This process is described in great detail in this paper. The abstract says:

  To assess the reliability of these results, we implemented a two-stage
  imaging procedure. In the first stage, four teams, each blind to the
  others' work, produced images of M87 using both an established method
  (CLEAN) and a newer technique (regularized maximum likelihood). This
  stage allowed us to avoid shared human bias and to assess common
  features among independent reconstructions.

They then generated a large number of "plausible" images, reconstructed the radio signals they would have given and ran them through the image analysis pipeline to see if it showed any biases.

  In the second stage, we reconstructed synthetic data from a large survey of imaging parameters and then compared the results with the corresponding ground truth images. This stage allowed us to select parameters objectively to use when reconstructing images of M87. Across all tests in both stages, the ring diameter and asymmetry remained stable, insensitive to the choice of imaging technique.

Incidentally, it's not really appropriate to think of the rejected data as image data in the sense you would get from a camera. It is (almost) recordings of the exact electromagnetic waveforms received by the telescopes. The image arises from subtle correlations between these datasets. So most of what is thrown away (apart from noise) is information about the exact phase of the radio signal on a nanosecond-by-nanosecond basis, which is not really interesting.


Is there likely other significant objects we haven’t discovered in the solar system?

Moons.  We've explored the planets up to Saturn with probes that remained in orbit for long periods and found moons down to quite small sizes.  While it is certain that there are more small undiscovered moons, these won't be large spherical objects.  Uranus and Neptune have not been visited by orbiting probes, and so there are surely lots of undiscovered moons. Nothing very large, or we could see it from Earth but relatively substantial moons could be orbiting Neptune but be undetected. However there are no planned missions to either Uranus or Neptune in the next 10 years.

Dwarf Planets.  There are probably spherical bodies (and hence "dwarf planets") in the Kuiper belt, in the region of space around Pluto. There could well be further discoveries of dwarf planets here. We are certain there are no more dwarf planets in the asteroid belt. On the other hand we discover new asteroids all the time.

Major Planets.  There is the intriguing suggestion that the orbits of the known Kuiper belt objects are correlated with each other in a way that could be due to a large, Neptune sized planet in orbit in the outer part of the solar system. It is far from certain if such a body exists, but if it does, it could be discovered within a decade.

Habitable zones.  The only Habitable zone is the one which is approximately 150,000,000 km from the sun.  If you go closer, water boils. If you go further it freezes. Liquid water surely exists under the ice of some of the outer solar system moons, but I would not call them "habitable".

Unknown Unknowns.  If you look back 10 years most of the new discoveries could not have been predicted.


Does the gravity of the planets affect the orbit of other planets in our solar system?

It does - although the term 'disrupt' may be a bit too strong to describe the effect; personally, I think 'influence' would fit better.

An interesting consequence of such iterations is something called orbital resonance; after long periods of time - and remember that the current estimate for our planet's existence is 4.54 billion years - the ebb and flow of tiny gravitational pulls cause nearby celestial bodies to develop an interlocked behavior. It's a double-edged sword, though; it may de-estabilize a system, or lock it into stability.

Quoting the Wikipedia entry,

  Orbital resonances greatly enhance the mutual gravitational influence
  of the bodies, i.e., their ability to alter or constrain each other's
  orbits.

Another gravity-related effect (although, as pointed out by Dieudonné, present only on our solar system between bodies that have very close orbits like the Earth-Moon and Sun-Mercury systems) is known as Tidal locking, or captured rotation.

More about orbital resonance on this ASP Conference Series paper: Renu Malhotra, Orbital Resonances and Chaos in the Solar System.


Why is the metallicity of elliptical galaxies low?

It isn't true that elliptical (or early-type) galaxies are metal-poor. This sounds like a pop-sci myth being propagated. The first line of the abstract of Pipino & Matteucci (2006) states that "Elliptical galaxies probably host the most metal rich stellar populations in the Universe".  Thus your suspicion of the idea that "ellipticals are metal-poor" is well-founded.

Measurements of galaxy metallicities and metallicity gradients have been around for decades and show that ellipticals obey fairly simple scaling relationships between mass, luminosity, velocity dispersion and metallicity.

In terms of metallicity, massive ellipticals with bigger velocity dispersions are a bit more metal-rich than the Sun, whilst smaller galaxies may get down half the solar metallicity. The range is much larger for spiral galaxies. In addition, there is a tendency for ellipticals that are older to have higher metallicities. Examples of these trends can be seen in Graves & Faber (2010) and Li et al. (2018).

The central plot here (taken from Graves & Faber 2010) shows how the metallicity of galaxies varies with their stellar velocity dispersion (an indicator of their mass). [Z/H] is a logarithmic base 10 scale where 0.0 means the metallicity of the Sun and -1 would a tenth the metallicity of the Sun. You can see that the elliptical galaxies (red) have a relatively narrow range of metallicities, that most are as metal-rich as the Sun, that less massive ellipticals have lower metallicity and that spirals (blue) have a much bigger range of metallicity.

The reasons behind these trends are still debated. But as you rightly point out, ellipticals likely form from mergers of objects that have already been actively forming stars. A low metallicity is therefore not necessarily expected. Much depends on how much enriched gas from the ejecta of previous (higher mass) stellar populations is retained within the galaxy. Higher mass galaxies with deeper gravitational potentials are able to retain more of that enriched gas.


Exactly the difference between binary and double stars.

Trinary System

Three stars gravitationally bound to each other.

Here are some examples

Triple Star System

Three stars that appear to be together but any one of them is gravitationally unbound.

Here are some examples.


Why is the Earth-Moon distance not the same at each perigee/apogee?

Isn't the Moon's orbit a fixed ellipse with Earth at one of the foci?

No, it's not. This isn't even true for the planets' orbits about the Sun. Each planet perturbs the orbits of the other planets, making Kepler's ellipses approximately correct rather than exact. The Moon's orbit is strongly perturbed by the Sun in a number of ways. The Moon's orbit deviates from being a fixed ellipse in a number of ways. One result of these solar perturbations (and to a much lesser extent, perturbations from Venus and Jupiter, and to an even lesser extent, from the other planets) is that the Moon's orbit precesses in a number of ways.

One such precession is the apsidal precession. The line from the Earth to the point at which the Moon reaches perigee does not point to a fixed position in space. It instead precesses with a period of about 8.85 years. This is what results in the so-called supermoons, which occur when the Moon's orbit is close to perigee when the Moon is full.

Another such precession is the nodal precession. The line of nodes (where the Moon crosses from above to below the ecliptic, and vice versa), also precesses, but with a period of about 18.6 years. We only get eclipses when the Moon is very close to a node at a syzygy (either a full Moon, resulting in a lunar eclipse, or a new Moon, resulting in a solar eclipse).


The terms refer to the viewing angle, i.e. from which direction do we observe a (disk) galaxy. If we happen to be located roughly in its plane of rotation, we see it from the "edge", whereas if we are more or less above or below its plane, we see its "face".

To make a more quantitative statement, the orientation is described by the inclination angle $i$, which is the angle away from "face-on". That is, the inclination of a galaxy that is observed face-on is $i=0$, and the inclination of a galaxy observed edge-on is $i = 90º$. For randomly oriented galaxies, the mean inclination angle will be around $i = 60º$.

Wrt. your second question, you're right that the text is not correct. The three small constellations lie north of hydra. "North" means "toward higher values of declination". Perhaps it's more easily seen in this image (from Wikipedia):


If my eyes were really big, would I see radio waves?

No.  But the reasons are biological, not physical.  Your eyes work by the interaction of electromagnetic radiation with certain molecules ( rhodopsin  which consists of the protein opsin linked to 11-cis-retinal, a prosthetic group.)
These molecules are tuned to detect light of particular wavelengths.  But they couldn't be tuned to detect radio-waves, since radio waves are much larger than molecules.  To detect radio waves, you need an antenna, not a light sensitive molecule.


What could be the name of a huge asteroid I watched some years ago?

Meteors don't move slowly.

Large fireballs do occur, rarely, but they move pretty fast, a few seconds at most. Meteors that are bright enough to be seen in daylight are even rarer. When a meteor is falling you can't see the "rock".  Even a large meteoroid is only a few metres across, and they are 50-100km up in the atmosphere, and surrounded by glowing plasma.  Take a look at some of the videos of fireballs. Very few are during the day (because even bright fireballs are not clearly visible) and the few which are visible during the day are very well know (the 1972 daylight fireball or Cheblynsk)

What you saw cannot be a meteor. Because meteors don't look like that.

Re-entering space debris or satellites move more slowly, and can have more of a "burning" appearance than a meteor, but there is no mention of a satellite re-entry in the database: http://www.satobs.org/reentry/Visually_Observed_Natural_Re-entries_latest_draft.pdf  For example on Jan 2007 the rocket for Corot-r (a space telescope launched by Russia) re-entered over the USA and generated many reports.  That would not have been visible from Venezuela (and was in the middle of the night).

If you had an exact date then it would be possible to search more deeply, without it, this will probably remain a mystery.   (I pondered briefly about a rocket launch from the Guiana site, but they always launch to the East, over the sea (for safety and efficiency) so that can also be excluded.


What does "O I" mean in the context of detected molecules?

This is the standard astronomer notation for neutral oxygen. Where chemists would write
$\mathrm{O}$ (or $\mathrm{O}^0$),
$\mathrm{O}^{+}$,
$\mathrm{O}^{++}$, and
$\mathrm{O}^{+++}$ (or $\mathrm{O}^{3+}$),
for neutral, once-, twice-, and thrice-ionized oxygen, respectively, astronomers almost exclusively write
$\mathrm{O}\,\mathrm{I}$,
$\mathrm{O}\,\mathrm{II}$,
$\mathrm{O}\,\mathrm{III}$,
$\mathrm{O}\,\mathrm{IV}$, etc.

The same is of course the case for other elements. For instance, you may have an $\mathrm{H}\,\mathrm{II}$ region (i.e. ionized hydrogen) surrounding hot stars, in a larger $\mathrm{H}\,\mathrm{I}$ (i.e. neutral) envelope.

In other words, we start at 1, where chemists start at 0.


