Why are ice giants not considered suitable for life?

A basic approach would be to estimate the air-pressure at the top of the mantle.

Uranus' surface gravity is a bit less than Earths, but, somewhat counter-intuitively, it's atmosphere is light enough that it's gravity increases as you move inside the planet towards the top of the mantle.  (the same is true for Earth).   For ballpark estimate, lets say the gravity works out to be about the same as Earths, because I don't want to calculate, but it might be a bit more than that.

The Mass of Uranus' atmosphere, by Wikipedia is about 0.5 times the mass of Earth.  It's surface area is about 16 times Earth (4 times the Radius), but the surface area of the Mantle would be less, about 80% of the radius which works out about 10.2 times the surface area.

If you spread 0.5 Earth masses over 10.2 times Earth's surface area, you get a pressure between 50,000 and 60,000 atm.   Some 50 times the upper limit you provided in your question.   That's probably a limiting factor for most "ice giants".   There's simply too much hydrogen and helium atmosphere to allow liquid water at 1,100 ATM.   Ice giants might create their own hydrogen by heat and pressure on Methane, creating diamond like structures and free hydrogen.

The heat below all that atmosphere (about 20% of Uranus' radius) would be outside your criteria too.  I couldn't find any good estimates for Uranus' mantle's temperature.   There's too many unknowns about why Uranus is as cold as it is on it's surface, so estimates on it's mantle temperature seem pretty tentative and unavailable, but what little I know about how temperature rises inside a planet, I don't see any way the top of Uranus' mantle would come close to the temperature criteria you gave.  Not at a depth of 20% of Uranus' radius.

Consider this chart from Uranus' wikipedia at 200,000 mbar Uranus temperature is approaching 340 K.    200,000 mbar is about 200 atm.   The mantle begins, by my ballpark estimate at about 50,000 atm.   The temperature wouldn't be close unless the planet had a very long time to cool down, perhaps tens of billions of years.

I could run ballpark estimates on Neptune too, but I suspect the results would be similar.

For an ice giant to meet your criteria, it would need to have comparatively very little hydrogen and helium compared to Uranus.  Perhaps if an ice giant was close enough to it's star that it had a hot enough surface temperature to lose much of it's hydrogen and helium through Jeans escape, and then, it had a long time to cool down, it might get the right criteria of liquid water at the proper range of temperature and pressure.  I don't see any other scenario where it would work.


Why haven't asteroid belts turned into new large bodies?

From Wikipedia:
http://en.wikipedia.org/wiki/Asteroid_belt#Formation

Planetesimals within the region which would become the asteroid belt were too strongly perturbed by Jupiter's gravity to form a planet. Instead they continued to orbit the Sun as before, and occasionally colliding.[27] In regions where the average velocity of the collisions was too high, the shattering of planetesimals tended to dominate over accretion,[28] preventing the formation of planet-sized bodies. Orbital resonances occurred where the orbital period of an object in the belt formed an integer fraction of the orbital period of Jupiter, perturbing the object into a different orbit; the region lying between the orbits of Mars and Jupiter contains many such orbital resonances. As Jupiter migrated inward following its formation, these resonances would have swept across the asteroid belt, dynamically exciting the region's population and increasing their velocities relative to each other.[29]


Why don't black holes quickly eat up the galaxy/universe?

Your rough analysis is basically correct; everything may well end up in black holes eventually. But the timescale for that is very long indeed.
The cross-section for direct interaction as you describe is vanishingly small. The diameter of a ten solar mass black hole is 60 km. The rate at which it scoops up interstellar material from a cylinder of diameter 60 km is totally negligible. e.g. A speed of 10 km/s with respect to interstellar gas and a typical gas density of $10^6$ H atoms per cubic metre, would result in an accretion rate of about 1.5 kg/year.
The chances of direct star to star collisions in a galaxy is exceedingly rare, so the chances of a wandering star crashing directly into a black hole are even smaller.
With supermassive black holes, the issue with your analysis is that they sit in the centres of galaxies and stars do not encounter them by chance. The stars and gas in a galaxy have angular momentum that must be conserved. This takes them on orbits which do not go near the black hole unless perturbed in some way. These supermassive black holes feed while they can, but when all the easily available matter has been consumed, they become dormant. This is one of the reasons that the peak of quasar activity was about 10-12 billion years ago.


What are the next planned space telescopes?

The James Webb Telescope is the next one on the launchpad that you might be familiar with.

Although there are a few differences that one ought to be aware of.  NASA has an entire program of telescopes to observe the universe, and many of them are designed for different wavelengths of light.  The James Webb is primarily designed for the infrared part of the spectrum.  While many people view the Webb as the successor to the Hubble, it would be more fair to say that it's really a successor to the Spitzer Telescope.  Hubble and Spitzer were both part of the "Great Observatories" program.  The Webb telescope is barely hanging on to funding, but so far the US hasn't managed to entirely abandon the program, science, or reality.


What is the relative velocity between planets in different solar systems?

Tens of kilometres per second for nearby stars. Hundreds of km/s for stars in distant parts of the galaxy to thousands of km/s for stars in nearby galaxies. Anything from "much more" to "it's complicated" for stars in very distant galaxies.

The Earth is orbiting the sun at 30km/s and the orbital velocity of exoplanets is on the same order of magnitude. The stars are also moving relative to the sun. We can measure the transverse velocity for nearby stars by seeing them slowly change position (relative to the distant background stars) and we can measure the radial velocity by the blue- or red-shifting of the light.

We can't directly measure the transverse velocity of distant stars, but their general motion is around the galaxy with a velocity, relative to the sun, in the 100s of km/s

Nearby galaxies are moving at similar velocities. Again, we can currently only measure the radial velocity, but their transverse velocities would be of the same order of magnitude, and in the 100s to 1000s of km/s.

Distant galaxies are part of the Hubble flow, and the expansion of space, so defining relative velocity or relative distance becomes tricky and needs to be done in the context of General Relativity, at which point things become more problematic, and beyond what is required for space opera.

For context, spacecraft in orbit around the Earth travel at less than 8 km/s and our fastest spacecraft (which are needed to travel close to the sun) have never reached 30km/s.


Can a telescope image be amplified to see more by adding distant lenses or mirrors?

Speaking in general, telescopes are designed as complete systems from the beginning. It's not generally feasible to improve performance by just adding yet another lens or mirror to the assembly. The reasons for this are not simple to explain, since this is basically the whole theory of instrument design.

Usually you start with a set of requirements (limiting magnitude, angular resolution, linear resolution in prime focus, etc) and with a set of constraints (cost, size, weight, materials).

You settle on a given architecture (newtonian reflector, Ritchey-Chrétien, doublet / triplet / quadruplet refractor, etc), and then start to work out the main parameters: aperture, focal length, etc.

There's an iterative process involving computer aided design to work out the details of each optical element; at each step you need to solve problems such as eliminating internal reflections, etc.

In the end, what you have is the design of the whole optical stack: say a Dall-Kirkham reflector, 1 meter aperture, f/8, with mirrors of whatever radiuses of curvature, etc.

Once that's set in stone, it's hard to change in a major way. Minor upgrades might be possible (add a corrector lens, or a different secondary mirror), but not huge modifications.

TLDR: No. Start with a bigger instrument from scratch if that's what you want.


How is dark matter different from ordinary (visible) matter?

There is certainly one and possibly two types of dark matter to solve two dark matter problems.

The first dark matter problem is that most gravitating matter is dark. Most of the inferred mass that appears to be responsible for the observed gravitational dynamics of large scale structures like galaxies and clusters of galaxies cannot be seen (i.e. is dark). That is, we can't detect it through visible light or any other electromagnetic waves.

Now it is possible that you could make up this mass out of very low mass stars,   cold white dwarfs, planets, black holes, lost golf balls or indeed any other form of cold, non-luminous matter. Such objects (bar the golf balls perhaps) certainly do exist, are known as baryonic dark matter (because they are made of "normal" protons and neutrons - baryons), but there is unlikely to be enough of this stuff to explain the observations.

The second problem is that most of the dark matter must be non-baryonic. This is deduced from the lack of sufficient plausible baryonic dark matter to explain the dynamics of galaxies and clusters, but also from estimates of the primordial abundances of helium, deuterium and lithium. Big bang nucleosynthesis depends on the density of baryonic matter compared to the universe as a whole and it looks like baryonic matter can only be responsible for a fifth of the dark matter.

Thus the bulk of the dark matter is supposed to be some non-baryonic stuff, probably in the form of particles that do not interact via the electromagnetic interaction and therefore do not emit or absorb light.


Requesting book references for a non-expert person with math background or just a non-expert person

I personally found Stephen Hawking's "A Breif History of Time" to be very interesting and informative.  It is a little bit advanced, but is put forward in plain English.  Here is a link to it on Amazon, and you can "look inside" to see part of the book and see what you think.

http://www.amazon.com/gp/product/0553380168/ref=pd_lpo_sbs_dp_ss_1?pf_rd_p=1535523722&pf_rd_s=lpo-top-stripe-1&pf_rd_t=201&pf_rd_i=055371449X&pf_rd_m=ATVPDKIKX0DER&pf_rd_r=1Z0JYCKHWXEFPK3TS6TM

After reading your comments / clarifications, you might like "A Field Guide to Stars and Planets."  It appears to be a good introduction into astronomy.

http://www.amazon.com/Field-Guide-Planets-Peterson-Guides/dp/0395934311/ref=sr_1_1?s=books&ie=UTF8&qid=1411293771&sr=1-1&keywords=stars+and+planets

As far as covering astronomy from a more mathematical perspective, you might like "A Student's Guide to the Mathematics of Astronomy."

http://www.amazon.com/Students-Guide-Mathematics-Astronomy/dp/1107610214/ref=sr_1_1?s=books&ie=UTF8&qid=1411294172&sr=1-1&keywords=math+of+astronomy

You can take a look inside each of these books to see what you think.


How to determine what directly imaged objects are?

Simple really. You observe again and see if the pair are moving together. If the star plus "planet" are quite widely separated (e.g. tens of au or more), then you would expect negligible orbital motion in the course of say a year and their relative position should remain quite constant.

On the other hand, if the "planet" is a much more distant star then it is quite likely to have a very different proper motion on the sky to the foreground star and so they will show an appreciable difference in their relative positions second time around.

