Has the Earth's wobble around the Earth-Moon barycenter ever been observed by a spacecraft?

Absolutely yes if you can count the wobbling motion of the spacecraft that tracks Earth's wobble about the Earth-Moon barycenter.
DSCOVR sits in a heliocentric orbit which is in a 1:1 resonance with Earth, otherwise known as a "Lissajous orbit associated with Sun-Earth L1".
Its EPIC camera is pointed carefully at Earth all the time in order to keep a 0.45° to 0.53° Earth nicely framed in its only slightly larger 0.61° field of view.
If the Earth-Moon average distance is 385,000 km and they have a mass ratio of 81:1, Earth's wobble should be roughly 4,800 km or about 38% of its diameter. If DSCOVR didn't track this wobble the Earth would appear to move +/- 0.18° in EPIC's field of view such that parts of Earth would disappear outside the frame around the times of first and third quarter of the Moon as seen from Earth.
One can download roughly daily data dumps from EPIC which contain data fields for the pointing vectors of the camera. One can subtract a smoothly varying rotation of $2 \pi$ per year and see both Earth's eccentricity and a monthly wobble due to center-of-mass motion of the Earth. I've looked at EPIC metadata in the past, and while I haven't done this specific analysis this can't not be true!

Why is DSCOVR in a Lissajous orbit? Wouldn't a halo orbit completely avoid the Sun exclusion zone?
Where are post-2017 EPIC images of Earth and the DSCOVR spacecraft coordinates available for download?

From this answer:

above: Image from NASA/NOAA from Bad Astronomy


highspeed black holes or neutron stars on (almost) head-on collision course and kinetic energy

I think it's a fun question, and I can answer some of it.

  If their event horizons don't "touch" I'd assume their courses will
  change considerably but it would be a normal flyby - right?

While the relativistic speeds and time dilation might change things a bit, a flyby is still a flyby. They would either fly past each other, the dense objects would change direction as a result, but the combined kinetic energy and momentum would remain constant.

Second option is they enter into orbit around each other.   Pretty much the same with any two massive objects passing near each other.

  But what happens if their event horizons "touch", if their event
  horizons "overlap" just a few meters?

With classic (non Kerr) black holes, if the event horizons touch the black holes have no option but to merge.  The event horizon is directly proportional to the radius, so when the two event horizons touch it, as that happens, no longer two black holes but one black hole.   The mass of the two objects combined has an event horizon that circles both of them.

For Kerr black holes this question is more complicated and I'm not sure but I think it's the same. Once the event horizons touch it's no longer two separate black holes but one bigger one.

  On the one hand both objects have a massive kinetic energy with
  opposite directional vectors. And on the other hand nothing can escape
  below the gravitational radius.

When you're talking about event horizons, gravity always wins.  Kinetic energy cannot exceed the speed of light, so it loses.

  Where does all the kinetic energy go ? Would the two black holes
  extremely deform, like creating a long string before they collapse
  into spherical form?

A long string doesn't make any sense.  A temporarily non-round event horizon as the holes merge seems possible, but think of a merger of two stretched balls, not a string.   Each black hole is centered around it's "singularity", those shapes might warp some as the holes collide or spiral into each other.  As an FYI, direct collisions would be rare.  Spiraling into each other is much more common.

The kinetic energy gets absorbed inside the black hole.  It can't, by definition of the black hole, escape.   Total momentum is likely preserved.

  And what if in a similar scenario one of the two objects is a neutron
  star. Would the neutron star be ripped apart when it scrapes the black
  hole on high-speed flyby ?

A few points to consider regarding a black hole. Anything that falls within a black hole's photon-sphere is unlikely to ever escape, so one of the magical distances to consider is 1.5 times the radius of the black hole where all orbits are destined to fall into the black hole.   If the black hole and neutron star are of similar size and the neutron star passes close to the event horizon and relativistic speed, then there's an interesting scenario where some of the neutron star (at close to the speed of light) would be outside the photon-sphere and have the momentum to escape and some of it would not.   Neutron stars are bound together very tightly, but the tidal forces would be very strong and it's possible that some of the neutron star material could get torn off in this hypothetical scenario.  Would the entire neutron star be torn apart?  Maybe.  It's hard to say.  The math in that hypothetical is way above my pay-grade.

If the neutron star is much smaller than the black hole, then it's likely inside the photon-sphere and it will spiral into the black hole, perhaps breaking apart in the process into an accretion disk, perhaps not, but it will get absorbed into the black hole if it's entirely within the photon-sphere, even at nearly the speed of light.

The problem with this hypothetical is that it's pretty much impossible.  Objects the size of stars don't accelerate at nearly the speed of light relative to other near by stellar mass objects.  What actually happens is described in the first answer to your question, the two dense objects would either have sufficient relative velocity to fly past each other or they would get caught into a mutual orbit that would likely lead to a merger.  In the actual universe, a close to the event horizon fly-by between a neutron star and a black hole would result in a merger of the two.  In theory, with enough kinetic energy the neutron star could fly close to and still fly past the black hole, but that much relative velocity probably never happens.

People smarter than me should feel free to correct anything I got wrong, but I thought this one was a fun one to answer.


Will JWST be as durable as the hubble telescope?

Hubble was in low earth orbit, and was always intended to be serviceable. In fact, the original plan for Hubble was to have the space shuttle carry it down from orbit and take it back up, but they decided that was too risky compared to servicing in orbit.

JWST, on the other hand, will be at the Earth/Sun L2 Lagrange point, like WMAP and Planck before it. The distance to that point is 0.01 astronomical units, which is about 1.5 million kilometers from Earth's center. The Moon's orbit around the Earth is about $1/4^{\mathrm{th}}$ of that. So, a manned servicing mission would be an unprecedented project (beyond Apollo), and an unmanned mission, also unprecedented, would face automation challenges from the 5 second light travel time (10 second round trip).

Because the L2 Lagrange point is only meta-stable (stable to offsets in two directions, unstable along the Earth-Sun line), JWST will have to expend fuel on station-keeping just to stay there. It needs to stay there because it can only be as cool as it needs to be by using the Earth as a sun-shade - those tennis court sized shades built in are actually to shade JWST from the thermal radiation of Earth & the moon.

Put simply, it would be cheaper to build and launch another JWST than to service the existing one.


What coordinate system is at rest relative to the center of the galaxy?

Galactic coordinates are sun centred with a zero longitude chosen to align with the centre of the galaxy when the measurements were made in 1958.

As defined, the galactic coordinates don't rotate, and so the position of Sag A* is very slowly drifting. Currently Sag A* is about 4 arc-minutes from the defined zero longitude, but decreasing by 5.7 milli-arcseconds per year (a tiny amount)

The task is ill defined, You can convert between different coordinate systems of the sky, but they all must be centred on the solar system.


When was the diameter of Titan first measured?

Discovery of Titan, 1655: Unknown diameter.
Dollfus, 1970: 4,850$\pm$300km (1). Measured by Filar micrometer (2) and diskmeter / double-image micrometer (3).  (Apparently a summary of earlier measurements, currently trying to find print copy)
NASA SP-340, 1974: Summary of above techniques, propose settling on 5,000km diameter until it can be measured by stellar occultation (the same process used to determine the size and shape of Ultima Thule). (4)
Elliot, 1975: Measured the diameter by limb darkening / lunar occultation (i.e. by how Titan passed by Saturn).
Calculated diameter somewhere between 5,132$\pm$47km and 5,832$\pm$53km.  (5)
Pioneer 11, 1979: Only 5,800km found in references, appears to still be using the 1975 calculation. (6)
Voyager 1, 1980: Same as above.
Cassini, 2004: The Cassini probe had radar which was able to "see" right through Titan's atmosphere.  Although the first flyby was on July 2, 2004, it appears that the radar was not used until the second of the 45 planned flybys (7,8).
This paper from 2009 combined data from all of Cassini's passes at that point.  This allowed the authors to build a model of the actual surface over enough of Titan to determine its size (~5150km) and that it is slightly oblate, like the Earth.  In terms of actually measuring the surface, this is probably the best answer.
(1): Surfaces and interiors of planets and satellites, Dollfus, Audouin, 1970.  Page 129 is apparently the specific table.
(2): Lunar Occultation of Saturn. I. The Diameters of Tethys, Dione, Rhea, Titan, and lapetus, Elliot, J.L., 1975.


Why do we need the "mean" in mean free path?

In every context we talk about "mean free path", we talk about particles. Particles usually bounce around a lot, from collision to collision. In between two bounces a particle moves in a straight line. That is the expected behaviour from Newton's laws of motion.

Between one collision and the next, a particle therefore moves a distance. That is the "free path". So why do we need to include "mean"? The thing is, you can never be sure about how far the particle moves after a collision. It might as well immediately bounce into another particle, as well as move a considerable distance. However, if you measure really many particles, their average is going to be pretty stable and predictable. That gives a nice, usable number. The "mean" here is for average. (In case you wonder, yes, this applies to optics too, but the concept requires a little more abstraction).


Can a natural satellite exist in a geostationary orbit?

Yes. Charon is in Pluto synchronous orbit. Pluto and Charon are mutually tide locked.


Would the (orbitting) speed of the Earth change if suddenly the sun disappear and the Earth is thrown into space?

Velocity is a vector quantity with both a magnitude and direction. The magnitude of velocity is speed. During the Earth's nearly circular orbit, the speed changes little, but the direction of motion changes, and this change of direction is the acceleration. The Earth is always accelerating towards the Sun.

When we speak of velocity one should mention the frame of reference: we measure the velocity of the Earth wrt the the Sun. The sun is therefore by definition "at rest" (we should actually use the centre of mass of the Earth-Sun system, an also include the gravitational effect of the other planets, I pass over these details) This frame approximates an inertial frame of reference

Mars doesn't stand still relative to either the sun or the Earth. We can use a frame of reference in which Mars is at rest, but this is then a non-inertial frame, it rotates, so in this frame both the Earth and Sun are moving, and accelerating due to the non-inertal motion of the frame.

If we (magically) remove the sun (but not the frame of reference in which the sun is at rest) then there will be no force on the Earth, and it will move in a straight line at constant velocity, relative to the frame in which the sun was at rest. It would move at the same speed as its orbit, about 100,000 km/h (in that frame of reference)


Why do the Pleiades look clearer when viewed indirectly?

