Why did astronomers believe most or all stellar black holes had masses no greater than 15 solar masses?

The so-called 'mass gaps' for black holes, according to theoretical models, are between 2-5 solar masses and 50 to 150 solar masses. (Actually, I have read that there is no good theoretical reason for the lower, 2 to 5 solar-mass gap....)

The lower mass-gap is suspected observationally because we have yet to observe a neutron star with mass greater than about 2 M$_{\odot}$ (the error bars on such measurements vary greatly depending on the method of observation, and its controversial currently as to what neutron star is the most massive currently known), and because we've also not discovered a black hole in an X-ray binary whose mass is lower than about 6 M$_{\odot}$. However, a binary neutron-star merger can produce a binary black-hole (BBH) which is likely to be in this lower mass-gap. Thus, in reality, the gap will probably be populated by high mass neutron stars (if they rotate very fast and have inclination, etc...) and by low mass black holes that result from binary neutron star mergers.
The current nomenclature for the masses of observed black holes reflects the uncertainty inherent in the field. Right now, although the limits are arbitrary:
stellar mass means $\lesssim 100$ M$_\odot$, intermediate mass means $\sim 1000 - 10^5$ M$_\odot$, supermassive means $\gtrsim 10^6$ M$_\odot$. There is no upper limit on the mass of the black hole from general relativity, but astrophysical and cosmological considerations, though model dependent, can yield an upper limit of about $10^{11}$, but it could be larger. These have been called "stupendously big" and "ultramassive."
Theoretically, the bounds on this lower mass-gap are motivated from below by stellar evolution, e.g. neutron stars are expected to only form up to a certain mass (depending on the metallicity), and astrophysical processes such as accretion. Very importantly, the bounds depend on the equation of state of nuclear material, which is uncertain and one of the "holy grails" that gravitational-wave (GW) astronomers are after.
So people often just write a range of $2 - 5$ for simplicity when writing in pop sci articles. The upper mass-gap attributed to pair-instability supernova is generally thought to be around $50 - 150$ M$_{\odot}$ (this is the black hole mass), but these limits are uncertain because they depend on the uncertainties of stellar evolution models and supernova physics. Just like with the lower mass-gap, this gap could, in practice, be populated by black holes that form from hierarchical mergers in dense stellar clusters (dynamical channel), or from isolated binary-star evolution that can in principle produce high stellar-mass binaries like GW190521 that survive failed supernova, although it is uncertain because this depends on stellar winds and core mass calculations.
EDIT: As Rob explains below, the history of stellar wind mass-loss theories is an active field. The progenitors of stellar-mass black holes are high mass stars (i.e. $m_{\rm ZAMS} \gtrsim 30$ M$_{\odot}$). For example, Wolf-Rayet stars have been viewed as the end-point of stellar nuclear evolution of high-mass stars since the 1980s. WR stars have strong winds for their size, which are line-driven, similar to the O-type stars that they likely evolve from. The mass-loss rate from the stellar surface is typically modeled as a power law of the star's luminosity and metallicity, i.e. higher mass and higher metallicity both imply more mass loss. So an initially high-mass star will experience strong winds, even more so if it has high metallicity (as Rob points out, low-metallicity environments might be loci of black holes near/in our galaxy). However, as you might hear/read in pop-sci and scientific articles, this does not preclude the existence of high-mass black holes! The most massive stars that we have observed to date are WR stars, implying they might have evolved from even more massive stars, and in the extreme case, supermassive stars are thought to be the "seeds" of supermassive black holes.

But, I have also read that astrophysicists were surprised to find black holes larger than 15 solar masses using the LIGO gravitational-wave Observatory....

I suspect you've read this from pop science articles, which often do not properly characterize the sentiment of a scientific article, but it's a difficult thing to do. The reason that anybody would be "surprised" to find black holes with masses greater than 15 M$_{\odot}$, is that before the discovery of BBHs (from the direct detections of gravitational waves) we only had suggestive evidence for the existence of stellar-mass black holes from observations of X-ray binaries. See here and here for reviews of known stellar-mass black holes in X-ray binaries.

“These observations are very hard to explain with astrophysical scenarios and they are quite easily explained with primordial black holes,” says Sébastien Clesse, a cosmologist at the University of Brussels in Belgium."

Again, this just depends on who you are asking. Everyone wants to say that their work is the most relevant formation channel for what LIGO/Virgo are seeing, and this is okay because the number of known sources is still too small to make rigorous claims about formation of BBH sources from GW detections. Also, since the signal-to-noise that the current detectors can achieve is so low (e.g. rarely above ~20), it is very hard to constrain the possible formation of even a single event. So many events that are on the edge of the current theoretical uncertainties are called "difficult to explain." This quote that you provided is perfect demonstration of this, since there actually is no evidence that priomordial black holes even exist to begin with!!!
For the event that you specifically asked about, GW190814, it is difficult mainly because of the very low mass of the secondary. But, this could be explained by, for example, isolated evolution of a binary star with an initial very low mass ratio, where the less massive component could evolve into a neutron star or a black hole. There are people who study isolated binary evolution and they claim that all LIGO sources can be explained by their channel, but they tune their simulations to produce the LIGO results. There are people who study dynamical channel evolution and claim that there is strong evidence for this or that source to be from this channel. People have to write papers and fund their students, and they do a lot of the ground-work that will make more rigorous investigations plausible once we do have larger samples of known sources. In the coming decades, it is expected that the third generation of GW detectors will provide on the order of MILLIONS of detections and then population-level constraints on formation will be much more convincing, and these detectors will have very high signal-to-noise system and constraining formation of individual events will also be more convincing.

I read elsewhere that, before LIGO, no stellar-black-holes had been detected above about 15.65 solar masses, but that reference did not say that none were expected above that mass....

I hope I've clarified this by now! If you read papers leading up to 2015 from the GW community, for example this review, they were certainly expecting to see binary black holes with masses greater than 15 M$_\odot$.


How to choose portable, yet powerful telescope

I was reluctant to do an answer because I do not own a NexStar 8 SE, so I can't speak from direct experience. But this information may prove valuable, so here it is.

The smaller NexStar scopes (like the 4) can be used tabletop-style. The scope is small enough so that the base is sitting just fine on a flat surface.

The 8 might be a little too big for that - I'm not 100% sure. However, if that's the case, there's a simple remedy. Take a piece of plywood or plastic, bigger than the base, thick enough to be rigid, and improvise a way to attach it to the scope base (use bolts / nuts with the same thread as the ones on the tripod). The wider area should stabilize it, and the extra piece of material is flat and easy to carry.

At your local home improvement store they surely carry plywood conveniently cut into round pieces of various diameters. Just get one of those, get the appropriate size bolts and washers, and that should be it. Maybe put 3 pads on the bottom of the plywood so it doesn't sit on the bolt heads. Get a spray can of shellac or urethane and give it a nice protective coating when you're done.

The NexStar mount should withstand the challenges of travel pretty well. It's the optics I'm more worried about.

As others have said, most tables are too wobbly for astronomy. Use a sturdy table, or some surface that doesn't move around, like the wide edge of a concrete balcony or something.

Another option is a collapsible dobsonian. Not the pseudo-collapsible ones like the big SkyQuest series, but a true collapsible scope like the Sumerian dobs. A 16" dob built this way folds down into the volume of a regular travel suitcase. A 10" collapsible dob is like a small suitcase. These are built for the specific purpose of being taken on long trips away from home.

http://www.sumerianoptics.com/

The 8" Alkaid Sumerian scope, at 800 mm focal length, is a pretty small dob already. The poles are probably pretty short. Should be easy to transport, and doesn't require much space when fully unfolded; you may actually have to place it on a stool or something.

Obsession Telescopes also makes dobsonians of all kinds. Their Classic series are bulky when packed, but the Ultra Compact series is basically a true collapsible dob which might be worth considering:

http://www.obsessiontelescopes.com/

You don't need a table for these scopes, since they are sitting on the floor. Table wobbliness issues are eliminated. :)

There are equatorial platforms made for all these scopes, so you could even use them for photography.

I do not own Sumerian or Obsession telescopes, but the feedback I hear through the grapevine is good. These are quality instruments that could be used as a general-purpose telescope even at home.

You could also take the DIY route and build a collapsible dob yourself, if you're so inclined. Many have done this, with great results.


To start, if I'm reading your question correctly you've got the general galaxy evolution model backwards, at least as far as morphology goes. At a very high level the picture goes like this: most if not all (large) galaxies form as spirals, then at varying points in their lives they merge with other large galaxies (either ellipticals or other spirals) and the product becomes an elliptical galaxy. This happens faster and more often in dense environments like galaxy clusters, and takes longer to occur for more isolated galaxies like the Milky Way (which will merge with Andromeda in a few billion years).
Question on isolated spirals
If what you were asking about is isolated spirals, well no I don't think an isolated spiral would ever turn into an elliptical galaxy by itself (at least not on life-age-of-the-Universe timescales). It may eventually start to resemble an elliptical galaxy (in that its stellar halo may grow) depending on how much it gets harassed by other small dwarf galaxies. No galaxy is truly isolated, there is always a hierarchy of smaller galaxies nearby and they gravitationally interact with the host galaxy. For example the Milky Way has the Sagittarius dwarf galaxy, and the Large and Small Magellanic clouds which will merge with us soon (it's basically happening now for Sagittarius, and the Magellanic clouds will be a few billion years). None of these events will be impactful enough to turn the Milky Way into a true elliptical, that will only occur with it merges with Andromeda.
Question on morphology versus internal dynamics
To specifically answer your other question on how morphology relates to the internal dynamics of a spiral galaxy: it most definitely does, but the relationship is complicated. When you see a spiral galaxy with a central bar, or spiral arms, these features are being generated by dynamical instabilities within the galaxy. This could happen thanks to a number of causes, which may operate from within the galaxy itself, or from outside the galaxy. One of the most common means of exciting a bar or spiral arms in a disk galaxy is by interactions with smaller dwarf satellite galaxies that orbit around the larger spiral galaxy and gravitationally perturb it. If this part of the answer sounds like a cop out, well it sort of is. Exactly how disk perturbations like bars and spiral arms are started, how they work, what the heck they do to the galaxy, these are all truly bleeding edge research topics that astronomers are working on right now.
Hope that helps!


Can you see man-made lights on the dark side of the Earth from the surface of the moon with the naked eye?

