What would the night sky look like if Earth orbited an intergalactic star?

Since Andromeda is already visible to the naked eye, to a civilization located at half the distance from the Milky Way, Andromeda would be still be visible. Its total brightness would be four times higher, but since its area would grow by the same factor, its surface brightness would stay constant. The Milky is less bright by a factor of ~2.5, but also smaller, so the surface brightness and hence visibility would be similar.

Individual stars would not be visible except if 1) one exploded as a supernova or 2) the civilization has evolved into beings with extremely large pupils, so that more light is being recorded by their retinas. In fact, we can calculate how large:

According to this list, the brightest star known in the Milky Way is called WR 25, and has an absolute magnitude of $M = -12.25$. Half of the distance from the MW to Andromeda is $d \sim 390,000$ pc, so the distance modulus is
$$\mu = 5\log d - 5 = 22.95,$$
and the apparent magnitude of WR 25 would be
$$m = M + \mu = 10.7.$$

Under the darkest conditions, humans are able to see objects down to an apparent magnitude of ~6, i.e. a factor of
$$f = 2.5^{10.7 - 6} = 75$$
smaller. Thus, to be able to see WR 25, the diameter of the aliens' pupils would have to be a factor of $\sqrt{75} \sim 9$ larger. Since human pupils are roughly 6 mm wide, their pupils would be 5 cm wide, and the eyeballs somewhat larger, incidentally roughly the size of a teacup.

Thus, it seems that my compatriot H. C. Andersen predicted the existence of these beings in his fairytale The Tinderbox.


The James Webb Space Telescope (JWST) will be Hubble 2.0, and, my personal favorite, it could get a better look at planets in other solar-systems than we've ever had.   Scheduled for launch in 2018.  (Footnote, as Rob Jeffries points out, it's infra-red through long-wave visible light only).   Hence, no Blue/Indigo/Violet or Green, but some yellow.   So, it's not quite like Hubble, but it's still often called Hubble's replacement.

Chandra observes X-rays.

Spitzer observes infra-red.    They both followed Hubble, but they're not the same type of telescope.


A couple of points on this.  While faint rings can exist outside the Roche Limit (like Saturn's G and E rings) - photo is too big to link but click here.  rings of any significant mass are only going to reside inside the Roche Limit, so the practical limit of the mass of a planet's ring is about the mass of the largest moon a planet is likely to have, roughly speaking.

A ring around a planet can form in a couple of ways, by giant impact, by a moon passing inside the Roche limit and breaking apart or by accretion, like how Enceladus feeds Saturn's faint e-ring.

Also, rings don't have to be particularly massive to be impressive.  The mass of Saturn's rings estimated here (and there's some uncertainty to that estimate, but close enough for our purposes)

  Based on Voyager observations, the total mass of the rings was
  estimated to be about 3 x 10^19 kg. This is a small fraction of the
  total mass of Saturn (about 50 ppb) and is just a little less than the
  moon Mimas

So the mass of Saturn's rings is the mass of a pretty small moon, a little under 200 km in radius, but that wasn't your question, so, moving on.

A planet with rings 200 times the size of Saturn (40,000 times the surface area) was observed and if we assume the same density as Saturn's rings (which is a very bad assumption to make, but this is just an approximation), 1.2 x 10^24, or about twice the mass of Mars.   Now we're getting somewhere, but there's a few points to make on our friend J1407B

First, it may not be a planet at all.  It's mass estimate is 10-40 jupiters which suggests it's in the brown dwarf category.   2nd, it's a young solar system, just 16 million years old, so the ring system may still be in the process of forming a moon system and not be a permanent ring at all.  Source.   Also, same article, it mentions that parts of the ring system block 95% of the sun's light, so it's a far denser system than Saturn and there's a moon in there thought to be 0.8 times the mass of earth, but I still don't think that should counts since it may be an accretion disk remaining from the formation of that solar-system.

If we set the planetary mass limit to be about 13 Jupiter masses, which is roughly the cut-off between heavy Jupiter and brown dwarf, and we take some estimates of the size of moons to planets, during formation, it's unlikely that moons can be more than maybe a ratio of one to several hundred, due to limitations in angular momentum and likely size of formation.   In our solar system, the largest moon to planet ratio, not counting giant impacts or captures is about 1 to 4,200 (Saturn to Titan).  Jupiter to Ganymede is 1 to about 12,800 and Tritan to Neptune (and Tritan may be a captured moon), about 1 to 4,800.   If we assume a heavy Jupiter of about 13 Jupiter masses and a moon in a decaying orbit that breaks up with a mass ratio of 1 to 400 - and needless to say, that's very ballpark, then this theoretical estimate works out to a practical limit around a heavy jupiter to have a moon system of roughly 10 earth masses.  (1/400th the mass of a 13 jupiter mass planet).

Now, there are creative ways to increase it, like, lets say there are two moons, perhaps in Trojan points to each other, and they both spiral in gradually.   Or, lets say a passing object of greater mass flies past, but the problem with a passing object is that, due to escape velocity speed, such an object, if it passes within the Roche Limit, it would have a highly elliptical orbit, which wouldn't lend itself to a ring system, since the momentum of the elliptical orbit would only pass through and not stay inside the Roche sphere.  What you need for a good sized ring system to form is a slowly decaying circular orbit, not a recent capture of a large passing object.

See Shoemaker Levy 9 which passed through but quickly went back outside Jupiter's Roche limit.

Now if we imagine a scenario of a double planet, which is probably a rare scenario but not impossible, and over time due to proximity and tidal forces, the two planets form a decaying circular orbit around each other, in this theoretical, lets say we have two heavy Jupiter orbiting each other, one of 12 Jupiter masses and the other of 6 Jupiter masses (purely theoretical) and the 6 Jupiter mass planet begins to break up as it enters the Roche limit of the 12 Jupiter mass one.   There's 2 problems with assuming you'd get a ring with half the mass of the planet.   One is, hydrogen and helium gas are likely at least 80% of the mass of the planet that's breaking apart, perhaps 90% and hydrogen and helium don't easily bind into ice particles, so over time, much of that gas is likely to get blown away by the solar wind.  You're left with a rocky and icy mass of quite a bit less than 6 Jupiter masses, probably less than 1.  The other problem, with an orbiting object that large, it doesn't break apart all at once, but it's denser core stays together and spirals closer to the planet while the lighter ices around the surface which do begin to form the ring, can get spun off and flung either into the planet or flung outside the Roche limit, as the denser core of the smaller of the two planet moved slowly closer, there would be a vacuum cleaner effect wrecking havoc on the very same ring that planet was forming.  For smaller moons, this effect can be seen around saturn as tiny moons create tiny breaks in Saturns' rings.  With a much more massive core of the planet that was forming a moon, you'd get orbital chaos and it wouldn't be a good scenario to form a ring, even with all that available ring material.   You'd still probably end up with a ring, but I think only a small percentage of the original smaller planet's mass would survive the process.

A similar problem arises with giant impacts.  If the impact is too large, the planet breaks apart.   If it's not quite that large, like the giant impact on earth was thought to be, 4.4 billion years ago, then debris is blown outside the Roche Limit and forms a moon, in our case 1/81st the mass of the earth, though at the time it might have been a bit more, maybe 1/72-74 or so, and there were probably two moons initially not one, but I digress.   The point is, to have debris blown off the planet but staying inside the Roche limit so it forms a ring, not a moon, you need a smaller impact than the one that formed our moon, and a smaller impact implies less debris, so with Rocky worlds, what does that mean, 1/100th?   1/200th?  In there somewhere.  With gas giants the impact method is even worse due to the outer layer of gas giants being so much hydrogen that you don't get good material for a dense permanent ring structure from giant impact.

So, there are practical limits to the size of a ring system you're likely to see (if we ignore young solar-systems still information like J1407b.   It's a very ballpark just barely educated guess, but I think the practical limit for a ring system is probably pretty tiny compared to the mass of the planet, like a ratio of 1 to 200 or so, which if you have a planet the mass of 13 Jupiters, that's still quite a bit of mass.   If we use the 1 to 200 ratio and 13 jupiters, that's the mass of 20 earths.  I have a very hard time seeing how a ring system around a planet could get much more massive than that.

Now, in theory, like, lets say we want to build a ring system, and we design this enormous snow blower and we blow ice crystals around a planet, just for fun.   A ring system would probably remain stable well beyond that mass.   I don't have the means to calculate where instability might come into play, but if you build your own ring system around a planet, you might be able to get the ring system as massive as 1/10th the mass of the planet, maybe a bit more before some kind of gravitational instability took over.   I don't think you're ever likely to see that in a natural situation, but theoretically I think it could be done.

Now, to answer your questions:

  does it depend on the mass of the planet?

absolutely.  The mass of the planet (and it's density) determines the size of it's Roche limit, which defines how far the ring system can extend.   As pointed out above, planets can have rings beyond the Roche limit but only faint ones.   Dense/Thick rings only form inside the Roche Limit and the more massive the planet, the larger the Roche Limit.

  presence/absence of moons?

Small moons don't matter much.  Saturn has small moons inside it's rings and they create small breaks in it's ring system.  Large moons, particularly if they are close to the planet, can create gravitational disruptions and wouldn't be good for a permanent ring system.

  distance to the star or other parameters?

Beyond the frost line is best, cause that's where ices don't melt.  Ices make up more mass of our solar system and probably most solar systems than rocky material and they break apart more easily, so statistically speaking, a ring system should do better far away from the sun, in our case, at least Jupiter Distance.   Jupiter's enormous magnetic field might also not be long term ring friendly, so in addition to far from the sun you'd want a planet without too strong a magnetic field that's shooting high velocity charged particles through the ring.  Here's a fun article about how Saturn's magnetic field erodes it's ring.  Jupiter would melt any ice ring it had much faster than Saturn does.

Hope that wasn't too long and it's more my trying to work it out than a definitive answer, but until we get much better telescopes, there might not be a definitive answer to this one.


Is it possible to calculate the focal length of a refactor I built?

The focal length of a telescope is the focal length of the lens at the front of the scope. In your case it's either 50, 30, or 20 cm.

To magnify the image formed by one of those objective lenses, you need an eyepiece. None of the lenses you listed will work as an eyepiece unless you want a magnification of 2.5 (50/20).

Eye pieces generally have a focal length of 25 to 30 mm or less. To get a quality image, you should buy a used eyepiece.


