Can eternal inflation and gravity explain dark energy?

We need to cover some basic concepts: inflation in our universe ended 13 billion years ago. Inflation occurred for just a short period of time after the big bang. Eternal inflation is different. It hypothesises that spacetime is infinite and keeps on inflating. In some regions of spacetime, the inflation stops and the energy from this inflation is converted into energy and matter. This region where inflation stopped becomes sort of like a 'bubble' in the ever inflating spacetime. This bubble is then a universe. This there can be multiple universes accelerating to this hypothesis.
Okay. Our current theory of gravity is General Relativity, and for now all you need to know is that it seats that gratuity thanks time to propogate and has only a finite reach unlike Newton's description of gravity.
So even if our neighboring galaxies have a slightly different observable universe, the gravity from galaxies far apart don't even affecter them at all. In fact, that is why or universe is expansion by without any noticable obstruction from gravity.
Gravity takes time to propagate and has a limited access, so far away galaxies don't even affecter each other. That is why expansion is actually happening.
Another thing is I think your understanding of dark energy is somewhat misleading. You say that dark energy pushes the universe from the inside. That is the exclamation it the universe. The expansion does not happen because of dark energy. Dark energy accelerates the expansion. We have observed that expansion of the universe is getting faster and faster as time passes. For this you need energy. That energy is dark energy, failed so because we haven't detected it yet.


Do other G-type stars follow an 11-year cycle of activity?

The answer is yes. Early work showed that just over half of sun-like stars exhibit magnetic activity cycles similar to the Sun. These cycles can be monitored by looking at chromospheric activity, measured in the emission line cores of their Ca II H and K lines; by looking at spot modulation of their light curves; or for a few stars, looking at their coronal X-ray emission over time.

Subsequent work has found cycles in most stars where there is good data (Olah et al. 2016). The cycles vary in length from star-to-star, where they have been found. Some stars show multiple cycle periods (e.g.  Olah et al. 2009). There is obviously an observational bias towards finding shorter periods, but the Sun's cycle is towards the longer period end of what has been found. There seems to be a rough correlation between cycle period and rotational period - the cycle period is 100-300 times as long as the rotation period.

An excellent short review is given by Korhonen (2015).


Rudolf Minkowski.

Its not likely to be the mathematician, Hermann. Nor the physiologist Oskar.  There's only one Minkowski who was an observational astronomer.

The paper describing M1-92 was https://iopscience.iop.org/article/10.1086/125855

The 2006 paper on Minkowski's object (Croft et al) cites Minkowski, R. 1958, PASP, 70, 143


What's the distance between Venus and Jupiter?

The furthest distance Venus is from the Sun is 0.73 AU

The furthest distance Jupiter is from the Sun is 5.46 AU

The closest distance Jupiter is fro the Sun is 4.95 AU

So if we assume the orbits are concentric circles (they're not !) then :

Minimum distance = 4.95-0.73 = 4.22 AU
Maximum distance = 5.46+0.73 = 6.19 AU

And a simple average of these gets you an average distance of about 5.2 AU.

The problems with that simple estimate are :

The orbits are not circles, but ellipses
The orbits are at an inclination
A more detailed calculation for circular concentric orbits produces a tricky double integral with no easy solution
Over time (the only way to do a true average) the orbits would change from even "simple" ellipses.

A number of online sources quote the average distance as 4.48 AU.  This appears to be nothing more than the semi-major axis of Venus's orbit subtracted from the semi-major axis of Jupiter, which is probably an under-estimate of an average separation, as it neglects that they can be at opposite sides of the Sun sometimes and the same side at others.

Minor update :

For those interested in seeing the effect of a more detailed calculation I tried treating the orbits as concentric circles and doing the full integral on WolframAlpha.com .

The required integral in this case is (units are AU) :

$$L = \frac {5.2044}{4\pi^2} \int_0^{2\pi}\int_0^{2\pi} \left[ ( sin(v)-0.139 cos(u) )^2 + ( cos(v)-0.139 cos(v) )^2 \right]^{1/2}du\, dv$$

where $0.139$ is the ratio of Venus's orbital radius to Jupiter's and $5.2044$ AU is the mean radius of Jupiter's orbit.