Did atoms in human body indeed come from stars?

The chemical elements in our bodies are inherited from the Earth. The Earth was formed in a disc of gas and dust swirling around the protosun 4.5 billion years ago. The material that formed the Earth was a selection of the material from that protostellar nebula that was itself once part of a larger molecular cloud.

So the atoms in our body were once part of this molecular cloud, so we need to understand how they got there.

After the first ten minutes or so, the universe contained mainly hydrogen, helium and some traces of lithium, deuterium and tritium - and that's all. No oxygen, iron, carbon etc.

Almost all of the heavier chemical elements are made inside stars. We could stop there - the atoms of carbon, oxygen, calcium etc. in our bodies must have been made in stars, and since these atoms/nuclei are stable, they must survive unchanged (you could argue about whether their electrons get swapped about in chemical reactions etc., but since electrons are indistinguishable this hardly matters).

But how do they get into a molecular cloud and what sort of stars make these elements? A couple of answers correctly identify massive stars that explode as supernovae as important. But they are by no means the only contributor, or even the most important contributor for some elements.

If we take carbon and nitrogen, these are manufactured in nuclear reactions inside stars of even a bit less than a solar mass during the horizontal branch and asymptotic giant branch stages. These stars may be less massive and produce less C and N than massive stars, but there are many more of them. The central material is mixed to the surface during thermal pulses and the outer envelope, enriched in a variety of chemical elements, is gradually lost into space via a slow wind. This is a major source of carbon, nitrogen, fluorine, lithium and a number of heavy elements - Ba, La, Zr, Sr, Pb and many others - produced in the s-process. About 50% of the elements heavier than iron are made in the s-process, which can occur in both massive stars that explode (mainly isotopes with $A<90$) and the less massive AGB stars with slow, massive winds (elements up to lead and bismuth).

Iron, nickel and many other elements such as sulphur and silicon are also produced during type Ia supernovae. This is the detonation of a white dwarf, the end stage of a low-mass star, after mass transfer or merger. Milder novae explosions caused by the ignition of material accreted onto a white dwarf also enrich the interstellar medium.

All these different processes produce distinctive patterns of element abundances.

The enriched material is swept up by neighbouring supernova explosions, by interactions with spiral arms and other molecular clouds. It cools, condenses and collapses to form a new generation of stars.

Analysis of "presolar grains" found inside meteorites tells us what our solar system formed from. These analyses tell us that all of the above processes were important in making the chemical elements that made up the Earth and hence those in our bodies.

[Further details on the production of elements heavier than iron (including supernovae, low-mass AGB stars, colliding neutron stars etc.) can be found in my Physics SE answer to this question. ]


When one asks the question, "How is a year defined?", there are actually a few different answers one can provide. What you're implicitly asking, even if you don't know, is how one defines a Sidereal Year. In effect, the Sidereal Year is the time necessary to complete one, full, 360 degree revolution around the Sun. This is opposed to the Tropical Year (or else the Solar Year) which is the time necessary to travel just slightly less than a full 360 degrees (due to things like precession and the definition of the Tropical Year).

So your question boils down to, how is a Sidereal year determined? In practice, this is often done by choosing a specific starting point and noting where the Sun is with respect to the stars in the Sky. Then, it will be an entire sidereal year (and a 360 degree revolution) before the Sun returns to the same position as previously observed. Once you've observed the Sun in the exact same position you know a full revolution has occurred of the Earth around the Sun. It should be noted that the motion of the stars over a year is almost completely inconsequential. The stars are so immensely far away compared to their own motion, that their apparent motion (otherwise known as the proper motion) is basically nothing over a year. In this way, we consider the stars "fixed".


Why the forbidden lines of [OIII],[NII] and [SII],[OII] are sensitive to electron temperature and electron density, respectively?

Forbidden lines like these arise in thin gases, where collisional de-excitation is unlikely. Forbidden lines become "quenched" by collisional de-excitation when the densities reach levels that depend on the detailed quantum physics of each transition (i.e. different for every line/transition).

If you study the ratio of a pair of lines at densities well below their "quenching thresholds", the ratio will depend only on the relative occupation of their upper levels, which in turn depends on the temperature, but is insensitive to density.

If on the other hand you choose a pair of lines with very similar upper energy levels, the ratio will be insensitive to temperature. However, if one transition is around or even above its "quenching threshold", then the ratio becomes very sensitive to density.


What caused these strange craters on Hyperion?

According to Thomas et al. (2007), these craters are not unusually deep:

It is unlikely that unusual crater depths significantly enhance the sponge-like appearance. Crater depth-to-diameter ratios for the 13 examples that can be measured reliably using shadow lengths average 0.21 ± 0.05 (s.d.). These ratios are similar to values for fresh lunar craters, and are slightly greater than for some small rocky objects. They are slightly larger than that for craters on large icy satellites, 0.14.

What is unusual and gives a "sponge-like" appearance to Hyperion is the high density of craters and lack of intercrater plains:

For craters with diameters between 2 km and 11 km, Hyperion has a cumulative number of craters per unit area twice that on Phoebe; at similar resolution these objects are strikingly different.

To explain this, they proposed that ballistic ejecta from impact events, which usually cover older craters, are lost to space instead.
A more recent study (Howard et al., 2012) suggested that mass wasting processes (landslides) and CO$_2$ sublimation also account for this special morphology, especially for the dark-floored craters. This has been confirmed by Dalton et al. (2012), who called these craters "suncups":

The irregular shapes of the sublimation pits (suncups) support the
hypothesis of extensive mass wasting and sublimation degradation
(Moore et al., 1996, Howard et al., 2011, Howard et al., 2012) largely
associated with the loss of volatiles (including water ice). When the
ices evaporate, entrained low-albedo grains accumulate in lag
deposits, which are subject to solar heating. Warm grains accelerate
both the local evaporation of ices and their own spatial
concentration.


This is almost certainly an example of a collisional ring galaxy. It's listed as such (under the alternate name AM 0417-391) in the catalog of Madore et al. 2009; here is a panel showing the galaxy from one of their figures:

Image (probably from the Digitized Sky Survey) of AM 0417-391 = LEDA 14884), taken from Figure 5 of Madore et al. (2009). "RN" = "ring nucleus"; "Cn" = possible "companion" galaxies -- though some of these are in this case parts of the ring.
These are thought to be the result of a smaller galaxy colliding more or less head-on with a larger disk galaxy. The smaller galaxy passes through (so these are sometimes called "drop through" collisions), but leaves behind a pronounced "ripple" in the gas, which becomes concentrated in an expanding ring and forms stars at a high rate; this shows up as a ring of gas and bright, young stars. Since most such collisions will be off-center (i.e., the smaller galaxy won't pass through the exact center of the bigger galaxy), the ring will be asymmetric, as is clearly the case for LEDA 14884.
These are very rare systems; Madore et al. (2009) quote an estimated frequency of 0.01% of galaxies in the local universe being collisional ring galaxies.


Why are relativistic jets perpendicular to the accretion disk?

We can think of the magnetic field lines as part of its host gas in the jet, in the sense that when the gas moves, the magnetic lines of force must move with it - and vice versa. Magnetic field lines passing through an accretion disc are thus forced to rotate with the disc. The particles of ionised gas above and below the disc are then urged to rotate with the field lines and consequently feel a centrifugal force that flings them outwards. Beyond a certain distance, however, the magnetic field is not strong enough to force the particles to rotate with it. The gas particles lag behind and bend the magnetic field lines backwards. As the field lines become more and more twisted around the axis of rotation, they begin to exert a strong squeezing force that may be responsible for the initial focusing of the beam. This tendency of a helical magnetic field to contract towards its axis may explain why jets are so tightly focused. In some circumstances, the screwed magnetic field can also act like a compressed spring and help to accelerate the gas even further.

The energy source for a jet is probably the accretion disk itself. As the infalling material gains speed, it begins to grow into a disc of matter that revolves around the black hole. In the accretion disk, the density becomes greater and greater the closer the matter gets to the black hole. This results in increasingly violent collisions between the particles. The enormous increase in particle collisions suddenly heats up the gas so much that it explodes and drives the matter back out. However, this material cannot shoot through the surrounding material in the disc. It can only escape through the central hole of the donut-shaped accretion disc: Two symmetrical jets form that shoot outwards, which are focused by the aforementioned "magnetic field tunnels" This mechanism would be suitable for radio galaxies; however, their jets only have a speed of about a quarter of the speed of light.
The jets of black holes reach speed of close to the speed of light (99% of c).
Jets in AGN, however, reach speeds far above the maximum orbital velocity of an accretion disc: they are accelerated to a fraction of a percent of the speed of light. Such a flow can therefore not only be generated with gravitational energy released by the infalling matter. But there is another source of energy that can be tapped: the rotational energy of a rotating black hole.
The surface area of a black hole can never decrease. Since a rotating hole has a smaller surface area than a non-rotating hole of the same mass, there is no principled reason why rotational energy cannot be extracted. The mechanisms that explain this process are based on the high similarity between a black hole and an ordinary electrical conductor. This analogy can be illustrated by imagining a charge being brought closer and closer to a non-rotating black hole. When the charge is far from the hole, the electric field lines protrude radially from the charge. But as the charge moves closer to the horizon, the field lines become more distorted: They "wrap" around the hole, so that a distant observer would see the electric field as arising at the surface of the horizon. It is as if the charge has spread across the surface of the hole, as it would if the "surface" were an electrical conductor.

A rotating black hole behaves like a rotating conductor. Let us assume that the gas approaching a rotating black hole has a magnetic field. When a charged particle, such as an electron, moves in a magnetic field, it is acted upon by a force that is perpendicular to its direction of motion and perpendicular to the lines of the magnetic field.  The force causes the electron to deviate from the direction of motion. The electrons in an accretion disc are free because the gas in it is in a plasma state in which the electrons are no longer bound to their atomic nuclei. The accretion disc rotates with its axis parallel to the lines of the magnetic field. The free electrons in the disk will tend to move either towards the centre, or away from the centre depending on the orientation of the magnetic field and the direction of rotation of the disk. In either case, the accumulation of electrons in one part of the disc and a corresponding minority of electrons in another part of the disc will create an electric potential. Conducting gas bound to the magnetic field lines transfers this electric potential to a circuit between the poles of the black hole and its equator. The jets use this circuit, which resembles a battery, as an energy source. The energy that is extracted from the hole accelerates charged particles and ultimately drives the jets. This creates a drag on the rotation of the black hole, which is thus slowed down.

