The Kuiper Belt is fairly large (with inner and outer radii $\sim30\text{ AU}$ and $\sim50\text{ AU}$), but it does not contain much mass. The total mass is likely somewhere between $0.01$ and $0.1$ Earth masses[1], [2], which is essentially a mass range from five times the mass of Pluto to the mass of Mars. Now, it was likely much more massive in the past - its origin and mass loss are tied to the evolution of the orbits of ice giants in the early Solar System (see in particular the Nice model) - but we're talking about the formation of planets in the future. At present, there simply isn't enough mass to go around.

Let's talk about "clearing the neighborhood". The IAU's definition of a planet merely states, as one of its three criteria, that a planet must have

  cleared the neighbourhood around its orbit.

Wikipedia gives several quantitative measures of whether or not an object has cleared its neighborhood. The quantity $\Pi$ (Margot (2015) is given by
$$\Pi(m,a)=\frac{m}{M^{5/2}a^{9/8}}k$$
where $m$ is the mass of the body (Earth masses), $M$ is the mass of the star (solar masses), $a$ is the radius of its semimajor axis (AU), and $k\simeq807$. A body is a planet if $\Pi>1$.  If the mass of this hypothetical planet, $m_H$, is $0.01$ Earth masses, then no orbit in the range $30\text{ AU}<a<40\text{ AU}$ gives me $\Pi(.01M_{\oplus},a)>1$. If $m_H$ is $0.1$ Earth masses, then almost all $a$ in the same range give me $\Pi(0.1M_{\oplus},a)>1$. That said, it is extremely unlikely that a body could amass even a small fraction of the total mass in the Kuiper Belt - remember the inner and outer boundaries!

The quantity $\Lambda$ (Stern & Levison (2002)) is given by
$$\Lambda(m,a)=\frac{\mu^2}{a^{3/2}}\times\left[\text{Terms accounting for the properties of the body's orbit}\right]$$
where $\mu$ is the ratio of the mass of the body to the mass of the star. The author's list a value on the order of $\Lambda=3\times10^{-3}$ for Pluto; if we assume that the orbital terms are about the same for this body, we find $\Lambda(0.01M_{\oplus},a)\sim0.075$ and $\Lambda(0.1M_{\oplus},a)\sim7.5$, with the cutoff being $\Lambda=1$. Again, it is not possible for the body to accumulate the total mass of the Kuiper Belt, so we can safely assume that $\Lambda\ll1$.


can moon rise before sun and set after it in the same day

It's possible, but it's not very impressive.
This month (August 2020 New Moon) from, say central Asia (I used Aral for my simulation on Stellarium) The moon rises at 05:48 local time (UTC+5) and the sun a few minutes later at 06:01
Of course you can't see the moon rise, it is the day of the new moon and it is invisible.
Then in the evening, the sun sets at 19:51 and the moon sets at 20:28.  Again, it is not visible at this time.  (I've not accounted for atmospheric refraction here, which will change the times but not the example.)
This doesn't happen from all locations every month, it just depends on whether the sun passes the moon during the day or during the night, I'd venture a guess that in any particular month, half the world will have the moon rise before, and set after the sun on the date of the new moon.
So this isn't valid for all locations.  If you were in America, then the moon would pass the sun during the night: on one day the moon would rise and set before the sun, the next day it would rise and set after the sun.
Its is also not valid for all dates at a single location.  It is possible that the New moon in September might pass the sun during the day when viewed from America, but not from Asia (though I haven't checked)


What has GAIA found out about the dwarf galaxy heritage of stars in the Milky Way, if anything?

Malhan et al. (2018) summarise progress so far (the Gaia astrometry is improving all the time as more data are collected) using Gaia DR2 data. In particular they produce "panoramic maps" of streams in the Galaxy halo, which I believe is exactly what you are looking for (see below).

Coherent tangential velocity structures, a.k.a. "streams", seen in the southern Galactic hemisphere by Gaia DR2 - from Malhan et al. (2018).


Why is Earth's rotation considered fast?

The interactions of the Earth and its moon have slowed the Earth's rotation significantly over the billions of years since their formation.  Wikipedia cites a paper that gives a figure of approximately five hours for rotational period in the past.  We don't know of a similar method to slow down Mars.

This implies that the Earth must have had a much faster rotation than Mars in the past.


How much does the sky change in a few thousand years?

Here's part of the sky in the year 1

It is part of the sky you may know well,  Orion and the dogs.  I've marked the current positions of Sirius, Procyon and Betelgeuse, with green markers so you can see how their positions have changed over 2000 years.  It's not a lot.
The first thing that the Babylonian astronomer might notice is that there is a pole star, one that is fixed. 2000 years ago, Polaris was more than 10 degrees from the True North. Now it is less than 1 degree.  It isn't immediately obvious from a casual glance at the sky, but it would be noticeable and surprising to Babylonian astronomer.
If they started looking at the position of stars at a particular time and a particular date, they would find that things are in the "wrong" position.  Of course "on a particular date" assumes a calendar, and the Babylonian calendar was not the Gregorian calendar. With more measurement they would find that the point of Aries is now in Pisces. Again, its not something you would notice at a glance, but it is within the ability of a Babylonian astronomer.
Now the stars have actually moved since year 1.  But it would not be casually noticeable.  It would be within the ability of our Babylonian astronomer to measure, if we asked them to.  The relative positions of Sirius and Betelgeuse have changed.  The Babylonians could measure angles in the sky precisely enough.  But we don't know if they did, and if they did we don't know if they wrote them down.
So our Babylonian would not look up and immediately say "something's wrong".  But if we 1. Prepared him by telling them to take careful measurements of Polaris, Sirius and so on. and 2. Gave them some prompts on what to look for in terms of proper motion and precession,  then it would be possible for them to observe some changes.


No sir/madam. An astronaut or a cosmonaut cannot walk on Mercury's surface. The reason is plainly simple. During daytime, its temperature surges up to 430 degree C making it 2nd hottest planet after Venus. Whereas, night time temperatures are as low as -180 degree C.

Despite of having ice caps at its polar regions which might seem attractive, its totally not feasible to land there and then mine the frozen ice to make up a base.

For a much more descriptive read, kindly refer this link : What it would be to live on Mercury


Why Do Planets Revolve Faster When they Are Closer to Their Parent Star?

Velocity is a form of kinetic energy, while height within a gravity well is a form of potential energy. For an orbiting body, conservation of energy will keep the total energy constant.

So as a planet moves away from the parent star, it loses velocity and gains potential energy. As it moves closer, it trades the potential energy back for velocity. The point of lowest potential has the highest velocity and the point of highest potential has the lowest velocity.


Can black holes collide with the Sun? In that case what will happen?

Yes, in theory. But;

No, this is very unlikely to happen during the life of the sun. If it did happen it would be Bad.

Many black holes are much more massive than the sun, even though they are much smaller. If a black hole were to enter the sun, some the matter that forms the sun would form into a accretion disc around the black hole and eventually fall into the black hole. The rest of the sun would be utterly disrupted by the massive release of energy as matter falls towards the black hole.

The solar system would be wrecked. The Earth, even if it survived intact, would not be able to support life.

Fortunately black holes are rare, and space is BIG. The chances of a black hole coming anywhere near the sun in the next billion years or so is very very small. There are more pressing things to worry about.


Can information leave the observable Universe?

Good question! I will need to go into some cosmology to answer it, my main source being Davis & Lineweaver (2003).

Nothing traveling through space should be able to leave the observable universe. However, information can and will inevitably leave our observable universe. How is this possible?

The Universe is expanding.

If we lived in a static universe, no objects would leave our observable universe. However, in our accelerating universe, space is being created between objects, with farther objects drifting away faster than closer ones.

Einstein said nothing can travel through space faster than light. However, space is literally being created between objects, so they can be pushed away from each other faster than light. So while the observable universe is expanding at the speed of light, distant objects are moving away from us even faster.

And the result? Take a look at this image:

This describes how the observable universe will evolve. The Hubble sphere defines the area in which objects are receding from us at or below the speed of light. Beyond this, objects are far enough to recede from us faster than light.

The light cone is the region of the Universe capable of influencing us today. Essentially, it is the observable universe, and its boundary is clearly marked. It is, as you mentioned, growing at the speed of light.

The light cone does not contain the entire Hubble sphere, since photons from many objects receding faster than light cannot reach us yet. Conversely, the Hubble sphere doesn't contain the entire light cone. This is because photons emitted by objects outside of the Hubble sphere can crawl into regions of space that are receding from us a bit slower than the one from which they were emitted. They can then crawl into regions that expand slower than the last one, and so on. Eventually, the photons will reach regions of space that aren't expanding faster than light, and they can reach us.

Now, there is a distance beyond which space expands faster than the light can traverse it. This distance is called the cosmic event horizon. Light from beyond the event horizon will never reach us. Our observable universe grows because more photons come into our view. However, the observable universe will infinitely approach the event horizon, since light from beyond there can never reach us.

Thus, after an incomprehensibly long time, the observable universe would stop growing. At this point, it may finally be possible to travel through space beyond the observable universe.


Helium rain inside metallic hydrogen within Jupiter, conditions and experiments?

Different sources give significantly different values for the conditions under which hydrogen shifts from molecular to metallic. See the discussion in the comments associated with this question. Pressures from about 20 to 300 GPa and temperatures very broadly around 10 000 K seem to be the most common.

Experiments can only just reach these pressures, at lower temperatures, for tiny volumes of material, and the question of whether anyone has made metallic hydrogen at all remains controversial. Certainly no one can make it repeatably. (wikipedia has a reasonable summary of the history) or make enough of it and keep it for long enough to check to what extent, if any, helium is soluble in it.

That said, analysing from theory whether it will be soluble shouldn't be that hard, since both hydrogen and helium are relatively simple atoms. An early paper concludes that the hydrogen would be at or close to the limit of how much helium could be dissolved. I'm sure there is more accurate recent work, but I couldn't find it quickly.


Predicting accurate time of local sunrise in mountainous area?