Did a piece of Halley's comet strike the Earth 1,500 years ago?

Little pieces of Halley's comet hit the Earth every spring. It's called the Eta Aquarid meteor shower, but I think you are thinking of something bigger.

Dallas Abbot's paper notes that at the 533-540 point on the GISP2 ice core there are Sn-rich particles, Ni-rich particles and cosmic spherules. These she identifies as evidence of a minor cometary impact. The resolution of the GISP2 core is sufficient to place this impact in the late spring, consistent with the Eta Aquarid meteor shower, which is known to be produced by Halleys comet.

Dallas notes that an increase in cometary dust in the atmosphere could partly explain the extreme weather events of 535-536, in combination with a minor volcanic eruption. Although there are other, competing hypotheses, and this is not considered to be settled science.


It would depend on what time the Earth started to spin around Michigan. There is no way to answer this without saying what time, time of year, and which century (or at least millennium) it happens.

Edit: Hi Dayna. Thanks for the the time of year the Keweenaw Peninsula becomes the North Pole. One could be omniscient except for that and not know this. But we also still haven't ruled out a single star since your OP. With the time though we could tell you exactly what star. The decade is not important. What century or part of history it happens, yes but not the decade. (yes, astronomy is weird).


Why has Venus's atmosphere not been stripped away by solar wind?

Venus has a strong ionosphere that protects it against violent solar winds. So, even though Venus has no intrisic magnetic field, it has an effective, induced magnetic field due to the interaction between the solar winds and the atmosphere, that protects it against solar winds.

Venus atmosphere is thick enough to have a consequent ionosphere, that would be the difference between Mars and Venus (and Venus was able to keep a thicker atmosphere due to its greater mass, contrary to Mars).

Sources:

Luhmann & Russell (1997),  on Venus
Luhmann & Russell (1997),  on Mars


Does anything orbit the Sun faster than Mercury?

The recently discovered asteroids 2019 LF6 and 2020 AV2, each taking 151 days to orbit the Sun, have the shortest periods currently listed in the JPL Small Body Database.
Vulcanoids are difficult to detect from Earth; none are known yet.
To remain in such an orbit, Evans and Tabachnik 1999 estimate a minimum diameter of 100 m and a semimajor axis between 0.09 and 0.21 au.
There are various competing definitions for a planet's neighborhood.
Soter 2006 says:

Two bodies share an "orbital zone" if their orbits cross a common radial distance from the primary and their periods are nonresonant and differ by less than an order of magnitude.

A Sun-grazing asteroid with perihelion 0.01 au and aphelion 0.31 au (Mercury's perihelion) would orbit in 23 days, well above the 9 day minimum for that definition.


Do some galaxies collide faster than the speed of light?

If some galaxies expand away from each other faster than the speed of light then do they collide faster than the speed of light?

The short answer is "no".
The expansion of the distant universe at apparent speeds faster than light is due to the expansion of space.  But the velocity of that expansion depends on how far apart the observer and object are.  If there are two objects A and B which we see apparently moving away from us at faster than the speed of light due to the expansion of spacetime then that is not how the objects will see each other.  They will be close to each other so from their reference frames the expansion of spacetime is negligible (just as it is for us and our "local" galaxies).
At some future time we think the universe may expand even faster at "small" ranges but in that case the two objects could not collide.  Once they are able to collide, they are close enough so that the effects of the expansion of the universe are not going to make them move apart faster than light.  Once objects move apart faster than light they cannot collide (unless the entire universe starts to contract).
If they are expanding away from each other faster than light (due to the expansion of space-time) then there is no way for them to collide.  No attraction could overcome the expansion of spacetime if it has already reached the point where they are moving apart faster than light.  It would require the entire universe to start contracting to do that.
So if it's going to collide with you, an object cannot be moving faster than light relative to you.

How long does it take for galaxies to collide?

This depends on how fast they are approaching each other.  In theory they cannot approach faster than light, so the upper limit is (simplistically) the distance divided by the speed of light.
A more realistic example would be Andromeda and the Milky Way which are approaching each other at a relative speed of 300 km/s.   Andromeda is about 2.5 million light years away and current estimates are for a collision in about 4.5 billion years, so that's moving much, much slower than the speed of light.


Has "GHz-spinning dust" ever been demonstrated in the laboratory?

This poster suggests an answer of yes and no. Essentially the poster reports a demonstration of some of the proposed mechanisms, but the rotation speeds produced are in the $kHz$ not $GHz$ range because the dust partcles are bigger, the vacuum less good and so on.

To quote the summary:

  First laboratory measurements of rotation of the analogs of interstellar
  dust grains induced by radiative torques have been made
  The measurements were made on irregularly shaped SiC particles
  particles of ~ 0.17 to 8.2 $\mu m$ effective radii levitated in an electrodynamic
  balance and illuminated with laser light at a wavelength of 5320 Å, at
  pressures of $10^{-3}$ to $10^{-5}$ torr.
  The rotation rates with radiation intensities of ~ $4-30 Wcm^{-2}$ were
  observed in the range of 1,000-22,000 rot/sec.
  The grain rotation rates were observed to follow the expected functional
  relationships, being directly proportional to the incident radiation
  intensities and inversely proportional to the drag represented by the
  ambient pressure.


Why do the solar system planets go rock-gas-ice instead of rock-ice-gas when moving away from the sun?

Prelude
It is now generally accepted in the planet formation community that planets form as a side-product of the star formation process in so-called protoplanetary discs.
Protoplanetary discs have initial masses of few to tens of percent of their stellar host masses, are relatively cold (T<150K in about 95% or more of their mass, which is outside the water iceline for a standard MMSN model) and are hence mostly detected in the infrared. The radiating infrared component is the 'dusty' component (first published detection and confirmation via the IRAS satellite in 1984-1985) making up about 1% of the mass, the other 99% being H/He gas.
Those discs are accretion discs, i.e. they loose angular momentum through various processes, which leads to mass infall into their host star. The dust settles into the midplane. For the case of turbulent accretion, dust and gas will be well-mixed and accrete relatively uniformly into the star, while in the case of disc-wind-driven accretion, H/He in the upper layers of the disc flows over the midplane and provides the accretion rate. Disc accretion rates can be too much for what the star can actually accrete and the excess mass is ejected vertically in jets that can exist throughout the disc lifetime, their mass decretion rates correspond typically to 1-50% of the disc accretion rate.
I mention the water iceline solely as point of reference, as its exact effect on the physics of planet formation is heavily debated, it currently cannot be observed, and icelines in several other molecules such as $\rm CO, CO_2, N_2,...$ could be playing roles too.
Planet formation
Our solar system originated very probably in one of those protoplanetary discs. We cannot follow the formation process over the disc lifetime, as this takes between 1-20 Myears (median value 3-5 Myrs, depending on the survey) and hence as often in astrophysics, we rely on snapshots and exoplanet statistics to try and puzzle together the physics.
50% of all exoplanetary systems harbour several rocky super-earths at radii interior to the water iceline. 6-10% of all stars possesses cold gas giant planets (giant planets at semi-major axes >0.5 AU) and 0.5-1% possesses hot gas giants (giant planets at semi-major axes <0.1 AU).
While from this our solar system seems to have an unusually low mass in the terrestrial planet zone, nonetheless physics seems to prefer to build rocky planets interior to the water-iceline. Those processes must happen in the protoplanetary disc phase and possibly shortly after gas removal (<100Myrs, it is poorly constrained which fraction of its final mass Earth possessed at disc dispersal).
Rocky planets are also thought to form beyond the water iceline. However in those regions of the protoplanetary disc, the mass reservoir is enormous and rocky planets can achieve runaway gas-accretion before their parent disc disperses.
Reaching runaway gas-accretion consists of two steps: First, after the rocky, multi-earth-mass planet is formed, it aquires an atmosphere that is hydrostatically connected to the disc via its own gravitation. This atmosphere cools slowly via Kelvin-Helmholtz cooling. The contraction allows for more mass to flow into the planetary domain, forming a massive atmosphere. Should this atmosphere reach a mass important enough for self-gravity to help the contraction further, the planet accretes more the more it cools and it cools more the more it accretes, hence runaway accretion is reached.
The architecture of the solar system
With all this, we can formulate the standard explanation for the architecture of the solar system:
Jupiter and Saturn are standard cold gas-giants that underwent a phase of rapid core-assembly and subsequent runaway gas-accretion. Uranus and Neptune grew far out in regions of low disc gas density (or small dust populations, increasing the core-assembly and cooling time) and hence were stuck in the hydrostatic gas accretion phase until the disc dispersed. The "ice" in ice giants hence refers to the solid component making up 60-80% of their mass, and not that they missed runaway accretion, which would make for a clearer name.
Now the other question is, why planets at small radii seemed to have evaded runaway gas accretion, in our solar system and at least 50% of exoplanetary systems. A candidate mechanism is "gas recycling", i.e. replenishment of entropy to protoplanetary atmospheres which prevents their contraction. This is possible close to the star because the gas is very dense, replacing cooling with advection as dominant entropy transport mechanism.
Summarizing
The broad strokes of the solar system architecture can be understood in terms of physical mechanisms which have been shown to work in simulations. However, when applying those same mechanisms in order to form synthetic populations of planets, those synthetic populations are usually inconsistent with the observed ones.
This is work in progress and needs future missions to the ice giants in order to measure their detailed heavy element abundances and use the latter to distinguish between competing formation scenarios, of which I have presented merely one.
The physics presented here is therefore very different from a simple 'heavy elements sink in solar wind' picture, which to the extent of my knowledge, was never considered as candidate for a planet formation model. Merely Laplace in the 18th century considered a similar sounding model of yours, of an extended solar atmosphere that centrifugally breaks into rings in order to form the planets. With my prelude as above, this model is now known to be incorrect however.


If the space is expanding near the speed of light or the speed of light, why Andromeda is approaching or coming near to our galaxy (Milky way)?

The Andromeda galaxy is about 2.5 Mly away, less than 1 MPc, so if the Hubble flow were relevant, it would have Andromeda moving at about 55 km/s away from us. Andromeda is actually approaching at about 300 km/s, or almost 6X that calculated due to the expansion of space. However, Andromeda and the Milky Way are graivitationally bound along with the rest of the galaxies in our local cluster. The individual (so called, peculiar) velocities of these local galaxies are the result of orbital motion within groups and around the local cluster barycenter.