What you describe is a technique called averted vision and takes advantage of the arrangement of cones and rods, two types of light sensitive elements within the eye:

  The retina contains two types of photoreceptors, rods and cones. The
  rods are more numerous, some 120 million, and are more sensitive than
  the cones. However, they are not sensitive to color. The 6 to 7
  million cones provide the eye's color sensitivity and they are much
  more concentrated in the central yellow spot known as the macula. In
  the center of that region is the " fovea centralis ", a 0.3 mm
  diameter rod-free area with very thin, densely packed cones.

Shorter, the cones are sensitive to color and work best in bright light, while the rods are most effective in low light conditions. And since the rods are concentrated off center of retina, our night vision is best off center. This averted vision technique can be used with a naked eye, as well as with any optical equipment like binoculars or telescopes, although it can be a bit of a bother with eyeglasses and contact lenses. Especially with the latter, using this advert vision technique can cause additional irritation of the cornea.

As per "beating the system" part, with a bit of practice, you could train yourself to view dim objects better with the corner of your eyes at will, and not merely by chance or for only short periods of time. It might seem hard with first few tries, but it really isn't and you soon get used to not looking directly towards the observed object, countering what we tend to do naturally, direct our eyes towards the object of our observations. Practice makes perfect, but there is no long term and unwanted side effects to doing this, and you'll still be able to direct your eyes towards whatever you'll want to see clearer, just as you did before.

There is one other thing to add to techniques to observing the night skies. Namely, our eyes need some time to adapt to darkness, and with our more frequent staring in computer monitors, television and other illuminated sources, our pupils need even longer to dilate and let the maximum of light to fall on the retina. So for best results, rest your eyes and do not expose them to anything except red light for 30 minutes prior to observing. This allows your pupils to expand to their maximum diameter and build up the levels of optical pigments, which are rapidly lost if exposed to bright light.

It is also important to observe with both of your eyes open to avoid fatigue. If you find this difficult, you may try covering the other eye with your hand or an eye patch. It should also be mentioned that both smoking and drinking cause deterioration in your sight. Your blood-sugar levels are also important and a low level of blood sugar causes lowering of sensitivity of eyes.

Lastly, your eyes will tire using this technique, just as they would by focusing for longer periods of times as you normally would. Look away every now and then, preferably not towards too strong sources of light, but still at least slightly brighter than those you previously observed. Also stretch your legs and blink fast a few times. Stretch your back too, and relax your bottom. You might find it funny, but this technique of relaxing and contracting muscles of that part of our physique has been used by trained spies to cheat polygraph tests that also measure pupil dilation. How's that for "beating the system", huh? It is by far the fastest way of relaxing your eyes, at least that I know of.


There exists probably about one supermassive black hole per galaxy, hence about 100 billion ($10^{11}$) in the observable universe.

If we take 1% of the stars ending as a black hole as a very rough order of magnitude estimate, we'll get about 1 billion stellar black holes per galaxy, hence about $10^{20}$ stellar black holes in the observable universe, give or take one order of magnitude.

Stars of more than about 15 stellar masses (O class stars) will probably end up as black holes. About 0.00003% of the stars are O-class. The lifespan of a star with 15 solar masses is about 11.5 million years. So, we need to multiply the 0.00003% at least by a factor 1000, since the age of the universe is more than 1000-times the life of an O-class star, resulting in 0.03% of all stars over time should have been O-class stars.
(Most stars are red dwarfs, and have a longer life-span than the current age of the universe.) Star formation, especially of heavy stars, in the early universe has probably been higher than today.
Therefore I'm estimating the number of stellar black holes as about 1% the number of all stars.

Intermediate-mass black holes as presumed centers of globular clusters are harder to estimate, since there isn't much observational evidence. If the hypotheses turns out to be correct, and we take 1000 globular clusters per galaxy, we get about $10^{14}$ IMBHs.

There is no observational evidence for the existence of small black holes (smaller than a few solar masses).


What orbital period would produce one New Moon (and one Full Moon) each year? What other effects would this produce?

This would naturally be achieved by a quasi-satellite. These are objects with a 1:1 orbital resonance with a different eccentricity to the planet. From the perspective of the planet, this leads to the quasi-satellite appearing to travel around the planet once a year in a retrograde direction. Starting with the quasi-satellite at its maximum distance from the Sun, this would be at full phase. Six months later, the quasi-satellite would be between Earth and the Sun, so would be at new phase. The diagrams below show an idealised version of the configuration based on Keplerian orbits in a non-rotating frame and in a frame where the Earth and the Sun are kept in a fixed position.

Real systems will evolve dynamically due to the gravitational interaction between the planet and the quasi-satellite, leading to precession and oscillations of the alignment of the orbits.
Earth currently has several known quasi-satellites, the closest of which is (469219) Kamoʻoalewa. The known quasi-satellites in our Solar System do not have long lifetimes: they are relatively recent captures into the quasi-satellite orbit that will eventually escape. Nevertheless, there are some conditions under which quasi-satellites can be stable over long periods, and perhaps some exoplanetary systems have more favourable conditions for quasi-satellites than ours does.


Estimates for "undetectable" planets in extra-solar systems

This is a partial answer. It is partial because this is an active area of research, with this very topic as one of the prime questions, second I've only chased down the estimates from one method; microlensing, because this method is not biased toward large (in size and mass) planets in close orbits around small stars.

The microlensing method is biased toward more massive planets in Jupiter like orbits i.e. cold planets. It is biased toward distant planets about stars much closer toward the galactic center. This is important because stars there have lower concentrations of heavy elements need to make planets. It is equally sensitive to planets round all sorts of stars, and even free floating planets. It can detect down to Earth mass planets.

Here is a link to a technical summary of the method. The part you are most interested in is section 4.1.3 pages 23. Microlensing Searches for Exoplanets Yiannis Tsapras
This poster paper has a summary plot Fig 3 of planet occurrence rates from various methods Poster: Planet frequency from microlensing observations Arnaund Caasan
There is also a link to an archive of papers on the subject: NASA archive: Planet Occurrence Rate Papers

I do a quick summary of the main results: (microlensing perspective).
Estimating the biases, combined with actual detections it is possible to estimate (or put limits on) the numbers of planets and how theoccurence depends on mass, distance from star.
Snow line == distance from star where water remains frozen. In our solar system it is between Mars and Jupiter. Jupiter's moons have lots of ice. Dimmer stars this is closer in.
1 AU = Earth-Sun distance

Number of planets beyond snow line (Jupiter and beyond in our solar system) is 7 times greater than close in hot planets. In the distance range 0.5 to 10 Au of 20% of stars have Jupiters. 50% have Neptunes, 60% have super earths. There are many stars with more than one planet. So there is about one planet for every star in the galaxy. At least!. About 1/6th of stars with planets have a solar analog (massive planets like Jupiter and Saturn with room for small rocky planets close in).


Does the magnetic field of earth restore?

Remember that a magnetic field is just that:  a field . There's no mass there.   Its shape depends on the electromagnetic properties of Earth -- primarily the behavior of the outer core.
When there's a significant Solar ejection,  the  electrically active particles in that mass cause local changes in the magnetic field pattern (equivalent to isobars , or lines of constant field strength, etc.)   So long as the solar ejecta do not damage the Earth in any significant way, once the particles are gone, the field lines will revert to their default layout.


Can a planet's changing tilt be aligned with the rotation around its star?

No, or at least not for a planet of any size. To change the absolute direction of the axis of rotation of a planet needs a large force acting in some way asymmetrically. In the case of the Earth this comes from the gravity of the sun and the moon acting on the Earth's equatorial bulge and it takes 26000 years to rotate the Earth's axis around a cone 46 degrees wide. To rotate the axis in a full circle (a cone 180 degrees wide) in a single year would only be imaginable for a very irregular planet (which means a small one, since big ones collapse into spheres under their own gravity) probably with a very large and close moon to pull on it.


There are several major types of black hole, and each forms differently:

Quantum black holes - these aren't currently proven to exist, but many scientists think they are likely. These are microscopic black holes, which are believed to have formed in the Big Bang, though they can be created wherever there is enough energy.
Stellar-mass black holes. These are the remains of dead stars, and the main type of black hole people talk about. They form when very large stars collapse to the point where even light cannot escape at the end of their lives.
Supermassive black holes. These, as their name suggests, are extremely massive - with masses of millions to billions of times that of the Sun. Scientists aren't quite sure as to how these formed, but the main theory at the moment is that they started as several stellar-mass black holes, left behind by the first generation of stars, which fell into the centre of galaxies that were forming around them. As they did so, they came together, to form one large black hole. Another theory is that they started life as 'dark stars', a theoretical type of star held together by dark matter, which could be far larger than normal stars. These would have been extremely short-lived, and collapsed to create a supermassive black hole.


Finding the number of stars n light-years from the sun

Wikipedia, quoting Erik Gregersen. The Milky Way and beyond. gives a local value of 0.004 stars per cubic light year, or 0.14 stars per cubic parsec.

So within n light years you would expect $0.004\times 4\pi n^3/3\approx 0.016n^3$ stars.

This predicts the nearest star to be about 4 light-years away, consistent with observation.


How long until we cannot see any stars from other galaxies?

First of all, we can't see stars from other galaxies (with a few exceptions, Cepheid variable stars for example are regularly used to determine distances to nearby galaxies). As it currently stands, we can only see stars from our own Milky Way (in which I'm including the large and small Magellanic Clouds). Type Ia supernovae aside, it might be possible to see Cepheid variables in the Andromeda galaxy, but anything further is almost certainly not possible.

The relevant number when it comes to the distance one could possibly see in the universe is the comoving distance to the cosmological horizon. This horizon defines the boundary between what can be seen and what cannot be seen, simply because the universe is not old enough for particles to have traveled that far (yes, even at the speed of light; in fact, if we're talking about what we can observe, the photon is the particle we care about).

