Does Universal Time really track mean solar time?

This question asks in effect: Is there a discrepancy, perhaps progressive, between the official time-measure UT1, and some other credible measure of mean solar time? This is harder to answer than it might first seem.

UT1 is the current official representative of mean solar time: it is doubtful that any independent measure of mean solar time (other than direct relatives of UT1 such as UT0 or UT2) has been recently defined or computed to any precision. (The method of the equation of time as normally carried out is only roughly approximate, because it ignores the many solar perturbations discovered since the early 18th-c. and its results may vary from accurate mean solar time by up to about 3 seconds: see Hughes et al. (1989), "The equation of time",  http://adsabs.harvard.edu/abs/1989MNRAS.238.1529H .)  But as the question indicates, one reason for questioning UT1 is that solar time naturally depends on two independent variables, related to the angular rates of the earth's axial rotation and of the earth's orbital motion around the sun:  In contrast, the official UT1 calculation appears to condense these into a single constant ratio, thus losing one of the independent variables of the natural physical model. This may well raise suspicion that UT1 deviates or will deviate from a more physically-defined mean solar time.

There is in fact a (no-longer-official) alternative and physical basis for calculating mean solar time. This can be gathered from the history of 19th-century mean solar time definitions and calculations. At the same time, there is also some documentation from the history to indicate the origin and nature of some of the steps taken in the 20th/21st-century, to change the old 19th-century physical model to the current (and very non-transparent!) basis of calculation for UT1, which is described in: (N Capitaine et al. (2003), "Expressions to implement the IAU 2000 definition of UT1", Astronomy & Astrophysics 406, 1135-1149, http://adsabs.harvard.edu/abs/2003A%26A...406.1135C ).

Classical method for determining mean solar time (Nautical Almanac, 19th-c.):  (Not discussed here is the method of the equation of time, which grew more cumbersome as perturbations of the apparent solar motion were successively discovered during the 18th-c.; all of them had to be included in precise calculations.)  A simpler equivalent method for mean solar time, to a suitable astronomical precision for its period, was provided by Nevil Maskelyne's tables of 1774 (found at http://docs.lib.noaa.gov/rescue/rarebooks_1600-1800/QB12M31774.pdf), especially his table of the Sun's mean longitude converted from degrees to hours and given to a precision of 0.01 time-second. His instructions showed how to use this to convert observed sidereal time to mean solar time. It may be unclear whether Maskelyne intended this method to make a standard out of Greenwich mean time, but that seems to be what happened. Thus initially, according to J J Lalande's 'Astronomie' (3 editions, Paris, 1764, 1771, 1792) methods for mean time still relied usually on the equation of time, but Lalande's third edition of 1792 (vol.1, https://archive.org/details/astronomielalande01lala , see art.1014-5, p.361) added a mention of Maskelyne's table, commenting that 'in England' it was used to find the mean time of an observation. The Nautical Almanac eventually began (from 1833) to tabulate the relation between sidereal and mean solar time (Sidereal time at (Greenwich) mean noon), and during the rest of the 19th-c. such tabulations provided what became standard observatory methods of finding mean solar from sidereal time (see J C Adams 1884, "On the Definition of Mean Solar Time", Observatory 7, 42-44, http://adsabs.harvard.edu/abs/1884Obs.....7...42A ).

The method has a simple physical basis in spherical astronomy, using a division into two parts of the arc of the celestial equator that extends from the equinoctial point eastwards to the point culminating at the meridian -- shown in the diagram below.

(The upper part of the diagram represents the celestial equator above the horizon (marked E W) for an observer in the northern hemisphere looking south. M is the culminating point where the equator intersects the (south) meridian.  Q is the equinoctial point 'first point of Aries'. QS represents the sun's mean longitude reckoned from the equinox: it has been (in concept) transferred from the ecliptic to the equator where it now represents the mean right ascension of the fictitious mean sun. For convenience of arrangement, the diagram represents a position in early afternoon (S somewhat west of M) in about June (S is nearly 90° east of Q). (N) is below the horizon.)

The time-definitions and explanations of J-J de Lalande (1792, 'Astronomie', 3rd ed. vol.1, art.1014, p.361, https://archive.org/details/astronomielalande01lala) state clearly (expressing right ascensions in degrees, according to Lalande's custom):-

"The mean longitude of the sun, plus the mean time converted to degrees, gives the right ascension of the mid-heaven."   That translates, in terms of the diagram here, to:  QS + MS = QM .

Lalande also pointed out (art.1015) that "The Sun's right ascension, or that of the star used to find the right ascension of the midheaven, should be counted from mean equinox like the mean longitude of the sun ....". Modern readers have to note further that (a) observed and tabulated 'sidereal time' up to about the end of the 19th century always included the effect of nutation in right ascension, i.e. it was 'apparent' sidereal time (see explanation in Nautical Almanac 1864, pp.515, https://babel.hathitrust.org/cgi/pt?id=mdp.39015068159329 ), and (b) also in those days tabular mean longitudes of the sun still included the effect of mean aberration: Newcomb's tables of 1895-8 were the first to attempt to remove it.

Lalande (art.1015) continued:  "it follows that the right ascension of the mid-heaven" {i.e. sidereal time} "added to the complement to 24h of the Sun's mean longitude gives the mean time", and "this method is used in England to find the time of an observation: Mr Maskelyne has given in his tables the Sun's movement in time to hundredths of a second"."  In terms of the diagram, this translates as:   mean time MS = QM + (24h - QS) = QM - QS .

A modern version of the foregoing classical method for defining mean solar time: During the period when methods like the foregoing were current, sources of the solar tables used to compute the annual almanacs were occasionally updated, when newer and more correct theories and tables became available.  Thus, for the (UK) Nautical Almanac, Delambre's mean elements for the Sun were used with minor corrections down to 1833, then Bessel's mean elements from 1834 to 1863, and Leverrier's from 1864 to the end of the century. The principle remained unchanged although slight discontinuities in the tabulations occurred at the points of changeover.  Accordingly, a modernized version of the same method can be implemented using recent accurate data for the sun's mean longitudes. Two modern sources of mean elements/longitudes for the sun (both in terms of ephemeris/dynamical time TT) are available: in J L Simon et al. (1994) "Numerical expressions for precession formulae and mean elements for the Moon and the planets" (derived from the JPL ephemeris DE200 which was the official basis of the Astronomical Almanac from 1984-2002), Astronomy & Astrophysics 282, 663-683, http://adsabs.harvard.edu/abs/1994A%26A...282..663S ; and in J Chapront et al., "A new determination of lunar orbital parameters ..." (derived from the JPL successor ephemeris DE405 in official use 2003-14), Astronomy & Astrophysics 387, 700-709, http://adsabs.harvard.edu/abs/2002A%26A...387..700C .  The solar positions in those two ephemerides hardly differ from each other for present purposes.  Mean solar longitudes reckoned from mean equinox of date are given in Simon et al. The data of Chapront et al. are for a fixed equinox of J2000 to which general precession in longitude has to be applied. Both sets of modern mean longitudes are geometric, leaving mean aberration to be applied as -20".4955 in arc.  With that adjustment they can then be used like the old mean longitudes, converted to time, treated as right ascensions of the fictitious mean sun, and applied in the formulae given above to relate mean solar time with mean sidereal time.  The results may be compared with current values for UT1. {Some example calculations will be added if requested.}

Modern IERS zero meridian: It is worth noting that it appears no allowance is needed for any difference between time at the old transit instrument at Greenwich observatory and time at the new IERS reference (zero) meridian. The ground track of the IERS reference meridian is located about 102m to the east of the meridian line through the old transit instrument. A paper of S Malys et al. (2015), "Why the Greenwich meridian moved" (J Geodesy 89, 1263-1272, at http://adsabs.harvard.edu/abs/2015JGeod..89.1263M ) accounts for the difference of ground track as an effect of local gravitational deflection of the vertical. Thus, while the ground track of the IERS zero meridian is appreciably east of the old Airy transit instrument, by a distance which would make a rotation of about 5.3" arc in longitude or about 0.35 time-seconds if it represented a geocentric rotation on a spheroidal earth, there has in fact been no rotation of the reference frame (within measurement error): the deflection of the vertical makes the difference a parallel displacement.  With the nominally meridian (but off-geocentric) plane of the old transit instrument practically parallel to a geocentric meridian plane through the more easterly ground track of the IERS reference meridian, a transit instrument mounted in the IERS geocentric meridian plane would 'see' stellar transits at the same times as those occurring at the Airy transit instrument.

Modern changes in mean solar time/UT1:  The 19th-century Nautical Almanac standard for mean solar time was altered in several successive ways towards the current standard for UT1. (Not discussed here are the corrections for observed polar wandering and seasonal effects leading to the differences between UT0, UT1 and UT2.)  But there was a history of 19th-c. controversy about one of the changeovers mentioned above from older to newer solar data, worth brief description because it foreshadows and may possibly have influenced some of the 20th/21st-c. departures from the use of mean solar longitudes.

At one of the changeovers (from Bessel's data down to the end of 1863, to LeVerrier's data used from 1864 on), a discontinuity of a little over half a second was generated in the tabular relation between sidereal and mean solar time  (shown e.g. by the daily tabular values of sidereal time at mean noon in the Nautical Almanac for '1863 Dec 32' and 1864 Jan 1) (data for Greenwich mean noon 1864 Jan 1 in the 1864 volume, at https://babel.hathitrust.org/cgi/pt?id=mdp.39015068159329 , and for the same physical day in the 1863 volume where it was designated '1863 Dec 32',  https://babel.hathitrust.org/cgi/pt?id=mdp.39015068159006 ).

There was a brief explanation and apology in the 1864 preface: the step-change arose because LeVerrier's theory had identified new long-term solar periodic perturbations: as long as these remained undiscovered, their effect had been implicit in the old mean elements, and so they had to be removed from the new; the largest of them due to Mars and Jupiter was near its peak at the start of 1864 and responsible for nearly all the time-discrepancy of the changeover.

