Overhead (galactic North) view of the stars in and around the Vela Constellation with distance

note: I'm about to practice Astronomy without a license, if I've said something wrong please feel free to comment or just edit and correct it, thanks!

Wikipedia's Vela (constellation) lists its brightest stars and links to each star's Wikipedia page. Here are the first five of them:

https://en.wikipedia.org/wiki/Gamma_Velorum
https://en.wikipedia.org/wiki/Delta_Velorum
https://en.wikipedia.org/wiki/Kappa_Velorum
https://en.wikipedia.org/wiki/Lambda_Velorum
https://en.wikipedia.org/wiki/AH_Velorum

On each page at the bottom of the data box on the right side there is a link to the star's page in Symbad. For Gamma Velorum (HD 68243) for example, the page is http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HD+68243

On that page you can see the galactic coordinates (GAL coor) are about 262.806, -07.697 and since you only want a 2D map use the azimuth (262.806) to draw a line from our position in the solar system, which is the origin of galactic coordinates (for more on that see answers to How was the galactic plane established? and also can you translate a GPS coordinate to a Galactic coordinate.

Now how far away is the star from us on that line? Use the parallax value for that. For the Gamma Velorum example it is about 3.556 milliarcseconds. the distance to an object (measured in parsecs) is the reciprocal of the parallax (measured in arcseconds). So the distance here is 1/0.003556 or 281.2 parsecs which is about 917.2 light years.

For your 2D projection on to a flat map, you'll need to foreshorten that distance, so multiply by cos(altitude): $\cos(-07.697)=0.9910$ so you put your dot along the azimuth line at 917.2 times 0.9910 = 908.2 light years.

You'll have to decide what to do if one of the stars is far away from the plane of the galaxy, because the projected distance won't accurately reflect the actual distance.


What is the difference between gas and dust in astronomy?

Yes, metals and other elements and molecules can exist in gaseous form under the right conditions of temperature and pressure. A "gas" is simply one of the fundamental states of matter, as in solid, liquid, or gas (and a few other states outside the scope of this question). But as a gas, these substances exist entirely as either individual atoms, individual elemental molecules, or individual compound molecules of multiple atoms (e.g. carbon dioxide).

Dust, on the other hand, is comprised of tiny particulate matter that has undergone the stronger intermolecular bonds to create substances like ice, silicates, and carbon compounds that float around in varying densities between the stars and between the galaxies. Since these particles are still extremely small (typically a fraction of a micron across), they can appear to be a gas, but these tiny, irregularly-shaped objects still exist individually in a solid or liquid state.


How does a telescope measure parallax angle?

No, the telescope doesn't measure the parallax. A sextant or any other angle measuring device fit on the telescope does.

And, we don't(can't) directly measure the parallax angle. Instead, we just track the position of the star/object throughout the year. A little bit of spherical astronomy math shows us that the path of a star in the celestial sphere defined by a fixed reference through the year is an ellipse around it's mean postion, called the parallactic ellipse

Here is a link which does the rough math involved.

The semi-major axis of this ellipse is equal to the parllax angle P, while the semi-minor axis is equal to P*sin(b), where b is the stars ecliptic latitude.

EXTRA NOTE: These parallactic ellipses are often rotated and parameters modified as the abberational ellipse is superposed on it. In practice, things get nigh complicated.


Angular diameter distance is the reduced circumference of the circle, centered at our location, on which the object was located when it emitted the light (or the reduced area of the sphere if you prefer). If the universe is spatially flat then this is the same as the metric radius of the circle. In general it's related to the radius by the function $S_k$ defined here, where $k$ is the Gaussian curvature at that time.
Transverse comoving distance is the same thing scaled up by $1{+}z$, i.e., the reduced circumference of the circle on which the same object is located at the present cosmological time, if it had no net peculiar motion in the mean time. It's therefore equal to (radial) comoving distance if the universe is spatially flat.


why there is a circle of light surrounding Sun and Moon?

Most probable you have been experiencing a weather phenomena: Ice crystals in the sky reflect or bend the light to form a halo.

So, this depends on the temperatures and moisture content of the atmosphere and cannot be seen every day.


How certain are we about the universe's flatness?

It's entirely possible that our seemingly flat observable universe sits inside a much larger and very curved universe. Since there is, by definition, no way to ever measure anything beyond the observable universe, such possibilities are essentially non-scientific musings.


Julian dates offer one nice feature: All astronomical observations recorded by humankind have a positive timestamp. A key downside of Julian dates is that current timestamps on computers that use 64 bit IEEE Standard for Floating-Point Arithmetic (IEEE 754) have a resolution of 40 microseconds. Almost all computers use the IEEE 754 floating point standard. This lack of precision can be a problem with some modern astronomical observations.
One way around this limitation is to use a pair double precision numbers to represent time, the sum of which logically represent the Julian date. This is the approach used by the Standards of Fundamental Astronomy (SOFA). Another approach is to use an offset.
One example is Modified Julian Date, subtracts 2400000.5 from the Julian date. This slices off the leading 24 which is always present on any recent Julian date (JD 2400000.5 was midnight on 17 November 1858, and JD 2500000.0 will occur at noon on 31 August 2132). The additional half of a day in the MJD offset effectively sets the start of a day at midnight rather than at noon.
Subtracting 2450000.0 is equivalent the slicing off the leading 245 from recent Julian dates. This simple slicing technique is valid from noon 9 October 1995 (inclusive) to noon 24 February 2023 (exclusive). The precision using IEEE 754 doubles at the latter date will be two picoseconds, which is more than precise enough for timestamping any astronomical observation for the past two decades and the next decade.


If a black hole pulls every object in the universe, why there is a light around it?

Well, you're partially correct here. A black hole traps everything inside its event horizon. Outside of the event horizon, matter (and energy) can still escape from its gravitational pull. This is due to the fact that the escape velocity of a black hole beyond its event horizon becomes greater than the speed of light itself.

So, the light you see around a black hole is actually just the matter and energy that is outside the event horizon. In other words, you're seeing the light and matter that is not actually inside the black hole. This "stuff" usually orbits the black hole at large fractions of the speed of light.


Is there a theory / equation showing whether or not two passing bodies will go into orbit around each other?

If there are only two bodies, then they will never enter a mutual orbit. For two objects initially gravitationally unbound, in order to become gravitationally bound you must remove energy from the system. With only two bodies (that don't collide), this does not happen. They will accelerate toward each other, change directions according to how close they get, and then leave each other again with exactly the same total energy and momentum as before, but in general shared in some other ratio (for instance, if a small body encounters a large body, the smaller will gain energy and leave with a larger velocity).

On the other hand, if you have three (or more) bodies, one may get slung out with high velocity, thus extracting energy from the two others, which can then go in orbit. But alas, there's no equation for this; the so-called N-body problem has no analytic solution, and must in general be solved numerically.


How can we use hypervelocity stars to determine the origins of the Universe?

First and foremost, I should clear something up: our knowledge of the Big Bang is incredibly limited, and so we don't even know the origins of the Universe today. General relativity and quantum mechanics both break down as you get closer and closer to $t_0=0$. All we really know is that something happened 13.8 billion years ago. However, if we look at the expanding universe, we can extrapolate that, 13.8 billion years ago, space was much less dilated than it is now.

General relativity suggests that the Universe would be packed into a singularity. This is not a good approach, since GR can't properly explain such conditions (QM is more fit to do so, but that alone can't explain how gravity acted), so we really haven't the slightest clue as to what happened 13.8 billion years ago. That said, we still know when the Big Bang (whatever it may have been) occurred, and that's what future astronomers may be able to find out.

The idea the article brings up is this: by one trillion years from now, the expansion of space will have driven away any other galaxies from our line of view, and CMB radiation will have become too redshifted to be detectable.

However, hypervelocity stars will likely be detected outside of the Milky Way (or rather, Milkomeda). Once these stars get far enough from the Milkomeda's gravitational influence, we can observe their acceleration due to the expansion of space, just as we do for many galaxies today. The 4-page paper the article is based on, Loeb (2011), actually makes this clear in a graph:

As the stars get farther and farther from Milkomeda, the gravity's pull on the stars will weaken and they will be accelerated by the expansion of space. This will really become noticeable once they are about 2.3 megaparsecs from the galaxy. Of course, these stars will not be easily found, but Loeb mentions that future astronomers will likely have advanced their tools and instruments by that time.

Loeb explains that, once we observe their redshifts as their distances increase, we can confirm Hubble's law. After this, we just need to measure the necessary parameters, and finally we can calculate the age of the Universe using the Friedmann equation:
$$t_0 = \frac{1}{H_0} F(\Omega_r,\Omega_m,\Omega_\Lambda,\dots)$$

That's the gist of it. The hypervelocity stars will be the only extragalactic objects detectable by then, and thus we will rely on them to confirm our measurements.

Interestingly, Loeb mentions that observing and analyzing the expansion of the Universe at that time would answer some of the uncertainties we have today:

  The working assumption so far was that the vacuum energy density remains constant and does not decay to a lower energy state over hundreds of e-folding times. Measuring the dynamics of hypervelocity stars at late cosmic
  times provides the added benefit of testing whether the cosmological constant is truely [sic] constant. Indications to the contrary might reveal new physics, such as that considered by evolving dark energy models.

Thus, even though much of the Universe will not be observable by that time, we will ironically have more answers about cosmology than we do today, simply by observing the few extragalactic objects that will be visible.

This is the general explanation of how we can find the time since the Big Bang. Remember that even now, we really have no idea about the origins of the Universe — it's a major unsolved problem in physics. All we know is something happened 13.8 billion years ago. Using hypervelocity stars, future humans would be able to determine the time since the Big Bang as well. Don't get your hopes up on humans surviving for 1 trillion years, though.


Stellar Mass black holes are (generally) black holes which form from collapsed stars. They get their name because the mass of the black hole is in the order of the mass of stars. I don't believe they would fit your definition of "Low Mass," as they are believed to be fairly standard in size.