The PeakFinder app1 can do most of what you're looking for.  It's available for Android and IOS as well as through a web browser.  The app shows the paths across the sky for both the sun and moon2 and displays rise/set times taking into account the local topography and viewpoint elevation3.  At least the Android version works without a network connection (nice for hiking).  IIRC, it uses SRTM data4 for elevations so it's not super-accurate but does a good job over all.

I plugged in the location, elevation (ground + 1m), and date you gave and got sunrise at 05:09 local time, reaching the ridgeline at 08:37 local time.  Here's a link to the page; screenshot is below.

1 I'm not affiliated with PeakFinder, just a satisfied user.

2 If using a web browser, click on the "sun" and "moon" icons on the page's toolbar.

3 It also shows what the surrounding terrain looks like and the names of visible peaks.

4 As does OpenStreetMap, IIRC.

Source: Cropped screenshot of PeakFinder site


What is the upper bound of number of stable interacting stars in a star system?

There is no true upper bound. You could argue that a galaxy is a vast multiple star system, with hundreds of billions of stars. Natural galaxies are not perfectly stable (random encounters eject stars over very long timescales), but one can construct stable systems by taking pairs of similar-mass binaries and placing them into a remote circular orbit around each other. This can then be repeated indefinitely, with $2^n$ stars forming a vast system.
In practice such neat systems are unlikely to form naturally (4 levels seem to be the max observed and often involves very unequal masses) and very wide systems will encounter other stars often enough to be destabilized. Eccentricities tend to have a triangular $f(e)=2e$ isothermal distribution, making things less stable. See (Duchêne & Kraus 2013) for more statistics.
(As an aside, Jaron Lanier and Piet Hut have suggested constructing such vast hierarchical systems as a clear indicator that there is intelligent life present.)


When stars explode why can we still see them?

Because we are not moving at the speed of light.

From our point of view, light travels at the speed of light, and so events on a distant star are not visible until the light reaches us.

From the point of view of the photon, the universe is 2 dimensional (it is flattened by infinite Lorentz contraction) and there is no passing of time (infinite time dialation). The star and you are at the same point. Photons have a weird point of view.

Fortunately we don't have to worry about that, because we have mass and so we are not moving at the speeed of light. We have a 3d universe and we have a dimension of time. Thank Higgs we have mass!


What happens to a star after it has become a red giant?

It will become a planetary nebula like e.g. the Cat's Eye nebula that was formed by the death of a star with about the same mass as the Sun ~ $1\ M_\odot $:

              Composite image using optical images from the HST and X-ray data from the Chandra X-ray Observatory

The Sun, and any red dwarfs above about 0.25 solar masses, will expand into what's called a red giant, a late stage of stellar evolution. At this stage, the star starts to fuse different elements, and eventually throws off its layers as a planetary nebula, leaving behind a white dwarf made of carbon and oxygen.

A red dwarf that is too small to become a red giant will not turn into a planetary nebula: its fusion processes will eventually cease and it will probably produce a white dwarf made mostly of helium. But the main-sequence lifetime of these very small stars is longer than the age of the universe, so this has never actually happened yet. The helium white dwarfs that do exist were formed in binary star systems (and were formed during their complex dynamics).

Red dwarfs are small main sequence stars - the smallest and dimmest that are still able to fuse hydrogen in their core. Brown dwarfs are even smaller (in terms of mass), but cannot even fuse hydrogen - they are thought to be able to fuse deuterium and lithium. Stars smaller than a brown dwarf turn out to be gas giants like our Jupiter, Saturn, Uranus and Neptune.


Time dilation due to the expansion of the universe

Hubble's law
$$
v=H_0 d,$$
relates the recession velocity $v$ of a distant object to it's physical distance $d$. Today, the physical distance coincides, by definiton, to the comoving distance $\chi \equiv d/(1+z)$, but if you want to know how fast two galaxies at a redshift of, say, $z=1$, you must plug in their physical distance at that time (and use $H(z)$, the Hubble parameter at that time, rather than $H_0$). Depending on how you measure their mutual distance, there are various ways to calculate this.
When we observe a galaxy at redshift $z$, general relativity predicts that we see its time dilated by a factor $1+z$. That is, some physical process which takes a proper time $t_0$, will be observed to take a time
$$
\boxed{t = t_0(1+z),}
$$
which is the formula you're after.

A good example of this is the time it takes for the brightness of a supernova to decline; a supernova at $z=1$ declines half as fast as a local (i.e. $z=0$) supernova (see e.g. Goldhaber et al. 2001).

Calculating the distance to that supernova, or the age of the Universe at the time it's observed, is a bit more complicated, involving a model of the expansion rate of the Universe (see e.g. this question about the Friedmann model).
Taking inflation to end when the Universe was $t\sim10^{-32}\,\mathrm{s}$ old, the corresponding redshift was of the order$^\dagger$ $z\sim4\times10^{25}$, so if we could observe physical processes there, they'd be dilated by a factor $\sim4\times10^{25}$, so 1 second would, at this dilation, take a 3 billion years. However, already 1 second after inflation, the expansion rate of the Universe had already slowed down so much that the corresponding redshift, and hence dilation factor, was "just" $\sim4\times10^9$. So the next second would only, from our point of view, seem to take $\sim140$ years.

Note that these considerations don't really have anything to do with the accelerated expansion, but is just a consequence of the expansion.

Note also that special relativity predicts the same time dilation; that is, if galaxies were flying through space rather than just being carried along with expanding space, we would observe the same relation between redshift and time dilation. However, the special relativistic interpretation of redshifts is incompatible with the relation between redshifts and magnitudes of supernova, and can be ruled out at $23\sigma$ (Davis & Lineweaver 2004).

$^\dagger$This can be calculated as follows: Until the Universe was $t_\mathrm{eq}\sim50\,000\,\mathrm{yr}$ old, corresponding to a redshift of $z_\mathrm{eq}\simeq3400$ (Planck Collaboration et al. 2018)
 its dynamics was dominated by the energy density of radiation, causing the size (scale factor $a=1/(1+z)$) to evolve with time as $a\propto t^{1/2}$. Calculating backward from $t_\mathrm{eq}$ yields $a\sim10^{-26}$ at $t\sim10^{-32}\,\mathrm{s}$, corresponding to $z\sim10^{26}$.


What are possible methods of discerning a starburst AGN composite galaxy?

For starbursts, their star formation rates are best estimated by a combination of their ultraviolet and infrared emission. [O II] and [O III] as well as Hydrogen recombination lines can be used to estimate star formation, but like the UV, these suffer from extinction.


What is the theoretical ages for Pop I stars

Since Pop III stars are the oldest stars, then they could have theoretically began to form in the Cosmic Dark Age (CDA), see section 5, since the oldest galaxy we've observed so far (in infrared) is about half way through the CDA (~$480$ million years). As the article states,

  Stars and galaxies are formed when dense regions of gas form due to the action of gravity, and this takes a long time within a near-uniform density of gas and on the scale required, so it is estimated that stars did not exist for perhaps hundreds of millions of years after recombination.

So not only must the temperature of the universe be sufficiently cool, gravity must have acted for a long enough amount of time for baryonic structure to begin to form.


Is it true that black holes emit only photons, electrons, and positrons in the form of Hawking Radiation?

When you look at Hawking's 1974 Nature paper (https://www.nature.com/articles/248030a0), it is very clear that like thermodynamic blackbodies, black holes can emit anything they want. Or to be more precise, the thermal radiation that is produced by gravitational vacuum polarization in the vicinity of the black hole event horizon is not restricted to photons.
But what else could it be?
Photons are there, obviously. And their peak wavelength is determined by the temperature of the black hole or blackbody. Unless that temperature is several billion degrees kelvin, there just isn't enough energy to form electron-positron pairs in meaningful quantities. (Astrophysical black holes have temperatures in the nanokelvin range.) Neutrino-antineutrino pairs perhaps, but the extreme short range of the weak interaction effectively limits their radiation. Gravitational thermal radiation (assuming gravity is quantized and gravitons exist - see Dyson's Poincaré prize lecture, section 5 in particular: https://publications.ias.edu/sites/default/files/poincare2012.pdf ) might also be present but again, close to twenty orders of magnitude less intense than photons.
If the temperature is high enough for electron-positron pairs to form, sure, those can also come out of the black hole. As well as proton-antiproton or neutron-antineutron pairs. But these hot black holes would be so small (just a few billion metric tons, with Schwarzschild radii in the subatomic range), they would have to be either primordial black holes or astrophysical black holes near the end of their evaporation, in the extreme distant future.
What else? There are no other stable particles for the black hole to emit. If a sufficiently hot black hole produces, say, a W+/W- pair, those would very rapidly decay the usual way into positrons (electrons) and electron (antielectron) neutrinos, or their muon/tau flavor counterparts (but those, too, would decay.)
So in the end, just like in the case of a thermodynamic blackbody, we find that although in principle, a black hole can emit other things, at non-extreme temperatures, the only thing it can emit with any efficiency is photons.


How fast are we moving relative to the CMB?

Yes, our (i.e. the Sun's) motion in the "global", or comoving, reference frame can be measured accurately from the dipole of the cosmic microwave background. The latest results from the Planck Collaboration et al. (2018) yielded a velocity of
$$369.82\pm0.11\,\mathrm{km}\,\mathrm{s}^{-1}
$$
in the direction
$$
\begin{array}{rcl}
\ell & = & 264.021º\pm0.011º\\
b    & = & 48.253º\pm0.005º
\end{array}
$$
(in Galactic coordinates).

Since Earth orbits the Sun with some $30\,\mathrm{km}\,\mathrm{s}^{-1}$, there's a small, biannual correction to this result. On much larger timescales ($\sim100\,\mathrm{Myr}$) our motion round the Milky Way alters our comoving velocity with the order of $\sim100\,\mathrm{km}\,\mathrm{s}^{-1}$.


Would the solar corona be visible from the distance of the outer planets?

If the sun is ever perfectly eclipsed as it is on Earth, then yes, the corona would be visible. The only reason we can't usually see the corona is because it is way dimmer than the rest of the sun. If the entire sun other than the corona is ever covered fully by another object, then the corona will be visible.


How can I calculate when the new moon will be observed from a particular meridian?

You need an Orrery. Specifically, one that includes Earth’s rotation.