The huge voltage drop in this circuit could actually be responsible for the creation of the matter that flows in a jet. At such a large voltage, electrons are accelerated so strongly that particles of matter and antimatter - mostly electrons and positrons - are created from the vacuum of space.
Mainly took this from here. Went on a little bit of a tangent but I hope I answered your question. I myself am actually not sure about why these magnetic field lines diverge from a straight course and "spread" themself across the surface of a black hole, but maybe someone in the comments can explain that.


Question about the formation of elements

It's very likely, that we don't have discovered every non man-made element. For some elements there exist only very short-lived isotopes.

Plutonium...

  is the heaviest primordial element by virtue of its most stable isotope, plutonium-244, whose half-life of about 80 million years is just long enough for the element to be found in trace quantities in nature

More about naturally occurring nuclides.

This paper mentions the detection of a curium ion in cosmic rays.
The process which forms curium can also form elements beyond curium, hence it's likely that there exist short-lived isotopes of elements in nature, which have not yet been detected. Research in this field is going on.


If the Moon were impacted by a suitably sized meteor, how long would it take to impact the Earth?

As several people have said, this is incredibly unlikely. Part of the reason why is that the "circling the drain" effect you describe doesn't really happen for solid objects much less dense than black holes. Orbits are not "precarious" in that way.

So, suppose something large enough and fast enough to change its velocity noticeably, but not large enough or fast enough to shatter it, did hit the Moon. The effect would be to shift the Moon from its present almost circular orbit around the Earth, into an elliptical one. Depending on the direction of the impact, it would either get a bit nearer to the Earth than it is now, once per orbit, or a bit further away (it also might swing North and South a bit). What is important though, is that this elliptical track is stable at least for a while. Suppose it gets knocked into an orbit that is 220000 miles from the Earth at its closest and 240000 miles at its furthest, that is where it will stay. It will not "spiral in".

Over a long enough period the gravity of the Sun also comes into play and things may shift a bit, but that is a relatively small effect.

Now, suppose that the impact was really big, or perhaps there were a long series of impacts (starting to look like enemy action..) so that the innermost point of the ellipse was eventually driven down to within a few thousand miles of the Earth, somehow miraculously not smashing the Moon to fragments in the process. At this distance it starts to matter that the near side of the Moon is closer to Earth than the far side, so that Earth's gravity pulls on it more strongly. If it orbited closer than about 3000km to the surface of the Earth for long (the Roche limit) these forces would eventually pull it to pieces, and Earth would probably have a pretty set of rings for a short time before internal collisions between the bits caused them to rain down on Earth and kill everyone.

Finally suppose the impact(s) was(were) so big that they actually put the Moon into an elliptical orbit whose innermost point was so close to Earth that the Earth and Moon touched. This is manifestly impossible without shattering the Moon, but in that case, the Moon would indeed hit the Earth. The time for the impact would be about 1/4 of the Moons current orbital period, which is to say about a week.


What happens to the information on the event horizons of two merging black holes?

How black holes encode or destroy information is an open question, which RichS touched on.  However, while he provided an answer consistent with the No Hair Theorem, I will provide an answer derived from the holographic principle.  I'd like to stress that both are equally valid, since we (as of yet) do not know enough about black holes.

In terms of the holographic principle, the information about the matter that formed the black hole is encoded in some manner (micro gravitational fluctuations maybe?) on the two-dimensional 'surface' of the event horizon.  It has been theorized that one could reconstruct this information by measuring the outgoing Hawking radiation, since this process decreases the radius of the black hole, hence the surface area of the event horizon, and subsequently the amount of information.  I would liken this to burning a book and then trying to reconstruct the book by measuring the properties of the ashes and radiated light.

When two black holes merge, they form a black hole of smaller mass than the combined masses.  Once more the total surface area of the event horizons has decreased to that of the event horizon of the new black hole, so information must have been 'radiated'.  Black hole-black hole mergers are not believed to have any optical counterpart (burst of light for sake of simplicity) and the merging process is derived from relativity, so Hawking radiation is not a component of consideration.  Where could the information have gone?  Well, luckily enough these mergers do radiate gravitational waves, now proven by LIGO's recent discovery.  Thus the information, if it does in fact encode itself on the event horizon surfaces, could be radiated/lost by the gravitational waves created during the merger.

EDIT: The above description sounds very 'hand-wavey', so I will expand on proposed theoretical method behind it.

Gravitational radiation is generated by the changing quadrupole moment, caused by the two in-spiraling black holes.  However, just as for the case of electromagnetic radiation emitted by oscillating charges, the quadrupole moment contribution is only one part of the greater multipole expansion of the oscillating masses.  Oscillations of the the event horizon would then cause deviations of the system from the simple quadrupole approximation, and result in gravitational radiation from the high-order multipole terms.  This radiation falls off with distance faster as one moves to higher-orders, making measurement of these contributions much more difficult.

Of course this is just one proposed solution to the black hole information paradox.


Finding galaxy spectra in SDSS Catalog by object name

One thing you might try doing is to use the SkyServer Navigate interface to see if the object was imaged by SDSS. Enter the name in the "name" box in the upper left and then click on the "Resolve" button. If an image with the galaxy shows up, click on the "Object with spectra" checkbox in the "Drawing options" panel on the left, and red squares will appear for objects with spectra. If there's a red square associated with the center of your galaxy, then there's a spectrum. (You can click on the links on the right-hand side to find out more; there may even be a little graphic of the spectrum you can click on.)
Here's an example for the elliptical galaxy NGC 6066, which does have an SDSS spectrum (as indicated by the spectrum-plot graphic to the lower right):

Something you should be aware of: in many cases, very nearby and luminous galaxies like your three examples simply don't have SDSS spectra. (This is true for your three galaxies, in fact.) Since such galaxies were known to already have plenty of spectra (including redshift measurements), there was little need to get SDSS spectra of them, so they were ignored. So you may end up needing to focus on more distant elliptical galaxies (NGC 6066 has a redshift of 0.034, while your example galaxies all have redshifts $< 0.005$.)


Are there any predictions of what galaxies exist in the Norma cluster/ Abell 3267?

VizieR lists two catalogues of Norma cluster members:

J/MNRAS/383/445 Woudt, P.A. et al. (2008) "Radial velocities in the Norma cluster (A3627)"
J/MNRAS/396/2367 Skelton, R.E. et al. (2009) "NIR Ks photometry of Norma cluster (A3627)" (this one contains angular diameter information)

The paper associated with the first of these, "The Norma Cluster (ACO 3627): I. A Dynamical Analysis of the Most Massive Cluster in the Great Attractor" describes the methodology used for estimating the cluster mass from their measurements of the radial velocities (in particular their dispersion) of the galaxies. From the paper:

For the determination of the dynamical mass of the Norma cluster, we have used both the virial theorem ($M_{\rm VT}$) and the projected
mass estimator ($M_{\rm PME}$), see equations (21) and (22) of Pinkney et al. (1996). The use of the biweight velocity centroid and scale (Beers et al. 1990) in the virial theorem (instead of the velocity mean and
standard deviation) leads to a more robust mass estimate ($M_{\rm RVT}$).
The latter is more robust against the effects of contamination by the
inclusion of possible non-members in the analysis. The projected
mass estimator (Bird 1995), on the other hand, is sensitive to the
presence of (spatially separated) subclusters due to its proportionality to the projected distance between galaxy $i$ and the cluster centroid
($R_{\rm \perp,i}$) (see equation 22 in Pinkney et al. 1996).

They also note that the masses they derive are consistent with masses estimated from the X-ray emission by Böhringer et al. (1996) and Tamura et al. (1998).
As regards what types of galaxy these are, unfortunately neither of the catalogues provide morphological types directly. On the other hand, they do include for many of the objects a reference to the WKK98 catalogue (Woudt, P.A. & Kraan-Korteweg, R.C. 1998).
WKK98 is also available on VizieR as J/A+A/380/441 and does include the morphological type.
You can use the VizieR query interface to join the tables (or you can download the tables in their entirety via the FTP page on VizieR and do the joining on your local machine, e.g. via the various lookup functions in your spreadsheet software of choice), which will give you a list of the morphological types of the objects.


My understanding is that the matter is not settled, definitively. There is a chain of logic that goes something like this:

Why is any galaxy red? Because it isn't forming any new, high mass, short lived, blue stars.
Why would a galaxy not form stars? Because the gas in the galaxy is either not dense enough or not cold enough to gravitationally collapse into new stars.

There are two main explanations put forward for the inability of the gas in really large galaxies to collapse and form stars. First, the most massive galaxies tend to be found in clusters of galaxies, and the gas in the galaxy clusters is too hot to collapse and form stars. Second, when large galaxies of comparable mass merge simulations suggest that the merger process heats up the gas in the galaxies, cutting off star formation.


I saw 3 consecutive object disappearing in the sky at same nearby location

You may have seen some artificial satellites passing into the Earth's shadow.
If so, Heavens-Above could help you identify them.
The reddish color just before disappearance would be due to the atmosphere scattering shorter wavelengths of sunlight.


Please help explain this unknown light in the sky. Not ISS. Not a star or comet. Too slow

That sounds like a high altitude nonfunctioning satellite to me. For example, a rocket body or upper stage that is tumbling. I have seen these types of satellites flash  many times before.

The pattern is irregular because of the spinning and changing angle to the Sun.


Check this paper: http://adsabs.harvard.edu/abs/2001ASPC..229..287M

LargerNi-56 directly implies brighter at peak, since it is the dominant source of radiation energy at early times.

The broadness of a light curve depends on the light curve timescale, which depends on ejecta mass, opacity, and kinetic energy. Since SNe Ia are from WD progenitors, there are not much different in ejecta mass. From simulations, kinetic energy is roughly the same across broad parameters. This leaves opacity to be the origin of the broadness. Since Fe-group elements have higher opacity than intermediate mass elements (IME), and more Ni-56 implies more Fe-group/IME, it is to be expected that high Ni-56 implies high opacity that implies broader light curve.