More than two observations might be needed to be sure. The bigger the proper motion of the foreground star, the more decisive the test.


Are there any natural satellites in retrograde Sun orbit?

Yes, there are, but they are very rare.

Wikipedia has a list of retrograde asteroids, and mentions:

  As of September 2016, of the more than 700,000 minor planets known, there are only 95 known retrograde minor planets.

That means only 0.014% of known minor planets are in a retrograde orbit.

The asteroid with the highest known inclination is 2013 LA2, with an inclination of reportedly (175.189±0.007)° according to the JPL small-body database browser.


How does the Sun orbit the Sun-Jupiter Barycenter?

The barycenter of two point masses M1 and m2 -- or their 'center of gravity' (which to avoid any doubt is not in itself a center of gravitational attraction) -- lies on the straight line joining them, and divides the distance between them in the proportion m2:M1 -- in classical/Newtonian terms. That is, the barycenter is separated from center M1 by the fraction m2/(M1+m2) of the total distance M1 to m2, and by the complementary fraction M1/(M1+m2) from the center of m2.

In a two-body system, the positions and motions of each body relative to their barycenter are similar in the way that two triangles can be similar: the corresponding angles for each are the same, the distances are in a constant proportion. (See also https://en.wikipedia.org/wiki/Barycenter and https://en.wikipedia.org/wiki/Two-body_problem .)

So at least part of the answer to the question about the Sun and Jupiter -- imagined as a 2-body system -- is that the angles including the mean anomaly are indeed the same for each.

However, this only applies approximately to the figures quoted in the question. These are not 'one-body' or 'two-body' orbit figures, they are heliocentric approximate mean elements for Jupiter in the n-body solar system as at J2000 (i.e. 2000 Jan 1.5), from a derivation applied to data over the interval 1800-2050. The figures appear in a GSFC/NASA Jupiter fact-sheet, and also on p.316 of 'Explanatory Supplement to the Astronomical Almanac' (1992, ed. K P Seidelmann et al.), where the error in Jupiter's position computed from the elements is stated as up to 300" in right ascension.

Mean elements such as those quoted in the question can be derived from the time-varying osculating elements obtained over a long period. Such elements may be obtained for the chosen period in different ways, for example

(a) by deriving them from results of a long-term combined numerical integration of the orbits of a set of solar-system bodies -- usually involving all of the major planets along with either some of the minor planets as further individual bodies and/or an approximate allowance made for the effects of those not included individually;  or

(b) by deriving them from an analytical theory (a method used for example to obtain the results in J L Simon et al., 'Precession formulae and mean elements for the Moon and the planets', Astronomy & Astrophysics 282 (1994) 663-683).

In either of these cases, subtraction of the periodic terms leaves the mean elements as non-periodic polynomials as a function of time-interval from a chosen epoch associated with the data.

A third and approximate approach, which seems to have been the one adopted for the figures quoted in the question, involves fitting the osculating elements to a chosen polynomial function over a chosen interval. The website http://www.met.rdg.ac.uk/~ross/Astronomy/Planets.html suggests that the figures here resulted from a 250-year least squares fit of data from the JPL DE200 numerically integrated planetary ephemeris to a Keplerian orbit where each element was allowed to vary linearly with time.  The mean elements for a date such as J2000 would then represent the value of the linear function at the chosen date.

If it is wished to take the approximate mean elements for Jupiter at J2000 to represent 1-body elements on that date, then that is perhaps about as close an approximation of the kind as can be made: and the angles will then be similar as already described, and the barycentric distances will be obtained by the fractional split already mentioned.

A more precise way of tracking the relation between the Sun and the solar-system barycenter is outlined in a 1983 paper from JPL on solar-system numerical integration by XX Newhall et al.). This takes into account all of the planets and their masses, and the result is a complex path that has perhaps never been represented by any trigonometric series, but only numerically and diagrammatically -- e.g. at https://en.wikipedia.org/wiki/File:Solar_system_barycenter.svg, thus:


Accuracy of predicting an asteroid strike in 1987

There are problems with this:

What happens when an asteroid hits the atmosphere

The asteroid will be travelling fast, more than 10 km per second. In front of it the atmosphere will be compressed and heat up, this causes the "meteor" (a streak of light in the sky). It also causes puts a lot of stress on the asteroid.

If the asteroid is very small (sand grain to pebble sized), it gets destroyed completely.  If it is larger it can explosively break up. But several chunks can continue at terminal velocity (ie no faster than a stone dropped from a high building) and don't form craters. These chunks can later be picked up as meteorites.

If the body is very large it can survive the atmosphere, hit the Earth at hypervelocity (several km/s) and form a larger crater.

If you want the body to survive, but only form a small crater you need either a very strong body (a solid iron asteroid) or an odd set of events, such as at the Carancas impact. In these cases a small crater or crater field can form. The size of the impactor is about 50cm.

Detecting and forecasting impact

A 50cm asteroid is not currently detectable until very close to the Earth.  We have seen asteroids that are a couple of metres across, but they need to be close to the Earth (and we need to be lucky).

In 2008 we did get lucky and saw a small object before it hit Earth. In that case we had a few hours of observations and by about an hour before impact we were able to narrow the impact zone down to "Northern Sudan".  Now (North) Sudan is huge, an area about 3 times the size of Texas. And in 2008 we already had an active Near Earth Object search program, something that we did not have in 1987.  Even if the asteroid was noticed by lucky amateur, before the internet it would have been very hard to coordinate the observations needed even to predict "Northern Sudan". This meteorite exploded at altitude and did not form a crater.

To conclude, most meteorites will not take out a single house and leave neighbours standing.  Such meteorites that could are not currently detectable before impact.  Even with current technology and lots of luck we could not predict the impact location better than "somewhere between New Mexico and Florida" but not a single address.

There are other things that can destroy a house, perhaps a sinkhole? Perhaps these are more predictable (with luck) than an asteroid impact.


Why are wavelengths shorter than visible light neglected by new telescopes?

There are some technological issues to solve with putting any large telescope into space - and a space telescope is required at UV wavelengths. It is not possible to optimise such an instrument to work at both UV and IR wavelengths because of issues like cooling, mirror coatings and such-like. The simple angular resolution limit of a telescope goes $\lambda/D$, so on the face of it, to get equivalent resolution to an optical telescope, a UV telescope can be smaller. However, you also have to have optics that are good to a small fraction of a wavelength, so much better than the visible/IR. At even shorter wavelengths then conventional "optics" doesn't work because photons get absorbed and you move to the grazing incidence technologies of X-ray telescopes, which is a whole different game and much harder to achieve a given angular resolution.

Given all that, back in the 80s/90s I would guess that a decision was taken about the wavelength range to be covered by the successor to HST (ie JWST at a cost of approximately 10 billion USD) The real reason that no major UV successor to HST or IUE is ready to go now is simply that it is considered that the most important science priorities are achievable at near and mid-IR wavelengths. These are: observing the high redshift universe (essentially no UV light is detected from galaxies beyond a redshift of 3), observing star and planet formation (mostly in dusty environments where UV light cannot emerge and protoplanetary discs emit mostly at IR wavelengths) and doing exoplanetary science (planets are cooler than stars and emit mostly in the IR).

Thus, I don't think there are any technological showstoppers to a big UV telescope (at least the equivalent of JWST), it just comes down to science prorities.


Astrophysics of Supernovae - Energy over Distance?

In short, there are no nice standard formulas for this. One can make some order-of-magnitude calculations, though.

The key formula you need is the inverse-square law: the intensity of a spherical energy source falls off with the inverse square of distance. $$I(r)=\frac{I}{r^2}.$$ The useful thing is that if you know that some source with intensity $I_1$ has certain effects at distance $r_1$ (say, a supernova being dangerous at 50 light years) then you get the same effects from a source with intensity $I_2$ at a distance $$r_2=\sqrt{\frac{I_2}{I_1}}r_1.$$ So a hypernova 10 times more luminious than a supernova would be dangerous out to about 158 lightyears (assuming the initial 50 lightyear range - this is where different papers and models might quibble a fair bit). Notice that you need a fairly large luminosity to get a vast range in this case.

Hypernovas are likely progenitors of gamma-ray bursts that release most of the energy along fairly narrow jets, 2-20 degrees across. That would amp up the range significantly. If the jet covers a fraction $f$ of the sky, that means the range is now $$r_{2 GRB}=\sqrt{\frac{I_2}{f I_1}}r_1$$ where the intensities are the true energy releases. For an opening-half angle $\theta$, $f=1-\cos(\theta)$ so we should expect $f$ to be between 0.00015 and 0.015, basically increasing the range by a factor 81 to 8.1 if one is unlucky enough to be in the beam.

The exact effect on different planets depends on many complex factors. Planetary evaporation is not a simple process unless the star delivers far more energy than the gravitational binding energy. The absorbed energy $\pi \epsilon R^2 I(r)$ where $0<\epsilon<1$ is the absorption efficiency must be much larger than $3GM^2/5R$, or $$ I(r) \gg \frac{3GM^2}{5\pi\epsilon R^3} $$ This scales proportional to mass and density, so to evaporate a ten times more massive planet you need ten times more radiation, assuming the same density.

Merely melting the surface or blowing off the atmosphere would depend on a lot of geophysical details. $\epsilon$ depends a lot on wavelengths, whether there is a plasma layer forming and similar messy issues. The light curve timecourse also matters, since some of the processes will be hydrodynamic flows rather than instant shockwaves or long-term heating.

A crude model might be that the absorbed energy heats up the atmosphere initially as $$\Delta T = \frac{\epsilon I(r)}{ C_P M_a}$$ where $C_P$ is the specific heat and $M_a$ the mass of a one-square meter column of atmosphere. For Earth, $M_a=10300$ kg and $C_P=1.00$ kJ/kg.K. A $10^{47}$ J hypernova at a distance of 50 ly radiating evenly would deposit $10^{47}/4\pi (50 \text{ly})^2\approx 35\cdot 10^9$ W/m$^2$, heating atmosphere if we assume $\epsilon=0.1$ by 345 K - enough to broil us, but not enough to evaporate the oceans (there $M$ is 1000 kg per meter depth, and $C_P=4.2$ kJ/kg.K). Were it just a supernova 100 times less intensive the effect is a balmy 3.45 K. On the other hand, with a focused GRB beam we can get a hundred to 10,000 times more heating (at which point the above formula stops being valid since the air ionises and becomes plasma).