Primordial black holes are theoretical black holes which were believed to have been created during the early moments of the universe. My understanding is that these would be a subset of quantum black holes, which also seem to be called micro black holes.

See this related question, in which the answer covers types of blackholes.

I'm not quite sure what the lower-end masses of observed black holes are, but in theory we could say that the lowest mass of a stellar black hole is just beyond the upper limit of mass for neutron stars, as above that limit they would be collapse into a stellar black hole. This is known as the TOV limit, and is approximated to be about 1.5-3 solar masses1, so we could say that the smallest stellar black holes could have a mass of about 1.5-3 times that of the Sun.

Note however, that this doesn't mean that the Sun is nearly massive enough to become a black hole. The mass of the resultant stellar black hole is only a small portion of the original mass of the star which created it.

1 I. Bombaci (1996). "The Maximum Mass of a Neutron Star". Astronomy and Astrophysics 305: 871–877.


Will Earth rotate backwards in the future?

No, the Earth will not start to rotate in the opposite direction. Ever.
The reason Earth maintains its direction of rotation is conservation of angular momentum. Just like a moving body resists changes in velocity because it has linear momentum, a rotating body will resist forces that try to change its rotation state. Angular momentum can be moved around in a composite system (consider a figure skater pulling in her arms while spinning), but it doesn't disappear unless there is a lot of friction. Which space lacks.
There is a grain of truth in the confused articles that led to the question: Earth is losing some angular momentum over time. Most of this is due to tidal interactions with the moon. When the moon formed Earth rotated much faster and the moon was close. Mutual gravitation caused tides that made some of the angular momentum transfer from Earth to the moon, slowing the rotation of both (to the extent the Moon now only turns one face towards the Earth, tidal locking) and giving the moon more angular momentum in its orbit, making it move outwards at a very slow rate. There are also tides from the sun (and other planets) but they are too small to matter here.
In reality, the sun will become a red giant in 5 billion years and expand, ending the story of the Earth and moon (ironically in this context, likely by tides dragging in Earth into the sun when it gets close enough to the surface). But if we ignore that, what would happen?
As the Earth's rotation is slowed by lunar tides, eventually it would become tidally locked to the moon in about 50 billion years. The moon would always stay over the same spot, visible only from one hemisphere. But Earth, like the moon today, would also continue to rotate around its axis. It is just that the "day" would be more than a month (about 47 days), corresponding to one full rotation of the Earth-moon system. But the direction would be the same as now, since the moon is moving around the Earth in the same direction as Earth is rotating.
At this point there would still be some loss of angular momentum due to solar tides, so over very long time periods (about 50 billion years more) the Earth and moon would slowly orbit closer and closer. Eventually they would merge into an (again) somewhat fast spinning body (still moving in the same direction!)
In the really long run tidal forces would slow this rotation so the big planet would be tidally locked to the sun. But again, the one-year rotation of the planet would be in the same direction as now.
There is a lot of nonsense claimed on the Internet. Learn the basic laws of physics like conservation of energy and momentum, get used to do rough calculations, and you can usually see through it quickly.


How can I map the zenith for an observer at a given time to equatorial coordinates (as would be measured in J2000)?

Zenith right ascension = sidereal time.

Zenith declination = geographic latitude.


Is the Universe really expanding at an increasing rate?

I don't take this at face value because we should expect more distant
  objects to have higher observed speeds and therefore higher observed
  red-shifts.

That's true.   That was the original Hubble discovery - the farther away things were, the faster they were moving away from us.

  Here's why.  Let's start with a model where the Universe expanded very
  fast early on, but has been slowing down ever since due to gravity, as
  one would normally expect.

Yes - that's what everybody thought following Hubble's discovery.

  Remember that, the farther away a cosmic object is, the farther back
  in the past we are observing it. An object 1,000 light years away, if
  it's light is reaching us now, is being observed in its state that
  existed 1,000 years ago. We are effectively looking through a time
  machine.

This is not lost on Astrophysicists.

  So if we observe a more distant object, we're observing an older state
  of that object. Therefore, we are observing it at a time when the
  Universe was expanding faster than it is now, so it has higher
  red-shifts.

OK, 2 points.   2nd point first.   The red-shift has to do with relative velocity, not speeding up or slowing down.    Something can be more red-shifted and slowing down and something can be less red-shifted and speeding up, especially since the acceleration/deceleration is comparatively slow compared to the relative velocity.

and other point - lets keep in mind, we don't know what a galaxy 3 billion light years away is doing now.  We can guess and we can run models, but we can only see what it's doing 3 billion years ago.

  And isn't that what we observe today? The more distant the galaxy, the
  higher its red-shift? This is not inconsistent with a "normal" model
  where the expansion is slowing down due to gravity.

Yes, the more distant the galaxy the higher it's red-shift.    But no, that's not inconsistent with expansion.  That's what you'd see, expansion or contraction, because red-shift is just relative velocity.

  What am I missing? Why are scientists trying to explain such things
  with weird dark matter and dark energy that otherwise have never been
  detected or found evidence of and aren't needed for any other model,
  and in fact get in the way of our models of physics and quantum
  dynamics?

A lot of these ideas are confusing.   They're confusing to scientists too, especially when they were first discovered - so you're not alone.

Dark matter was observed because galaxies were behaving strangely.   The stars in the outer arms of the galaxy were observed to be moving much too fast and faster than the stars more towards the middle of the galaxy and that made no sense.   The galaxies also weighed too much and the only way to explain this was extra mass in kind of a halo around the galaxy, but this extra mass, also, didn't interact with electromagnetic waves like the mass here on earth does - so they called this extra mass (and there's a lot of it, more than there is regular mass), but since it's invisible, they called it "dark matter" and it's not dark like dirt or coal, it's dark as in - invisible.   It's completely transparent to light, but it has mass and they still don't know what it is.  They have some OK theories, but nothing definite.

Now, dark energy - think about the big bang and all matter flying apart - the galaxies twice as far are moving away twice as fast, BUT, as you said, because of gravity, we should see the galaxies that are twice as far moving away more than twice as fast, cause the nearer the galaxy, the more time it's had to slow down - aha, they thought, if we can compare the speed of the galaxies 4 billion light years away to the speed of the galaxies 2 billion light years away to the speed 1 billion light years - etc, etc and measure it all carefully, we can measure the rate at which gravity is slowing down the universe. - that makes sense right.

And with careful measurement of Type 1A supernovas, which temporarily outshine entire galaxies - with remarkable consistency (what they call a standard candle - a very bright standard candle, but a standard candle all the same) - with that, they thought they could measure the gravitational slow down of expansion - exactly what you're talking about.

The problem was, the measurements told them the opposite of what they expected to find.   The measurements told them that the galaxies 2 billion light years away were traveling slightly more than half as fast as the galaxies 4 billion light years away, and so on.  They checked this, cause it had to be wrong, then they re-checked it, and re-checked again and the only conclusion was, stuff out there is speeding up, not slowing down - cause that's what the telescopes tell us.

Dark energy wasn't a hair-brained scheme that mad scientists thunk up.  It was an observed reality that nobody expected (well, cept just maybe for Einstein and his cosmological constant, but that's another story).

Dark energy's just a name anyway.   They have to call it something, even if they're not sure what it is or how it works.


Bounded Observable Universe and General Relativity

The universe may be bounded (or it might not be) and parts of it are receding faster than the speed of light (in some sense) but these two notions are independent.  It would be possible to have a universe that unbounded and expanding, or bounded and stationary etc.

Similarly it seems that "connected" means connected in the topological sense: the only simultaneously open and closed non-empty subspace of the universe is the whole universe. This is independent of any physical laws such as "information can't travel faster than the speed of light" which may make parts of the manifold inaccessible to us.

The universe could be said to be the disjoint union of the observable and non-observable universe, but these sets are not both open in the topological sense.


What is the apparent magnitude limit for the naked eye?

The apparent magnitude classification was done arbitrarily by Ptolemy. His thought was to set the 20 brightest stars to the first position, the less bright stars to the second position and so on up to the faintest stars which were given the sixth position. After the use of Pogson's law we were able to give to stars not only natural numbers, but also numbers with decimal digits. When we started to use the first telescopes, we realised that the numbers which are greater than 6 have meaning. So the natural limit of naked eye is the apparent magnitude of +6. Here you can find a nice work from the italian Light Pollution Science and Technology Institute which shows as the typical limit of naked eye compared to the light pollution of each area.


The expansion of the universe and how stars and galaxies still collide

The expansion is overridden on a small scale by gravity. Our galaxy is not expanding, and the stars are bound together.  In fact, a whole group of galaxies don't notice the overall expansion.


Conflicting information about the star Delta Pavonis?

To answer a question, we do need the source of information (wikipedia just isn't good enough).

According to the SIMBAD database, delta Pavonis is a G8 subgiant, with a temperature of 5512K and a $\log g$ of 4.23 (in cgs units) (Gray et al. 2006), which looks correct for a subgiant classification. The metallicity is given as $+0.13$ in the same source.

There are a large number of other spectroscopic determinations of the star's parameters listed in the SIMBAD measurements. All seem to agree on the subgiant nature, but for example Bensby et al. (2014) makes the star a little warmer (5635 K) and more metal rich (+0.37).

So what appears to be at odds with this is the age/mass combination that you quote.

The wikipedia entry and information you quote appears to be taking information from Takeda et al. (2007). They estimate the age and mass of the star from the temperature and gravity using a Bayesian fitting method and the YREC stellar evolution models.

If this is really important to you, then you need to read the Takeda et al. paper. They show that the posterior probability distribution of age and mass can have multiple peaks. In the case of delta Pavonis it appears that wikipedia has got things wrong (what a surprise). The age that is quoted is actually for a secondary peak in the posterior probability distribution. The secondary peak in the mass distribution is for $M= 1.101M_{\odot}$. I think that age, mass pair makes much more sense.

