detector used to take the picture.


Continuum at 230 GHz what does it mean 'The observations were made in the CO J=2–1 line and continuum at 230 GHz (1.3 mm)'?
"The observations of the continuum were also made with a bandwidth of 320 MHz. "
http://articles.adsabs.harvard.edu/pdf/2000A%26A...353L..25C

"Continuum at 230 GHz" seems like an example of oxymoron phrase?

This is the frequency that the observations were made at. These are sub-mm (the wavelength is given as 1.3mm) observations (between the far infrared and the radio parts of the spectrum) and so it is more usual to talk in terms of frequency rather than wavelength (although there are exception e.g. the 21cm line of neutral hydrogen which is in the radio at 1427 MHz).

In the optical and infrared , we would normally say "observations were made at 550nm" but through the relationship $c=\frac{f}{\lambda}$ (where $f$ is the frequency, $\lambda$ the wavelength and $c$ is the speed of light), we could also say "these observations were made at 545077 GHz (545 THz)" although almost nobody ever would.

The "bandwidth of 320 MHz." part refers to how much of the spectrum is being sampled and digitized around the parts of interest, the CO J=2-1 line at 230.538 GHz (molecular spectra in the radio reference) and the continuum on side. The continuum is measured to get the strength of the background "underneath" the line and is then subtracted to get the true strength of the CO line.


Modeling egg shaped stars I am well aware of one-dimensional stellar models:

The simplest commonly used model of stellar structure is the spherically symmetric quasi-static model, which assumes that a star is in a steady state and that it is spherically symmetric.  It contains four basic first-order differential equations: two represent how matter and pressure vary with radius; two represent how temperature and luminosity vary with radius.

But what if we moved from spherical symmetry to cylindrical symmetry? Did somebody already set up all equations and solve them for general rotational symmetric ellipsoid?
What changes, if we would assume a lemon-shaped or (most interestingly) an egg-shaped star?
What would be the (intutive) results of such a stellar model? I am sure, somebody solved the equations already and I am just missing the appropriate search terms.
References

The mathematics of egg shape gives a brief mathematical background about one of my favorite mathematical objects

Cylindrical symmetry is not as hypothetical as it might sound:

Ashley Strickland wrote for CNN about "Unusual tear-drop shaped, half-pulsating star discovered by amateur astronomers"
WASP-12b is is reviewed by NASA as An egg-shaped planet.

The pre-print by E.C. & L.V. Nolan On isotropic cylindrically symmetric stellar models seems to cover the topic, but is not too intuitive.
Related

Can a donut-shape planet or star be formed?

Diclaimer: This is not (yet) an answer! To attract answers, I decided to start an answer draft which can be expanded by others.
Cylindrical coordinates
Every point in our cylindrical coordinate system is defined by a tuple $(r,\varphi,z)$ where $r$ is the distance from the rotational axis. We also define $Z$ as the height of our solid of revolution, i.e. $0 \leq z \leq Z$. The shape of the body is defined by shape function $s(z)$.
The volume $V$ of the object is then given by $$V= \pi \int_0^Z \left( s(z) \right)^2 {\rm d}z$$
Mass conservation
The mass density $\rho(r,z)$ does not depend on $\varphi$.
to be continued
Specific shape curves
Up to now, all maths has been performed for a general shape function $s(z)$, so let us now look at some specific ones
Egg as rotational body
For an egg with $z$ being the distance from the symmetry axis, we could for instance a formula by Narushin:
$$s(z) = 1.5396 \cdot \frac{B}{Z} \cdot\sqrt{ \sqrt{Z}\cdot z^{\frac{3}{2}}-z^2}$$
In this formula, $B$ is the maximum breadth and $Z$ is the height of the egg.


Number of stars & planets in Milky Way Galaxy? Is there any specific numbers on stars & planets count on Milky Way Galaxy ?

A consensus number is that there roughly $10^{11}$ stars in our Galaxy (though this number is certainly uncertain by a factor of at least two, because it is based on extrapolating what we know about stellar populations in our vicinity). Most of these stars are of lower mass and are much less luminous than the Sun.

The number of planets is even more uncertain. It now seems probable that most stars like the Sun have at least one planet, but we really don't know that much about the diversity of planetary systems, how typical something like the solar system is, or how planetary systems might change as a function of position in the Galaxy. We also don't know that much about planetary systems around the dominant (by number) very low-mass stars. They certainly can have planetary systems, but the fraction that do is still a work in progress.


Can moons exist inside a planet's roche limit? Saturn's rings are inside its own Roche Limit - the limit beyond which bodies start disintegrating due to the tidal forces of the parent body. But as every rock in the ring can be regarded as a satellite of Saturn, why are they not disintegrated to smaller constituents? Also, is it possible for a body to hold itself together even within the Roche Limit (maybe due to some cohesive forces holding it together)? I thought it may be the reason for the rocks in Saturn's rings holding themselves together- but then again, I have no reason to believe it may be the case any way. Can someone provide an estimate for what the cohesive forces may be for a metallic rock as a fraction of the internal gravitational forces in it?