It's highly doubtful you could see any normal light source on the surface of the earth.
Using $$\text{brightness} = \frac{\text{luminosity}}{4 \pi \times \text{distance}^2}$$
(with brightness in watts, and luminosity in watts per square meter.
and distance to moon of $3.84 \times 10^8$ meters.)

Try a hypothetical light source 100 megawatts output, all visible light, no heat.

$$\text{brightness} = \frac{100  \times 10^6}{4 \pi \times 1.474 \times 10^{17}}$$

$$\text{brightness} = 5.4 \times 10^{-11} \text{ watts per square meter}$$ at the lunar surface.

That's pretty dim. By, contrast sunlight at earth's surface runs about 1300 watts per square meter.
In reality it'd take about a gigawatt to produce 100 megawatts of light in the visible range.
That's about what it takes to power a city of a million homes.
Cities also bounce most of the light they do produce off the ground, which'll have an albedo of somewhere around 0.3. So with ordinary city lights it'll take over 3 gigawatts to reach $5.4 \times 10^{-11}$ watts per square meter on the lunar surface.

You might fare better with a big laser. The Apache Point Observatory Lunar Laser-ranging Operation picks up multi-photon signals from the Apollo retroreflectors using only a 1 gigawatt laser and a 3.5 meter telescope. As the article states, the laser beam only expands to 9.3 miles in diameter on the way to the moon, so you might see it wink at you.

At 10 parsecs, the sun has a magnitude of 4.83. It'd be visible on an average night. That magnitude corresponds to a brightness of 3X10e-10 watts per square meter, about 5.6 fold brighter than our hypothetical earth based light source. That puts our light at magnitude 6.5 to 7. Naked eye visibility runs to about 6.0


During an eclipse, does the size of the moon and sun match perfectly?

To expand on what taupunkt commented, on average, according to the article Solar Eclipse: What is a Total Solar Eclipse & When is the Next One? (Rao, 2014), it is a unique quirk of nature that

  The sun's 864,000-mile diameter is fully 400 times greater than that of our puny moon, which measures just 2,160 miles.  But the moon also happens to be about 400 times closer to the sun than the Earth

This is often stated in middle and high school science textbooks.

However, the orbits of the Earth around the sun and the moon around the Earth are elliptical, as illustrated in the diagram below:

Image source: NOAA

This results in the exact ratio mentioned above not occurring resulting in an annular eclipse, where the moon is too 'small' to fully cover the solar disc.

A comparison between a total and annular eclipse is shown below:

Image source: ScienceJedi

NASA's Eclipse data page, in particular, their Eclipses and moon orbit information page discuss in detail (with considerable data), the

  interaction and harmonics of the synodic, anomalistic, and draconic months not only determine how frequently eclipses occur, but they also control the geometric characteristics and classification of each eclipse.

Within the NASA link, data about how much of the sun's disk that would be obscured by the moon has been calculated (from observations and predicted). For example, using the data for solar eclipses from 2011 to 2020, The average fraction of the sun's disk that is and is expected to be obscured (this is referred to as 'Eclipse magnitude') in an annular eclipse varies from about 0.95-0.99 (95% to 99%) for that time period.

A catalog of eclipses up to the year 3000 is provide on this NASA page.

The greatest difference would occur if the moon is at its furthest from the Earth (apogee) and the Earth closest to the sun (perihelion).


The energy density of DM (to be fluid,state parameter=-2)increases with time. Does it violate the concept of accelerating universe?

For phantom energy model we have $$H^2 = H_0^2\Omega_pa^{-3-3w_p}$$ such that $w_p < -1$

$$\dot{a}^2 = H_0^2\Omega_pa^{-1-3w_p}$$

$$\dot{a} = H_0 \sqrt{\Omega_p} a^{-1/2(1+3w_p)}$$

$$\int a^{1/2(1+3w_p)}da = \int H_0 \sqrt{\Omega_p}dt$$

Now at this part we should define the integrals upper and lower bounds. When we set time from $t → t_0$

$$\int_a^{1} a^{1/2(1+3w_p)}da = \int_{t}^{t_0} H_0 \sqrt{\Omega_p}dt$$

We get something like,

$$\frac{2} {3(1+w_p)}[ 1 - a^{3/2(1+w_p)}] = H_0 \sqrt{\Omega_p}(t_0 - t)$$

$$a^{3/2(1+w_p)} =1 -  \frac{3(1+w_p)}{2} H_0 \sqrt{\Omega_p}(t_0 - t) $$

Let us take $w_p = -2$

$$a^{-3/2} =1 +  \frac{3}{2} H_0 \sqrt{\Omega_p}(t_0 - t) $$

So

$$a =(1 +  \frac{3}{2} H_0 \sqrt{\Omega_p}(t_0 - t))^{-2/3} $$

here $y$ axis is the scale factor $(a)$ and $x$ axis is the time $(t)$

For $C = H_0 \sqrt{\Omega_p}$ and $a=t_0$

As you can see scale factor goes to infinity

For the acceleration equation

$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3c^2}\epsilon_p(1+3w_p)$$

For $w_p=-2$

$$\frac{\ddot{a}}{a} = \frac{4\pi G}{3c^2}5\epsilon_p$$

From the graph and also from here its clear that $\ddot{a} > 0$.


Explaining the Drake Equation on a smaller scale

No need to make it complicated: what about this...

Just scribble a rectangle on a piece of paper, and say "there are 100 billion stars in our galaxy"....

Then, color off (let's say) 1/3 of the rectangle, and say "only one third of those are the sort of star that could have life, so that's blah billion"

Then, color off (say) 9/10ths of that box, and say "we believe about 90% of those have planets - so that's blah billion"

Then, color off (say) 1/20th of that box, and say "of those with planets, it seems that about 1 in 20 have Earth-like planets. Now we're down to blah billion..."

and so on.

(Note: the Drake equation has a number of fairly silly terms relating to "nuclear war!", which were added as political sops in that era; suggest ignore these unless you want to sound 90 years old!)

So just scribble a box or draw a line on a piece of paper ...  or maybe use "a bag of marbles" as the other answer suggests.

Just BTW there is in fact an entire documentary (I noticed it on "Netflix") called "The Drake Equation" which does exactly what you say...

.. it is not really very good as I remember.  I think the guy simply draws a line in the ground, to do the "fractions" demo, you know?  (ie, they just erase more and more of the line). It doesn't need to be more complicated than that.

It's worth noting that the Drake equation simply points out:

(i) if you multiply those three or four fractions together, you get the number of civilizations in the galaxy. Which is self-evident.

but, the whole point is

(ii) we have utterly no clue - not even vaguely - what most of the fractions are,

You could say it's a written formula, which, helps clarify our thinking on, something we are utterly clueless about. So rather than just vaguely saying "we're utterly clueless," we can speak more clearly about the nature of our cluelessness!

although interestingly,

(iii) very admirably, the issue of "How many stars have planets?" ... one could say that issue has been somewhat settled these very years, as we speak - that's great.


Why doesn't Earth's axis change during the year?

My understanding is that the Earth's axis points in the same direction in space during its entire orbit around the sun. And this is what causes our seasons.

The second statement is correct. The axial tilt is the primary driver of the seasons.

The first statement is not exactly correct. There is a small but persistent change in the orientation of the Earth's axis. The change over the course of a year is small. The change over the course of 13,000 years is large. In 13,000 years the Earth's axis will be tilted by about 47 degrees compared to the current orientation. In yet another 13,000 years the Earth will be oriented very similarly to how it is oriented now. I'll write more about this later.

  My question is why doesn't the axis follow the orbital path (kind of like a car driving around a circle)?

That is not a good mental model. The only contact between your car and the ground are the bottoms of the four wheels on your car. The forces at these four small contact areas are what change your car's orientation and change your car's velocity vector. This means that the torque on your car as you turn the steering wheel and the centripetal force on your car that makes your care negotiate a turn are intimately connected. Turn the steering wheel gently and the heavy front end of your car follows suit. Assuming a gentle turn, the light rear end of you car soon follows suit. If you turn too hard, your car starts to skid instead of turn. Turn even harder yet and you risk having your car flip and roll. The motion of your car depends on that intimate coupling between force and torque.

Gravitation doesn't work like that. It instead acts on each and every bit of the Earth, all together, all at once, all the time. There is a slight variation in the gravitationally-induced acceleration across the Earth due to variations in distance to the Moon (or Sun), but these variations are small. The net gravitational force on the Earth is very much decoupled from the small net gravity gradient torque on the Earth.

The principal result of the gravity gradient torque exerted by the Moon and the Sun on the Earth is a very slow but large precession of the Earth's rotation axis (for more, read this wikipedia article on axial precession). This precession makes the Earth's rotation axis rotate about the normal to the ecliptic at a rate of one revolution per 25772 years. There are lesser effects collectively called nutations and polar motion that result from gravity gradient torque. These lesser effects are much, much smaller in magnitude but have much, much shorter periods. The largest of these lesser effects results from the 18.61 year cyclical variations in the lunar node. Even this largest component of nutation is small, two orders of magnitude smaller than the large but slow precession.


what is the percentage of stars with planetary systems?

For the purposes of the Drake equation you may as well assume that every star has a solar system.

At present, the exact fraction is unknown since the search techniques are limited to finding planets with certain characteristics. For example, transit searches tend to find close-in, giant planets; doppler shift surveys are also most sensitive to massive planets with short orbital periods, and so-on.

One can try and account for this incompleteness, but although some examples have been found, we are in the dark about the fraction of stars that have an Earth-sized planet at distances further than the Earth is from the Sun, or the fraction of stars that have any kind of planet orbiting beyond where Saturn is in our solar system.

For the Drake equation, you need to know the fraction of stars with planetary systems times the number of "habitable" planets per system is. There have been attempts to estimate this from Kepler data.

Possibly the best estimate at present is from Petigura et al. (2013), who estimate that 22% of sun-like stars have an earth-sized (1-2 Earth radii) planet that receives between 0.25-4 times the radiative flux of the Earth.

Obviously this is a lower limit because it doesn't include planets smaller than the Earth. It also doesn't include the moons of giant planets, which may also be habitable.

We still know very little about this fraction when it comes to M-dwarf stars, which are the most common type of star in the Galaxy...


Neutron stars do indeed shine and some like this one have been observed in visible light. Like white dwarf stars, if they were scaled up to be the size of the Sun, they would outshine the Sun and most other stars in the galaxy. It is only their small size that make them hard to observe.