This changeover and discontinuity attracted little attention for nearly twenty years; but then E J Stone, an astronomer of some repute and director of a university observatory, was trying in the 1880s to account for growing discrepancies between observed places of the moon and positions calculated by Hansen's 1857 'Tables de la Lune'. Stone seized (quite mistakenly), as the cause of the lunar errors, on the 1864 discontinuity of about half a second in calculated mean solar time arising from the changeover from Bessel's mean solar elements to those of Leverrier (E J Stone, 1883 MNRAS 43, 335-345 and 401-407).  The two effects were unrelated and Stone had also made a mistake that exaggerated errors by a factor of about 365. Stone's mistakes were pointed out in print by no less a quartet than John Couch Adams (1883 MNRAS 44, 43-47; 1884 MNRAS 44, 82-84), Arthur Cayley (1883 MNRAS 44, 47-49; 1884 MNRAS 44, 84-85), Simon Newcomb (1884 MNRAS 44, 234-5; 1884 MNRAS 44, 381-3; 1894 MNRAS 54, 286-8) and G B Airy (1883 Observatory 6, 184-5), but Stone stubbornly stuck to his position.  He also wrote a vehement paper for the Royal Society, based on his faulty analysis of the lunar errors, in which he insisted that no revision should ever again be allowed in future to create any discontinuity in the tabular measure of mean solar time (E J Stone (1883), Proc. R. Soc. Lond.  35, 135-7, http://rspl.royalsocietypublishing.org/content/35/224-226/135 ). This was of course a demand incompatible with the use of corrected solar mean longitudes whenever new theories and observations would show the need for correction. But strangely enough, Stone's principle of requiring continuity has actually been followed in the alterations to UT1 made explicitly in 1984 and 2003 (and implicitly in 1960). It is hard to tell whether or not this arose from some descendant of Stone's influence.

There have been four main occasions of alteration away from the 19th-c. practice described above: in 1900, with Newcomb's formula for time quantities; in 1960, and the introduction of Ephemeris Time into many calculations but not into the time quantities; in 1984 with a fresh official definition of UT1 adopted from Aoki et al (1982), "The new definition of Universal Time", http://adsabs.harvard.edu/abs/1982A%26A...105..359A ; and in 2003 with the introduction of the current basis of UT1, described briefly in IAU 2000 resolution B1.8 and more fully in N Capitaine (2003) "Expressions to implement the IAU 2000 definition of UT1", http://adsabs.harvard.edu/abs/2003A%26A...406.1135C .

1900:  Newcomb's time formula made a very small alteration in the time-equivalent of the sun's mean longitude, by increasing the quadratic term. The motivation and effect go back to the work and practice of LeVerrier, summarised by A Gaillot in 1886: "Sur la mesure du temps", Bulletin Astronomique, Ser. I, v.3, 221-232, http://adsabs.harvard.edu/abs/1886BuAsI...3..221G .  LeVerrier's analysis of precession supposed that the earth's axial rotation was perfectly uniform. He also found that his theory of the precession of the equinoxes in right ascension led to a quadratic term discordant with the quadratic term for the sun's secular acceleration in longitude. By incorporating an increment in the quadratic term in the sun's longitude to remove the 2nd-order discrepancy, the effect was to make increments in mean time closely proportional to those in sidereal time (and to transmit the supposed uniformity of earth rotation into a proportional uniformity of the derived (but altered) calculation for mean solar time).  Newcomb mentioned the point briefly in his book "The elements of the four inner planets and the fundamental constants of astronomy" (1895, at  https://archive.org/details/cihm_16774 , see p.188). This appears to be the historical origin of the still-current practice of making the measure of mean solar time an exact linear function of earth rotation, a point specifically raised by the current question. No-one now supposes that earth- rotation is uniform.

1960:  With the introduction of Ephemeris Time (https://en.wikipedia.org/wiki/Ephemeris_time) into the official almanacs from 1960, the solar data was effectively corrected, not as had been usual by change of numbers, but by time-shifting it by a displacement DeltaT, the difference (Ephemeris Time - Universal Time). The numbers were maintained unchanged, but they related now to a different timescale. No alteration at all was made in the calculation of nominally mean solar time. The effect was to retain the buildup of the time errors constituted by the difference between the new best estimate of the sun's mean position and the old.

1984: The definition of UT1 was revised in a way that maintained continuity of value and rate at the moment of changeover, the start of 1984 (see S Aoki et al., (1982), link above).

2003: A further revision to UT1 was made, maintaining continuity at the moment of changeover, the start of 2003 (see N Capitaine et al., (2003), link above).

Summary/conclusion:  The description above shows something of how the current calculations for UT1 have departed in principle from classical methods used to define and calculate mean solar time. But it seems no longer to be claimed that UT1 is mean solar time: thus the IAU 2006 NFA glossary calls UT only "a measure of time that conforms, within a close approximation, to the mean diurnal motion of the Sun". Current values of UT1 can be compared with results of a modernized version of an old standard method for mean solar time from sidereal time. It seems that differences between this and current UT1 are still well under a second, though they can be expected to increase with increasing DeltaT, the difference in seconds between ((dynamical) TT - UT1). Thus the differences probably are remaining less than allowable differences for example between UT1 and UTC. The IAU gloss seems broadly justified, although it may be of continuing interest to have a method of calculating a more traditional form of mean solar time to check on any accumulating discrepancies.

{Notice of any mistakes in all of this will be gratefully appreciated!}


Your question is based on a false premise:

there is no way to see

On the contrary, the distant universe is being observed to infer information about cosmology and observational cosmology is a field of active research. Like all scientists, cosmologists make hypotheses and test them by making observations. Luckily for cosmologists, light travels at a finite speed, so by looking at objects very far away, you see how they were a long time ago.
A better question might be "How do cosmologists know what they do?".
Here are a couple examples of things that have been observed to gain insight on cosmology.

Measuring the redshift of galaxies shows that the universe is expanding. Observations of type Ia supernova even indicate that this expansion is accelerating.
Observing the cosmic microwave background disproved some hypotheses about the early universe.
Comparing the amount of stars and gas in galaxies to the speed at which they spin shows that something is amiss, so more observations are being made to try to figure out what is going on.

Now, of course, we have to make some assumptions to make this all work. For instance, cosmologists generally assume that the universe is homogeneous and isotropic at large scales.


What might cause a planet to have a significant tilt in their orbit?

Orbital tilts are almost certainly an effect of the formation and evolution of the protoplanetary disk (at least in the majority of cases). They could also be captured rogue planets.

Consider the eccentricity of Pluto. Like Pluto's eccentric orbit, it's probable that its current inclination is the result of orbital resonances with Neptune. This paper (unpublished grad student research) describes some such scenarios.

It is likely these resonances are between the Jupiter-induced precession frequencies of the ascending nodes of Pluto and Neptune. Pluto's inclination could either be relatively stable now, or it may still oscillate on the timescale of several million years.

I wish I could give a more definitive and detailed explanation - it's just one of these cases where no one knows for sure.


We have not yet discovered any planets that are certainly capable of supporting life, (except Earth) and the few rogue planets that we do believe that we have found have been gas giants, no good for life.

It is unlikely that life could exist on a rogue planet. The surface of a terrestrial planet that was not in orbit around a star would be exceedingly cold, as cold as Pluto.

The only place for life to start would be in a layer of liquid water, similar to that on Enceladus. If the core were sufficiently warm, it could maintain a liquid water layer, under a thick layer of ices.

No such rogue planet is known to exist.


What's this red stuff in some astronomical photos, e.g. Centaurus A?

"Clean the image", how apt, because most of what you can see is caused by dust.

It is a bit difficult to comment specifically because I think you have chosen an example which is a combination of images and not all the images are in the visible part of the spectrum and so have had a "false colour" applied. In fact it is a combination of optical, X-ray and sub-mm images.

In general terms red is present in false colour astronomical images, usually used to representing the coolest or longest wavelength light in the image.

Red is also present in true-colour images of nebulae and is caused by light emitted due to transitions of atoms and ions between excited states separated by particular energies. Red light is predominantly due to Hydrogen alpha and ionised nitrogen emission.

The coolest stars are also red in true colour images, simply because their pseudo-blackbody radiation spectra peak in the red part of the spectrum for temperatures lower than 5000 K. The integrated light from old stellar populations (predominantly cool main sequence stars and red giants) would therefore have a reddish tinge - for example in elliptical galaxies.

As I said, it is hard to say in the case of this particular composite image, but the band across the centre of the galaxy is dust. It obscures (makes dark), the optical image, but warm dust glows in the sub-mm image. The structures coming out of the galaxy at right angles are jets from the active galactic nucleus traced in the X-ray.


Question regarding the Milky Way when calculating galactic space velocities for galaxies

v$_r$ and μ are radial velocities and proper motion as seen from the Sun (ie heliocentric)  and the equations in the appendix convert to v in the $S_o$ frame which is the center of the Galaxy.  So, the motion of the MW is (0,0,0).  If you want to, you can solve for v$_r$ and μ for the MW by plugging v=(0,0,0) into the equation, and find the heliocentric proper motion and radial velocity of the Galactic center. But it is not needed.


How to convert magnitudes to bolometric luminosity?

Quoting from Wikipedia,

  The bolometric magnitude Mbol, takes into account electromagnetic
  radiation at all wavelengths. It includes those unobserved due to
  instrumental pass-band, the Earth's atmospheric absorption, and
  extinction by interstellar dust. It is defined based on the luminosity
  of the stars. In the case of stars with few observations, it must be
  computed assuming an effective temperature. Classically, the
  difference in bolometric magnitude is related to the luminosity ratio
  according to:

$ M_{bol,*} - M_{bol,sun} = -2.5log_{10}(\frac{L_*}{L_{sun}})$

  In August 2015, the International Astronomical Union passed Resolution
  B2[7] defining the zero points of the absolute and apparent bolometric
  magnitude scales in SI units for power (watts) and irradiance (W/m2),
  respectively. Although bolometric magnitudes had been used by
  astronomers for many decades, there had been systematic differences in
  the absolute magnitude-luminosity scales presented in various
  astronomical references, and no international standardization. This
  led to systematic differences in bolometric corrections scales, which
  when combined with incorrect assumed absolute bolometric magnitudes
  for the Sun could lead to systematic errors in estimated stellar
  luminosities (and stellar properties calculated which rely on stellar
  luminosity, such as radii, ages, and so on).

[leading to the accepted definition of]
$ M_{bol} = -2.5log_{10}(L_*) + 71.1974... $ , where the constant term is the zero-point luminosity $L_0$ .

Dunno if this helps, other than that you have to determine the spectral luminosity of the star in question.


Is remote measurement of isotope ratios possible, or must a sample be aquired

Yes, isotopic ratios can be readily established by spectroscopy.

Molecules where the constituent atoms differ by one mass unit have different vibrational and rotational energy levels and the gaps between them depend on the reduced mass of the molecule.

For example, in a diatomic molecule
$$ \mu = \frac{m_1 m_2}{m_1 + m_2}$$
and the vibrational frequencies depend on $\mu^{-1/2}$.