And the result of this tedious integral $L=5.2293$ AU, which is not exactly much of an improvement on the much simpler $5.2$ AU given earlier.  But it at least dispels then notion on some online pages that it could be $4.48$ AU.

Last (?) minor update :

The above calculation just did a geometric average over the two orbit shapes, but made no attempt to consider the motion.  Well it occurred to me to revise that and allow for two bodies in a circular motion in the same plane.  Guess what ?  Same answer !

Well, almost.  In this case it's $5.2297$, which is an insignificant difference as we're already ignoring inclination and using circular orbits.  I'm inclined to put down the difference to a numerical calculation margin of error (the integrals do not have simple closed forms and would easily propagate errors, I think).

The integral boils to an average over the synodic period and is a simpler single integral which reduces to :

$$L = \sqrt{a^2+b^2} \frac 1 {2\pi} \int_0^{2\pi} \sqrt{1-2\frac {ab}{a^2+b^2} cos\theta}\, d\theta$$

I do prefer this result myself as it's taking account of orbital motion and over time, rather than purely geometric.


Asteroids hit the surface of the moon at speeds of about 25 km per sec, so they are mostly vaporised on impact. If a geologist inspected a crater on foot and looked hard enough, he probably would find a few relatively small fragments, particularly if the meteorite was of the iron/nickel variety. A few such fragments were recovered from the Barringer Crater in the United States. It is unlikely an impact would leave no residue at all. but the residue might be difficult to identify. When trying to visualise the impact of, say, a 100-ton meteorite, it is helpful to imagine that it turned into 100 tons of high explosive on impact.

Yes, craters can form for other reasons. Most craters on the moon are impact craters, but a small number are volcanic. Underground collapse sometimes occurs in the vicinity of volcanoes; the resulting holes are usually elongated, and are called graben. Lava tubes sometimes collapse too, leaving a  long, winding  valley like a river bed. These indentations are called rills. Objects left on the moon tend to survive for centuries in good condition, because there is no weather worth talking about, just solar wind (charged particles), cosmic rays and electromagnetic radiation.


Can an Earth-like planet survive if our Sun went Supernova?

One way of estimating this is to look at how much energy could be received by the planet. At distance $d$ it takes up $r^2/4 d^2$ of the sky as seen by the supernova. So for a $10^{44}$ J blast that is about  $4.5444\times 10^{34}$ J.

The gravitational binding energy of Earth is about $2\times 10^{32}$ J. So we have about 227 times as much energy as is needed to separate all pieces of the planet to infinity. It is also a few thousand times the energy needed to heat up an Earth-mass of iron from 0K to vaporisation. So, yes, it looks like it could well obliterate the planet.

Whether it actually does so is a complex question of how protected the evaporating planet is by its plasma sheath during the explosion. Given that actual terrestrial exoplanets in close orbits at 2000K may lose an Earth mass per gigayear it looks plausible that at least slower scenarios can vaporise planets.


Companion Sun the cause of comet impacts with Earth

Ahh, I see you've met our friend, the hypothetical star known as Nemesis. I don't know which scientist(s) you're referring to, but the idea goes back quite a while. The original theory was created by Raup and Sepkoski way back in 1984. Their paper analyzed mass extinctions in the past and concluded that there was a pattern. Note, though, that the paper does not specifically say that a star orbiting the Sun is the cause; it merely favors general impacts of "extraterrestrial" origin.

However, two more teams analyzed the data and posited that there was a celestial body responsible. Whitmire and Jackson and Davis, Hut, and Muller (Regrettably, you must pay to see the papers, which I have not done) suggested that the Sun had a companion in the far reaches of the solar system, and it was perturbing comets and other bodies, which came towards Earth.