Expansion dominates on larger spatial scales, such as distances between clusters and superclusters of galaxies.


Why is the difference from a perfect sphere the same for the Earth and the Moon?

It isn't, as soon as you look one more digit further.

From NASA's Earth Fact Sheet and Moon Fact Sheet which I found here

                             Moon (km)   Earth (km)      Ratio     (rounded)
Mean radius (volumetric)     1737.4       6371.000      0.27270      0.273
Equatorial radius            1738.1       6378.137      0.27251      0.273
Polar radius                 1736.0       6356.752      0.27310      0.273

If you would like to compare the Moon and Earth parameters further, see the Moon Fact Sheet:


Length of a tidal day or by how much is the Moon late each day?

As you point out, the mean moon conjucts (and actually eclipses)
the mean Sun every M = 29.5305891203704 days. If the moon's sydonic
period were exactly 29 days, the gain would be clearly 1440/29 or
49.6552 minutes. If the moon's sydonic period were exactly 30 days,
the gain would be clearly 1440/30 or 48 minutes. Thus, your
calculation seems to be correct here.

2, 3, 4. I think this calculation is incorrect because you're
confusing sidereal days with solar days, and your attempt to correct
for this is in the wrong direction:

Suppose the moon culminates at noon at a given location (which
technically means the sun also culminates, and we have an eclipse,
as above).
Twenty four hours later, the sun culminates again, and the moon
has moved 13.17636 degrees along the ecliptic in its orbit.
However, the sun has also moved (about 1 degree) along the
ecliptic in the same direction.
In other words, the moon has only "gained" about 12.17636 degrees
with respect to the Sun (it's actually a little more, since the Sun
moves less than one degree along the ecliptic per solar day).
Since we measure time by the sun, we only need to rotate an extra
12.17636 degrees for the moon to culminate.
That's less time than we'd need otherwise, not more.
I'm too lazy, but if you do the calculations keeping the above in
mind, you should get the same answer.


How does the cepheid variables help in finding distance

Why not read his original paper? It's fairly short by today's standards, despite being such a monumental turning point in our understanding of the size and nature of the Universe.

In short though, Hubble measured the distance to Cepheid's the same way anyone does. There is a distinct relation between the luminosity oscillation period of the star and the star's luminosity. If he can measure the star's oscillation period with good accuracy, he can derive the luminosity. If he can also then measure the brightness (i.e., how bright it appears to be to him) then he can calculate the distance (since luminosity, distance, and brightness are all related).


Does time dilation increase within overlapping gravitational fields?

Gravitational time dilation depends on the observers' relative positions in the gravitational potential.

For relatively weak gravitational fields the rate at which a clock ticks for observer A compared to the clock rate for observer B is given by
$$\frac{d \tau_{\rm A}}{d\tau_{\rm B}} \simeq \sqrt{1 + \frac{2\Delta \Phi_{\rm AB}}{c^2}} \sim 1 + \frac{\Delta \Phi_{\rm AB}}{c^2},$$
where $\Delta \Phi_{\rm AB}$ is the difference in gravitational potential between observers A and B.

Gravitational potential is a scalar quantity, so the basic answer to your question is yes - the effects of the two neutron stars would be summative. However, I would hesitate to use the approximation above when dealing with the strong gravity near neutron stars.


Why are the high density regions in outer space black holes?

Making a major edit to this answer, as it would seem that there is a rather significant misunderstanding in the formula.

Initially describing r as the radius may have been a mistake.  While we may normally describe r as the radius of Earth when describing Earth, this is because we are describing the escape velocity of Earth from the surface, which means that the difference between our location and the center of gravity of the Earth is roughly equal to the radius of Earth.

When determining escape velocity from an arbitrary distance, r should be the distance between the escaping body and the center of gravity of the body being escaped. For objects at a normal distance, changing the volume, and thus the density of the object, will not affect nearby objects, as the mass stays the same.

However, this changes when you get close, when you compress the object this allows you to get closer to the center of gravity of the object. As you know, gravity is a weak force which deteriorates quickly with distance.  As such, when you decrease the volume of the body, a body can then become closer to its center of gravity, increasing the potential maximum escape velocity.

The body becomes a black hole when the physical volume of the body becomes small enough that the Schwarzschild Radius extends beyond it. Because we take the same mass, and decrease the volume, the result is a much denser area of spacetime, but this isn't generally the result of only a tiny volume, but also a large amount of mass.


Just to put in a simple answer:

if you mean (A) for astronauts just in orbit (at the space station and so on) or on the surface of the moon.  Yes, it is incredibly bright.  Just like on Earth in the Sahara at noon!

A point of confusion is that THE >> SKY << IS BLACK, but it is incredibly bright.  On Earth in daytime, the sky happens to be blue, and it is incredibly bright.  On Moon in daytime, the sky happens to be >> black <<, and it is incredibly bright.  (Same deal in orbit as on the Moon.)

(If you are, say, explaining to a young person "is there daytime/nighttime on the moon" that's the salient point to explain: on the moon, confusingly, the sky is black during both daytime and nighttime.  But everything's just as bright on the moon in daytime as on Earth in daytime, there's the same amount of light, i.e. you can trivially read a book or see the ground, just the same as on Earth.)

If your question is about (B) in the solar system - say "on Uranus" or the like.  There is remarkably less light there than on Earth, the sun is much smaller.  Many young people don't realize this; everything's basically dark all the time from about Saturn outwards!

Outstanding page explaining this:

http://www.astronoo.com/en/children/sun-apparent-size.html

Finally if your question is about either (C) inside our galaxy but not near a star, or, (D) midway between the galaxies:

That's too difficult.


Under General Relativity (GR) alone, a Black Hole's (BH's) event horizon is a point of no return -- anything that passes through the event horizon is lost and gone forever, and nothing comes out. Hence, under GR alone, BHs are utterly black and don't have a temperature at all.

This is why the absorption of radiation (or anything else) by a BH doesn't raise its temperature -- it just gets swallowed up and lost. (It's mass, angular momentum and charge do remain, but that's all -- see the No Hair Theorem.)

(Note: The accretion disk that surrounds a BH can be very hot indeed, but that's another thing entirely.)

Stephen Hawking discovered that applying quantum mechanics to BHs showed that BHs would emit a random spray of radiation, and that that radiation was precisely what a black body would emit -- black body radiation.  This is called Hawking radiation.

Black body radiation is simply the thermal emission of a perfect absorber of radiation, and leads to the inescapable conclusion that a BH does have a non-zero temperature.  Interestingly, Hawking's analysis showed that the effective temperature of the BH is inversely proportional to its mass and that solar-mass BHs (which are the smallest for which we have actual evidence) would have a temperature of about 0.00000006 K. Kinda cold, but still not zero.

Note that, unintuitively, a solar mass BH get colder as it absorbs radiation. Because any radiation (or anything else) it absorbs increases its mass, and since higher mass BHs are colder, the more energy you dump into one, the colder it gets!


How close would a Sunlike star have to be in order to be dangerous to look at it through a spyglass?

This is more a question of "what is the brightest light I can look at", and subsequently this answer is more physiology, geometry than astronomy which only gives us the distances.
According to the wikipedia page on laser safety it is around 1mW/cm². A binocular - or any optics - again change the equation as so far that they accumulate all light falling onto its entry pupil into your eye.
Stars are sufficiently far away so that they are point sources (that's why they wobble around in the viewfinder due to atmospheric disturbance, unlike solar-system objects).
Energy emitted by a star is obtained by a simple black-body assumption $E=4\pi R^2_\star \sigma T^4$ and energy $W$ per on an area in distance $r$ thus is $W_\star = \frac{4\pi R^2_\star}{4\pi r^2}\sigma T^4 = \frac{L_\star}{4\pi r^2}$ (using the stellar luminosity $L_\star$ further on). Let's assume that your pupil has 0.5cm opening, thus the area there is $A_e = \pi 0.25^2 = 0.2cm^2 = 2\cdot 10^{-5}m^2$. Let's assume a binocular with 5cm aperture, thus the area there is $A_b=\pi\cdot 0.05^2m^2 = 8\cdot 10^{-3} m^2$, thus the ratio $A_e / A_b = 2.5\cdot 10^{-3}$. Thus the light at the entry of the binocular needs to be $W_{max} = 1mW/cm^2 \cdot 2.5\cdot 10^{-3} = 2.5\mu W/cm^2 = 2.5\cdot 10^{-2} W/m^2$ at most to not become dangerous. So, what distance does that imply? Let's equate:
$W_{max} = W_\star $ and solve for $r$: $r = \sqrt{\frac{L_\star}{4\pi W_{max}}}$. So for a sun-like star ($L_\star = 4\cdot 10^{26} W$) we get a minimum distance to safely look at its disk of roughly r=240 AU or 0.004 light years.
You can modify that equation for other stars with higher or lower energy output. Alpha centauri is not much brighter than our sun. But for a star like Vega ($L=40L_{sun}$), the minimum safe distance would already be more than 6 times as distant - yet still very near for any distances when considering distances to other stars. Thus any star becoming dangerous to look at with the naked eye or a binocular would pose a threat to our solar system.
And even if you consider not binoculars, but telescopes, say 1m diameter, thus 400x the light gathering power compared to the considered binocular, we would start to get distances of the order of a light year... even then all stars are further away.
(this still sounds crazy... is there an error in the estimates? But likely this errs on the side of caution. While you should not look at the sun even with the unaided eye, you're not immediately blind - and it's intensity is $1400W/m^2$ compared to the assumed safe power of $0.025W/m^2$, thus there's a difference of $10^5$)


Different types of stars are responsible for creating different groups of elements via nucleosynthesis. Most of the "lighter" metals (Na, Al, Ga, Ge, Se), were formed by exploding massive stars, and most of the "heavier" metals (I, Ir, Ag, Au, Pt, U, Pu) were created by merging neutron stars. Exploding white dwarfs are partially responsible for elements like Ti, V, Cr, Fe, Co, Ni. Dying low mass stars were involved with the creation of some of the elements Sr, Y, Zr.

The following periodic table of elements (from Wikipedia), shows the elements color coded according to their source of creation.

All the elements on Earth were created by various stars that seeded our region of the universe prior to the formation of our solar system.


Is there any proof of space being created?