You've answered you own question. If there are cohesive forces beyond that of simple self-gravitation, then objects can survive intact inside the self-gravitating Roche limit - as does every solid item on the surface of the Earth for example.
Saturn's rings are made of ice, not rocks.
The tensile strength of ice is about $10^6$ N m$^{-2}$.
For a self-gravitating body, the Roche limiting radius is about
$$ R_L \simeq 1.4R_p\left(\frac{\rho_p}{\rho_s}\right)^{1/3},$$
where $R_p$ is the radius of the planet and $\rho_p$ and $\rho_s$ are the densities of the planet and satellite respectively. This is for a rigid object. The leading constant would be bit higher at about 2.4, for a fluid.
Thus the self-gravitating Roche limit does not depend on the size of the satellite, it just depends weakly on its density.
However this limit becomes much larger if the cohesive forces holding a body together exceed the gravitational forces. This becomes true for smaller objects because the gravity at the surface of an object does depend on its size, whereas the cohesive forces do not.
A critical size will depend on the geometry, density and tensile strength. Roughly speaking, gravity and the cohesive forces will be equivalent for an satellite of radius
$$ r \sim \frac{T}{ \rho_s  g},$$
where $T$ is the tensile strength and $g$ the surface gravity. But
$$ g \simeq GM_s/r^2 = \frac{4\pi G}{3}\rho_s r$$
Thus
$$r = \left(\frac{3T}{4\pi G}\right)^{1/2}\rho_s^{-1}$$
For ice of density $1000$ kg m$^{-3}$ and $T\sim 10^6$ N m$^{-2}$, then $r = 60$ km.
This rough and ready analysis suggests that ice objects smaller than 60 km could survive inside the self-gravitating Roche limit.
FYI, the innermost named moon of Saturn, Pan, has a radius of about 15 km, an icy composition and is inside the Roche limit. There are other satellites in the ring system (Epithemius, Pandora) with radii of 30-60 km. Whereas the largest inner moon, Mimas with a radius of 200 km, is in an orbit at just over 3 Saturnian radii and outside the Roche limit.


What is the 2D elliptical projection of the Celestial Sphere called, and how can I make one? I see this kind of projection of the celestial sphere into 2D frequently, usually with an image of the milky way galaxy or cosmic microwave background. This particular image shows the path of visible transits of the inner solar system planets across the sun as seen from locations outside of the solar system from the Motherboard article At Least 9 Exoplanets Could See Earth With Present-Day Human Technology which links to: R. Wells, K. Poppenhaeger, C. A. Watson, R. Heller; Transit Visibility Zones of the Solar System Planets, Monthly Notices of the Royal Astronomical Society, stx2077, https://doi.org/10.1093/mnras/stx2077

Question: What is the 2D elliptical projection of the Celestial Sphere called, and how can I make one? Is there a python-friendly package, but more importantly, is there a mathematical expression for this transform?

I want to say that's a Mollweide projection. I know that they're pretty common in astronomy; many images of the CMB use them. I was actually working with one recently. Given latitude $\varphi$ and longitude $\lambda$, the $x$ and $y$ coordinates of an object are
$$x=R\frac{2\sqrt{2}}{\pi}(\lambda-\lambda_0)\cos\theta,\quad y=R\sqrt{2}\sin\theta$$
for projecting from a sphere of radius $R$. $\theta$ is the solution to
$$2\theta+\sin2\theta=\pi\sin\varphi$$
Similar images can be found here (labeled as Mollweide projections), which are of the same shape and proportions:

Matplotlib will let you do Mollweide projections; a nice tutorial can be found here. Basemap is a possibility, although I haven't used it before. It has quite a few different built-in projections. There are some interesting features available in the different matplotlib implementations, and using the ephem package, you can use astronomical data. That said, I've found that the transformations aren't that hard to implement by hand; I was able to get some basic projections done from scratch.


Why are the high density regions in outer space black holes? The escape velocity of an object is given by the following formula.

$$v_e=\sqrt{\frac{2GM}{r}}=\sqrt{\frac{2 \mu}{r}}= \sqrt{2gr}$$

As far as I am aware, the mass of the star which formed the black hole is the same as the black hole itself, the only difference is the black hole's density.

Now, the escape velocity of the star which formed the black hole is not as high because light can escape the star. Considering the equation above, $G$ is constant, $M$ is constant, however, $r$ changes resulting in a much smaller region as compared to the star. Consequently, it's the radius which determines the high escape velocity of a black hole, not its density. Then why is it that I always hear that black holes behave the way they do because they are highly dense regions in spacetime? What am I getting wrong? Thanks.

Making a major edit to this answer, as it would seem that there is a rather significant misunderstanding in the formula.

Initially describing r as the radius may have been a mistake.  While we may normally describe r as the radius of Earth when describing Earth, this is because we are describing the escape velocity of Earth from the surface, which means that the difference between our location and the center of gravity of the Earth is roughly equal to the radius of Earth.