As for neutrinos, it seems they do too. This link might help explain neutrino activity in / from neutron stars further.


Which stars, identifiable with the naked eye, have exoplanets?

Stars with an apparent magnitude of 7.0 or greater are typically not visible to the naked eye. So if you take a list of exoplanetary host stars, such as this, and you sort by apparent magnitude, in ascending order, then you will see a list of identifiable stars known to have exoplanets, until you reach an apparent magnitude of 7.0 on the list.

Finding the star can be trickier, but Wikipedia will usually show you what constellation you can find the star in. With a little work, and maybe the help of some other posts on this site, you could probably identify which stars are visible in your area and time.

Most exoplanets have not been given formal names themselves, but instead a lowercase letter is appended to the end of the name or designation of the star, for example "Beta Geminorum b (abbreviated β Geminorum b, β Gem b) and HD 62509 b" for Pollux b, the first planet discovered around Pollux. Incidentally in 2014, the International Astronomical Union did release a process for giving proper names to exoplanets, and Pollux b was given the name Thestias. If you look up other exoplanets on Wikipedia, you can find out whether they have been given proper names as well.


How do scientists know that distant parts of the universe obey the physical laws exactly as we observe around us?

We don't know in general but to the extent we can measure, the laws seem to be the same, even if conditions are not.
For example radioactive decay: We know how fast various elements decay, and we can observe the results of radioactive decay in distant supernovae.  The conclusion is that, for at least some elements, the rate of radioactive decay is the same on Earth as it is in distant supernovae.
After accounting for redshift, spectral emission lines remain unchanged by distance. This implies that the fine-structure constant is indeed constant.
Distant galaxies have gravitational fields, and interactions between galaxies proceeds in the same way in distant galaxies as it does in local ones. Eventually, the justification is philosophical:  There is no observational reason to believe gravity behaves differently in distant parts of the universe, and so we believe that it does not,
In the extreme conditions of the early universe, some physical laws were different.  For example, instead of distinct electromagnetic and weak fields, there was a single Electroweak field.  But this can be described as single "law" with the electromagnetic and weak interactions being just the low energy approximation of the electroweak interaction.
So if it were discovered that Gravity (for example) was working differently in distant parts of the universe, but that there was a consistent pattern or rule for how it varied, then that would simply become the new theory of gravity (with general relativity becoming only the local approximation to this new law).
There is a more fundamental assumption: that the behaviour of matter and energy in the universe can be modelled by "laws".  There are no angels dancing on pinheads. The justification for this is strictly in the realm of philosophy.


First of all it has to be stated that it is difficult to have a scientific discussion about multiple universes. The problem with multiple universes is that it can be logically deducted from assumptions and hypotheses, but not tested nor empirically verified. This makes discussions around this topic, like string theory, highly "religious".

There are different multiverse theories coming from cosmology, physics and philosophy. One of the theories comes from the cosmological theory of cosmic inflation. Cosmic inflation states that there are always fluctuations on the microscopic level and this results in energy waves that can form into matter and anti-matter pairs. The cosmic inflation now states that at the beginning of our universe there was a cosmological constant or a big vacuum that is by nature unstable and caused the microscopic fluctuations to happen on a cosmic size and thus resulted in a rapid expansion of the early universe. This cosmic inflation is a variation of the (hot) big bang theory, that explains a number of observations in the universe that are not explained by the (hot) big bang theory. What it doesn't explain is that if I have an energy fluctuation at microscopic level, it can happen that this energy wave splits into two matter pairs or an anti-matter/matter pair. Later these pairs annihilate each other and what has happened at microscopic level evens out on macroscopic level. However, when particles appear at microscopic level, what needs to happen to let these particles exist for a longer time to have stars evolve from this matter.

Now back to the multiple universes. The going theory now is that if such a thing can happen to our universe, such a thing can happen in the endless vastness that contains our universe and multiple universes can start and exists next to each other. These multiple universes might also explain that when I have a cosmic fluctuation, I have two universes that come into existence. One with the matter and one with the nati-matter. Bear in mind!!!! These are all philosophical hypotheses and none of it can be verified, falsified or empirically tested.

If I now have made it logically possible to have multiple universes coming from different cosmic fluctuations, I can now have different universes. For instance, I can have for each matter universe a corresponding anti-matter universe to get the energy balance evened out. Now if I have several universes with each different compositions, I can also have universes where, at least, the laws of nature apply differently. In a universe without dark matter, laws of physics apply differently. Whether the "laws of nature" are totally different in other universes, is difficult to scientifically proof. Especially with theories that are build on our laws of nature.

When you think we're already on thin ice here, philosophers can make it even more unreal. If I have cosmic fluctuations that can happen under different  circumstances, which result in universes with different compositions that then have a different application of the laws of nature, I can equally well argue that there will be universes where certain laws of nature don't exists, while there are some particles that don't exist in that universe.

So, to answer your question briefly, there are theories that say there are other universes that have different laws of nature. However, multiverse theories are highly speculative and are difficult to proof. So, whether these theories are right, we'll probably will never know.


How did the Hulse-Taylor Pulsar Provide Confirming Evidence of General Relativity?

One of the predictions of general relativity is that certain objects can give off energy in the form of gravitational radiation. This means that over time, the orbits of the two neutron stars should decay, and they should come closer to each other. Using relativistic formulae, it is possible to predict the changes in the orbits and the energy of the emitted gravitational waves; analysis of the system showed that general relativity's predictions were correct.

Studying gravitational waves requires linearizing the Einstein field equations. However, these can be reduced to certain simpler formulae when it comes to detecting waves from a binary system, as was the case with the Hulse-Taylor binary.


How much do the mascons on the moon affect surface gravity?

Upon some google-ing and wiki-ing I found this image of a gravity map of the moon:

That scale up the top is measured in milli-Gal which is thousandths of a cm/s^2. For scale gravity is ~9.81m/s^2 which equals ~981000mGal. The difference between gravity at sea level and the top of mount everest is 2Gal, or 2000mGal, which is 0.2% of average gravity. On the moon, the variation seems to be about 1000mGal, with the average being 162000mGal, which is about 0.6% of average gravity.

In conclusion, on the surface, there is no noticeable difference. Not to a human anyway.


Can a tidally-locked planet maintain an obliquity

Well, I can't do the math, but if an intuitive answer is OK:

  If a planet is tidally-locked, can it have an obliquity?

I'm going to cheat and give you an answer using moons, because many moons are are tidally locked to their planets.

Io is a funny one, it's slightly eccentric orbit and proximity to Jupiter causes it to resurface pretty regularly.   It's axial tilt isn't listed and might not be known.

Europa.  Axial tilt  0.1°

Ganymede.   Axial tilt 0–0.33°

Callisto.  Axial tilt zero

Titan, to Saturn,  Axial tilt zero

Triton, to Neptune,  Axial tilt zero

Our Moon is a bit of an oddball, it has an Axial Tilt that's 1.5424° to it's orbit.   click here for picture.   Why our Moon varies a bit more than the others, I'm not sure, but it might be the proximity to the sun which keeps the Moon from orbiting more directly over Earth's equator and causes some imperfection in it's tidal lock.  (Kind of a guess, I'm not sure on this part)

But, simple answer, Tidal locking with solid bodies by definition, the same side of the planet or moon faces the object it orbits all of the time.   Obliquity doesn't make much sense if the planet's primary rotation is in sync with it's orbit around an object.   (Tidal Locked planets do in fact rotate).  So, it seems counter intuitive that a planet could have a high obliquity and at the same time, be tidally locked to the object it orbits.  Mercury and Venus, work as examples for this common sense argument too.  Mercury is, in a sense tidally locked but due to it's eccentric orbit it's in a 3:2 ratio.  Mercury has 0 obliquity and Venus appears to be on it's way to being tidally locked to the sun and it has a 177.36 degree obliquity, but that's essentially 2.64 degrees, if Venus wasn't rotating in the opposite direction of the other planets.   As Venus moves into being tidally locked to the sun, it's rotation should approach 0 (or 180) obliquity, the slowing of it's rotation in comparison to it's rotation matching it's orbit suggests that this is the only likely outcome.

If you want to check the axial tilt of other moons, more listed here.

Now, there may be some factors that could, just maybe, lead to exceptions, perhaps A planet with very high inclination might be a wild card.  Or a rotation that's in a 2:1 resonance with the tidal locking.  A binary star system could have some odd orbits too (See Pluto's and Charon's, satellite Nix here.  Now, Nix isn't spherical, so it's prone to much more pronounced oddities, but a planet around a binary star could exhibit some strange orbital resonances, but I suspect using a binary star system to find an exception isn't what you had in mind.

I'm 90%-95% confident in saying that obliquity should tend towards zero as the planet becomes tidally locked to a single star that it orbits.

  A long way in the future it will be tidally locked. As it slows, will
  the obliquity also be eroded or will other forces (such as the orbit
  of the moon) allow it to maintain an obliquity?

This is a tricky one because if the sun wasn't going to go Red Giant, the Earth would eventually become tidally locked to the Moon (in 10-50 billion years), and not the sun.   If the Earth didn't have a moon, then it could get tidally locked to the sun, in, oh, I don't know, 100-300 billion years as a guess.   The Moon is the dominant tidal force on the Earth, so in it's current set-up, the Earth can't get tidally locked to the sun.   If the Moon was smaller, then maybe somebody could run the numbers on this question.   If the Moon was 1/3rd it's current mass then the Lunar and Solar tidal forces would be similar, and . . . then it becomes an interesting question.

Curious sidebar, while the Moon currently stabilizes Earth's obliquity, that's not always going to be the case.  As the Moon moves further from the Earth, the Earth's obliquity is expected to show much more pronounced fluctuations, but that doesn't have anything to do with the Earth being tidally locked to the sun.

Hope that helps.  Corrections welcome.


Does Saturn raise tides on Titan's lakes?

Sort of...  Titan is tidally locked to Saturn, so points the same side toward Saturn all the time.  The tide caused by Saturn on Titan should be nearly unchanging, except that Titan has an elliptical orbit, and is at times closer.  This cases a significant tidal effect, but it doesn't travel like on Earth.  See this article for more information.
This article states that this tide deforms Titan about 33 feet.  I do not know how much the lakes themselves would change though.
The sun should also cause a very minor tide because it is further away.


Aren't there any rock or similar firm material on/in the gas giant planets?

The inside of a giant planet is not like regular gas.