My noting the meridian on the model Earth, you can see how the Earth may be rotated some amount each time the Moon passes on a line between the Earth and Sun.

The wikipedia page gives a formula.  Take the result (given in UT) and apply the timezone offset for the desired longitude.

The same page discusses Lunar calindars and may have the info you are really interested in.


Measure an angle in the sky from a photo?

A plate solver matches the stars in an image to a catalogue of star positions.
Astrometry.net provides a web service and a software API for this.
Given only another user's cell phone photo, Astrometry.net determined that the image scale was 112 arcseconds per pixel.
It also estimated an image size of 72 by 54 degrees, apparently using an approximation which is only valid for small angles.

Assuming low distortion (not a fisheye lens), the image is a gnomonic projection of part of the celestial sphere.
If $r$ is the distance from the image center in pixels, $p$ is the pixel scale in radians per pixel, and $\theta$ is the angular separation from the center in radians, then

$$rp = \tan \theta$$

For example, I used the measurement tools in GIMP and Stellarium on two bright stars near the center to verify the central image scale of 112 arcsec/pixel or 0.54 mrad/pixel.
The edges of the 2320x1740 image are 1160 and 870 pixels away from the center.
Solving for $\theta$, the edges are 0.56 and 0.44 radian away from the center, so the image's angular dimensions are 64 by 50 degrees from edge to edge.

Daytime images would of course have the same pixel scale and angular dimensions as astrophotos taken with the same equipment.


It could start a fire if the screen is at the focal point of the optical system. That is how you light fires with a magnifying glass.
Here, the blackboard is likely away from the focal point, so you can see the shape of the eclipse (and you get a bigger image) without setting things on fire.
Although this is fairly safe, there are a few things to pay attention to:

If you do this, make sure nobody can walk between the telescope and the screen, because if they go near the focal point, they could get very hot.
Doing this will cause your telescope to heat up. If there are any plastic parts, they can melt.
The telescope in the picture seems to have a small opening. Don't do this with a big telescope. You don't need to collect a lot of light.

Not an answer to the question, but an important note: Observing the Sun is the most hazardous thing you can do in astronomy. Make sure you know what you are doing before you try.


How would astronomical seeing on Mars differ from that on Earth?

As far as I know, "seeing" (or rather the effects influencing optical wave propagation) is caused by turbulence in the atmosphere.

Using the Reynolds number Number $ Re = \dfrac{\rho L v}{\mu}$ as a measure for turbulence:

density $\rho$ drops due to the reduced pressure (about 1/100 earth pressure), additionally the gravity is smaller than on earth
characteristic length $L$ will stay similar
average wind speed $v$ is about 2 times higher than on earth
the dynamic viscosity $\mu$ of an (ideal) gas is independent on the pressure, and the temperature dependency can be approximated with $\sqrt{\dfrac{T_{mars}}{T_{earth}}} \approx 0.9$

So for an average martian day an atmospheric Reynolds number would be much smaller than on earth and I would expect a far better seeing due to less turbulence

Additonally, I would guess that the effects of a thinner and dryer atmosphere would help as well, since the index of refraction is dependent on the gas pressure as well.


Is it correct to say that only binary and solitary star systems are stable?

No.
It isn't proven that any real orbital systems are stable on sufficiently long time scales.  Even our (single star) Solar System is not stable.  In their paper: On the Dynamical Stability of the Solar System, Batygin and Laughlin, use the numerical techniques pioneered by Laskar to show scenarios in which various planets fall into the Sun or are ejected from our Solar System on a timescale of 20Gyr.
The area of math that dominates long term orbital stability questions is Chaos Theory, which studies the types of systems of differential equations for which very tiny changes in the initial conditions cause very different resultant solutions.  Over long time scales, tidal forces, stellar radiation, and other energy sources cause orbits to evolve.  When resonances are reached, orbits can begin to de-stabilize.  Examples of this can be found in the Kirkwood Gaps of the asteroid belt.
Given the overwhelmingly high populations of binary star systems in the Milky Way (table below from wikipedia), we could probably call binary and single star systems stable at least at 10-100Gyr time scales.

However,Correa-Otto and Gil-Hutton point out that merely being in a galaxy causes binary orbits to be perturbed:

The main influence of the Galaxy on the WBS [Wide Binary Star systems] are
twofold: first, the disruption of the binary system, which produces
gaps in the orbital elements distributions of the three sets; and
second, the changes in the orbital configuration of the surviving WBS,
which modify the orbital element distributions of the population in
each set.

Given the paucity of discovered n-star systems where n is greater than 2, I think we can safely say these systems are mostly unstable at timelines on the order of 1-10Gyrs. In fact, many of them are unstable on much shorter timelines.


Moons with curlicue paths around our Sun?

A moon will describe a path like this if its orbital speed relative to its parent planet is greater than the parent planet's orbital speed about the Sun.  (Assuming the moon's orbit about the planet and the planet's orbit about the sun are roughly coplanar;  I'll ignore Uranus for the remainder of this discussion.)  Going through the planets:

Neptune:  All moons from Proteus inwards (7 total) orbit Neptune sufficiently fast.  However, they're all pretty small;  in particular, all are irregularly shaped, since they're all too small to form into spheres under their own gravity.
Uranus:  (ignored because the paths of the moons are just too baroque)
Saturn:  All moons from Dione inwards (20 total), along with Dione's trojan moons Helene & Polydeuces, orbit Saturn sufficiently fast.  This includes a few large & charismatic "pre-Voyager" moons: Mimas, Tethys, Enceladus, and Dione.
Jupiter: All moons from Europa inwards (6 known), including Io, orbit Jupiter sufficiently fast.

It's possible that there are some satellites of dwarf planets that might also make this list, but I didn't spot any immediately.  In general, a satellite of a dwarf planet would have to be orbiting its parent extremely closely to make this happen, since the orbital speed of a satellite is roughly proportional to $\sqrt{M/r}$, with $M$ the mass of the parent and $r$ the radius of the satellite's orbit.


There are many reasons why nothing has been detected. We could be too far away, no one is listening, listened at wrong time, and many more. But this does not mean there is no life on other planets. We just haven't discovered them yet.


Why did the then PhD student Didier Queloz get the Nobel prize?

He shared the 2019 Nobel Prize "for the discovery of an exoplanet orbiting a solar-type star".  It was shared with Michel Mayor. Maybe Queloz did much of the work and Mayor thought he rightfully deserved some of the credit. Queloz only got a fourth of the money because the prize was split between Mayor-Queloz and James Peebles. Peebles got his half for "for theoretical discoveries in physical cosmology."

This was a very unique case because it involved no new physics really just the discovery of what most astronomers knew existed.


How small a star can provide Sun-level illumination to its planets?

I consider small stars to mean main sequence stars (or brown dwarfs) with mass less than the Sun, and then also consider compact stellar remnants - white dwarfs and neutron stars.

Main sequence stars and brown dwarfs

What you need is a mass-luminosity relation combined with an expression for the tidal radius in terms of the stellar mass. The latter also depends on the mass of the planet, so when you say earth-sized, I'll assume that means mass and radius.

So going through the calculation.

Flux at the planet is $L/4\pi r^2$, where $L$ is the luminosity and $r$ the orbital radius (assumed circular).

Let's next use an approximate relation that
$$\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^3$$
A more accurate numerical relationship could be obtained from evolutionary model calculations. There is a complication that the luminosity of very low mass stars and brown dwarfs does depend on age.

For a rigid body the tidal radius (Roche limit) is
$$ r_t\simeq 1.4 R_E\left(\frac{M_{\odot}}{M_E}\right)^{1/3},$$
where $R_E$ and $M_E$ are the radius and mass of the Earth.

So setting $r=r_t$, the flux to $1400$ W m$^{-2}$ and replacing $L$ in terms of mass, we find
$$\frac{M}{M_{\odot}} = \left[\frac{4\pi \times 1400}{L_{\odot}} (1.4 R_E)^2 \left(\frac{M_{\odot}}{M_E}\right)^{2/3}\right]^{1/3}.$$

All that remains is to put the numbers in and I get $M = 0.026 M_{\odot}$.

So, a small brown dwarf - but there are caveats. First, and most importantly, as I said, the luminosity-mass relation is a bit rough and ready, and is certainly age dependent for something as low mass as this. Second, the expression for the Roche limit depends a bit on structural properties of the planet, but I think this introduces relatively little uncertainty.

If you use that mass and the tidal radius formula, you find that the orbital radius is only $5R_E$. As the minimum size of a brown dwarf is about a Jupiter radius, it appears that tidal breakup would not be the limiting factor for an Earth-like planet. Also note that the spectrum of radiation from a brown dwarf might contain the same power per unit area as sunlight, but it has a very different spectrum; mostly infrared with almost nothing in the visible band.

Compact stellar remnants

Smaller "stars" are possible in the form of the collapsed remnants known as white dwarfs or neutron stars. Planets have been detected around neutron stars.

White dwarf stars are "born" very hot, cool rapidly and then more slowly. To answer your question requires an addendum, which is a minimum time you want this illumination for. A newborn WD has about a solar luminosity and has a similar diameter to the Earth. So you planet could be 1 au away from this and still receive a sun-like power per unit area. BUT, it would be short-lived and the spectrum of the light would be nothing like the Sun. The WD would have a temperature in excess of 50,000K and most of the light would be in the UV.

Cooling curves (and observations) suggest they cool to $10^{-4}$ solar luminosities in less than a billion years and have more sun-like temperatures, but to receive sun-like illumination at this point, your planet would need to be only 0.01 au from the white dwarf. White dwarfs have a mass a little less than the Sun. Using the same tidal radius formula as above we get $r_t \sim 0.004$au. So I conclude that yes, it might be possible to have a long-lived planet with Earth-like illumination around a white dwarf. But note (a) conditions change rapidly (compared with the Sun) and (b) you have to arrange to get the planet that close, since an orbit of that size would have been inside the red giant progenitor of the white dwarf.

Neutron stars are born in the cores of supernovae explosions and have a radius of about 10 km and mass about 1.5 times that of the Sun. If your planet can survive the supernova (or be constructed afterwards) it faces a similar set of problems to a planet around a white dwarf.