This technique is used for example to look at carbon monoxide feature in the infrared and establish the ratio of $^{13}$C/$^{12}$C.

The isotopic shift for atoms is much smaller in general, but is still detectable. The different mass of the nucleus changes the reduced mass of the electron (a little) and this changes the wavelength of the characteristic transition.

An interesting example of this has been the search for $^6$Li in the spectra of population II stars (e.g. Asplund et al. 2006). This isotope of Li should be produced in the big bang along with the more common $^7$Li but is more readily burned in stellar interiors.

The main optical resonance line of lithium occurs at 670.8 nm. The difference between the  wavelength of this line (actually, it is an unresolved doublet) is 0.015nm. This separation cannot be resolved in stellar spectra and instead the shape of the absorption line has to be modelled to estimate the $^6$Li/$^7$Li ratio.

The isotopic shifts become smaller for heavier elements.


How did the lighter elements end up in the center of the solar system? Solar System Formation

The solar system contains very little of elements heavier than Helium - less than 2% by mass.

This is reflected in the chemical abundances measured in the photosphere of the Sun. i.e. The Sun does contain heavier elements.

Your question is the wrong way around; it is not that the heavier elements have not sunk into the middle, it is that the vast majority of hydrogen and helium that was in the same place as the planets when they formed, did not end up as part of the planets. In fact, even this is only partially true. The mass of planetary material in the solar system is also dominated by the hydrogen and helium in the gas giants.

So the conundrum is only why the smaller planets don't have a similar composition to the Sun. The answer to that is temperature and gravity. A small, hot planet just doesn't have the gravity to retain fast moving hydrogen and helium atoms, unless they are trapped in some compound (like water!).

Thus, the small planets close to the Sun are depleted of light elements.


What is the size of the largest asteroid orbiting the Earth, other than the Moon?

There are no other known natural satellites orbiting the earth. Most of the objects orbiting the earth are man-made and largest object other than the Moon orbiting the earth is the International Space Station which orbits the earth at a height of about 400km from the surface and is about 100m wide.

It may be odd to you that there are no other natural satellites orbiting the Earth. We are not sure but it maybe that the probability of an object ending up in a stable orbit around the Earth is not that high. There are roughly 3 different categories of natural satellites - Large moons in roughly circular orbit like the Moon, small rocks in low orbits formed as a result of collision and nearby asteroids that a planet catches in its gravitational well.

It maybe that there are Earth is not massive enough to capture fast moving asteroids in orbit. Or the Moon which is quite large relative to the earth destabilizes a satellite's orbit around the Earth and there are not many stable orbits that a random natural object can get itself in.


What is this "Table of astronomy" about?

Fig. 1 and many others on that page seems to be diagrams of various trigonometric relations, used e.g. to convert between coordinates of celestial objects. But I'm not sure about this particular one.

Fig. 21 is a armillary sphere, i.e. a physical model showing the position of objects in the sky. Because this particular model has Earth in its center, it's also called a Ptolemaic sphere. Had it been centered on the Sun, it would have been a Copernican sphere, seen in Fig. 22.

Fig. 53 is a quadrant, which is  an instrument used to measure angles, e.g. between stars, or longitude and latitude.

Fig. 56 is a horary quadrant, which is used for finding the time of the day, using the Sun. This particular version was made of wood, paper, and brass by Henry Sutton around 1658.

Additionally,

Fig. 2 shows a diagram of the diurnal motion, i.e. the apparent motion of celestial objects around Earth.

Fig. 3 is titled "Aspect Trine Tetragon". Aspect is the astrological term for the angle between planets. This diagram shows a triangle ("Trine"), a square ("Tetragon"), and a pentagon.

Figs. 11 and 12 are micrometers (not to be confused with the length unit $\mu\mathrm{m}$), which is a device used to measure lengths very accurately. It was invented by the British astronomer William Gascoigne.

Figs. 13–20 are diagrams of the phases of the Moon and drawings of its surface. It seems the person who drew these diagrams accidentally swapped "1" and "8" in Fig. 18, since the figures run like 13, 14, 15, 16, 17, 81, 19, 20.

Fig. 34–38 illustrate how Lunar eclipses occur.

Figs. 43, 44, and 45 show the positions and orbits of the planets, Earth, and the Sun as depicted in the Ptolemaic (geocentric), Copernican (heliocentric), and Tychonic system, respectively. The latter used the mathematics of the Copernican system, but held Earth in its center since that's what Tycho Brahe believed.

Fig. 57 shows a gnomon, which is the part of a sundial casting the shadow.

Fig. 58 shows a globe. Fig. 59 seems to show how the spherical sky can be mapped onto a flat map by dividing it into 12 elongated stripes, each going from the south pole to the north pole, with a central width of 2 hours, or 30º. One of the stripes (around R.A. ~ 15 h, it seems) is shown in Fig. 60 with the constellations in that region.


Do heavier elements breakdown during supernova?

The answer is that in a pre-supernova star, most of its mass is still in the form of hydrogen and helium. It is only the central core where the primordial H and He has fused to heavier elements.

This picture of onion layers is typically what you see in elementary text books. It is completely misleading in a quantitative sense. It schematically represents what's at the centre of the pre-supernova star, but in terms of the mass that is in each shell (it is obviously a 3d object) you get completely the wrong idea, because this diagram is only about 1 Earth diameter across, compared with the actual stellar radius of something like the distance between the Earth and the Sun!

Here is a more sophisticated plot taken from a paper by Fuller et al. (2015). It shows time until the supernova along the x-axis and the y-axis represents a radial assay of the chemical composition from the centre of the star to the outside. The initial total stellar mass is $12M_{\odot}$. As you move leftwards towards the supernova explosion, notice how what is at the core changes - from being H dominated, to He dominated, to C/O dominated then Si and finally Fe (actually iron-peak elements). Note how much mass is contained within these core region for each stage of nuclear burning. The edge of the "helium core" encloses the central $4M_{\odot}$ of the star. The subsequent heavier element cores inside the onion ring structure enclose significantly less mass, until the iron core is around $1.3M_{\odot}$ just prior to the explosion. Blue shading indicates regions that are thoroughly mixed and homogenised by convection.

After the explosion, the neutron star that is produced will also have a mass of around $1.3-1.4M_{\odot}$. In other words most of the rest of the star (about $10M_{\odot}$ just prior to the explosion) gets blown out in the supernova. But of the $8.6M_{\odot}$ that makes into the interstellar medium, well over half is still in the form of hydrogen and helium; the minority will be carbon, oxygen, neon, silicon, iron etc., and only a very small fraction of that will have been transformed (by the r-process) into elements heavier than iron and nickel.

Thus although the material injected back into the interstellar medium is enriched with heavier elements, there is still plenty of hydrogen to start a new generation of stars. It is also the case that star formation is an inefficient process, so the material from which the supernova progenitor formed will still mostly be around in the interstellar medium. The picture you should have is of a gradual enrichment with heavy elements, especially as the interstellar medium gets churned up and mixed through a variety of processes (including supernova explosions!).

EDIT: Here is an even more awesome picture. The lower plot shows the relative mass fraction of each element as a function of enclosed mass as you work your way out from a 15 solar mass star (it has shed 2 solar masses during its evolution). The really awesome thing is that it is animated, so it shows you the first few moments after the core-collapse and how things start to change. Note that the outer $5M_{\odot}$ of the envelope is about half H and half He by mass prior to the core collapse. Lots of He and then O in the layers below that. The upper plot show how the density temperature and outward velocity are behaving. The image is from the website of Woosley and Heger (2007), a canonical work on the subject.

Hmm. I can't upload the animated gif. Here it is; well worth a look.


What is the Maximum Speed that can be acheived Because of Acceleration Due to Gravity?

It all depends on your frame of reference, but the biggest speeds caused by gravity will involve black holes or perhaps neutron stars.

From the point of view of a far external observer, an infalling object (from infinity) reaches a maximum inward speed of $0.385c$ (where $c$ is the speed of light) towards the black hole at 3 times the Schwarzschild radius. Thereafter, the object will appear to slow down and tend towards being stationary at the event horizon.

From the point of view of a "shell observer", that is somehow held stationary in the gravitational field of the black hole, the infalling object would have a speed that tended towards $c$ as the shell observer tended towards the event horizon at the Schwarzschild radius. It is meaningless to try to extend this inside the event horizon since (stationary) shell observers cannot exist inside the event horizon.


An image of the whole sky (360 degrees) centred on Sagittarius and rotated so the galaxy is in the middle:

This image from the "2MASS infrared survey" actually uses infra-red light, as it penetrates the interstellar dust better than visible light. It shows that any apparent curvature is an illusion. The milky way goes straight across the sky.

It also shows that when we look towards Sagittarius we look towards the galactic centre, and when we look away from it (towards Auriga) we are looking through to stars in the outer spiral arms. The night side of the Earth faces the galactic centre during July, and faces outwards in January.

Also visible in the image are the two Magellanic clouds (small but nearby galaxies outside the milky way), the clear bulge around the centre of the galaxy and the dark bands of dust along the milky way. The central bulge is much brighter in infrared than the spiral arms. Using infrared makes these features more visible.


Why does the Supernova 2006cm give a very different value for the Hubble constant? Why doesn't it increase error bars for the Hubble constant?

At a distance of $d = 87\,\mathrm{Mpc}$, with a Hubble constant of roughly $H_0 = 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ cosmological expansion should make the host galaxy UGC 11723 recede at $v=H_0 \,d\simeq6100 \,\mathrm{km}\,\mathrm{s}^{-1}$.

However, galaxies also move through space, at typical velocities from several
$100\,\mathrm{km}\,\mathrm{s}^{-1}$ in galaxy groups (e.g Carlberg et al. 2000) to some $1000\,\mathrm{km}\,\mathrm{s}^{-1}$ in rich clusters (e.g Girardi et al. 1993; Karachentsev et al. 2006).

The observed velocity of $4900\,\mathrm{km}\,\mathrm{s}^{-1}$ (Falco et al. 2006) is hence $\sim1200\,\mathrm{km}\,\mathrm{s}^{-1}$ smaller than the Hubble flow, but quite consistent with what may be expected.

This is why supernovae at such small distances cannot be used to deduce the Hubble constant, unless a very large number is observed such that statistical errors cancel out.


If fusion only happens near the center of the sun, what's with all the "fire" on the surface?

99% of nuclear fusion of hydrogen nuclei into heavy nuclei happens at the core of the Sun. This is because only at the core is the temperature high enough to overcome the coulomb barrier caused by the electrostatic repulsion between the nuclei. The solar core produces a lot of energy $(3.828×10^{26} W)$ by fusion and since all that energy has to go somewhere, it eventually reaches the surface making the surface of the Sun hot (6000K).