What is minimal aperture/magnification to recognize phases of Galilean moons?

From Earth the resolution of is limited by the atmosphere.  Turbulence in the atmosphere make it impossible to resolve the moons of Jupiter to more than a few pixels wide.

Amateur refractor telescopes tend to be small and less powerful than reflector telescopes. I'm not aware of any amateur refactor telescope that could resolve a moon of Jupiter to more than a point of light. The eye is unable to resolve an object that is less than about 60 arcseconds and it would need to be closer to 1000 arcseconds before many surface details became visible. The Moon, for comparison, is 1800 arcseconds.

Professional equipment can partially overcome the atmosphere by using adaptive optics, but this is not a feature that is generally available to amateurs.

Jupiter and its moons orbit outside the orbit of Earth, so from our point of view Jupiter is never partially lit less than 99%. The moons are sometimes eclipsed by Jupiter, but they too are never partially lit less than 99%.  With amateur equipment, Jupiter and its moons are never anything other than "full".

Even with the Hubble telescope, the resolution is not sufficient to identify phases of the moons.

What Galileo saw was Jupiter, resolved as a disc and 4 moons orbiting it which appeared to him as "stars" (he called them the Medicean stars). The four moons were not always visible, sometimes a moon was hidden in the shadow of Jupiter, or behind Jupiter from our point of view. At other times a moon would be too close to Jupiter to be visible, or outside the field of view of his telescope.

What he realised was that these were bodies in orbit around Jupiter, and that they obeyed the same rule for orbits that Kepler had proposed for the planets.


In a planetary system close to the galactic core, would it be possible to see the supermassive black hole?

If the black hole is active, meaning it is still capturing matter from its surrounding, it will have a large accretion disk all around, which is the only way to dissipate angular momentum for the matter falling into it.

As a result of this dissipation all the matter will warm up and emit radiation. This disc will be fairly large and thus clearly visible as a bright object in the sky.

This page shows an image captured by Hubble of such a disk:

Strictly speaking then, one cannot see directly the black hole, as its view will be covered by the bright emission of the disk.

However, the presence of the disk will allow to observe the black hole thanks to its gravitational effect.


How does gravity interact with a photon?

It is simply not true that gravity can only interact with mass. Rather, any long-range spin-2 force interacts with all energy-momentum equally, and it source is the stress-energy-momentum tensor. That is one way to state the equivalence principle.

Note that a massive object in its own rest frame has an associated energy $E = mc^2$, which under ordinary conditions is usually much higher than any stress terms (including pressure) or momentum (outside the rest frame). Thus one can usually pretend that gravity couples to mass, but it isn't so--rather, the gravitational charge is energy, and the entire stress-energy-momentum tensor couples to the gravitational field. This is analogous to how for electromagnetism, there is electric charge, but the electric currents also make a difference.

In the weak-field limit of general relativity, one can consider the fact that light is gravitationally deflected twice as much as a naive Newtonian prediction would be (for an object at the speed of light under Newtonian mechanics, that is) as the fault of light putting a pressure in the direction of its propagation equal to its energy density. But in general, the motion of a test particle is determined by its four-velocity and the geometry of spacetime only, and the light would just be a special case of having a 'lightlike' four-velocity.


Why do we only see the clouds directly infront of the moon glow from moon light?

This is an experiment that you can do at home:

Put a paper fixed in front of a light bulb
Move your head around and see how the bright spot on the paper accompany your moves.

But, why is that? How is it that not all the paper is lighted the same?

When light travels through a translucent material some of it gets reflected back, some bounces inside of it and some goes through. As always the shortest path is the easiest, so the light coming straight "bulb-paper-eye" gets through easily (brightest spot) and as you go further from the center you are not seeing direct light anymore, but diffused light, so it gets dimmer and dimmer because to get to your eye it has to bounce in the right direction. That doesn't imply that the rest of the paper is not getting light, it is just not bouncing in your direction and the same happens with the clouds.


What are the units of distance in this subtended angle calculation?

The units of radians are 'dropped' because unlike most units, they are dimensionless. Recall that the definition of radian angle measure is the ratio of the length of a circular arc to its radius. Thus, radian angle measures have units of $[\mathrm{Length}]/[\mathrm{Length}] \equiv 1$, i.e. dimensionless.

Radians are units in the sense that they give information about what standard the angle quantity is measured by. Thus, the convention of writing $\mathrm{rad}$ is useful in that it distinguishes it from other ways of measuring angles (e.g., in your question, arc-seconds), but in terms of dimensional analysis, $\mathrm{rad} \equiv 1$, so $\mathrm{m}/\mathrm{rad} = \mathrm{m}$.


Why do planets tend to rotate in the same direction although they have formed from tumbling asteroids?

You are right that the tilt of the asteroids are distributed in very random way, and that the rotation of the Solar nebula is a minor contributor to that tilt, and only skews it a little.

However, you are not right that randomness simply adds up. The randomness does in fact cancel out more and more when you combine a large amount of asteroids, until the rotation of the nebula becomes the dominant factor. This is related to the Law of large numbers.

For instance, throw a dice. The outcome is random. Throw 10 dices, calculate their sum, and divide by 10. Not so far from the average any more? You can do the same thing with thousands of dices, or millions of asteroids. When the number of asteroids that form an object is really high, the tilt is not going to be far from the average value, determined by the nebula's rotation.

The same argument goes for inclination, and the fact that even though the planets' orbits are elliptical, they are not as far from circular that a random orbit would be.


Where might astrometric observations of the Roadster spacecraft be catalogued?

The Minor Planet Center inconspicuously puts this under "Distant Artificial Satellites Observation."
DASO Circulars 561-566 have observations of the roadster 2018-017A.
It's also on the short list of artificial objects for which MPC can generate ephemerides.


We don't need to see things to understand them.

It is now generally understood that gold is formed in neutron star collisions.  The evidence for this partly theoretical: Models of neutron star collisions suggest they would produce large amounts of heavy elements, and regular supernovae would not produce gold in the right amounts to explain the concentrations found in the solar system.

Secondly, we have observational evidence from the collision of two neutron stars, detected by gravitational waves, which allowed for scientists to analyse the light coming from this collision in detail. We can't see gold, but we can see radiation that indicates the presence of radioactive isotopes that will decay to form gold.

This gold would float in the interstellar space for a while until it forms part of the dust around a newly forming star. That dust gathers together to form planets. Most of the actual gold we find came to the Earth carried in by meteorites, as the original gold will have fallen to the Earth's core when the Earth was molten.

So we have a good model for explaining how gold can be formed in neutron star collisions and evidence for this actually happening. That is enough for this to be the generally accepted theory.


How did the Mars satellites "take cover" from the recent comet flyby?

I've written an answer or two dealing with this within the last week, and I can't believe I never thought about this. Great question.

  These spacecraft are in orbit around Mars, constantly in motion, so they can't very well just pack up and move to the other side of Mars to avoid getting hit with the dust.

Believe it or not, they can just pack up and move. All they have to do is alter their orbital speed - i.e. either slow down, speed up, or change their orbit so they will automatically go a different speed, and so be in a different place. They execute a series of orbital maneuvers to ensure that they are in the part of their orbit that temporarily carries them behind Mars. Orbital maneuvers aren't hard; practically any spacecraft can modify its orbit slightly, if need be. According to this,

  The orbiters will be repositioned to a “safe zone,” where Mars will actually help shield the satellites from the cloud of dust particles.

In other words, the spacecraft will be moved so they are on the opposite side of Mars as the comet. They can't be there for long - after all, they orbit Mars extremely quickly. But, fortunately, they don't have to be there for long. As this says,

  The satellites will remain in this region for a period of up to 40 minutes, until the threat has subsided.

Why do short a period? Well, the comet will only be in the area for a short time: NASA estimates the worst part will last roughly 20 minutes. After that, the danger will have subsided. Any particles that could damage the spacecraft will be long gone.

  Did these orbital modifications affect the quality of research these satellites will be able to perform in relation to Mars?

I can't find a source for the answer to this, but I would expect there wouldn't be any problems. You should be able to undo any maneuvers that would change the orbit of the craft.


There are many such online services, most of them free and a Google search away, so I won't make this a list answer, and since you didn't specify what data in particular you're hoping to find, here's one of my favorite satellite trackers with fairly complete position data, ground track display, map options and so on: NASA iSAT (Interactive Satellite Viewer). You might want to allow it to use your location for more convenient display, but it isn't necessary to enable that.

That's for a desktop computer, but if you're hoping to spot satellites (a.k.a. tracking) on the go, then I'd personally suggest installing Sky Map + on your smartphone / tablet (it is free, but there are of course others), because it includes a lot more useful information and has both compass / gyroscope / GPS positioning for easier identification of spotted satellites (it pans automatically) as well as manual panning mode. You can tap on screen to display the toolbar and select objects that you want displayed, or switch to ruby & black monochromatic night mode, if default display is too bright. Satellite information can be updated in settings via an additional download, and all of it is free.

For downloadable data alone (to perhaps include in your own software or import into your database), one of such sites is CelesTrak that includes TLE (Two-Line Element) sets and a Space Track TLE Retriever 3 tool to download Space Track data sets.


I know you didn't specifically ask this, but I feel it answers the spirit of your question. I'm just going to go through the list of features and specifications and explain what they mean.

Features

Maximum magnification: 675x

This should hopefully be obvious, but this claims to provide images that are 675 times larger that they would appear to the naked eye. However, don't get too excited about this number. When it comes to telescopes, the magnification power is not the most important number. Telescopes for amateurs will try to play off this ignorance though and very often talk about how good that telescope's magnification to make it sound impressive.

Large aperture offers clearly illuminated views

The aperture is the physical size of the collecting mirror/lens. Usually aperture size is specified as the diameter of the main mirror/lens and larger is better. A larger aperture means the telescope can collect more light and thus you can see fainter objects you couldn't normally see with your naked eye. The phrase "clearly illuminated views" doesn't really mean much though, its just marketing.

Stable construction provides solid support and control

I guess they're just telling you the telescope won't wobble on it's stand. Which is good. It would be hard to look at a fixed object in the sky while your telescope is wobbling all around.