First of all, it is hot. You are away of how it gets hot inside the Earth (causing volcanos). It is also hot inside Jupiter, but since Jupiter is bigger, it is hotter. It would be hot enough to vaporise rock under "normal conditions".

The pressure is immense, and things stop behaving as you are used to when the pressure is very high. For example, all the atoms in a gas are pushed so close together that they are touching (like they would be in a liquid). This means that there is no boiling point and no real difference between the gas and liquid phases. At extreme pressures, hydrogen atoms are pushed so close together that their electrons can start to flow from atom to atom, forming a fluid metal.

Pressure that can push atoms so close that they can turn hydrogen to a metal mean that you can't send a probe to this region. You can't swim in a metallic hydrogen sea.

Nevertheless, heavier atoms like iron, silicon, carbon, oxygen, will tend to fall towards the centre of the planet, and so there may be a since rocks are made mostly of atoms like iron, silicon, carbon, oxygen, you could describe this as a "rocky" core.  But don't think this is a kind of solid surface that you could stand on.


This appears to be one of Sergii Tsololo's more convincing efforts in the synthetic "space" image genre.
It is listed on various stock photo sites as
"Blue and purple space galaxy,"
or with added 4-way spikes,
"Blue space starfield with flares."
A real astrophoto would identify the subject and not be tagged "abstract" or "fantasy."

To check it another way, I submitted the image to
Astrometry.net,
whose plate solver failed to match the field within their 10 minute time limit.
When I submitted an amateur photo of the
Trifid Nebula
as a control, Astrometry.net found a solution in less than 1 second.


What are the positions of the Earth, Moon, and Sun at a time when there is the least difference between high and low tides?

'Neap' tides when there is the least difference between high and low tide, generally occur around the time of the first and third quarter moons.
'Spring' tides when there is the largest difference between high and low tide, generally occur around the time of the New and Full moons.
The exact timing is complicated by your position on earth, the shape / size and depth of the ocean, coastlines also the moons distance from the earth and earths distance from the sun.


Do space objects stay in orbit once they get sucked in? Can they escape?

If it can get captured, it can escape: we know this because the law of gravity is time-symmetric. If there is a trajectory that leads to a capture, then by running time backwards, there is a trajectory that leads to escape.
What happens with small objects like 2020CD3, is that they pass into the space between the Earth and moon, and it is the interaction between the Earth and Moon (and sun) that slows them down and allows them to enter into a long looping orbit.
But this orbit is too close to the moon to be really stable.  Usually they go around in a complex orbit that changes each time, as they are affected by the gravity of the Earth, moon, (and sun).  Usually after a few orbits they will have another interaction with the moon, but this time it will speed the object up, and eject it from the Earth's gravity well.  Interactions that lead to a permanent capture are rare.
There are no objects in Earth orbit that have been captured.  Asteroids are rather rare in the inner solar system, and the moon probably does a good job of ejecting any that do get into an orbit around the Earth.  There are two captured asteroids around Mars, and many of Jupiter and Saturn's moons may have started out as asteroids.  Triton is also probably a captured body. So it is possible for a body to get into a stable orbit from capture, but it is rare.
There is no sucking effect.  Instead the orbit of the asteroid around the sun gets tangled up in the gravity field of the Earth and moon for a while, but eventually it will almost always escape.


How could the universe be a few light years across just 1 second after the big bang?

Your question hinges on the misconception that nothing can travel faster than light. No physical object can move faster than light, but plenty of non-physical things can. For example, take the straight line $y = mx + c$. If one varies $c$, the x-intercept of the line also changes. It's possible that this x-intercept moves faster than light for a suitably small $m$. The caveat is that this intercept is a mathematical point, and it cannot be used to transfer information. Relativity isn't violated.

In the case of the universe's expansion, the physical objects in the universe aren't moving faster than light, but they can appear to do so because space is expanding. Let's take Alice and Bob, separated by one meter, and let the space between them expand at some rate. There's no speed-of-light limit on how fast this rate can be. If space expands by a billion meters per second, then one second later, Alice and Bob would be separated by a billion and one meters, but they wouldn't have traveled faster than the speed of light since from their own point of view, they're both stationary.

In the same way, our best guess of the current universe is that the expansion is speeding up. A very far away object might eventually appear to recede from us faster than the speed of light. Relativity remains unviolated, although it means that we'll eventually be causally disconnected from these faraway objects (that is, we cannot affect or even observe them, and vice versa). Something of this sort - exponentially-increasing expansion - happened in the very early universe.


There is a lot of scope to provide a very detailed answer here. The rate depends on what sort of a galaxy you are considering and when, what its star formation rate is (or was) and what its total stellar mass is.

A good reference is the Annual Review of Astrophysics article by Maoz et al. (2019). This says that for a Sbc galaxy like the Milky Way, the specific rate ( the rate per unit of stellar mass in the galaxy) of type Ia supernovae is $10^{-13}$ per year per solar mass of stars. For a Milky Way stellar mass of $\sim 6\times 10^{10} M_\odot$, this means a rate of 0.006 per year, or one type Ia SNe every $\sim 200$ years.

In the local universe this rate scales as about $M^{-0.5}$, where $M$ is the galaxy mass (valid over about 5 orders of magnitude for $M>10^7 M_\odot$; Brown et al. 2018). So the specific rate of type Ia supernovae is larger in smaller galaxies.

The rate also decays after a galaxy is formed, or after a burst of star formation as something like $t^{-1}$.


Would the heliosphere disappears if our sun becomes a white dwarf?

Yes the heliosphere will disappear. As you know, the heliosphere exists due to the solar wind, which consists of charged particles that follow the magnetic field lines starting from the Sun and extending far into the universe (of course at some point they return to the Sun, as the loops need to be closed, but it's so far away that we usually assume it to be at infinity). Now when the Sun goes Red Giant and then loses its outer shells (planetary nebula), it will become a white dwarf. Then there won't be any wind originating from its surface and therefore there won't be any "magnetic bubble" left.
The heliosphere stays roughly the same in size as long as the Sun remains in its Main Sequence stage, as a star remains relatively stable during this time of its life.


What is the formula to figure out how far away a star is?

There is no single formula that tells you how far away a star is. In general, it is difficult to find the exact distance to a star just using its brightness since the brightness of stars vary.

For close stars parallax is used to accurately tell the distance to the star. For stars farther away more complex ways of finding the distance to the star must be used based on how far away the star is, these methods make up the cosmic distance ladder.


What's the mass of the interstellar neighborhood of the Sun?

You don't have to guesstimate to come up with the answer.

What you do is look at the dynamics of stars with respect to the Galactic plane - in particular, the velocity dispersions of stars with known distances from the plane, combined with a reasonable assessment of where the Sun is with respect to the plane (close), yields an almost model-independent assessment of the mass density in the disk of our Galaxy in the vicinity of the Sun.

This work was done using the Hipparcos cataogue by Creze et al. (1998). They found that the local mass-density (all forms of matter) in the Galactic disk was $0.076 \pm 0.015 M_{\odot}/pc^3$.

Thus within 50 light years (15.3 pc), because the Galactic disk has an exponential scale height much bigger than this (about 100-200 pc), we can assume a constant density and derive a total mass in the solar neighborhood of $1100\pm 200\ M_{\odot}$.

This mass is almost compatible with the total mass estimated to be in stars, white dwarfs, neutron stars, brown dwarfs and the interstellar medium gas. For instance, according to Chabrier (2001), the total mass density is almost accounted for in the form of 60% main sequence stars plus white dwarfs and neutrons stars, about 5% brown dwarfs and 30% gas, and so the contribution of dark matter to the local disk density is probably very small. Dark matter is not distributed in the same way as normal matter; it did not collapse to a disk, because it is dissipationless and probably exists in a spherical halo. So this consistency between the local baryonic matter density and the total mass density is not an argument against the dark matter hypothesis.

EDIT: I found a few estimates of the local dark matter density that use the kinematics of local stars as a constraint. The range of densities quoted is from 0.008 to 0.02 $M_{\odot}$/pc$^3$ (Bovey & Tremaine 2012; Garbari et al. 2012) i.e. small compared with the contribution from stars and gas.


Is there an IRAM satellite that measures thermal radiation at 250 GHz, or was this a ground-based instrument?

The Nature paper by Bertoldi et al. (2006) says:

Our millimetre observations were performed with the Max-Planck Millimeter Bolometer (MAMBO-2) array detector at the IRAM 30 m telescope on Pico Veleta, Spain.

This is a ground-based telescope in the Spanish Sierra Nevada mountains at 2850m to try and get above as much of the precipitable water vapor, which has a very large effect on sub-mm observations, as possible. Its website is here
Since it's common to refer to the Spitzer and its predecessor, the IRAS mission, when discussing far-infrared observations, I suspect the authors typed "IRAM", thought "IRAS (satellite)" and made a simple typo that was not caught before publication.


Are there accurate equinox and solstice predictions for the distant past?

I wrote
https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-astro-13008.c
to find historical solstices and equinoxes. The full results are at:

https://github.com/barrycarter/bcapps/blob/master/ASTRO/solstices-and-equinoxes.txt.bz2

Accuracy:

I'm using the EARTH_IAU_1976 precession model and the
EARTH_IAU_1980 nutation and obliquity models, which are
the only models supported by the CSPICE library I'm
using:

http://hesperia.gsfc.nasa.gov/ssw/stereo/gen/exe/icy64/doc/frames.req

This is the same library and the same models NASA uses to
do its own calculations, but the models are known to have
limited accuracy. Quoting "http://aa.usno.navy.mil/faq/docs/SpringPhenom.php":

The times given in the tables are accurate to within two or
three hours for 25 to 5 BCE, and one or two hours for 4 BCE
to 38 CE. The uncertainty in these times arises from the
stochastic, that is unpredictable, change in the Earth's
rotation rate.

The accuracy gets worse the further back OR the further
forward you go: not only don't we know precisely about the
Earth's rotation in the past, but we can't even predict the
Earth's rotation precisely for the future.