When determining escape velocity from an arbitrary distance, r should be the distance between the escaping body and the center of gravity of the body being escaped. For objects at a normal distance, changing the volume, and thus the density of the object, will not affect nearby objects, as the mass stays the same.

However, this changes when you get close, when you compress the object this allows you to get closer to the center of gravity of the object. As you know, gravity is a weak force which deteriorates quickly with distance.  As such, when you decrease the volume of the body, a body can then become closer to its center of gravity, increasing the potential maximum escape velocity.

The body becomes a black hole when the physical volume of the body becomes small enough that the Schwarzschild Radius extends beyond it. Because we take the same mass, and decrease the volume, the result is a much denser area of spacetime, but this isn't generally the result of only a tiny volume, but also a large amount of mass.


Given that a color index ($B-V$) of star A is lower than that of Star B, What can we say about the stars? The color index $B-V$ of the star is defined as
$$B-V\equiv M_B-M_V$$
Let us suppose it's given
$$(B-V)^{(A)}<(B-V)^{(B)}\Rightarrow M_B^A-M_V^A>M^B_B-M^B_V$$
The notations are a bit confusing, but I think you would pardon me.
Now we know that Stellar magnitudes decrease with increasing brightness which means a star with absolute magnitude $-5$ will be brighter than a star with absolute magnitude $-2$.
I'm reading Introduction to Modern Astrophysics By Carroll and Ostlie. The  author says,

A star with a smaller $B-V$ color index is bluer than a star with a larger value of $B-V$.

I don't understand, How??
Star A is bluer than Star B means
$$M_B^A<M_B^B$$
Given four numbers with relation
$$x-y<a-b\not\Rightarrow x<a$$
Then how we can say so??

Color is a difference, not an absolute value
Being "blue" doesn't (necessarily) mean that a light source has a large flux in the $B$ band. A color is not an absolute value; it is the ratio between two fluxes or, equivalently, the difference between two magnitudes.
Being blue means "More flux in some short wavelength band (e.g. $B$) than in a longer-wavelength band (e.g. $V$)". Because of the stupid magnitude system, this then means a lower value in $B$ than in $V$, i.e. $B-V<0$.
Hence, the smaller $B-V$ value a star has, the bluer it is.
Intuitive example
So when you write

Star A is bluer than Star B means $M_B^A<M_B^B$.

it's actually not true. This mathematical expression only means that Star A is brighter (in the $B$ band) than star B; it says nothing about the color.
For instance, if you compare the flux of the Sun with the flux of a blue laser pointer, the Sun most definitely emits more blue light than the laser pointer, i.e. $B_\odot < B_\mathrm{laser}$. But that doesn't mean that the Sun is blue, because it also emits much more red light.
Astronomers are astronoming
In the above, I've followed the sloppy astronomer notation of using $B$ and $V$ both for the filters/bands/bandpasses, and for the absolute magnitudes. More clearly, we can also write the magnitudes as $M_B$ and $M_V$.
Not also that, in astronomy, the term "bluer" is often used interchangeably with "shorter wavelength", irregardless of the actual wavelength. Similarly, the term "redder" is used for longer wavelengths.
For instance, we could say that a red photon is bluer than an infrared photons. Similarly, a UV photon can be redshifted as it travels through the expanding Universe. If it doesn't travel too long, it might be observed as "blue", in the sense that its wavelength is around 4400 Å, but although it's not red, we still says that it has been redshifted.
Relation of color to temperature
For blackbodies, hotter objects are bluer and more luminous. This is a corollary of Planck's law, which relates the brightness $B$ per wavelength ("spectral density") of a blackbody to its temperature $T$.
If you integrate Planck's law you can show that the total power $P$ increases tremendously with temperature ($P\propto T^4$; Stefan-Boltzmann's law), while if you differentiate it you can show that the peak wavelength $\lambda_\mathrm{peak}$ is inversely proportional to the temperature ($\lambda_\mathrm{peak}\propto1/T$; Wien's displacement law).
In other words, 1) hotter objects emit more flux (i.e. are brighter), and 2) hotter objects emit a larger fraction of their flux at short wavelengths (i.e. are bluer).
The figure below (an annotated version of this) shows an example of the colors of a $5000\,\mathrm{K}$ (blue solid line) and a $4000\,\mathrm{K}$ (dark green solid line), respectively. The $B$ and $V$ fluxes are measured at $4400\,\mathrm{Å}$ (purple dashed line) and $5500\,\mathrm{Å}$ (light green dashed line), respectively.
The flux ratio between the $B$ and $V$ wavelengths are larger for the $5000\,\mathrm{K}$ object, so its magnitude difference is smaller.
In other words, it's bluer.
(I've omitted a factor 2.5)

Reddening
Stars are rather close to being blackbodies. However, a hot star could be reddened, e.g. by dust which preferentially absorbs blue wavelengths. Less hot stars may be  blanketed by metals absorbing light and re-emitting at lower wavelengths, effectively also reddening the star. Or an object could simply not be a blackbody, e.g. a blue laser pointer.
Comparison to velocities
If, for some reason, you don't know the value of $B-V$, but you know that "object A has a smaller $B-V$ value than object B", then you can't really say anything about their "true" color; you can just say that "A is bluer than B", although neither of these objects need be blue.
You can compare this to velocities. All velocities are relative, so "$v = 10\,\mathrm{km/h}$" doesn't mean anything, unless you specify a reference frame. But even given a reference frame, the statement $v_\mathrm{A} > v_\mathrm{B}$ doesn't tell you how fast A and B are; it just tells you that A is faster than B.