Honestly, I (and quite a few others) don't think there's much truth to the idea. Why? Because we haven't found any evidence of its existence. As Wikipedia explains,

  Searches for Nemesis in the infrared are important because cooler stars shine in infrared light. The University of California's Leuschner Observatory failed to discover Nemesis by 1986. The Infrared Astronomical Satellite (IRAS) failed to discover Nemesis in the 1980s. The 2MASS astronomical survey, which ran from 1997 to 2001, failed to detect a star, or brown dwarf, in the Solar System.

and, later on,

  In particular, if Nemesis is a red dwarf star or a brown dwarf, the WISE mission (an infrared sky survey that covered most of our solar neighborhood in movement-verifying parallax measurements) was expected to be able to find it. WISE can detect 150 kelvin brown dwarfs out to 10 light-years. But the closer a brown dwarf is the easier it is to detect. Preliminary results of the WISE survey were released on April 14, 2011. On March 14, 2012, the entire catalog of the WISE mission was released. In 2014 WISE data ruled out a Saturn or larger-sized body in the Oort cloud out to ten thousand AU.

In fact,

  According to NASA, "recent scientific analysis no longer supports the idea that extinctions on Earth happen at regular, repeating intervals, and thus, the Nemesis hypothesis is no longer needed." And, indeed, a recent sky survey by NASA's WISE mission found no star or brown dwarf orbiting the Sun.

NASA's conclusions are laid out in this page, which talks about how WISE found no brown dwarf in the Oort Cloud . . . but found other brown dwarfs further out.


What is the current theory for the formation of the Earth's Moon?

The current accepted theory is known as the Giant Impact Hypothesis, where according to this NASA webpage "Origin of the Earth and Moon" (Taylor) a Mars sized object collided into the early Earth.

This theory allows explanations of (from the link above):

The chemical makeup:

  The giant impact hypothesis is consistent with our ideas for how planets were assembled and explains some important features of the Earth-Moon system, such as why the Moon has only a tiny metallic core.

and in terms of orbits:

  To account for the amount of angular momentum in the Earth-Moon system, Cameron estimated that the object would need to be about 10% the mass of Earth, about the size of Mars. (Angular momentum is the measure of motion of objects in curved paths, including both rotation and orbital motion. For the Earth and Moon this means the spin of each planet plus the orbital motion of the Moon around the Earth.)

Image source

However, recently, there has been some revisions to this theory.

According to NASA's page "NASA Lunar Scientists Develop New Theory on Earth and Moon Formation", based on concerns that the Mars-sized would likely to have had a different composition to the Earth (inconsistent with the current similarities in geochemistry).

The new hypothesis is (from the link above):

  After colliding, the two similar-sized bodies then re-collided, forming an early Earth surrounded by a disk of material that combined to form the moon. The re-collision and subsequent merger left the two bodies with the similar chemical compositions seen today.

This is also discussed in "Huge Moon-Forming Collision Theory Gets New Spin" (Wall, 2012), basing the revised hypothesis on the rotation rate, which is theorised to have been very fast, from the article:

  Earth's day had been just two to three hours long at the time of the impact, Cuk and Stewart calculate, the planet could well have thrown off enough material to form the moon (which is 1.2 percent as massive as Earth).

Further discussion, based on the geochemical makeup is discussed in "Geochemical Constraints on the
Origin of the Earth and Moon" (Jones and Palme) conclude that:

  Although none of these observations actually disproves
  the giant impact hypothesis, we find it disquieting that the
  obvious consequences expected of a giant impact are not
  observed to be fulfilled.

So, there is still some question as to precisely how the Moon was formed.


You seriously cannot expect the sun to have a layer that would contain some of its harmful radiation. Sun consists of a plasma and is not solid. Learn more about its structure here. Magnetic field is a kind of layer you might think of apart from all gases and energy it radiates. But that is harmful and not protective for us. It is highly unlike earth which has a solid crust. Any layer around sun wouldn't actually protect us as it would have to contain a huge amount of energy and that would make it more unstable.

Earth just happens to be at the right distance from the sun that was conducive enough to harbor a suitable atmosphere with just the right combination of the right elements and compounds (especially oxygen for breathing, ozone for protection and nitrogen for maintaining atmospheric balance). This is why life has been possible here.
Earth also happens to have a magnetic field strong enough to protect us from the solar wind, as described in details here. It just diverts the harmful energy away from us. Containment of huge amount of energy is the major concern here and nobody is doing that.