This is an intriguing proposition, but I would ask how your hypothesis explains that the universe appears to be flat? That is with $\Omega_M + \Omega_\Lambda = 1$. The evidence for this comes from measurements of the cosmic microwave background, yet if we sum up all the matter (including dark matter), we only arrive at $\Omega_M \sim 0.3$.
I do not think that your proposed arrangement both allows an accelerating expansion of the visible part of the universe (which note, can only be gravitationally influenced by "outside regions" that are within causal contact with it) and for it to be geometrically flat.
In addition, I note you are attempting to explain the expansion in some sort of Newtonian sense - I suppose your clumps of far-away dark matter are meant to be "stretching" the observable universe? But that isn't what happens in GR and any Newtonian interpretation would have to counter the problem that if the matter outside the observable universe was uniformly distributed (i.e. in a spherical shell), then it would exert exactly zero force on everything inside it (the shell theorem); but if it were not symmetric then you have to explain why the expansion of the universe appears to be (very close to) isotropic observationally.


How do we know if an object is redshifted?

There are two methods, one more reliable than the other (though both are pretty good.)

Key point: The brighter a star is, the more detail we can see in its spectrum -- you can think of it as being able to magnify the spectrum more so as to be able to see finer details.  This also allows us to see fainter lines (not all spectral lines are equally intense.)

We have very extensive reference spectra of all the elements under a wide variety of conditions, so when a star is bright enough we see many spectral lines and can match them up to reference spectra.  No two elements have similar spectra, so if you can see many lines, you can't easily confuse one for another.

When you can account for nearly all of the lines in a star's spectrum by a plausible set of elements in plausible abundances and all with the same red shift, you have a very, very reliable match and accurately know the object's red shift.

But distant stars (and galaxies) are very faint and so the spectra we get are very low resolution and show only a few of the strongest spectral lines. They can frequently be matched because hydrogen is so common that the strongest lines seen will be hydrogen lines and even if you can only spot two or three of them, if they bear the same relative positions as bright hydrogen lines, you can safely identify them and read off the red shift.  This is less reliable, but still reliable enough that it rarely is a problem.

In the very dimmest cases you may only see a single line. You have a very rough idea of the redshift from the object's brightness and all you can do is assume it's the strongest hydrogen line and see if it's in roughly the right position given the distance deduced from the object's brightness and type.  This works too, but is still less reliable.  (Much of Hubble's original work relied on just a single line for the more distance objects, but proved to be correct when observed later with dramatically better equipment.)


Could Sgr A* be a core of dark matter instead of a supermassive black hole?

I'm going to say, "probably not", because (as tfb points out) we have very good evidence for SMBHs in other galaxies -- in particular galaxies with active nuclei where the phenomenon associated with accretion disks around BHs are seen. The most extreme case is probably NGC 4395, where a Seyfert 1 nucleus and associated radio jets exist. The best estimates for accretion-disk+SMBH models suggest a SMBH mass of only a few hundred thousand solar masses; a recent direct dynamical measurement (den Brok et al. 2015) indicate a central "dark object" gravitational mass of $M_{\rm BH} \sim 400,000$ solar masses. There are numerous other galaxies with active nuclei where the estimated or measured central-object masses are $M_{\rm BH} \sim 10^{6}$--$10^{7} M_{\odot}$.
This is rather difficult to square with the claim that "above a critical mass, a dark matter clump could gravitationally collapse into a supermassive black hole" because if you follow up the associated reference to Argüelles et al. (2021), it ends up predicting a critical mass of $\sim 10^{8} M_{\odot}$. This is consistent with something like M87 ($M_{\rm BH} \sim 6 \times 10^{9} M_{\odot}$), but not with all the evidence for accreting SMBHs in galaxy centers with $M_{\rm BH} < 10^{7} M_{\odot}$.
The "flares observed from Sgr A* in recent years" are, I believe, generally supposed to be evidence for a very weak, transient accretion disk around the SMBH (e.g., the discussion on this web page), and so would be difficult to explain with this model.


Any advice for figuring out whether that is a real celestial object or man-made?

Yes - in almost all cases, you can readily identify a celestial object by going out the next clear night, and looking for it in the same position.  (Very few celestial objects move far over a few days.)

In response to comments about geostationary satellites:
I doubt it is a geostationary satellite, answers to this question indicates these are usually too faint to be seen without special equipment. Also they don't move, whereas nearby stars will move a little over the space of half an hour or more.


Could there be a M-type asteroid beyond Jupiter?

According to the Grand Tack Hypothesis, in the early solar system, Jupiter and Saturn migrated inwards and then outwards, scattering asteroids as they did so.

In this hypothesis it seems inevitable that some m-type asteroids will have ended up in orbits with large semi-major axes.


Can the moon's gravity influence be observed on the space station's orbit?

According to NASA, satellites in Earth orbit don't experience strong enough interference from the Moon to "whisk a spacecraft out of the game". To a satellite in Earth orbit, the gravitational pull of the Sun is actually ~160 times stronger than influence from the Moon. Source: A New Paradigm for Lunar Orbits (see EDIT note below for details).

However, gravitational pull from the Sun, the Moon, planets such as Jupiter (as well as other factors such as solar radiation pressure, atmospheric drag, etc.) do cause enough interference to slightly alter the orbital path of any satellite orbiting Earth. To compensate for the deviations all satellites must periodically fire their thrusters for brief periods of time in order to maintain the correct desired orbit. If a satellite has no fuel to perform these "orbital stationkeeping" maneuvers it essentially becomes space junk. Source: Is the ISS, and all other satellites, affected by the gravitational pull of the moon?

EDIT: Consider a satellite with mass of 3,000 kg rotating around Earth on geostationary (GEO) orbit of 42,164 km radius. Gravitational pull from the Earth is constant throughout the orbit and equals ~673 N (Newtons). As the satellite rotates around the Earth it gets closer or further away from the Moon as well as the Sun and, therefore, their gravitational pulls change. The pull from the Sun is about 17 Newtons but changes only by 0.02 N. The gravitational pull from the Moon varies between 0.12 N (when the satellite is between the Earth and the Moon) and 0.08 Newtons (when the satellite is furthest away from the Moon over the opposite side of the Earth). So even though the pull from the Moon is much smaller than from the Sun, the variation of the Moon's perturbation is about twice as large than from the Sun.

Note: the numbers above are computed under a simplified assumption that all orbits are on the same plane. In reality, due to the inclinations of orbits, the numbers can be somewhat lower, but the overall picture remains about the same.


Standard Big Bang model and space curvature

What exactly did the book say? It all depends on what you mean/define as "curvature". What you describe appears to be a description of the behaviour of $\Omega$. Inflation does indeed drive $\Omega$ towards unity and simultaneously flattens space because the radius of curvature grows exponentially bigger.

If $\Omega < 1$ at some early epoch then it should decrease quickly with time such that $\Omega << 1$ in the present day - this means that the universe has negative curvature, but does not mean it is becoming more curved.

In the Friedmann equation, the curvature parameter $k$ is a constant $(1,0,-1)$
$$H^2 = \frac{8\pi G\rho}{3} - \frac{kc^2}{a^2}$$

Here, the spatial curvature is $k/a^2$ and the radius of curvature is $a$ if $k=+1$. Thus as the universe expands and $a$ gets larger, any curvature becomes smaller.

In a little more detail - one can write the above equation in terms of the density parameter $\Omega$, the ratio of density to the critical density $3H^2/8\pi G$:
$$(\Omega ^{-1}-1)\rho a^{2}={\frac {-3kc^{2}}{8\pi G}}$$

During inflation, the energy density $\rho c^2$ remains constant as $a$ grows exponentially. In order to keep the left hand side equal to the right hand side (which is just a collection of constants), then $\Omega$ must be driven very close to unity, while $k/a^2$ will tend towards zero.

After inflation then $\rho$ will vary with $a$ depending on whether the expansion is dominated by matter ($a^{-3}$) or radiation $(a^{-4})$. In both these cases $\rho a^2$ will decrease as the universe expands, such that if $k \ne 0$, then $(\Omega^{-1} -1)$ must increase, which means that $\Omega$ must either grow or shrink away from unity. But $k/a^2$ continues to get smaller as the universe expands.


When we say a galaxy is 200 million light years away, does this account for the expansion of space in the time it took it's light to reach us?

Usually the distance data doesn't take the expansion of the universe fully into account. That's not important for a distance of 200 million lightyears, but as we get close to the border of the visible universe, it makes a relevant difference. Therefore astronomers prefer to talk of redshift instead of distance.

To be a little more precise: The light travelling from a galaxy to Earth takes only that part of the expansion of the space between the galaxy and the Earth into account, which it still has to traverse.

For a proper distance measurement in the everyday's sense, we would have to send light to the galaxy and reflect it back to Earth, measure the time, divide it by the speed of light, and take the half of the resulting distance.

If this experiment would be performed, some of the now visible objects would never reflect light back to Earth, because they are already too far away.

For a more detailed discussion about the various notions of distance, read Wikipedia's article on distance measures in cosmology.
The usual distance data is provided in the (one-way) light travel distance.


If the Wow! signal was not directed at us, then where?

In the opposite direction (RA= 07h 25m 31s ± 10s, +26° 57′ ± 20′, J2000.0, seen from Earth) of the positive horn of the Wow!-signal (RA= 19h 25m 31s ± 10s, −26° 57′ ± 20′, in J2000.0 coordinates), Iota Geminorum, aka Propus, (RA= 07h 25m 43.6s, +27° 47' 53") is within a 3-sigma error ellipse (1.26 sigma in RA, 2.54 sigma in Dec, 2.84 sigma total error).

However, this doesn't reveal, whether the signal pointed to anywhere.


Seeking advice regarding buying a telescope

All the magnifications listed above 84x will be useless on a 60mm telescope.

Still, such a telescope can be useful if it is well made (particularly a steady tripod/mount) and you have reasonable expectations.

It will show lots of detail on the moon.  You will be able to see the phases of Venus and Mercury (which is pretty much all anyone can see of those from Earth).  You will be able to see Mars as a disk and maybe glimpse an ice cap or a little surface detail under excellent seeing with a favorable apparition.  You will be able to see Saturn's rings and Jupiter's moons and maybe with good conditions a band or two on Jupiter.

It will only show a few of the very brightest nebulae, and only as a faint smudge.