In theory, is there anywhere in the universe where velocity=0? Earth is traveling at a certain velocity.  Earth orbits the Sun at velocity X.  The sun is orbiting the center of our galaxy at velocity Y.  The galaxy is orbiting (something?) at velocity Z. On and on...

In relation to zero, Earth has a velocity of X + Y + Z + (?).

Is it theoretically possible that there would be a point in space where nothing is moving and velocity is essentially 0?

If so, what would time look like at this point in space?

Yes.

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

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

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

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

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


Why isn't everything yellow(ish)? I understand that - although appearing yellow - the sun is white. And that this is because the other light colours are scattered by the atmosphere. For example:

Google image search for "The Sun from Space"

That was difficult for me to come to terms with.

But if yellow light is what's getting through the atmosphere - why doesn't everything have a yellow tinge to it?

But if yellow light is what's getting through the atmosphere - why doesn't everything have a yellow tinge to it?

Everything does have a yellowish tinge to it, but only for roughly the hour after sunrise and the hour before sunset. These a photographer's golden hours:

As the Sun rises, its color changes from reddish to orangish to yellowish -- and then to nearly white, even when the Sun never does get very high.

You must not (and cannot) look at the Sun, even for the most fleeting of moments, once the Sun gets above 10 to 15 or so degrees above the horizon. The only time you can look at the Sun, even briefly, is when the Sun is low on the horizon and air mass is high. We think the Sun is yellow (or in some civilizations, orange or even red) because we cannot see the Sun for what it truly is, which is nearly white.

What bit of off-white color sunlight does have when the Sun is 10 to 15 or so degrees above the horizon is offset by the blue light from the sky, aka "diffuse sky radiation". The color of the light from the Sun and the sky when the Sun is high in the sky is about as white as light can get.

There's a bit of physiology going on here as well. To those who have lived their most of their lives inside buildings, bright sunlight is a bit yellowish and a bit too harsh. To those who spend much of their time outdoors, the fluorescent lighting that now dominates the urban landscape is a bit bluish and a bit too harsh. What city dwellers who rarely ventures outdoors thinks of as white light is not very white to those who spends most of their waking hours outdoors, and vice versa.


Mars night sky - how many stars visible? How would the night sky look like from the Martian surface?

Would there be more stars visible or less (compared to Earth)?

Would they appear brighter/dimmer/same?

I'll add to Wayfaring Stranger's comments. In fact most of the time you would be able to see fewer stars in the night sky of Mars, than in a good dark night sky on Earth, because of dust obscuration.

Even in favourable conditions, the optical depth of the Martian atmosphere is usally somewhere between 0.5 and 1 per airmass. (Petrova et al. 2012; Lemmon et al. 2014) and is nearly wavelength independent. This corresponds to a reduction in flux to between 37% and 60% of its value above the atmosphere. This compares to a typical optical (V-band) extinction of around 0.1 magnitude at a good site on Earth, which allows 90% of the flux through.  There are times when it can be much worse than this on Mars.

This means that the limiting magnitude on Mars would be somewhere between 0.44 and 0.96 magnitudes brighter than it is on a dark site on Earth and that all the stars would be fainter by these values.

As a rough guide to the effect this has I took the Hipparcos catalogue and made a frequency histogram of V-magnitude of the stars. If we define a Hipparcos magnitude of 6 as the limit you can see from a good dark site on Earth, then you would see 4559 stars (that's over the whole of the sky in both hemispheres). If that limit was reduced to 5.56 or 5.04, then this number decreases to 2745 or 1560 respectively. Thus there would be a reduction in the number of stars you could see by typically between a factor of 1.66 and 2.92.


Can redshift be measured using fourier? Are there pictures that contain high enough dynamic width, so that amateurs can use fourier spectrum analysis to observe the redshift themselves?

I know pictures usually are encoded using RGB and/or CMYK and therefore lose a lot of color information from a physical perspective, but I'm wondering if there exists a picture format that could store that much information, and if so, could amateurs then use fourier spectrum analysis to actually see the red-/blueshift?

No, amateur astronomy equipment is not capable of doing such an analysis. Most color CCDs use something called a Bayer filter to detect color. Essentially, certain pixels detect only red photons, other pixels detect only green, and still others detect only blue. Each filter has a high transmission over some band of wavelengths (say, for red, 600 - 700 nm). Photons outside of this band are absorbed. The CCD pixels behind the red Bayer filter then count how many photons they see (they actually measure charge, but that's a detail). The point is, the CCD can't measure a photon's precise wavelength; it can only count how many it sees over some wavelength band.