So life has been possible only because the way earth has been created, well of course all energy comes from the sun. But the sun just provides energy and doesn't save us from itself.


Galaxies move though space independently of the orientation of their axis of rotation. That this is true can be appreciated from the fact that their direction through space is relative; that is, in the reference frame of an observer that is passing the galaxy in its plane, it is moving like a frisbee, whereas in the reference frame of an observer who is moving along its axis of rotation, it is moving like an American football:

Galaxies form by accreting gas which often lies in filaments of gas, and this is thought to have a tendency to align their spins with the filaments. Higher-mass galaxies grow by mergers along filaments, which should align their spins perpendicular to the filament. I'm unsure if it has actually been confirmed observationally, but it is seen in simulations (e.g. Krolewski et al. 2019). Thus, a merger between a massive and a low-mass galaxy will have a tendency to make them meet with their spins at right angles of each other. But this alignment is only in a broad, statistical sense, and mergers at all angles do indeed occur.


Can a dying, swelling, star be rejuvinated by enveloping orbiting gas giants?

Stars turn into Red Giants not because they're running out of fuel, but because they're accumulating material they can't use for fusion (yet) in the core. The star isn't so much dying of starvation as it is wallowing in its own muck.

Red giants form when the fusion is no longer taking place in the centre of the star, but instead in a shell around the centre. The centre is filled up with the products of fusion, which themselves cannot fuse until temperatures and pressures get high enough. The exact reason why this results in red giant formation eludes me still, but you'll find plenty of questions asking about it on this SE.

Furthermore, most stars that are larger than red dwarfs never fully use up their fuel. There is often plenty of hydrogen in the outer reaches of the star, way more than you'll find in Jupiter (which, as zeta-band mentions, is only about 1/1000th the mass of a sun-mass star).


Whats the deal with black holes and "no information from inside the event horizon can leave"?

You are right in that, if you could send a message of any kind from inside a black hole then there is no magic required rebuild you from the information about you. If you could send even a single bit of information, then the black hole wouldn't be so black. You want to do this using the rotation of the black hole.

This doesn't work, for various reasons.

  Information has left the black hole; its mass, rotation, location, and a few other properties

The mass etc is a property of the black hole, not information travelling from inside the black hole. The reason is that you can't manipulate them in the way you describe, so they can't be used to send a message.

  I then enter the black hole and construct another station

Here is your first problem. From the frame of reference outside the black hole, there is infinite time dilation at the event horizon. Hence you never quite reach the event horizon, nor enter the black hole.

From the frame of reference of the infalling person, the event horizon is crossed. So let's somehow pretend that you are now inside the event horizon.

  Its purpose is to manipulate and sustain a small black hole "orbiting" the inside of the bigger black hole.

You can't orbit, you can only fall. The orbital speed is higher than the speed of light, so you will reach the singularity, as surely as you will reach next Tuesday. Your time has now been infinitely dilated, so events that take happen in finite time for you, take an infinite amount of time in the frame of the outside observer, so that makes sending a message using variation in rotation impossible.

Finally, from inside the black hole you can't change the total mass or rotation. The angular momentum of a closed system is conserved, so if you use your thrusters to change the direction of the small black hole you are carrying, the total angular momentum of the black hole remains constant. You can't change the total angular momentum with thrusters if you include both the angular momentum of the ship and the angular momentum of the gas ejected by the thrusters.

Similarly you can't change the total mass or location of the black hole from inside the black hole. And you can't do anything in a finite time from the perspective of an outside observer.

Thus no information can be encoded in the observed properties of the black hole.


I think your question covers a lot of ground cause you're asking it from 2 different perspectives.

  How were the orbits of planets first mapped out?

and

  How do we know what position in space a planet will be at certain time
  so that spacecraft trips to planets could be planned etc.

These should probably be separate questions, as they're quite different.

First one first, here's a bit of history:  Using parallax Hipparchus was able to get a pretty good estimate of the distance to the Moon about 200 BC.  The distance to the sun was much more difficult but Aristarchus was the first to come up with a method and others were able to improve upon it to get more accurate measurements.