The definition of the comoving distance is:

$$\chi = \frac{c}{H_{0}} \int_{z=0}^{z=z_{Hor}} \frac{dz^{'}}{E(z^{'})}$$

where the function $$ E(z^{'}) $$ contains your choice of cosmology (the following is for flat universes only):

$$E(z^{'}) = \sqrt{ \Omega_{\gamma} (1+z)^{4} + \Omega_{m} (1+z)^{3} + \Omega_{\Lambda} }$$

This distance can be extrapolated out into the future (given your choice of cosmology), and ultimately will tell you what the distance to the cosmological event horizon will asymptotically reach. This is the distance at which no object beyond could ever come into causal contact with you.

Another simpler way to calculate the particle horizon is to calculate the conformal time that has passed from the beginning of the universe, and multiply it by the speed of light $c$, where the conformal time is defined in the following way:

$$\eta = \int_{0}^{t} \frac{dt^{'}}{a(t^{'})}$$

where $a(t^{'})$ is the scale factor of the universe, and its relationship to time depends specifically on your choice of cosmology. Again, if the ultimate value of this horizon is desired, you would need to integrate this to infinity.

SUMMARY: I'm having a hard time finding the exact number of the comoving size of the cosmological event horizon, though I'll keep searching. If I can't find it I suppose I'll have to do the calculation when I have some free time. Almost certainly it is larger than the size of the local group of galaxies, making our night sky largely safe from the fate of the universe. Anything further than this horizon, however, will monotonically become both fainter and redder. I also think that it does asymptotically reach some value, rather than eventually increasing infinitely or decreasing passed some point in time. I'll have to get back to you on that.

Here's a nice review of distance measurements in cosmology.


Why does Wikipedia state that the Solar System has one unconfirmed planet?

That was a rather recent (16 January 2021) change to that wiki page made by a user named Mariofeds, who does not have a wikipedia user page and who has just one contribution to wikipedia.
It may well have been vandalism.
The associated talk page explicitly excludes adding the hypothetical Planet Nine as an unconfirmed planet. In line with this discussion, the number of planets listed for the solar system had remained as eight confirmed, none unconfirmed for a long time.


Is there any data source for the rotational axes of bodies in the solar system?

Collecting and reporting the requested information is the job of the IAU Working Group on Cartographic Coordinates and Rotational Elements. They issue reports every few years; the most recent report was in 2009. The report is currently available at usno.navy.mil and astrogeology.usgs.gov, and elsewhere. Table 1 provides data for the Sun and planets (including Pluto), table 2 for satellites, and table 3 for other solar system bodies.

The question asks specifically about the Moon. The Moon's orientation is hideously complex. I won't even begin to replicate the expression in table 2 of the cited report. The report specifically addresses the Moon's orientation, admitting that the expression in table 2 is approximate. The best source is an ephemeris such as JPL's Development Ephemeris. These data are available in the form of SPICE kernels.

References:

Archinal, Brent Allen, et al. "Report of the IAU Working Group on Cartographic Coordinates and Rotational Elements: 2009." Celestial Mechanics and Dynamical Astronomy 109.2 (2011): 101-135. DOI 10.1007/s10569-010-9320-4


What should the mass of a planet be in-order for its escape velocity to be near the speed of light?

As @KeithMcClary says in his comment, escape velocity depends on both mass and radius. The smaller the radius for a given mass the higher the escape velocity. So if you could somehow compress the Earth until it was just a few centimeters across, its escape velocity would approach the speed of light.
At the other extreme, if you filled a space twice the diameter of Earth's orbit with copies of the Earth, without doing any compressing at all, it would already be a black hole.
If you just pile matter onto a planet and let its own gravity compress it, you need about 2 solar masses before it is close to being a black hole (at which point it is 10-20 km across).


What Constellation Is This? (Photo Included)

If it was rather small, I suppose you were looking at the Pleiades, which is an open star cluster (not a constellation).
Take a look at this web application: https://stellarium-web.org/
You can set a time and location and find constellations and other objects really easily.


If months are based on the moon, then why are the months longer in the Gregorian calendar than lunation?

Ancient Egyptians and Mesopotamians came up with “administrative” calendars of 30 days, that were easier to calculate than “real” lunar months of sometimes 29, sometimes 30 days. At the end of the 12 months, they added five days of religious observances, sometimes positive, sometimes negative.
When Julius Caesar conquered Egypt, he knew about the very regular Egyptian calendar and appointed his mathematicians and astronomers (if we can really call them that, as Roman science wasn’t very developed) to discuss with their Egyptian peers and devise a better calendar for the Roman Republic (reminder that Caesar was not the Emperor; the first one was Augustus, his [adoptive] son, and not right away).
Eventually, it was decided in Rome that, in order to follow with the seasons, the already-existing Republican Roman calendar was to be modified. It contained 10 months of 29 and 30 days plus a “winter” of 59 days, which became divided in January and February. As Julius Caesar was the one who brought this change, he was “honored” by having a month renamed from Quintilis (“Fifth”) to Julius (our July)—and that month was given 31 days, as well as a few others. Also, the beginning of the year was shifted (at this point or later; it’s not clear) to January, making the “numbered months” off with their new rank—e.g. “October” is not eighth (“octo”) anymore.
Caesar’s reform was not well applied, and eventually, Augustus had to enforce the new calendar, and he was also “honored” with a 31-day month, right after Julius’, so that’s how Sextilis (“Sixth”; and it had 30 days) became Augustus (with 31 days).
The rest is basically adjusting here and there to make it a total of 365 days at the end of the year, someone sometime having the “brilliant” idea of alternating 30- and 31-day months (a little like nature “alternates” between 29- and 30-day lunar months).
Side note… In some languages, the word for Moon and the word for month is the same—it’s already quite similar in English! For example, Romanian has “Luna” for both, and Korean has ”dal.”


What would we see if all stars disappeared simultaneously?

The farthest object that we can see with our naked eyes as of now is the Andromeda Galaxy (2.6 million light years away), although the M83 (14.7 million light years away) has also been reported to be seen with the naked eye. (You may want to click here for more information on what the farthest objects we can see with our naked eyes, from Universe Today or see this answer from this same site.)

Assuming that just before the moment all the stars disappear, these two farthest objects are still emitting light, then we can know judging by their distance from us, that this light will only reach our eyes 2.6m years later (because 1 light year = light travelling for 1 year) from the Andromeda Galaxy, and 14.7m years from M83.

Therefore to answer your question more directly:

We would still see the stars (assuming this is what you are referring to by "anything") for up to 2.6~14.7m years, depending on what really is the farthest thing we can see in the sky.


Did I see a mentionable event in Germany early morning on 6th of February 2019?

Fireballs are real but rare events. Meteors are formed when small rocks enter the atomsphere from space. Larger rocks or small asteroids will produce brighter meteors. The brightest meteors are called fireballs or bolides. It is not uncommon for parts of the rock that made the fireball to survive and can later be found as meteorites.
A couple of things can be confused with fireballs.  When satellites re-enter, they can burn up. Satellites move much more slowly than true fireballs, most fireballs last less than 10 seconds.
A comet is an astronomical, not atmospheric, object. It looks like a fuzzy star with a tail. It remains almost motionless in the sky, sometimes for months. Really bright comets are rare, and even the brightest comets are much much dimmer than a fireball.
There are several organisations that collect reports of fireballs:

https://fireball.amsmeteors.org/members/imo/report_intro
https://ukmeteornetwork.co.uk/fireball-report/

The advice given states:

Please, don't report sighting that lasted more than 30 seconds: the vast majority of fireballs are only visible for few seconds.
Please, don't report recurring events: seeing a fireball is extremely rare and often a once in a lifetime event.
Please, don't report slow blinking objects or lights crossing the sky going by 2 or 3: a fireball looks like a big shooting star.


Why must all magnitude systems have a reference point?

Because you can't take the logarithm of something with dimensions only the log of a number.
Your magnitude would depend on what you measured f in. In other words, by choosing a set of units, you choose a zero point! Whether that unit is Jy, W/m$^2$ or the flux of some other object.


What object could I have seen at late June on the Sky?

Checked on Sky Map on my phone for late June, 2016 in Kolkata - and Saturn and Antares were close to each other around then in the Eastern sky, so these may be the two objects you saw.  The colour change sounds like atmospheric effects affecting the seeing that night.  I don't know if June in Kolkata is particularly worse than usual for atmospheric conditions, but even on a seemingly clear night there can be a layer of turbulent air above you that will diffract the light from the sources and produce unsteady images.  This is usually worse closer to the horizon as well (as these two objects were).
Planets don't usually twinkle, so if Saturn was one of the objects, then it must have been a strong effect.
Nothing to freak out about, and keep looking up! :)


How to calculate the S/N of IFU or long-slit galaxy observations

Unless you in fact find a "standard" definition for the mean or collective SNR, there is no way to answer this.  As you noted, the signal strength varies with view angle (as well as wavelength), so there's really no single value.  If you have a defined algorithm which processes all your spectral data to produce a final image or analysis value, then you can use standard statistical techniques to calculate the SNR of the output as a function of the SNR of all the contributing factors. (Here "factor" means each independent input variable, i.e. spectral line).  I recommend getting a copy of Bevington to learn how to do this.   Note- that appears to be a not-necessarily-legal-copy; you can purchase the real thing anywhere, e.g. amazon


Eyepiece diameter for small focal length planetary viewing on a large dobsonian?

I would recommend a quality 1¼" 9 mm or 12 mm eyepiece as the optimal solution.

You already have a 19 mm so something close to that such as 15 mm would bring small difference, so take a step or two down to a 9 mm or 12 mm Plössl lens.

While a 2" eyepiece usually offers better eye relief (from bigger lens) the cost is also substantially more, and if you compromise on quality by buying a cheaper eyepiece the results will be less than satisfactory - lack of clarity, distortions from imperfections in the glass, etc.

Extra magnification can also be obtained by using a Barlow, but you are also adding extra lenses to the optical path which will slightly reduce the amount of light you are getting at the eyepiece, but also any imperfections will be magnified.

Also, with Dobsonians (usually) not having any tracking mechanisms you will be manually tracking your object of interest, and extra magnification means that it will be moving more quickly across your field of view as well as making it a little more difficult to keep it smoothly in the eyepiece. These are also arguments in favour of a 2" eyepiece over a 1¼" eyepiece, but either way, good eyepieces can be kept and re-used when/if you upgrade your telescope.


What would the view be list from TRAPPIST-1g, looking back towards TRAPPIST-1?