The slide, that someone provided in your comment, explains the same idea here (with more references).


How can I calculate how far through the day the prime meridian is of different planets

NASA's Navigation Ancillary Facility (NAIF) publishes planetary constants kernels (PCK) which are basically text files containing pole orientations for the largest known bodies.

PCKs includes the parametric orientation of their prime meridian in agremeent with IAU standards (meaning datum is J2000.0).

The latest PCK dates from 2011 and is actively used by most operational interplanetary flight projects across the world.

Using the NAIF SPICE library (available in Fortran, C, Matlab and IDL) you can load this kernel and read the orientation of the prime meridian for your body and date of interest.

Time offset between bodies can be reconstructed as angular offset between meridians. But you may find some time-related functions within SPICE which can make your task easier.


Difference between luminosity, luminous flux, and radiant flux?

Photometric units can often seem complex at first sight.

Luminosity

Luminosity refers to the total energy released by an object per unit time. This quantity is dimensionally equivalent to power, and is therefore measured in joules per second, or watts $(\mathrm{W})$. Another common unit is the solar luminosity $\mathrm{L}_{\odot} = 3.846 \times 10^{26}$ $\mathrm{W}$.

Luminous Intensity

Luminous intensity refers to the wavelength-weighted power emitted by a source in a particular direction. Wavelength-weighted is important here. Because the human eye is more sensitive to some wavelengths than others, this weighting is designed to set a standard for perception. This is set at $555$ $\mathrm{nm}$, the wavelength to which the human eye is most sensitive. This unit is called the candela $(\mathrm{cd})$, and is equivalent to a source emitting at $555$ $\mathrm{nm}$ with a power of $\frac{1}{683}$ $\mathrm{W}\cdot \mathrm{sr}^{-1}$. Essentially this is taking into account the fact that concentrating the same amount of light into a smaller solid angle will cause it to appear brighter. Finding the luminous intensity of a source with multiple wavelengths is a little more complicated. Let us define the following:

$I_v$ is the luminous intensity in candelas,

$I_e$ is the radiant intensity in watts per steradian, and

$\bar{y}(\lambda)$ is the standard luminosity function, which does the wavelength-weighting.

Then we can define the luminosity intensity of a multi-wavelength source to be:

$I_v = 683 \int_{0}^{\infty} \bar{y}(\lambda) \cdot \frac{dI_e(\lambda)}{d\lambda}d\lambda$

Luminous Flux

The luminous flux is equivalent to the total wavelength-weighted power emitted by a source. It is similar to luminosity. It is measured in the lumen $\mathrm{lm}$, which is equivalent to the candela-steradian $(\mathrm{cd}\cdot\mathrm{sr})$. One can imagine this as the human-useful amount of light emitted by a source.

Radiometric Units

The radiometric units are the same, except they refer to the true emitted power, with no wavelength-weighting.


The Sloan Digital Sky Survey website has a good page with lots of links on it to retrieve spectra in different ways, you can access it here. The most useful of these is probably this one. This is just one of many places you can search but its a good starting point if your just curious about analysing spectra and not too picky about the source.


Can there be planets, stars and galaxies made of dark matter or antimatter?

Dark matter galaxies are possible but very speculative. On a theoretical level, they are hard to form because dark matter interacts only gravitationally (see Anders Sandberg's answer), which makes it hard to lose energy and become bound structures. On an observational level, they would be hard to detect. Gravitational lensing can do something, but since one cannot actually see the galaxy, it's also hard to say where the dark galaxy is -- if there is one at all.
Still, people have studied the idea, so it's not impossible.
Antimatter galaxies: At some level the idea that there are antimatter galaxies out here is appealing. First it can solve the baryon asymmetry problem at a stroke. It's also the case that an antimatter star would shine. From long distance, it would also be practically indistinguishable from a "normal" star.
However, there are strong reasons to believe that there are no antimatter galaxies. That's because antimatter annihilates with normal matter, which leaves experimental signatures. If any part of the Earth were made of antimatter, it would immediately vanish in a flash, so we can be sure that the Earth is mostly matter. Similarly, if the Sun were made of antimatter, we would be quickly annihilated (thanks to the antimatter solar wind radiating from the anti-Sun), so we can be sure the Sun is also mostly matter. Similar arguments allow us to conclude that the Milky Way is almost entirely matter, the Local Group is almost entirely matter, etc, all the way up to the largest structures in the sky.
If antimatter galaxies exist, they are probably outside our observable universe, at which point some will argue it's no longer science.


Do stars made by universe getting smaller or larger in mass?

The size of stars depends on the amount of non-hydrogen elements ("metals") affecting the opacity of the gas that forms them. Metals make it easier to radiate away heat when a gas cloud is collapsing, making it easier for a large amount of gas to accrete (although it may mainly act by increasing the minimum mass of stars that form). There is a fair disagreement about whether the first stars ("population III") were very massive (hundreds of solar masses) or actually small, but I think the rough consensus is that very massive stars were likely. Once they emerged and died, they added metals to the interstellar medium and star formation eventually became more like the present.
The expansion of the universe does not have much effect once gas become trapped in dark matter halos and start to form galaxies.
Once the star has formed opacity and mass determines its structure and luminosity. It turns out that for increasing metallicity the maximum stable mass of a star decreases (section B.2). For the eventual metallicity at the end of the stelliferous era Adams and Laughlin estimate a maximum mass star as only 30 solar masses.


How do astrophysicists handle the topic 'boundaries'?

The literal shape of space became a topic with Einstein's theory of general relativity. Before that, there had certainly been arguments about whether space was unbounded or had some limit, but they were mostly philosophical. In general relativity space-time is a mathematical manifold whose geometry is linked to observable features such as gravity, the expansion of the universe - and also lightcones, event horizons and singularities. This meant that (1) the theory now included various forms of boundaries set by the geometry, and (2) that the theory clearly leaves certain things "outside".

The first category of boundary are benign in the sense that they denote either regions that cannot influence each other but have the same kind of physics (we can never observe beyond a certain distance, we can never see the interior of a black hole from the outside), or represent parts of physics where the theory clearly breaks down (and "knows" it) like gravitational singularities or the big bang singularity. How to resolve those breakdowns is of course a rich frontier for theoretical physics, astrophysics and cosmology.

The second category is subtler. Einsteinian gravity is strongly based on the profound insight in Gauss' Theorema Egregium that the geometry of a manifold can be described only using properties found in the manifold, with no recourse to embedding it into a bigger outside space. The perennial question "what is outside the universe" is simply removed: general relativity does not need an outside, doesn't talk about an outside, and doesn't benefit from having an outside. That doesn't mean one cannot construct theories where space-time manifolds are embedded in other spaces but they are theories that need to be motivated rather strongly - there is no obvious evidence making them necessary, so Ockham's razor wants to shave them off.

Similarly quantum mechanics also adds boundaries but in the information domain: certain things are measurable only in a partial sense, and while the part that cannot be measured can still be described nicely mathematically the theory shows that we cannot know its state (in a pretty specific way).

To sum up, there are some things we know we cannot know or interact with but are not particularly worrisome: in a large accelerating expanding universe there are visible galaxies we can never reach, and there are galaxies that will forever be beyond our horizon. We however expect astrophysics to work the same over there. There are also limits to knowledge that are more annoying: the interior of black holes, what is beyond the CMB radiation. Then there are the theoretical boundaries where quantum gravity or other new insights we do not have yet are needed to continue inquiry. And finally there are the "boundaries" where we simply have no information, like "outside the universe" or some multiverse theories. Different people work on different things - philosophers and theoretical physicists are having fun in the last category (and arguing about how one can rigorously argue about such things), while most astrophysicists prefer to work in the data-rich regions and use it to extrapolate that is likely true for the other regions.

(James K. cited Wittgenstein's adage about being silent, but it is worth noting that this is usually interpreted more as a challenge than advice by many theoreticians.)


What is the period of Jupiter being visible in the sky?

You can make a ballpark estimate of Jupiter's synodic period if you know that Jupiter orbits the sun once every twelve years, while Earth orbits in the same direction in one year.  So when your Earth calendar has completed a year, Jupiter has moved approximately one-twelfth of the way around its orbit, and Earth has to move approximately another month to catch up.  That gives you a synodic period of about thirteen months, consistent with the more careful calculation by Mike G.
The ballpark estimate doesn't work very well if the periods are too similar (e.g. for Mars or Venus), but it's fine for the outer planets.
As to when you can see it again, Jupiter's next solar conjunction is on 2021-01-28.  So if it disappears into your evening twilight at the end of December, you should expect to find it in the morning twilight around the beginning of March.


What I can observe with a 80/900 refractor telescope? What can I study?

There is a wonderful intersection of Physics, Mathematics, Applied Math (numerical analysis) and Observational Astronomy in a systematic analysis of the movement of four bright Galilean moons of Jupiter!

They are easy to see in a small telescope, they have well defined orbits around Jupiter, and they have some trick up their sleeves that you can analyze.

Here's an exercise you can do right now. When Jupiter is anywhere near quadrature, it's shadow extends behind it by an angle of almost 11 degrees, so the moons will often be eclipsed. That means instead of appearing or disappearing into Jupiter's overwhelming brightness, they can "pop" in or out of visibility as they pass out of or into Jupiter's shadow.

                a (km)
Io             421,800
Europa         671,100
Ganymede     1,070,400
Castillo     1,882,700

Jupiter radius  71,492

You can predict when these happen, and then watch to see them and confirm your calculations. You can start timing them to get more accurate information.

From a mathematical perspective you can check the periods of their orbits (from your measurements, or from a table) to see if there are any surprising numerical relationships!

You can check the inclination of their orbital plane to see if the eclipses always happen or sometimes don't.

You can read the answers to When will the next series of mutual eclipses of Jupiter's moons begin? and start planning for 2021 when these begin again!

You can learn to do numerical integration of solar system bodies to simulate their orbits, or use the Python package Skyfield or the JPL Horizons website for high accuracy positional information.