Most objects like a red hot iron or tungsten filament of a bulb glow when heated. An idealized version of this is called black body radiation. The colour at which the body glows or the frequency of the electromagnetic radiation depends on the temperature. Since the surface of the Sun is hot, it glows. This is not combustion but natural heat glow. Moreover glowing hot plasma rises up in columns while cooler denser gases settle at the surface of the Sun at speeds upto 7km/s in a process called solar granulation causing the Sun to look like a fierce fire ball.


I'm not an astrophysicist but I'll give it a shot. I'm taking this mostly from Hawking's "A brief history of time" and my own personal studies.

First, the universe is not like the earth. As far as I know this analogy only works when one considers the concept of imaginary time. According to Hawking, when you consider the theory of relativity with a imaginary time (which mathematically is pretty much only considering time as a imaginary number, rather than a real one) the equations that describe space are mathematically similar to those that describe a sphere. This is a trick used to try to make a quantum mechanics concept (Feynman's sum over histories) work with relativity.

This, however, does not mean that we can experience space (or the universe) like a sphere, because we are not capable of experiencing imaginary time. We experience only real time. If we were to experience time as it is defined by imaginary time, we would be able to to travel to the future and past as easily as we would travel from latitude -25º to +25º, and we all know that is not the case.

And in this answer, I'm not even taking into account the expansion of the universe, which would complicate things even further for someone trying to go around the universe. Just for the sake of the argument, imagine going around a sphere that is growing more and more. Even if you take the shortest path possible, depending on the rate of expansion the task becomes either impossible of very hard to say the least.

Hope this helps.


Why Free-Free emission is regarded as Thermal emission?

Thermal radiation $\neq$ blackbody radiation.

Thermal radiation is radiation that comes from a system where an equilibrium has been reached, where the various energy states are occupied according to the Boltzmann distribution and the particle velocity distributions are Maxwellian at some given temperature.

That does not necessarily imply that the radiation field is in equilibrium with the matter at the same temperature. It is this latter equilibrium that is required in order for the radiation field to approximate to a blackbody.

So the following statements are true:

Blackbody radiation is thermal radiation. Thermal radiation is not necessarily blackbody radiation.

An example of thermal, but non-blackbody, radiation would be  thermal bremsstrahlung. The particle velocities in the gas follow a Maxwellian distribution at a characteristic temperature and this determines the spectrum of the emitted free-free radiation. However, the bremsstrahlung radiation is able to escape from the gas without further interaction - in other words the gas is "optically thin". The result is a thermal bremsstrahlung spectrum that is not at all like a blackbody spectrum, especially at frequencies $<kT/h$.

On the other hand, the bremsstrahlung process is purely caused by the interaction of free electrons and ions. If you allow there to be atoms and partially ionised atoms in the gas, then it is feasible for those to be in equilibrium at the same temperature. The population of the various energy states in the atoms and ions is determined by this temperature, and transitions between these levels produces spectral emission lines in a completely predictable, temperature-dependent, way.


What does the active area on a typical CCD based detector look like?

An astronomical CCD is in principle sensitive across its entire active surface; there are no physical gaps between pixels. The pixels themselves are controlled and maintained by the underlying electronic circuitry, which sets up potential barriers between individual pixels. In practice, there are variations in the actual sensitivity within individual pixels: sensitivity is highest in the center and falls off towards the edges, although it never goes to zero anywhere within the pixel.

This paper (PDF) is a recent analysis of a (frontside-illuminated) CCD; Figure 13 shows the measured pixel response functions (how sensitive an individual pixel is across its surface), and Figures 14 and 15 show the photometric sensitivity map (what the sensitivity is for uniform illumination). In the latter case, you can see that the sensitivity never varies more than about 20% or so, depending on the wavelength of the light. Some of the variation corresponds directly to wiring within the CCD absorbing some of the light, but this has a very minor effect on the sensitivity.

The comparison made in Florin Andrei's answer to a consumer-camera-style CMOS sensor, as shown in the image, is misleading, both because CMOS pixels are physically somewhat more discrete in ways that CCD pixels are not, and because astronomical CCDs do not have per-pixel filters glued on top of them.

(The argument that "there has to be a narrow slice of silicon in between pixels to prevent short-circuits" is not correct for CCDs; the electrical separation between pixels is maintained by the underlying electronics, not by any material barrier. In fact, you have to be able to create temporary "short-circuits" between adjacent pixels in a column in order to transfer the accumulated electrons during readout. That's what the "charger-transfer" in "charger-transfer device" means.)


Does parallax measurement depend on position in the sky

It's all about basic geometry.
Base for the parallax measurements is Earth's orbit around the sun, giving you a maximum of 300 Mio km. With a given base dimension, you get the best precision when the base is orthogonal to the star direction. (In the other extreme, you get no parallax at all if the base is in line with the star).
For stars close to the ecliptic you get this optimum base angle only using two specific dates, half a year apart from each other (those when the star appears 90 degrees distant from the sun).
For stars nearly perpendicular to the ecliptic, you can choose any two dates half a year apart, giving you more chances to contribute max-precision measurements.
If doing a continuous observation of the star over e.g. a year, the difference should amount to a factor of sqrt(2), if the other parameters are comparable.


How to correct observed flux densities for redshift

You are familiar with the relationship between flux and luminosity, right? This is usually taken as the definition of luminosity distance in extragalactic astronomy,
$$ F = \frac{L}{4\pi D_L^2}.$$
The quickest way to convert this into a relationship with spectral variables is to use differential 1-forms. So, $L \rightarrow L_\lambda \,\mathrm{d}\lambda_e$ and $F \rightarrow F_\lambda \,\mathrm{d}\lambda_o$, with $\lambda_e$ the photon wavelength at emission (also called the rest frame wavelength) and $\lambda_o$ the observed wavelength at $z=0$ (also called the comoving wavelength). Thus
\begin{align}
  F_\lambda \,\mathrm{d}\lambda_o & = \frac{L_\lambda \,\mathrm{d}\lambda_e}{4\pi D_L^2} \Rightarrow\\
  F_\lambda & = \frac{L_\lambda}{4\pi D_L^2} \times \frac{\mathrm{d}\lambda_e}{\mathrm{d}\lambda_o}.
\end{align}
Now you just need to take your derivative using the definition of redshift, $\lambda_o = (1+z)\lambda_e$, and use some way of calculating the luminosity distance in your cosmology.

Formally, a lot of people like to use $\nu F_\nu = \lambda F_\lambda$ as though it were a total flux, but it's not. The derivation will work, but only because redshifting scales all frequencies/wavelengths by a single constant. $\nu F_\nu$ isn't a measure of the flux under any part of the spectrum, it's a spectral flux using a different coordinate system. Consider the following
\begin{align}
   F & = F_\nu \,\mathrm{d}\nu \\
    &= \nu F_\nu \frac{\mathrm{d}\nu}{\nu} \\
    &= F_{\ln \nu} \,\mathrm{d}\ln \nu.
\end{align}
In other words, $\nu F_\nu$ is still a spectral flux. If $F_\nu$ is in $\mathrm{ergs}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$ then $\nu F_\nu$ will be in $\mathrm{ergs}\,\mathrm{cm}^{-2}\,(e\text{-fold})^{-1}$. The $e\text{-fold}$ is a pseudo-unit, like radians, dex, octave, decibel, or magnitudes. In principle it can be omitted, just like how angular frequencies can be in radians per second, or just $\mathrm{s}^{-1}$.

Fun math exercise: show that $\ln(10) \nu F_\nu$ has units of $\mathrm{ergs}\,\mathrm{cm}^{-2}\,\mathrm{dex}^{-1}$.


Is there an advantage to the equatorial region of the far side of the moon for a radio telescope or would any crater on the far side work?

tl;dr: The less complicated you want to make your dish, the closer to the equator you want to put it.

Since you are asking about single craters I assume this question is about using a crater as a natural pre-form for a single, large dish antenna like Arecibo Observatory and FAST.
On the Moon with no atmosphere or ionosphere nearly the whole sky is available for good signal, but as you move away from the current zenith position (i.e. "straight up") there are several problems. One is that the effective area of the telescope may be reduced for geometrical reasons at steep angles, and the other is that the optics of getting a good focus is increase.
Below you can see that Arecibo can't look much beyond 18° away from the zenith, and while the much newer and more complex FAST dish with it's changing dish shape articulated by moving cables and actuators can go as far as 60° away, it loses useful collection area beyond 26° not even counting $\cos(\theta)$ losses.
You will also notice that each telescope's maximum zenith angle attempts to be larger than the geographical latitude of its location, so that it can see the ecliptic.
Moving your lunar dish near the poles means you will lose access to a lot of the sky and be restricted to a "cap" of declinations either near +90° or -90°.
So the less complicated you want to make your dish, the closer to the equator you want to put it.
Arecibo
From Overview of the Arecibo Observatory; ALFALFA, Undergraduate Workshop , Gregory Hallenbeck January 14, 2013

Latitude: 18° 20′ 39″ N
Can move dome to zenith angle position of 19.7°, But only to ~18° with good performance

FAST
From Wikipedia:

Latitude: 25° 39′ 11″ N
The maximum zenith angle is 60-degree when the effective illuminated aperture is reduced to 200 m, while it is 26.4-degree when the effective illuminated aperture is 300 m without loss.33,3
Although the reflector diameter is 500 metres (1,600 ft), only a circle of 300 m diameter is used (held in the correct parabolic shape and "illuminated" by the receiver) at any one time.22 Thus, the name is a misnomer: the aperture is not 500 m, nor is it spherical.
3Project FAST
22The Five-Hundred-Meter Aperture Spherical Radio Telescope (FAST) Project
33The optics of the Five-hundred-meter Aperture
Spherical radio Telescope


Is the motion of the Sun around the Galaxy a result of gravitational pull?

I would say the answer to your question is yes and no. Yes that the orbit of our Sun in the Milky Way is due solely to gravitational forces as you suggest, but no in that they are not all radial towards the center of the galaxy.

The motion of the Sun around the center of our galaxy can be broken into two motions. The first is an "azimuthal" orbital motion around the center of the galaxy due to radial gravitational forces from mass interior to our orbit (be it stars, gas/dust, dark matter, dragons, etc.). The other type of motion is an oscillation into and out of the plane of the milky way. This is also caused by gravitational forces, albeit not radial forces from the center of the milky way. As the sun moves "above" the plane of the Milky Way, there is more mass below than above and it gets pulled "down". When it is in the plane it has residual energy from "falling down" and continues to fall "below" the plane where it then gets pulled back "up". This type of oscillatory motion is constant and analogous to a mass on a spring.