Now, this is NOT necessarily the case when it comes to professional astronomy. To my knowledge, there do not currently exist optical CCDs that can do spectroscopy without a diffraction grating. For one thing, optical diffraction gratings work really well, so there is no reason. The other reasons are that optical photons don't generate a lot of charge in a CCD, so it would be hard to measure, and there are generally a LOT of optical photons, so you'd get a lot of pile up.

By contrast, X-ray CCDs often do spectroscopy, albeit not with Fourier analysis (for example, the ACIS CCDs on Chandra). There are a lot fewer X-rays than optical photons, and they deposit a lot more energy on a CCD. Unfortunately, you can't do X-ray astronomy from the ground, so even if an amateur astronomer could somehow afford one, it wouldn't do them any good. Even in the case of X-ray astronomy, you generally use an x-ray diffraction grating to get a spectrum if that is what you are interested in.


The systematic review of past observations in the hope of finding things that may have been missed I seem to remember a TV science programme which re-examined the slides that led to the discovery of Pluto and then took a fresh look at earlier slides taken of the space which Pluto was passing through. These slides did indeed feature Pluto but were probably not part of Lowell’s search for the planet. Maybe the old data/images(if still on record) could be revisited with the advantage of later discoveries to reveal things of interest like say orphan planets?

Recently there was this case:-

https://www.extremetech.com/extreme/234037-astronomers-observed-a-nova-explosion-for-the-first-time-by-accident

Which inadvertently did this but are there any programmes/projects retrospectively doing this intentionally?’

So, it is the systematic review of past observations in the hope of finding things that may have been missed or even unknown at the time.

The first step for most of these projects is to make the old data available. Just about all of the old data is on photographic plates, so there are projects at many of the great observatories to scan their archives.

A good example is the project at Smithsonian Astrophysical Observatory (located at Harvard) to digitize the entire plate collection, ranging from ultra-early photographs to sky surveys that Harvard Observatory did during a large part of the 1900s. The plates are scanned, and they are matched with star atlases to determine their exact coordinates and their limiting magnitude and the filters used, if any.

This data -- and lots more from all over, including the epochal Palomar Sky Survey -- is available (or soon will be) using special software such as Microsoft's Worldwide Telescope.

So the old data is being collected and curated and being made available to everyone.

(Note that the old data, being photographic, is limited in how deep they go, so many objects are simply not visible in them, including most minor planets.)

And lots of projects make use of it.  The ones I'm familiar with tend to use the data to look for prediscovery images or to extend light curves into the past.

New surveys like the Large Synoptic Telescope will generate cubic shittons of data and are being deliberately designed to separate the collection of data from the use -- it'll collect deep full-sky images every few days and then people will mine that data for things that the people who planned the survey hadn't thought of.  It will revolutionize astronomy.  Key point: All this old data is sure to be made available in the same way and the very bright people developing ways to find interesting stuff in the LSST will also find good stuff in old photographic plates from the 20s.


Could Earth have been seeded with life? The earliest life on Earth is thought to have began shortly after its formation (a few hundred million years after, by fossil evidence, and this while the late heavy bombardment was raging). With the next discernible leap (eukaryotes) happening billions of years later, That seems like a very short timeframe for life to spontaneously develop. Is it possible that life was seeded on an early Earth by a supernova of an earlier, life-bearing star, blasting living material across the cosmos?

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


What's the correct distance to Betelgeuse? Reading [1], [2] and this smartphone app[3], I'm getting different answers to the question.
The smartphone app is stating 427ly. [1] states approx. 700 ly,  and [2] states approx. 724ly.
Even if I take [1] as the correct value, why does [2] state 724ly and the app stating 427ly?
Thanks
[1] - https://en.wikipedia.org/wiki/Betelgeuse
[2] - https://earthsky.org/astronomy-essentials/how-far-is-betelgeuse
[3] - https://www.escapistgames.com/sc.html