Let's take a look at the star and the planet's characteristics first. We have the star TRAPPIST-1:

$M = 0.082 M_{\odot}$
$R = 0.117 R_{\odot}$

The planets are all roughly in the scale of $0.7 R_{\oplus}$ to $1.2 R_{\oplus}$ with their semi-major axes of:

b: $1.66 \times 10^6 \; \mathrm{km}$
c: $2.88 \times 10^6 \; \mathrm{km}$
d: $3.14 \times 10^6 \; \mathrm{km}$
e: $4.19 \times 10^6 \; \mathrm{km}$
f: $5.57 \times 10^6 \; \mathrm{km}$
g: $6.73 \times 10^6 \; \mathrm{km}$

So if we would stand on the surface of TRAPPIST-1g and look towards the star, we would be at $1/25$ the distance to the star compared to the sun. The star however is roughly $1/8 R_{\odot}$. That would make the star appear about at about $3$ times the diameter compared to our Sun from the Earth.

TRAPPIST-1f would every few days pass TRAPPIST-1g with a closest approach of $1.16 \times 10^6 \; \mathrm{km} \approx 3 \times r_{\mathrm{Earth-Moon}}$ which is roughly three times the distance to the Moon.

Since $R_{\mathrm{TRAPPIST-1f}} \approx R_{\oplus}$ and $R_{\oplus}/R_{\mathrm{Moon}} = 3.67$ it would seem that TRAPPIST-1f would appear roughly the size of the Moon on closest approach.

The other planets are similar in radius but their distances are $2$ to $5$ times further away and would thus appear smaller by that factor.


What would happen to black holes during the Big Rip?

The universe won't expand fast enough to rip atoms apart, but if dark energy turns out to be a fact it will expand fast enough to avoid gravitational collapse. Black holes won't be ripped apart, they will just carry on as normal, as will the galaxies they belong to. But as space expands the galaxies will move further and further apart, going their own separate ways with not enough gravity to keep them together, eventually fading away into the black nothingness of space. Further research may mean that dark energy will have to be revised or reinterpreted, so at the present time a collapsing, blue shifted universe can't be completely ruled out.


How can I calculate the time between sunset and actual darkness?

One may define three forms of twilight on Earth.  Although the actual amount of light depends on weather, topography, and land cover, they are defined as:

Civil twilight: Solar angle > -6°
Nautical twilight: Solar angle > -12°
Astronomical twilight: Solar angle > -18°

Human eyes see logarithmically so % of sunlight left is not a very useful measure, unless you are interested in incoming shortwave radiation for solar panel or other energy calculations.

To answer your question, you need to calculate at what time the Sun will be at those elevations below the horizon; this is calculated in a similar fashion to sunrise and sunset times.  However, on top of that you also need to consider atmospheric refraction.  See position of the Sun and sunrise equation.  The formulae are quite complicated, but you may be able to simplify them if you're only interested in the duration, not the absolute times.

The time after sunset at which those occur are a function of latitude and time of year.  On this carpet plot from Wikipedia you can read the length of different forms of twilight at 70°N (north of the Arctic circle):


How much radiation would something need to output to sufficiently affect the CMB?

It is not very clear what you mean. There are many cosmic sources of microwave radiation that emit so much energy and/or are so close to us that they confuse measurements of the cosmic microwave background. For example "compact sources" such as giant molecular clouds, supernova remnants, H II regions; "diffuse" sources due to dust in the galaxy and in our solar system. In order to get a clean signal, these foreground sources have to be removed from the data. There are also more subtle cosmological effects caused by lensing and gravitational time dilation caused by the large scale structure in the universe.

There are many ways and techniques that are used to characterise and remove these backgrounds (e.g. see Ade et al. 2013).

The most confusing background source would be one that is diffuse, spread over the whole sky almost uniformly and with a spectrum that is very similar to the cosmic microwave background. The source that most closely matches these properties is dust in our own solar system. An imperfect subtraction of this may have led to controversy over possible large scale anisotropies in the CMB- for example see Schwarz et al. (2015); who discuss the problems of removing the signatures of solar system dust.


What it would look like to observe people with a different time flows?

Normal case:

The effect would not be very noticeable unless the large planet was extremely large / high gravity (e.g. it would look "almost" normal) looking at one another.  If the planet was that large, it would probably not be able to support life, but to see what would happen, lets assume it can support life.

Relativistic case, from the large planet, observing the small planet:

If the large planet were extremely "heavy / massive" (think more like a neutron star), the people on the large planet would observe the people on the small planet moving much faster.  It would be like watching a movie with "fast forward" on.  However, light would also be blue shifted, which would produce other interesting visual effects (e.g. what was red might now be blue, and what was in invisible "infrared" might now be red depending on the amount of the shift.)  The amount of blue shift (light wave compression) and "fast forwarding" would increase if the planet were made bigger.  The rest of the universe would have similar effects when observed from the large planet.

Relativistic case, from the smaller planet, observing the large planet:

The people on the smaller planet would observe the people on the extremely large planet moving in "slow motion".  It would be like watching a movie in slow motion.  The people on the large planet would also be "red shifted".  For example, something that was blue might appear red, and something in the ultraviolet range might now appear blue.  The "slow motion" effect and red shift (light wave expansion) would increase if the extremely large planet were made bigger.


How do we distinguish old craters from new ones on the Moon?

Aside from the excellent points made in James K's answer, there are other ways to date craters.  For example, when the rays of one crater overlay those of another, we know that the former is younger than the latter.

We can also estimate ages of large craters by counting subsequent craters inside the crater floor.  Crater counting is one of the more common ways of estimating the age of lunar surface features.

We can also determine ages of some craters by looking at whether it was affected by known ancient lunar processes.

Then there is radiometric dating.  Large impacts spread materials across the surface of the moon, so the Apollo mission samples contained ejecta from many craters, some of which are identifiable by things like mineral composition unique to that part of the moon.

Once you have a good date for one crater, you can use that data to help determine the age of others.  For example, if the floor of a crater has X smaller craters  in it, and you know how old that crater is,  you can then use those counts to estimate ages of other craters with similar counts.   With enough varied data sets, you can generate some pretty good functions for estimating absolute ages.

For example,  have a look at the image of the Aristarchus Plateau below:

NASA (image by Lunar Reconnaissance Orbiter) [Public domain]

Notice the two craters at the bottom of the plateau.  The one on the left is the crater Herodotus, and the one on the right is Aristarchus.

Looking at the differences between the two of them, the first thing you notice is that Herodotus is filled in with the same lava that fills the basin around the plateau, but Aristarchus does not have that fill.  So right there we know that Herodotus pre-dates the event that filled in Oceanus Procellarum, while Aristarchus came later.  This makes Herodotus at least 1.2 billion years old, which is the current best estimate for the last time lava flowed on the moon.

Now, we can do better than that with sampling.  If we get a sample of the basalt inside Herodotus, we can use radiation dating to determine when it crystallized, and that would give us a very good upper limit of the age of the crater.  And in fact we did get such samples from Apollo, and that, plus crater counting, plus looking at other features around it with known ages, plus the amount of degradation in the crater walls and the coloration of the material, we can tell that Herodotus dates to the Imbrium era, or about 3.8 billion years ago.

As for Aristarchus crater, you can see lots of evidence that it is young.  For one thing, its interior and ejecta are incredibly bright - it's the brightest crater on the moon.  We know from radiological dating and crater counting that similarly bright craters are young.  The walls of the crater have minimal slumping, and the ejecta around the crater is still well defined.

Aristarchus is one of the 'Copernican' craters, so named because they all formed during the Copernican period, which starts 1.2 billion years ago and goes right up to the present.  Other large copernican craters include Copernicus, Tycho, and Kepler, among others.

Now look at this wider, high contrast image of the Tycho region.  You can see Copernicus and Kepler on the right, and Tycho lower center.  Note that the crater rays overlap each other, and by careful examination we can tell the order of the impacts.  That helps us further refine the relative ages of the craters,  and if we can narrow down the age of one with sampling or other techniques, it helps narrow down the ranges of the others.

Btw,  the order of creation from oldest to newest Copernicus, Kepler, Aristarchus, Tycho.

Here's a pretty good reference describing a number of techniques for dating features on the moon:

Relative Crater Age Determination

This is an iterative process - each time you narrow down the age of a crater, that can be used as data to further refine the ages of others,   But it's not a precise science - age estimates of various craters can still vary by hundreds of millions of years.  There is still a lot we don't know about how the Moon formed and evolved.


Will the UK be able to witness the tetrad in April 2014?

It is great that you have an interest in astronomy.

Unfortunately, the UK will be on the wrong side of the Earth to see the lunar eclipse on April 15. If you you want to experiment with various places on earth that will see it, and what they will see, get a planetarium application like Stellarium or SkySafari or TheSkyX.

A lunar eclipse is always easy to spot - it is where the full moon is. Also, because it is the result of the Earth getting in the way of the Sun shining on the Moon, it is always after the sun sets and after the full moon rises in the East. That is, if you're going to be in a place to see the whole sequence, not just a part of it.

Just so you know, the tetrad is not one but a series of lunar eclipses over 2 years, this is just the first lunar eclipse in the series. The last tetrad was 2003-2004, so you were around, but probably not paying as much attention at that time :-)

Unfortunately, Cambridge UK will not see any of the first 3 eclipses in this tetrad. However the 28 Sept 2015 eclipse will be visible from that location.

Incidentally, one of the better places to see this lunar eclipse will be Easter Island (Rapa Nui) which will see the whole eclipse, and with the totality occuring high in the sky, near the meridian. Also, not much light pollution.


The tidal field of the Galaxy does lead to the oscillation of the plane of very wide binaries.  The mechanism of this oscillation is identical to the Kozai-Lidov mechanism (the only difference is that in the case of KL oscillations the tidal field is generated by the averaged orbit of a tertiary stellar companion).

However, if you run the numbers, the timescale for these oscillations is extremely long for any of the planets (I forget how long exactly, but much longer than the age of the Universe).  It's somewhat shorter for comets in the Oort Cloud, only a few billion years, because their orbits are much larger and so would tend to experience larger a larger tidal effect.