They do not provide detailed information/discussion on individual cases. Like you, I cannot see how they arrive at 1 solar mass for a star of any age, unless they have picked up a pre main sequence solution in the HR diagram - for which the age corresponding to 1 solar mass would be very small. However, they claim that their models start at the ZAMS. The fact that they list no primary peak in the age posterior pdf suggest to me that the pdf look like a declining function from zero (the ZAMS), with a secondary peak at the 7Gyr (quoted by wikipedia). This then suggests that this secondary solution is also not a brilliant fit and that the fitting process would really have liked to make this younger than ZAMS - i.e. a PMS star (which it isn't). In turn then, this may suggests that the gravity used by Takeda et al. might be too low.

Takeda et al. used spectroscopically derived parameters of temperature 5590 K, $\log g= 4.31$ and $[M/H]=+0.26$ from Valenti & Fischer (2005). These authors also did their own isochrone fits to these parameters finding $M=1.045 \pm 0.058 M_{\odot}$ and an age of 5.3-6.8 Gyr. They say that the mass at the best-fitting isochronal age is $1.07M_{\odot}$.

I think part of what is puzzling you is that you have to remember that having a metal content that is twice the Sun's does affect the evolutionary tracks considerably. It is not that the evolutionary timescales are massively different, but the effective temperature of the star is changed considerably for a given mass and age.


Do planetary rings have geometric bounds?

The Saturn's Phoebe ring that has an orbital inclination 173° to the ecliptic so it is actually in a retrograde orbit and is tilted 27° to Saturn's inner rings, shows that there clearly isn't any limit to orbital inclination. It is feeding off the Saturn's moon Phoebe (possibly due to micrometeorite impacts) indicating that as long as the planetary ring has a source of its materials with already inclined orbit, the ring will follow it (preserve the angular momentum of its originating body). That of course doesn't exclude polar orbits, but we have yet to see any rings like that. Theoretically though, their orbit is just as stable as any other as long as it doesn't intersect other gravitational disturbances, like Lagrange points, or paths of other celestials that would stop it forming a complete ring with accretion (something that is only partially done by Iapetus intersecting the Phoebe ring). Majority of planetary rings might be made of planet's own materials though, and would then naturally follow its own rotation, more or less (depending on their formation) and again preserving angular momentum.

Phebe ring is also extremely huge, stretching from calculated 59 to 300 Saturn's radii (observed by NASA's Spitzer Space Telescope in the infrared range between 128 and 207 Saturn's radii, more can be read in this Emily Lakdawalla's blog post), attesting that as long as the ring's materials would not reach escape velocity or be gravitationally attracted more by another celestial body (beyond the L1 point) there isn't much of a limit in their size either. In theory, if we for example imagine some rogue brown dwarf that isn't gravitationally bound to any solar system and floats freely in the interstellar medium not close to any stars, the ring size would only be limited by the interstellar medium's own pressure. At least until the brown dwarf in questions roams closer to some stronger gravitational influence and loses its ring, that is.

So the maximum size of the ring would in theory indeed be limited only to the L1 point (where gravitational attractions of two massive bodies cancel each other). As for the minimum distance, that depends on the size of particles that the ring is made of, stretch of planet's atmosphere, and strength of its radiation pressure that would be negating gravitational attraction. So it is difficult to give any fair minimum value. If we take a fast spinning planet (e.g. dwarf planet Haumea) as an example, one that would spin so fast that it would actually lose some of its materials due to it reaching escape velocity at the planet's equator, and if we assume this process can be sustained for long enough period, then there isn't any minimum as the planetary disk would be essentially touching the planet's surface. Eventually, the planet would lose much of its own rotation, similar like a fast spinning figure skater would by stretching her arms, by shifting some of its mass into larger radius, and the planet would cease replenishing disk's materials.


Forms of stellar orbits around the galactic center

I'm not sure what the focus is on Sgr A*? Only the stars that are very close to the Galactic center can be said to be "orbiting Sgr A*", the rest of the stars in the Galaxy orbit in the the entire Galactic potential, to which Sgr A* is a minor contributor.
The stars near Sgr A* have a wide variety of orbital eccentricities. This is shown clearly in the picture below (credit: Cmglee, Wikimedia commons).

Stars in the Galaxy at large also show a variety of orbital eccentricities in their non-Keplerian orbits. The orbits are non-Keplerian because they are not orbiting a point-like source of gravity and the orbits do not form closed shapes.
Many, in fact most, of the stars near the Sun are in the Galactic disc population and have roughly circular orbits, confined close to the plane of the Galaxy. About 1% of nearby stars are part of the old halo population which have a wide dispersion of eccentricities and orbital inclinations.
The Sun is currently in a mildly "elliptical orbit" (it is not a closed ellipse as discussed above and it wobbles above and below the Galactic plane), since its motion is not entirely tangential to the Galactic centre - it has components of 5-10 km/s towards the Galactic centre and upwards out of the Galactic plane, compared with a tangential orbital velocity of around 240 km/s. See https://astronomy.stackexchange.com/a/12311/2531 and How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?


How feasible is it that we could see the Central Black Hole Sgr A* occlude one of it's closely orbiting stars?

I'm sure someone will write a longer/better answer, so I'll keep this short and just mention that when a black hole passes between the observer an a star, we usually say that it becomes gravitationally lensed rather than simply occulted. This can potentially provide much richer data than an occultation but requires more complex analysis.

Optical/Infrared

It also takes heroic efforts to try to image near Sgr A* optically. You need to use infrared wavelengths because of dust and an array of telescopes using adaptive optics, combined interferometrically. See How did they make a video of the center of the galaxy, and what is it exactly that's flashing there?

If there was a dedicated, second Very Large Telescope array pointed at SgrA* all night every night, there is certainly the possibility that a gravitational microlensing event could occur, and an otherwise invisible star could suddenly brighten substantially as it passed directly behind Sgr A*.

Are there candidates?

As far as I know there are no currently known stars closely orbiting Sgr A* that are likely to pass directly behind it. However, I suppose that there could be one so dim that it has not yet been observed, and a lensing event would be the first time it gets noticed, so this is certainly a possibility.

But lensing can brighten any background star, it would not have to be in a close orbit around Sgr A* at all.

Radio

Gravitational microlensing works essentially the same at all electromagnetic wavelengths, so it can be observed in radio as well as it can optically.

However, individual stars tend to radiate most of their energy optically and are not so bright at radio wavelengths. It would be hard to pick out the radio radiation of a background star above the strong emission of heated plasma in SgrA*'s accretion disk.

To this end I have just asked How far have individual stars been seen by radio telescopes?

I think (but I'm not sure) that you'd have to pick a time of extreme quiet (little/no infall) to even try.

above: homemade GIF from ESA video ESOcast 173: First Successful Test of Einstein’s General Relativity Near Supermassive Black Hole


What does the "a" at the end of a spectral type mean?

If present, an a (or b or ab) do not refer to spectral peculiarities but are part of the luminosity class definition explained further up on the page.

Occasionally, letters a and b are applied to luminosity
classes other than supergiants; for example, a giant star
slightly less luminous than typical may be given a
luminosity class of IIIb, while a luminosity class IIIa
indicates a star slightly brighter than a typical giant.[27]


As the front image on the wikipedia-page already indicates, a total solar eclipse is not always total.
Earth's orbit is slightly elliptic, and so is the Moons orbit around the Earth.
Now take the Moon's slight orbital inclination into account and far from all total eclipses are really total.
Unlike stated usually.
In fact wiki states "On average, the Moon appears to be slightly smaller than the Sun as seen from the Earth, so the majority (about 60%) of central eclipses are annular."

On a sidenote: It's important to understand that if the inclination and eccentricity of the orbit would be zero, then we'd have a perfect total solar eclipse on earth every month!
But yeah you can look it up, that this is not the case. There are, after all, not every month news reports from ppl staring into the sun, are there ;)

On other note, we know that the Moon is increasing its distance to earth. This happens because Orbital angular momentum is converted into internal heat for both bodies through tidal forces.
This situation will evolve over time, so yes, it's a coincidence.

And no, I'm not aware of the Moon's distance having any strong implications for life.
Tell me, if this is not clarified enough from the above.


How was the time of equinox measured in ancient times?

The "specified value" would be the latitude of the observer, and would be the average of the maximum and minimum values observed at the solstice.

The altitude of the sun above the horizon is measured when it is due South. This is what "meridian elevation" means. So no clock is required. However this method can at best determine the date of the equinox. Precise observations over several years could be combined to obtain an time of equinox.

The position of the sun above the horizon can be simply determined by measuring the length of a shadow. However Ptolemy had access to astrolabes and quadrants for more accurate measurement of the sun's position.


Going from the second moments of an object to its ellipticities and half-light-radius

You simply cannot obtain any information on the half-light radius from the quadrupole moment only of a galaxy, without knowing the galaxy profile.  Consider for example two (extreme) light profiles:

A galaxy consisting of a uniform disk of radius $R$ – that is, its surface brightness would be something like $I(r) \propto H(R_1 - r)$, where $H$ is the Heaviside function.
A galaxy consisting of a point source + a uniform disk – that is $I(r) \propto \delta(r) + H(R_2 - r)$.

You can easily find a combination of radii $R_1$ and $R_2$ such that the quadrupole moments of the two galaxies are identical.  However, clearly their half-light radii cannot be (for the second galaxy, in particular, the half-light radius vanishes).

From what I can see, the relation you wrote holds for a Gaussian profile, and even then it is not correct: your $r$ is not the half-light radius but the standard deviation of the Gaussian profile.  For different profiles there is no guarantee that your relation will work (in general it will not): in general, it will only give you something proportional to the square of the half-light radius, but the proportionality constant will depend on the specific light profile.


Gravitational field released during supernovae

What you are missing is that the shell theorem, which says that for a spherically symmetric mass distribution (we can count the pre- and post-supernova states as this for the sake of argument), that the gravitational field at some distance $d$ from the mass is the same as if all the mass were concentrated at the centre of the distribution.

Therefore the gravitational field due to the pre-supernova star is unchanged after the supernova at a distance $d$ until some of the mass (or equivalently in General Relativity, energy) as travelled beyond a distance $d$.