These can also help you calculate the eclipse times by Jupiter, and the mutual eclipses of Jupiter's moons.

Finally, you can try to measure the speed of light by seeing that the eclipses come sooner or later than average when Jupiter is closer or farther than average from the Earth. This is one of the early ways that the finite speed of light was verified.


How to convert a gravitational force to speed and direction

You must apply Newton's law
$$ m\frac{\mathrm{d}\boldsymbol{v}}{\mathrm{d}t} = \boldsymbol{F}$$
which related the force $\boldsymbol{F}$ to the acceleration (=change of velocity). Note that positions, velocities, and forces are all vector quantities. Also note that as the objects move (change position), their mutual forces change (both direction and strength).

There is an enormous amound of literature on the resulting orbits for celestial bodies, going all the way back to Kepler and Newton. I suggest you look into elementary books or start with some Wikipedia articles.


What is the predominant element in the dust of the Sombrero Galaxy?

I think that first, we have to properly appreciate the size of the Sombrero Galaxy. It is roughly 50,000 light years (15 kilo parsecs) in diameter. That might be only half of the diameter of our own Milky Way galaxy, but still makes each and every pixel on the photograph you're attaching in your question stretch more than 100 light years in distance.

That's a huge distance, so we can probably agree that clouds made of nearly anything and at least as dense as what you'd on average expect the interstellar medium to be comprised of, would obscure even the strongests light sources within that galaxy. This is important to realize, so we somewhat lower your expectations; These areas aren't tightly packed with anything, we're just talking of such vast distances that the light has to travel through, that even almost nothing will be enough to stop it from shining through.

Now to your question what it is comprised of. It is in essence matter from the interstellar medium that stops this light from reaching us, so nothing in particular that you wouldn't expect also in similar areas between stars within our Milky Way, or indeed any other galaxy that is roughly in the same age period of its development. Since the Sombrero Galaxy isn't even all that far away in cosmological sense, merely around 29 million light years, it has an apparent magnitude of ~ 9 and is a rather good target for observations of even hobbyist astronomers with enthusiast telescopes. More importantly though, you aren't looking that far into the past. Again, in the cosmological sense. For comparison, our Sun is roughly 4.5 billion years old, and we're observing the Sombrero Galaxy as it was 29 million years ago.

So what is this interstellar medium made of? On average, it is comprised of mostly gas clouds of hydrogen and helium forming during the primordial nucleosynthesis (read: Big Bang) and at most a few percent of heavier trace elements, some also in gas form and an itsy-bitsy percentage of these would combine into even heavier dust particles, mostly made of carbon, silicon, and oxygen (source for the interplanetary dust composition: NASA APOD). Remember, when I say clouds and dust, they are incredibly thin. So thin, you'd have to be traveling at a significant portion of C to even notice it much, unless you're looking for it with extremely sensitive equipment, like Voyager 1 and 2 do, for example. But even with such low particle density, some regions would be slightly denser, and some slightly thinner. And that is what creates that majestic dust lane of the Sombrero Galaxy.

They are interstellar medium, mostly made of primordial hydrogen and helium, and dust made of carbon, silicon, oxygen and other trace elements, and all of it hundreds of light years in length and obscuring light from the galaxy's stars to penetrate through at varying levels, depending on background light strength, proximity, and local interstellar medium cloud density. The color of the cloud says nothing of its composition. It is mostly black with parts of it shifting towards red hues due to molecules of gas and dust in the interstellar medium scattering blue light from the source of light more than they scatter red light. I.e. the same reason why the Sun or the Moon appear more red when they're low on the horizon.


How can I convolve a template spectrum with a photometric filter response spectrum?

Your feeling is right: You shouldn't convolve the spectrum and the filter, you should only multiply so that flux outside the bandpass is suppressed. Subsequently you integrate the resulting function over wavelength, so that flux density (in energy/time/area/wavelength) becomes flux (in energy/time/area).

Simply setting the flux to 0 outside $\lambda_1$ and $\lambda_2$ (or, equivalently, just integrating from $\lambda_1$ to $\lambda_2$) corresponds to a "top-hat filter". Most realistic filters are more smooth.

So, something like (untested):

import numpy as np
from scipy.integrate import simps

lamS,spec = np.loadtxt('spectrum.dat',unpack=True) #Two columns with wavelength and flux density
lamF,filt = np.loadtxt('filter.dat'  ,unpack=True) #Two columns with wavelength and response in the range [0,1]
filt_int  = np.interp(lamS,lamF,filt)              #Interpolate to common wavelength axis
filtSpec  = filt_int * spec                        #Calculate throughput
flux      = simps(filtSpec,lamS)                   #Integrate over wavelength
print 'Total flux is {0:8.3e}'.format(flux)

Magnitude

I'm wondering, however, if "the total luminosity within the bandpass" is really what you're interested in. This quantity doesn't really bear any physical significance for the source, as it depends on the particular filter. Usually, one would normalize to the filter, thus getting the magnitude, which is independent of the filter shape.

In the AB magnitude system, the magnitude $m_\mathrm{AB}$ of a source is (Oke & Gunn 1983; note that they make a sign error in their own definition):
$$
m_\mathrm{AB} = -2.5 \log f_\nu -48.6,
$$
where the average $f_\nu$ in $\mathrm{erg} \,\mathrm{s}^{-1} \,\mathrm{cm}^{-2} \,\mathrm{Hz}^{-1}$ is given by
$$
f_\nu = \frac{1}{c}\frac{\int ST\lambda\,d\lambda}{\int T / \lambda\,d\lambda},
$$
where $S$ is the spectrum with units $\mathrm{erg} \,\mathrm{s}^{-1} \,\mathrm{cm}^{-2}$ Å$^{-1}$, $T$ is the filter curve (un-normalized as above), $\lambda$ is the wavelength in Ångström, and $c$ is the speed of light in Å $\mathrm{s}^{-1}$. That is, a flat spectrum (in $\nu$) of height $f_\nu$ would have the same integrated flux over $T$ as $S$ would.

So:

import numpy as np
from scipy.integrate import simps
c_AAs     = 2.99792458e18                          # Speed of light in Angstrom/s
lamS,spec = np.loadtxt('spectrum.dat',unpack=True) #Two columns with wavelength (in Angstrom) and flux density (in erg/s/cm2/AA)
lamF,filt = np.loadtxt('filter.dat'  ,unpack=True) #Two columns with wavelength and response in the range [0,1]
filt_int  = np.interp(lamS,lamF,filt)              #Interpolate to common wavelength axis
I1        = simps(S*T*lam,lam)                     #Denominator
I2        = simps(  T/lam,lam)                     #Numerator
fnu       = I1/I2 / c_AAs                          #Average flux density
mAB       = -2.5*np.log10(fnu) - 48.6              #AB magnitude


Why did the edge of Olympus Mons collapse?

This source from NASA says:

Most scientists think the the cliffs formed by landslides. This collapse is driven by the weight of the huge volcano exceeding the strength of the rocks it is built of.

In other words, one theory is as you have surmised: The outer 7km shell of the volcano appears to have slid down, all around the mountain, forming the cliffs.  This paper has more information.
Another theory says:

The part of the Olympus Mons edifice surrounded by cliffs is morphologically similar, albeit at a much greater scale, to the volcanic seamounts on Earth like
the Canaries and the Hawaii.

This says that the Hawaiian volcanoes have a similar cliff around them, so perhaps Olympus Mons was formed in an ancient sea.  This paper by the same author gives the reason for why the cliffs don't go all the way around as more evidence for a sea on the portion where the cliffs did form.  A third paper by this author states that this also explains the aureole extending out so far.
A third theory is that as the lava and ash spread out from the volcano, eventually they reached a point where rather than forming rock they more formed sand dunes - which eventually started to blow away.
So, in summation, it could be because the mountain is too tall; because the bottom 7km or so was underwater as the volcano was forming the mountain; or because the outer portion of the mountain was only formed as ash.


How do ground-based telescopes deal with earthquakes?

Earthquakes are not uncommon in Hawaii and Chile, where several observatories are built. I only have personal experience with the observatory in La Silla, where a minor earthquake once occurred while I was observing. This was an old telescope, and the image was ruined. Modern telescopes may be able to withstand small seismic activities, but since we felt the ground shaking, I think the image would've been ruined on any telescope.

Luckily, images are not taken in one, long exposure. Rather, they're composed of multiple exposures that are added subsequently. You typically don't expose for more than 30 minutes. A faint object that requires, say, 5 hours of exposure, is then given by the median of 10 half-hour exposures. Why the median and not the mean? Although earthquakes are, after all, rare, cosmic rays quite often ruin a small region on the image, but when you take the median, such outliers are removed$^\dagger \!\!\!$.

In other words, the whole night isn't ruined, only the one exposure that was taken when the earthquake occurred.

Of greater concern are more violent earthquakes, which may in principle damage the telescope. Modern telescopes are prepared to deal with such seismic activity. For instance,

  …ESO's VLT has a complex mechanical system that is automatically activated when
  it experiences a strong earthquake of magnitude 7 or higher. There are
  a series of clamps around the edges of the mirror, which can — at a
  moment's notice — lift the entire 23-tonne mirror off the actuators
  and secure it to the telescope’s support structure.

In 2006, an earthquake close to Hawaii shifted slightly the position of various things on the Subaru telescope, apparently resulting in people there working with pre- and post-2006 numbers, according to former telescope operator Dan Birchall.

Source: ESO, where you can also find more about how to prevent damage in case of earthquakes.

$^\dagger$For instance, if the values in a given pixel of 10 images are $[21,25,22,19,22,27,23,65536,18,24]$, then obviously something went wrong in the 8th exposure, and the median $(22.5)$ is more characteristic of the "true" value than the mean $(6573.7)$.


What is the physical geometry of this apparent "eclipse" of a tiny moon of Asteroid Florence?

The vertical axis is signal delay, and therefore distance from the radar antenna; this is more or less true spatial information. As the caption indicates, delay is increasing from top to bottom in the image.