Sample output:

EQU 511720269.432607 A.D.   2016-03-20 04:30:01
SOL 519734120.174820 A.D.   2016-06-20 22:34:11
EQU 527826138.004142 A.D.   2016-09-22 14:21:09
SOL 535589116.137776 A.D.   2016-12-21 10:44:07

Format:

The first column indicates whether this is a solstice or an equinox.
The second column is the ephemeris time of the
solstice/equinox. If you're doing serious astronomical
work, this is the column you should use. The approximate definition is the number of seconds since the J2000 epoch. The precise definition as used in CSPICE is: https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/FORTRAN/req/time.html#Ephemeris%20Time%20(ET)
The remaining columns indicate the UTC time of the
solstice/equinox in a more human-readable format:

The CSPICE libraries assume that the Gregorian
calendar reformation occurred on 4 October 1582
(meaning the day after 4 October 1582 was 15 October
1582). Looking at these lines:

SOL -13191695511.794357 A.D.   1581-12-11 20:07:27
EQU -13183992131.003845 A.D.   1582-03-10 23:57:07
SOL -13175951250.170158 A.D.   1582-06-12 01:31:48
EQU -13167875920.223862 A.D.   1582-09-13 12:40:38
SOL -13160138634.917915 A.D.   1582-12-22 01:55:23
EQU -13152434815.793312 A.D.   1583-03-21 05:52:23
SOL -13144394485.035870 A.D.   1583-06-22 07:17:53
EQU -13136319216.460808 A.D.   1583-09-23 18:25:42

you can see that the dates of the solstices/equinoxes jump
ahead 10 days per the reformation.

The CSPICE libraries use the Julian calendar prior to
4 October 1582. In reality, the Julian calendar was
introduced in 46 BCE:
https://en.wikipedia.org/wiki/Julian_calendar
Prior to 46 BCE, there were other calendar systems in
use, but the CSPICE libraries assume the Julian
calendar goes back indefinitely:

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

My calculations go back to 13201 BCE (the limits of DE431,
the ephemeris I'm using), and it's possible human beings
weren't even using calendars regularly at the time: quoting
"https://en.wikipedia.org/wiki/History_of_calendars#Prehistory":

A mesolithic arrangement of twelve pits and an arc found in
Warren Field, Aberdeenshire, Scotland, dated to roughly
10,000 years ago, has been described as a lunar calendar
and dubbed the "world's oldest known calendar" in 2013.

Notes:

It takes the Earth 365.256363004 days to revolve around
the Sun with respect to the fixed stars, but the time
between vernal equinoxes is slightly less (365.242190402
days) because the position of the vernal equinox moves
(precesses) with respect to the stars. Source:
http://hpiers.obspm.fr/eop-pc/models/constants.html
The Gregorian calendar's average day length of 365.2425
is much closer to 365.242190402 days than the Julian
calendar's 365.25 average day length, but it's still not
perfect. As noted in
Do solstices and equinoxes shift over time?
if we continue using the Gregorian calendar in the far
future, the equinoxes and solstices will drift
backwards. By 17090 (the limit of DE431), they will look
like this:

EQU 476198945887.238159 A.D.  17090-02-22 08:56:59
SOL 476207018540.040894 A.D.  17090-05-26 19:21:11
EQU 476214808218.146362 A.D.  17090-08-24 23:09:09
SOL 476222655067.609985 A.D.  17090-11-23 18:49:59

about a month behind their "regular" times.

MY EARLIER PARTIAL ANSWER FOR REFERENCE:

Since HORIZONS (http://ssd.jpl.nasa.gov/?horizons) and SPICE (http://naif.jpl.nasa.gov/naif/tutorials.html) can compute the ecliptic and solar position back that far, it should be possible to compute equinoxes and solstices with reasonable accuracy. However, I haven't been able to find a site that actually lists these dates (I'm pretty sure USNO did this at one point, but I can't find their list). Other possibly helpful sources/questions:

How did Meeus calculate equinox and solstice dates?
http://ns1763.ca/equinox/eqindex.html
http://data.giss.nasa.gov/ar5/srvernal.html
http://www.usno.navy.mil/USNO/astronomical-applications/data-services/spring-phenom (appears to be down at the moment)


Is the "mean orbit" centered on ellipse center or ellipse focus?

But where is this orbit centered?

The true anomaly is the angle as measured from the central body between periapsis passage and the object's current location. The orbit is an ellipse with one of the two foci at the central body. This concept is central to Kepler's laws and Newtonian mechanics.
The eccentric anomaly is the angle as measured from the center of the ellipse between periapsis passage and the projection along the minor axis of the point in question onto the minimal bounding circle that encompasses the elliptical orbit. That minimal bounding circle is necessarily centered on the center of the orbit rather than the central body.
Mean anomaly is a fictitious angle. Unlike true anomaly and eccentric anomaly, mean anomaly does not indicate where the object is. Mean anomaly instead is the angular displacement that a fictitious object in a circular orbit about the central body with the same semi-major axis length as the object in question would have passed through in the time since periapsis passage.

Aside: The image from wikipedia is flat-out wrong. The center is correct, but the size is incorrect. The circular orbit should intersect the ellipse, twice. It should be outside of the ellipse at periapsis and inside the ellipse at apoapsis.


How would the pocket cellular clock work?

The device that you saw is called a nocturnal. It calculates the local time based on the month of the year and the position of Polaris and one or more other stars.

Since stars change position throughout the night, they can be used to determine time; but the positions of stars change throughout the year, so the input of the month of the year is needed. Polaris is used as a relatively fixed reference point.

More information:

Nocturnal (instrument) - Wikipedia


Change of time in vernal equinox by altitude from the Sea level

In other way, does the time of vernal equinox change with altitude?

No.

The equinoxes are defined as the points in time at which the Earth's equatorial plane intersects the center of the Sun. From an Earth-centric perspective, the Sun appears to crosses the equatorial plane from south to north at the March equinox, and from north to south at the September equinox.

Altitude, latitude, and longitude, and length of day do not come into play. In fact, days are slightly longer than are nights almost everywhere on the Earth on the dates of the equinoxes. This is a result of the Sun's angular diameter and atmospheric refraction, both of which mean we can see the Sun when the center of the Sun is a bit (0.83 degrees) below the horizon.


Why is the difference between Apollo and Aten asteroids considered meaningful?

The first point to make is that categorization isn't an exact science.  It's done for convenience and practicality and sometimes, familiarity.  It's not going to be perfect.

The second point to make is that scientists like categorizing things.  Take the Taxonomic ranking of living creatures.  Everything from a paramecium  to an elephant has a species, a genus, a family, an order, a class, a phylum, a kingdom and a domain.   Categories are everywhere in science and they're useful.   Taxonomy classifications aren't perfect, but classifications don't need to be perfect to be useful and provide distinction and quick and familiar identification.   The same is true for asteroid classification.

Both Atens and Apollos are already lumped into one category called Earth-crossing asteroids, so, when needed, they can be discussed together.   It's more common to separate them by the two types, but it's not required.

Earth-crossing asteroids or Earth-Crossers are a sub-category of Near Earth Objects, and Near Earth Objects or NEOs are of particular interest because every so often, an undesirably large object hits the Earth.

NEOs can be divided into Atens, Apollos, Amors, Comets, and Atiras, sometimes called Internal Earth Objects or IEOs, but there are relatively few Atiras and they're sometimes included with the Atens, though technically, Interior Earth Objects aren't Atens if their aphelion is too far inside the Earth's orbit.

NEOs can be classified as potentially hazardous (PHOs) if they are over 140 meters across, per Wikipedia (NASA gives a different definition of PHOs below - so take that definition with a grain of salt).

Because of the infrequent but "could happen at any time" impact possibility, a well funded effort is underway to catalog, track and calculate the trajectory of every NEO larger than 100 meters, called the Near Earth Object Project.  Because so much attention is given to NEOs, the distinction and naming of Atens and Apollos makes perfect sense to me and doesn't seem unreasonable at all, but as pointed out above, there aren't any clear-cut rules when it comes to categorization.

What's most relevant to the NEO project, in addition to size is Minimum Orbit intersection distance or MOID.   PHO's by the NASA article, as opposed to Wikipedia, are categorized by Earth-MOID of less than 0.05 AU and an absolute magnitude (which is related to surface area, which is related to size) of 22 or brighter.

Not all crossing asteroids (determined by aphelion and perihelion) have orbits that come close.   Because orbits take place in 3 dimensions, orbits can be crossing, but never get close.  Pluto, for example, is a Neptune-crosser but the orbits never get close.

I added MOID cause I thought it was interesting, but back to categorization.

The other 7 planets have their own versions of Atens and Apollos, called inner and outer grazers.  See Jupiter crossing objects.

The 7 planets also potentially have libration asteroids, distinct from Crossers.   Libration asteroids are (captured, in horeshoe orbits, or Trojan).   The 4 inner planets have very few libration asteroids.   Jupiter has tens of thousands.  The outer 3 known planets have several known libration objects each.

There's also resonance objects (Jupiter's Hildas, Neptunes Plutinos).   Objects can also be defined by outer orbits (Trans) or inner orbits (Cis), and by prograde or retrograde, which isn't the same that planets appear to do, but orbits relative to the sun can move opposite to the 7 planets.   In the inner solar-system, there are very few retrograde orbiting objects but out beyond the 8 known planets, retrograde is more common.

The dwarf-planet Pluto, for example, is a Plutino, which is a subset of a Neptune-resonant-object, and a subset of a Trans-Neptunian object, which is a subset of a Kuiper-Belt object (KBO).  I mention that only to point out how many orbital categories there are.   Scientists like categories.

There's other categories include Centaurs, Scattered Disk Objects, Oort Cloud Objects and, of-course, Asteroid belt objects and lets not forget comets, which are distinct from Asteroids, though the distinction gets a little fuzzy for dead comets.

Another categorization example, Jupiter has so many trojans, that it's two camps are separated into Greeks and Trojans, which is odd, because Jupiter's Trojan camp is a subset of Jupiter's Trojan asteroids.  That's a naming system that stuck, but which I think is less than ideal.  You can also call them L4 Trojans (Greek camp) or leading the planet's orbit and L5 Trojans (Trojan camp), following the planet's orbit.

The overall point is that there's many of categories of asteroids.  Atens and Apollos seem to me, fine and relevant, not arbitrary.   (want an arbitrary definition, how about when a moon-planet system becomes a duel planet system - that's pretty darn arbitrary).  Many definitions are arbitrary.

(too long?)


Nebulas are not static objects - they evolve.

Initially a nebula might contain mostly (as an example) hydrogen, helium and some other elements in a cloud, but over time that material will clump together to form larger and/or denser material.  Eventually parts of some nebula can even condense enough to form a star and it's planets.  Our solar system was (probably) once part of a nebula.