After a core-collapse supernova, if the pre-supernova star was say $15 M_{\odot}$, then what might happen is that $1.4 M_{\odot}$ is left as a neutron star remnant, whilst $10^{46}$ J of energy is released - mostly in the form of neutrinos travelling at almost the speed of light and the envelope of the star expanding outwards at tens of thousands of km/s and a kinetic energy of about $10^{44}$ J.

The neutrinos carry an equivalent rest mass of a mere $0.05M_{\odot}$, so if you were on a (indestructable) planet in orbit around the supernova, then yes, after the main neutrino pulse has passed there would be a small decrease in the gravitational force felt towards the centre of the planet's orbit (not instantaneously, the neutrino pulse lasts some tens of seconds) that would result in an outward acceleration and  the orbit widening slightly.

Sometime later (the orbital radius divided by the supernova ejecta speed) the main mass loss from the supernova would pass by and this would result in a drastic decrease in the gravitational force and a drastic widening of the orbit.


Great project! and welcome to Stack Exchange. I'll post a short answer but I think someone can add a more detailed, thorough and insightful answer.
I think that website is not well suited, so I'll answer based on you switching to Horizons. If you like Python then it's more fun to use Skyfield.
If you want apply an equation based on a Kepler orbit model, you'll need to use data where the Sun stays in one place and Mars orbits around it. That would be Heliocentric with the Sun at (0, 0, 0).
That there are three zeros raises the issue of the number of dimensions; proper Kepler orbits are sort-of in 3D i.e. they have an orbital plane that can be tilted to a reference plane, but the orbits are planar. Two problems; your equation assumes 2D flat orbit because of the way $\varphi$ is defined. Ideally you'd like data in the plane of Mars' orbit and you may need to transform NASA/JPL Horizons data into Mars' orbital plane yourself because there are only two main "official" planes, no real planet remains perfectly in a plane.
So what you do depends on how far down the rabbit hole of pretending orbits are planes that you want to go.
Zeroth order approximation
Go to Horizons
Use this tutorial and set it up to match the following:
Current Settings
Ephemeris Type:      VECTORS
Target Body:         Mars [499]
Coordinate Origin:   Sun (body center) [500@10]
Time Span:           Start=2020-10-04, Stop=2020-10-05, Step=1 d
Table Settings:      quantities code=2; output units=KM-S; CSV format=YES
Display/Output:      default (formatted HTML)
             -- OR --
Display/Output:      download/save (plain text file)

Here's a sample line for Mars for today using the Sun as origin (I've truncated some decimal digits). You see right away that Mars is about 201 million km from the Sun, it is also about 4 million km below the J2000.0 ecliptic.
2459126.500, A.D. 2020-Oct-04 00:00:00.00,  2.036231544E+08,  5.355405115E+07, -3.872888712E+06...

From here you can approximate
$$r = \sqrt{x^2 + y^2 + z^2}$$
and
$$\varphi = \arctan2(y, x) - \text{286.502°}$$
Since you are going through all four quadrants it's better to use a computer's arctan2(y, x) or atan2(y, x) with two arguments, not $\arctan(y/x)$ which only works in two quadrants (i.e. 1/7 = -1/-7).
First order approximation
You see right away that Mars is about 201 million km from the Sun, it is also about 4 million km below the J2000.0 ecliptic.
If you want to correct for Mars' orbit's tilt with respect to the ecliptic, you can just find the best plane fit to one Martian year's of data and make your own Mars ecliptic.
But I recommend you do the zeroth order first and see how well or poorly it works, then you can decide if you want to tilt.


What caused this mysterious stellar occultation on July 10, 2017 from something ~100 km away from 486958 Arrokoth?

There were three attempts to measure Arrokoth by occultation, and the June 3rd attempt didn't detect anything.  The July 10th attempt had a tiny blip, that appeared to be in the "wrong place", well away from the location that astrometry had predicted. The July 17th occultation was successful, it determined the shape and location well.
Some thought that the July 10th "blip" might have been evidence of a moon.
It was not, it turned out that there was an error in the software that was used to produce the astronometry (the author of the software describes one as "a case of me failing to read my own documentation, another one a sign error") If those bugs are fixed, then the blip on July 10th is moved to the same place as the blips on July 17th.  That blip is Arrokoth.


How can I make Stellarium match the magnitude of stars visible in London, UK?

Ok go on Stellarium then hover over the left side of the screen afterwards click on the location and select London, England (City Of London is the original walled city), then go and hover over the left side of the screen and select sky and viewing options and in the sky section and select the air pollution level from location database then it will automatically set the air pollution depending on where you are on Earth, cool right?


What is the next planned mission that can discover life on another planet?

The Exomars rover of ESA and Roscosmos would be the obvious answer. To be launched in 2018. It will drill 2 meters deep and as far I know is the only mission since the Vikings in the 1970's, to explicitly be equipped to find biosignatures, signs of life. But the Russians, who will land the rover, have had a very poor Mars mission success rate, ESA cooperates with them since NASA abandoned the project a few years ago, and ESA's only own landing attempt on Mars failed too, Beagle 2. And some biologists think it is more challenging to detect sparse exotic microbial life than what that rover is capable of. It weights about 1/3 of MSL Curiosity, and underground life can maybe be very local.

I want to recommend the blogger Robert Walker who writes at great length and well informed, still interestingly speculative, about possibilities for life on Mars, what one maybe should be looking for.

I should add that since 1960 SETI uses telescopes to pick up evidence of interstellar life which is powerful enough to somehow change its environment to make itself astronomically detectable. That's the other main potential except for probes in the Solar system. But not even SETI people sit up waiting for it any more.


Based on your comments, I think your confusion comes from having seen the classic rugby ball-shaped image of the CMB.
The CMB we observe is not from the whole cosmos, but only from a thin and completely spherical shell centered on us, with a radius $R_\mathrm{CMB}\simeq45.6\,\mathrm{Gly}$ (using a Planck 2015 cosmology). Just as the shell of Earth can be projected onto a rugby-shaped figure using a Mollweide projection, so can the shell of the CMB. Here's a figure from Universe Adventure that can help visualize:

Although we (still) can't see beyond the CMB, the observable Universe is all the way out to redshift $z=\infty$, while the CMB comes from $z\simeq1100$, but the difference is not big; $R_\mathrm{obs.Uni.} \simeq 47\,\mathrm{Gly}$. Thus, the volume of the observable Universe is
$$
V = \frac{4\pi}{3}R_\mathrm{obs.Uni.}^3 = 435,\!000\,\mathrm{Gly}^3.
$$

The total Universe is probably much, much larger, and may in fact easily be infinite.


Does lightning on Venus affect the chemical composition of the clouds?

Observations by many spacecraft that have visited Venus over the last
40 years appear to confirm the presence of lightning storms in the
Venus atmosphere. Recent observations by Venus Express indicate that
lightning frequency and power is similar to that on Earth. While
storms are occurring, energy deposition by lightning into Venus
atmospheric constituents will immediately dissociate molecules into
atoms, ions and plasma from the high temperatures in the lightning
column (>30,000 K) and the associated shock waves and heating, after
which these atom and ion fragments of C,O,S,N,H-containing molecules
will recombine during cooldown to form new sets of molecules. Spark
and discharge experiments suggest that lightning
effects on the main atmospheric molecules CO2, N2, SO2, H2SO4 and H2O
will yield carbon oxides and suboxides (COm, CnOm), sulfur oxides
(SnO, SnOm), oxygen (O2), elemental sulfur (Sn), nitrogen oxides (NO,
N2O, NO2), sulfuric acid clusters (HnSmOx−.aHnSmOx e.g. HSO4−.mH2SO4),
polysulfur oxides, carbon soot and other exotic species. While the
amounts generated in lightning storms would be much less than that
derived from photochemistry, during storms these species can build up
in a small area and so their local concentrations may increase
significantly. [...]

Source: M.L. Delitsky, K.H. Baines, Storms on Venus: Lightning-induced chemistry and predicted products, Planetary and Space Science, Volumes 113–114, 2015,Pages 184-192, 10.1016/j.pss.2014.12.005


Is there sufficient evidence to confirm additional Planet in solar system

This question already has muliple answers on this site...

However : Yes, astronomers DO have enough data to speculate about a possible Ninth Planet.

The orbits of six KBO are correlated, and a possible ninth planet could be the reason for those peculiar orbits. See image below for the computed results of the possible orbit of the ninth planet.:

However, since the Planet has not yet been SEEN or DETECTED, its existence is only supposed. We have nothing to prove it really exists. But if it exists, we know where it is :p


Yes. Stellarium is a very nice package that meets your needs. I use it on my mac and it works great. There are versions for Windows and Linux as well. It is a free package. You can enter your location and see the night sky in any direction. It has zoom capability and you can rotate sky to look in any direction. Their website describes it as a planetarium for your computer. It has a database of over 600,000 stars. The website is here


Powerful-enough telescopes to observe the Lunar Roving Vehicle from Earth?

I don't think so. In order to resolve an object one metre across (as a single pixel) at a distance of 384,400km (the diameter of the Moon's orbit), you would need a telescope mirror about 200 metres in diameter, and to achieve a resolution of 10cm (which you would probably need to form a recognisable image), you would need a telescope mirror with a 2km diameter. Currently the largest Earth based telescope has a 30-metre mirror.


No, it cannot. A black hole is something vastly different from a star. It's vastly different from anything else. It's not a thing, really, but more like a portion of very distorted spacetime. Nothing escapes from it simply because there is no way out - spacetime is distorted in such a way that all trajectories lead to the center.

Now, there is a mechanism where radiation is generated just outside the black hole, sucking energy from its gravitational field and therefore from its mass. This is called Hawking radiation. It does not come from inside the BH, but the way it's created it leeches the BH's energy / mass. In time, this would diminish the mass of the BH.