The relative orbits of the planets wasn't hard to calculate once you had one of them, more on that here.

We can measure the length of a year on Mercury or Venus quite easily, as well as their relative angle to the sun (though Mercury's orbit is quite eccentric, you'd want to take several measurements).  Even the wacky Copernican model with just circles on top of other circles made pretty good distance calculations with the sun in the center and the planets orbiting the sun.

Kepler was the first to recognize that the planets orbited at slightly different orbital planes, and from what I've read, Kepler was prouder of that discovery than the discovery of the elliptical orbits that he's most often remembered for.  The combination of slightly different orbital planes and elliptical orbits and Brache's improved accuracy to his measurements was a huge step forward in planet mapping.  Kepler still didn't have the distance to the sun well plotted out, but he figured out the model for planetary orbits.   After that, it was just a matter of improving the distance measurements.

As to how NASA does it now.   I'm not an expert, so somebody else might be able to answer that one better.   It gets pretty mathematical, 3 position vectors, 3 direction vectors and velocity changes based on position, but we have nearly pinpoint laser measurements now.

  I'd also like to know the most primitive way that this could be done
  and if there is anyway this could be done using only maths and simple
  measuring tools assuming no previous knowledge about the planets
  positions.

It depends what you mean by primitive.    Galileo's telescope enabled him to see phases of Venus, and that isn't possible without a telescope.  The telescope also made it possible for Galileo to see 4 of Jupiter's moons, which orbited Jupiter and not Earth.

Aristarchus tried to eyeball the difference in shadow from the sun between the earth and the moon and he estimated the Sun was some 18-20 times more distant than the moon.  Details here.  To his credit, he successfully deduced that the Sun is much further than the moon, and he estimated the angle ratio to be 87 degrees when it was 89.83 degrees.  Despite his efforts, his estimate was off by a factor of about 20, in distance though he was only off by 2.83 degrees.

Brahe used the best equipment available to him at the time, much of it made specifically for his observations.  More on that here.  Brahe also had the advantage of having accurate time pieces.

Brahe believed in the Ptolemaic model but he was unable to get it to match his data and he urged Kepler to figure it out when he was old and knew he wasn't going to figure it out himself.  (There are rumors that Kepler poisoned Brahe to get more credit for himself, but I don't think that actually happened, Brahe's body was even exhumed to test that theory some time later but no traces of poison were found).

Kepler, unlike Brahe, liked the Copernican model but he couldn't get that to work because he believed in Copernican circles.   He tried ellipses, more out of desperation because he couldn't get circles to work and the ellipse model matched the data.   Kepler had not only Brahe's but years of Copernicus' records to work with as well as observations preceding them, some going all the way back to ancient Greece.

I'm not sure it would be possible to figure out that planets have elliptical orbits and how far they were from Earth without good equipment.  Kepler had a lot to work with and a determination to find a simple explanation.

Hope that's not too wordy.  I can try to clean it up a bit later if you like.


What type of star does theory predict should be the most massive?

There is no particular type. Yes Wolf Rayets were once more massive and have suffered extensive mass loss. Other examples of evolved massive stars are very luminous red supergiants and luminous blue variables (LBVs). The evolutionary tracks through these processes are somewhat poorly understood, as are the exact magnitudes and durations of the various mass loss phases.

However, they all started out as massive main sequence stars with very short hydrogen burning lifetimes. There are a number of examples of O3V main sequence stars in the Galaxy which probably have masses of around 100 Suns.

Hence the most massive stage of a massive star's life is (briefly) as a main sequence star of type early-O.


Calculate eclipse days and duration of satellite

I found a convenient formula to determine Eclipse (satellite in shadow). From Ismail, et al., Equation 10 here.

Image from same article:

So time in shadow = ((Theta-E)/360)*Orbital Period.