Neutron stars are born with surface temperatures of billions of degrees, emitting copious gamma ray radiation. They cool very rapidly - within a million years the surface temperature is down to an X-ray emitting million degrees and perhaps to UV-emitting temperatures of tens of thousands within 100 million years. It may be that they never cool much below this, since they are heated by their rotational spin-down, the decay of their strong magnetic fields and accretion of gas from the interstellar material. Even at 50,000K, because of its small size, the neutron star has a luminosity of only a millionth that of the Sun. That means your planet would have to be moved within 0.001 au of the neutron star to receive the same power and this would be well inside the tidal breakup radius of $\sim 0.005$ au.

I conclude that there really is no long-term possibility for sun-like illumination from a neutron star. Note also, that I have ignored magnetospheric and pulsar-like emission from the neutron star. These are also short-lived phenomena only seen in the first million years or so of a neutron stars existence.


I hope the following doesn't come across as self-promotion, but you can have a look at the source code for my near-space 3D map:

http://beyondproxima.com/

Just View Page Source.

It's not perfect - the 3D projection is not geometrically accurate, and some brighter stars are duplicated. But it might give you a starting point?


how to measure temperature of the distant star

To measure the surface temperature of a star it is used is it's black body spectrum. You would have to get the light curve of the light from tat star and then by checking it's peak you could guess its temperature.

I really don't know which instruments to use to get the light curve from a distant star but you could use a tool like this one for the final conversion: http://astro.unl.edu/naap/blackbody/animations/blackbody.html


Does Oxygen-burning process produce Neon?

The pages you’re looking at describe fusion of two similar nuclei with each other (e.g. oxygen with oxygen).  But fusion doesn’t have to involve identical nuclei, and reactions involving more abundant nuclei will in general occur at a higher rate (though this can vary based on other factors, too).
Long before oxygen-oxygen or neon-neon reactions become important, reactions involving helium nuclei (also called “alpha particles”) take place.  That is what the text in your book is referring to: C + He gives O, and O + He gives Ne. See the list in this Wikipedia article on the alpha process.


There is a useful model of spacetime as a rubber sheet that is bent by masses laying on it. But it should be remembered that this is an analogy (Obligatory xkcd) and most analogies fail if pushed too far. Spacetime isn't made of something that can rip.

A rotating black hole, a "Kerr black hole" is stranger than a static one, as it pulls spacetime around it and can accelerate objects passing close to it. Even so there are no rips, no singularities outside the event horizon.

However in another way, at the centre of every black hole there is a singularity, and a singularity is a single point "rip" in spacetime. General relativity can be used to predict what happens to spacetime around the black hole, and even inside the event horizon. But at the singularity the gravity becomes infinite, and at that one point, spacetime doesn't exist.

It is suggested that a "naked" singularity, not surrounded by an event horizon, cannot exist.


Will comet 21P that just passed Earth on Sept 10 produce Draconid storm on Oct 9?

Earthsky discusses this.

There isn't a storm every perihelion, and predicting meteors is tricky, as we can't observe the meteoroids in space, we don't know if the Earth will pass through a dense patch or not.

Key quote "How many Draconid meteors will you be able to catch? No one expects a Draconid storm this year, but one can always hope!"


In googling I stumbled across the following list of Perihelion dates and times back through the beginning of the twentieth century, and forward to then end of the twenty-first.

http://cococubed.asu.edu/data/perihelion_earth_1900-2100.dat

This seems to agree with the values generated by the US Naval Observatory's Perihelion and Aphelion page, which accepts individual years as values. (Site is kind of fickle on Mondays for some reason; perhaps scheduled downtime.)

https://aa.usno.navy.mil/data/docs/EarthSeasons.php

Mean Anomaly, for an elliptical orbit, can be thought of as the quasi-angular parameter representing the angle swept since the time of periapsis of a hypothetical object in a circular orbit with the same radius as the semimajor axis of the elliptical orbit. As such, yes, it will always be 0 when your actual object is at periapsis.

Argument of Periapsis, on the other hand, is the angle measured in the plane of the orbit, from the Ascending Node to the Periapsis of the orbit, in the direction of travel around the orbit.  It remains the same regardless of the position of the object in the orbit. As such, it will only be 0 at periapsis when the Ascending Node and the Periapsis are the same point.


After Oumuamua passage, has an "urgency" telescope time request procedure been set up?

For observing "predictable" transient events (e.g. Gamma Ray burst or gravitational wave events), observers apply (i.e. write a proposal for peer review) for "target of opportunity" (ToO)time in advance. Typically, you might say we expect LIGO to give us 2 reasonably located, strong GW events dutring the next semester and we would like 3 hours of telescope time to follow each one of these up.

If the proposal is approved, then the ToO time can be "triggered" by the proposers at any time (up to their maximum of 2 events per semester, or whatever they got approved for), by contacting the observers who are currently at the telescope and asking them to execute the ToO observing that they asked for. The current observers programme is interrupted (and they may not be compensated for this, other than perhaps co-authorship on a paper arising from the ToO).

Totally unpredictable events like Oumuamua or a new Galactic Supernova have to rely on director's discretionary time (DDT). This is a fast track, way of getting some observing time. It is still peer-reviewed to some extent and only meritorious proposals will get any time allocated. Often a DDT proposal could be approved within 48 hours.

If something truly extraordinary happened (and I would not classify Oumuamua in this way) that required observing right now, such as for example a local supernova, then I imagine that any observer at a telescope would be happy to break off what they are doing and point whatever instruments they had at it. It should be noted though that there is a risk here. At many observatories (for example ESO) you are only supposed to observe what you said you were going to observe. Departures from this can get you into trouble. A wise observer would seek permission from observatory staff/astronomers before chasing up some object somebody phoned up and asked them to observe.

See https://www.eso.org/sci/facilities/paranal/sciops/ToO_policies.html for some more information on how these policies and procedures work at ESO.


From the introduction of the source paper, we can see that what is described is a particular bump in the extinction curve of the galaxy centered at 2175 Ångström (a unit of length used for wavelength). The strength of this bump is large compared to the Large Magellanic Cloud, and is rarer toward the Small Magellanic Cloud.

As mentioned in the comments, the main candidates for the origin of the feature are non-graphitic carbon or polycyclic aromatic hydrocarbons (PAHs). The absence in other environments is attributed to low metallicities or different radiative environments.

Basically, this is a case where we lose information in science reporting. "Very high content of carbonaceous dust" was a translation of a specific strength at a specific wavelength, and "other galaxies" was a translation of surrounding galaxies.

Source:

The 2175 Å extinction feature in the optical afterglow spectrum of GRB 180325A at z=2.25 - Zafar, et al.


What defines the plane of an accretion disk around a black hole?

Let me give an answer based on my intuitive understanding of the thing...

 Initial conditions

The initial accretion disk movement is defined by the objects falling into it, the black hole's mass and its spin of course.

If you take the situation where a single star falls on a quiet black-hole, the accretion will appear along the orbit of the infalling object.

Credit: NASA, S. Gezari (The Johns Hopkins University, Baltimore, Md.), A. Rest (Space Telescope Science Institute, Baltimore, Md.), and R. Chornock (Harvard-Smithsonian Center for Astrophysics, Cambridge, Ma.) and live on youtube...

 Equilibrium in an active galactic nucleus

Now on the long run, if additional matter is constantly coming in, at first thought I would think

the accretion disk momentum will be exchanged with the one of the black hole through the Penrose process,
Infalling matter momentum will reflect the overall host galaxy momentum - especially if the galaxy is a spiral


Mapping illumination to moon phase (font) icon

While you have "illumination" it would seem to be easier just to map "moon age" (in days) to a character
The average synodic month is 29.53 days, and you have 28 characters, so calculating age*28/29.53 and then rounding to give a number between 1 and 28 (or 0 and 27). Then it is just a matter of adding 97 if that number is in the range 0-13, or 36 if that number is 14... etc.
Slightly better would be to find out the actual length of the synodic month (if you can calculate moon age and illumination this should be possible) and use it in place of 29.53.  The actual length varies from about 29.2 to 29.8 days depending mostly on the position of the Earth's orbit


I occasionally view the Earth illuminated side of the new and old moons with my Mallincam electronic eyepieces. Unlike the sun illuminated side there are no shadows in the view but the features do stand out quite clearly.

There is little if any contrast across the view that the eye can perceive. Other than looking with the eye I have never attempted any processing of an image so anything is possible.

I think I understand the concept. The earthshine should be different based on what part of the earth is facing the moon.

Unfortunately the Earth's cloud cover will probably have more effect on the earthshine than the surface itself.

There are thousands of astro images on the Mallincam.io forum. Ask the group if there are any earthshines in the files. If not someone might provide you with a few.

https://groups.io/g/MallinCam


Why does the sun rise north of east between the vernal and autumnal equinox?

You know, the first time someone told me this, I was absolutely certain they were confused, ignorant, or otherwise mistaken, and I told him he had his facts wrong. To be precise, we were talking about an observer on the equator. I maintained that the Sun would rise due east and set due west, all year round. He said no.

What gave me pause, though, is that this person was far from ignorant, so maybe the other two appellations didn't fit either. I went home and tried to construct a convincing proof of my side, and in the middle of the night a very simple and convincing proof did come to mind. Except that it was a proof that I was wrong.

Imagine that you're on equator at sunrise on the summer solstice. Due east is along the equator. But the plane of the equator is tilted relative to the Earth's orbit, and the Sun is well above (north) of it. You cannot see it due east at sunrise. It has to be $23.5^\circ$ north of east. (Remember, this is for an observer on the equator.)

By the same token, the Sun must rise $23.5^\circ$ south of east on the winter solstice, and gradually transition between these extremes during the year.


What cosmic disaster is most likely to happen?

Asteroid impacts are a possibility, but a killer asteroid typically affects the Earth every few dozen million years. This would probably destroy all life as we know it, but life will probably nonetheless persist. Statistically, by 500,000 years from now, we should have been hit by an asteroid 1 km in diameter, and 100,000,000 years from now, we should have been hit by one that is the size of the one that killed the dinosaurs. Statistically it's very unlikely we will get one that completely obliterates the Earth, causing it to be completely and utterly destroyed.