Best approximation for Sun's trajectory around galactic center?

The Sun orbits in the Galactic potential. The motion is complex; it takes about 230 million years to make a circuit with an orbital speed of around 220 km/s, but at the same time it oscillates up and down with respect to the Galactic plane every $\sim 70$ million years and also wobbles in and out every $\sim 150$ million years (this is called epicyclic motion). The spatial amplitudes of these oscillations are around 100 pc vertically and 300 pc in the radial direction inwards and outwards around an average orbital radius (I am unable to locate a precise figure for the latter).

This is established by measuring how the Sun moves with respect to the average motion of the stars in the solar vicinity - the so-called local standard of rest. Estimates of the Sun's motion with respect to the LSR vary a little. According to Dehnen & Binney (1998) the Sun currently moves at 10 km/s inwards, at about 5 km/s faster than the average star tangentially to the Galactic centre and at about 7 km/s upwards out of the Galactic plane. Though as I mentioned previously, these motions are of an oscillatory nature around the mean value. The Sun is currently about 8.5 kpc from the Galactic centre and about 20 pc "north" of the Galactic plane.

I leave it as an exercise for the reader to visualise this orbit. From a side-view it would look like a pseudo-sinusoidal oscillation. From above the disk it would look like a slightly eccentric ellipse with a very large precession.

All of the above assumes that the Sun moves in a smooth Galactic potential. In fact the presence of spiral arms, a Galactic bar and the clumpy presence of giant molecular clouds can and will perturb the Sun's Galactic orbit around the regular behaviour described above, so predicting the Galactic location of the Sun in a billion years or so is not necessarily going to be very accurate.


How many planets have we discovered that can support human life?

There is currently only one planet known to be capable of supporting human life, and you're on it.

Several planets have been found in the region in which we expect water to be liquid on much of the planet. Of these, only one fits the criteria of being Earth-sized and well placed in the habitable zone: Kepler 186-f

However we know nothing about it's atmosphere (or lack of one). The star is a red dwarf, so it could be subject to dramatic solar flares. The planet is rather colder than earth, so could be in a perpetual "snowball world" state, depending on the composition of the atmosphere and the strength of the greenhouse effect. The atmosphere would be very unlikely to be even close to breathable, and it is nearly 500 light-years from Earth, so could not be reached in a reasonable amount of time, even with much more advanced propulsion.

At the moment we can't usually detect most Earth-like planets in the habitable zone of brighter stars like the sun, though the probably do exist and may be common.


Why are stars more metallic closer as you move closer to the galactic bulge?

It has to do with the formation of the Milky Way.

At the beginning, the Milky Way was much more spherical than it is now - perhaps closer to what an elliptical galaxy is like than a spiral galaxy. Population III stars would have formed first, then quickly died out. Next came Population II stars. They formed when the galaxy was still somewhat spherical, and so they tend to inhabit the galactic spheroid/halo.

Eventually, the rotation of the Milky Way flattened out much of the remaining gas and dust, and some of the stars. When younger stars formed, they formed in the flatter disk, nearer to the center. The disk itself became smaller that the spheroid/halo. Thus, the younger Population I stars are found in the galactic disk, and are closer in.

No more stars will form in globular clusters; they are relatively dust-free and contain old, Population II stars. The same is true for the galactic halo.

Source: Populations & Components of the Milky Way

The galactic bulge itself contains several populations of stars. Some may have come from the halo and thick disk (thus being metal-poor) while others may have formed together more recently from the thin disk itself (thus being metal-rich).


How do you mount these binoculars and telescope on this tripod?

What you want to attach the telescope to a tripod is something like this:

https://www.firstlightoptics.com/dovetail-bars/baader-vixen-style-dovetail-clamp.html

Which is a clamp that holds the standard vixen dovetail used by most small telescopes.  The important thing with the one linked to is that it has a threaded hole to allow you to attach it to a standard 1/4" photo tripod screw. (Not all do - some of them are designed to bolt to a mount, so only have countersunk holes).

Similarly, the tripod adapter that came with the binoculars should have a threaded hole on the bottom part to attach it to the tripod. The captive screw in the adapter attaches to a threaded socket on the front of the
binocular hinge - there's likely to be a circular cover that unscrews or clips off to reveal the socket.

Unfortunately, judging from the picture of the tripod, it looks as though you may be missing the camera platform bit. Normal photo tripod heads will either have a platform with a captive 1/4" whitworth tripod thread thumbscrew, or a platform with a quick release holder for a removable plate which has the screw (the quick release versions make it easier to quickly attach or remove the camera (with the QR plate attached) from the tripod.

From the picture, it looks to me as though that platform part is missing.

If the tripod head was designed to be removable, it will usually be screwed onto a 3/8" bolt attached to the leg section. There may also be some extra
smaller screws on the underside of the platform the head is attached to to
prevent it from unscrewing by accident. If it is removable, you may be able to replace the entire tripod head unit (This is fairly common with more expensive photo tripods, where you can buy the legs and heads separately).

But you might find it cheaper to look for a secondhand tripod.

For supporting a telescope, you usually want a fairly solid and stable tripod. That can often mean heavy, large and expensive. Most of the smalller /  cheaper tripods are designed for lightweight cameras and can be shaky with
a heavier load.

The problem with using a photo tripod for normal binoculars is that when you're observing near the zenith (looking straight up) the tripod gets in the
way. You can also get things called parallelogram mounts that put the binoculars out on a pivoting arm with a counterweight on the other side - since the binoculars aren't over the tripod you can point them straight up without the tripod getting in the way.

[Edit - now that you've added the manual pictures, it looks as though you may just be able to attach the scope directly to a standard tripod, without needing a dovetail clamp, assuming at least one of the two holes in the bottom of the telescope dovetail is threaded. Some tripod heads have a rectractible or spring loaded pin, as well as the main screw, mainly designed for use with movie/video cameras. If the spacing on that matches up with the other hole in the scope dovetail, then the pin helps prevent the scope accidentally unscrewing (since it stops the scope turning round on the platform).]


Cherenkov radiation (Wikipedia)

  Cherenkov radiation, also known as Vavilov–Cherenkov radiation,[a] is electromagnetic radiation emitted when a charged particle (such as an electron) passes through a dielectric medium at a speed greater than the phase velocity of light in that medium. The characteristic blue glow of an underwater nuclear reactor is due to Cherenkov radiation. It is named after Soviet scientist Pavel Alekseyevich Cherenkov, the 1958 Nobel Prize winner who was the first to detect it experimentally.1 A theory of this effect was later developed within the framework of Einstein's special relativity theory by Igor Tamm and Ilya Frank, who also shared the Nobel Prize. Cherenkov radiation had been theoretically predicted by the English polymath Oliver Heaviside in papers published in 1888–89.


How does radio bandwidth restrict the choice and use of science payload on interplanetary probes?

Sample returns are expensive and risky.  Radio is simple and reliable.  Radio bandwidth on a particular mission is determined by several engineering tradeoffs.  Higher bandwidth requires either larger antennas, more output power, or more complicated processing.

Spacecraft are typically limited in mass, power and compute capabilities. What most missions have plenty of, is time.  Lower bitrate takes longer, but allows more mass and power to be used by other experiments.  This is often a highly desirable tradeoff.

Missions like New Horizons that have a fairly limited time to record lots of observations do so to a data storage device, and then spend months or years relaying the data back to Earth during the long coast to the next target.


How to scale down solar system data to simulatable values

As barrycarter mentioned in his comment, you should be more concerned with units and less with scale.

Generally, it's best to stick to conventional units that people recognize. (It will keep your head on straight and make it easier for others to review your work.) In astronomy, these are a little different than the standard SI units, since things are—shall we say—big, and numbers get out of hand quickly (as you have noted). Here are some units suggested in the Astronomical system of units Wikipedia page:

Time: Day. This is probably not too important, so feel free to use seconds here. (After all, it's only a 5 order of magnitude difference.)
Mass: There are several conventions, but the one I've personally seen most is solar masses. Since you're modeling our solar system, it makes a convenient reference.
Length: I would definitely use astronomical units. (What an appropriate name!)

If you stick to these units, you should be able to limit error due to rounding and reduce the computational load of large numbers.


Why would the Perseids meteor rate fall off after maximum faster than the increase before maximum?

A very interesting question. Neither Bruce McClure at EarthSky nor I knew the answer, despite the fact that the observed gradual rise in numbers for the Perseids - and their quick drop-off - is something we've both observed for decades. So we asked a true expert on meteor showers, Robert Lunsford of the American Meteor Society. He also said: "Good question!" And, he said: "It seems that the particle density of the meteor stream [what you're calling here a 'fuzzy tube'] is greater prior to maximum than after. Perhaps the core of the stream, which produces maximum activity, is not quite centered in the stream. This is true for some other major showers such as the Geminids. The core of the Geminids must be really off-center as the display nearly disappears 48 hours after maximum. Gravitational effects from the sun may also play a role in this shift."

So here, as so often in astronomy, the answer suggests our knowledge of nature is - at present - inexact. Are any of you old enough to remember when the Voyager spacecraft first encountered Jupiter? Nowadays, we're used to seeing the very detailed images of Jupiter's cloud bands, but when we first saw that detail - via the Voyagers in the late 1970s - we were blown away! The situation with meteor streams in space is likely exactly the same. We picture them as symmetrical only because we don't have instruments refined enough to reveal their structure. All we have are the long-observed rates of meteors in annual showers - such as the Perseids and the Geminids - suggesting that meteor streams do indeed have a structure.

Hope this helps!


Have there been successful attempts at spotting Neptune with the naked eye?

Well, having seen Neptune and identifying Neptune are two totally different things.

Let's tackle this one step at a time. Your link says:

  From these observations it would appear that, if seen upon a perfectly black background, a star of magnitude approximately $8.5$ would be at the limit of unusually good vision.

The problem with this is that there are stars, diffused sunlight and airglow that prevent the night sky from being near-perfectly black. This has always been a problem, and it's only gotten worse with air pollution — meaning that we would be less likely to see Neptune unaided nowadays.

My guess is that in the past, there have been a few people who have seen Neptune, but none that would take note of it. Do you know how many objects of magnitude less than $7.8$ are there in the sky? The answer is, approximately 30 thousand! Good luck finding Neptune.