So there's no single answer to your question.  They can contain most forms of matter.


What is CMB radiation doing to the universe?

It is said that due to the CMB radiation our universe has been expanding?

No. It's the the other way round. The CMB is the result of the expansion of space. In the initial stages after the birth of the universe, there was hot plasma everywhere and light could not travel much farther. The universe was opaque at that time. But slowly as the Universe cooled and atoms began to form, light could travel farther. This light got redshifted due to the expansion of space and can be now detected as the CMB radiation. It's in the microwave region of the spectrum and just above absolute zero(about 2.7 K).

  What do we mean by Dark Energy?

After Hubble's discovery, we came to the conclusion that the Universe is expanding. We expected the expansion to slow down as time progresses but what we observed as that the expansion was accelerating and is still doing so. Physicists and cosmologists did not have any explanations for this and so it was named "Dark Energy". Dark in the sense that it is unknown.


Can you simulate artificial gravity by using an electromagnetic field?

There are two aspects to this question: first, creating a field that can exert a force and second, having a body that is coupled to that field and "feels" the acceleration.

Electric field: You probably want a uniform field of acceleration, like on earth, where the same amount of gravity is present everywhere in our daily life (close to the surface of earth).  This can be achieved by putting sources into floor and ceiling of your ... space station.  With various setups this could create a homogeneous electric field (like a parallel plate capacitor).

Now, use the field by putting charged bodies into it.  I could imagine a space suit with an eletrically charged layer, insulated on both sides.  I don't have an idea for a suit in the magnetic field though, because there are no magnetic charges.

Quick calculation: Assume a spherical human with a radius $r$ of half a meter.  The suit is a charged conducting sphere and charging it with $V_\text{suit}=1\mathrm{MV}$, which is a lot, even for Marx generators, then the charge stored in the suit is

$$Q = V_\text{suit} 4\pi \epsilon_0 r = 6\cdot10^{-5}\,\mathrm{C}\,.$$

This doesn't look like a lot in the exponent, but Coulomb is an insanely large unit.  Now charging up the capacitor formed by floor and ceiling to some voltage $V_\text{deck}$, the total force on the suit is

$$F = E \cdot Q = \frac{V_\text{deck}}{d}\, Q$$

And the acceleration of the body should be one g:

$$F = m a = m g$$

At a mass of lets say $m=70\,\mathrm{kg}$ this requires a voltage of:

$$V_\text{deck} = \frac{m g d}{Q} = 2\cdot10^7\,\mathrm{V}$$

Good luck with that :-)  The typical dielectric strength of air is $3\mathrm{MV}/\mathrm{m}$ so you would instantly get an electrical arc between floor and ceiling.

Magnetic field:  Magnetic fields behave a bit differently and the best option is probably a linearly varying magnetic field $\mathbf{B}$ from floor to ceiling (use Helmholtz coils).  If you manage to build a suit with some magnetic dipole moment $\mathbf{m}$ (for example using a permanent magnet, or running a current through the suit), the force is

$$\mathbf{F} = \nabla (\mathbf{m} \cdot \mathbf{B})$$

The dipole moment of a normal neodymium magnet is given to be about $1\,\mathrm{Am}^2$ here.  For the same $m=70\,\mathrm{kg}$ body the magnetic field has to have a gradient of 7 Tesla per centimeter ...  or a difference of whopping 1400 Tesla from floor to ceiling (give or take one magnitude for not scaling the magnet to human weight).  As a comparison, the magnets at LHC generate fields of about 10 Tesla.

Bottom line: Earth is big, therefore gravity at earth's surface is strong, because of that it's hard to reproduce it using electric or magnetic fields.

Graviton: The graviton is still a hypothetical particle.  Even if we confirm its existence, its behaviour should be (for the large part) very similar to gravity as we know it.  "Bombarding" something with gravitons is then equivalent to having a large mass nearby.  Understanding the theory will most probably not enable us to build graviton guns that work analogous to lamps producing photons.  Speaking of large masses, how about lining the floor with neutron star matter?

Ferrofluid: A paramagnetic material can be used to couple to the magnetic field, but I believe you need active stabilisation (circuits in the floor and ceiling).  There actually seems to be a theorem about that: Earnshaw's theorem.  There are other implications to drinking ferrofluids, besides the chemical toxicity, because it might accumulate in some spots of your body and more of the artificial gravity will attack at that spot.

I guess, we are still better off with gravity-like acceleration by rotation or propulsion ...


Celestron Astromaster 114eq Star pointer finder scope help required

For an astronomical object, a 1000 mm focal length objective forms an image 1000 mm away. However, the image of a terrestrial object focuses farther out: 1010 mm for a target 100 m away, 1020 mm for a target 50 m away. When switching from a terrestrial target to an astronomical one, the eyepiece must move closer in.

Aligning the finder on a terrestrial target is bound to produce some parallax error. After focusing the telescope on a bright planet or the Moon, aim the telescope at Polaris and fine-tune the finder.

Finally, even with good finder alignment, sometimes you will have to sweep to locate your target. It gets easier with practice.


How often are supernovas visible by the naked eye from Earth?

One way to answer this is to look at how often naked-eye supernovae were seen in the past. In the last 2000 years, there have been (at least) eight such supernovae recorded: in 185, 393, 1006, 1054, 1181, 1572, 1604, and 1987. Seven of these were in our galaxy, while the last (SN 1987A) was in the Large Magellanic Cloud.
It's possible that supernovae in the far southern sky would not have been visible to Chinese astronomers (the most reliable and comprehensive records we have), so some additional supernovae visible from the far south may have been missed. We could round the total up to ten, which gives us a rate of about one every 200 years. If you want to worry about some being visible but not recorded (e.g., because they happened during a particularly chaotic time in China), then perhaps the rate would be a little higher -- maybe one every 150 years?


What is the furthest star for which we can measure the diameter?

It's neither the angular diameter or prallax precision that is the limiting factor, but the fact that it is difficult to get the interferometric measurements for faint stars.

State-of-the-art angular diameters are measured by infrared interferometry (e.g. with the CHARA array - Gordon et al. 2019). The most precise measurements of angular diameters have an uncertainty of about 17 microarcseconds in that paper. That means the smallest angular diameters that can yield
 a solid detection of the disc of a star are about $\theta = 50$ microarcseconds. But such measurements can only be achieved for stars down to about $V \sim 6$.

We can write down an equation for the relationship between $\theta$, the radius of the star $R$ and its distance $D$.
$$ D = \frac{2R}{\theta} = 186\left(\frac{R}{R_{\odot}}\right) \left(\frac{\theta}{50\ \mu{\rm as}}\right)^{-1}\ {\rm pc}$$

This means that if a $1R_{\odot}$ star has its distance known more precisely than the $\sim 30$% error in the angular diameter, then it's radius can be measured out to distances of about 200 pc (with a precision of about 30%). This is easily achieved - a $1R_{\odot}$ solar-type star at a distance of 200 pc has a brightness of $V \sim 11$ and has a parallax uncertainty in the Gaia catalogue of only about 0.1 mas (mostly a systematic error at present), meaning the distance is known to about 2%. Thus, parallax uncertainty is not the limiting factor. However, a 6th magnitude solar-type star would be at $D \sim 20$ pc. Much further away than this and the source is just not bright enough to get an angular diameter measurement with CHARA.

On the other hand, a large red giant star, with $R \sim 200 R_{\odot}$ could in principle have its radius measured to 30% out to a distance of 37 kpc with the angular resolution available. Such a star would not however have a measurable parallax at present. The parallax would be 27 microarcseconds, which should just be resolvable in the final Gaia data reduction in a few years time. However, the star would also have $V \sim 14$ and is way too faint for the current interferometric capabilities of CHARA. A 6th magnitude red giant is at $D \sim 500$ pc, a distance at which both its angular diameter and parallax are precisely measurable.

So the answer is that the limitation is not the angular resolution, or the parallax. It is the brightness limit for the interferometers that are used to do the measurements. This limitation means that you can measure the diamters of solar-type dwarfs to a few 10s of pc, and the diameters of the largest red giants to about 1 kpc.

Update: There are other interferometers that can be used. The VLTI/Amber interferometer in Chile utilises larger telescopes than CHARA and can in principle work to fainter magnitudes. A paper by Chesneau et al. (2014) measures an angular diameter for HR5171A (a red hypergiant) to be $3.39 \pm 0.02$ mas, at a distance of 3.6 kpc.
https://www.aanda.org/articles/aa/full_html/2014/03/aa22421-13/aa22421-13.html

Although much further away, this huge star is bright enough ($V \sim 6.5$) that an angular diameter measurement can still be achieved. Note though that the limitation here is still the brightness of the object, not its distance or angular size.


Do you mean the time it takes for the exoplanet to rotate around its axis? The transit method is used to detect the presence of an exoplanet as it passes in front of the star (as seen from Earth). After several transits you can get the orbital period, i. e. the time it takes for the planet to orbit around the star.

There are methods to measure the rotation period of astronomical objects (exoplanets included). You can use the Doppler effect to measure the change of wavelengths of the planet's light as it rotates. This technique has been used to measure the rotation period of exoplanet Beta Pictoris b.

Suppose the planet (as seen from our point of view) rotates from right to left. If so the right-hand edge of the planet's disc will be moving towards us, while the left-hand edge will be moving away from us. Therefore, the wavelength of the light coming from the right will be slightly shorter (i. e. it will be blue-shifted) than that coming from the left (which will be red-shifted). The greater the difference in wavelengths, the faster the planet is rotating. The method cited in the article above seems to be a bit more complicated, but that's the basics of it.

Alternatively, if you can image the planet with enough resolution, you might be able to spot different terrains or patterns on its surface, and by measuring the time it takes for those patterns to appear again in the same apparent position you might be able to calculate the rotation period. But this is probably unfeasible for most exoplanets, at least with the instruments we have right now.


The rate at which a clock ticks is a local property. There is no universal rate of time which could speed up, so the answer is no.

A clock stationary in the same frame of reference, and the same gravitational field as me will tick at the same rate, (one second every second) It is only clocks that are moving, or in different gravitational fields that that tick at a different rate. If I am falling into a black hole, and I stop the check the time, I wouldn't see my watch slow down.