Metal tripod with slow motion control rod

Not sure it matters that the tripod is metal or not. But they do say it has a "slow motion control rod". Basically, these are rods you can turn to move the telescope very slightly rather than swinging it with your hand. This just provides the ability to finely move the telescope to help point it precisely.

Strong performance in both high and low magnification

Pretty obvious statement, albeit a vacuous one. What counts as "strong performance"?

Easy sky focusing with finderscope

The telescope comes with a finderscope. You can see that as the small "mini" telescope attached to the main telescope. When trying to point your telescope at a specific star, it can be hard to know if you're pointing at the right one. The finder scope provides some basic ability to view a larger region of the sky and try to get your telescope generally pointed in the right direction. From there, you can look through the main telescope and use the "slow motion control rod" to get it pointed exactly right.

Specifications

Aperture: 60mm (2.36inch)

This is the aperture I mentioned above. They're telling you the diameter of the main lens. Bigger is better here because it means the telescope can collect more light and see fainter objects. That's why professional astronomical telescopes are constantly getting bigger and bigger. I'll note that they're probably playing mental games by saying its 60 mm when 6 cm would have sufficed. More than likely, 60 sounds more impressive than 6 which is why they chose to use mm.

Focal Length: 900mm.f/15

This tells you two pieces of information. First, that the focal length is 900 mm. This telescope is a Refraction telescope. The light enters the telescope and immediately hits the main lens. The main lens is curved and bends the light, focusing it to a point. This focused light it sent to the eyepiece which will then direct it right to your eye for viewing. They're telling you that the telescope focuses the light to the eyepiece over a distance of 900 mm.

The other quantity is the f-number. It is the ratio of the focal length to the aperture size. You can hopefully easily see that 900/60 = 15 and thus the f-number is written as f/15. The "/" here does not imply division, it's simply how the f-number is written.

To be honest, neither of these two quantities will really matter much for you. I think they're just quoted because they're common "optics" numbers you'd see with higher end cameras and they want this telescope to sound like a serious product. The only thing about this that affects anything for you is the field of view, that is how much of the sky you can see at one time. This will only matter when looking at extended objects like the Moon. A longer focal length means a larger field of view. The f-number is virtually useless for astronomical applications. It only really applies in photography.

Finderscope: 5x24

This is again the finderscope I mentioned above. It's used to get your telescope pointed more or less in the right direction so you can then look through your telescope and do fine pointing. The number 5x24 signifies that it magnifies by 5x and has a 24 mm diameter aperture. This is going to be cheap and may not focus very well so don't expect anything amazing. You'll only use it to point the telescope though so if you're fine with that then so be it.

Diagonal Prisms: 90degree

This is just the prism at the end of your telescope, and before the eyepiece, that let's the eyepiece be 90 degrees turned from the main telescope shaft. This is useful because you can sit next to your telescope and view through the lens on the side rather than having you align your eye directly with the telescope's main barrel. It just provides easier and more comfortable viewing.

Slow Motion Control Rod For Easy Vertial Micro Adjustment

Already described this above.

Height: 128cm

Just giving you the height of the telescope. Although it's unclear precisely what height this is. Certainly the height changes as you move the telescope around on the mount. Is this the maximum height? Is it the height when the telescope is level? Who knows.

Standar: 0.965” Acceddoness Include

I actually have no idea what this means. Clearly they misspelled Standard. Other than that, I don't know what an "Acceddoness" is.

Eyepiece: SR4mm H12.5mm H20mm

Just providing you with descriptions of the three eye pieces that come with this telescope. The measurements are the eyepiece aperture sizes. A larger size generally means higher magnification and higher field of view.

3X Barlow Lens, 1.5X Erector

There are two additional optical pieces provided. The first is a "Barlow lens". This will sit in front of your other eyepieces and serves to provide an additional 3x magnification. The second is the Erector lens. It will provide an additional 1.5 magnification but it also will "erect" your image. The optics of your telescope will flip your image upside down. This is just how optics works (and it's the same reason you appear upside down in a spoon). In truth, when looking at astronomical objects, it doesn't really matter if you're looking at it upside down or not, but cheaper telescopes like to provide the illusion of having fancy gadgetry like erector lenses. There's not much of a reason to flip the image to be "correct".

Telescopic Aluminum Tripod

It's just a basic metal tripod that lets you sit your telescope on a stable mount. The legs can adjust to different heights and are thus "telescopic" which has nothing to do with the fact that it is a tripod for a telescope.


So, what exactly is an 'ultra-cool' dwarf star?

Breaking the phrase down:

Dwarf star - a term I will never understand - is used to describe relatively small, dim stars. Unfortunately, this encompasses most main-sequence stars, which are indeed dwarfs compared to large giants and supergiants.
Ultra-cool, as called2voyage already discussed, means that the star has an

  effective temperatures of less than 2,700 kelvin.

This is about half of the Sun's surface temperature. Spectroscopically, it means that these stars are of class M7-L8, i.e. really, really cool stars that just barely reach the threshold for nuclear fusion. Actually, some L dwarfs will never fuse hydrogen as normal stars do, and will become brown dwarfs. Therefore, "ultra-cool dwarf" doesn't necessarily refer to just main sequence stars.

Other characteristics in some cases (from Cruz et al.):

Lithium in some L-type objects, indicative of low temperatures, low masses, and young ages in those particular objects.
Very low surface gravities in late M-type dwarfs (though likely not as much in an M7 dwarf like TRAPPIST-1).
Very low metallicity in certain other L-type objects.

TRAPPIST-1 actually appears to be a rather massive ultra-cool dwarf, then, in comparison to some of these other objects.

Here's an annotated Hertzsprung-Russell diagram, with a box for ultra-cool dwarfs and a circle for where TRAPPIST-1 is, approximately:

Image from Wikipedia courtesy of user Saibo, under the Creative Commons Attribution-Share Alike 2.5 Generic license.


How fast does Venus move as seen from the earth?

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.


Where can I find the positions of the planets, stars, moons, artificial satellites, etc. and visualize them?

There are many resources online, so this is a community wiki
answer. Please feel free to add to it!

If you want to visualize the stars/planets/etc (as viewed from Earth
or another location), you are looking for planetarium software:
https://en.wikipedia.org/wiki/Planetarium_software

If you want accurate positions for stars/planets/etc, you are looking
for horizons:

https://en.wikipedia.org/wiki/JPL_Horizons_On-Line_Ephemeris_System
http://ssd.jpl.nasa.gov/?horizons

If you want to compute star/planet positions yourself, you have
several options:

SPICE (http://naif.jpl.nasa.gov/naif/tutorials.html) will give you
results that match HORIZONS very closely. You can also use some of
SPICE's functionality online at
http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation

SPK (Spice kernel) files are available at https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/ -- while these are primarily intended for use with CSPICE, the format is documented and you can use it directly. It's described at https://github.com/skyfielders/python-skyfield/issues/19 and implemented in Python at https://github.com/brandon-rhodes/python-jplephem/tree/master/jplephem

If you want to do the SPICE computations yourself (numerically solving the differential equations), see https://astronomy.stackexchange.com/a/13491/21

You might also want to use an n-body simulator to do the computations yourself: https://physics.stackexchange.com/questions/25241/what-open-source-n-body-codes-are-available-and-what-are-their-features

IAU SOFA (http://www.iausofa.org/) will give you the International
Astronomical Union's "official" libraries to compute positions.

VSOP theory (https://en.wikipedia.org/wiki/VSOP_%28planets%29) is
another option.

PyEphem (https://en.wikipedia.org/wiki/PyEphem), which derives
from MIT's libastro library
(https://stuff.mit.edu/afs/athena/project/xephem/src/xephem-3.5.2/libastro/)

Skyfield: https://github.com/skyfielders/python-skyfield

libnova (http://libnova.sourceforge.net/)

If you want to treat planetary orbits as simple ellipses and ignore perturbations, you can find orbital elements at https://ssd.jpl.nasa.gov/txt/p_elem_t1.txt but you'll probably want to visit https://ssd.jpl.nasa.gov/?bodies#elem first

Note that these computational libraries don't always agree with each
other or with planetarium software:
https://stackoverflow.com/questions/16293146

If you want a catalog of stars/etc, NOMAD
(http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad) and GAIA (http://gea.esac.esa.int/archive/) are
the largest such catalogs (about 1 billion entries each, lots of overlap). The NOMAD link also provides references to smaller catalogs.

For high-resolution astrophotography try ALADIN (http://aladin.u-strasbg.fr/#AladinLite)

If you want to find artificial satellites and spacecraft, start at
https://en.wikipedia.org/wiki/Two-line_element_set

Transits of Mercury and Venus across the Sun? http://eclipse.gsfc.nasa.gov/transit/transit.html

Planetary conjunctions?
http://laps.noaa.gov/albers/ast/conjun/conjun.html

Occultations of stars by asteroids? The International Occultation
Timing Association (IOTA) at http://occultations.org/

Create your own Sun-Earth-Moon model by using a ~17 inches (~44 cm) beach ball or box as the Sun, 4-5 mm object like a pencil eraser as the Earth and a 1 mm sized drill bit or other object to represent the Moon. Then, find a 150-foot string or tape measure. The sun and earth are ~154 feet (~46m) apart. The earth and moon are ~4.7 inches (12 cm) apart.


Which things can LIGO see that LISA can't, and vice-versa?

Gravitational wave detectors have a frequency range that they are sensitive to.

In the case of LIGO it is about 10Hz to 1kHz. The lower limit is imposed by seismic noise, the upper limit by "shot noise" (basically not having enough photons to sample the interferometer path difference at high frequencies).

LISA is in space and doesn't have the problem with seismic noise. However, there is still an upper limit determined by shot noise. As I understand it, LISA will not/can not use the same sort of resonant cavity arrangement that LIGO uses to boost the effective laser power and hence number of photons in the apparatus. Thus the upper frequency limit for LISA is more like 1Hz.

So now to your question. What LIGO can see that LISA can not, are gravitational waves with frequencies of 10-1000 Hz. The astronomical phenomena that lead to such waves are the merger of compact binary systems with stellar masses, rapidly rotating, asymmetric pulsars and perhaps supernova explosions.