Minimum Mass Solar Nebulae Scaling Factors

Attempt to do an intuitive explanation:

The Sun was formed from a cloud of matter. A small amount of that matter remained around the Sun, and had a similar composition. This was the solar nebulae.

But the composition of the planets are now not the same as the Sun. This is explained by that some of the matter was collected by the Sun, and some of it was blown out of the system.

That was only the light stuff! like hydrogen and helium. Most iron and other metals are conjectured to have remained in orbit, forming the planets.

As the Sun is representative for the original composition (partly at least, stellar fusion has skewed the hydrogen/helium ratio). So if iron is 0.14% of the Sun, that is also the original abundance in the nebulae. So all the iron in the planetary bodies is 0.14% of the original nebulae mass.

From that you can calculate the original mass. It is a 'minimum' because some of the iron may have escaped as well.


What am I looking at here at Google Sky?

It's a star with pretty strong diffraction spikes.
To find out which one it is, you can just look up its coordinates in databases like SIMBAD.
Then you find that it is the star Xi Cygni.
You can also find some information on it on Wikipedia.


Is **Voyager I’s** reduced data transmission rate as described in [this article][1] because of the distance or because its transmitter getting slower?

Transmitters don't slow down as they age. Red shift isn't involved, at all. As a transmitter gets further away the signal at the receiver gets weaker, so it's deliberately slowed down. Data channel capacity depends on signal to noise ratio and bandwidth. As the signal gets weaker the data rate can be slowed to reduce the necessary bandwidth.  Noise doesn't increase: it's just a loss of signal strength at the receiver. Narrowing the receiver bandwidth reduces the noise received, but also limits the information bandwidth.


Is it accurate to say that we have a gravitational attraction towards all objects in the known universe?

Yes

The gravitational field from an object does extend infinitely far out into space, if there is an infinite space for it to fill. There isn't any sort of limit to how far an object's gravitational influence can travel. However, there is a catch: Gravity travels at a finite speed (the speed of light). If gravity propagated infinitely fast, any two objects would be guaranteed to be affected by each other. However, because it is not infinitely fast, two  objects that have never been less than 13.8 billion light-years apart (the universe is only 13.8 billion years old) could only now be feeling each other's influence. If the Sun were to suddenly explode, it might not feel the influence of an object 13.8 billion that has never been less than 13.8 billion light-years away.

  Is the gravity blocked by the same way light is blocked by an object?

An gravitational field extending infinitely (eventually) in all directions means that one object should always have an influence on another. But the gravitational influence of one object can sort of be "blocked" by another object if its field in one spot is strong enough compared to the  other. For example, we still feel the force of Mars' gravity, but, to us, it is largely overwhelmed by that of the Earth.

We can show this easily using Newtonian gravity. Newton's law of universal gravitation is formulated
$$F=G\frac{m_1m_2}{r_{12}^2}$$
where $m_1$ and $m_2$ are the masses of the two objects - we'll say $m_1$ is the mass of a human, $m_h$, and $m_2$ is the mass of Earth, $m_E$. A human on Earth's surface is roughly $6,371 \text { km}$ from it's center, or $6,371,000 \text { meters}$. Mars, on the other hand, is, at its closest, $0.3814 \text { AU}$ from Earth - or roughly $57,210,000 \text { km}$, or $57,210,000,000 \text { meters}$. Care to do the math? Well, what with this distance being squared ($r^2$), you can tell at a glance that the force from Mars is many orders of magnitude less than the force from Earth.


HD? Perhaps start with: VizeR. You can search for any star you like, regardless of catalogue.


How is the Sun tilted to its orbit around the Milky Way's center?

The wikipedia page says

  Obliquity 7.25°[2] (to the ecliptic)
  67.23° (to the galactic plane)

Interesting fact: the sun's layers rotate at different speeds, as well as different rates depending on latitude.


Is there any habitable exoplanet around Tau Ceti?

TLDR version: probably not, and claims for the habitability of planets in this system are on shaky ground.
Long version follows.
Planets
So as of Feng et al. (2017), there are four planet candidates around Tau Ceti:

Planet
Minimum mass ($M_\odot$)
Semimajor axis (AU)

Tau Ceti g
$1.75^{+0.25}_{-0.40}$
$0.133^{+0.001}_{-0.002}$

Tau Ceti h
$1.83^{+0.68}_{-0.26}$
$0.243^{+0.003}_{-0.003}$

Tau Ceti e
$3.93^{+0.83}_{-0.64}$
$0.538^{+0.006}_{-0.006}$

Tau Ceti f
$3.93^{+1.05}_{-1.37}$
$1.334^{+0.017}_{-0.044}$

Note that the designations Tau Ceti b, c and d refer to planet candidates that are no longer thought to exist. The error bars refer to the 1% and 99% percentiles. $M_\oplus$ is the mass of the Earth.
The Feng et al. (2017) paper also notes that the system is dynamically packed, which does not bode well for the prospects for additional planets between the known planet candidates (note that their figure 17 shows the regions where the planets would interfere with each other, not the regions of stability for an additional planet).
The habitable zone
The conclusion of the paper gives the luminosity of Tau Ceti as 0.52 times solar and the effective temperature as 5344 K. Using these values, the habitable zone limits can be calculated from Kopparapu et al. (2013), which assumes that habitable conditions are maintained by the carbonate-silicate cycle with carbon dioxide as the main (non-condensible) greenhouse gas.
Inner boundaries

Boundary
Distance (AU)

Recent Venus
0.551

Runaway greenhouse
0.723

Moist greenhouse
0.729

The moist greenhouse limit is the most conservative inner boundary, it occurs where sufficient water vapour enters the upper atmosphere that water loss begins to occur from the planet. In our solar system, Earth is located close to this limit in the inner part of the most conservative habitable zone.
The runaway greenhouse limit occurs where the positive feedback from water vapour overwhelms the stabilising negative feedback from the silicate-carbonate cycle, driving further evaporation of the oceans and higher temperatures. This is thought to have occurred on Venus, leaving the planet in the state it is in today.
The recent Venus limit is based on the possibility that Venus may have retained oceans for several billion years. This is not known for certain as our knowledge of Venus's evolution is rather incomplete and the conditions on the planet's surface are not favourable for driving rovers around investigating the geology.
From this we see that Tau Ceti e is located close to the recent Venus limit and is closer to the star than the runaway greenhouse limit. This suggests that any oceans that may once have existed would likely have boiled off, leaving the planet in a Venus-like state.
Planets g and h are too close to the star.
Outer boundaries

Boundary
Distance (AU)

Maximum greenhouse
1.279

Early Mars
1.330

The maximum greenhouse limit is the furthest distance from the star that a cloud-free carbon dioxide atmosphere can support conditions compatible with liquid water. Beyond this, the increased scattering leads to increased reflectivity of the planet and the CO2 would begin to condense, removing it from the atmosphere and leading to runaway cooling. This is the most conservative outer habitable zone boundary. Note that by this point, the planet would require several bars of CO2 which would make it toxic for humans.
The early Mars limit is based on the observation that Mars managed to maintain surface water (e.g. various rivers, and a possible northern ocean) in the early solar system when the Sun was significantly fainter than it is today. Tau Ceti f is located right at this limit.
Extensions to the habitable zone
None of the planets fall into the most conservative habitable zone, and Tau Ceti e and f are at the boundaries of the most optimistic estimates for the habitable zone boundaries. There are nevertheless options for extending the habitable zone.
At the inner boundary, a runaway greenhouse effect could be avoided on dry planets, where there simply isn't enough water to evaporate to drive the positive feedback, see Zsom et al. (2013). It isn't clear to me that such a planet can be described as habitable, since such planets may lack the hydrothermal systems that could act as the sites for abiogenesis. Their geological evolution would likely be substantially different to Earth's without water to lubricate plate tectonics.
Another possibility is on slowly-rotating planets, where substantial cloud layers can build up on the day side of the planet and increase the reflectivity, as noted by Yang et al. (2014). On the other hand, Scholz et al. (2018) have noted that there appears to be a universal mass-spin relationship that extends from planets to brown dwarfs. This predicts that super-Earths would likely spin too fast for this mechanism to work unless they have been spun down by stellar tides or a large moon.
On the outer boundary, adding additional greenhouse gases such as methane may work to extend the outer habitable zone, see for example Ramirez & Kaltenegger (2018). This has been suggested as the mechanism for allowing surface water on Mars, which would suggest that the "Early Mars" limit is an observed data point within the methane habitable zone. Another possibility is that a dense hydrogen atmosphere could maintain liquid water, e.g. Pierrehumbert & Gaidos (2011) though the pressure of such an atmosphere may well have implications for the geology of the planet and hence the potential for abiogenesis.
Planets whose climates are stabilised by something other than the carbonate-silicate cycle, or have substantially different atmospheric compositions would have different habitable zone boundaries (if subsurface oceans on icy worlds are habitable, there may be interesting prospects for dwarf planets in the outer debris belt), but this is already getting speculative enough, besides there is another possible objection to the habitability of these planets...
Planetary masses
A limitation of the radial velocity method is that only the minimum masses can be derived. With Tau Ceti, we have a possible means to estimate the true masses: the star is surrounded by a debris disc (this would likely provide a source of impactors onto the planets, how bad the situation is depends on how much material is being perturbed into the inner system). Using Herschel observations, Lawler et al. (2014) give an inclination of 35±10 degrees. Assuming that the planets lie in the same plane as the disc, the true masses would therefore be approximately 1.74 times greater than the minimum masses.
Under this assumption, the true masses of the planets e and f both come out as about 6.85 Earth masses. Taking the 99% lower limit on the minimum mass error bars and a 45° orbital inclination as a low estimate, these would be 4.65 Earth masses for e and 3.62 Earth masses for f.
The nature of the planets
According to Rogers (2014), the transition between rocky and Neptune-like planets is somewhere in the region of 1.4 to 1.6 Earth radii. Using the mass-radius relationship from Zeng et al. (2016) and their core mass fraction of 0.26 for typical terrestrial planets, these radius limits correspond to terrestrial planets of roughly 3.3 to 5.4 Earth masses.
This suggests that Tau Ceti e and f are fairly likely to be sub-Neptunes rather than rocky planets, although the caveats are that in the optimistic case they can have masses below the rocky/Neptune-like transition, and that there do seem to be a few cases of rocky planets above the transition (most of those are likely to be evaporated cores of Neptune-like planets, which wouldn't apply to Tau Ceti e and f as they have much lower levels of stellar irradiation).
Conclusion
Given the current state of knowledge, Tau Ceti does not look like a good prospect for habitable planets. Tau Ceti e and f are rather marginal in terms of their location within the habitable zone, and their masses are sufficiently high that there is a good chance that they are sub-Neptunes rather than rocky planets. The dynamical packing of the system makes it unlikely that there can be a smaller, temperate planet in the habitable zone between the known planets.