The horizontal axis is frequency shift (due to the Doppler effect). To the left are signals with a lower frequency relative to what was sent from the radar antenna, i.e. reflections from points moving away from the antenna; to the right, signals with a higher frequency, coming from points moving towards it.¹ This axis is more difficult to interpret.

For a single rigid body, which Florence presumably is, the only way it can have points moving in different directions at the same time is by rotating. On one edge of the body, points that have just come into view of the radar are moving quickly towards it. This relative movement slows down as the points approach the centre of the radar-visible disk; then, they move away increasingly fast until they disappear from view on the opposite edge. (The speed change isn’t linear, which means the body will be somewhat warped on the radar image.)

Only this effect allows the radar to distinguish between points that are at the same distance. If the asteroid wasn’t rotating at all, or if we happened to look exactly onto one of its poles, it would appear only as a thin vertical line in the image. For a rotating body, note that the radar cannot tell to which direction the rotation is: A mirror asteroid rotating the other way would look exactly the same to it.

Now for the moons. They are not rigidly connected to Florence, and apparently not tidally locked to it either, so they will not follow its rotation. With that, the starting point “moving in different directions at the same time = rotation” which worked on the main asteroid breaks down. When one of the moons is moving towards the radar, that tells us exactly nothing about its relative position to Florence; the visualization is misleading in that respect. Therefore, it is absolutely possible for a moon to be straight behind Florence even when it appears to its side in this particular visualization.

Emily Lakdawalla has a nice blog post on this topic, with animated drawings: How radio telescopes get “images” of asteroids

¹ After correcting for the movement of Florence – or the whole Florence-and-moons system – relative to Earth, of course.


how steady over time are right ascension and declination values?

The values are more steady for objects close to Ecliptic poles, but in genenral, they change by a few arcseconds every year.

Yes, we lack an absolutely fixed frame of reference.


Based on the smallest number of photons that can deliver information about a star, what is the largest size of the universe?

So, assuming my assumptions aren't assinine (they very well could be) ...

Your words, not mine. Your "very well could be" is the case.

Assume the universe is infinite

The universe might well be infinitely large, but the observable universe is all that we can possibly see. By all signs, the universe is 13.8 billion years old. We can't see the stuff that is beyond the observable universe because not enough time has passed for light to have traveled that far.

To make matters worse, the early universe was very hot and dense. Photons traveled but a short distance before being absorbed. We can't see those very early photons because they no longer exist. This changed about 379,000 years after the universe formed, at which time the universe had sufficiently cooled so as to allow photons to roam free. The photons let loose by that clearing of the universe form the cosmic microwave background radiation.

Assume it is possible to receive a single photon of light from any star within it that has a line-of-site to the Earth

Per the big bang theory, the first stars formed about 100 to 200 million years after the formation of the universe. Stars that formed at that time beyond the observable universe: We can't see them. Due to the continuing (and possibly accelerating; google "dark energy") expansion of the universe, we will never see those stars.

There's a famous paradox related to your question: Why aren't we bathed night and day by light from an infinite number of stars if the universe is uniform and is infinitely old> This is Olber's paradox. The solution is simple: The universe is not infinitely old.

Assume that the photon has travelled an infinite distance, then such a photon, if enough information could be extracted from it, would prove the universe is infinitely large

The only information that can be extracted from a single photon is its frequency, and even that is a bit fuzzy due to quantum mechanics. A photon from a remote star is indistinguishable from a photon of the same frequency emitted a nearby lightbulb.


Are white dwarf stars supported by proton degeneracy as well?

Straightforwardly no.

For a start there are almost no free protons inside a white dwarf. They are all safely locked away in the nuclei of carbon and oxygen nuclei (which are bosonic).
There are a few protons near the surface, but not in sufficient numbers to be degenerate.

Let us assume though that you were able to build a hydrogen white dwarf that had equal numbers of free protons and electrons.

The density at which the electrons become degenerate is set by the requirement that their Fermi (kinetic) energy exceeds $kT$. The Fermi energy is given by
$$E_F = \frac{p_F^{2}}{2m} = \left(\frac{3}{8\pi}\right)^{2/3} n^{2/3} \left(\frac{h^2}{2m}\right),$$
where $n$ is the number density (which would be the same for protons and electrons), but $m$ is the mass of a proton or electron, which is different by a factor 1800.

Thus for a given white dwarf temperature, the electrons become degenerate at number densities a factor of $(m_p/m_e)^{3/2} = 78,600$ times lower than do protons.

Even if we were to compress a hypothetical hydrogen white dwarf to the point where the protons were also degenerate (which for a typical white dwarf interior temperature of $10^{7}$ K, would need mass densities considerably in excess of $10^{12}$ kg/m$^{3}$), the (ideal) degeneracy pressures would then be given by
$$ P = \frac{h^2}{20m}\left(\frac{3}{\pi}\right)^{2/3} n^{5/3}$$
and so we immediately see that the degeneracy pressure due to protons would be $\sim 1800$ times less than that due to the same number density of electrons.

If we push up to higher densities then both the protons and electrons will become relativistically degenerate. In this case the pressure becomes independent of the particle mass. However, the Fermi energies of the particles would now be high enough ($>1$ GeV !) to block beta decay and instigate neutronisation. The protons and electrons start combining to form neutrons and an n,p,e fluid is formed where the neutrons vastly outnumber the protons and electrons. This in fact prevents the protons ever becoming relativistic, even at neutron star densities, and the contributions of electrons to the pressure is always orders of magnitude higher than for the protons.

A handwaving way of understanding this is that the degeneracy pressure depends on the product of the momentum and velocity of the particles. In turn, the momentum of the fermions depends on how much they are compressed through the uncertainty principle. For a given particle number density, the separation $\Delta x$ is the same for protons and electrons and therefore the uncertainty principle says that the $\Delta p \sim \hbar/\Delta x$ of the momentum is also the same. This is another way of saying that the Fermi momentum does not depend on the mass of the fermion; however, for a given fermion momentum, the velocity clearly does! Therefore the degeneracy pressure must be lower by approximately the mass ratio of the fermions


How to convert between AB and Vega magnitudes?

A few things:

The conversion between magnitudes is dependent on the filters being used, you can see it on the tables from the website you linked. That being said, the conversions available in there will not be useful for you if you are trying to replicate Stern et al 2005, as they use redder filters (in the website with the conversions, the longest wavelength available is 2.19 microns, whereas the shortest in Stern et al 2005 is 3.6 microns). If you can find the proper relationships, you should get the expected results.
The paper states that they used IRAC data, so it is a good idea to start looking over there. In fact, you can find some relations in point 7 here and you could also take a look at point 3, might be necessary (I do not know what data you are working with). If you are not using IRAC data, you might as well look for a different source for the conversions.
Are you perhaps using the UltraVISTA catalog? If not, it might also be a good idea to check it out. Their README contains some details about the photometry and references about how they acquired the data (including IRAC), which could give you some more hints.


Which spiral arm of the Milky Way is Kepler-62 in?

The Kepler space telescope had a field of view along the so called Orion arm, or spur, of the galaxy. The same structure which we ourselves are inside. It basically looks along a line where the star density is the greatest within the distance that it could detect transiting planets. Less than 3% of the galactic diameter, I think. Note that the illustration your question linked to seems to overestimate that distance. One link and another link. Kepler-62 and all another Kepler stars are certainly in that FOV. It would not be able to see transiting planets beyond the local arm. However, if the Kepler K2 mission gets going, it will use several different field of views.


Are there naturally radiation shielded areas on Mars? (shadowed by terrain formations)

If you want protection from radiation on Mars, you'd have to either live in caves, or else build some kind of thick dome overhead, or live behind thick walls/ceiling. Cosmic radiation will reach you anyway as long as the sky is visible, so deep valleys can only help you so much.

The only way to stop hard radiation is to place a lot of "stuff" (thick mass) between you and the source. A pile of dirt would suffice in many cases. The geologic layers on top of a cave would work well. I guess a dome could also work if it's thick enough. Habitats with thick ceilings (and walls) could also be an option. Have a large water reservoir? Put it on the roof of the habitat: voila, instant radiation shield.

The main concern is the stuff coming in from directly overhead; near horizon even Mars' atmosphere is thick enough to reduce radiation a little. So prioritize the ceiling.

I guess at the base of a hill with some overhang above your head you'd get some shielding, but again - is the sky visible? That's the main concern. Also, areas under an overhang may or may not be safe (stable) long term.

Find a very large boulder (the size of a house or bigger), much taller than your habitat, and place the habitat right next to it - you've cut 50% of radiation right there.

For a shorter term stay, a simple habitat with thin panels would be enough - not very different from the interplanetary vehicle after all.


No, it seems that the formation of a supermassive black hole is a consequence of galaxy formation, not a cause of it. The mass of the black hole is a very small proportion of the mass of the matter and dark matter in the galaxy. The details are not yet worked out but it is the distribution of dark matter that forms the galaxy, and then galaxy mergers contribute to the growth of a large black hole in the centre.
As indicated in the answer to your previous question, "sucked" is entirely the wrong metaphor for how black holes work. Black holes have lots of mass in a small space, so they "pull" gravitationally.

Now gravity gets weird when it is very strong, as it affects the time. From the perspective of a person outside the black hole, a clock on a body falling towards the horizon will slow down, and will appear to stop as the body reaches the black hole, This means that we will never see something fall into the black hole.  However the mass of the body that falls towards the black hole will increase the effective mass of the black hole, and so increase the radius of the event horizon.

For objects further from the black hole, the acceleration towards the black hole is proportional to its mass. So it is not the case that an object will "start to feel the pull", but instead the pull will increase.  Note that mass that is orbiting close to the black hole will act almost like part of the black hole from the point of view of someone further out. This is not a special property of black holes, just ordinary gravity.
Black holes form an accretion disk around themselves, a disk of matter orbiting the black hole at very high speeds and getting very very hot. They also produce high energy jets of particles moving close to the speed of light from their poles. These are produced from the accretion disk, not from the black hole itself. Objects that fall towards the black hole don't come back.


Hydrogen Line observation - Interference from trees

The 21 cm line is from atomic hydrogen which is going to be free, neutral atoms in vacuum, where the electron is bound only to the proton.