It turns out, the smaller the BH, the stronger the Hawking radiation. This in turn makes the BH lose weight even faster, so the HR gets even stronger, and so on. It's a vicious cycle. If the black hole is small enough, the process turns into an explosion - HR gets so intense, the whole mass of the black hole is converted into radiation at once. It's not a supernova, not as big as that, but it is a powerful explosion.

The only issue here is - HR is very weak when BHs are big. So regular BHs will take an incredibly long time to slim down, if ever, to the tiny size where HR spirals out of control. Basically this never happens in reality.

TLDR: Not a supernova, just a huge bomb, but not very likely.


What's the proportion of Mars' surface compared to Earth's surface uncovered of water?

The surface area of Mars is 144798500 km2

The land area of Earth is 148940000 km2 (figures from wikipedia)

Thus, the land area of Earth is slightly (about 3%) larger than the land area of Mars. However much of the Earth's surface is suitable for human habitation, but none of Mars is suitable for humans.


Of the order 10000–100000 years.  As an example of an old Supernova remnant, you can see the Vela SNR, estimated at 11,000–12,300 years old.
Nasa describes the evolution of a remnant
Initially, the shock wave expands freely, As it grows it sweeps up material from the interstellar medium which slows its expansion. Expansion causes adiabatic cooling (or the Sedov-Taylor phase) until the point that electrons start to recombine with nuclei. This then allows for more effective thermal emission of UV radiation, which cools the remnant further, and it "disperses into the surrounding medium over the course of the next 10000 years."
A paper (mentioned by Rob in comments) describes this in greater detail, and offers this graph of the radio luminosity of a supernova remnant.

 Illustration of a model radio light curve for Type Ia and CC
SNR, evolving through the ejecta-dominated and Sedov-Taylor phases. The
dashed line marks the transition to the radiative phase, where we assume the
synchrotron emission becomes inefficient,


The American Association of Variable Star Observers tracks observations of a number of variable stars, including Betelgeuse. There are typically many such observations each day, primarily at visible wavelengths. It's quite easy to generate a light curve in the band(s) and date range of your choice.

I'm also aware of a Twitter bot that tweets light curves from AAVSO data several times a day, including long-term projections, but I assume you'd be more interested in looking at the individual measurements yourself.


What other “planetary formation mechanisms” exist beyond “gravitational collapse”?

"Gravitational collapse" is not the generally used name of any planet formation model. Gravitational disk instability may be what is meant.

In the context of the answer you refer to, the other, and probably dominant planet formation is known as the core accretion model.

Of course gravity plays a crucial role here too, but the distinction between this and direct collapse of an instability in a gas disc (what is perhaps being referred to as gravitational collapse) is that the major phase of gas accretion takes place after the initial formation of a rocky/icy core. This may account for the correlation between planet frequency and host star metallicity.


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


How clearly would I be able to see a galaxy with the naked eye if viewed from a "close" distance?

The surface brightness of an extended source in independent of distance. So for the surface brightness of a large spiral go look at M31 (The Great Galaxy in Andromeda) . What will change with distance is the apparent size, but the surface brightness will remain the same.

The surface brightness of a near edge-on galaxy will be greater than one face-on due to the greater number of stars close to any sight-line, and the integrated brightness will of course vary with distance.


Are there any good images of Sirius B at apastron in the Sirius system?

HST took pictures close to periastron in 1997, when the separation of the stars was about 3.7 arcsecs. See Bond et al. (2017). There are no pictures exactly at periastron.

HST took pictures close to apastron in August 2016, when the separation was 10.7 arcsecs (see the same reference). There are no pictures from HST closer to apastron but some may be taken over the next year or so.

Here is a raw HST image from August 2016, taken from the HST archive  (freely available and easy to find).

There are many photographs of the binary taken at around 11 arcseconds separation. Table~3 of the same reference above lists a stack of observations taken between 1970 and 1975 when the separation was around 11 arcseconds. One of these images is shown below; Sirius B is the faint object below and to the right of Sirius A. The other symmetrically spaced blobs are image artefacts.


Is it possible to have a trinary star system with a planet that orbits their barycenter?

This configuration should be possible if the planet is located sufficiently far from the barycentre (i.e. sufficiently far outside the outermost orbit of the triple): close-in orbits will end up being destabilised.
So far there are no confirmed exoplanets orbiting the common barycentre of three (or more) stars: while planets are known in triple star systems, they are usually found orbiting just one of the stars. There might be a triple star containing a circumbinary planet but can't find it right now (there's definitely a quadruple with a circumbinary planet though).
If you're willing to consider candidate planets, and to relax your requirement that the stars have the same masses and that the planet is as close to the barycentre as possible, Phuong et al. (2020) suggest the presence of a protoplanet located outside the dust ring of the system GG Tauri A, at a separation of ~290 au. GG Tauri A is a T Tauri triple system, comprising the primary GG Tau Aa (~0.6 solar masses) separated at ~35 au from a close (~4.5 au) binary GG Tau Ab1/Ab2 (~0.38 and ~0.3 solar masses).
The configuration is as follows (apologies to users of assistive technology):
            ┌ GG Tau Aa (~0.6 Msun)
            │
      ┌35 au┤
      │     │      ┌ GG Tau Ab1 (~0.38 Msun)
      │     └4.5 au┤
      │            └ GG Tau Ab2 (~0.3 Msun)
290 au┤
      │
      │
      │
      └ GG Tau Ac (candidate protoplanet)

Masses for Aa, Ab1 and Ab2 from Di Folco et al. (2014)
The protoplanet is inferred from spiral structure observed in the disc. The paper suggests that other spiral structures may be induced by additional protoplanets (GG Tau Ad and Ae) located even further out. As mentioned before, these protoplanets are not currently considered confirmed.
GG Tauri A is itself a member of a quintuple system, the binary GG Tauri B is located at a projected separation of ~1500 au from GG Tau A. GG Tau Ba is a ~0.12 solar mass star, while GG Tau Bb is a ~46 Jupiter mass brown dwarf (White et al. 1999).


Are these real paths of planets as traced from Earth?

From this answer here are views of the inner and outer planets. I used the python package Skyfield which uses the NASA JPL ephemerides to generate positions of all the planets, then I subtracted the Earth's position to generate the geocentric positions.
I agree with your guess that the first image is too perfect.
The paths will not be closed, because there is not a rational number relationship (e.g. 5:9, 2:3, 1:1) between the periods of the planets.
The ones in the image are reminiscent of the Spirograph drawings I used to make a very long time ago.

...plotting using the Python package Skyfield.
The "inner" plotted bodies (from outer to inner) are Mars, (Earth), Moon, Venus, Mercury, The Sun. The data is for 2000 days.
The "outer" plotted bodies are Uranus, Saturn, Jupiter, Earth, and The Sun. The data is for 2000 months.
Distances are in AU.


Firstly, Big Bang didn't happen at a point in space, away from which we are traveling. Big Bang was the creation of space. This space has been expanding ever since, so that the distances between everything increases, but Big Bang happened right where you are, where the Andromeda galaxy is, where GN-z11 is, and so on.

The Universe has evolved ever since. If we want to know how galaxies look 13.8 billion years (Gyr) after Big Bang, we can just look around in our neighborhood. If we look too far away, we don't see 13.8 Gyr old galaxies, because the light has taken some time to reach us. Thus, if we want to see galaxies that are 12.8 Gyr old, we simply look 1 billion lightyears away; if we want to se 10 Gyr old galaxies, we look (roughly) 3.8 billion lightyears away, and so on.$^\dagger$ In this way, we look back in time.

If we look away far enough, we would in principle be able to look 13.8 Gyr back in time. However, we face a problem in that until the Universe was 380,000 years old, it was opaque to light (why this is so is another story). It wasn't opaque to gravitational waves, however. And since lots of GWs are thought to have originated during the epoch called inflation, which were responsible for the expansion of the Universe, and which took place a fraction of a second after the creation of space, we say that GWs offer the possibility of looking all the way back to Big Bang.

$^\dagger$This is somewhat imprecise, since galaxies weren't created instantly after Big Bang, and since all galaxies weren't created at the same time. But for the sake of the arument, let's pretend they were.


Is it possible for galaxy clusters to interact?

Yes.

Theoretically, structure is expected to form first on small scales (stars and stellar clusters), and later on increasingly larger scales — galaxies, groups, and eventually galaxy clusters (see e.g. Longair 2006). This is confirmed, at least to some extent, observationally. For instance, galaxies have been detected out to a redshift of $z=11.2$ (400 million years after Big Bang; Oesch et al. 2016), while cluster have only been detected out to $z=2.5$ (2.6 billion years after Big Bang; Wang et al. 2016).

Thus, when clusters formed, the Universe had already expanded so much that interaction between them is rather rare.

It does happen, however. One of the most important examples is the Bullet Cluster, which consists of two colliding galaxy clusters. The reason I mention this is that it really beautifully confirmed the existence of dark matter. The image below (from NASA's Astronomy Picture of the Day) shows the two clusters after the collision. The stars and galaxies are so far apart that collisions between galaxies are rare, and collisions between stars virtually never happen. Thus, they have just passed right through each other, as seen in the image. The gas between the galaxies, however, collides, slows down, and is separated from the galaxies. This heats the gas to millions of degrees, emitting X-rays (seen in red). The blue stuff is a map of the mass distribution, made using gravitational lensing. This mass is clearly separate from the gas, but coincides with the galaxies, and gives a much higher mass than the visible mass, roughly 5.5 times more, which is exactly what is found for the ratio between dark and normal matter using other methods.


Would an object shot from earth fall into the sun?

Assume that a spacecraft is instantaneously accelerated at the Earth's surface (disregarding the atmosphere for simplicity). We'll consider this from the Sun's reference frame; in other words, the Sun is stationary and the Earth is moving around it.

The spacecraft is accelerated to a velocity which is precisely equal and opposite to the orbital velocity of the Earth around the Sun, making it completely stationary in the instant after the acceleration.

What happens next? Well, we can consider the forces acting on the spacecraft:

The Earth's gravity causes a force in the direction of Earth.
The Sun's gravity causes a force in the direction of the Sun.