Can we calculate the average temperature of the heliosphere?

The heliosphere is mainly defined by the region dominated by solar wind against the interstellar medium.

"The solar wind is divided into two components, respectively termed the slow solar wind and the fast solar wind. The slow solar wind has a velocity of about 400 km/s, a temperature of $1.4–1.6×10^6 K$ and a composition that is a close match to the corona. By contrast, the fast solar wind has a typical velocity of 750 km/s, a temperature of $8×10^5 K$ and it nearly matches the composition of the Sun's photosphere."

Hence with 1 million Kelvin we get a good estimate of the average temperature of the heliosphere.

The temperature is the freeze-in temperature derived from carbon ion charge states measured by SOHO. Hence it's not the black-body temperature, but a reasonable definition of temperature for solar wind.

"Coronal temperatures are inferred from density ratios of adjacent charge states. The freeze-in temperature derived from a given density ratio is the electron temperature that reproduces this ratio in a static situation. For the analyzed iron charge states we use the ionization and recombination rates of Arnaud and Raymond [1992]". More details about SOHO's temperature measurement here.

If the heliosphere would be in the thermodynamic equilibrium, according to Planck's law or Wien's law, you would get the peak at about 3nm, in the soft x-ray range. But the equilibrium assumption probably doesn't hold.

For the hot interstellar medium extreme ultraviolet radiation measurements have been tried (CHIPSat), some successful (EUVE).
Hence, in an ideal environment we would measure the radiation intensity across the electromagnetic spectrum, look for the peak intensity, apply Wien's law, and retrieve the  effective temperature, which would be identical to the black body temperature in an ideal world.

The plasma of the heliosphere is very rare, therefore almost transparent. That's why we don't feel that heat of about 1 milion degrees. Microwaves play a minor role in that temparature range.

A temperature profile of Sun's corona and heliosphere is shown e.g. on pages 42 and 47 of this survey.


Internet Resource which register all Ultra-long Gamma-Ray Bursts

I guess one of them you're referring to was GRB 130427A (not in December of 2010 or 2011 tho, but in April 2013)? But yes, there are several such catalogs, as are different designators and phenomenological classification methods for these gamma-ray bursts (GBR). For example, GRB 130427A is a simple nomenclature that consists of a date of discovery (130427 denotes 27th of April, 2013 and suffix A is just the first letter from the alphabet, increased when there's more than a single such event in a day), but it doesn't indicate e.g. its type or origin. The same GRB event is so cataloged under designator CSS130502:113233+274156 in the Catalina Real-time Transient Survey catalog, SDSS J113232.84+274155.4 in Sloan Digital Sky Survey catalog, and so on.

So, as you can see, it would help if you could clarify which designators you saw, so we can identify the catalog easier. But not wanting to leave you empty-handed, one such catalog that is frequently used and is relatively easily searchable is the Goddard Space Flight Center's GRBCAT (Gamma Ray Burst CATalog) (on top of those already mentioned).

Post update to the question, I've managed to identify SWIFT pages for the two GRB events in question:

101225A
111209A

Catalog numbers, references and data are available on linked pages. And here is a complete SWIFT GRB catalog (note, it's rather long, listing all GRB in a single page).


I'm not far from you (Israel) and I am also waiting to see comet ISON.

Comets are very unpredictable, and comet ISON has not yet reached the critical part of its journey which will determine visibility for us. Comet visibility is usually due to reflection of the Sun's rays on the coma and tails of the comet, as the nucleus is to small to see directly. So ISON's visibility (and all comets for that matter) depend on how much gas, water, and other materials that it outgasses. This in turn depends on how much volatiles it has, how much the Sun melts them, how much the slingshot around the sun and tidal forces fracture the nucleus, and other factors. So the actual visibility that will occur is very difficult to determine in advance. That said, ISON has all the markings of this being its first trip around the Sun. Thus, we assume that it has many volatiles to release and we assume that ISON will be exceptionally bright.

In any case, ISON will be most visible when it is closest to Earth, near the end of December. It should be most visible in the early morning sky, I'm guessing at about 4:00 AM or so. This is because ISON will be close to the Sun (thus visible close to the time that the Sun becomes visible). Like all celestial bodies, it will rise in the East. Each day the comet will move farther from the Sun and thus will be visible earlier and earlier in the night, until it disappears from view probably sometime in mid January.

You are welcome to visit Beersheba and have a peak through my binoculars!


What should I consider in order to accurately calculate the moon phase?

The moon's phase isn't easy to calculate accurately. Fortunately, the 4 main phases (new moon, first quarter, full moon, last quarter) are a bit more stable than the intermediate phases. Good main phase algorithms use a dozen or so sine calculations for the new or full moon, and a similar number for the first or last quarter. Those two algorithms you linked use very simple approximations so their results are rather crude.
The period of the moon's phase cycle (the synodic month) is stable over the long term, but it drifts considerably over the short term. So algorithms like those in your links are usable over a rather long time span, and they can be useful first approximations to feed to better algorithms, but by themselves they only yield precision results by accident. ;)
The guru of astronomical calculations for bodies in the solar system is Jean Meeus, so if you want quality algorithms make sure they reference Meuss. :)
For seriously accurate ephemerides, especially if you need to cover a wide time span, consider using the work of JPL. You can access their ephemerides at the Horizons Web page; a telnet and an email-based interface are also provided.
If you want to do precise astronomical calculations (eg for eclipse prediction) you'll need to learn about the various time systems used in astronomy, in particular Terrestrial Time and its forerunner, Ephemeris Time.
I see from your SO profile that you're familiar with Python. Good stand-alone Python moon phase code isn't easy to find online, but eventually I did find keturn's py-moon-phase, which is a Python port of the core code of John Walker's rather ancient Moontool, which uses algorithms from Meeus and also from Peter Duffett-Smith. I haven't actually tested John Walker / keturn's code, but it does look rather good (although it uses UTC instead of Ephemeris Time or Terrestrial Time). I have very similar (but rather messy) C code that's generally accurate to half a minute or better over the 20th century.
However, there's a package called PyEphem that can do all sorts of astronomical calculations. James K notes in the comments that PyEphem is now discouraged for new projects. The PyEphem developers recommend Skyfield instead. Astropy can also do a lot of what PyEphem does, and much more.


Did this paper just argue that there is surely life on Titan?

The underlying assumption in the paper is that the planet has water in liquid form.

The paper suggests that finding both CO2 and CH4 in the atmosphere of a rocky planet in the "habitable zone" (in which water can exist in liquid form) would be hard to explain, without life. Thus one doesn't need to detect atmospheric oxygen. If you can detect both CO2 and CH4, this should be taken as evidence of anoxic life (probably in the form of microbes)

Titan has CH4, but it doens't have much CO2 and there is no liquid water.
Titan is not in the habitable zone. On Titan, H2O is a forms part of the "rocks", and the oceans are made of hydrocarbons. Titan is not a good location to look for life as it doesn't have free liquid water. We have a photo from Titan's surface and it isn't covered in Microbes


Yes, indeed! Many nebulae are visible from Earth in a small and cheap telescope, and even to the naked eye (if you are standing in a sufficiently dark place).
In fact, yesterday I was watching the Orion Nebula with my 4.5" telescope (which is worth $200 or so) from my apartment in the middle of Copenhagen.
The term "nebula" is a bit of a… well, nebulous term, as it covers several distint phenonomena, for instance:

Galaxies,           e.g. the Andromeda Galaxy (though this is an old expression)
(Open) stellar clusters,   e.g. the Eagle Nebula
Globular clusters,  e.g. 47 Tucanae
Planetary nebulae,  e.g. the Eskimo Nebula
Supernova remnants, e.g. the Crab Nebula

However, you should be aware that even in a large and expensive telescope, the nebulae do not look anything like the beautiful images you find on the internet. With your eye, you will merely see diffuse, whitish clouds. The beautiful colors arise only in images taken through telescopes with long exposure times. If you want to make such pictures, you will need a telescope with a motor (since the telescope must follow the rotation of the sky) and the possibility of attaching a camera to it.
People are often disappointed when shown a nebula through a telescope. I think the beauty of it lies in knowing what it is that you are seeing, knowing for instance that it lies 600 lightyears away, that you are looking 600 years back in time, and that stars are right now being formed there.
Googling something like "nebulae visible in a small telescope" should get you started. Good luck!


A day in earth, a thousand years somewhere else

Assuming you mean a day in terms of an Earth day in both cases, the phenomenon I believe you're referring to is called time dilation, which affects the speed at which time is experienced by one body relative to another. This rate is affected by velocity and gravity, each of which can cause the bending of spacetime that results in the discrepancy between objects. Both gravity and velocity slow one's time as they increase.

The situation you describe is probably not naturally occurring, however is not outside the realm of reality.  As referenced by Wiki from Calder's book Magic Universe, Calder claims that an acceleration of a constant 1G (what we feel on Earth all the time) would result in the affects of time dilation allowing you to traverse the entirety of the known universe in a single human lifetime (for the traveler). Conversely, during your trip, the rest of the universe would be aging 'normally'. There are plenty of technical problems with the feasibility of achieving this, but is conceptually possible.

I'm not particularly qualified to explain the equations behind it, so I'll leave that to our more capable members. Coincidentally, time dilation is the excuse I give when asked why I walk so quickly.


What's the slit and slot on EIS(EUV Imaging Spectrometer) instrument of Hinode?

You can go back and reference the paper published on this telescope. Specifically it is Culhane et al. 2007. This paper talks about the overall telescope, however the paper describing the optics is Korendyke et al. (2006). Unfortunately this optics paper is behind a paywall.

Within the Culhane paper, in section 3, the describe the slit mechanism as

  Exposure times are controlled by a rotating shutter while a slit exchange mechanism can allow selection of four possible apertures — two spectral slits and two spectral imaging slots.

That's about as much detail as they give, except to note in table 1 that the slit and slot widths are 1, 2, 40, and 266 arcsec.

In the Korendyke optics paper, they give a bit more detail.