So there is no standard clock rate. Clocks in the early universe (or at least things that depend on time, such as nuclear decay) ran at the same rate, in their local gravitational field (at one second per second)

If we were to observe a clock from the early universe, it would be receding at great velocity (and so it would be redshifted and time dilated)

Now, the gravitational time dilation does not depend on the amount of mass directly, but on the intensity of the gravitational field. A galaxy has huge mass, but the only place in which the gravitational field is significant (from a GR perspective) is in the neighbourhood of Neutron stars and black holes.

To get gravitational time dialtion you need an intense gravitational field. Just having a lot of mass is not enough. After the period of inflation, the universe was homogeneous, and there were no significant "lumps" to give a net gravitational field. As the gas in the universe collapsed into galaxies and stars, and eventually black holes regions of intense gravity formed, in which clocks would run slowly.

However there is no general speeding up of clocks in the universe.


Specialty hardware will be able to do it, as might clever programming if you have a commercial, computerized mount and feel like hacking away. Either way, an ordinary mount will be able to follow targets in the asteroid belt easily -- their motion across our sky isn't actually very fast. What I'm about to write really only applies to NEOs.

With NEOs, the problem isn't just the apparent speed of the objects across our sky, it's that the objects are usually in irregular orbits while moving at that speed. This is in addition to the apparent motion due to the rotation of the earth.

To track these objects you'd have to project the position of the object onto our sky, using the object's ephemeris data as your starting point, then add the apparent motion due to the rotation of the earth. That position could be calculated by the mount or calculated by a computer tethered to it, which would send updated pointing directions to the mount. The specifics on how to do this depend on your hardware. It's not a small project.


Is Limb Darkening because at centre we are directly looking core, whereas at limbs we are seeing only photospheric part?

We can't see the core, but we can see deeper into the photosphere of the sun. The deeper layers of the photosphere are hotter and brighter, and this is the cause of limb darkening.


Why do post main sequence stars enter the red giants branch?

If your question is why the star starts moving up the red giant branch, it's in essence because of the behaviour of the surface opacity and the development of a substantial convective envelope in order to meet the surface boundary condition. It's basically the Hayashi track in reverse.

We can say this because if you create a model of a star and artificially suppress convection in the equations, then the model just keeps getting colder and colder at roughly constant luminosity, instead of ascending the giant branch.  See e.g. Fig. 2 of Stancliffe et al. (2009). Note that the star still becomes large in terms of radius, just not bright.


Does creation of galaxy depend on black holes?

Most of the galaxies are believed to host a super massive black hole. The black hole in return has a dramatic impact on the evolution of the galaxy, an effect usually called "AGN feedback". Quoting W. Ishibashi, A. C. Fabian:

  Active Galactic Nuclei (AGN), regions of extremely high luminosity at the centers of some galaxies, are thought to radiate due to accretion onto a supermassive black hole.  This emitted radiation, or feedback, sends energy into its environment and influences the evolution of the host galaxy.  The details of this feedback are not well understood.  We do, however, observe several relationships between properties of the host galaxy and the central black hole — the most prominent of which is the M-sigma relation.  This empirical correlation states that the black hole mass (M) scales as the fourth power of the velocity dispersion (sigma) of stars in the galactic bulge.  Clearly, there is some connection between the evolution of the black hole and that of its host galaxy.

The topic of understanding how the supermassive black hole form in the first place, evolve and interact with their galaxy is still in debate among the scientific community, so it is impossible to fully answer your question yet.


How to calculate the period of a lunar mansion?

If I understand the concept of "mansion" correct, it is similar to an astrological sign, and astrononomically simply refering to a certain zone in which a certain object (here: the Moon) can be seen on the sky.
More specifically, a Lunar station - [Manazilu ʾl-Qamar منازل القمر] according to Islamic astronomical system seems a certain sector of the sky and the sky is divided into 28 of theses sectors. This defines a celestial coordinate system with $x$-axis being the ecliptic, the celestial longitude. The longitude is measured in degrees east of the vernal equinox (the ascending intersection of the ecliptic and equator):
An algorithm to determine the Lunar station at a given time would have as input parameter only the position of the Moon on the sky (actually only the Ecliptic longitude $\lambda$ of the Moon) and check in which interval $\lambda$ lies. The lower boundary of each mansion is given in the 3rd column in the Wikipedia-table shown in the question in a rather inconvenient format, namely in degrees and minutes from a start of a zodiac symbol.
In a first step, you would have to convert the Wikipedia list to a intervals with real celestial longitudes. You would end up with 28 if-clauses in your algorithm to determine a string $s$ containing the name of the lunar mansion:
$s_{\rm lunar~mansion}(\lambda) = \left\{ \substack{ {An-Nath} {\rm ~if~}
  0 \leq \lambda < 12^\circ 51' \\
\vdots \\
Butnu 'l-Hut {\rm ~if~} 0 \leq \lambda < 347^\circ9' = 330^\circ +17^\circ 9'
}\right.$
Here we used the fact that Pisces starts at $330^\circ$ celestial longitude.
Since you are after the begin and end dates for a each lunar mansion you have to calculate at which time the Moon celestial longitude $\lambda$ equals to the boundaries of the 28 intervals. I would do it vice versa and calculate for each day  of the year at 00:00:01 local time the longitude of Moon and then use $s(\lambda)$. You will end up with a table of 365 or 366 date- string-pairs. In the last step, you discard all lines of the table where a string repeats and reduce the list to 28 entries, with the start dates in the first column.


Yes, binary comets do exist; 8P/Tuttle is an example of one. This is an area that is still being heavily researched, but it is suspected that binary comets can "form from collision, mutual capture, or fission."

In recent news, the comet 67P/Churyumov-Gerasimenko has also been revealed to be a contact-binary by the European Space Agency's Rosetta spacecraft.

Sources:

Radar observations of 8P/Tuttle: A contact-binary comet - Harmon, Nolan, Giorgini, Howell
Rosetta's target is 'double' comet - BBC


Why do we use filters in telescopes for astronomical imaging?

In general, the CCDs used to capture images do not register the energy (therefore, colour) of the incident photons on them - they just count the number of photons observed by each pixel (or a value proportional to the number of photons, as the are not 100% efficient). So, they essentially just show overall brightness variations across the image.

If you want to capture colour information you therefore have to use filters. I.e., if you want to get information on the number of red photons (e.g., the intensity of red light incident on each pixel of your image), you use a filter to block out all other light. You can do this with multiple different filters to build up a full colour image.

In most commercial digital colour cameras the CCDs have a filter mask over four pixel patches: two filtering for green light, one for blue light, and one for red light. The outputs from these pixel is used to build the full colour image.


This page about a commercial product goes into some detail about their algorithm.  It does the triangle matching you describe, with something like simulated annealing to get a more optimal solution.

The accepted answer to this closely related question recommends Hugin panorama software; it's open source, so you should be able to glean the algorithms used.


Are there observable changes in a star about to become supernova, minutes or hours before the explosion?

I think your best bet would be detecting neutrinos generated by nuclear burning inside the star (as we do for the Sun). Once the star hits the carbon-burning stage, it's actually putting out more energy in neutrinos than in photons. During the silicon-burning phase, which lasts for a few days and is what creates the degenerate iron core (that collapses once it is massive enough), the neutrino flux increases to about 1047 erg/s a few seconds before core collapse. (The peak flux during core collapse is about 1052 to 1053 erg/s). This paper by Asakura et al. estimates that the Japanese KamLAND detector could detect the pre-supernova neutrino flux for stars at distances of several hundred parsecs, and provide advance warning of a core-collapse supernova several hours or even days in advance. Since your characters are in the same system as the star, they'd hardly need a large underground detector to pick up the neutrinos.

This plot shows an example of neutrino luminosity (for anti-electron neutrinos) versus time for a pre-supernova star (from Asakura et al. 2016, based on Odrzywolek & Heger 2010 and Nakazato et al. 2013); core collapse begins at t = 0s.

By measuring the spectrum of energies for different types of neutrinos and their time evolution, you could probably get a very good idea of how far along the star was, particularly as we can probably assume your characters have much better models for stellar evolution than we currently do. (They'd also want to get accurate measurements of the star's mass, rotation rate, maybe internal structure via astroseismology, etc., in order to fine-tune the stellar-evolution model; these are all things they could do pretty easily.)

The core collapse itself would be signalled by the enormous increase in neutrino flux.

This "What If" article by Randall Munroe estimates that the neutrino flux from a core-collapse supernova would be lethal to a human being at a distance of around 2 AU. Which, as he points out, could actually be inside of a supergiant star, so your characters would probably be a bit further away than that. But it does show that the neutrino flux would be easily detectable, and that your characters might well get radiation poisoning from it if they were closer than 10 AU. (Of course, you'd want to detect it more directly than just waiting around till you started to feel sick, since that might take longer than the shock wave takes to reach the surface of the star.) This is just to bring home the fact that they wouldn't have any problem detecting the neutrinos....


Is there a way to estimate the age of M dwarf stars?

Really difficult.

Very young ($<50$ Myr) M dwarfs may still exhibit lithium in their photospheres. Any older and it will have been depleted.

For older dwarfs you then move on to looking at rotation and activity. Both decline with age, but in an M dwarf as cool as this, the decay timescale is many Gyr, so it is only weakly constraining and not very well calibrated for such low mass stars. A very slow rotation rate or very low level of magnetic activity might indicate it was older than 5 Gyr.

The lack of flares and the lack of light curve modulation by starspots may indicate that it has spun down, is magnetically inactive and is therefore older than 5 Gyr ( I cannot immediately locate the source of this Wikipedia claim). The ratio of X-ray to bolometric luminosity would be more definitive and could probably be derived from the ROSAT all-sky survey (or a constraining upper limit would be found).

A useful paper is by Stelzer et al. (2013). http://arxiv.org/abs/1302.1061
She finds various activity diagnostics for nearby stars, including this one. Its projected rotation velocity is 1.5 km/s, which suggests a rotation period of $\sim 10$ days for a star of this size. This could not be less constraining on the age! The X-ray to bolometric flux ratio is $10^{-4.45}$ which suggests it is neither very young or very old - say between 1 and 5 Gyr for a star of this mass.

Another possibility is to look at the Galactic space motion and position. Older stars tend to have higher velocities, especially perpendicular to the Galactic plane, and also tend to be found further from the Galactic plane. The velocities for this object do not look large, suggesting it is not very old, ie younger than perhaps 10 Gyr and likely a young disk object with age of $<5$ Gyr.