Can orbital migration cause a planet to crash into the sun?

Definitely. In the early solar system, there were probably several more planets than there are today. Gravitational interactions between the planets caused some to be flung into the sun, while others were sent spiralling into the sun until it settled out into the more or less stable configuration we have today. With the Kepler spacecraft and other exoplanet finding programs, we've come to realize just how big a factor migration can be. The best example of this is what are called 'hot Jupiters'. These are gas giant planets up to several times the mass of Jupiter that are in orbits that would be well within Mercury's orbit in our own solar system. These planets could not have formed where they are now, as when their parent stars ignited fusion they would have blown away all the gas in their immediate vicinity. This means that these planets must have migrated all the way in towards their sun. For each one we see in a stable orbit tight around its star or spiralling its way in, there were probably several more that collided with their parents. We just see the 'lucky' ones.

Just to assuage any fears: This sort of thing mostly happens on such large scales in the first few million years of a solar system's history. A major migration isn't going to happen in our system. Our planets' orbits aren't entirely stable, but they'll last longer than the Sun will.


Shall we say now that Pluto is "larger than Mercury" ?

The implication of the question is that this extra 1000 miles should be added to Pluto's radius. The answer is no. For all of the solid planets, it's that solid surface (or solid+liquid surface in the case of the Earth) that counts, not the outer reaches of the atmosphere. The surface is a clear-cut, non-arbitrary boundary. The atmosphere? They can extend a long way out. A non-arbitrary boundary is always going to be preferred over an arbitrary one.

That's not possible in the case of the Sun and the giant planets. A somewhat arbitrary boundary is needed for those bodies. That somewhat arbitrary boundary explicitly excludes the upper reaches of the atmosphere, which extends out for many thousands of kilometers. For the giant planets, some use a tenth of the Earth's atmosphere as defining the arbitrary "surface", others, one atmosphere, yet others, ten atmospheres. That factor of 100 variance in pressure amounts to about a hundred kilometers. That's not that much considering how big the gas giants are.


What is the difference between the way light travels in a gravitational field and the way it travels through an expanding universe?

If like me you think the answer is C, can you explain why light should travel differently through a gravitational field to how it travels through an expanding universe? In other words, why in one case does the speed of the photon vary, while in the other case it doesn't?

I think the reason is, we can take the Cosmological redshift as a Doppler redshift, where the speed of the photon does not change but only the wavelength changes due to the expansion of the universe.

Meanwhile, in Gravitational redshift, its the "redshift" is purely caused by GR and it's possible that the light loses its speed due to this effect.

Here is the link why I think cosmological redshift can be thought as Doppler redshift.

Here


Will the shape of the constellations change if we observe from another planet, say Mars?

There will be no change observable by naked eye from any planet in the Solar System.

It's easy to estimate the distances for which change is visible.  There are two key points. First, changes in the constellations are due to parallax.  Second, by definition, a baseline of 2 AU (astronomical units) produces a parallax of 1 second of arc at a parsec's distance (wihc turns out to be about 3 light years) and half an arc second at two parsecs, and so forth.  Double the displacement to four AU and you double the parallax.

The nearest visible star to the Sun (Alpha Centauri) is just over a parsec away.  For it to move a barely noticeable distance (say a tenth of a degree or 1/5 the diameter of the Moon) in the night sky, the baseline would have to be 360 times as long, since 0.1 degree is 360 arc seconds.  So the closest star in the night sky would have moved a just-perceptible amount if observed from 500 AU away from Earth.

The brighter stars which make up the constellations range from the distance of Alpha Centauri out to thousands of light years, but many of them are within 100 light years (30 parsecs).  While a movement of 0.1 degree is perceptible, for the constellations to be obviously distorted, you'd probably want many stars to move by more than a degree (two full Moons).

So: 30 parsecs and 1 degree requires a movement of 1AU * 30 * 3600 (arc second/degree) or about 100,000 AU.  That's about 1.5 light years.

If you voyaged away from the Sun, a keen observer would notice an slight effect somewhere out past the Kuiper Belt, and a few stars would be massively displaced by the time you were out in the Oort cloud, but for wholesale disruption of the constellations, you'd have to go about a light year.

Two points: First, at a light year's distance from the Sun, the constellations would still be somewhat recognizable, though there'd be no doubt at all that things were scrambled.

Secondly, the shifts would be largest in the direction perpendicular to your line of travel (the baseline for parallax) and much smaller fore and aft.


How big are the rocky/icy cores of Jovian planets?

10-30 Earth masses is an estimate from this abstract, but 12-45 is a more accurate answer.


How do we calculate the image circle of a telescope at prime focus?

Image Circle will vary with scope design. Just about all scopes will cover the field of a 1.25" eyepiece barrel, and scopes with 2" focusers will usually cover the field of a 2" barrel (There are some exceptions where the field is bigger than 1.25" but doesn't cover the full 2"). Basically, you need to check the spec for the scope - and in any case the size of the focuser usally sets an upper limit. (With smaller SCTs and Maks, the size of the baffle tube can also limit the image circle).

Most small refractors for astrophotography will cover an APS sized DSLR sensor.
If you're looking for something to cover a full frame DSLR sensor, the choice gets more restricted - quite a few of the field flatteners/reducers for smaller refractors are only designed for APS coverage.

If you want to go for an unusually large sensor, then your choice is even more restricted - something like Takahashi's FSQ106ED astrograph has an 88mm diameter image circle in native mode, large enough for some medium format sized sensors, or 44mm with the reducer or extender - but there's a high price tag to go with it.

However, if you're looking at large format astonomical CCDs, they're usually very expensive too. Most of the more affordable astro cameras have smaller sensors, so are less demanding on image circle size.


With this scope you will be able to see quite a few things. Unfortunately, you won't be able to resolve much detail with some objects.

At 50mm aperture, the light gathering ability is not fantastic. That being said, some brighter objects will be easily seen. Your max power (zoom) is approximately 2*aperture, and the power is your objective focal length (360) / eyepiece focal length. Say you have a 6mm eyepiece in, your power (zoom) will be 360/6 = 60

PLANETS

Mercury and Venus will be easily seen with your telescope, at the highest power, you may be able to resolve some of the phases of Venus.

Mars will be visible as a reddish round blob, with hazy edges.

Jupiter will be visible along with the 4 Galilean moons (Io, Ganymeade, Europa, Callisto), although cloud bands will be very hard to resolve. Perhaps if you are at a dark sky site with perfect seeing, you may be able to make out a faint cloud band, but you will likely be looking at a large bright dot, with 4 smaller ones near it. A handy app for your phone called 'Moons of Jupiter' will allow you to tell which ones you are looking at.

Saturn will be a nice sight as you will be able to see the rings. This is always a spectacular sight for newcomers to astronomy. Although it will be rather small (even at maximum power), it will still be a nice sight.

Uranus and Neptune will likely be too dim for you to see. In a dark sky site in great conditions, you may be able to make out a tiny dim dot for each, but that is unlikely.

DSO's (Deep Sky Objects)

For DSO's, your best bet is to go for clusters and nebula at around mag +6.5-+7 max. Any dimmer than that and you will struggle. This means you will be able to see such globular clusters as the great cluster in Hercules, the Andromeda galaxy and the Orion nebula (these 3 were picked as they are the brightest in their categories) although they will look like just fain smudges.

Stars

Some of the brighter Carbon stars are a nice sight as they are often pleasing shades of red, which distinguish them from background stars. Binary stars are also a popular sight. There are a lot of good lists of binary stars online, although you will want to go for ones with a large separation between them, so that you will be able to see both. Mizar and Alcor are a good starting point for binary stars, these are easy to locate in the handle of the 'Big Dipper' in Ursa Major. If you want a bit more from your binary stars, you could look for some colour contrasting binaries, such as Albiero in Cygnus, which is a lovely binary star with one blue and one gold star.

Some open star clusters may be nice to go for, such as the Beehive cluster in Cancer, or the Pleiades in Taurus.

The best thing to do is to download a few apps, get yourself a star chart and get to a local dark sky sight and see what you can see. There are hundreds of objects you will be able to see. You could also look for a local Astronomy group based in your location for further guidance and also a chance to look through some other telescopes too!

Oh and let us not forget the Moon! That is always a beautiful sight!


What happens to the Gas Pressure when working out the Eddington Luminosity?

The Eddington luminosity is defined in this way - but I guess that is not the answer you are looking for!

However, if we ask what isotropic luminosity is required in order to balance the inward gravitational force on a lump of gas then yes we could consider gas pressure as well.

The correct formulation is that the pressure gradient due to gas plus radiation balances the density times local gravity.

The justification for neglecting gas pressure would depend on the application. For example, when using the mass limit to justify an approximate upper mass limit for stars, it is reasonably easy to show that the ratio of radiation pressure to gas pressure  $\sim 0.1 (M/M_{\odot})^{2}$, where $M$ is the stellar mass in solar units. Thus for stars $>3M_{\odot}$, radiation pressure dominates, and because $L_{edd} \simeq 3\times10^{4} (M/M_{\odot})\ L_{\odot}$ and $L \simeq (M/M_{\odot})^{3.5}\ L_{\odot}$ for a main sequence star, then $L<L_{edd}$ means that
$$ (M/M_{\odot})^{3.5} < 3\times10^{4} (M/M_{\odot})$$
and hence $M < 61 M_{\odot}$. At this mass the radiation pressure will be nearly 400 times the gas pressure, justifying the use of a "gas pressure-free" Eddington limit.

I suspect, though I can't give chapter and verse, that for the intense radiation fields normally associated with the Eddington limit this will always be the case.