Distance measurements for Betelgeuse are a bit of a mess. Solutions based on parallax would be ideal, but Betelgeuse has a rather large angular diameter at most wavelengths thanks to its extended envelope; optical and infrared observations usually fall in the 40-60 mas range (see Dolan et al. 2016 for a recent review), while radio observations show a disk of emission roughly twice that size (O'Gorman et al. 2017). The parallax is expected to be much smaller that the angular diameter, on the order of ~5 mas, and so it depends strongly on the choice of the center of emission.
One of the first decent parallax results was obtained by the Hipparcos satellite in 1997, whose astrometric measurements allowed comparatively precise measurements of location, parallax and proper motion for over 100,000 stars. Hipparcos measured a parallax for Betelgeuse of $\pi=7.63\pm1.64$ mas, corresponding to a distance of $131\pm30$ pc$^{\dagger}$. This is that 427 light-year number the app cited. The Hipparcos-only result was subsequently improved upon significantly by van Leeuwen 2007, who found $\pi=6.56\pm0.83$ mas, cutting the old uncertainty in half; this would correspond to a distance of 152 parsecs. If you're going to quote a Hipparcos result, this is the one to pick.
More recent results indicate that this value is likely too low. Combining Hipparocs data with multiple multifrequency radio measurements using the Very Large Array, ALMA and e-MERLIN (Harper et al. 2008, Harper et al. 2017) gives derived values of $197\pm45$ pc and $222^{+48}_{-34}$ pc, with the former barely consistent with the purely optical results (the latter value converts to 724 light-years). These groups note that the Hipparcos stochastic astrometric solution required the addition of so-called "cosmic error" or "cosmic noise" terms to individual position measurements.
The papers above note that the photocenter at both and optical radio wavelengths does not coincide with the barycenter, and may change on timescales of months to years. Therefore, extended, long-term observations would be necessary to reduce any photospheric "jitter" or other variations which could lead to variations in emission and thus astrometric fitting. Harper et al. 2017 proposed joint ALMA and Expanded VLA/Jansky VLA observations and mm and sub-mm bands over a period several years, but also suggested that this would require "a Herculean effort" to achieve, for logistical reasons (how many telescope committees would be inclined to commit that much time up front?). Perhaps interest in Betelgeuse's recent luminosity dip could motivate this sort of observation.

$^{\dagger}$Rob Jeffries makes the point that given the pretty terrible signal to noise ratios of many of these parallaxes (Hipparcos was in particular was bad, but none of the observations are amazing), it's not clear that you can really get good, meaningful uncertainties on the distance measurement from them via $d=1/p$. I agree; it's safe to say that the jury is still out on those, and any source that claims a distance and an error should give make that quite clear.


How was Earth's "quasi-satellite" 2016 HO3 "first spotted" and it's orbit determined? Under a question I asked a few days ago Have there been any documented mini-moons since 2006 RH120? @Hobbes mentioned the recent news about 2016 HO3 - a near-Earth asteroid that stays near Earth because its orbit around the sun is coupled to Earth. Thus the term quasi-satellite.
https://www.youtube.com/watch?v=SbbAnVU4rmY
In this NASA JPL video (above) the view is rotating around the sun following Earth. You can see the earth move slightly closer and farther from the sun since the Earth's orbit is not quite circular.
https://www.youtube.com/watch?v=2mVfE_qmQAg
This video (above) from http://arksky.org/calendar/alerts/714-what-is-it-the-strange-new-object-2016-ho3 shows a projection of 2016 HO3's motion against the stars as seen from Earth's location, but in a fixed direction. You can see the the sun and planets tend to follow the ecliptic, while 2016 HO3 does a figure-eight every year.
The NASA JPL news brief states:

The asteroid's orbit also undergoes a slow, back-and-forth twist over multiple decades. "The asteroid's loops around Earth drift a little ahead or behind from year to year, but when they drift too far forward or backward, Earth's gravity is just strong enough to reverse the drift and hold onto the asteroid so that it never wanders farther away than about 100 times the distance of the moon," said Chodas. "The same effect also prevents the asteroid from approaching much closer than about 38 times the distance of the moon. In effect, this small asteroid is caught in a little dance with Earth."
note: Paul Chodas is the manager of NASA's Center for Near-Earth Object (NEO) Studies

The item goes on to say:

Asteroid 2016 HO3 was first spotted on April 27, 2016, by the Pan-STARRS 1 asteroid survey telescope on Haleakala, Hawaii, operated by the University of Hawaii's Institute for Astronomy and funded by NASA's Planetary Defense Coordination Office. The size of this object has not yet been firmly established, but it is likely larger than 120 feet (40 meters) and smaller than 300 feet (100 meters).

First, I think it's quite cool that there even exists a Planetary Defense Office. It must be fun to go to work every day knowing that you are defending the planet!
I am interested in the story behind "...first spotted...".
The Pan STARRS telescope PS1 (http://pan-starrs.ifa.hawaii.edu/public/home.html) is managed by the PS1 Consortium. The original plan was for four telescopes, the status of PS2 is unclear. PS1 is a 1.8 meter telescope with a Gigapixel camera at the relatively fast f/4.4 focus, corrected for a very large 3° FOV, which means it's of the "primary with giant hole" variety (e.g. LSST). Typically it records an image every 30-60 seconds, and can record the entire night sky every three days.
I looked at the entry for 2016 HO3 (SPK-ID: 3752445) in the NASA JPL Small Body Data Browser and saw this:

The solution (shown on June 18) was calculated on June 11 2016 by an "O. Matic" :) using data going back to 2004.
As I said above, I am interested in the story behind "...first spotted...".
What were the sequence of events? Does "spotted" mean identified/flagged as an unknown object? Did it trigger a search of a database of earlier survey images from other telescopes? Is there a list available of the 80 total observations used? I'm curious if a twelve year old "first known image of 2016 HO3" exists.