  Light passing through the slit is dispersed and stigmatically reimaged by the toroidal grating onto two 1024 x 2048 pixel CCD detectors. The focus of the grating can be remotely adjusted with a mechanism. In flight, the mirror will be rotated in ~0.125 arcsec increments about the Y axis [solar north–south (N-S)] to sample different structures of the solar surface with the slit. High-resolution spectroheliograms (raster images) are formed by steadily moving the solar image in fine increments on the spectrograph slit and taking re-peated exposures. An interchange mechanism allows selection among two slits (1 and 2 arc sec wide) and two slots (40 or 266 arc sec wide). The slot observations of the solar disk produce spectrally pure images
  of large areas of the disk in bright solar emission lines with each exposure. The slot images exhibit a modest spatial blur along the dispersion direction.

For completion, they also include the diagram below of the optics.

From reading all of this, it appears to me that the various slit/slot widths are designed for taking different types of images/spectra, either of very focused, single-instance areas (such as with the slits) or else of large regions (using the slots).


Would we detect a spacecraft before it landed on Earth?

Some thoughts about Earth-based detection systems:

We are looking at asteroids using radar: For instance the former Arecibo telescope could actively send out radio waves and calculate object position and size. However, we might be able to detect an incoming spaceship using this technology, assuming that the object approaches Earth within the radio beam, if we we are currently studying another object. In other words: Detecting an incoming object would be purely chance.

An all sky camera network for meteorite detection is another outward-looking system worth mentioning, but that would be very very briefly before landing of a hypothetical extraterrestrial space ship. I am sure you are not after this option since we would be seeing the object already with naked eyes if it would not be incoming at a very remote location. However, this option is to my knowledge working 365/24/7 and covers large portions of the sky, I believe.

Edit suggested by uhoh: There are also initiatives like NASA's Space Fence or ESA's Space Surveillance and Tracking (SST) segment monitoring space debris within the geospace which could possibly detect other objects as well, see also https://space.stackexchange.com/q/45144/12102

Concerning satellite-based observations:
If the objects emits significant radiation e.g. in parts of the spectrum of communication satellites, we might detect anomalities or malfunctions in one or more satellire and by chance figure out the extraterrestrial source of the distortion. This would of course only hold for slow approaching objects.


When are the planets occulted by the sun?

The search for sungrazing comets requires looking very close to the Sun for faint objects. This requires a chronograph and for faint objects is better done above the atmosphere. The SOHO spacecraft near the Sun-Earth L1 point does this nicely.
Here is a GIF I made for this answer. In addition to some comets you can see Venus on its approach to occultation by the Sun in 2016. Seeing the Pleides so close to the Sun is exciting as well!

These LASCO C3 images from SOHO were downloaded sohodata.nascom.nasa.gov/cgi-bin/data_query. The square frame is about 15.9 degrees wide.

You can see many trajectories for 2016 in http://sungrazer.nrl.navy.mil/index.php?p=transits/transits_2016 (shown below) and replacing 2016 with other years works as well.
The problem with this page is that while it may flag likely occultations, it does not unambiguously predict them, so you may need to use another source for accurate predictions.

Here's a quick calculation for 2016 using the Python package Skyfield. It seems both Mercury and Venus were able to hide behind the Sun.

Jackpot!  Mercury, Jupiter, Saturn, Mercury

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos, EarthSatellite

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180
Rsun              = 696392. # https://en.wikipedia.org/wiki/Sun

load    = Loader('~/Documents/fishing/SkyData')
ts      = load.timescale()
data    = load('de421.bsp')

names   = ('Mercury', 'Venus', 'Mars', 'Jupiter', 'Saturn')
objects = [data[name + ' barycenter'] for name in names]
earth   = data['Earth']
sun     = data['Sun']

ts      = load.timescale()
days    = np.arange(1, 366, 0.1)
times   = ts.utc(2016, 1, days)

observations = [earth.at(times).observe(obj) for obj in objects]
sunobs       =  earth.at(times).observe(sun)
sundist      = sunobs.distance().km
separations  = [obs.separation_from(sunobs) for obs in observations]

sunhalfangledegs     = degs*np.arctan2(Rsun, sundist)
sunhalfangledegsmin  = sunhalfangledegs.min()

if True:
    plt.figure()
    for i, (name, sep) in enumerate(zip(names, separations)):
        plt.subplot(5, 1, i+1)
        angle = degs*sep.radians
        minangle = angle.min()
        plt.plot(days, angle)
        plt.ylim(0, None)
        ymin, ymax = plt.ylim()
        ymax = 5*max(ymin, minangle, sunhalfangledegsmin)
        plt.ylim(0, ymax)
        plt.plot(days, sunhalfangledegs, '-r', linewidth=0.5)
        plt.title('observed Sun-Earth-' + name + ' angle (degs)')
    plt.show()


Pluto is actually smaller in diameter than Triton, and is also farther away, meaning that Triton covers roughly 1.4x (according to WA) the angle that Pluto does, making occultation that much more probable apriori, ignoring their actual orbits.

In addition to the above, New Horizons recently observed Pluto's atmosphere with far more detail than what we can do from here on Earth, so it makes sense to measure a different target.


Why the more-luminous star has larger surface area when two stars have the same temperature?

I'd say your problem is that you are imagining the light rays emanating out from the center of the star.  Try it again with light emanating from each square area on the surface, and you will see that being at the same temperature means each square on the surface sends out the same number of rays, but there are just more such squares.  Then you will see a lot more light rays coming from the larger star, in proportion to its surface area.


What might the surface of Jupiter's rocky core look like?

There's not going to be any surface features.

First of all, let's assume you could look through a thick soup of metallic hydrogen. This material has densities starting at that of water, going to the densities of solid rock and even above, while still remaining liquid.
On a sidenote: No space probe will ever be able to look into that mess. Only gravitational measurments will help.

In recent years, it has been shown that under such conditions typical minerals are thermodynamically not stable, and thus will dissolve into the hydrogen-helium mix.
Thus, the core will not have any well-defined boundary, but it's going to look more like the following sketch (emphasis on sketch!):

which I took from this author's website, which has also a lot of follow-up in-depth research material linked.

In short: Essentially any topographical features can be ruled out. That thing down there will be a liquefied, high-pressure mess.

It is of course correct to say that Jupiter might not even have a core, but to build a planet with 1 Jupiter mass and one Jupiter radius one usually requires a certain amount of high-Z material in the core, which is why the core-idea is very attractive.
Also, going around on conferences this year, it seems that JUNO does find a 5-25 Earth masses core, but there is no publication on this yet.


I can attempt to address the second part of your initial question (*"Is it a particle, a wave,...?") Einstein's theory of general relativity states that mass and energy bend space-time. Space-time, in turn, tells matter how to move (John Wheeler put this more elegantly).

This concept is completely different from the theories of the other three fundamental forces (electromagnetism, and the strong and weak nuclear forces). In these quantum theories, forces are mediated by particles called gauge bosons. Electromagnetism is carried by photons, the strong nuclear force by gluons, and the weak nuclear force by W+, W-, and Z bosons. There have been many attempts to find a quantum theory of gravity - that is, to use quantum principles to construct a field theory of gravity. In these theories, gravity would indeed be mediated by a particle, dubbed the graviton. String theory is one example of these theories; the physics community is divided about it.

You probably have heard other terms tossed around that embody interesting concepts. A gravitational wave is essentially a ripple in space-time emitted by an object or system of objects. There are strict restrictions over what sort of objects can emit these waves; binary neutron stars are one consistently cited example. These waves should not be confused with the aforesaid hypothesized gravitons; while gravitational waves carry energy, they do not "mediate" gravity. They should also not be confused with the unrelated gravity waves.

So, basically, the predominant view in the physics community is that general relativity is the best description of gravity; at the moment, gravity is considered to be the bending of space-time, so it is indeed "something else entirely." Many theories of quantum gravity, including string theory, however, attempt to create particles called gravitons as force-carrying bosons. If evidence is found relating to these theories, then we may well learn whether or not gravitons do exist. One more thing about waves: Because of the quantum concept of wave-particle duality, any particle can be described as a wave, which has a wave function. So if gravity is carried by a particle, then it is also due carried by a wave!


How much higher is the radiation around Jupiter than at Chernobyl, after the explosion?

Let's first quickly estimate whether the numbers by HBO's TV series make sense:
According to gesundheit.ch/strahlung the liquidiators were exposed to $2 \ldots 20 {\rm Sv}$ which compares to a life dose of $\approx 100 {\rm mSv}$ for usual people. I remember from my education that each worker was supposed to spent only ${\rm 40s}$ on the roof of the reactor. Assuming a dose of ${2 \rm Sv}$ in ${\rm 40s}$ corresponds to ${\rm 180Sv/h}$, so the value of ${\rm 12\,000R/h=120Sv/h}$ makes sense.
On the other hand, you give an absorbed dose of $20 \cdot 10^6 {\rm rad}=2 \cdot 10^5{\rm Gy}$ which is a physical quantity. However, the dose equivalent in Sievert quantifies the biological effect (of the deposit of a joule of radiation energy in a kilogram of human tissue), see https://www.ncbi.nlm.nih.gov/books/NBK230653/

${\rm 1 Sv}$ is the amount of radiation necessary to produce the same effect on living tissue as ${\rm 1Gy}$ of high-penetration ${\rm 200 keV}$ x-ray

Only if we incorrectly assume that the composition of the radiation on the roof Chernobyl in April 1986 and on Juno's flight path are comparable, we could proceed with cross-multiplication by setting ${\rm 1 Sv= 1 Gy}$.
But this (wrong) assumption is not enough, we also need a time frame for your Gedankenexperiment: If we assume that the radiation dose would remain constant (aka reactor core being exposed) at the lower bound of ${\rm 120 Sv/h}$, the maths gives $2 \cdot 10^5 \frac{\rm J}{\rm kg}/ (120 \frac{\rm J}{\rm kg \cdot h}) = 1\,666.7 {\rm h}$ which is a bit more than 69 days of survival for the Juno probe on the roof of a burning nuclear reactor. Realistically, I assume the probe would stop working much earlier.
Further reading
You might also find the presentation Charged Particle Fluxes and Radiation Doses in Earth-Jupiter-Europa Spacecraft’s Trajectory by M. V. Podzolko et al. interesting which gives quite a few estimates on radiation dose around Jupiter.