The stationary spacecraft is therefore going to accelerate towards the Earth and towards the Sun. Since the Earth is moving away quickly on its orbital path, the gravitational force is not enough to pull the spacecraft back into an Earth orbit; however, it will nudge the spacecraft into an elliptical orbit.

To demonstrate the situation, I have created a small simulation which can be viewed in a desktop browser. Click here to try the simulation. (You can click "View this program" to check the code, and refresh the page to restart the simulation.)

The simulation is physically accurate (ignoring the effects of other planets), but the spheres have been enlarged for easy interpretation. The Earth is represented as green, while the Sun is orange and the spacecraft is white. Note that, while the spheres representing the spacecraft and Sun intersect, the distance between the two physical objects is always larger than 3.35 solar radii.

This screenshot shows how the spacecraft has been pulled into an elliptical orbit by the Earth:

Finally, we could consider a more realistic scenario where the spacecraft is accelerated until it reaches zero velocity (again, in the Sun's reference frame) at a certain distance from the Earth. At the instant it reaches zero velocity, the engine is stopped.

In this case, the result is essentially the same: there are still forces exerted by the Earth and the Sun, so an elliptical orbit will result. The further the rocket is from the Earth when it reaches zero velocity, the more elliptical the orbit. If the Earth is so far away that its gravity is negligible, the spacecraft will fall directly towards the Sun.


My email to Paul Schlyter and his reply, posted with permission:

  On http://www.stjarnhimlen.se/comp/tutorial.html you note that
  Mercury's ascending node is given as:

  N =  48.3313_deg + 3.24587E-5_deg   * d

  According to https://ssd.jpl.nasa.gov/txt/p_elem_t1.txt Mercury's
  ascending node precesses in the negative direction.

  As Where to get rate of change for calculating ephemeris from JPL Horizons notes, NASA's
  value matches the negative of your value almost exactly.

  Are you using a different sign convention here, or is this an error?

Reply:

  It's neither - instead I use a somewhat different reference system.

  I use the "epoch of the day" instead of the fixed epoch of, say,
  J2000.0. So add the rate of precession to the negative rate of change
  you found at NASA's site, and you should get my rate of change.

  Since I use the "epoch of the day", the positions I get are suitable
  to compute e.g. the rise, transit, and set times for the planets. Of
  course they are also good for computing the positions of several
  planets relative to one another. But if you need the positions to plot
  on a star map drawn for some fixed epoch, you need to apply a
  correction for precession to these positions.

  Yes, it can be confusing to distinguish those different astronomical
  coordinate systems from one another. But that's a consequence of
  living on the planet Earth which orbits the Sun, rotates, and wobbles.


Were there any images of a transitary event of Jupiter or Saturn as imaged by a deep probe mission

Cassini is supposed to have used its VIMS spectrometer to observe a Venus transit from Saturn in December 2012. NASA said it was a first. They were more interested in an absorption spectrum than an image.

Celestia can model such events.
Here is a video of a simulated grazing Jupiter transit from Saturn.
There are similar videos of Saturn transits from Uranus and from Neptune.

Transits are rare events because two planets must be near the intersection of their orbital planes at the same time.
A numerical integration study by John Walker found only 1413 transits of Jupiter and 201 of Saturn from outer planets in 250000 years.
A similar, smaller study could determine whether a given spacecraft could have caught a transit on its way through the interplanetary void, but I would not expect a positive result.


How do we find the exact temperature of a star?

This question is very broad - there are very many techniques for estimating temperatures, so I will stick to a few principles and examples. When we talk about measuring the temperature of a star, the only stars we can actually resolve and measure are in the local universe; they do not have appreciable redshifts and so this is rarely of any concern. Stars do of course have line of sight velocities which give their spectrum a redshift (or blueshift). It is a reasonably simple procedure to correct for the line of sight velocity of a star, because the redshift (or blueshift) applies to all wavelengths equally and we can simply shift the wavelength axis to account for this. i.e. We put the star back into the rest-frame before analysing its spectrum.

Gerald has talked about the blackbody spectrum - indeed the wavelength of the peak of a blackbody spectrum is inversely dependent of temperature through Wien's law. This method could be used to estimate the temperatures of objects that have spectra which closely approximate blackbodies and for which flux-calibrated spectra are available that properly sample the peak. Both of these conditions are hard to satisfy in practice: stars are in general not blackbodies, though their effective temperatures - which is usually what is quoted, are defined as the temperature of a blackbody with the same radius and luminosity of the star.

The effective temperature of a star is most accurately measured by (i) estimating the total flux of light from the star; (ii) getting an accurate distance from a parallax; (iii) combining these to give the luminosity; (iv) measuring the radius of the star using interferometry; (v) this gives the effective temperature from Stefan's law:
$$ L = 4\pi R^2 \sigma T_{eff}^4,$$
where $\sigma$ is the Stefan-Boltzmann constant. Unfortunately the limiting factor here is that it is difficult to measure the radii of all but the largest or nearest stars. So measurements exist for a few giants and a few dozen nearby main sequence stars; but these are the fundamental calibrators against which other techniques are compared and calibrated.

A second major secondary technique is a detailed analysis of the spectrum of a star. To understand how this works we need to realise that (i) atoms/ions have different energy levels; (ii) the way that these levels are populated depends on temperature (higher levels are occupied at higher temperatures); (iii) transitions between levels can result in the emission or absorption of light at a particular wavelength that depends on the energy difference between the levels.

To use these properties we construct a model of the atmosphere of a star. In general a star is hotter on the inside and cooler on the outside. The radiation coming out from the centre of the star is absorbed by the cooler, overlying layers, but this happens preferentially at the wavelengths corresponding to energy level differences in the atoms that are absorbing the radiation. This produces absorption lines in the spectrum. A spectrum analysis consists of measuring the strengths of these absorption lines for many different chemical elements and different wavelengths. The strength of an absorption line depends primarily on (i) the temperature of the star and (ii) the amount of a particular chemical element, but also on several other parameters (gravity, turbulence, atmospheric structure). By measuring lots of lines you isolate these dependencies and emerge with a solution for the temperature of the star - often with a precision as good as +/-50 Kelvin.

Where you don't have a good spectrum, the next best solution is to use the colour of the star to estimate its temperature. This works because hot stars are blue and cool stars are red. The colour-temperature relationship is calibrated using the measured colours of the fundamental calibrator stars. Typical accuracies of this method are +/- 100-200 K (poorer for cooler stars).


Wikipedia claims that Saturn's rings may be as old as Saturn itself; NASA says that they could be only a few hundred million years old.

The 4.something-billion-years-old fits in with one of the two theories of the rings' formation - that Saturn swept up material from the protoplanetary disk. The date makes sense, because there's no way the disk would last for billions of years.

The newer date does fit in with the moon-torn-apart hypothesis, which is perhaps the dominant hypothesis. The hypothetical progenitor moon could have been torn apart at any point between Saturn's birth and the present day. "A few hundred million years ago" fits in reasonably well.

I should add that all the theories surrounding the rings' formation have reasonable evidence in their favor, though there's quite a lot for the 4.something-billion-years theory (see here, here and here). Thank you, Cassini!


Why are retrograde orbits more stable than prograde ones?

In any system of orbiting bodies, the length of time two bodies spend in proximity determines the mutual gravitational perturbation of each body on the other.

If both bodies are orbiting in the same direction, the inner body (closer to the central mass [primary star, for instance]) will overtake the outer one, because it is orbiting at a faster angular velocity. Because both bodies are orbiting in the same direction, they will have a great deal of time close to one another, and more time mutually to disturb one another. This is especially pronounced when the two orbits are adjacent and close.

On the other hand, if two bodies are orbiting in opposite directions, they spend very little time in close proximity, so gravitational perturbation is minimized. Passing one another happens very quickly.

By analogy, compare the amount of time two vehicles spend close to one another on a highway when one is passing the other and going in the same direction. A passenger in one could have time to study the other vehicle in great detail. Then, compare two vehicles moving in opposite directions; any passenger could only gain only a brief glimpse of the other vehicle up close. The amount of time for a passenger to study the other vehicle is analogous to the amount of gravitational disturbance one orbiting body will have on another.

A similar problem occurs when the orbital periods of two bodies are simple ratios of one another (orbital resonance). The repeated tugging of both bodies on one another at the same place in their orbits creates an additive distortion at that location in their orbits. This problem would occur in both situations -- orbiting in the same direction and opposing directions.

Though the reasoning is simple, the exact amount of perturbation is quite complex. A "simple" 3-body problem requires the use of advanced mathematics (celestial mechanics).


Yes, there should be dark matter within the Earth, but at very low densities - a few $10^{-22}$ kg/m$^3$ (Bovy & Tremaine 2012), something like one hundredth of the density of the interplanetary medium. The whole of the Earth would contain few hundred grammes!

Of course, this density averaged over a sphere a lot larger than the Galaxy adds up to a lot of mass, but unlike luminous matter, dark matter should not be particularly concentrated in the plane of the Galaxy (and observations of the dynamics of local stars confirm this).

Neither do we expect dark matter to be especially concentrated within the Earth. Although it does interact gravitationally, the kinetic energy it receives from falling in to the Earth's gravitational potential will be exactly sufficient for it to escape again, unless it suffers some inelastic interaction inside the Earth. Some researchers do think that it may be possible that weak interactions may trap dark matter in the core of the Sun (e.g. Vincent et al. 2015) or even more likely, in neutron stars (Guver et al. 2012). But the Earth is not large or dense enough for this to be a likely scenario.


In rocky planets,does fast rotation cause flatting or low flatting imply slow rotation?

The flattening of a planet is a function of both its spin rate and its structure. But for a series of planets of homologous structure the flattening depends on the spin rate. the faster it spins the greater the flattening (for spin rates typical of planets any way). So if a planet is spinning slowly it will display little polar flattening, similarly if it does not show flattening it should be spinning slowly.


Why is there free magnesium in the ecliptic (if there is) and why would some be orbiting within the ecliptic plane but retrograde?