In water, or any other hydrogen containing molecule, the electron's orbit will be dramatically modified (or even "stolen") by the atom to which it is attached, and so the transition will no longer exist as a narrow line.

Certainly there might be all kinds of RF interference from objects encroaching on your antenna's reception pattern, which may have side-lobes as well as the main beam pointing upwards, so like @Kozaky's comment suggests, you should to some experimentation if possible to understand contributions to background.

Your sensitivity will depend on the bandwidth you choose for your receiver, as well as the thermal and Doppler broadening of of the source that passes by, so make sure you have an understanding of what the bandwidth should be to see a peak, and also if there are significant offsets in frequency due to a Doppler shift.

above: From here

  Ground state hyperfine levels of hydrogen (parallel and antiparallel) with the spin-flip transition, emitting radiation at 1420 MHz. The corresponding wavelength is 21 cm. (21-cm line, hydrogen line)


If rotation has nothing to do with gravity, why do astronomers always link swirling dust and gas to stellar evolution?

Any time you have a large cloud, different parts of it will randomly be moving in different directions. If you pick any line through the middle to consider, slightly more particles might be moving (say) clockwise than counterclockwise around that line, just by chance. And there has to be some axis where that random difference is the largest.
Over time, the gas and dust may start to contract under its own gravity. Like a spinning skater pulling in their arms, the rotation will speed up, and there will be collisions so that over time there will be a consistent rotation in a certain direction, just arising from the original random motions.
Instead of producing the gravity, it kind of does the opposite thing. The objects really want to move in straight lines, and gravity is the only thing that keeps pulling their paths around the middle. If there were no motion, the objects would just fall straight down towards the middle, so in that sense, the rotation is acting against the gravity. Explanations of the formation of the solar system mention the rotation of the initial cloud because it was the precursor to the revolution of the planets about the sun, along with the rotations of the planets and of the sun itself. It is not because the rotation somehow creates the gravity.
When you spin an object, things inside it swing out towards the edge. That's really just their momentum carrying them in a straight line until (say) the wall of the can gets in the way. But if you're in a spinning room and the wall is constantly keeping you from flying off in a straight line, it kind of feels like gravity is pushing you against the wall. But it's not gravity, just the wall pushing inward on you and steering you in a circle. The inward force of the wall is called centripetal, and the outward force that you perceive is technically not a true force, but it is a meaningful concept called centrifugal force. But, again, it is a very different thing from gravity.


Can we see the past image of Earth because of curving light by massive objects?

Light can be made to do this around black holes, sort of. At a certain distance from a black hole you have what's called a photon sphere. At this distance, determined by the mass of the black hole, photons travel in orbits because of the space-time curvature.

So to answer your question, maybe.

Firstly, a massive entity would need to be located at some point in space in order to influence the photons into an orbit - okay yes, possible. Secondly, this massive object would need to have a mechanism of instantaneously disappearing from space-time in order for the photos to tangentially come out of orbit and back towards earth.

I read on a previous question on Physics.SE that there is no such solution to the Einstein Field Equations which allows for a massive object to just disappear - the closest thing would be for the density distribution of the object to change but I can't say if that would yield the result you are asking about.

Rob Jeffries mentioned that images do exist which suggest that photons experience such a change in trajectory - this could happen on paths which are very close to a photon sphere-like one, but deviate enough from it to allow photos to escape.


"Really dense" is a bit of an understatement for small black holes.

A Black hole the mass of the Earth would have a diameter of about 18mm.

And it doesn't get any easier if you start smaller. The density of small black holes is bigger than large ones:

A black hole the mass of a human would have a diameter that is no only smaller than a proton, but a billion times smaller than a proton. Consider the difficulties in crushing something that small.

A black hole the mass of a proton...

The notion that you can make a black hole by "crushing matter very very hard" is many magnitudes beyond what is realistic.

It is, perhaps, possible that tiny black holes could be created in very high energy particle collisions.  We're fairly sure that no such black holes have been created at CERN.  Particle collions of much higher energies occur when cosmic particles hit the atmosphere, and again there is no evidence of black holes being created but we are looking: https://arxiv.org/abs/hep-ph/0112247


What if a white dwarf is less massive than her partner?

If they are close enough to exchange mass, the Red dwarf will always lose mass to the white dwarf.  That doesn't mean there won't be some exchange going back the other way, but the white dwarf will always win the exchange rate and it won't even be close.   It makes no difference whether it's heavier or lighter.   What matters primarily, is how tightly held it's outer mass is to each star and how strong the tidal forces are.

The white dwarf's mass is very tightly bound to it by gravity and being so small, it's not going to experience strong tidal variation.  The red dwarf's outer layers are far less gravitationally bound to it, and being much larger, it experiences greater tidal bulging.

How much mass is exchanged would depend on how close they are, but around any star, and if they are close enough, the white dwarf will always steal mass from the main sequence star.

A curious thing happens with main sequence stars.   As their mass goes up, their surface gravity goes down.   A star with 0.25 times the mass of our sun, extrapolating from this chart would have a radius about 32-33% of our Sun, and being 1/3rd the size, using the inverse square rule, it's surface gravity would be about 9/4ths or 2.25 times the gravity on the surface of our sun - roughly speaking.  There would be other factors like rotation rate and solar storms, and, as mentioned above, how close they are, but there's no way that a main sequence star of any size could steal more mass than it gives up from a densely packed White dwarf.


Why is the sunset at different times at the same longitude at equinox?

The only idea I have that it might be due to the definition of sunset.

The instance of sunrise occurs when the first bit of the Sun becomes visible at the horizon, while the instance of sunset occurs when the last bit of the Sun disappears below the horizon. This means the actual position of the center of the Sun is about 50 arc minutes below the horizon at sunrise / sunset.
At sunrise and sunset on the days of the equinoxes, the angle between the Sun's path and the horizon is roughly equal to 90° minus the absolute value of an observer's latitude. The Sun's path is much more oblique at your more northern location.
It takes the Sun about 200 seconds (3.333 minutes) to traverse those 50 arc minutes at the equator on the equinoxes. At some latitude $\lambda$, it takes about $200 \sec(\lambda)$ seconds for the Sun to go from 50 arc minutes below the horizon to the horizon on the days of the equinoxes. That's 4.89 minutes at 47° latitude and 8.90 minutes at 68° latitude, a difference of four minutes.
The remaining minute could easily be due to rounding. Web sites typically publish sunrise and sunset times rounded to the minute because the difference between actual and predicted times varies by over a minute due to due to variations in the atmospheric state.


Regarding the original picture from the question

No, it is not feasible to get a picture like this. According to this webpage the photograph you posted was taken with the Hubble Space Telescope's Wide-Field Planetary Camera 2.

The telescope itself has a 2.4 m (7.9 ft) mirror and the focal length of 57.6 m (189 ft) and is placed outside the Earth's atmosphere. You'd need a large instrument with adaptive optics to get near this quality (assuming you can't afford to place a telescope in orbit).

That said, it is possible for you to capture the polar ice caps on Mars using an amateur telescope. The quality of these will largely depend on the equipment you have and the location (atmospheric turbulence is very unforgiving if you use large magnifications) as well as the imaging technique.

Here's a gallery of planet pictures taken by amateurs. You can find some photos of Mars in which the polar caps are visible.

Regarding the Mars pictures you found on the Sky & Telescope website

To answer your questions about this image, I see no reason to think it wasn't taken with an 8" telescope. However, there must have been more to it than just snapping a picture.

According to the author, the photographs were taken using a Celestron 8" SCT telescope with a Meade 3x Telextender and Expanse 120 Camera.

We're talking about a focal length of 2000 mm with 3x extender, which gives us 6000 mm. Not bad.

The Expanse 120 is a CCD camera. Now, I'm not sure that's the case with these specific pictures but one way to reduce the impact of atmospheric turbulence is to record a video of a planet and stack the individual frames. Many of them will look very bad due to turbulence but occasionally, you'll get a better quality image. Selecting the best frames from a video and combining them into a single image can help you get a lot more detail.

Have a look at this guide explaining how consumer-grade computer webcams can be used. It contains a neat comparison of a single frame and a stacked image.

Please note that I'm merely speculating as to which techiniques could be used to obtain such images. I can see that the author posted their email on the S&T site and that there's a comments section as well. Feel free to ask them yourself.


The defintion of star/planetary/solar system

I think it's largely a matter of personal preference. Some people will want you to be technically correct and only use "solar system" to refer to the Sun and the objects that are under its gravitational influence. Wikipedia defines it as such:

  The Solar System comprises the Sun and the objects that orbit it, whether they orbit it directly or by orbiting other objects that orbit it directly.

NASA says that the term is more general:

  A solar system refers to a star and all the objects that travel in orbit around it. Our solar system consists of the sun - our star - eight planets and their natural satellites (such as our moon); dwarf planets; asteroids and comets.

As an astronomy enthusiast, I do differentiate between "stellar system"/"star system" and "planetary system". I'd use "planetary system" to refer solely to the planets in a system, as well as their moons, while I'd use "stellar system" to refer to the whole thing. That's just my opinion, though, and I'm unsure what professionals use. In casual conversation (especially with a non-expert), you can probably use whatever term you want. If you're using it formally, you should probably do some more research. And while you're at it, check whether or not you should capitalize "Solar" and "System" in "Solar/solar System/system". If people nag you about terminology, they'll nag you about capitalization.


There is evidence that both black holes and their event horizons exist.

The primary evidence for stellar-mass black holes arises from observations of the dynamics of binary systems. What has been found, for at least 20 binary systems, is that the optically visible star has a dark companion, that is usually more massive, and more massive than can possibly be supported as a neutron star or any other degenerate configuration ($>3M_{\odot}$ and mostly $>5M_{\odot}$ - see Narayan & McClintock (2014) for an excellent, accessible review).

Supermassive black holes at the centres of galaxies can also be "weighed" using the dynamics of gas, or in the case of our Galaxy, actually watching the stars orbiting something that must have a mass of $4.4\times10^{6}M_{\odot}$, yet is remarkably compact and emits little or no radiation.