A star could become a planet, but this almost never happens. It would probably have to lose almost all of its mass early in its life, perhaps by a black hole. Then, the exposed core would have to be stopped from becoming degenerate and further mass loss would have to continue. It is possible, but it is not realistic and  will not happen in a human lifetime, even if you observed every star in the universe (including white dwarfs).
A brown dwarf definitely can become a planet. A brown dwarf early in its life acts like a star, fusing deuterium and reaching temperatures of almost 2000 K. But as it ages it cools down and stops fusing deuterium. Then it acts like a planet.
In theory, a brown dwarf directly could become a planetary-mass object through mass loss. This could happen from a giant impact or another body accreting from it.
Contrarily, a planet (brown dwarf) can become a star if it accretes enough mass. For example, the most massive brown dwarfs have masses of about ~80 Jupiters, while the least massive stars have a mass of about ~60 Jupiters (there is an overlap). Therefore, a "planet" could become a star under special circumstances.
As @User123 said in a comment, all stars start as a "planet," without burning hydrogen. While it accretes more material, fusion ignites in its core and it becomes a star.


What do astronomers do different now that they know about fusion of the Sun?

How has knowledge of fusion of the Sun changed astronomy?

TL;DR: It changed everything.
Up until the 1930s, physicists and astronomers argued for a rather young universe while geologists and biologists argued for a markedly older Earth. The problem was that physicists and astronomers couldn't image any known processes that would possibly allow the Earth to be as old as geologists and biologists claimed. Note well: geologists and biologists of that time were not arguing for an Earth that is 4.55 billion years old. They were wanting an Earth that was at least several hundred million years old.
Hans Bethe's 1938 paper, along with a few papers that immediately preceded Bethe's paper, paved the way for addressing that discrepancy. It wasn't until the 1950s when Clair Cameron Patterson showed that the Earth was much older than geologists and biologists thought. The 1950s also were when the apparently ridiculous claims of an expanding universe finally become accepted.


How Does Cosmology Constrain the Number of Neutrino Species?

The original research paper, p.43, section 6.3.2.

I leave it to someone else to find comprehensible words to describe the connection between CMB, redshift and effective neutrino number.


Do magnetic fields affect planetary rings?

(I will assume that these planets orbit a star.) Planetary rings are formed by the gravitational capture of small objects, mostly ice and dust. These particles can be magnetic and in such cases would be affected by the planetary magnetic field. It is also to be pointed out that the magnetic field is not really uniform, but let us assume that it is. The presence of a magnetic field would be important for the sustenance of a ring, but the stability of the ring is an entirely different issue. If the ring does consist of magnetically sensitive materials, then large enough fluctuations in the magnetic field can in principle cause the ring to become unstable. So depending on the answer of stability, I choose to say yes, it does depend on the magnetic field. To calculate the stability would be a complex n-body problem but can be averaged over.


death of a red dwarf star / minimum mass needed for a white dwarf?

Stars that have a mass lower than about $0.5 M_{\odot}$ will not ignite helium in their cores, in an analogous fashion to the way that stars with $M<8M_{\odot}$ have insufficiently massive cores that never reach high enough temperatures to ignite carbon.

The cause in both cases is the onset of electron degeneracy pressure, which is independent of temperature and allows the core to cool at constant pressure and radius. [A normal gas would contract and become hotter as it loses energy!]

The end result for a $0.5M_{\odot}$ star will be a helium white dwarf with a mass (depending on uncertain details of the mass-loss process) of around $0.2M_{\odot}$. Such things do exist in nature now, but only because they have undergone some kind of mass transfer event in a binary system that has accelerated their evolution. The collapse to a degenerate state would be inevitable even for the lowest mass stars (which would of course then be very low-mass white dwarfs). As an inert core contracts it loses heat and cools - a higher density and lower temperate eventually lead to degenerate conditions that allow the core to cool without losing pressure.

The lowest mass stars ($<0.3 M_{\odot}$) do get there via a slightly different route - they are fully convective, so the "core" doesn't exist really, it is always mixed with the envelope. They do not develop into red giants and thus I guess will suffer much less mass loss.

The remnant would be a white dwarf in either case and is fundamentally different from a brown dwarf both in terms of size and structure, because it would be made of helium rather than (mostly) hydrogen. This should have an effect in two ways. For the same mass, the brown dwarf should end up bigger because the number of mass units per electron is smaller (1 vs 2) and also because the effects of a finite temperature are larger in material with fewer mass units per particle - i.e. its outer, non-degenerate layer would be more "puffed up". NB: The brown dwarfs we see today are Jupiter-sized, but are still cooling. They will get a bit smaller and more degenerate.

A simple size calculation could use the approximation of an ideal, cold, degenerate gas. A bit of simple physics using the virial theorem gives you
$$ \left(\frac{R}{R_{\odot}}\right) \simeq 0.013\left(\frac{\mu_e}{2}\right)^{-5/3} \left(\frac{M}{M_{\odot}}\right)^{-1/3},$$
where $\mu_e$ is the number of atomic mass units per electron.
Putting in appropriate numbers I get $0.32\ R_{Jup}$ for a $0.07M_{\odot}$ Helium white dwarf versus $1.01\ R_{Jup}$ for a $0.07M_{\odot}$ completely degenerate Hydrogen brown dwarf (in practice it would be a bit smaller because it isn't all hydrogen).

However, it would be interesting to see some realistic calculations of what happens to a $0.07M_{\odot}$ brown dwarf versus a $0.08M_{\odot}$ star in a trillion years or so. I will update the answer if I come across such a study.

EDIT: I knew I'd seen something on this. Check out Laughlin et al. (1997), which studies the long-term evolution of very low-mass stars. Low-mass stars do not pass through a red giant phase, remain fully convective and can thus convert almost all their hydrogen into helium over the course of $10^{13}$ years and end up cooling as degenerate He white dwarfs.


What is the actual black hole merger speed?

It varies per object, and throughout the merger (and doesn't neatly increase linearly).
An object with a lot mass can not come close to the speed of light, that would require almost infinite energy. Extremely tiny objects, particles, can come close to the speed of light but not actually reach it. Light can go the speed of light, only in a vacuum. Gravity always travels at the speed of light, unlike everything else which goes slower.
The first animation shows what happens when two large objects approach each other, miss, and then don't slingshot away from each other - they become trapped in an extreme elliptical orbit.
The second animation shows what happens when the orbits are relatively close, > ~0.01 parsecs - instead of an elliptical orbit they have a somewhat circular-like orbit, around the barycenter.
There are many different trajectories and resulting orbits, the binaries are also affected by: inflow, outflow, and other objects entering and leaving the area of their orbits. No one answer, no two are alike.
The paper "Observation of Gravitational Waves from a Binary Black Hole Merger" (Feb 11 2016), by B. P. Abbott et al. offers one study of one merger, they provide this illustration:

Top: Estimated gravitational-wave strain amplitude from GW150914 projected onto H1. This shows the full bandwidth of the waveforms, without the filtering used for Fig. 1. The inset images show numerical relativity models of the black hole horizons as the black holes coalesce. Bottom: The Keplerian effective black hole separation in units of Schwarzschild radii ($R_S=2GM/c^2$) and the effective relative velocity given by the post-Newtonian parameter $v/c=(GMπf/c^3)^{1/3}$, where $f$ is the gravitational-wave frequency calculated with numerical relativity and $M$ is the total mass (value from Table 1).

In-between those distances occurs a situation referred to as the final parsec problem.

Final parsec problem
When two galaxies collide, the supermassive black holes at their centers do not hit head-on, but would shoot past each other if some mechanism did not bring them together. The most important mechanism is dynamical friction, which brings the black holes to within a few parsecs of each other. At this distance, they form a bound, binary system. The binary system must lose orbital energy somehow, for the black holes to orbit more closely or merge.
Initially, the explanation is easy. The black holes transfer energy to gas and stars between them, ejecting matter at high speed via a gravitational slingshot and thereby losing energy. However, the volume of space subject to this effect shrinks as the orbits do, and when the black holes reach a separation of about one parsec, there is so little matter left between them that it would take billions of years to orbit closely enough to merge - more than the age of the universe. Gravitational waves can be a significant contributor, but not until the separation shrinks to a much smaller value, roughly 0.01–0.001 parsec.
Nonetheless, supermassive black holes appear to have merged, and what appears to be a pair in this intermediate range has been observed, in PKS 1302-102. The question of how this happens is the "final parsec problem".
A number of solutions to the final parsec problem have been proposed. Most involve the interaction of the massive binary with surrounding matter, either stars or gas, which can extract energy from the binary and cause it to shrink. For instance, if enough stars pass close by to the orbiting pair, their gravitational ejection can bring the two black holes together much quicker than would otherwise be the case.

See the "final parsec problem" link for the whole webpage.

"... the gravitational time dilation ..."

Means: In the black hole's timeframe time moves forward normally for both the black hole and the observer in their own frames. In each other's frames the outside observer see the black hole's time seemingly come to a standstill while the black hole observer sees the outside observer's time pass very rapidly.


Could a planet form without a molten state?

If you accrete a mass of $M$ the total binding energy will be $3GM^2/5R$, which will at least partially turn into heat. If the melting temperature is $T$ and specific heat capacity $C$ we get the condition $$\frac{3GM^2}{5R}=CTM,$$ or $M/R<5CT/3G$. So if we use a low silicate rock melting point of 600 C and basalt thermal capacity then $M/R<1.8313\cdot 10^{16}$ kg/m, while the upper range 1200 C gives $3.0899\cdot 10^{16}$ kg/m. So if we consider a basalt planetoid it will start to have some melt when cohering into one ball of radius larger than 1000 km. So, not much of a planet.


Quadrantids Meteor Shower 2020 viewing times

Quick answer because the shower is tonight.

For any given meteor shower in a given year (e.g. Quadrantids, now) where can I see a rough prediction of how many days it will last?

The PDF linked on the page linked in the answer there says the shower is unusually short (FWHM  of 4 hours) and the maximum is at 08:00 UTC which is 4 PM where I am and must be similar there, so its already getting weaker by nightfall, so you should start watching BEFORE the radiant rises because while their trails all appear to "radiate" away from the radiant, they go in all directions. I'd say start checking around midnight looking in the direction where the radiant will rise later, which I suppose must be at least roughly EAST.

Try to find a viewing location with an unobstructed and dark eastern sky.

Keep watching all night of course, as the radiant rises more could be seen  geometrically even though the number may be decreasing for other reasons, and after all that's only a prediction and not guaranteed.

At least it's not as short as this one! What makes some meteor showers continue for days, while the "Unicorn shower" can be shorter than one hour?


What happens if I were to add angular momentum to a maximally spinning black hole?

How would you add angular momentum?

The only way to do that is to drop something with angular momentum (relative to the centre of mass of the BH and the thing, if you want to get picky) into the BH. Could be a spinning object, or an object fired in at a oblique angle, or circularly polarised light or even suitably polarised gravity waves.