Lower frequency GWs can only be seen with LISA. This would include stellar binary systems with orbital periods longer than about 10 seconds, merging supermassive black holes and maybe GWs from the big bang.

What could move from being observable in LISA to being observable in LIGO? In principle, the merger of any binary system results in a gradually increasing frequency and amplitude. Stellar mass, neutron star or black hole binaries can get to frequencies around 1kHz before they merge, but sweep through lower frequencies before that, but with much lower amplitudes. Perhaps the best bet would be the merger of intermediate mass black holes with masses of $10^3$ to $10^4$ solar masses, which would have merger frequencies of tens of Hz, but a significant amplitude at much lower frequencies prior to merger.

I suppose the implication of the quote you gave is that this newly discovered inspiralling white dwarf binary might be another possibility. I don't think that is correct. It has a 7 minute orbital period, which puts it in LISAs frequency domain (the GWs emitted have twice the orbital frequency), but because white dwarfs are physically larger than neutron stars, the peak frequency at merger (some time in the future) will only be a few Hz and not visible to LIGO.


What type of telescope can show cliffs on the moon surface from a city location?

The main things to look for are:

Decent optics (nearly anything except those with plastic lenses).
A steady mount that points where you want to, and moves smoothly. An altitude-azimuth mount is fine because with practice you can guide at high powers.
The eyepieces should be 1 1/4 inch size or more expensive 2 inch. The older 0.96 inch are very hard to find. This lets you upgrade eyepieces later.

The larger the aperture the better. Don't worry about magnification.

Hint: find your local astronomical society, ask there, go to one of their "star nights", they may even have telescopes on loan.


Can someone help me to identify this section of the night sky?

I've just submitted a cropped section to Astrometry.net (that's astrometry, not astronomy). Click the "web" to upload an image and start the search, then take some time to read How (the heck) does Astrometry.net work? while waiting for it to compute...

No quick solution. That doesn't mean that this isn't real, it's just that the first attempt didn't succeed.
better luck next year!


Consider you have a capsule with water at $T=300$ K. Water will start boiling right after you remove containing pressure. Evaporating water will take out the heat, required for phase transition into vapor, so remaining water will cool off. It will boil until it will reach $T_0=273$ K temperature. Then part of water will start to transit into ice. The remains are ice and vapors. Solving equation $c_p m_i + c_p m_i (T-T_0) - c_L m_v = 0$, where $m_i$ is mass of ice, $m_v$ is mass of vapor, $c_L,c_p$ --- energy required for phase transitions, you can obtain fraction of ice which will remain. Whole process takes up to few minutes (depending on what size is the hole of container). Quantity of water does not change the rate of evaporation, which mostly depends on the heat transfer. The heat transfer in this case is the exchange between vapor and ice, i.e. between water and water itself, which does not depend on quantity or the size of free surface. Water will boil in every point, over all volume at once.

If the hole of container is small, then process will be slower and remaining water will solidify into a solid brick of ice. If the hole of container is large, then all water will jump out immediately and break down into small drops. Similar to liqiod nitrogen breaking down when you pour it on table.

Atmosphere is the exact reason for oceans to exist on Earth, so without it the oceans have no meaning. Jupiter and Saturn have oceans of their own, but those are hidden beneath very thick atmospheres.


Has great eyesight been necessary for astronomers?

Johannes Kepler
Wikipedia:

"However, childhood smallpox left him with weak vision and crippled
hands, limiting his ability in the observational aspects of
astronomy."

He made great use of Tycho Brahes great systematic observations in his theoretical work. He did not need exceptional eyesight for his developments in optics and telescopes.


Here is a link to a diffraction grating that can do what you want. It is mounted in a 1.25" filter ring that attaches to an eyepiece, or to most astro cameras. I believe they also sell adapters for other cameras and software to extract spectra from the photos.

Here is a link to a spectra of Vega that I took with a 120 mm refractor using  the Star Analyser grating.


Your confusion may stem from the way we sometimes draw the Sun (not to
scale) as though it's close to the Earth.

In reality, as you note, the lines of vision to the Sun are nearly parallel:

Note that the angles (measured from the red line, the horizon, to the
arrow) are still different, but for entirely different reasons.

I've actually stolen @Andy's answer (and @Tanenthor's comment), but
I'm hoping the diagram makes it easier to understand.

Re the parallel lines, the 11810x60 image below may help explain: even
though the lines eventually converge (on an object about 10 times the
distance to the moon, still nowhere near the distance to the Sun),
they are nearly parallel when leaving Earth.


While objects blocking the sun may have an effect on the twinkle if viewed on another planet, a twinkle is mainly caused by the effect the atmosphere of a planet has on the light, or something interfering with the light that is passing to your eyes.
From Wikipedia's article on Twinkling:

Twinkling, or scintillation, is a generic term for variations in apparent brightness or position of a distant luminous object viewed through a medium. If the object lies outside the Earth's atmosphere, as in the case of stars and planets, the phenomenon is termed astronomical scintillation; within the atmosphere, the phenomenon is termed terrestrial scintillation.[2] As one of the three principal factors governing astronomical seeing (the others being light pollution and cloud cover), atmospheric twinkling is defined as variations in illuminance only.
Stars twinkle because they are so far from Earth that they appear as point sources of light easily disturbed by Earth's atmospheric turbulence, which acts like lenses and prisms diverting the light's path. Large astronomical objects closer to Earth, like the Moon and other planets, encompass many points in space and can be resolved as objects with observable diameters. With multiple observed points of light traversing the atmosphere, their light's deviations average out and the viewer perceives less variation in light coming from them.


Your reasoning is at least true in part.
First off.  We have only one example of life.  It is based on carbon chemistry and water solutions. As we only have one example, we don't know if fundamentally different chemistries are possible. We just don't know if the Earth is unique or special, or whether life is commonplace throughout the universe.
If you head over to https://worldbuilding.stackexchange.com/ you will find lots of people speculating of different biologies, including those with a sulphur and sulphuric acid based chemistry
However, given our understanding of chemistry, it does seem likely that most life in the universe will be carbon/water based.  Carbon is the only element that produces complex structures, and is also one of the most common elements.  Oxygen is another very common element, but it is very reactive and so will tend to react with hydrogen to form water.  Water also happens to be one of the very best solvents.  So carbon seems to be the best element to build biological compounds and water is the best solvent to carry those around.
This means that is is a reasonable hypothesis that life will be found most often on planets that have surface temperatures and pressures that allow for liquid water.  That is a pretty narrow range.
We are adapted to Earth.  We have had 4 billion years of evolution to adapt to Earth, even as Earth changed.  For example we have adapted to an oxygen-rich atmosphere.  We could have adapted to Sulphuric acid (at least to dilute sulphuric acid).  Oxygen is as toxic as acid to those bacteria that haven't evolved to tolerate it.
Whether the Earth has adapted to us is more controversial.  This is the Gaia hypothesis of James Lovelock "that living organisms interact with their inorganic surroundings on Earth to form a synergistic and self-regulating, complex system that helps to maintain and perpetuate the conditions for life on the planet." (wikipedia)  However "the Gaia hypothesis lacks unambiguous observational support and has significant theoretical difficulties" ( David Waltham. Lucky Planet).
We don't know if the Earth is unique, nor even if it is special.  Our current ability to detect planets would struggle to detect the Earth, if it orbited another star. So we don't know how planets there are that have liquid water on the surface. We also don't know how many of them harbour life in any form. We don't know if life is possible in alternate chemistries. Even if life turns out to be common, we might still be special in that we have oxygen-breathing, multicellular, complex and intelligent life.
If it turns out that life is rare... we don't know why it would be. Life started pretty soon on Earth.  If it turns out that life is actually rare we would need to explain that. But we don't have enough data to do so yet.


How do I build a stargazing aficionado's Sun funnel to observe our nearest star safely?

As for projecting the Sun onto a screen at a low cost, I would recommend starting with a ~50-200$ sunspotter box, which is basically a lens mounted on a wooden box, that projects the sun onto a white piece of card. The advantage of using a telescope is that it can be programmed to track the Sun, so that if you want to trace sunspots, for instance, you can do so at leisure, instead of having to race to finish before the Sun moves off of the page.

As for observing coronal features- this is a very difficult task. A H-alpha filter (expensive) can be purchased for a home telescope outfitted for solar observing. This would allow you to see filaments/prominences, which is low T (10^4 K) chromospheric material that extends into the corona, suspended by magnetic field and buoyancy forces.

To observe coronal material from Earth, you would need a coronagraph, which blocks out light from the central disk of the Sun. This is because coronal features are extremely dim, as compared with the photosphere. From Earth, coronal features are best observed with during a total eclipse. Using a telescope mounted coronagraph at home is not ideal (and probably not feasible) because a lot of light from the Sun would leak into your field-of-view, due to both diffraction, as well as sky brightness. The University of Hawaii takes these sorts of observations, but even with all of their expensive equipment, the space-based LASCO instrument is able to do far better.


Why does Pluto's Orbit underlap Neptune's Orbit

There are a few peculiarities of Pluto's orbit. These are:

Its high orbital eccentricity (e = 0.25) causes Pluto's perihelion to be ever so slightly smaller than Neptune's perihelion.
Pluto has an orbital resonance with Neptune. Its orbital period is exactly 3/2 of Neptune's. This orbital resonance is the cause of its orbital eccentricity that you ask about.
Pluto has an orbital inclination of 17 degrees to the ecliptic. No one knows why, we can only guess.

  The origin of Pluto's eccentricity explained:

During the later stages of planet formation, in this early Solar system, the planets were still forming. Neptune exchanged angular momentum with the remaining planetesimals, and its orbit expanded outwards. If Pluto were in a near-circular orbit larger than Neptune's (at about 33 AU), there is a high chance that Neptune could capture Pluto and lock it an orbital resonance (this probably happened when Neptune was at about 25 AU). As Neptune's orbit continued to expand outwards, this expansion drew out the eccentricity of Pluto's orbit.

Source and further reading: The Origin of Pluto's Orbit: Implications for the Solar System Beyond Neptune


Will I be able to see the "Beads of Baily" at 99%+ totality?

No, at 99% a narrow sliver of sun will be visible, the width of that sliver will be about 1.5% of the diameter of the sun at its thickest point.