You can also look at the metallicity. This is difficult to measure in M-dwarfs. Maldonado et al. (2015) give [Fe/H]$=-0.05 \pm 0.03$. This is very close to the solar value and does not constrain the age beyond suggesting the star is younger than 10Gyr.

Beyond this, it is basically impossible. Even if you had a very precise distance and could place the star on a HR diagram, low mass M dwarfs evolve so little in luminosity and temperature over many Gyr that this offers no constraint.

So my conclusion - the age is between 1 and 5 Gyr, with the probability peaking somewhere in the middle.


Why is the boundary of friends-of-friends (FOF) halo corresponding to iso-density contour?

If you have enough particles to sample your field, then a given (local) density corresponds to the same mean distance between your particles. The FOF algorithm keeps linking particles to a halo until a particle is farther away than the linking length. That means that the halo boundary will be ~exactly on the border between regions of interparticle distances smaller and larger than the linking length, corresponding to a boundary between larger and smaller densities, respectively. That is, an iso-density surface.
In contrast, a halo finder using the spherical overdensity (SO) algorithm will always result in a spherical boundary, which will only be an iso-density surface in the case of a completely symmetrical halo.
This is the reason that FOF halo finders tend to include more filaments outside the region identified by SO halo finder. This may or may not be desired.


Instead of using flexible mirrors, could software be used for adaptive optics?

Short answer: no.

Why not?  because the mirror is placed at a position in the optical train such that it controls the wavefront phase and tilt of each subsection, "subaperture" of the incoming beam.  Once an optical system has formed an image, all phase information is lost.  There  are some software methods which can correct for simple aberrations such as astigmatism or coma, but those require a near-constant type of error across the input field.  The problem with atmospheric turbulence is that the errors are essentially of a localized and random nature.

Note that even a plenoptic camera which, by tricky use of sublenses at the focal plane, record enough information to change focus, cannot correct for general phase aberrations.


SWAN seems to be on a parabolic orbit. This means that it is the first time that it is interacting with the Sun, making it a "new" comet. New comets are covered in a layer of very volatile elements that will vaporize pretty fast when the comet is still far away from the Sun. This causes a surge in brightness. But once those very volatile elements are gone, it stops getting brighter and the comet's brightness can even decrease.

In addition to this, it is mostly losing gas, so its tail isn't as visible as if it were a tail of dust.

(From this article, in French)


How many Kuiper Belt objects have moons? How do we know this?

"Moons" is a misnomer here, the majority of Kuiper Belt Objects (KBOs) with companions are binaries.
Now there are several populations in the Kuiper Belt, characterized by their eccentricities and inclinations. The cold classicals are hereby the most important ones, having mostly circular, low inclination orbits and a high binarity fraction of about 30% (Noll et al. 2008).
Hot classicals, with eccentric and inclined orbits possess a binarity fraction of about or less than 1 in 40.
The other important KBO populations, i.e. the scattered disc, the resonant ones and the centaurs possess negligible binarity. In terms of numbers, the classicals make up about 2/3 of the entire KBO population by numbers, and the cold classicals are about 50-60% of the total classical population.
So in total you'd get a fraction of at least $f_{\rm Moon}\approx 0.66*0.5*0.3\approx10\%$ of KBOs having 'moons'.
In order to find and characterize binary orbits, strange light curves pose a first hint, but only high spatial resolution actually helps to discern the two binary components. As most KBOs are low-mass objects, their binary orbital times are long. Prolonged, 5-15 year snapshot observations with the HST allowed recently to characterize the spin direction, for example (Grundy et al. 2019).


Why are there no gamma-ray bursts detected in our galaxy?

All models of gamma-ray bursts involve extremely energetic phenomena: particular types of supernovae, the coalescence of binary compact objects, strong magnetar flares, or tidal disruption events. It turns out that these events are quite rare - so rare, in fact, that GRBs would be expected to occur in a low-redshift Milky Way-like galaxy at a rate of only one every few million or tens of millions of years (Zhang & Meszaros 2003). If the Milky Way was undergoing a period of intense star formation, yielding more massive stars and therefore more supernovae, then this rate would go up, but still not significantly. Bear in mind that the Galactic supernova rate, for example, is believed to be only several supernovae per century, and the vast majority of supernovae don't lead to gamma-ray bursts.
We know of many extragalactic GRBs for several reasons. It helps that we can observe large numbers of galaxies (thanks to how bright the bursts are), and if we could look at millions of Milky Way-like galaxies, it wouldn't be surprising if we could detect $\sim$1 event per year. (There's also the advantage that star formation peaked at a redshift $z\sim2$, and so high-redshift objects would be more likely to produce more GRBs!)


Spacetime curvature illustration accuracy

Newtonian gravity of a point-source can described by a potential $\Phi = -\mu/r$. If we suppress one spatial dimension and use it to graph the value of this potential instead, we get something that looks very close to this illustration, and is indeed infinitely deep at the center--at least, in the idealization of a point-mass. And farther away from the center, it goes flat, just as many illustrations like this have it.

Illustrations like this are fairly common, and I'm guessing that they're ultimately inspired by the Newtonian potential, because they have almost nothing to do with with the spacetime curvature.

Here's an isometric embedding of the Schwarzschild geometry at an instant of Schwarzschild time, again with one dimension supressed:

Above the horizon (red circle), the surface is a piece of a paraboloid (the Flamm paraboloid). Unlike the potential, it does not go flat at large distances.

Being isometric means that it correctly represents the spatial distances above the horizon at an instant of Schwarzschild time. Below the horizon, the embedding isn't technically accurate because the Schwarzschild radial coordinate does not represent space there, but time. Although if we pretend it's spacelike below the horizon, that would be the correct embedding. Picture the below-horizon part as having one-directional flow into the singularity.

Since we've only represented space and not time, the embedding is not enough to reconstruct the trajectories of particles in this spacetime. Still, it is a more accurate representation of a part of the spacetime curvature of point-source--specifically the spatial part.

  The velocity of the object from this perspective, would seem to increase, until a point - where the velocity in x,y coordinates starts to decrease due to most of the motion happening "down" the time dimension. Is this also correct? Would a photon seem to slow down when moving down the well, if seen from above?

The above is an embedding of a slice of spatial geometry, and is not a gravity well. The mathematical form of the paraboloid above the horizon is best described in cylindrical coordinates as
$$r = 2M + \frac{z^2}{8M}\text{.}$$
Here the vertical $z$ coordinate doesn't mean anything physically. It's purely an artifact of creating a surface of the same intrinsic curvature in Euclidean $3$-space as the $2$-dimensional spatial slice of Schwarzschild geometry.

For the Schwarzschild spacetime, radial freefall is actually exactly Newtonian in the Schwarzschild radial coordinate and proper time, i.e. time experienced by the freefalling object, rather than Schwarzschild time. So the Newtonian gravity well isn't actually a bad picture for the physics--it's just not the geometry and so is not a good representation of how any part of spacetime is curved. For non-radial orbits, the effective potential is somewhat different that than the Newtonian one, but ignoring the effects of angular momentum gets us the Newtonian form.

In Schwarzschild time, yes, a photon (or anything else) does slow down as gets near the horizon. In fact, in Schwarzschild time it never reaches the horizon, which is one indication that the Schwarzschild coordinates are badly behaved at the horizon. The coordinate acceleration actually becomes repulsive close to the horizon--and for a fast enough infalling object, is always repulsive. This can be understood as the particle moving to places with more and more gravitational time dilation. In proper time of any infalling observer, however, close to the horizon the acceleration is always attractive.


Earth is bound gravitationally to the Sun. So it is orbiting together with the Sun around the center of the Milky Way.

This takes one galactic year.

Hence your presumption is correct.

Seen from a non-rotating system relative to the Milky Way barycenter, Earth's orbit isn't an ellipse, but ressembles a distorted helical trajectory. Distorted, since the orbit around the Sun isn't perpendicular to the orbit around the center of the Milky Way.


Are sunspots vertically displaced from the surrounding photospheric plasma?

Yes.
The idea that sunspots are depressed slightly came as a possible explanation for the Wilson effect. The Wilson effect was discovered as the shape of sunspots as viewed from Earth changed as the Sun rotates, in a way consistent with the change in perspective looking onto a slightly depressed region. While this isn't the only explanation for the effect, it's certainly the most prevalent.
More specifically, as Solanki (2003) writes, the depressions indicate a lowering of the layer where the optical depth $\tau=1$ (keep in mind that the bottom of the photosphere is the layer where $\tau=2/3$). There are two causes mentioned: lower temperature and magnetic effects.
Sunspots are cooler than the surrounding areas, as is well known, and thus appear darker. We see the same thing apparent at the boundaries of solar granules: Cooler, darker gas sinks and lets the hotter gas (which is less dense) move upward. Additionally, the opacity $\kappa$ is temperature-dependent, which may impact how far one can see into the star.
Not only is temperature a factor, but so is the magnetic field. Sunspots are, at heart, a magnetic phenomenon, and thus the radial force equation is substantially different. Normally, in a star, the equation of hydrostatic equilibrium is
$$\frac{\mathrm{d}P}{\mathrm{d}r}=-\rho g$$
for pressure $P$, density $\rho$ and gravitational acceleration $g$. However, when the magnetic field becomes important in a sunspot on the solar surface, the force balance is
$$\frac{\mathrm{d}P}{\mathrm{d}r}=\frac{B_z}{4\pi}\left(\frac{\mathrm{d}B_r}{\mathrm{d}z}-\frac{\mathrm{d}B_z}{\mathrm{d}r}\right)$$
where $r$ and $z$ are the radial and vertical coordinates (note the change of coordinate system - $r$ is along the surface, and $z$ is perpendicular to it!). The force from the magnetic field implies a lower gas pressure and a greater depression.


Direction of Earth’s poles relative to the Sun’s movement around the Galaxy

There is no difference in the number of cosmic particles.

You seem to imagine that there are static clouds of gas in the universe that the sun is moving through.  This is not the case. The gas in the galaxy is orbiting at roughly the same speed as the sun.

The sun does have some proper motion relative to the interstellar medium, but the interstellar medium doesn't penetrate the bubble formed by the solar wind.

There is no dragging. The gravity of the galaxy affects both sun and planets, we all move around the galaxy together. There is no shifting of the plane at all.

Interstellar particles are found in ice cores, a paper suggests that variation in the rate of cosmic radiation may cause hydromagnetic shocks that could transport metal rich dust into the inner solar system. No 30 million year periodicity is mentioned.