The evidence for event horizons is more circumstantial, but not absent. In any case, absence of evidence would not be evidence of absence - event horizons are difficult to prove; they are small and far away, with elusive observational signatures. The lack of an event horizon could only be accommodated by inventing something even more bizarre than a black hole.

The most compelling argument for an event horizon is found in sources that accrete from their surroundings, either in a binary system or at the centres of galaxies (see Narayan & McClintock 2008; read particularly section 4). In some circumstances, the accretion flow can become radiatively inefficient (dark), in which case one expects that if the accretion flow reaches a "surface", it will radiate its kinetic energy strongly as it thermalises on impact. On the other hand, a black hole event horizon can swallow such flows without trace. The prediction is, that in their low-states, transient black hole accretors will be much less luminous than their neutron star counterparts, and this is what is found (by a factor of 100).

Other evidence is that quiescent neutron star binaries show hot, thermal emission, presumed to be from the neutron star surface. None has been seen from the black hole binaries. Type I X-ray bursts are seen in neutron star binaries, caused by an accumulation of matter and subsequent nuclear ignition on their surface. The black hole candidates do not show these bursts. The accretion rate onto the supermassive black hole in the centre of the Galaxy should produce thermal radiation at any "surface" that would be quite visible in the infrared part of the spectrum. No such detection has been made.

All of these observations are best explained if the black hole has no surface and can simply make the accreted energy "disappear" inside its event horizon.
Narayan & McClintock conclude that these lines of evidence are "impervious to counterarguments that invoke strong gravity or exotic stars".


Oxygen in a methane atmosphere akin to methane in oxygen atmosphere?

Yes, it would work.

The difference between "fuel" and "oxidizer" is rather made for convenience, but it's not a fundamental distinction at the macroscopic level. The only differences are at the microscopic level, related to what the electrons do during the reaction. All such processes are redox reactions (reduction-oxidation), where one participant is the reducer, and the other is the oxidizer. That simply means the reducing agent loses electrons, while the oxidizing agent gains electrons; once the electron balance is final, the two components are bound together and the reaction is complete.

The oxidizer doesn't even have to be oxygen, it can be many elements or compounds hungry for electrons.

The bottom line is - you can burn either gas in an atmosphere made of the other. The gases don't care, as long as they react with each other. There are videos online where you can see chemists burning oxygen in an atmosphere of hydrogen, or burning hydrogen in an atmosphere of chlorine, etc. All these are redox reactions.

As to what's the minimum percentage of methane in the atmosphere that could sustain an oxygen flame, I am unsure. But it must be similar to the situation on Earth, where a low-enough oxygen contents in the air will eventually extinguish any flame. I would guess that 20% methane would work just fine for burning oxygen, but that's just a guess, so take it for what it's worth.

As others have pointed out, a methane atmosphere would indicate a very different chemistry at the planetary level, which may or may not be able to sustain life in general. It would definitely not sustain life as we know it.

Also, the Goldilocks zone for a methane atmosphere is different from the same zone for an oxygen atmosphere - it's further away from the star, because methane retains infrared better than oxygen (methane is a "greenhouse gas").

Fire and Flame 19 - Oxygen Flame in Hydrogen (Royal Society of Chemistry)
Fire and Flame 21 - Oxygen Balloon in Hydrogen (Royal Society of Chemistry)
Fire and Flame 22 - The Planet Jupiter (Royal Society of Chemistry)
Fluorine - Periodic Table of Videos
Exploding Hydrogen Bubbles (Periodic Table of Videos)


Determine full Moon position into sky from Sun position

The Sun's and Moon's hour angles at those times will be similar (both about 48$^\circ$ east of the meridian), but their declinations will be opposite (Sun 19$^\circ$ north of the equator, Moon 19$^\circ$ south).
The Sun in the morning will be over 30$^\circ$ higher than the Moon in the evening.

As the Moon rises and sets about 50 minutes later each day, a better predictor for the Moon's position at 21:30 on July 27 is the Moon's position at 20:40 on July 26, about 1$^\circ$ lower than the eclipse position if observing from a mid-northern latitude.


what would a captured planet do if it was in a retrograde orbit?

Totally dependent on the details.  A rogue planet could be any mass from Pluto to 10 Jupiter masses - quite different gravitationally.

And exactly what it's trajectory was makes a huge difference.  If it came barreling into the inner system it could cause chaos, particularly a large body.

It would certainly have to pass thought the Oort cloud and possibly the Kuiper Belt.  Any appreciable mass doing would disturb many, many objects which have the potential to fall into the inner solar system (or not).  So that alone could be dangerous for us (or not).  It's a dice throw - quite random from our point of view.

Stable is relative.  If it disturbed orbits in such a way that in a hundred thousand years Earth would become uninhabitable that might be not too bad - a lot can happen to a species in that time scale, especially one armed with nukes. :-)  So this isn't a question with a single answer.


When did the concept of "celestial sphere" fall?

Friedrich Bessel's 1838 observation of stellar parallax disproved the notion that all stars were equally distant.

However, by this time, the idea that the stars were on a sphere surrounding the solar system had already been discounted (but not disproven) by the application of the Copernican principle (The Earth does not have a privileged position) and Occam's Razor (Adopt the model that has the fewest assumptions)

Giordano Bruno is particularly noted with the idea (c. 1580) that the sun is just one star among many which are distributed throughout the universe.


What are some night sky objects I could see with my Celestron UpClose 20x50 Porro Binocular?

Depending on the season you can try different objects. Try large star clusters and nebulae:

The Pleiades
Orion nebula
Andromeda galaxy
M13 globular cluster

Also double stars like epsilon Lyrae.

It helps to have a foto tripod and an adaptor for the binoculars. Gives much better view when the binoculars are resting on the tripod.


Rings are made up of tiny (and not so tiny) pieces of rock and ice that are in some way the bits "left over" from the formation of the planet. The theory involves the Roche limit - and is that particles that are already within this limit can't accrete into a larger body because of the tidal forces involved.

Another theory is that they are formed when a moon comes closer to a planet than the Roche limit, the tidal forces cause it to break up and form a ring. Though the presence of "shepherd" moons in the rings of Saturn does hint that this may not be major source of material.

Both explanations, to me, imply that you'd only get major ring systems around larger (gas giant) planets, though it doesn't preclude rings around smaller (rocky) planets. This seems to be borne out by our solar system where the gas giants have rings whereas the rocky planets don't.

Source

As to being able to see them, you should be able to see the rings of Saturn with an amateur telescope which has a 50 - 100 powers of magnification. With binoculars you'll probably see a misshapen blob.

Source


I think there are no deep sky objects (or extended objects) that show any color to the naked eye. They are too faint to activate the color sensors, so they appear white. I think you saw something nonastronomical.

You will have a better idea the next time you see it; get a better estimate of its location so that you can find it on a star chart. Use binoculars if you have them.


Precession and the location of the sun on the horizon at equinox

The fountain will always line up with the Sun during the equinox. The precession of the Earth's axis will not change this fact. What it will change is exactly when the equinox happens. However, the fountain doesn't care about when the equinox occurs, only that it does and when it does happen, it will be pointed at the Sun.

The equinox is simply the precise point in time when the Earth's axis is neither leaning towards or away from the Sun, but rather leaning along it's direction of travel. There are many analgous definitions of the equinox, but the main point here is that the equinox is simply a specific orientation of the Earth's tilt with respect to the Sun. So long as the Earth is in that specific state, the equinox will occur and the fountain will line up with the Sun. As I said though, the precession may cause the date of the equinox to drift over time.


Is Higgs Boson mass equal to the missing mass of dark matter?

To answer this question, we need to understand what dark matter and a Higgs Boson are.

Higgs Boson

A Higgs Boson is an elementary particle in the standard field of quantum physics. It was theorised by a few people, namely Peter Higgs. The Higgs Field is a field which we can now believe exists, as we can detect it through its excitations, which become Higgs Bosons. The quantum excitations of the Higgs Field create Higgs Bosons.

A Higgs Boson is a particle with no spin, infact, it is the only particle with no spin, electric charge or colour charge. It's also extremely unstable and decays into other particles almost instantly.

Dark matter

Dark matter is a heavily researched topic in physics at the moment. Out of all matter in the universe, dark matter makes up about 80% of it!

One example of a theorised candidate for a dark matter particle is a WIMP. These are abbreviations of 'Weakly Interacting Massive Particles'. These particles are not normal matter, therefor we can't detect them. Our efforts to detect them include trying to look at the annihilation of WIMP particles.

The main characteristics of WIMPS are that they only interact through the weak nuclear force or gravity. They also larger than standard particles (a dark matter particle with mass less than a electronvolt is classified as light dark matter)

Wimps move slowly and are therefor cold, making them one of the main candidates for cold dark matter.

Think of dark matter as a 'skeleton' for normal matter.

Conclusion

From looking at both types of particles, we can clearly tell that Higgs Bosons are completely different from dark matter. We are still trying to find out about them both, as both of them can answer fundamental questions in physics.


How to make projection from altitude and azimuth to screen with screen coordinate system?

The example sky map appears to be a stereographic projection; circles of right ascension and declination are projected as circular arcs, and their intersections are projected as right angles.
The example has a 195° field of view, and the eastern and western horizons are mislabeled.

Let the celestial sphere be centered at the origin with radius 1, zenith at (0, 0, 1), northern horizon at (0, 1, 0), and eastern horizon at (1, 0, 0).
Assuming azimuth 0° is north:
$$\begin{align}
x &= \sin \mathrm{Az} \cos \mathrm{Alt} \\
y &= \cos \mathrm{Az} \cos \mathrm{Alt} \\
z &= \sin \mathrm{Alt}
\end{align}$$

To project from the nadir (0, 0, -1) onto the plane z = 0:
$$ (x', y') = \left( \frac{x}{z + 1}, \frac{y}{z + 1} \right) $$

To fit this between (0, 0) and (2R, 2R) on a canvas with northern horizon at (R, 0) and eastern horizon at (0, R), scale by -R and translate by (R, R):
$$ (X, Y) = (R(1 - x'), R(1 - y')) $$

For a sky map centered overhead, if north is up, east should be on the left.