But how do you calculate Theta-E? Well, for my part, I used the law of similar triangles. You will have two similar isosceles triangles - one from the Earth to the point of the umbra cone and one from the Sun to the point of the umbra cone. Cut the isosceles triangles in half and use the right triangles. The sides of similar triangles are proportional and using the distance of 1 AU and the radii of the Sun and Earth, you can find the distance to the point of the umbra cone with a little algebra. Then use the arc-tangent to find the half-angle to the distance at GEO, then doubled it to find Theta-E.

You should be aware, however, that you are dealing with (1) circular orbits with (2) 0 inclination. The time of year also can change things.

My result was very close to that of James K. with only a few seconds difference.


Meteorites have distinct compositions and so meteorites with similar compositions can be grouped into families.  There are, for example the iron meteorites, meteorites with lots of carbon and so on.
One family are called HED meteorites.  They don't contain chondrules and show evidence of igneous processing (the rocks have been melted). The are differentiated.  This suggests that they didn't form from the raw materials in the solar system, but formed on a larger body, and then were ejected into space. The look very like igneous rocks on Earth.
It is suspected that these come from Vesta. They don't come from the Moon or Mars (we know about the composition of these worlds and they don't match). Vesta is big enough to be differentiated, and in the right place to allow for pieces of rock ejected by impacts to be perturbed into Earth-crossing orbits. Observations by the Dawn probe are consistent with Vesta being the source of HED meteorites (By contrast, Ceres has a different composition and so is not a likely source), and there is a large impact crater (the Rheasilvia crater) that could have ejected a huge amount of rock into orbit around the sun.  Some of the debris from this impact would still be falling to Earth, even after a billion years.
So Vesta fits the requirements.  That isn't quite proof that these HED meteorites did come from Vesta, but it is a reasonable belief.


Could the Earth use gravitational lensing / bending of light to see it's own bottom?

Whimsical question. Similarly, you could reason that someone on a planet far, far away would be holding up a mirror, such that we could see our own reflected past again.

Unfortunately, both options are practically infeasible, even if the light would be undistorted and redirected perfectly towards us. Remember that (1) Earth is small, (2) the reflected signal is weak as it is, (3) any object capable of reflecting this back towards us would be a black hole, the closest of which is further away than most known exoplanets, and the signal falls of with $r^{2}$ with $r$ twice (!) the distance to the object doing this lensing.

Now, if we somehow were able to observe it anyway, it would be straightforward enough to estimate the look back time - the distance to the lens would presumably be fairly well known.

How bright it would have to be is not really a sensible question - there are too many variables in play for that (which (hypothetical future) telescope are we using, what are the observing conditions, how far away is the source, what kind of source are we talking about...). And at the end of the day, the brightness of your source is not even the biggest problem; finding a suitable lens is.


When the moon eclipses the sun what direction is darkened first: left or right?

Relative to the Sun, stars, and Earth's shadow, the Moon always appears to move from west to east.
How this looks relative to the horizon depends on your latitude and time of day.
If a solar eclipse occurs in the middle of the day, the Moon appears to move from right to left if you're in a northern latitude, or from left to right if you're in a southern latitude.
If you're in a tropical latitude, the Moon appears to move downward if the eclipse occurs in the morning or upward if in the afternoon.


Total solar eclipse, supermoon, and spring equinox all happening at the same time: anything special about this?

There wouldn't be a direct correlation between the equinox and the moon's orbit, so there really isn't any special - it's just a coincidence.


Can planets have orbits other than elliptical or circular orbits?

Orbits are conic sections therefore can be either circular, elliptical, parabolic or hyperbolic.
Of these 4, only first two form a closed curve under 2 body hypothesis, while the later two extend to infinity.
If you talk about planet,  by definition it has to orbit a star which would require it to have a closed orbit hence circular or elliptical, with respect to the star. For any other kind of orbit the body will just fly away to infinity never to return back.  Infact comets are considered to be parabolic, but in theory they have highly elliptical orbits with their aphelion lying near the edge of the solar system.

But it is possible for a planet to have other kind of orbits if we  consider their motion from a different reference frame such as with respect to another planet.
So for an inertial frame of reference a planet will have a circular or elliptical orbit, even Pluto.


What are the possible solutions to the Red Supergiant problem?