Gamma-ray bursts may irradiate the Earth to the point that complex life will no longer be able to survive, and only very resilient organisms like waterbears may live. At some point in the next 300,000 years, WR 104 is expected to go supernova, and if the poles are somewhere around 12 degrees or lower  we could be in the path of the supernova. We aren't 100% sure what its axis is angled at though. Either way, it won't actually destroy the Earth, but just irradiate it.

In roughly 7.5 billion years, the Earth will probably have been destroyed by falling into the Sun, as it engulfs the inner planets. This is nearly certain, but if somehow, the Earth-Moon system were to survive, then 50 billion years from now, the Earth and Moon will have become tidally locked. Then the Sun would extract angular momentum from the system as a white dwarf, and will cause the Earth's spin in accelerate and the Moon's orbit to decay. If this happens, then roughly 65 billion years from now the Moon will crash into the Earth, which would probably destroy it.

Keep in mind that there are many more ways that life as we know it could end, but these are not cosmic disasters, but are geologic and biological disasters, so they are not included. Generally speaking, it's very hard to completely destroy the Earth.


Will Triton have a thicker atmosphere as the Solar System ages?

Triton's orbit is decaying due to tidal interactions, it is predicted to reach Neptune's Roche limit in about 3.6 billion years (Chyba et al., 1989) where it would likely be disrupted, possibly forming a ring system. This timescale is shorter than the timescale for the Sun to become a red giant.
Even before that, Triton's orbital decay will bring it into the region of the inner prograde satellites. There's a possibility that it could experience one or more major collisions (since Triton is in a retrograde orbit, the collision velocities for an impact with a prograde satellite will generally be higher). Such collisions would likely have an effect on the inventory of volatiles on the surviving objects.
This makes prediction of the future evolution of conditions on Triton extremely difficult.
Even if this weren't the case, there would still be a question of whether a world as small as Triton could maintain liquid water. Arnscheidt et al. (2019) provide a conservative estimate for the lower gravity limit for waterworlds to maintain a stable atmosphere at around 1.48 m/s2. Triton's gravity of 0.78 m/s2 is below this threshold, putting it in the non-habitable "comet-like" regime. They also note that deglaciation of an icy world into a long-lived habitable state is unlikely: except at very low albedos, the required stellar flux is sufficiently high that the moon transitions directly to a short-lived, high temperature state.


Is it plausible for a planet that is positioned in the habitable area of a solar system to have little extractable metals?

Interesting question, though probably a better fit for  world-building.

Firstly "metals" is often used by astronomers to refer to any element other than hydrogen or helium. That is clearly not the intended use here. Terminus has a solid surface at temperatures that humans can stand, and a breathable atmosphere (perhaps after terraforming). So, what can it be made of?

The most obvious option is silicate rock (mainly aluminium and magnesium silicates) similar to what most of the outer layers of the Earth are made of. It is certainly imaginable that a planet might be more thoroughly differentiated than Earth, so that all the iron etc. ended up in the core, and there were no, or almost no, useful ore bodies in the crust.

On the one hand, this clearly fails, since magnesium and aluminium are metals. On the other hand, it is quite possible in context that what is meant is "easily extractable metals" and/or "the metals required for a specific range of technologies", which might include iron, copper, etc. (given that Asimov was writing in the '30s that is probably what he was thinking of).

In fact, with enough energy you can extract aluminium or magnesium from these rocks. We don't do this, because it's cheaper to find rare compounds of those elements from which the metal is more easily extracted, but the Foundation could certainly have had aluminium and magnesium in this scenario assuming they had enough power.

A stranger possibility would be carbon. Several gas giants are believed to have carbon-rich cores. Suppose one of those lost its outer layers in some cataclysm, the resulting smaller planet could differentiate to have a thick diamondoid crust over a rocky mantle and iron core. Although if diamonds were literally as common as dirt on Terminus, I can't help feeling someone would have remarked on it.

A similar but slightly less exotic option would be silica -- so a planet with a mainly or entirely quartz crust. I can't see a scenario for that to form -- one where you'd have silicon and oxygen and not much else around when a planet was forming.

Now we're definitely into worldbuilding territory, but I wonder is ice is a contender? If you found a world with an icy crust a few hundred km thick, is there a way to terraform it without melting all (or even most) of the ice? YOu could imagine covering the ice in a layer of vacuum-filled diamond honeycomb and then dumping a few meters of imported or manufactured soil over it, creating an atmosphere somehow and then putting a heavy greenhouse layer (CO2 and SF6 and so on) to warm the surface. Could the first-empire scientists have done something like that?   Taking this back to the original question -- an ice world could exist at the outer edge of the habitable zone (or just beyond it, if you like) and then be made habitable by adding a greenhouse layer and changing its albedo. If you can avoid melting the ice, that would meet the requirements.


It depends on the size of the black hole and the size of the bit of material you are trying to pull apart, as well as what it is made of. What is important is the difference in gravitational pull between one end of your sample and the other. This difference force grows as the inverse of distance cubed as one moves in towards the black hole.

However, nothing special marks crossing event horizon and if we were talking about a supermassive black hole (a million solar masses or more), you might not even notice it had happened, since the force difference, sometimes called the tidal force, would be quite small. At the event horizon of a smaller black hole, the tidal force would be much larger. To see this, we note that the difference in force between two ends of a bar of length $\Delta r$ is approximately
$$ F = \frac{8GM}{r^3}\Delta r,$$
where $M$ is the black hole mass. At the event horizon $r = 2GM/c^2$, so
$$ F = \frac{c^6}{G^2M^2}\Delta r.$$

The spaghettification force continues to increase as the object heads towards $r=0$, which will occur in a finite proper time for the falling object. So it will always increase sufficiently to rip something apart, no matter what it is made of; the question is merely at what radius does this happen.

It is of interest however to calculate whether it would happen to an object outside the event horizon, simply because then an external observer (at a safe distance) might witness it. This in turn depends on the material making up the object, it's size and the mass of the black hole.

Consider a material of tensile strength $T$, size $\Delta r$ and cross-sectional area $\sim (\Delta r)^2$. For this to break up before reaching the event horizon then the tidal force at the event horizon must exceed $T (\Delta r)^2$ and thus
$$ \frac{c^6}{G^2M^2}\Delta r > T (\Delta r)^2 $$
$$ M < 2\times10^5 \left({T \Delta r}\right)^{-1/2}\ M_{\odot}$$

Diamond has $T\sim 10^{11}$ N/m$^2$. If it was of size 1 cm, then a black hole would need to be larger than 6 solar masses for this chunk of diamond to survive being ripped apart outside the event horizon. Other materials or larger bits of material are easier to break and would not survive outside the horizon of even more massive black holes. e.g. The tensile strength of bone is about $10^{8}$ N/m$^2$, so a 0.1m piece of bone would be pulled apart prior to falling through the event horizon if $M< 60M_{\odot}$.


What's the difference between a crater and a basin?

The French version of the Wikipedia article has a "nomenclature" section, absent from the English version. It says:

Lorsque l'énergie cinétique de l'impacteur est suffisante pour
atteindre le manteau à travers la croûte et provoquer des épanchements
magmatiques, on parle de bassin d'impact et non plus de cratère
d'impact.

Which roughly translates to:

When the kinetic energy of the impacting body is sufficient to reach
the mantle through the crust and trigger magmatic outpourings, the
term impact basin is used instead of impact crater.

However, there is no reference to back this claim, so I would take this with a pinch of salt...


How do they know how far out Farout is? (2018 VG18)

It is a process of orbit determination.

The apparent position of any solar system body will change from night to night as a result of the combination of its actual motion around the sun and the motion of the Earth. A very distant body such as "Farout" will be moving very slowly.

To determine an orbit, a minimum of three observations is needed. From three observations it should be possible to fit a Keplerian elliptical orbit to the observations and from the orbit one can tell the object's position at any time. The method takes into account the effect of parallax as the position of the Earth is used in the determination of the orbit.

However there will be observational errors; the exact location of the body can't be seen, only a white blob on a photograph. This means that there can be errors in the orbit determined from the minimum of three observations. To counter this, further observations over a longer period of time can be taken. The process of orbit determination then becomes a process of finding the best fit orbit to the observations. The more spread out (in time) the observations are, the better is the orbit determined.


What would be the obstacles to building a 6.5km space telescope besides cost?

The basic issue with a large optical telescope is to keep all of the mirror in an extremely stable position relative to the rest of the mirror. Depending on the wavelength of your light you probably want to limit deviations from your perfect shape to less than about 100nm, so, you need to account for and deal with somehow any force which might act on one part of your mirror differently from another with enough force to bend the whole mirror by more than this tiny amount. At a glance, therefore you will need to deal with:

tidal forces from Earth, Moon and Sun (I think you can ignore tides from Venus and Jupiter, but I'm not certain)
Orbital perturbations arising because Earth is not spherical
temperature variations from moving in and out of Earth's shadow at different times and from sunlight and Earthlight (and maybe moonlight) hitting different parts of the structure differently
interactions of any conducting components of the mirror with Earth's magnetic field
flexing and oscillations of the structure arising from thrusts used to point the mirror

and probably a lot more that I haven't thought of.  These problems would become somewhat simpler if you put the telescope much further from the Earth, or even better much further from the Sun (say a very distant orbit around Neptune) but that would make assembling it by spacewalking impossible.


What would be the outcome for life in our galaxy if the merger of the Milky Way and Andromeda creates a Quasar?

Quasars have very pointed jets perpendicular to the rotational axis of the accretion disc around the central black hole. As long as this jet doesn't happen to point into the galactic plane, the cosmic radiation of these jets won't be a too serious problem.

But a collision of two galaxies can become an eventual problem in a different way. A merger of galaxies triggers star formation and sweeps most of the galaxy free of interstellar dust and gas. So, after a short (hundreds of millions of years) period of time, we'll get lots of new stars. But they'll eventually exceed their main sequence lifetime. Due to lack of interstellar matter only very few new stars will form afterwards. That's a difficult settings for formation of planets and life.