Baily's beads are caused by the sun shining through valleys on the edge of the moon. Such a valley would have to be about 50km deep (or surrounded by hills 50km high).  You do get some beads at the thin edge of the sliver but these would be too bright to see against the sliver of sun (which would still be blinding bright)

To get Bailey's beads you need totality, with the sun shining through valleys. Baily's which is why they are normally only visible for a few seconds before true totality, usually a couple of seconds after the predicted start of totality.

You can see the worksheet I used to make these estimations at https://ggbm.at/sZ3EYp7z however it is undocumented.


what is the net effect of nodal and apsidal lunar precession?

(Hope I’m not too late, two and a half years later… Just saw this post now—and I wasn’t a member of Astronomy SE back then…)
The short answer is no. These two effects don’t occur in the same plane. Imagine the lunar orbit as an elongated ellipse—much more elongated than it is in real life. Imagine the Earth as one of the foci, closer to one tip than to the other. Imagine the plane of the ecliptic as, if you wish, parallel to your floor or your desk. Now, the Moon’s orbit is tilted—by a small amount, about 5.14°—to the ecliptic (NOT to the Earth’s equatorial plane, like other planetary satellites, but that’s a different story). So imagine your first ellipse tilted with respect to your floor or desk.
The first movement, the precession of the perigee, makes it so the ellipse pivots in is own plane around the Earth. So its plane doesn’t change at that point.
The nodal precession, though, changes the orientation of the plane. The movement is still around the Earth, but parallel to the plane of the floor or the desk.
The closest everyday parallel I can find is if you own a ventilator (fan) that pivots on itself to aerate in different directions. The movement of the fan blades would represent the apsidal precession (the first you mention), while the swiveling of the fan head left and right would represent the nodal precession.
Hope this helps!


Why do we not find larger quantities of helium in the Sun's atmosphere?

Helium is burnt in Sun's core, and it is heavier (specific weight) than hydrogen.
Hence most of the helium stays in Sun's core.
(More details)


How fast will the sun grow when it becomes a red giant?

Yes, the time is very long, but exactly how long depends on when you "start the clock" for the expansion to a red giant.  If you take the end of the "main sequence" (when the core hydrogen runs out) as the start of the expansion phase, it's more like a billion years to complete the process of becoming a maximally expanded red giant.  But the answer that gets 5 million years focuses only on the phase of most rapid expansion (which begins after the star is already quite a bit larger than the Sun).  Either way, our lives a the smallest blink of an eye to a star!


And also there are some terms related to black holes, 3D shape, ball
  etc. Apart from those complexity, my intention is simply to be
  clarified, why we call our universe flat? And by word flat, what I
  meant "having a level surface; without raised areas or indentations".

Imho in order to accept the answer, you have to realize the contradiction in the sentence above: You mean level surface, but we call the universe flat :) "Flat" in general relativity does not refer to "having a level surface; without raised areas or indentations". Please accept this first.

Now let's think of a ball (yes, a ball; it's not complexity). Would you call its surface flat or curved? I'd call it curved. Let's think of a pan. I'd call its surface flat.

Both the surface of a ball and a pan have two dimensions. The ball itself has three dimensions obviously, but its surface has two dimensions. I hope this is straightforward.

Now, everything described above about surfaces also applies to volumes. Volumes can be flat, and they can be curved. The issue is that humans cannot visualize things with three dimensions bending, so they cannot visualize a curved volume. But this is okay: We can still grasp the meaning of flat and curved, just by seeing it on surfaces, and understand that it can be extended to volumes.

However it's not just visualization and understanding: Flat and curved volumes (i.e. spaces) have very different physical and mathematical properties. A usual example is that in a curved space the sum of the angles in a triangle is not 180 degrees. There are also other properties and those enable astrophysicists to measure the actual curvature of the universe. I hope this was helpful.


Why are O III lines so prominent in the spectra of emission nebulae when the amount of oxygen relative to hydrogen is a million times smaller?

This is an excellent question. Think about the way in which emission happens. $\text{H}\alpha$ emission happens when an electron makes a transition from the third energy level to the second, emitting a photon in the process with the energy equivalent to the difference of the energies of the two states, roughly $1.9\text{ eV}$. There's a lot of hydrogen in emission nebulae - and in nebulae in general - and so it plays an important role in the emission spectrum.

Emission from $[\text{O III}]$ comes from a different process: collisions. Free electrons with energies of $\sim1\text{ eV}$ - about the amount of the oxygen's low excitation potential - collide with oxygen atoms. This produces forbidden lines that would otherwise have been unlikely to be created, as the ionizing photons typically would not have the right energies. Therefore, doubly-ionized oxygen (in particular) becomes extremely important in spectra, generally second only in intensity to $\text{H}\alpha$.

Given that $[\text{O III}]$ emission is collision-based, it makes sense that there is a temperature dependence - which there is, as is the case with $[\text{N II}]$ lines and is responsible for their high intensities. Conditions in many hot nebulae (in this case, $>10,000\text{ K}$) are right for such collisions and thus forbidden line emission. Hot host stars can sometimes help achieve high electron temperatures.


Given that Kepler's Third Law as stated on most websites doesn't include mass, how does it work for planets in other star systems?

Though the question may be closed as a duplicate, here's an answer that shows how to calculate periods using Kepler's $\color{blue}{\text{3rd}}$ law but with all the constants and units explained.

Here's the equation from Wikipedia's Orbital_period; Small body orbiting a central body. The idea here is that the the size difference is so big that we can assume the central body doesn't move. In reality there is always at least a tiny motion. The Sun-Earth system rotates around a point about 450 kilometers from the Center of the Sun, and outer planets (especially Jupiter and Neptune) move the Sun so much that the center of mass is sometimes completely outside the Sun!

$$T = 2 \pi \sqrt{\frac{a^3}{GM}}$$

Which is Kepler's $\color{blue}{\text{3rd}}$ law:

$$\frac{T^2}{a^3} = \frac{4 \pi^2 }{GM}$$

  What are the units?

While it is convenient to use AU and years, as you've found out that only works for orbits around the Sun. The reason is in that case you're really simplifying the equation to a ratio; period divided by another period around the same body, and semimajor axis divided by the semimajor axis of that same other body.

For general calculations I recommend you use meters, seconds and kilograms. You can use any set of units, but you have to be careful that everything is in the same units.

  What is $GM$?

It's the gravitational constant $G$ times the mass $M$ of the central body. You can look them up separately, but it turns out that you can find $G$ times $M$ listed as the standard gravitational parameter. These are more accurate than multiplying $G$ times $M$ because each one alone is known to much less accuracy. That's because we can't (easily) measure the mass of a planet by comparing to a standard kilogram. Instead we look at periods and distances of orbits and determine the product $GM$ together.

Here are a few of the numbers from the Wikipedia article, I've rounded them to three decimal places. You can see more decimal places and discussion in the question Where to find the best values for standard gravitational parameters of solar system bodies?.

body      GM (m^3/s^2)
-------   ------------
Sun       1.327E+20
Earth     3.986E+14
Moon      4.905E+12
Jupiter   1.267E+17

So to get the period of the Earth around the Sun:

$$T = 2 \pi \sqrt{\frac{(\text{1.496E+11})^3}{\text{1.327E+20}}} = \text{3.156E+07 sec} = \text{1 years.}$$

To get the period of Europa around Jupiter:

$$T = 2 \pi \sqrt{\frac{(\text{6.709E+08})^3}{\text{1.267E+17}}} = \text{3.067E+05 sec} = \text{3.550 days.}$$

Wikipedia gives 3.551 days, which is the same considering we're only using three decimal places here. If you take advantage of using $GM$ instead of multiplying $G$ times $M$ and use many more decimal places, you should get agreement to the known periods to many more decimal places as well.

  But why don't I get the right period for the Moon?

The Moon's mass is more than 1% that of Earth, so the approximation treating the central body as not moving doesn't work.


How do we have photos of galaxies so far away?

There are two reasons that often — but not always — light from galaxies millions and even billions of lightyears away make it through the Universe and down to us:

Particle number and particle size

First, the intergalactic medium (IGM) is extremely dilute. The number density of particles out there is of the order $n\sim10^{-7}\,\mathrm{cm}^{-3}$, or roughly 26 orders of magnitude lower that the air at sea level! That means that if you consider a tube from Andromeda to the Milky Way with cross-sectional area of $1\,\mathrm{cm}^{2}$, it will contain roughly one microgram of matter (thanks to Rob Jeffries for catching a factor $10^6$ error).
Second, even if a photon comes close to an atom, it will only be absorbed if its energy matches closely some transition in the atom. Since most of the atoms are ionized (and thus should be called plasma instead, but in astronomy the distinction if often not made), there are no electrons to absorb the photon. The photons are more likely to interact with the free electrons via Thomson scattering, but the Thomson cross section is immensely small $(\sim10^{-24}\,\mathrm{cm}^{2})$, so even if you consider the CMB photons — which have traveled through the Universe almost since the Big Bang — only around 5% of them have interacted with electrons on their way.

In other words: The amount of transmitted light depends on two factors: 1) The amount of matter along the line of sight, and 2) that matter's ability to absorb the light. In the IGM, both are tremendously small. When the light enters the interstellar medium (ISM) inside our galaxy, it may encounter denser clouds with atoms that are able to absorb the light. But usually (although not always) "dense" is still very dilute compared to Earth's atmosphere.

Mathematical expression

In general, if a beam of light traverses a region of particles, each with a cross section $\sigma$ (measured e.g. in cm$^2$), passing $N$ particles per area of the beam (measured e.g. in cm$^{-2}$), then the opacity of the medium is given by the optical depth $\tau$, defined by
$$
\tau \equiv N \, \sigma.
$$
The transmitted fraction $f$ of photons is then
$$
f = e^{-\tau}.
$$
In general $\sigma$ depends on the wavelength, and thus part of the spectrum may pass unhindered, while another part may be completely absorbed.

The figure below (from here) shows the spectrum of a quasar lying at a distance of 22 billion lightyears, i.e. $10\,000$ times farther away than Andromeda. You see that there are several thin absorption lines (caused by intervening hydrogen clouds whose densities are a factor of 10-100 higher than the IGM), but still most of the light makes it down to us.