Whatever you do, you increase both the mass and the angular momentum of the BH. Now the limit on the spin of a black hole is expressed in terms of a relation between these two quantities and the Schwarzschild radius of the black hole (which is determined by the mass) and it turns out that whatever you add, and however you add it (if you're firing something in obliquely, it has to actually hit the BH so it can't be too ar from the centre of mass) you cannot increase the angular momentum enough relative to the increase in mass to break that limit.

The equations you have to look at are on wikipedia see especially this section. One way of expressing the limit is given there as
$GM^2 < Jc$  where $J$ is the angular momentum. Suppose you have a black hole just under this limit and add an object of mass $m$ angular momentum just under $Gm^2/c$ velocoty just under $c$ and on a trajectory that just grazes the Schwarzchild radius of the hole at $\frac{2GM}{c^2}$. The angular moment of the result is just under $$GM^2/c + 2GMmc/c^2 + Gm^2/c$$ where the terms are respectively the original angular moment of the hole, the angular moment from the grazing impact and the angular momentum of the added body. This works out to be (just under) $$G(M+m)^2/c$$, the limit for the (now-enlarged) black hole.


What are "non-Keplerian" orbits? What are some familiar examples in our solar system, and can some still be closed?

What exactly are "non-Keplerian" orbits?

Orbits that don't follow Kepler's laws.
Strictly speaking, all orbits are non-Keplerian. In practice, one can model some orbits as basically Keplerian, but with perturbations. Sun synchronous satellites are one example of orbits that are close to Keplerian, but not quite so. The Earth's equatorial bulge makes satellites fail to orbit in a plane. Sun synchronous satellites take advantage of the precession induced by the Earth's equatorial bulge so that their orbits precess by 360° in one year. Connor Garcia's example of satellites in pseudo-orbits about the Sun-Earth linear Lagrange points form another set of examples.

What are some familiar examples of distinctly non-closed orbits in our solar system?

Every planet, every moon, every asteroid, ...

What are closed orbits?

Bound orbits are orbits that stay bound to some central object. Parabolic and hyperbolic trajectories are not bound. Closed orbits are bound orbits that repeat their path. Any attractive central force law can result in circular orbits. These are trivially closed. There are only two attractive central force laws that can result in closed noncircular orbits: An inverse square law (e.g., Newtonian gravitation), and a linear force law (e.g., a Hookean spring). This is Bertrand's Theorem.

What are some familiar examples of distinctly non-Keplerian orbits in our solar system?

Every planet, every moon, every asteroid, ...

Can some non-Keplerian orbits still be closed?

Other than a Hookean spring, no.


Can the energy transport by radiation occur in the convection region of a star?

Radiative energy transport continues. The point is that the radiative flux, which is proportional to $dT/dr$ can be overtaken when the temperature gradient achieves the adiabatic value and convection starts. Once convection is started, it is very efficient and the majority of energy flux will be transported by convection.
Details
Broadly speaking, radiative flux is ever-present and proportional to $T^3(dT/dr)/\kappa$, where $\kappa$ is the opacity.
Convective flux begins when the temperature gradient exceeds the adiabatic value and is proportional to the difference between these two to the power of $3/2$ and to the temperature, local gravity and other thermodynamic properties of the gas.
In most cases what happens is that convection is very efficient when it turns on. This means that most of the energy flux is transported by convection and the temperature gradient is almost equal to the adiabatic value. Thus you can estimate a fraction of the flux carried by radiation by assuming the adiabatic temperature gradient$^1$.
I think this fraction is small throughout the bulk of most convection zones.
There are however some circumstances where you need to be a bit more careful. These are where the density/temperature/gravity/opacities conspire to stop convection. So at the edges of convection zones there will be more complicated regions where the fraction of radiative flux grows and the temperature gradient can be superadiabatic. An obvious example is the outer parts of cool stars where the convective envelope changes into a photosphere dominated by radiative heat transport.
$^1$ An example:
In the convection zone
In the convection zone in the Sun at $r\sim 0.9R_{\odot}$, with an enclosed mass of $m\sim 1M_{\odot}$, the temperature $T \sim 10^6$ K, density $\rho \sim 20$ kg/m$^3$ and pressure $P \sim 10^{11}$ Pa.
The adiabatic temperature gradient
$$ \left(\frac{dT}{dr}\right)_{\rm ad} = \frac{T}{P}\left(\frac{\gamma-1}{\gamma}\right)\frac{Gm\rho}{r^2} \simeq  3\times 10^{-2}\ {\rm Km}^{-1}\ ,$$
where the ratio of specific heats $\gamma=5/3$.
The radiative heat flux is given by
$$F_{\rm rad} = \frac{16\sigma_{sb} T^3}{3\kappa \rho}\left(\frac{dT}{dr}\right)\ , $$
where $\kappa$ is the opacity. From a plot in Turck-Chieze & Couvidat (2011), I estimate
$\kappa \simeq 30$ m$^2$/kg at this radius.
Plugging the numbers in: $F_{\rm rad} \sim 1.5\times 10^7$ W m$^{-2}$.
This is to be compared with the actual energy flux of $L_\odot/4\pi r^2 = 8\times 10^7$ W m$^{-2}$.
Thus, the convective flux must be about a factor of 5 times larger than the radiative flux.
In the radiative zone
Then to check the methodology:
Consider somewhere well inside the radiative zone at $r=0.2R_\odot$, where $\rho \sim 4\times 10^4$ kg/m$^3$, $T\sim 9\times 10^6$ K, the enclosed mass $m \sim 0.35M_\odot$ and $\kappa \sim 0.2$ m$^2$/kg.
The adiabatic temperature gradient is $\sim 0.1$ K m$^{-1}$ and using this in the radiative flux formula gives $F_{\rm rad} \sim 3\times 10^9$ W m$^{-2}$ compared with the actual flux of $1.6\times 10^9$ W m$^{-2}$.  i.e. In this case the temperature gradient must be about half of the adiabatic gradient, such that all of the flux is carried by radiation.


I would suggest you check out SAOimage DS9, which is one of the most commonly used FITS viewers (probably the most commonly used one) among professional astronomers. It allows you to display images with a variety of different scalings and color maps (and, yes, there is an "Invert Colormap" setting under the "Color" menu).


A recent article examines if exoplanetary systems follow the Titus-Bode rule. They find that among the systems with three or more confirmed planets that they studied, the planetary distances follow a logarithmic law roughly 53% of the time. So it is possible that the Titus-Bode rule is more than a simple coincidence.
There is no solid explanation for this, but from Wikipedia:

it is possible that, given a combination of orbital resonance and shortage of degrees of freedom, any stable planetary system has a high probability of satisfying a Titius–Bode-type relationship.


Conversion between Astronomical Frames, ex. IRCF, FK5, FK4, etc…

I'm not sure this answers your question, but ICRF is a non-precessing
reference frame.

On J2000 (2000-01-01 12:00:00 UTC), the z axis of the ICRF frame was
exactly parallel to a line connecting the Earth's center to the
Earth's north pole.

Since then, the Earth's north pole has precessed, so the ICRF z axis
and the geocenter-north pole line are no longer parallel.

To see this, you can visit http://ssd.jpl.nasa.gov/horizons.cgi and
use these settings:

(this is actually looking from the north pole "down" to the Earth's
center, so we would expect a negative z value).

The interesting part of the results:

$$SOE
2445701.000000000, A.D. 1984-Jan-01 12:00:00.0000,  1.007696415090285E+01, -3.712508526111991E-02, -6.356744326939433E+03, -2.994380889331686E-07, -2.570088696418019E-07, -4.731801778858530E-10,  2.120384334099952E-02,  6.356752314245179E+03,  5.329972950848752E-25,
2451545.000000000, A.D. 2000-Jan-01 12:00:00.0000,  1.601081935638483E-01,  1.746805655485580E-01, -6.356752309828780E+03,  2.230286001476848E-07, -8.078442785945724E-07, -1.658173608778433E-11,  2.120384334099953E-02,  6.356752314245181E+03, -5.549834335071262E-23,
2457389.000000000, A.D. 2016-Jan-01 12:00:00.0000, -9.877267476902714E+00,  3.153970319756332E-01, -6.356744632653835E+03,  1.794887917287106E-07, -5.409368308792473E-07, -3.057333255094828E-10,  2.120384334099952E-02,  6.356752314245179E+03, -5.736383388350969E-23,
$$EOE

It's a little hard to notice because of the scientific notation (and
the embarrassing fact that x and y aren't exactly zero at J2000), but
the x and y values at J2000 (the middle entry) are nearly zero.

The y value is nearly zero for all three entries, but the x value
changes. In 1984, the x value is about +10m, and, in 2016, the x value
is almost -10m, showing the effects of precession.

One benefit of a non-precessing frame like ICRF is that star positions
are relatively constant (excluding proper motion and the solar system's own
motion). For example, the z axis will always point (very roughly) to
Polaris, even after precession has changed the north star to, for
example, Deneb.

By default, NASA now uses ICRF coordinates to indicate where planets
are, so, if you're looking to map planets on a J2000 starchart, you
should use the ICRF coordinates.

The only problem you'll have using ICRF/J2000 coordinates (instead of
J2016 coordinates) is that any calculations you make for
stellar/planetary azimuth and altitude will be off slightly, not
enough to really notice unless you are doing precision work.


Where can I find data on Mercury's orbit to fit a model to?

The available astrometric data for Mercury come from meridian transit timings (right ascension and declination) and radar or spacecraft ranging (distance).
NASA JPL has some
here,
and the French IMCCE has some
here.
To use those, you'll have to work out some geometry including the position of Earth in its own orbit.

For your purpose it may be easier to use numerical ephemerides via
JPL HORIZONS or SPICE,
or IMCCE Miriade or Multisat,
which can output Cartesian coordinates if requested.
Of course these are fit to observational data themselves.
The underlying data files (e.g. DE430, INPOP17a) come with papers detailing sources, methods, uncertainties, etc.


When you travel fast around a black hole, do you experience high speed yourself?

If you jump out an aeroplane, do you experience high speed? Many people would say you do, since your velocity relative to the ground increases.

Note the importance of the phrase "relative to the ground" since to describe speed you always need to say what you are measuring the velocity relative to.

A simple choice is to measure your velocity relative to a distant observer. In this case then, yes as you pass close to the black hole your velocity relative to the distant observer increases. As you approach the event horizon, you velocity relative to the observer will approach the speed of light.

Depending on how close you pass the the event horizon, you will experience significant time dialation, so that more time will pass in the observer's frame than in yours. This means that if you manage to escape the gravitational pull of the black hole, and return to the observer, you will have found that the observer has aged compared to you.