How to retrieve from NASA Horizons site the needed data to plot analemmas for other planets?

You did two things wrong, one minor, the other major. The minor thing that you did wrong was to chose the solar system barycenter as the target rather than the Sun. Use the Sun.
The major thing you did wrong was to use a step size of 1477 minutes. That is (to within a minute) the length of a Mars sidereal day. The Sun will march across the horizon if you use a sidereal day. What you want is a Mars solar day, which is 1479 minutes and 35 seconds long.
This presents a new problem. Ideally your step size should be 88775 seconds. But you cannot choose seconds as the unit of a step size. You're stuck with minutes, and that is suboptimal. There is another approach:

Choose the "equal intervals (unitless)" option for the step size,
Make the time difference between the start and end time be an integral multiple of 88775 seconds, and
Specify the step size as the integral multiple chosen to determine the time span.

For example, choosing 711 as the integral multiple and 2020-Aug-20 12:00:00 as the start time means the end time should be 2022-Aug-21 01:03:45. You might need to tweet the end time to get the analemma to close up properly.

Edit
I noticed that you used the center of Mars. You need a point on the surface to get a proper analemma. Horizons provides several predefined points. If you choose Viking 1 / Chryse @499, you'll have to modify the start and end times so that the Sun is above the horizon. Add 12 hours to the start and end times shown above appears to work very nicely for Viking 1's location.
Below is what I get plotting the Sun's elevation vs azimuth as seen from the Viking 1 / Chryse location, with 711 intervals (712 data points) sampled over the period 2020-Aug-21 00:00:00 to 2022-Aug-21 13:03:35 UT.


Is light emitted from a heavier body more redshifted than light from a lighter body?

Redshift is not related to the time it takes to travel from the source to the viewer, not to the distance from the source to the viewer.
But light is red-shifted by travelling out of a gravitational well.  One way to see this is to consider the equivalence principle: A person at the front of an accelerating spacecraft would see light shone from the back to the front of the craft become redshifted, and would say this is due to the fact that they are moving faster when the light reaches them than when the light was emitted.  If the same craft were held stationary in a gravitational field the same experiment must have the same effect, ie the light is observed to be red-shifted.
Alternately you can consider energy:  If you fire a bullet vertically up, it will slow down, and lose kinetic energy as it moves out of a gravitational well.  If it is travelling fast enough it might be able to escape the pull of gravity, but it will be going slower by the time it reaches a great distance.
Light, on the other hand, can't lose energy by going slower, but it does lose energy and momentum as it rises out of a gravitational well.  This is seen by the photons becoming red-shifted as they rise.
Redshift can be caused by a Doppler effect (of when an object is moving away from you) or by a cosmological effect (as the space between you and the object is stretched with the expansion of the universe) or by gravitational effects as above.  Redshift is not caused by the time it takes to travel from the source to the viewer.  But as more distant sources are receding faster,  the red-shift of more distant galaxies (that has been travelling for longer) tends to be greater.


Is there evidence for the speed of gravity?

It has long been one possibility that gravity does not work as in general relativity, or as in the Newtonian limit that is relevant to the gravity in a galaxy.  Thus, many different modifications to gravity have been tried.  But the reason they are generally not very successful is that dark matter seems to alter gravity on many different scales, from within a galaxy, to over galaxy clusters, to the expansion of the universe as a whole.  Thus it generally requires multiple different modifications to fit them all, and that makes for a very clunky model of gravity-- one can fit anything if given enough free parameters.  That sounds like it would be the problem for trying something based on the speed of gravity-- the timescale for propagation of changes would be very different over the three scales I just mentioned, yet dark matter has a similar influence on all three scales.


360 degree equirectangular projection of all naked eye stars with precessional motion for simulating

The "cylinder" projection of Stellarium does this:


Observing a point 13.82b ly away, 1b years ago

Time is kind of funny when you look at such distances. Let's imagine that you are running away from a person throwing a ball at you. The ball will travel further to hit you, based on your speed, than the distance that was present when you started to run.

The universe is expanding. The early Universe expanded very quickly, giving large deviations from the expected position today than what we see through a telescope. Something that we see that is 12 billion years old is actually quite a bit further away today, although we can't know exactly where it is. As I understand, this hypothetical object would be around 40 billion light years away today (See this question), given current theories.

But yes, if you looked at a galaxy that appears in today's distance to be 13 billion light years away, then the universe was 800 million year old, roughly speaking. Galaxies took about a billion years to form, however, so you probably wouldn't see such a galaxy.


You are forgetting parallax.

Something which is very distant and lies on the celestial equator will have a declination of 0°, but if it is nearby then its declination will only be 0° to an observer located along Earth's equator.

From your location in Zagreb, something like the ISS, even if it is crossing Earth's equator at a longitude of ~16° will appear to be in the southern sky, and your declination coordinate will likely be south of 0°.


What happens to oxygen produced on the Sun (or other stars)?

The Sun is a small main sequence star. It does not produce oxygen via fusion. It can't. The temperature and pressure in the Sun's core are too low. Fusion in the Sun is currently limited to production of helium. This will remain the case for several billion years.
That said, there is oxygen in the Sun, about 1% by mass. This oxygen was produced long ago by other stars at the end stages of their lives. Our Sun is a third generation (or more) star. Most of the Sun is far too hot for those oxygen atoms to combine chemically. One exception is sunspots, relatively cool areas on the Sun's photosphere. (Relatively cool means less than 4500 kelvins, so still quite hot.) Molecules can form at these lowish temperatures, and scientists do see signatures of many different molecules in the light coming from the Sun.

**Update, in response to edits to the question**
Molecules cannot form inside of a star. The temperatures are just too high. Molecules decompose (split apart) into their constituent parts at high temperatures. The Sun's photosphere is about 5800 kelvins, which is already too hot to sustain very many molecules. Temperature rises rapidly with increasing depth below the photosphere. The Sun's core temperature is about 15 million kelvins (27 million Fahrenheit), and the Sun is a small star. Larger stars have even higher core temperatures. At 15 million kelvins, there aren't even atoms, let alone molecules. There are instead atomic nuclei and electrons. Atoms are stripped of their electrons at those extreme temperatures.
In five to seven billion years, our Sun will have fused all of the hydrogen in the core into helium. That's when our Sun will become a red giant. Even then, it still will not produce oxygen. The first stage a one solar mass star experiences after leaving the main sequence is the red giant phase, where the core is an inert mass of helium surrounded by a shell of fusing hydrogen.
Eventually (after another billion years or so), the temperature of that helium core will rise to the point where the helium starts fusing into carbon, plus a little bit of oxygen via the first step on the alpha ladder. At this point, the Sun will leave the red giant phase and join the horizontal branch of the Hertzsprung–Russell diagram. This is a rather short-lived phase of a star's life. The carbon and oxygen produced by helium fusion quickly (in stellar timeframes) form an inert core. At that point, our sun will become an asymptotic red giant.
The red giant and asymptotic red giant phases are rather messy affairs, wracked by convulsions where the star expels lots of gas. Our Sun will lose about half its mass to such convulsions. Molecules do form when this expelled gas cools. This results in some of the prettiest pictures in astronomy, shown below.

(source: inspirehep.net)


That's a big asteroid!

Recent discoveries of objects of this size are rare, one example was 2013UQ4. We believe that all objects of this size in the main asteroid belt have been discovered. (The discovery of 2013US10, which appeared to be main belt object turned out to be an error, the combination of observations of two different objects) This object has a highly inclined and retrograde orbit. It turned out to be a comet not an asteroid.

For context US10 it was at magnitude 19 when discovered. It is unlikely that a large body would remain undiscovered until it was at mag. 10 or 9, especially if it was in the main belt. A very small object that is rapidly brightening could reach these magnitudes before discovery, but not a 10-20km sized object that your story implies.


How could a hobbyist astronomer determine apparent magnitude of a star?

The easiest way to determine the magnitude of a given star is probably to use the Pogson relation. The idea is to determine the magnitude of a star knowing the magnitude of a reference star; it is thus quite easy, using a well-known reference as Vega or Sirius.

The Pogson relation is given by:

$$m_1-m_2=-2.5\ log\ \left({\frac{E_1}{E_2}}\right)$$

where $m_1$ and $m_2$ are the magnitudes of star 1 and star 2 (your reference star), and $E_1$ and $E_2$ brightness (that could be in arbitrary units, which is a good news for you).

With a CCD detector the brightness is easy to determine since it's just the flux of a pixel. You don't have to convert it in physical units (as Watt per square meter) since you have a ratio $\left({\frac{E_1}{E_2}}\right)$, so the raw intensity is enough to determine the magnitude.

As for any other measurement with a CCD camera, you have of course to determine the bias, dark and flat-field and to reduce your data before estimating any magnitude. The precision will depend on the number of observations performed and, as always, it is better to perform several short exposure images than a single long exposure one.


Have we measured or found evidence of a detectable origin point of the big bang?

There is no single origin point.

Before the big bang, the entire universe was contained in an extremely dense singularity - there was no "space" for the universe to expand into, the very fabric of spacetime was contained entirely in the singularity. There is no single "origin point" in the universe where the big bang took place where we'd see the entire universe receding from us. Rather, from any point in the universe we can see everything else receding. The big bang happened everywhere, since the whole universe, including space itself, came from that singularity.


How could a neutron star cause repeated 'fast radio bursts' detected coming from outside of the galaxy?

The progenitors of FRBs have been tied to soft gamma-ray repeaters (SGRs), and are likely young magnetars displaying irregular atmospheric and magnetic activity. The earliest connection between the two - as far as I know - was made by Lyutikov (2002), who compared the outbursts to solar flares (the magnetar bursts are sometimes referred to as "giant pulses"), based on models by Thompson et al. (2002).

The theory is that currents dissipate into the neutron star's magnetosphere, at rates coupled to the star's crust. Slow deformation can allow dissipation to proceed at a constant rate, while "fractures" in the crust can lead to quick outbursts. The release of magnetic energy through reconnection may then be responsible for the radio outbursts. The evolution of the field is similar to the behavior of the Sun's magnetic field in the corona, leading to the analogy to solar flares.