Mg I does indeed mean neutral magnesium atoms, but May and his collaborators weren't observing magnesium atoms orbiting in the inner Solar System -- they were seeing absorption by magnesium atoms in the Sun's atmosphere.

In the optical and near-infrared, the zodiacal light is sunlight scattered/reflected by interplanetary dust particles. Not surprisingly, the spectrum is largely the same as that seen coming directly from the Sun, complete with all the Fraunhofer absorption lines. Their approach was to focus on a narrow region of the solar spectrum containing a strong Mg I absorption line, and measure how the the dust particles redshifted or blueshifted the reflected solar spectrum, since that would tell them the velocities of the dust particles relative to the Earth.

(They also sometimes detected Mg emission partly filling in the absorption line; since this was always at the rest wavelength of that transition and thus not Doppler shifted, they deduced that it was probably emission from Mg atoms in the Earth's upper atmosphere, likely deposited by vaporized meteoroids.)

As for possible retrograde motions: the zodiacal dust is thought to come from three main sources: occasional collisions between asteroids, sublimation from comets passing through the inner Solar System, and a possible contribution from interstellar dust grains as the Sun moves relative to the local interstellar medium. The second source -- comets -- can give you retrograde orbits because many long-period comets (coming originally from the Oort Cloud) have retrograde orbits.


What is a singularity? What is at the center of a black hole? Specifically regarding space-time

This is more of a question for the Physics stack, but I'll give it a shot, since it's fairly basic.

You need to understand something before we begin. The theoretical framework we have to gauge and answer this sort of thing is called General Relativity, which was proposed by Einstein in 1915. It describes things such as gravity, black holes, or just about any phenomena where large densities of mass or energy are involved.

There's another chapter in Physics called Quantum Mechanics. This describes, usually, what happens at very small scales - things that are super-tiny.

Both GR and QM are fine in their own way. Both are tested against reality and work very well. But they are not compatible with each other. Meaning: you cannot describe a phenomenon from a GR and a QM perspective, both at once. Or meaning: we don't have a coherent set of equations that we could write down, and then "extract" out of them either a GR-like view of reality, or a QM-like view.

The problem is, the center of a black hole is both very high mass density and very high gravity (and therefore right in the field of GR), and very small (and therefore "quantum-like"). To properly deal with it, we'd have to reconcile GR and QM and work with both at once. This is not possible with current physics.

We pretty much have to stick to GR only for now, when talking about black holes. This basically means that anything we say about the center of a black hole is probably incomplete, and subject to further revision.

  A star dies, collapses into a black hole, what is at the center? The
  star's mass compacted into the size of the plank length or something
  similarly small? Is there really nothing at the center of a black
  hole?, surely the core collapsed into something, just really small
  right?

According to General Relativity, it collapses all the way down to nothing. Not just "very small", but smaller and smaller until it's exactly zero in size. Density becomes infinite.

You can't say "Plank length" because, remember, we can't combine GR and QM, we just don't know how. All we have here is GR, and GR says it goes all the way down.

It is quite possible that the singularity is not physical, but just mathematical - in other words, whatever is at the center is not actually zero size. Quantum mechanics in particular would be offended by zero-size things. But we can't say for sure because our knowledge here is incomplete.

I'm using words such as "size" (which implies space) and "becomes" (which implies time). But both space and time in the context of a black hole are very seriously warped. The "becoming" of a black hole all the way down to the zero-size dot is a reality only for the unlucky observer that gets caught in it. But for a distant, external observer, this process is slowed and extended all the way to plus infinity (it's only complete after an infinitely long time). Both observers are correct, BTW.

EDIT:

So, when we are saying "density is infinite and size is zero at the singularity", this language applies to the unfortunate observer being dragged down in the middle of the initial collapse of the star.

But from the perspective of the distant observer, a black hole is still a chunk of mass (the original star) in a non-zero volume (the event horizon of the BH). To this observer, the density of that object is finite, and its size is definitely not zero. From this perspective, anything falling into the BH never quite finishes falling, but just slows down more and more.

Both observers are correct. So, keep in mind, when I talk about "infinite density", that's the inside observer point of view.

  What is a singularity? Is it just the warping of space time that makes
  it this way?

You get a singularity whenever there's a division by zero in the equations, or when the equations misbehave somehow at that point. There are many different kinds of singularities in science.

http://en.wikipedia.org/wiki/Mathematical_singularity

In the context of a black hole, the center is said to be a gravitational singularity, because density and gravity are suggested to become infinite, according to the GR equations.

GR says: when you have a lump of matter that's big enough, it starts to collapse into itself so hard, there's nothing to stop it. It keeps falling and falling into itself, with no limit whatsoever. Extrapolate this process, and it's easy to see that the size of it tends to zero, and density tends to an infinite value.

EDIT:

Put another way - if density becomes large enough, gravity is so huge, no other force is strong enough to resist it. It just crushes all barriers that matter raises to oppose further crushing. That lump of matter simply crushes itself, its own gravity pulls it together smaller and smaller... and smaller... and so on. According to current theories, there's nothing to stop it (QM might stop it, but we cannot prove it, because we don't have the math). So it just spirals down in a vicious cycle of ever-increasing gravity that increases itself.

Space and time are really pathologic inside the event horizon. If you are already inside, there's no way out. This is not because you can't move out fast enough, but because there's really no way out. No matter which way you turn, you're looking towards the central singularity - in both space and time. There is no conceivable trajectory that you could draw, starting from the inside of the event horizon, that leads outside. All trajectories point at the singularity. All your possible futures, if you're inside the event horizon, end at the central singularity.

So, why the center of a black hole is called a "singularity"? Because all sorts of discontinuities and divisions by zero jump out of the equations, when you push math to the limit, trying to describe the very center of a black hole, within a GR frame.

http://en.wikipedia.org/wiki/Gravitational_singularity

Speaking in general, physicists don't like singularities. In most cases, this is an indication that the mathematical apparatus has broken down, and some other calculations are necessary at that point. Or it might indicate that new physics are taking place there, superseding the old physics.

One last thing: just because we don't have a combined GR/QM theory to fully describe the center of black holes, that doesn't mean a pure GR research in this area is "wrong" or "useless". It doesn't mean one could imagine some arbitrary fantasy taking place inside a black hole.

Astronomers these days are starting to observe cosmic objects that are very much like black holes, and their observed properties are in very close accord with what GR predicts for such things. So research in this field must continue, because it's clearly on the right track, at least in the ways we can verify today in astronomy.


Is every single large solid body orbiting entirely outside of the frost line covered in ice?

No, they aren't all covered in water ice.
As an example, Io, a moon of Jupiter is covered in volcanically deposited sulfur and sulfur dioxide frost.

Solar system mechanisms are pretty complex and varied, so we wouldn't expect similar quantities of water as exist on Earth.  Also, the "frost line" for water in the Solar System isn't the same as the "frost line" for many other compounds like Ethane and Methane.  Saturn's moon Titan has lakes of the stuff:

Instead of water snow, Titan has benzene snow that dissolves as it falls in the lakes!  Delightful...


How can a supernova affect black hole in a binary system?

The likely result would either be a black hole-black hole binary system; a neutron star-black hole binary system, or the black hole and the compact remnant from the second supernova explosion would go their separate ways at reasonably high speeds.

You cannot disrupt a black hole in this way. In fact all that will happen to the original black hole is that it will likely get a bit more massive from accreting some of the supernova ejecta.


I answered this same question at physics.SE. I specifically joined this part of the SE network to address this duplicate question at this site.

The astronomy community faced two crises with regard to what constitutes a "planet", first in the mid 19th century, and more recently at the start of the 21st century. The first crisis involved the asteroids. The second involved trans-Neptunian objects. Both crises challenged astronomers to question what a "planet" was.

1 Ceres, 2 Pallas, 3 Juno, and 4 Vesta were discovered in quick succession during the first decade of the 19th century. There was no international astronomical organization at the time of these discoveries; the International Astronomical Union wouldn't be formed for another century. Instead, the designation of what constituted a "planet" fell on the major astronomical almanacs such as the Berliner Astronomisches Jahrbuch (BAJ). Those discoveries at the start of the 19th century were treated as newly discovered "planets". This situation remained static for about 40 years.

That changed in 1845 with the discovery of 5 Astraea. During the 1850s, the list of objects orbiting the Sun grew to 50, and during the 1860s, the list grew to over 100. The response of the BAJ and others was to demote Ceres, Pallas, Juno, and Vesta from planethood status to some lesser status, either minor planet or asteroid. Astronomers didn't have a clear-cut concept of what constituted a planet other than that they should somehow be large. Ceres, the largest of the bunch, is not very large. The end result of all of these discoveries starting in 1845 was that the first four discovered asteroids were demoted from planethood status.

The second crisis started in 1992 with the discovery of (15760) 1992 QB1. By 2006, the number of trans-Neptunian objects had grown significantly. Were these things "planets", or something else? Some astronomers, notably Alan Stern, wanted the term "planet" to be extremely inclusive. Most astronomers balked at this idea.

Paradoxically, it was Alan Stern himself, along with Harold Levison, who provided the key criterion of "clearing the neighborhood" that lies at the heart of what the IAU deems to constitute a "planet." Their paper, Stern and Levison, "Regarding the criteria for planethood and proposed planetary classification schemes," Highlights of Astronomy 12 (2002): 205-213 suggested splitting "planet" into two categories, "überplanet" (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune) and "unterplanet" (Pluto+Charon, Eris, Ceres, Sedna, and a host of others).

Stern is being quite hypocritical when he rants that there is no clear-cut boundary between "planets" and "dwarf planets." The boundary is huge, and Stern knows this. The ratio of the square of an object's mass to its orbital radius about the Sun is key in determining whether an object can clear most of the junk from the vicinity of the object's orbit. There is a five order of magnitude difference between the smallest of the planets and the largest of the dwarf planets in terms of this ratio. This five order of magnitude difference figures predominantly in that paper by Stern and Levison.