Even with telescopes, Neptune can be hard to point out. Galileo actually discovered Neptune 200 years before it was recognized as a planet, but he mistook it for a star. Point is, even aided eyes often cannot identify Neptune.

Anyway, the reason Curtis could see such dim objects is because he had exceptionally good vision and was aided by tools:

  Without using the diaphragms and apertures, Curtis was able to see stars of magnitude $6.5$ directly against the background of the sky. With their aid he easily observed stars in the magnitude range $7.2$ to $7.4$. Four stars between magnitudes $7.9$ and $8.1$ were usually seen with little difficulty...

Without these tools, he probably wouldn't have seen it at all. Not to mention, Neptune's brightness fluctuates — at its brightest, it is $7.8$; at its dimmest, it is $8$. That might not seem like much, but when finding Neptune, it could make a big difference.

Regardless, even in ideal conditions, it is generally agreed that Neptune is the only planet that cannot be seen with the naked eye. Most sources state that you need to at least use binoculars to see Neptune.


You're correct that "$p$-wave" in this context means that the charmonium has orbital angular momentum $L=1$. The principal observable effect is that two-particle states with even $L$ have even parity, while those with odd $L$ change sign under parity.

If you were to skim the Particle Data Group's list of charmonia (or the $c\bar c$ section of the short list) you would find that each particle has a value listed for spin, $J$, and parity, $P$. The lightest few states are

$\eta_c(1S)$, with $J^P = 0^-$. This is a charm-anticharm bound state where the two quarks have opposite spins and the angular momentum $L=0$. (The total parity is negative because the intrinsic parities of the $c$ and $\bar c$ have opposite sign.)
$J/\psi(1S)$, with $J^P=1^-$. This has the same parity as the spinless $\eta_c$, but a unit of spin. So it must be that here the two quarks have the same spin and $L=0$.
$\chi_{c0}(1P)$, with $J^P=0^+$. The change in parity suggests that this state must have a different $L$ than the $\eta_c$ and $J/\psi$, and it's the first with a $P$ in its name. There's also a $\chi_{c1}$ and $\chi_{c2}$ with $J^P=1^+, 2^+$, respectively. That suggests these are the three spin combinations of a charm-anticharm pair with spin $S=1$ and orbital angular momentum $L=1$.

There are also some charmonia with $(2S)$ and $(2P)$ in their names. There don't appear to be any definitively labeled $d$-wave charmonia.


The definition of eclipsing binary star systems

I assume that the diagram indicates what the observer sees (if they had a big enough telescope!). i.e. The viewpoint is nearly in the orbital plane but not quite. Why then are the eclipses asymmetric, with the secondary eclipse being shallower than the primary? Well probably because the surface brightness of the two stars is different - i.e. they have different surface temperatures. The secondary would be cooler, so when its surface is
obscured by the primary, then that results in less flux being eclipsed. Note that this would be the case whether you were looking along a line of sight exactly in the plane of the orbit or at a small angle to it.
There is no definition of an eclipsing binary that says the plane of the orbit must be in the line of sight from Earth. It has to be close to that in order to get eclipses, but it does not have to be exactly true.


Is there a possibility that a white dwarf can turn into a neutron star or a black hole?

The answer is: to a neutron star - possibly; to a black hole, no.

The process whereby a neutron star is formed is known as an accretion induced collapse and is being seriously debated, especially in the case of white dwarfs that are born at the upper end of the "natural mass range" for white dwarfs and then accrete more mass as part of a binary system. An excellent read are the introductory sections of Taurus et al. (2013), who go through the motivation, process and (limited) observational evidence. See also Schwab et al. (2015); Ruiter et al. (2018).

Explosion vs Collapse

A white dwarf can respond to the accretion of material by exploding or collapsing. It depends on the competition between energy released in fusion reactions and the energy being locked away by endothermic electron capture (also known as neutronisation) reactions.

If thermonuclear reactions are initiated then the likely outcome is a runaway nuclear reaction - the pressure inside the star does not rise rapidly enough to prevent the entire star undergoing fusion. The energy released exceeds the gravitational binding energy and the likely outcome is a type Ia supernova.

On the other hand, the white dwarf is supported by electron degeneracy. If neutronisation begins to occur in the core, then protons (in nuclei) capture electrons to form neutrons. This destabilises the star causing it to collapse. The collapse would proceed (quickly) in a similar way to a core collapse supernova. The nuclei would dissociate, neutronisation would run to near-completion and the collapse would be halted by the formation of a neutron star.

There is little possibility that a black hole could be formed by such a collapse. The collapsing object would be of order 1.4 solar masses and comfortably smaller than the maximum mass of observed neutron stars (at least 2 solar masses). Therefore the collapse will be halted at the neutron star phase.

White dwarfs of moderate mass

Most white dwarfs of moderate mass have a C/O composition. They will need to accrete a lot of mass to get to a density (at about $4\times 10^{13}$ kg/m$^3$, reached at $1.38M_{\odot}$ in a non-rotating WD) where neutronisation becomes energetically feasible. Before this happens, it is probable that fusion reactions are ignited (due to high density, rather than temperature). The threshold density for ignition is *lower * for nuclei with lower atomic number (He < C < O) because of lower Coulomb repulsion, and the ignition threshold densities for He and C are also lower than the neutronisation threshold for C.

This means that in a C/O WD that has accreted a lot of matter, ignition could take place in C at the core, or it could be triggered in He (at even lower densities) at the base of a deep accreted shell of material. The outcome would likely be runaway thermonuclear fusion and the complete destruction of the star.

More massive white dwarfs

O/Ne/Mg WDs are made as the final stages of more massive stars ($8-10M_{\odot}$) and are probably born as remnants with much higher mass $>1.2M_{\odot}$ than typical C/O WDs. More massive WDs are smaller, with higher density. The neutronisation thresholds for O, Ne and Mg are only $1.9\times10^{13}$, $6\times 10^{12}$ and $3\times 10^{12}$ kg/m$^3$ respectively (all lower than for C, especially for Ne and Mg). This means that a O/Ne/Mg WD may have to accrete very little mass to reach this central density, begin neutronisation, which leads to collapse. In addition if such densities are insufficient to trigger C burning in a C/O WD, then they certainly won't be high enough to trigger burning in O/Ne/Mg because of stronger coulomb repulsion. Further, if little mass is accreted, then there won't be a deep envelope of accreted material in which to ignite burning off-centre.

For all these reasons, O/Ne/Mg WDs may be more likely (Liu et al. 2018; but see also Wang 2018) to collapse than explode (the collapse would cause a type of core-collapse supernova though).

Does accretion induced collapse occur?

At present there is only indirect evidence. When we look at recently formed neutron stars - identified as fast-spinning pulsars - we see they generally have very high velocities. These velocities are thought to result from an asymmetric "kick" delivered by a type II core-collapse supernova. This in turn suggests it might be quite difficult to retain a neutron star in a binary system, but many neutron stars are observed to be in binary systems, and many of these, particularly the millisecond pulsars, are thought to have undergone significant mass transfer in the past.

Further evidence comes from the retention of a significant population of neutron stars inside globular clusters. Again, the kicks might have been expected to expel most of these. In addition, there are a number of examples which appear to be "young", in that the ratio of their spin periods to rate of spin decay indicates they were formed recently. Since there are no high mass stars in globular clusters, and so no possible progenitors for these objects via the core-collapse of massive stars, then accretion induced collapse of a high mass white dwarf is a possibility.


Should only spherical satellites be considered 'moons'?

The need to distinguish between these potato-shaped and spherical moons hasn't arisen. As such, there aren't two different words to designate these two types of moons. And as long as the community of astronomers who study natural satellites doesn't feel the need for there to be an extra category, there likely won't be one created preemptively.

Perhaps, one day, we will discover that potato-shape and spherical satellites form in very different ways, of have very different properties (not just esthetic). Then the case might be made to sort them into two different categories.

On a side-note, defining arbitrary categories is never trivial. How spherical is spherical? The boundary has to be set somewhere, but there always will be edge cases where an object could reasonably fall in either category.


Why does Saturn have both moons and rings?

You're right that density is the important thing here. The Roche limit is the distance from the main body $d$ such that
$$d=1.26R_M\left(\frac{\rho_M}{\rho_m}\right)^{\frac{1}{3}}$$
where $_M$ denotes the main body and $_m$ denotes the satellite.

As you can see from the chart on the Wikipedia page, Pandora and Prometheus are both at least one and a half times the Roche limit from Saturn. Therefore, they're in no danger of being ripped apart any time soon.


Why is twilight longer in summer than winter and shortest at the equinox

See this diagram,

Now, the slant appears to be the same for June and December Solstices, doesn't it? First thing, it is a wrong conception that Sun travels more "perpendicular" to horizon in the summers. It just reaches a higher point as you can see, not that the slant is more.

Where does the disparity come in?

Civil Sunrise begins when the Sun is 6° below the horizon, Nautical Sunrise begins when it is 12° below the horizon and 18° for Astronomical Sunrise.

So, the Civil/Astronomical Disparity is due to the extra 12° of altitude that the Sun has to cover. Now, the diagram I have posted is better to explain the sunset, so let it be that way. It is just the same for our purpose.

The motion of the sun is uniform along the circles that you see. But these angles which decide our sunrise are measured perpendicular to the horizon, from the zenith. Are you aware of what happens when you project circular motion along a straight line?

That's right. You get a sinusoidal curve. The special thing about this curve is that it changes fast at some points and slow at others.
Transit is the position when the Sun is highest in the sky. In the curve for equinox (as in Fig 1), if you assign 0° to that point, the Sun is at 90° at sunset and -90° at sunrise. Both positions give you the peak of a sine curve and the change is lowest at that point. That might explain why you get the least difference at Equinox.

Now, if you consider the curves for June/December solstice (as in Fig 1), the position for sunset/sunrise is a little off 90°, hence rate of change is faster there as compared to the equinox.

That should explain why the disparity is larger at solstices as compared to equinox. But what about the difference in June and Decemeber solstice?

Now, my dear, you are entering dangerous territories. Let me remind you that the line/plane along which we are making our degree measurements is perpendicular to horizon. But the plane of circular motion is still at an angle to it. So, the sinusoidal curve has to go through another set of projections before you get the exact thing.
On Equator, the disparity between the solstices vanish, yes, as the aforementioned two planes align.