Popov & Postner (2007) state that Lyutikov's model's predictions agree with several observed bursts. While the arXiv version of the Letter concerning FRB 121102  does not specifically delve into the model, it seems to support it, although there are doubts as to the feasibility of repeated giant flares.


Why is the Cosmic Microwave Background at the same distance no matter the direction we look?

The CMB is visible at a distance of 13.8 billion light years in all directions from Earth, leading scientists to determine that this is the true age of the Universe.

This is wrong in a few ways. First, we do have good reason to think that the CMB was produced around 13.8 billion years ago, but that doesn't mean it's 13.8 billion light years away. The light is right here (that's how we detect it), and the matter that emitted the light is roughly 47 billion light years away in the present era, if our cosmological model is correct.
Second, we don't work out the age of the universe from the distance to the CMB (or the matter that emitted it). We can't see how far away it is. We work out the age of the universe by fitting a cosmological model to a variety of evidence (including the detailed spectrum of the CMB), and then we say that the CMB was emitted 13.8 billion years ago because that's what the model implies.
Third, the CMB doesn't date back to the beginning of the universe – although the time from the big bang/end of inflation to the emission of the CMB (around 380,000 years) is so much smaller than the time from the big bang to now (around 13.8 billion years) that this makes little difference.
We aren't at the center of the universe. The universe is just homogeneous (the same everywhere), and it's large enough that light hasn't had enough time to cross it since the CMB was emitted. The CMB we see is from the same distance in every direction because it's all been traveling for the same amount of time. So your first scenario is the correct one.


How are the solar flares from May 23, 2021 extraordinary?

They weren't at all extraordinary. The Sun has been quiescent for the last three or so years. This quiescence is typical of the period between solar cycles. Solar cycle 25 is finally becoming a tiny bit active.

What is a "magnetic canopy of a sunspot"?

Bad prose. An active region on the Sun is an area with an especially strong magnetic field.. Normally the magnetic field lines surrounding an active region are closed; they loop back toward the Sun. I guess one could call this looping the "magnetic canopy of a sunspot." Occasionally the loops break. That break of the loop results in a solar flare or a coronal mass ejection.

Why is that solar flare so spectacular?

It wasn't. That active region resulted in 9 C class flares and 2 M class flares.
That's nothing compared to what we will see in a few years, and those in turn most likely will be nothing compared to the Carrington Event. The Sun is coming out of its solar minimum slumber.

How are sunspots named?

Sunspots are neither named nor numbered. Active regions on the Sun however are numbered, and have been since 05 January 1972. Active regions can have several sunspots, or none. The numbering of active regions is consecutive. The standard practice is to report only the last four digits and the date.

It is said that the "CME might hit Earth's magnetic field", which is probably not same as "it will hit Earth", or is it?

CMEs typically hit the Earth's magnetic field rather than the Earth. Only the very strongest CME's can punch through the Earth's magnetic field. People living at high latitudes might see some nice auroras (aurora borealis (northern lights), and aurora australis (southern lights) from this event.
When a CME hits the Earth's magnetic field, the field channels most of the particles around and away from the hitting the Earth. Some of the particles end up getting channeled toward the Earth's poles. This channeling of some of the particles toward the Earth's magnetic poles is why those auroras occur. Even something as strong as a Carrington-level event will have the greatest impact at high latitudes.


As @astromax observed, one of the primary factors that makes a space based telescope better than an equivalent telescope of the same size is scattering.

Along with scattering, there is also refraction which can be especially problematic when combined with atmospheric turbulence. In the modern era, this problem can be remedied to a certain extent using adaptive optics, but since the Hubble was designed, built, and launched before AO became practical in the 1990's, a space telescope represented the pinnacle of optical clarity for that time.

However, there is another important optical property of the atmosphere, absorption. Although the Hubble is primarily a visible light telescope, it does have instruments that cover both the near UV and near IR, both of which are absorbed by the atmosphere more than visible light.

Furthermore, there are practical advantages of space based telescopes. There is neither weather, nor light pollution in space.


Is there any difference between rotation axis 90 degrees and 270 degrees?

No, I don't believe there's any meaningful distinction.

A planet's axial tilt can be thought of as a 2-dimensional quantity: its angle relative to the ecliptic (or to whatever baseline you're using), and the direction in which it's tilted. When we talk about a planet's axial tilt as a number of degrees, we're only talking about the first component.

A planet whose rotation axis is perpendicular to the ecliptic would have a tilt of 0°. If its rotation axis is parallel to the ecliptic (rotating "on its side"), its tilt is 90°. If its axis is "vertical" but it's spinning backwards, its tilt is 180°. The range from 0° to 180° is enough to express all possible tilts, both prograde and retrograde.

Consider taking a planet with a 0° tilt and tilting its axis by 270°. The result is exactly the same as tilting it 90° in the opposite direction. Since the axial tilt is usually expressed as just the magnitude of the tilt, and not its direction, the distinction between 90° and 270° can be ignored.

And if we want to express the direction as well, we'll just say that it's tilted 90° in some specified direction.

We could have had a convention where the tilt ranges from 0° to just under 360°, with the direction specified more narrowly, but that would be less useful; the particular direction of a planet's axial tilt is less interesting than its magnitude. Also, the direction varies over time due to precession.

We could also have had a convention where the tilt ranges from 0° to 90°, and we also specify whether it's retrograde or not; then Venus would have a tilt of 3° rather than 177°.


Why can't we use the semi-minor axis in Kepler's third law?

The relationship between $a$, $b$ and the eccentricity of an orbit $e$ is
$$ a = b\left(\frac{1+e}{1-e}\right)\ .$$
It is clear therefore from a mathematical point of view that you cannot get a similar constant relationship by replacing $a^3$ by $b^3$ in Kepler's third law. You would have to use an expression involving both $b$ and $e$.
Another way of saying this is that you can have (infinitely) many orbits that have the same $a$ and $T$, but which have different $b, e$ combinations.
I guess from a fundamental point of view it boils down to being able to express the total energy of an orbit as
$$ E = -\frac{GMm}{2a}$$
which is independent of the shape of the ellipse.


Computation of mean sidereal time at Greenwich using solar position algorithm

The formula 28 from SPA that you linked is given with units and in a form that do not match official versions. It takes a little work to 'translate' it: also, it is needlessly susceptible to roundoff error (because all significant figures for both the date and time-of-day are strung together in one very long number).  The authors of SPA do not appear to have been striving for full official accuracy, probably their application did not require that.

I can't quite work out from the data in the question what might have happened to the calculations, online or otherwise, but maybe the following reference data could be of help.

GMST for 0h UT is still tabulated in the Astronomical Almanac, and there are also two rather easy formulae derived from the official definitions that you can use (see below), they make somewhat better preservation of end-figures than a calculation in the SPA form.

In the meantime, just for reference, the Astronomical Almanac for 2003 gave the GMST for 0h UT on 2003 Oct 17 as 1h 40m 21.4554s (1.6726265 h).  That is for 0h UT, and during the following 24h, the GMST for x hr UT is is easily found, it is very close to  $ [ (GMST for 0h)  +  x * 1.00273791 ] $ .

2003 was a changeover year for the official (IAU) calculations that relate GMST and UT1. Before 2003, the formula of Aoki et al (1982), "The new definition of universal time" (http://adsabs.harvard.edu/abs/1982A&A...105..359A) was the one adopted. From 2003 Jan 1 the formula of N Capitaine et al. (2003), "Expressions to implement the IAU 2000 definition of UT1", Astronomy & Astrophysics 406, 1135-1149 (http://adsabs.harvard.edu/abs/2003A%26A...406.1135C)  was (and still is) adopted. The difference between the results during 2003 was small (+0.0003 time-secs to get the result of the newer calculation), but it looks as if AA for 2003 was still using the older 1982 formula.

The Aoki et al (1982) formula gave :

GMST of 0h UT (in time-secs) =

$ 24110.54841 + 8640184.812866 *Tcy + 0.093104 *Tcy^2 -0.0000062*Tcy^3 $

plus a very small correction for DeltaT: this can be had from the version of the Aoki formula given in the Capitaine reference at B.3 on page 1149, (which seems to be the one that was used in official calculations) but it can practically be left out at the precision of the data in the question. Tcy means the time interval from J2000 (Julian Date 2451545.0) reduced to centuries of 36525 days. (The JD for 0h on 2003 Oct 17 0h UT was JD(UT) 2452929.5 days and DeltaT was close to +64.55 secs.)

The current formula from the Capitaine (2003) reference (most conveniently from formula B.2 on page 1149) gives

$ 24110.5493771 + 8640184.7945360 *Tcy + 0.0931118*Tcy^2 -0.0000062*Tcy^3 $

plus a very small fourth-order term (given in the reference but omitted here) and again a very small correction for the slight effect of DeltaT which can be left out or included if desired as

$ +307.4771600 * DeltaT/(86400.0*36525.0) $ (time-secs), where DeltaT is still in seconds.


Are gravitational waves part of dark energy?

Gravitational waves cannot be part of dark energy, because -- just like electromagnetic waves -- their energy density dilutes as the universe expands: both due to the simple expansion of space and due to the cosmological redshifting that the expansion produces.

Dark energy, on the other hand, has an energy density that remains constant as the universe expands (truly constant if it's the cosmological constant, very close to constant if it's something stranger like quintessence), so it can't be made up of GWs.

GWs probably don't contribute much to the energy density we attribute to dark matter, either, since dark matter is diluted by the cosmological expansion but not by redshift. So if the energy density attributed to DM ten billion years ago were due to GWs, then it would be significantly lower than the DM energy density we measure today. (Also, current cosmological observations indicate significant dark matter was present very early in the universe, before the star formation that was necessary to create binary black holes in the first place.)

The GW energy from a binary-black-hole merger is enormous, yes; but those are very rare events. The current estimate from LIGO observations is about 50 or 100 such mergers per year per cubic gigaparsec, which translates into a rate of less than one per million years per galaxy. This paper by the LIGO and Virgo Collaborations has estimates of the current GW energy density in units of the current energy density of the universe (their Figure 1):

The green curve shows the estimated energy density from binary-black-hole mergers as a function of the frequency of the GWs; the red curve shows the contribution from binary-neutron-star mergers. The combined peak is always less than $10^{-8}$ of the critical energy density, so even summing up over frequencies, we have an energy density much smaller than that due to ordinary matter ($\Omega \sim 5$%), so it's probably safe to ignore its effects.