The only difference between the proposal by Stern and Levison versus the voted-upon IAU resolution is that while Stern and Levison wanted to designate hundreds (and perhaps thousands) of objects into subcategories of "planet" ("überplanet" and "interplant"). On the other hand, the IAU chose to designate those objects as the mutually exclusive terms "planets" and "dwarf planets". This is consistent with how astronomers dealt with that first crisis. Planets should be "large." Stern and Levison provided the necessary ammunition to distinguish large from not so large.


Confused about the significance of sidereal time

Let's start with a description of sidereal time. From my knowledge, sidereal time at any location can be described as:
the right ascension currently at the zenith
Here are the definitions of zenith and right ascension, if you need them.
So, if the sidereal time at some location is 12 hours, 0 minutes, and 0 seconds, the right ascension directly above that location at that time will be 12:00:00. So, one sidereal day is the time it takes for the sidereal time at any location to "repeat itself" - i.e., the time between one 00:00:00 and the next 00:00:00. This means that one sidereal day is the time it takes for Earth to rotate 360 degrees.
The time that full rotation takes - i.e., one sidereal day - has been calculated to be roughly 23 hours and 56 minutes, or 1436 minutes. Since you're measuring your time in solar time, you would expect to see this 4-minute discrepancy in a lot of places, because, well, 23 hours and 56 minutes is 4 minutes less than a full solar day of 24 hours. From one day to the next, most objects (the Moon is a notable exception) will rise 4 minutes earlier.
So, yes, your analysis is correct - this daily 4 minute change is not really "significant" since it is a consequence of the Earth's rotation around the Sun, and you should expect to see that.

A side note:
Additionally, we can see that the time it takes Earth to rotate one full degree is:
$\dfrac{1436\: \rm{minutes}}{360 \: \rm{degrees}} = 3.99 \dfrac{\rm{minutes}}{\rm{degree}}$
So, this 4-minute daily change is also a nearly 1-degree daily change.


What is the difference between azimuth angle and hour angle?

Azimuth is measured along the local horizon: 0° is geographic north, 90° is east, and so on.
Lines of equal azimuth converge at the zenith and nadir.
Hour angle is measured along the celestial equator: 0h is the local meridian, 6h is in the western sky, etc.
Lines of equal hour angle converge at the celestial poles.
In other words, the two angles are measured around different axes.
The angle between those axes is the complement of the observer's latitude.


Are satellites orbiting around earth visible to the naked eye?

A lot of satellites are visible under the right conditions. Usually up to 2 hours after sunset and 2 hours before sunrise. This allows the sun to strike the satellite when you are on the dark side. Depending on the orbit, it will take between 1 and 5 minutes to traverse most of the sky. Usually, they will enter the shadow and you lose sight of them.


You are referring to what is known as proper motion, which is the apparent motion on the sky due to the relative motion of the solar system and the stars.

To measure an accurate parallax you have to observe for long enough that you can disentangle the parallax motion, which is of course modulated on a 1 year cycle, from the proper motion of a star, which produces an approximately linear, monotonic drift of the star's position with time.

This is perfectly illustrated in this graphic produced by HST/NASA, which shows the track on the sky taken by Proxima Centauri (a nearby star with a large parallax and a large proper motion). The small, periodic wiggles are caused by the parallax shift, whilst the linear trend in the path is due to proper motion. The two are easily separated if you observe for at least 18 months with several precise position measurements during that time.


Orbital speed is (vector) sum of tangential and normal speed?

"For any object moving through space, the velocity vector is tangent to the trajectory."(Citing https://en.m.wikipedia.org/wiki/Orbital_state_vectors).

Therefore, it's both of these components, there is no normal component1 (relative to the central body)... the orbital speed represents the intensity of the orbital velocity which is always tangent to the ellipse (or parabola/hyperbola) the body in orbit describes.

The Moon's orbital speed being 1 km/s means it moves along an ellipse around the Earth at a speed of 1 km/s (this speed is roughly constant, so that ellipse can sometimes be approximated to a circle).

1However you can separate orbital motion into two components normal to each other (the most popular such separation is to the radial component and one normal to it, but this is unnecessary for circular orbits where there is no radial velocity, only radial acceleration which causes the curvature of the trajectory), and in that case the orbital speed will indeed be the square root of sum of the squares of those two.

$$v^2 = v_{||}^2 + v_{\perp}^2$$

But in two-body Kepler orbits there is no velocity component normal to the plane of the orbit, which is the plane containing the $\mathbf{v_{||}}$ and $\mathbf{v_\perp}$, which is also the plane containing $\mathbf{r}$ and $\mathbf{v}$.


What is the exact time of sun stay in each zodiac every year on sidereal year duration?

A sidereal zodiac is not the same as the IAU constellations but corresponds more closely to them than the Western tropical zodiac does, e.g. the eastern edge of Virgo remains near Spica despite precession.

I assumed equal 30° blocks of ecliptic longitude with the J285.25 equinox as origin (Lahiri ayanamsa).
Your local tradition may differ slightly, but any difference should affect all signs equally and have little effect on Sun residence times.

I computed these statistics for the years 1970-2070 using Skyfield and the DE430 ephemeris:

     mean  stdev   mean
       seconds     days
 1  2667075 ±186  30.8689
 2  2702097 ±174  31.2743
 3  2717490 ±139  31.4524
 4  2708635 ±122  31.3499
 5  2678191 ±139  30.9976
 6  2635085 ±167  30.4987
 7  2591328 ±182  29.9922
 8  2558345 ±168  29.6105
 9  2544241 ±125  29.4472
10  2552351 ±101  29.5411
11  2580755 ±128  29.8698
12  2622569 ±167  30.3538

Gravitational interactions with other planets tend to make any given year slightly longer or shorter than the mean sidereal year of 31558150 seconds.
For the same 100 years as above, I computed a standard deviation of ±320 seconds in sidereal year length.


How did the Ancients understand the ecliptic using the Ptolemaic system

The ecliptic was understood to be the path of the sun relative to the sky.

Ptolemy believed the sky rotated daily around the Earth. All the planets, including the Moon and the Sun, shared in this motion.

However, the planets had secondary motions, relative to the sky. And in particular relative to the stars that they thought were fixed to the sky. This motion was discovered by careful observations of the sky, and was very well known, long before Ptolemy.

The motion of the stars was very simple: therefore when Ptolemy was describing his system, he only concerned himself with the motion of the planets (including the moon and sun) relative to the stars. It was clear that the planets all approximately followed a great circle relative to the stars. This great circle is the ecliptic. The Sun's motion was the simplest. It was exactly on the great circle, and it's position relative to the stars moved almost evenly. The other planets (including the moon) had orbits that were less even, but by introducing a system of Eccentrics, Deferents, Epicycles and Equants, Ptolemy was able to account for nearly all their motions.

The precession of the equinoxes was noticed because the Greeks used a calendar based on the heliacal rising (the date that stars are just visible before sunrise) of certain stars. And it was noticed that the heliacal rising was changing relative to the seasons and the equinox. We understand the reason for this to be the precession of the equinoxes.


Stars, especially larger stars, are hot enough at their core to produce significant amounts of heavier elements: Carbon, Nitrogen and Oxygen. When the stars die, some of these elements are returned to the interstellar medium.

The nebula from which stars form contains not only Hydrogen, but also all the other elements found on Earth. The nebula contains the Carbon that is the basis of life, the Nitrogen that forms the atmosphere, the Silicon that makes rocks, and the oxygen that reacts with hydrogen to form water.

Most of the water on Earth was actually brought to the Earth on asteroids after the Earth had formed, see How did water get on Earth


What is the abbreviation for solar time, hour angle of the sun?

Use SHA, solarHourAngle or some variant thereof, as James K said, if your programming language allows it. Longer variable names can be more cumbersome to type and have more possibility of mis-typing the name, but can save a lot of time by providing clarity of purpose when reviewing code at a later date.


Obtaining stars vector position in relation to the center of the galaxy at a given date

EDIT: It turns out a major problem with galaxy-center-based coordinates is that we don't actually know how far from us the center of the galaxy actually is. According to https://en.wikipedia.org/wiki/Galactic_Center estimates vary between 24,800 and 28,400 light years. You could arbitrarily choose a value, but the results would be off by at least 2,000 light years in some cases.

I was surprised that I couldn't find something like this
online directly. The closest I found is the file
"heasarc_tycho2.tdat.gz" on
http://heasarc.nasa.gov/FTP/heasarc/dbase/dump/

The format is explained on
http://heasarc.nasa.gov/W3Browse/all/tycho2.html and it
includes the galactic latitude and longitude which are
explained here:

https://en.wikipedia.org/wiki/Galactic_coordinate_system

Using this information, you'd still have to compute the
positions yourself, but at least all of the data is there.

Of course, as others have noted, you can take any star
catalog with any set of coordinates and convert it to the
format you need. The catalog I list above just makes it a
little easier, since they're precomputed some of the steps
for you.

As you mention above, SIMBAD also includes galactic
coordinates, but I couldn't find an easy way to download
the entire catalog.


Potential formation mechanisms of the interstellar asteroid 'Oumuamua

If you don't mind speculation, it seems to me that a massive impact - like the one that is thought to have formed our moon, but even larger, could eject shards like that.    That such a shard could escape it's solar-system and fly through space between solar-systems isn't unreasonable.   Asteroids (to my understanding) are somewhat loosely formed and an elongated shape like that is unlikely.

A small moon or larger asteroid near a super-nova might get turned into an elongated shape like that by the blast as another speculative answer.   A 3rd possibility might be a ejection from a high velocity and somewhat violent debris disk, perhaps from a planet that fell inside the Roche limit of a white dwarf star.

It's still an object that's strange for two completely separate reasons.  It being the first ever object observed to be passing through our solar-system and it's thought to be very unusual shape (footnote - better images would be nice).   Still, the intersection of two unrelated firsts scores high in statistical improbability, even if the shape can be explained, it still leaves open the question of why two never before seen events coincided with the first ever observed exo-asteroid.