Because the light we see from this quasar was emitted so long ago, the Universe was considerably smaller at that time, and thus the density was larger. Nonetheless, only a small fraction is absorbed. The farther away the light is emitted, the longer ago it was, which means smaller Universe, and higher density, and thus the more light is absorbed. If you consider this quasar (from here) which lies 27 billion lightyears away, you see that much more light is absorbed in part of the spectrum. Still, however, much light make it through to us.

The reason that it is only the short wavelengths that are absorbed is quite interesting — but that's another story.


Terminology: planet plus atmosphere vs planet and atmosphere as separate things

About the only time it could make any difference is when considering impacts on the planet. In that case, if a body enters the atmosphere, it is considered to have impacted the planet. For example the dust sized particles that form meteors don't reach the surface of the Earth, yet are still considered to have impacted the Earth.

On the other hand, quoted sizes of diameters normally don't include the atmosphere since the top of the atmosphere is fuzzy, and varies with temperature. For exoplanets, none of this matters since we don't have anything like enough resolution to determine atmospheric depth, or even the presence of an atmosphere in many cases.

Remember that gas giant planets don't have a solid surface. The cloud tops are then used as a proxy for the surface.


Why does Hellas Basin look white on some maps?

That map of Mars is a composite of many different images that were taken at different times during Mars' year.  Mars' arctic circles are very similar to Earth's, and during that hemisphere's winter, that terrain is in permanent shadow, allowing seasonal frost to form.
Even though Hellas is equatorward of the antarctic circle, it gets cold enough, days are short enough, and it is low enough in elevation so the atmosphere is thick enough to allow seasonal frost to form on Hellas.  It is highly variable year-to-year, but it typically covers much of the floor of Hellas.
So, what you are seeing in the first map is southern summer and northern summer for the poles (or probably late spring given the extent of the south polar ice), and then very early spring near Hellas.  I would guess that those were the best images they could find from Viking to make that particular mosaic.
That is in contrast with the 2nd picture, which probably better matched seasons for a more smooth mosaic and one that better represents late spring / early summer at most pixels.


Measurement uncertainty of the computed center of gravity from pixel image

Every pixel value $S_i$ on the detector at $\vec x_i$ has some error $N_i$: CCDs for example have a background noise $N_\text{bkg}$ from read-out electronics, thermal noise, and sky background, plus a Poissonian photon noise $N_S=\sqrt{S}$.  In many cases this noise follows a Gaussian distribution reasonably well.  After subtracting the background, a position measurement

$$ \vec x = \frac{\sum{S_i \vec x_i}}{\sum{S_i}} $$

has the uncertainty

$$ \operatorname{std} \vec x
= \left[\sum_i \left(\frac{\partial x}{\partial S_i} N_i\right)^2 \right]^{1/2}
= \frac{\left[\sum_i
    \left(
      \sum_j\frac{(x_i - x_j) S_j}{\sum_k S_k} N_i
    \right)^2
  \right]^{1/2}}{\sum_k S_k} $$

$$ N_i^2 = N_\text{bkg}^2 + N_{S_i}^2 $$

using Gaussian error propagation ... I hope I have run the math correctly.  The way I've written it is a bit weird, but shows one important property of this method: If you look at the sum over $j$, you can see how the noise of a pixel is basically weighted by the pixel's distance to the centroid result.  The same error on a pixel value has more impact if the pixel is farther away from the centroid.

Better methods account for the uncertainty $N_i$ on pixel values in the first equation already.  You can do this by introducing additional weights in your centroid or go to fitting models, which is the “usual” method in astrometry, as far as I have seen.

This more involved approach to measuring positions uses the point spread function $P(\vec x_i - \vec x_\text{obj})$ of the instrument under given observing conditions.  The model can be an approximation, e.g. a Moffat function, or an empirical model build from unconfused bright stars in the image, e.g. a spline interpolation.  For point sources, the typical least-squares fit of position and flux of the model to an image easily produces results close to the statistical optimum of parameter estimation in terms of uncertainty.  With our modern computing power the easiest way to get the uncertainty for a given model and data noise is often a bootstrapping algorithm.

Of course extended objects require a bit more work in the model, for example assumptions on their shape, as you have given in your question.


Was Newton the first to mention the orbital barycenter?

I'm pretty sure Newton was the first to apply the notion of the barycentre to celestial motion, although the concept of a centre of mass may predate him.
Before Newton, there was a very strong demarcation between celestial motion and terrestrial motion, inherited from the Greek philosophers like Aristotle. (Similar philosophies applied outside the Western world). Celestial bodies had a celestial nature, and moved according to celestial principles, eg, they were perpetually in motion, and that motion was composed of perfect circles. Bodies on the Earth obeyed a different set of laws appropriate to their earthly nature. The essential quality of the earth element is to move towards the centre of the universe and to not otherwise move, unless something else interferes. So terrestrial objects didn't move unless they fell, or were pushed in some way, and would soon stop moving.
Newton did a lot of great scientific and mathematical work, but his outstanding insight was that a single scheme could describe both terrestrial motion and celestial motion. And that is the "core" point of the anecdote about Newton seeing the falling apple. Prior to Newton, the very idea of attempting to apply mundane mechanics to the heavens would have seemed strange to most natural philosophers, if not downright blasphemous.


What is the extent of the "worldwide campaign" to observe near-earth object 2012 TC4?

The JPL press release mentions that more than 12 observatories, labs and universities will be involved but doesn't mention which.
https://www.jpl.nasa.gov/news/news.php?feature=6906

The University of Maryland has a website on the 2017 apparition of 2012 TC4
http://2012tc4.astro.umd.edu/index.shtml
Here's a snapshot of the list of planned observations shown there


Your question presumes that the ISS is beyond Earth's gravity, that it has escaped earth's gravitational pull. This is not correct. All objects with mass in the universe affect all other things with mass in the universe, the effect just gets weaker with distance. So the ISS is feeling the effect of gravity from Earth significantly more than the moon is.

The reason the ISS doesn't just fall to the earth, either directly or gradually spiralling towards the earth, is that it is travelling fast enough around the earth that it is continually "missing" earth. It is sometimes described as 'falling' constantly around earth.

If I am to be properly correct though, the reality is that ISS is in fact falling towards the earth, getting closer and closer to Earth all the time. It needs occasional boosts to push it further back out in it's orbit.

Just to blow your mind a little bit: The ISS is pulling on the Earth with the same force that the Earth is pulling on ISS.


The most successful methods are by gravitational microlensing and direct observation in the infrared or far infrared.

The transit method doesn't work well for rogue planets, because usually at least three transits are needed to confirm a planet.


How far apart are stars in a binary system?

Distances ($a$) between binary stars vary wildly, from the order of the radius of the stars, to more than a light-year! The plot below (from here) shows a compilation of several surveys, with the color indicating the method by which they've been detected.

Separations are roughly normally distributed in $\log a$, peaking at a typical separation of tens of AU. The corresponding periods have median values of a few hundred years (e.g. Raghavan et al. 2010; Duquennoy & Mayor 1991).

As commented by Oddthinking, if stars are too close we are no longer able to resolve them visually (green bars). But we can still detect them spectroscopically (yellow bars): If we observe the blended spectral lines of two stars, we see the lines shift back and forth as the two stars orbit each other and their light is Doppler shifted.

On the other hand, if stars are too far from each other, their orbital periods of several thousand years makes it impractical to observe them orbit each other (we have only observed double stars for some 200 years); instead they're detected because they simply follow the same path through space — i.e. they have a common proper motion (blue bars).

As commented by PM2Ring, the fact that the two $x$ axes suggest a 1:1 correspondence between orbital radius and period seems odd — by Kepler's law, this also depends on mass. My random guess is that, because low-mass stars (which are most abundant) are poorly constrained (difficult to observe) and high-mass stars are rare, then most observed binaries are of the order of a Solar mass. If they're not too far from each other in mass, then there is an approximate correspondence.


What does the word "compact" mean in astronomy/astrophysics?

Compact stars are stars of great density; stellar remnants in which the fermion constituents have become degenerate. i.e. white dwarfs and neutron stars (though sometimes black holes would be included).

A compact binary is a binary star system containing at least one compact object.


Oxygen is the third most common element in the universe after hydrogen and helium. Elemental oxygen is very volatile, but it's also highly reactive and will combine with almost anything it encounters.
The resulting oxides can be volatile in the case of CO (boiling point 81.6 K) or carbon dioxide (sublimates at 194.7 K), or extremely refractory: silicon dioxide has a boiling point of 3220 K, making it less volatile than iron (boiling point 3134 K). Magnesium oxide doesn't even melt until it reaches 3125 K, just a little short of the boiling point of iron. Rocky materials are almost entirely composed of oxides and silicates, and overall are about half oxygen by mass.
Carbon will readily react at high temperatures with the oxygen in many oxides to produce CO or CO2 (this being the basis of many smelting processes on Earth). Since oxygen is so common, most of the elemental carbon that ends up accreting with rocky materials can be expected to end up getting oxidized.
The moon not only appears to have formed by a process that allowed most of its volatiles to escape, that process also enriched it with light elements. As a result, it has a small iron core, but lots of silicates and metal oxides for carbon to react with, forming volatile CO or CO2 in the process.


Amateur radioastronomy: dish suggestions

The size of your dish determines two things:

Along with the temperature of your electronics, determines the signal-to-noise ratio of your telescope.
The size of your dish determines the angular resolution you can expect. This has an approximate relationship of $$ R = \lambda / D$$ where $R$ is your angular resolution, $\lambda$ is your wavelength of light and $D$ the width of your aperture (dish).

Instead of talking about dish size lets talk about interesting things to look at and see what the requirements are:
The Sun
The sun is a good source of radio waves in the 10cm wavelength. Since the sun is around half a degree, we would need a dish at least 11.5 meters before you could expect to see the sun as anything more than a point source.
Jupiter
Jupiter has some magnetic field effects that produce radio waves in the 10-100cm wavelength range. Of course Jupiter is 50 arcseconds across, and would need a dish 412 meters in diameter to do any amount of resolution.
Cosmic Microwave Background
The CMB was one of the earliest measurements of radio waves from space. The strongest peak of the CMB is in the 1mm wavelength. However it is very weak. The original investigators cooled their radio detector with liquid helium, and I imagine something similar would be required.
Terrestrial and Man-made sources
There are plenty of man-made radio sources. Airports have a radio wave emitter in the 10cm range for radar. There are plenty of geosynchronous satellites along the equatorial plane that use the X-band in the ~2-5cm range. These by definition have a strong enough signal to be seen from Earth with even a small dish.
Summary
Resolving power is likely out of the reach of an amateur radio astronomer since the dishes have to be either very large, or you must use a much more sophisticated radio interferometry method to achieve any angular resolution. You can however still see things as point sources in the sky if you are so inclined. An excellent example of what is likely accomplished by most people would be this project here.