Anyway, I was hoping to give you a qualitative description just so you get the gist of it. Here are the equations if you need a mathematical idea

$$\cos a = \sin \delta \sin \phi + \cos \delta \cos \phi \cos H$$,

$a$ is (90+6)° for civil sunrise/set, and (90+18)° for astronomical sunrise/set,

$\delta$ is declination of Sun on that date. Google it.

$\phi$ is latitude of the place

To get the disparity, you need to find $H$ from the formula for two values of $a$ and get their difference. Remember to divide by 15 to convert from degrees to hours.

Here is a graph showing disparity along declination(-23.5° to +23.5°)


In regards to the holographic principle, what exactly is information?

The answer here is deceptively simple, and has very little to do with gravity directly. You only really need to know a bit of quantum mechanics, and the answer comes almost for free.

In any quantum mechanical theory, we have a Hilbert space $\mathcal H$. (Actually, we really want a rigged Hilbert space, but this distinction isn't particularly relevant here.) This can be as simple as the space of a single q-bit, or it can be as complicated as you need it to be (e.g. the Fock spaces which arise in quantum field theory). The Hilbert space describes all the possible configurations of any system described by this theory; an individual vector is a specific configuration. We also have a linear operator $H$ on $\mathcal H$, called the Hamiltonian, which describes the dynamics of such a system via the Schrödinger equation.

The "information" is just knowing what state $| \psi \rangle \in \mathcal H$ your system is in. This state vector tells you the result of every possible measurement, and thus contains all the information of your system. If you really wanted to, it's possible to express any such state uniquely as a sequence of zeros and ones, but that's a very classical way of thinking, and we're dealing with quantum information, which means that the fundamental objects aren't zeros and ones, but state vectors.

So, when we talk about holography, what we're really saying is that we can determine the state $| \psi \rangle$ that our system (e.g. the universe) is in simply from knowing the results of experiments we perform on the boundary of the spacetime (e.g. infinitely far away). The holographic principle alone doesn't say how this reconstruction works, only that it is possible.

Since this is a bit broad, it may be helpful to see how it works in the only really well-understood example, the AdS/CFT correspondence. In this case, we have a theory of quantum gravity  in $d+1$ dimensions with Hilbert space $\mathcal H$ and Hamiltonian $H$, constrained to have spatially-asymptotically AdS$_{d+1}$ metric (AdS is just a special, maximally symmetric solution to the vacuum Einstein field equations which has nice properties that make this work). It turns out that, at least morally speaking (there is ongoing work to understand the full extent of this), when we collect all the observables in this theory and shuffle them around in well-defined ways, we can construct out of them vectors in a different Hilbert space $\mathcal H'$. That is, we have a linear map $T: \mathcal H \rightarrow \mathcal H'$. This map is 1-to-1, meaning that we don't lose information. Strictly speaking, it doesn't need to be onto, but this distinction isn't crucial at first pass, so you can think of $T$ as a 1-to-1 correspondence (i.e. a linear isomorphism) if you like. In addition, the Hamiltonian $H$ can be mapped to $H'$, which describes compatible dynamics to $H$ on $\mathcal H'$ (in the sense that one can first evolve in the original theory and then map to the new one, or first map and then evolve, and the results will be the same). When we look at the pair $\mathcal H', H'$, we recognize this not as a $d+1$-dimensional theory of quantum gravity, but as a $d$-dimensional theory without gravity, but with extra symmetries that turn it into a so-called "conformal field theory" (which come from the extra symmetries of the AdS spacetime). In some sense, this new theory can be thought of as living on the boundary of AdS. So when we say that all the information is contained on the boundary, we mean that given a state $| \psi' \rangle \in \mathcal H'$ which describes the boundary theory observables, we can reconstruct the full state in the quantum gravity theory simply by $T^{-1}(|\psi'\rangle)$. What I've given here is just the 2-minute summary; AdS/CFT is still an active area of research and everything I said above is only approximately and/or morally true.

You might wonder why we need to rely on quantum mechanics. It turns out that (at least in the AdS/CFT correspondence), we can't do it classically. Highly quantum mechanical behavior in one theory is recovered by the classical limit of the other, and vice-versa. This is both a blessing and a curse, but at any rate there's no real classical equivalent. This is suspected to be true in any nontrivial example of the holographic principle, so we're pretty stuck describing things in terms of quantum information.


Will the SKA's total bandwidth still exceed the Earth's internet's bandwidth?

Will the SKA's total bandwidth still exceed the Earth's Internet's bandwidth?

No.

First some conversions:

1024 Gigabytes = 1 Terabyte
1024 Terabytes = 1 Petabyte
1024 Petabytes = 1 Exabyte
1024 Exabytes = 1 Zettabyte

From the SKA Telescope website: "The Operational Model" (.PDF), on the webpage: "Info Sheets", this explanation and graphic is offered:

"The data journey

Basic data products will be produced and stored in Cape Town for SKA1-mid and Perth for SKA1-low. From there, they will be delivered to a global alliance of SKA Regional Centres for further processing and archiving and access by the user community.

New data models will need to be developed as current software packages for radio astronomy data reduction don’t have the capabilities to handle the SKA’s large bandwidths and Field of View dataset.".

"Major dates

SKA1 OPERATIONAL CONCEPT DOCUMENT - Operational Principles

late 2027 - Commencement of Key Science Projects (KSPs)

late 2026 - Commencement of PI-led programmes

2024–2027 - Key Science Project (KSP) planning & proposals

2022 - Start of Observatory & Science commissioning

2020 - Start of construction activities

2017–2018 - Prototypes deployed at the telescope sites"

See also: "SKA Community Briefing – 18 January 2017", page 12:

Nothing will flow to /dev/null.
Source: Cisco - Global - IP Traffic: https://www.cisco.com/c/m/en_us/solutions/service-provider/vni-forecast-highlights.html#

Globally, IP traffic will grow 3-fold from 2017 to 2022, a compound annual growth rate of 26%.

Globally, IP traffic will reach 396.0 EB per month by 2022, up from 122.4 EB per month in 2017.

Global IP networks will carry 13.0 EB per day by 2022, up from 4.0 EB per day in 2017.

...
Internet Traffic

Globally, Internet traffic will grow 3.7-fold from 2017 to 2022, a compound annual growth rate of 30%.

Globally, busy hour Internet traffic will grow 4.8-fold from 2017 to 2022, a compound annual growth rate of 37%.

Globally, Internet traffic will reach 350.8 EB per month by 2022, up from 96.0 EB per month in 2017.

Global Internet traffic will be 11.5 EB per day by 2022, up from 3.2 EB per day in 2017.

Global Internet traffic by 2022 will be equivalent to 1 trillion DVDs per year, 88 billion DVDs per month, or 120 million DVDs per hour.

By 2022, the gigabyte equivalent of all movies ever made will cross the Internet every 1 minutes.

Global Internet traffic by 2022 will be equivalent to 193x the volume of the entire Global Internet in 2005.

...

No, it's not even close.


Luminosities and Apparent Brightness of Two Stars of which one is 8 times the distance of the first star

Welcome to Astronomy Stack Exchange!
Let the star which is farther away be A, and the fainter one be B.Since this perceived brightness is something called flux density (sometimes referred to as just flux), which the following formula holds its own for: $F = \frac{L}{4\pi r^2}$
($L$ is the luminosity, and $r$ is the distance) it would mean the luminosity ratio can be calculated in the following manner:
$F_A = F_B$
$\frac{L_A}{4\pi r_A^2} = \frac{L_B}{4\pi r_B^2}$Substituting for $r_A = 8r_B$:
$\frac{L_A}{64r_B^2} = \frac{L_B}{r_B^2}$
$L_A = 64L_B$
The farther star is much more luminous than the closer one.
P.S. This is only if there is no extinction, if it is present, I cannot make an estimate without more details.


There are different ways to model something. From what you're asking, there are two main types of modeling: forward modeling and inverse modeling.

Forward Modeling

In this type of modeling, you have a specific model that defines the "current" state of your system. In the case of exoplanet atmospheres, it'd likely be something that defines the molecular content, ionization level, density, etc. of your exoplanet atmosphere. Then, you use the known physics/math of your system to decide how it will behave. In this setup, what you've created is a system for predicting system states from a predetermined physics model.

Such an example would be someone creating their own atmosphere of an exoplanet in a model and then saying, okay what happens when I shine light through this atmosphere. What observations might I record?

Inverse Modeling

In some sense this is the opposite of forward modeling, albeit it doesn't really mean you're running a model to see into the past. Instead, what happens with this setup is you know a particular state or result, and you want to construct a model of your system which can produce said state. Essentially, you want your model to arrive at a certain state when it is done calculating. If it does, you have a reasonable confidence that your model was some indication of what your system is actually like.

In this situation, you'd measure components of the atmosphere, e.g. the radius of the planet as a function of wavelength, and then create a model of the atmosphere which can hopefully reproduce your observations. If you can, then the hope is that the model accurately represents what your system is.


A question about the distribution of mass among the planets in the Solar System

First, Pluto isn't a planet, and hasn't been for about a decade. It's a dwarf planet, and is better grouped with other small, rocky objects beyond Neptune's orbit, such as Haumea. These objects, and other minor planets, have more in common with objects in the Kuiper Belt and Oort Cloud than the eight planets of the Solar System.1

Therefore, we see a different pattern: Four relatively low-mass planets (Mercury, Venus, Earth and Mars) in the inner Solar System (the four terrestrial planets) and four relatively high-mass planets (Jupiter, Saturn, Uranus, and Neptune) farther out (the four giant planets). The question now is why the two groups are arranged as they are, why the terrestrial planets are close to the Sun while the giant planets are further away.

I wrote about this in my answer to Why do the gas giants in the Solar System have comparatively large orbits compared to the inner planets?. It boils down to something called the frost line, a rough boundary before which ices of water, ammonia, methane, and other compounds cannot form. Beyond it, they can form ice crystals (not just water ice, of course), which made it easier to accrete large amounts of material from the protoplanetary disk. This leads to the more massive gas giants.

In general, this explains the mass distribution of the planets, although there are exceptions (e.g. Hot Jupiters), which often result form planet-planet interactions, rather than forming at distances inconsistent with the frost line.

1 I exclude Planet Nine from this list, as its existence has yet to be verified.


A Black Hole (BH) is an object of General Relativity (GR), not of Newtonian physics, so the answer involves both.