If we are talking about spherical accretion, then because the luminosity is constant at any radius, then the Eddington limit applies to material at any radius. But at greater radii from the central source, for a constant velocity inflow, the density will fall as $r^{-2}$, and the temperature will be lower at large radii. Thus the ratio of radiation to gas pressure will likely increase with radius and so at some point it will always be appropriate just to consider the radiation pressure.

What this implies is that reaching the Eddington limit could stop the accretion, but that even if it does not (e.g. clumpy accretion), then the increasing gas pressure at smaller radii could limit the accretion if the compressed gas is unable to cool.


Determine distance from the galactic center in DS9

You cannot estimate the distance to anything or from anything in 3-dimensional space using a 2-dimensional coordinate alone.

You may calculate an angular distance (as per Mike G's answer), then assume they are all at a common distance and translate that angular distance into a physical distance, but since they aren't all at the same distance, that won't work.

To do what you want requires an estimate of the distance to these sources.


I have some questions about star formation

I'll answer your questions in order 1-4:

The term "spiral density wave" refers to the arms of a spiral galaxy, which have a somewhat higher density than the medium in between (though not much higher). A gas-forming cloud traveling through such an arm experiences tidal forces, which perturbs the cloud and may cause local regions of the cloud to exceed its Jeans mass and collapse.
The term "knots" means clumps of higher density inside the cloud that has begun to collapse.
"Such clouds" refers to the cloud out of which a proto-star is forming. That is, it doesn't refer to the proto-star itself, but to the cloud around it. However, since the proto-star will accrete mass for a while, part of the cloud will eventually become part of the star.
The gas forms a disk around the proto-star, and loses angular momentum through heating+radiation, thus decreasing in distance to the proto-star and eventually accreting onto its surface. So yes, you can call it an accretion disk, but note that accretion disks also exist in other contexts, e.g. the accretion disk around a quasar. Also, once the central tempereture of the proto-star ges hot enough to ignite nuclear fusion and the proto-star becomes a "real" star, radiation pressure will tend to disrupt the disk, blowing the rest of it away (how much depends on the radiation field, which in turn depends on the mass of the star).


What are the last images from the Galileo orbiter before impacting Jupiter?

The wonderful article linked by Wayfaring Stranger in the comments alludes to the reason for not sending photographs, but does not answer the question directly. As with most spacecraft outside Earth or Lunar orbit, the data rate is extraordinarily low. This page gives data rates for the current Rosetta mission, which are on the order of a dozen or two KiB/s. So a 10 MiB uncompressed image file would take 15 minutes to send. Though it would be an undoubtedly interesting photograph, it would come at the opportunity cost of magnetic readings, temperature readings, radiation readings, and other important scientific data which would help plan our next probe to survive the environment.


Converting rectangular to polar coordinates

You miss the simple fact that a circle is 360° periodic. Your result is correct:
360°-14.66° = 345.34°.
Thus you miss that the arctan usually returns -180...+180° and you don't correct for the fact that the longitude is positive defined in the range of 0 ... 360°. Thus add +360° when your result is negative.


I decided to answer because the role of perception is not yet taken into account.

The way we see is not only dictated by the optics of our eyes as cameras. It depends on how the brain elaborate or interprete the signals.

It is well known but still surprising that a finger at arm length can accommodate something like seven to ten moons (if not ten to thirteen, I can't look right now). Still, it quite possible that asking a person (not aware of that) something like  "how many moon can you cover with a finger?" we get an answer such as "may be half" or " one? two? Tell me...".

Specifically to Moon:

https://en.m.wikipedia.org/wiki/Moon_illusion

Photographing the Moon with a smartphone or a point and shoot camera we get things like this picture (3.5 mm, aperture 2.6).

The moon looks much smaller in the photograph than as seen in reality. The result does not look too dissimilar from the pic posted by OP.
(Let me add that my photo was cropped a bit, so the native size of the moon is even smaller).

All this is to say that whatever is aimed at giving a feeling of realism should consider perception of reality rather then size of reality.

This is done in Stellarium and many others sky viewing apps where a "realistic" option is used to alter the angular size of the moon paradoxically away from reality. I.e. the moon is arbitrarily scaled up to fit our common perception.

I think this is what the OP is dealing with.

PS I am pretty sure that my pic has been uploaded. For some reason it does not appear to be the case. If necessary let me know. By the way it was a house with sky and a small than commonly perceived Moon ;)

Per this comment here is a cell phone image with a horizontal FOV of about 60°. However in this pic the moon is not that small. I suspect the brain still enlarges it as it is down at horizon and trees are close :)


Short answer: because Neptune's argument of the perifocus, the reference point for the true and mean anomalies jumps around a lot because Neptune's orbit is nearly circular.

If you run HORIZONS w/ the following parameters:

you'll notice Neptune's mean anomaly also decreases at times.

Both of these occur because Neptune's argument of the perifocus (from which both anomalies are measured) is itself jumping around quite a bit.

Except when Neptune is actually at perihelion, the argument of the perifocus is a theoretically computed point, not representing any actual physical quantity.

The function NASA uses to calculate the argument of the perifocus is oscelt, and specifically notes:

   6) If the eccentricy of the orbit is very close to but not
      equal to zero, the argument of periapse may not be accurately
      determined.

To summarize, using the true anomaly to determine's Neptune's position is a bad idea, unless you also account for the changing argument of the perifocus.

As you note, Neptune's ecliptic longitude behaves much better, because the zero point of the ecliptic, which, while based on the Earth's precession and thus not completely fixed, nonetheless moves very slowly, much slower than Neptune's orbit.

Of course, the "best" coordinate system to use is something like ICRF, which is unchanging with respect to the fixed stars.


Is the effect when two planets passing each other different, when they pass in the same or opposite direction?

From the any of the planet's point of view, the encounter is exactly the same regardless of direction if the relative speed and force is equal. This should be pretty clear for symmetry reasons. For any encounter, the law is $v_{in} = v_{out}$. However, if the encounter takes a long time, you would also have to consider the influence of the central body of the system. The problem then becomes a three-body problem and this case is not restricted enough for an analytical solutions. We can make a few observations though: As the velocity vectors of both objects before the encounter have no radial component, both orbits are going to get a higher eccentricity. Any other information depends on knowing the distance of the encounter and the relative masses.


Any known Triple Star Systems with this hierarchy?

Short Answer:  Star sytem HD 188753 seems to be an example of the type of star system asked for.
Long Answer:
Part One of Two:  The orbital characteristics.
In any multiple star system, all of the stars will orbit the combined barycenter of the system.  Any pairs will orbit around the barycenter of the pair, and the barycenter of each pair will orbit around the barycenter of the whole system.
If there is a close pair of stars plus a third more distant star in the system, the close pair will act like a single star with their combined mass.  If their combined mass is less than that of the single star, their barycenter will orbit farther from the barycenter of the system as a whole than the single star will.  If their combined mass is greater than that of the single star, their barycenter will orbit closer to the barycenter of the system as a whole than the single star will orbit.
According to the formula found at:  https://en.wikipedia.org/wiki/Barycenter[1] the barycenter of the pair of stars will be at a point closer to the more massive star than to the less massive star.  It will be 0.444444 of the distance between the two stars, while the less massive star will orbit the barycenter at a distance of 0.55555 of the total distance between the two stars.
the more massive star should orbit at a distance at least a few times as large as the distance between the stars in the pair.  Assume that if the distance between the two stars in the pair is 1, the distance to the more massive star is 5 or 5,000 as examples.
According to my calculations, if the third star orbits at a distance of 5, the distance between the barycenter of the pair and the the barycenter of the system as a whole should be 2 times the distance between the two stars in the pair.  The third star will orbit the system barycenter at distance of 3.
According to my calculations, if the third star orbits at a distance of 5,000, the distance between the barycent of the pair and the the barycenter of the system as a whole should be 2,000 times the distance between the two stars in the pair.  The third star will orbit the system barycenter at distance of 3,000 times the distance between the two stars in the pair.
So in star systems with a a pair of stars whith a total mass greater than that of the single star, the more massive pair of stars will orbit closer to the barycenter of the system tha tthe single star will, even if it is more massive than any one star in the pair.
So the question is are there any known trinary star systems with a pair of stars and asingle stars whose mass mass is slightly less than that of the pair, though slightly more than any one star of the pair.
I'm sure there are many trinary star systems where the single star is more massive than the pair of stars let alone one of hte stars in the pair.
I'm sure there are many trinary star systems where one or both stars in the pair, and thus the pair as a whole, are more massive than the third star.
The problem is finding a system with three stars of very similar masses.
Part Two of Two:  An Example.
Using the list of trinary stars at: https://en.wikipedia.org/wiki/Star_system[2]
I found that one of the six trinary stars listed has a pair slightly more massive than the single star.
HD 188753 is a trinary system with a main star HD 188753 A with a mass of  1.06 plus or minus 0.007 that of the Sun, and a pair of stars HD 188753 B & C.  B has amass 0.96 plus or minus 0.05 that of the Sun, and C has a mass of 0.67 plus or minus 0.05 that of the Sun.  So A should be a lot more massive than C, and slightly more massive that B, but the pairr B & C together should have about 1.53 the mass of A.   Although in this case the uncertainty in the masses of A & B makes a lsigh psosibility that B could be slightly more massive than A.
In a 1999 version of a catalog of multiple stars, 551 out of 728 systems listed were trinary stars.
https://en.wikipedia.org/wiki/Star_system#Multiple_star_systems[3]
So presumably seraching multiple star catalogs should find other examples of star systems satisfying the requirements of the question.


Why are Mars northern/southern hemisphere altitudes so different?

The placement of the center of the map is messed up, but it's nobody's mistake. From chapter 4 of The Surface of Mars:

The elevation difference between the two hemispheres
offsets the planet’s center of figure from its center of
mass by 2.986 km as measured along the polar axis

As explained in another answer, the altitude is defined based on Mars' center of mass.
Regarding the question that is in the title ("Why are the hemispheres so different?"), this sharp contrast between the Northern and Southern hemispheres is known as the Martian dichotomy.
The elevation on Mars follows a bimodal distribution. That is most of the land is either 4 km under the mean altitude or 1.5 km above the mean altitude.
But not only is the altitude different between the two hemispheres, but the crustal thickness and the number of impact craters are also very different.
There are several competing hypotheses to explain the dichotomy. Here are a few.
Geological phenomena
Part of the crust was overturned
Very early in the history of Mars, there was a magma ocean that crystallized. Solomon et al. (2005) suggest that the top the mantle ended up denser than what was underneath, and it flipped.
There used to be active margins
By analogy with the dichotomy seen on Earth between oceans and continents, Sleep (1994) suggests that the northern hemisphere may be the result of new crust being generated and spreading from the center. The limit between the two areas would have been an active margin, similar to the Andes where the oceanic crust drops under the continental crust.