It isn't usually an issue because most experiments are simply concerned with finding exoplanets. They are rarely designed in such a way that it is easy to estimate population statistics because of all sorts of biases that go into selecting the targets. Unfortunately the search for exoplanets has turned into a sport where discovery is everything.

If one assumes random orientation of orbits (and that is all it is, an assumption) then the probability of a transit scales roughly as
$$P \simeq \frac{R_p+ R_s}{a}$$
where $R_p$ and $R_s$ are the radius of the planet and hot star respectively and $a$ the planet's orbital radius (with small modifications for non-circular orbits). The larger this is, the more likely a transit is to occur. Hence large exoplanets orbiting close to large stars are more likely to transit. In principle then, this effect can be corrected for when calculating the statistics and frequency of exoplanets.

So how good is the random orbital inclination assumption? I honestly think nobody knows at the moment. I have done work on the possible alignment of spin axes within the low-mass stars of clusters (Jackson & Jeffries 2010) finding consistency with the random hypothesis. More recent work using asteroseismology suggests that there may be alignment for more massive stars (Corsaro et al. 2017). However, even if the spin axes (and therefore presumably the majority of planet orbits) of stars in clusters line up, there is no obvious reason why each cluster should have the same angular momentum vector When the clusters eventually disperse into the field then they would, presumably, form a pseudo-random distribution?

Except, what if the Galactic tides or a large-scale Galactic magnetic field played a role in shaping the angular momentum direction of the clouds that formed the clusters. Might it be possible for some alignment to persist to old age? Corsaro et al. argue that interactions within a cluster are not sufficient to "scramble" the angular momenta after star formation has finished. Close interactions between stars become much less likely after they emerge from a cluster into the field. An intriguing piece of work by Rees & Zijlstra (2013) found that there was evidence for a non-random distribution of orientation for bipolar planetary nebulae towards the Galactic bulge. This suggested that the orbital angular momenta of binary systems responsible for the bipolar shape of the nebulae were oriented in the Galactic plane. The result is highly statistically significant but as far as I know has not been followed up despite its obvious implications for estimations of transit yields from exoplanetary surveys.

I think that there will be a much better answer to this question once we have all-sky exoplanet searches of the quality of the Kepler satellite (the main Kepler survey was in one particular direction). It should become very obvious if there are changes in the planet yields as a function of position of the sky (although you also have to control for the types of star being observed) associated with any large-scale alignment. Maybe there is enough information in the Kepler K2 fields that are taken at positions around the ecliptic - I have not seen any analysis. However, such data will surely become available with the launch of NASA's all-sky TESS satellite in 2018.


The Wikipedia list given by @NeutronCat in a comment contains this external link. This compilation by Wm. Robert Johnston has not been updated since 2004, but contains both stellar and supermassive black holes, along with some parameters (coordinates, distance, mass...).


Are there any physical phenomena that destroy information?

This is one of the current questions of cosmology/physics.  So far all evidence points to the negative inside a black hole so far as quantum numbers go, but it's not clear what you're considering as "prior state."

A simple counter-example is a camera. Once the photons are absorbed by the film or solid-state sensor, all phase information is lost.  Is that what you were looking for?


If I travelled near a black hole, my time would progress slower relative to someone on Earth. This is clear enough.

Yes, no problem with the gravitational time dilation.

  However, what if we sent a probe with a camera to a black hole? When we watch the screen, would we see time through the camera's perspective — that is, would the Universe appear to progress faster as the probe got closer and closer to the black hole?

No. We'd see the universe progressing normally, because we aren't subject to that gravitational time dilation. (I presume we're at a safe distance). Let's suppose it was a TV camera that took a picture 25 times a second and sent it back to us, adequately catering for redshift. The camera starts off taking 25 pictures a second as measured by us. But after a while we notice we're only getting 24 pictures a second, then 23, and so on. We see events in the wider universe progressing at their normal rate, but eventually the movie starts getting jerky as the frame-rate reduces. In the end the frame-rate reduces to zero, and that's the end of the show.


The Sun orbits in the entire Galactic gravitational potential, which is very approximately similar to that provided by considering that all the mass interior to its Galactic orbit (about 100 billion solar masses) were concentrated at the Galactic centre.

The black hole at the centre has a mass of 4 million Suns, so contributes less than 0.01% of the mass required to keep the Sun in its present Galactic orbit of about 200 million years. Or to put it another way, if the Sun just orbited the black hole with a period of 200 million years, we would need to be about 800 light years from the Galactic centre, rather than the 30,000 light years we presently are.


First of all: "How gravity really works" is a deep question, and any serious scientist would quickly concede that all we have is an incomplete working model. You certainly have heard about  General Relativity; the first image on the page is your trampoline.

Our working model, General Relativity, is working because it explains a lot of observations very nicely. (Careful, here is another deep question lingering: "Explains" means that we can predict some observations from other observations with the model of gravity we have in our  mind. It does not necessarily mean that we understand the "real nature" of the underlying issues.) But we are very confident that the model is working over a wide range of observations. One of the last "first-time" observations which followed the predictions and thus gave us more confidence in the model was the two black holes colliding lately. Lately? Well, billions of years ago. We just learned about it lately.  Here is a link to a New York Times article with an impressive video. (I think one can still read a limited number of Times articles for free, so try it out.)

Our model of gravity is incomplete because it doesn't connect well to the model of nature we have for other things (elementary particles, quantum physics). For a while (like 70 years or so) it didn't connect at all; Einstein himself completely failed to connect the dots, which was probably not encouraging since he had received the Nobel Price for laying one of the foundations of quantum physics and was the obvious authority about gravity. If he couldn't do it, who could?

If I'm not mistaken, the physicists today are making progress, slowly. This connection between quantum physics and gravity is one of the main unresolved problems in modern physics.

Last, let me address your concern about the planets spiraling into the sun. This idea probably comes from actual balls on an actual trampoline spiraling in, I suppose. You probably know that the balls lose speed due to friction, much the same way you slow down on your bike when you stop pedaling. Some of the kinetic energy is transformed into heat.

And you know what? You are right. Given enough time, the planets would eventually fall into the sun. Low-flying satellites fall back to earth after a few years, because there are still traces of atmosphere slowing them down out there. The reason is that there is "friction" in the wider sense involved in all large-scale processes in the universe. That is actually one of the fundamental physical principles making up the world we know. It's just that the near-vacuum between the planets doesn't provide that much friction, and the planets are fairly massive bodies with an enormous mass and kinetic energy. It will take a long long time for them to lose enough energy that they'll be so close as to touch the sun. (Perhaps too long to happen at all.) In fact, over human life times the planets, moons and stuff are almost perfect examples for movement without friction. But in the astronomical time scale -- billions of years --, there certainly is friction. For example, the moon is showing us always the same side because friction slowed its rotation so that the rotation is now "locked" with its orbit.

Bottom line: The idea that gravity bends space and time "explains" all large-scale observations so far; the "trampoline" is a good model for a 2-dimensional "space", i.e. a surface, if you ignore friction.


What type of planetary-mass object would Planet Nine be?

It is a planet - either a gas giant core, a mini-Neptune, or a super-Earth.

First off, Mike Brown has stated outright

  “It is a planet—there’s virtually no doubt,” he said. “What we now call planets are objects that can gravitationally dominate their neighborhood. Pluto is a slave to the gravitational influence of Neptune. By area, Planet Nine dominates more of the solar system than any other known planet—it’s only because of this that we can infer its existence. And because of this we’re pretty sure it’s not a small object: it’s at least ten times more massive than Earth and five thousand times more massive than Pluto. In many ways, you could argue that this is more of a planet than anything else in the solar system.”

I wrote an answer to How was the hypothetical ninth planet kicked so far out of the Solar System? that discussed the possibility that the ninth planet is the hypothetical 5th gas giant proposed by Nesvorný (2011). The details can be found in that answer, but basically, Brown & Batygin's reference to previous papers by Morbidelli et al. suggest that they think that this is what remains of an ice giant, like Uranus and Neptune, that formed near Saturn and Jupiter but was ejected in a series of gravitational interactions that led it to the Kuiper Belt and beyond.

That said, Batygin's estimates place the ejection of this planet way before previous models of a 5th gas giant predict, so either we accept models with a  6th gas giant (also an ice giant) and say that both of these were ejected independently of one another - not impossible - or we go for some other option. The bottom line is that this is a planet, if we go with the above assumptions.

Two other guesses have been made based on the planet's mass, estimated by Brown & Batygin at about 10 Earth masses. One is that it is a very massive super-Earth, hanging around the proposed upper mass limit for those objects. This implies that the planet is rocky, but does not imply anything else. The other guess is that it is a mini-Neptune, with a predominantly gaseous composition but at about the same mass. The super-Earth hypothesis arose in part because of claims a long time ago about the ejection of a 5th terrestrial planet.

I've used the term "planet" throughout this entire answer, and that's because this object may be one - and most likely was in the past. Let's look at the IAU's three major criteria to be a planet:

It orbits the Sun. This is definitely true, if it exists. Also note that if this is a captured rogue planet, as some have proposed, then we can still call it a "planet", because it now orbits the Sun.
It is in hydrostatic equilibrium. This is most likely true, and is true if any of the three models I suggested above are correct.
It clears its neighborhood. This is a bit iffy. It clearly influences objects around it, but we don't know much about those influences besides observations of the six Trans-Neptunian Objects (TNOs) Batygin & Brown studied. That said, an object with a mass of 10 Earth masses will most likely clear the neighborhood around its orbit if it has a stable orbit, which should be the case here. See the quote at the start of this answer for more information.


Do the orbits of planets change sometimes?

The orbits of the planets and other bodies are changing all the time, because of the influence of other bodies.

Yes, planets can change orbit significantly because of the influence of other planets. There is a theory that has Neptune and Uranus swapping orbital positions a long time ago.