First Newton: As another answer notes, the acceleration due to gravity depends on the mass of the body divided by the square of the distance from it.  At a given distance (say, ten million miles) the acceleration due to gravity is simply proportional to the central body's mass.  It doesn't matter if the mass is iron, water, stone or chunky-style peanut butter. It's all exactly the same as long as you're far enough away to be outside it all.

(A GR sidelight: This is even true of BHs as long as you're more than a few BH radii away from the BH.  The gravitational field of a BH becomes Newtonian fairly quickly as you move away from it and, further out, a BH and a equal mass of chunky-style peanut butter have the same gravitational field.  If an object -- any object -- collapses into a BH without losing or gaining mass, its gravitational field at a reasonable distance away remains the same.)

Still Newton: The surface gravity of an object depends on the size of the object.  If an object is denser, it will be smaller for the same mass and its surface gravity will be higher. This is due to the division by the square of the distance. For a denser object, the surface is closer to its center and the surface gravity is accordingly higher.

(This is why the Moon's surface gravity is 1/6 Earth's in spite of the Moon being only 1.2% of the Earth's mass. It's also 1/4 the diameter of the Earth, so at the Lunar surface you are 4 times closer to the center and thus its mass affects you 16 times as much.)

Black Holes are General Relativistic objects, and the differences from old-fashioned Newtonian objects show up when you get within a few radii of the BH.  The effects that close are subtle and complex and not much like the kinds of forces we're familiar with.  Basically (handwaving really fast) the geometry of space-time becomes more and more distorted as you approach the BH's surface and at the Event Horizon, the geometry is such that there is no longer a direction out -- all directions lead in towards the center. (This is what defines the Event Horizon, actually.)

So the main thing that differentiates an ordinary mass from a BH, is that in a BH the object's mass is so concentrated that time and geometry themselves are distorted to the point that here is no way out.

This is fundamentally different from the simple increase in the gravitational field's strength that Newton's theory predicts.


As @usethedeathstar said, MIA satellite most likely means 'Missing In Action'. The following lyrics from the song 'MIA satellite' seems to solidify this:

  MIA Satellite-
  You've got no path in sight.
  You leave me hanging in the dead of night-
  In the dead of night.

Regarding the ability to identify artificial satellites, it is possible to identify some satellites. Some identification guides:

Iridium

Iridium satellites are usually faster than most satellites, but their tell-tale sign is that they go from dim to bright to dim as they rotate. This bright flash is known as an Iridium Flare.

ISS

This moves East to West and is brighter than the average satellite.


what is the time mentioned after date (1962-07-03+02:00) the +02:00?

As indicated in the comments, the +02:00 is the ISO 8601 syntax for the time zone. Specifically, it indicates a time zone 2 hours east of UTC.

whois dbpedia.org indicates that the site is hosted in Berlin, Germany, which is consistent with the +02:00 time zone.

The fact that the data includes the date and time zone but not the time is probably an error. This particular datum 1962-07-03+02:00 is Tom Cruise's birthday; in fact he was born on that day in Syracuse, New York.

A data representation that includes a date and time zone but no time is peculiar, and I'm not sure that it conforms to the ISO 8601 standard. The description in the Wikipedia article seems to imply that a time zone can only follow a time. I haven't checked the standard itself.

A representation of just the date: 1962-07-03 would make more sense, and would be more useful. This is probably a bug in the software that runs dbpedia.org.


How to build a powerful home made telescope

You can buy remarkably cheap reflecting telescopes these days on a mount called the Dobsonian mount. They don't (usually) have fancy GoTo automated pointing and tracking, but give fantastic bang for buck as the majority of their cost is invested into the biggest primary mirror possible.

Building a telescope would be a great hobby and learning experience, and there are many books which will show you how. Just don't expect to save a significant sum of money over a commercially built scope - it could even be more expensive to build your own.

Fancy GoTo tracking mounts are not usually worth the expense and hassle (in setting them up) for a first-timer, though the Orion Intelliscope series is an intriguing compromise. For an extra $100 or so, the telescope has no guiding motors, but tracks where you push the telescope and a hand controller will tell you how much further you have to go till you reach your target.

My recommendation? Get the biggest Dobsonian mount scope you can afford. With a mirror greater than 10" things begin to get large and unwieldy, so 10 inches (mirror diameter, not tube length) is my recommended maximum size for a first telescope. Just go for the biggest aperture you can buy and easily carry. Don't invest much in GoTo systems, as although they can be quite useful they are also a pain to set up if you don't have the scope permanently mounted.


You ask if

  there some property that a black hole has that is immune to the star's
  power?

The short answer is "yes" -- the property of being a black hole. Assuming the collision is at a reasonably low relative velocity, the black hole would consume most of the mass of the star, so that eventually you would be left with a much bigger black hole and a certain amount of matter and radiation heading away from the site where the collision took place. The details of how long this would take, how much matter and energy would escape and so on will vary depending on the exact structure of the star and the details of the collision.

Essentially you can't disrupt a black hole much by throwing matter into it (if the matter you throw in is VERY dense, like another black hole or a neutron star there will be some temporary disturbance to the event horizon, but it will settle down in a fraction of a second), all it does is absorb the matter and get bigger. Even if the matter is super-hot dense fusing plasma from the centre of giant star. In the course of falling in, the matter gets distorted and heated (much hotter than the interior of even a large star) and sometimes whirled around the black hole, and this can appear as a huge explosion, but in the middle of it, the black hole is sitting there absorbing matter, largely unperturbed.


Calculate planet's surface temperature by distance from star

Neat and tidy calculations for this aren't possible.

The math behind effective temperature of a planet can be found here and that's a straight forward calculation where you can input distance, solar energy, planet's albedo and get an average temperature.   Earth's effective temperature calculation comes to -21 °C, which obviously isn't accurate, but that's as far as the calculation can take you.

Feedback mechanisms:

With essentially no change to distance from the sun, and only relatively small changes in tilt and eccentricity, Earth's temperature can fluctuate from a current 15 °C global average to about 5 °C as it goes in and out of ice ages.

Feedback mechanisms like albedo (ice cover is the key driver, but also deserts, surface area of the oceans), CO2 capture, which increases with colder oceans, can amplify small changes, leading to significant changes in surface temperature.   An orbital change, which should account for maybe 1 degree C, or an increase in CO2, which, by itself, captures less than 1 degree C, through feedback mechanisms, can lead to much larger temperature changes.

More on feedback mechamisms here.

Land placement and Ocean circulation

5 or 10 million years ago, Earth's average temperature was about 18-20 °C and that change was driven, primarily, by the closing of the Isthmus of Panama which affected ocean circulation.  Just the closing of the Isthmus made ice ages possible.  That's a pretty small change in the grand scheme of things, whether your planet has a near equatorial passage where oceans can flow or whether it doesn't, but that one thing can mean more than 10 degrees.

As Antarctica drifted over the south pole (the last 30-35 million years or so) glaciers formed and the earth cooled, because as ice forms, more sunlight is reflected into space and oceans levels drop.  Oceans having the lowest albedo and ice the highest.  Also, as Antarctica driffs away from the south pole over the next 20-30 million years, Earth is expected to warm, unless another change occurs, like the continued absorption into land and the oceans of CO2.

Earth can only enter an ice age when there's land near the poles.   The glacial period of 440-460 million years ago most of the land was over the south pole.  The sun was about 4% less luminous and the earth had much more CO2 (though CO2 levels did drop during that time, they didn't drop close to where they are now) and the orbit of the earth, 440 million years ago has some unknowns, but a large land mass over the south pole is thought to have played a key role in making that glaciation possible.  The tipping point of ice formation at the poles is particularly sensitive and a significant variable.

So, even with very similar to earth like planets.  Just where the land is and how well the oceans circulate and whether the planet can form glaciers and sea ice, or not, can fluctuate perhaps 20 degrees C in global average temperature.  How much CO2, how much active volcanism, how much reflective material in the upper atmosphere - all variables.

Even with identical earths, the feedback mechanisms make it difficult to predict what Earth's temperature would be if you pushed Earth 1 million or 1/2 million miles further from the sun.  The calculations for surface temperature wouldn't follow a simple mathematical formula.   Now, you'd still get colder as you moved the planet away, but one push might give you 1-2 degree and the next push might give you 5-10 and the one after that 0-1.  There would be no way to do a clean formula.


Both of these sound like satellites reflecting the Sun, even the first one which didn't fade.  There are so many satellites now with unusual orbits that being close to midnight in time doesn't mean you won't see them anymore like it used to.  Years ago, if the local time was midnight you could almost be assured no satellites would be visible but that's not the case now.
I think your second event was also a satellite, or more specifically a pair of satellites.  Since they vanished near your zenith at that time of the night, my guess would be they entered the Earth's shadow.  Polar orbiting satellites can catch the Sun at unusual times due to the geometry as can a multitude of the new sats being launched constantly.
I forgot to add the the first one you saw, somewhat near M5, would be in the right position for that one to have been a geostationary sat (perhaps tumbling) but it also could have been another type, as well.


How open is China in sharing data from the Chang'e 4 mission?

Different space agencies and different governments have different policies about archiving their data.  One commonality that I have found, though, is that their systems are often difficult to find, they are often difficult to navigate or search and get the data, and then just as difficult sometimes to process.  Sometimes, they also require an account.
Doing an internet search for simply "Chang'e 4 data archive" linked me to this conference abstract as the first link, which was presented at the mid-2019 Data Users' Workshop in Flagstaff, AZ.  The abstract explains the different data types (such as level 0A, level 1, level 2C, etc.) and the different instruments, and it also explains their data archiving plan.
Perhaps more important for this question, it provides a link to China's "Data Release and Information Service System" within China's Lunar Exploration Program.
It has a lovely English translation button in the upper-left, so clicking on that (because I do not speak nor read any type of Chinese), and the news to the right shows they have Chang'e 4's 12th data release as their most recent posting.  You can follow through the links and find the data archived there, or, this is a direct link to the search engine.
Of course, as with any mission, agency, or government, you have to assume that they are publishing everything, but I see no specific reason to doubt that in this case.  Keep in mind, though, that there is always a waiting period between the time the data are taken and the time they are posted on public archives.  With NASA, that is usually 6 months, though at the start of mission data returns it can be more like 12 months.  So, just because data might not go up through today in these archives, that does not mean that they won't in 6–12 months.