However, this would result in the formation of magnetic stripes that are observed in the southern rather than in the northern hemisphere.
Thermal convection in the mantle
As suggested by Zhong and Zuber (2001), there might have been strong convection in the mantle with a single convection cell.
These models succeed in reproducing some differences between the two hemispheres.
Impacts
A giant impact in the northern hemisphere
Wilhelms and Squyres (1984) suggest that a giant impact very early in the history of Mars might have caused the northern depression. They also indicate the location of the impact.
However, no thinning of the crust has been found there, like what is observed for younger and smaller craters. The ring deposits that such an impact would have created are missing. The edge of the northern plains are very irregular, when impact craters tend to be somewhat circular.
A giant impact in the southern hemisphere
Leone et al. (2016) propose that the dichotomy is the result of an impact in the southern hemisphere and that the result of the impact in more crust, rather than missing crust.
A few not quite as giant impacts
Explaining the irregular boundary between the two areas is rather difficult with a single impact, so Frey and Schultz (1988) suggest that a few big impacts might have caused the peculiar geography of Mars.
However, this hypothesis doesn't completely explain why the impact basin isn't circular.

So the question of the martian dichotomy is still open and doesn't have a definitive answer yet.
Further reading:

Mysteries of Mars by De Blasio, chapter 2
The Surface of Mars by Carr, chapter 4
Origin of the Martian crustal dichotomy, George E.McGill and Steven W.Squyres


Mass accumulation rate and orbit-time of earth

OK. Let's do the numbers. The Earth presents an circular "profile" of radius about 6 million meters to the Sun so it's cross-sectional area is about $10^{14} m^2$. The solar irradiance at the top of the atmosphere is about 1400 $W/m^2$ so the total power received from the sun is about $10^{17}W$, by coincidence, more-or-less equal to $c^2$ in SI units. So the mass equivalent of the sunlight received by Earth is about $1 kg/s$ or about $3 \times 10^7 kg/year$. This is about one part in $10^{17}$ of the mass of the Earth per year, so even if were all absorbed and none of it re-radiated (which would be thermodynamically impossible) the increase in mass is so small that the orbital change would be unmeasurable. In fact as @James K has already pointed out in a comment, more or less the same amount of energy (and therefore mass) is radiated away as heat radiation, so even this tiny change does not really occur.


How to shift an AGN X-ray spectrum to rest frame

The frequency is related to the redshift by $$\frac{\nu_{\rm obs}}{\nu_{\rm emit}} = \frac{1}{1+z}$$

Another useful relation is the fact that, for any redshift,

$$\frac{I_{\nu}}{\nu^3} = {\rm constant}$$

where $I_{\nu}$ is the specific radiative intensity.

The specific count rate (counts per area per time per energy interval) is proportional to $I_{\nu}/\nu$. (Do you see why?)

That should give you enough information to derive how the measured count rate changes with $z.$ Of course, to know the actual value, not just the function of $z$, you'd need to know the actual count rate measured, not just the normalized value.


Is atmospheric turbulence irrelevant for ExoPlanetary transits and radial velocity measurments?

One can imagine turbulent eddies in atmosphere as very weak optical lenses which focus and defocus stellar radiation. This leads to image degradation (seeing) and fluctuations of flux being registered through some aperture. The latter effect is called scintillations. It is very prominent for naked-eye observations. For telescopes the averaging by large aperture reduces the magnitude of the scintillations. Nevertheless it is main limiting factor for high-precision photometry at telescopes larger than ca. 2 m. See e.g
http://adsabs.harvard.edu/abs/2012MNRAS.426..647K

As for spectrscopy, atmospheric turbulence does not affect the wavelength of radiation as long as in modern high-precision spectrographs (e.g. HARPS) actual detection takes place in vaccuum.


How does gravity affect pressure under water?

For a liquid, hydrostatic pressure is $\rho g h$ where $\rho$ is density (this is always the same for all water) g is gravitational acceleration and h is depth.

The gravitational acceleration on Europa is 1.3 $\text{m/s}{}^2$ (compared with 9.8 $\text{m/s}{}^2$ on Earth). But on Europa there is 20km of ice floating on the water.

As a rough estimate, the gravity in Europa is 1/10 that on Earth, so the pressure at the bottom of the ice/top of the water would be comparable to the pressure at 2km in the Earth's Oceans.

Since we can build submarines that can operate at 10km, we could reasonably suppose that we could operate submarines down to about 100km on Europa, which takes us to the rocky surface. However we have no way to pass through 20km of ice crust to reach the water layer. Building the submarine may be the easy part!


how does redshift prove expansion is accelerating?

You're right, a redshift will give you a measure of a velocity, not an acceleration. I'm assuming when you say acceleration, you're referring to the accelerating expansion of the universe so you're not looking at the particular acceleration of an object (which you could get from examining the force of gravity imparted on that object).

So in addition to a redshift, you need to find a way to independently measure the distance. To get out to an appreciable redshift (z~0.5), Type Ia supernova are used as "standard candles". They are considered standard because they're believed to be from white dwarfs that explode because they have reached their maximum possible mass, the Chandrasekhar limit, of 1.4 solar masses. Therefore, when they explode, they should all look about the same and any difference in brightness would be from how far they are away from us. So they measured the distance and the redshift of the galaxy the supernova originated from to determine the expansion of the universe appears to be accelerating over time.


First thing to mention is all our limited knowledge (best guesses) from back than only goes back to the fraction of time just after the big bang not the creation moment itself. Moreover you have to remember that spacetime itself was also created at that moment as well as all the stuff in it. This means that using the term singularity as we understand from the core of a blackhole, which is to say an infinitely dense and infinitely small point/puncture IN the fabric of space doesn't really marry up. The universe no matter how tightly packed conditions were at this early stage had no space to warp/puncture into.

The universe has no centre, it began everywhere, simultaneously and instantly with all matter distributed uniformly and in equal density. Without a central point all matter was pulling evenly on its self in all directions so wouldn't coalesce at a single point and collapse down as it does in a black hole.

And as to your mention of supermassive black holes sucking in parallel universes "they don't"

Hope this helps


3 years later, how strong is the evidence for planet 9?

It's worth noting that Siegel's article seems to extrapolate from one paper (Shankman et al., 2017) to "most scientists" without much justification, so it's not particularly clear what most scientists think anyway (and that's before we get into the question of whether the "most scientists" referred to includes scientists whose discipline is not astronomy-related). Usually in these kind of situations there's a lot of back-and-forth as new objects are found in the outer solar system and the analyses of systematic errors and biases in the various surveys are refined. If there is an actual detection of Planet 9, that would resolve things fairly definitively, the reverse situation is going to be rather more difficult to prove.

In response to the OSSOS paper (Shankman et al., 2017) Michael Brown and Konstantin Batygin released a new analysis of the clustering in the outer solar system, taking into account the OSSOS results that are the subject of Ethan Siegel's article. The paper can be downloaded from the arXiv. Brown also has a Twitter thread summarising the result. From the paper, their conclusion on the apparent absence of clustering in OSSOS:

  That is,
  the uncertainties in the measurement of clustering from
  the OSSOS data are so large that OSSOS would not be
  capable of confidently detecting the clustering seen in
  the larger data set even if it were real and present in the
  OSSOS data. Because of the limited survey region and
  small number of detected objects, OSSOS observations
  are equally consistent with being drawn from a uniform
  distribution of longitudes of perihelion and with being
  clustered in longitude of perihelion as strongly as seen
  in the ensemble data. No conclusions on clustering of
  longitude of perihelion observed in the complete dataset
  can be drawn from the OSSOS data.

They do note that this does not necessarily prove the existence of Planet 9, but in the absence of Planet 9 there would still need to be an explanation for the clustering.

Batygin et al. subsequently posted a review of the Planet Nine hypothesis on the arXiv, favouring a lower planetary mass and a less eccentric orbit than originally proposed. The new parameters do not favour Planet 9 as an explanation for the solar obliquity.

On the anti-Planet 9 side of things, it's worth looking into Kavelaars et al. 2019, a subsequent paper from the same authors as the study based on the OSSOS data, who note that there seems to be an absence of objects with perihelia ($q$) between 50 and 75 AU that would be expected if a planet was causing the alignment, which so far doesn't seem to be explainable by detection bias.

  A survey
  that might have detected a TNO at $a ∼ 500\ \mathrm{au}$
  and $q ∼ 75\ \mathrm{au}$ is actually more likely to have detected objects with similar $a$ but smaller values
  of $q$. The same is true of the other $q > 75\mathrm{au}$
  detections: the lower-$q$ but similar $a$ detections
  are always more likely. Thus, the lack of detections in the $50 < q < 75\ \mathrm{au}$ range may be indicating that there really is an absence of TNOs
  on orbits in this range. This strongly contradicts models of orbital evolution that include
  an additional planet, as the gravitational action of such an object would cause TNOs to
  be distributed across a range of $q$ values at any
  given moment (Figure 2; Shankman et al. 2017a;
  Lawler et al. 2017). Thus, if the lack of objects
  in the $50\ \mathrm{au} < q < 70\ \mathrm{au}$ range is real, the
  hypothesized external planet can be excluded.

It still is an active area of research, so it remains to be seen how the various studies will hold up.


Why can I see sometimes a horizontal half moon instead of a vertical one?

It occurs because the lit half of the moon points towards the sun, along a great circle in the sky.  The half-moon will be 90 degrees away from the sun in the sky. So by (say) 8pm (at equinox for simplicity) the half-moon will be visible in the South west, but the sun will be below the horizon.  The great circle from the moon to the sun will go "down" relative to the horizon.
The angle that it points down will depend on where you are on Earth. Only near the equator could the moon be completely horizontal,