Phil Plait's blog, Bad Astronomy, states (emphasis mine)

One of them was just discovered recently by astronomers: the asteroid 2016 HO3.
. . .
Here’s how that works. The asteroid was first seen in April 2016 in observations of the sky taken by the Pan-STARRS observatory, designed to look for asteroids and comets that get close to Earth. That’s recent enough that a really good orbit for it is hard to determine, but it turns out it was seen in older observations (those can be found by tracing the orbit backward and checking if any observations of it were archived), providing a much longer baseline and therefore a better orbit.

In other words, it seems like a basic approximation of the orbit was created, and the astronomers looked at images taken of areas near where the asteroid should have been, according to the calculations. The calculations could then be refined, and further previous sightings could be investigated.


What kind of radiation do supernova remnants emit? Do supernova remnants emit EM radiation? Moreover can neutrinos be emitted by these remnants?

Young supernova remnant release a substantial amount of radiation as they expand into the interstellar medium. In the free expansion phase, the outgoing shock wave heats up matter to $\sim10^6$ Kelvin, producing thermal x-ray emission. Throughout their lives, the remnants are responsible for quite a lot of non-thermal emission, including synchrotron radiation from relativistic electrons. This synchrotron emission stretches throughout the electromagnetic spectrum, including at radio frequencies. There is, additionally, non-thermal gamma ray emission through several mechanisms, including inverse Compton scattering and hadronic interactions.

Supernova remnants are also thought to be one of the main sources of galactic cosmic rays. Protons are accelerated through the shock front to energies of up to a few petaelectronvolts. Cosmic ray production may be associated with gamma ray emission if there is additional dense gas for the cosmic rays to interact with.

This brings us to the final point of your question: yes, supernova remnants are responsible for some neutrino production. Those cosmic ray/gas interactions produce (in addition to those gamma rays) neutrinos through hadronic channels and the decays of short-lived intermediate particles, including pions and kaons. This also means that probing the gamma ray spectrum of a supernova remnant can tell us something about its neutrino spectrum.


Data Analysis: Image Sharpening through Posterisation I had this image, a picture of Jupiter:

I wanted to measure the pixel diameter of Jupiter. As evident, the image is quite blurry, and thus, many value of equally justified pixel diameters can be obtained. To eliminate this I posterised the image (which is technically, quote from Gimp, "designed to intelligently weigh the pixel colors of the selection or active layer and reduce the number of colors while maintaining a semblance of the original image characteristics"):

With the above result, I obtained three pixel diameters of Jupiter (one from each of the coloured circles), and with a little bit of calculation, realised that the yellow circle gave rise to the most accurate value of Jupiter's diameter (which was surprisingly off by less than one percent!).

My questions are:

1) Why is my original image like that? Is it due to some form of light scattering?

2) Is there a better method of obtaining an accurate pixel diameter than posterising the image

3) Is the posterised result expected? Are there names that can be given to the three distinct spherical objects (which seem to resemble shadows, like the umbra, prenunbra, antumbra, etc.)? Is there a reason why the middle posterised object gives the most accurate value?

1) your original image is blurred and possibly overexposed. I can't tell if the blurring is due to poor focus or inaccurate tracking. I think you really need much better, sharper images that show the true boundary of the planet before you can start to measure diameters.

2) with a sharp image, there should be no need for posterising the image.

3) "Is there a reason why the middle posterised object gives the most accurate value" - well we could claim that it marks the "average" estimated diameter, but in my opinion the answer is: coincidence.  If you posterised it to give, say, four bands instead of three, which would you take as your "true" diameter? And without knowing the real diameter of Jupiter in advance, how would you ever know?

Edit: by the way, I noted your previous question on this subject. Did you read and use those answers? How sure your magnified image is really the size you think? You could try to verify this by photographing a double star of known separation...


Are other planets' gravity measurable on Earth? I understand that we can calculate the gravitational force  other planets exert on Earth, but can it be measured?

I think the questions you're asking is, "All of the planets exert a gravitation force on everything else in the Solar System, including the Earth. Can we make measurements of that force in a laboratory on the surface of the earth?  (I.e., not by observing events on the other planets or their orbits.)"

If I'm right, the answer is "No, the effects are too small."

First, calculate the gravitational acceleration we feel on Earth from, say, Jupiter.  As it happens, Earth is roughly 10,000 Jupiter radii away from Jupiter.  The acceleration due to gravity follows an inverse square law, so the acceleration due to Jupiter on Earth is about 10-10 that at the surface of Jupiter or about 3x10-10 g.  This is small, but perfectly measurable.

The problem is that we can't measure gravity directly. We're sitting on a freely-falling Earth and (as far as Jupiter is concerned) are freely falling with it. Whatever acceleration Jupiter is imparting, it imparting the same acceleration to the Earth, us and our laboratories and everything in them.  So there's no force to measure.

We can measure tidal forces directly, since they are in essence the differences in acceleration due to objects being a different distances from the gravitating object, but they drop off as the inverse cube of distance and are undetectable at this distance using current technology.