Typically it was visible a few weeks only, about 3...6 months at most. There is an excellent summary in Broughton (1979). In 1910 the peak visibility was in April, the last recorded sighting of Halley without instrumentation on 13th June.
Yeonan (1986) gives an overview including instrumentation; while it has positional and brightness data for dates in 1911, they are for a magnitude > 10, thus far beyond being visible with the naked eye.
So no, in 1911 Halley was not visible anymore without then bleeding-edge astronomical telescopes.


Will we be able to see our galaxy if we travel far enough?

In theory, yes; in practice probably not ...

i) It's perhaps helpful to take the Andromeda galaxy as a first working example. It is somewhat bigger than our own Milky Way galaxy but not so big as to be a problem.

When we look at Andromeda, it appears to occupies an area about 3 degrees by 1 degree. By comparison, the Moon appear about 0.5 degree across. Andromeda is a good target to see, a little faint but it is just about visible with the naked eye on a good night.

Andromeda is about 2.5 million light years away so, even travelling at the speed of light, it would take 2.5 million years to get far away enough from the Milky Way so we could look back and see our galaxy like we see Andromeda today.

( In Star Trek, the writers seemed broadly to limit themselves to adventures within our galaxy for the same reason: within the realms of sci-fi, perhaps they could justify getting around the Milky Way within a reasonable period of time - not so to travel between the galaxies )

ii) To reduce the apparent size of our galaxy to a dot would require travelling at least 10x further or probably more - depending on how you are observing. 50 million light years, anyone?


What is radiation pressure and how does it prevent a star from forming?

Radiation pressure is nothing but electromagnetic interaction.

Imagine a hydrogen atom hit by a stream of photons coming from the same direction. Although the atom as a whole is neutral, the electron and the proton are physically displaced, forming a dipole, i.e. a positive-negative charge couple. Some of the photons therefore scatter against the dipole transferring to it some momentum. So the atom start moving in the same direction as the photons. If the photons are in the ultraviolet, the electron can be exited to higher orbitals and possibly stripped from the atom. In this case the scattering is even more efficient.

Now imagine a star surrounded by a layer of hydrogen. Gravity attracts the layer towards the star. The photons emitted by the star try to push the hydrogen atoms away from it, through the electromagnetic force.

Very massive stars are very luminous and hot, which means that they emit a lot of ultraviolet photons. When the pressure transferred from the photons to the layer is larger than the gravitational attraction, then the layer begins expanding, effectively stopping growth of the star.

In the figure posted by the OP there is also dust. I don't know the details of photons-dust-gas interactions (we need a stellar atmosphere expert, I guess), but the basic principle is nonetheless the same.


Percent-illumination of crescent moon and its naked-eye visibility?

Any percentage at all would be visible, if the sky is dark enough. The surface on the illumined part is as bright as the street in front of your house during the day. So even a very tiny sliver would be clearly visible.

But you're running into a different problem here. As the illumined portion decreases, the apparent position of the Moon in the sky gets closer to the Sun. At some point it will be lost in the glare. It's a contrast issue.

Your best bet is a coincidence: if the Moon's apparent position is very close to the Sun's, while the Sun is setting, and you have clear visibility to a flat western horizon. In that case, the Moon's crescent could become very thin indeed, and still remain visible.

I've seen it like that, once. The Moon was very close to the Sun, several lunar diameters away. The Sun had just descended under the horizon. The Moon was very low in the sky, extremely thin, almost too thin to estimate the width of the illumined portion. The expression "razor edge" comes to mind. The thin curved line made less than half a circle (maybe 1/3?), with the two ends gradually tapering off in terms of brightness.

Due to the unusual shape, it took me a second or two to realize that that was actually the Moon. I showed it to my kids and their reaction was: "That's the Moon? NO! Wait... Really?" It was the Moon, really, confirmed with SkySafari on my phone.

It was not drawing attention to itself by any means, being almost lost in the sunset glare - not bright, just a thin, discreet, whitish curve in a yellow-red sky. But once you saw it, it was clearly there. It would have been invisible if the Sun was not below horizon, I think, or maybe not. Definitely invisible at mid-day, without any doubt.

I wish I had access to a telescope during that rare sight. :( Although, with the Moon at such low elevation, seeing must have been atrocious.


Opposite Constellation of Sagittarius A*

Per https://en.wikipedia.org/wiki/Galactic_Center, the galactic center
is at RA 17h 45m 40.04s, Dec -29° 00' 28.1" in the J2000 epoch.

The point opposite this would be RA 5h 45m 40.04s, Dec +29° 00' 28.1"
(J2000).

The nearest star I could find to this position is a relatively
uninteresting 16th magnitude star:

The nearest "interesting" star is HIP 27088, which is an 8th magnitude
double star:

My calculations show this point is in Auriga, not Gemini, albeit just barely:

A slightly more zoomed out view:

As @zephyr notes, the fastest way out of our galaxy would be "straight up" (perpendicular to the plane of the galaxy), not away from the center, since the Milky Way's "height" is much smaller than its length/width.

If someone wants to replicate what I did on Stellarium and use RA/DEC grids, be sure to set the date to 1999 Dec 31 at noon GMT: the grids Stellarium displays are based on the current date, not J2000.

NOTE: http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad lists even more stars than Stellarium (Stellarium uses only part of this catalog), so you might be able to find an even closer star to this point, but I was too lazy to do so.

EDIT: Newer versions of Stellarium let you overlay galactic coordinates. I'm too lazy to do this myself, but if someone else wants to, it would be a better answer.


Sunrise and sunset times when compared between the vernal and autumnal equinoxes

Instead of looking at sunset/sunrise times, it's easier to look at noon, which is right in the middle between sunset and sunrise: 11:50am in March and 11:35am in September.

The fact that noon doesn't always happen at the same time of the day (when measured with steady clocks) is expressed by the Equation of Time. Two factors play a role here:

The plane in which Earth rotates around the Sun (the ecliptic) is tilted with respect to the one it rotates around its axis (the equator)
The Earth's orbit around the Sun is an ellipse, not a circle, and it moves faster than average in its orbit from (roughly) October up to and including March, and slower in the other months


If you are referring to the Earth...

If the number is positive, the star is moving away from us. When the number is negative, the star is moving towards us.

Hope that helps!

Source with more info: https://cseligman.com/text/stars/radialvelocity.htm


How can one explain the apparent motion of the Sun from a heliocentric point of view?

The picture shows 3 paths of the sun over a single day, plotted on three particular days: Midsummer, midwinter, and an equinox. The apparent motion of the sun is due to the rotation of the Earth, not the motion of the Earth around the sun.

Why is the path of the sun a part of a circle?

Because the Earth is spinning, so you are moving in a circle around an axis that passes through the poles of the Earth

Why is this path in a plane?

The local surface of the Earth is modelled as a plane tangent to the sphere of the Earth. As the Earth spins, this plane rotates, and so relative to this plane, the sun appears to move. Because the plane is rotating in a single axis and at a uniform rate, the sun appears to move in a circle, across the sky. As it is a circle, it is in a plane.

Why is this plane tilted?

It is tilted because you are not at the poles, the surface of the Earth on which you stand is tilted with respect to the axis of the Earth

How to get this "tilt-angle"?

It is your latitude, or more exactly 90 - latitude. If you are at the poles the sun would travel in a horizontal plane. If you are at the equator the sun travels in a vertical plane.

The different paths (red green and blue in the picture) are due to the tilt of the Earth at 23.4 degrees, and so in Summer (in the north) the sun is higher at midday and further North at rise and set.

Why does this motion seem to be uniformly from a equatorial point of view.

The image appears to show the path of the sun from a location about 45 degrees North, for example Northern USA. Again, I may be misunderstanding the question.


Existence of a mass-volume astronomical chart?

Something like this?

https://arxiv.org/abs/astro-ph/9608168

But you can do literature search about the topic using NASA ADS for example, or astro-ph arxiv. They are pretty comprehensive databases of papers from astronomy/astrophysics field. If you dont find anything there you are pretty sure that no one did it. If you do find some results there (and I know for a fact that mass-size diagrams are explored/created for galaxies, clusters, etc) it might shed some light on your problem as well.

NASA ADS:
http://adsabs.harvard.edu/abstract_service.html

astro-ph arXiv:
https://arxiv.org/archive/astro-ph


Why is the number of Long Gamma Ray Burst (LGRB) event occurrences greater than the number of Short Gamma Ray Burst (SGRB) event occurrences?

Long and short GRBs are thought to arise from different types of event, involving different types of star. Therefore the question you should be asking is why are their event rates so similar?!
Long duration bursts are thought to be produced in the death throes of rapidly, rotating massive stars - the hypernova model. Short GRBs are thought to be produced during kilonova explosions, after the merger of two orbiting binary neutron stars. Short GRBs are less luminous than long GRBs, but they are also systematically closer, and so their observed gamma ray fluxes and hence detectability are similar.
However, these are quite different things, so there is no reason that they should share similar event rates.


What's the cheapest lens to buy for showing ISS to a flat earther?

I think you'd want a focal length of 700-1000 mm, and the cheapest way to get that would be a basic Newtonian reflector of 100 mm aperture.
Examples: Meade Polaris 114mm, Celestron PowerSeeker 114EQ, Orion SkyQuest XT4.5.
There are brackets available to mount a smartphone on any telescope.

Before trying to track ISS, practice acquiring easy targets such as the Moon, the Pleiades, etc.
Then use Heavens Above to predict ISS passes visible from your location.
After you collect your evidence and your hyper-skeptical acquaintance rejects it, you can still use the scope to enjoy the planets and a few dozen star clusters, nebulae, and galaxies.

On the other hand, maybe you don't need a telescope.
ISS is visible to the naked eye, is silent, and doesn't appear to slow down as much near the horizon as aircraft do.
Seeing it disappear as it enters Earth's shadow could be persuasive.