Effect of the absence of atmospheric refraction to sun rise and sunset times What is the effect of absence of atmospheric refraction on sin rise and sunset time?
Is it will be delayed or earlier for sunrise and sunset, and why?
Thank you:)

The simple answer is that atmospheric refraction makes the disk of the sun visible at sunrise or sunset, when it's actually half a degree or so below the horizon:

In reality, both the extent of refraction of the sun's disk and the refraction of the apparent horizon come into play. Both depend on the atmosphere's density and temperature profile; which changes irregularly. That makes it impossible to predict exact sunrise or sunset times when there is an atmosphere.


What's the exact story of jupiter and our earth, not becoming a super-earth? I've been watching the Cosmos documentaries by Neil DeGrass Tyson and the ones from Brian Cox, and in one episode (I don't remember from which series) there's something about the creation of the rocky planets in our solar system I don't understand. Let me first clarify what I understood from the explanation:
At the beginning of our solar system, there were lots of gasses and dust, all over the solar system. Some of those coalesced into planets, who became bigger and bigger by attracting more gasses and dust. (That was also what I learnt as a child)
However, when looking at exoplanets, there seem to be quite some "super earths": rocky planets, being quite larger than the rocky planets in our solar system. This created the question why the rocky planets in our solar system stopped growing, and then, a theory, involving Jupiter was formed:
Apparently, Jupiter, being a big gass giant, spiraled towards the Sun, and being so heavy, Jupiter worked like a vacuumcleaner, sucking up the gasses and dust from between the rocky planets, and like this stopping the growth of the rocky planets.
The spiraling movement of Jupiter was stopped, thanks to the formation of Saturn: that large planet pulled Jupiter back from further spiraling towards the sun, and like this, the planets in the solar system got, more or less, their current size.
When I first heard this, I thought "Ok, that's an interesting theory", but later, I remembered something from my years at university:
During my studies (I studied mathematics and mathematical physics), we have proven that the gravity, caused by a sphere with a mass, is exactly the same as the gravity, caused by a single point with the same mass.
This gives me the following idea (about the creation of Saturn and its impact on Jupiter): first there was a ring of gass, floating outside the orbit of Jupiter. Then, some of that gass coalesced and formed Saturn. The mass of the ring and of the resulting planet are the same.
So, the gravity, caused by that ring of gass, and the gravity, caused by Saturn as a coalesced planet, should be equal. Hence this could not have had any impact on Jupiter's movement, so the explanation of our rocky planets, not turning into super-earths because of Jupiter, spiraling towards them and being stopped by Saturn's creation, is wrong!
There are three possible solutions to this:

Either the theory is wrong indeed, but I would really be surprised that I, as a simple guy, am able to find an error in an official theory that simply.
Either I missed something in the explanation. In that case, can somebody tell me what the correct explanation is?
Either my reasoning is wrong: indeed a sphere and a point with the same mass have an identical gravitational impact, but my extrapolation towards a ring of gass and a coalesced planet is incorrect. In that case, can anybody point out the flaw in my reasoning?

Thanks a lot for reading and answering this (I realise it's a long one).

Your reasoning is wrong: you cannot extrapolate from the outside of a sphere of matter to the inside of an annulus or a shell of matter. The gravitational pull of a uniform annulus of matter is 0 inside that annulus.
See the wikipedia article for more details on that theorem.
So having a disc of matter around you is not the same as having a sphere of matter next to you.


How to calculate phase angle of a satellite? I'm making a program for predicting satellite passes. I'm trying to find out if the satellite is illuminated by the Sun and not in Earth's shadow. I need to know its phase angle: the angle between the observer on Earth, the satellite and the Sun.

Please explain it in simple terms if possible.

(Inserted by reviewer, extension to question, posted originally as answer)

I have TLE data for the satellite (contains right ascension). From that I got ECI position, azimuth, elevation, altitude. For observer I have latitude and longitude and for the sun: elevation and azimuth.

Draw the triangle Sun-Earth-Satellite, we will first find the angle Sun-Earth-Sat.

The angle between the Sun, the observer, and the satellite will be the angular separation between the Sun and the satellite on the observer's sphere. Here is how you could calculate that:

You have horizontal coordinates for both bodies so the easiest way to go from there would be to look at the spherical triangle Zenith-Sun-Sat, the angle at zenith will be the difference between azimuths, and the Zenith-Sun and Zenith-Sat lengths will be $90^{\circ}-h_{Sun}$ and $90^{\circ}-h_{Sat}$, respectively.
Now using the cosine formula for spherical triangles, one may obtain the following formula:

$$\cos^{−1}(\sin(h_{Sun})\sin(h_{Sat})+\cos(h_{Sun})\cos(h_{Sat})\cos(A_{Sun}−A_{Sat})).$$

Now use the sine theorem to find the angle E-Sun-Sat (the sine of this angle divided by the sine of the one we calculated will be equal to the ratio of distances from the Earth to the satellite, and from the satellite to the Sun, respectively), and to find the third angle, subtract the two from $180^{\circ}$.

Note: if you do not know the distance from the Sun to the satellite, I am certain you may use the distance from the Earth to the Sun as the error is probably negligible.