Solutions to the Red Supergiant Problem can be either observational or physical, and to date, both types have been proposed. Recent data and improved computer modeling have both helped make the later stages of red supergiant evolution clearer, leading to possible solutions to the problem.

Walmswell & Eldridge (2012) suggested an observational solution, namely, that circumstellar dust resulting from mass loss via the strong stellar winds of a red supergiant could lead to strong extinction in some cases, making it hard or impossible to detect the resulting supernovae. Based on data collected on 18 supernovae, they found an upper limit for the mass of a Type IIP  supernova progenitor (at 90% confidence) to be around $27 M_{\odot}$, which would explain the dearth of detections of progenitors in the $16$-$30 M_{\odot}$ range.

The observational difficulties are moot if there is a physical reason behind the problem, i.e. if there is some mechanism preventing these red supergiants from undergoing Type IIP supernovae. Yoon & Cantiello (2010) suggested that red supergiant stellar winds could be increased to "superwinds" by envelope pulsations. This would lead to periods of larger mass loss rates than expected, followed by periods of smaller mass loss rates. This would correspond to oscillating around states of instability due to the pulsations, and would be prominent only in red supergiants of masses greater than or equal to about $19M_{\odot}$. The result could be drastic, with stars losing well over half their masses. The star could then evolve toward a Type Ib or Type IIb supernova.

Smartt et al. did consider the possibility of envelope loss in the paper you cited, but dismissed it as being unfeasible for a star in the mass range. However, it appears that they did not consider the superwinds induced by pulsations, making this solution feasible once more. It's also worth mentioning that the authors suggested direct collapse to black holes with at the most a very faint supernova; this idea had been explored in the past.


Photosphere is relatively transparent. Is that right?

See the other answer by @RobJeffries for a more thorough explanation. Here I just address the the question of how something transparent can also be visible.
Transparent things can still emit bright light. Under some circumstances it can even approximate blackbody radiation. Think about red hot glass, and then imagine it is much bigger and the glass is now gas.

above: from Glowing Glass

above: from Wikipedia


Real-time position of a distant celestial body

The most robust, but generally not computationally feasible way would be a full N-body simulation of all the relevant interacting neighbours of the object in question.

On a more crude note, one could observe the eigen motion of the object (its velocity vector), its distance to us, and from that information (assuming the eigen motion is constant over the period between now and when the light was emitted from the object) one may infer the current position.

You can of course make a hybrid of the above methods: use the observed motion of the object, and use simulations to include the interaction with other objects.

Note that 'current position' is not unambiguously defined. Astronomers would often consider the observed position to be its current position, even though we are looking back in time. And, needless to say, on cosmological scales these concepts pose more problems.


The details are not known for certain, however it is generally thought to have a rocky core, formed of silicates (aluminium and magnesium silicate for example) and a large mantle of water ice, perhaps 50% by volume, 25% by mass. The surface crust is formed of various carbonaceous and iron minerals: siderite, dolomite, cronstedtite. There may be a small iron core.


Fastest recorded apparent motion of a comet or asteroid seen from Earth (degrees/day)?

This is more of a recent example than a record.
On 2019-07-25, asteroid 2019 OK passed about 65000 km from the Earth at a relative speed of 24.5 km/s.
The Minor Planet Center lists multiple
observations from sites in Italy and Armenia an hour before closest approach.
Using pairs of these (streak endpoints?) by the ISON-Castelgrande observatory, I compute motions of 29.8° per hour at 00:23 UT and 31.7° per hour at 00:26 UT.
The MPC ephemeris service
gives sky motions of 30.6°/hr and 32.5°/hr from site L28 at those times.
If someone had observed the asteroid from (25°S, 85°E) in the Indian Ocean at 01:22 UT, they would have seen it move 76° per hour.

Update: On 2020-08-16, tiny (H=29.8) asteroid 2020 QG passed by at 12.3 km/s, about 9320 km from the Earth's center or 2950 km from the surface.
A NASA news article said this was a new record for close approach distance.
An observer in the right place at 04:09 UT would have seen a maximum sky motion of
$$\mathrm{\frac{12.3~km/s}{2950~km} = 0.0042~radian/s = 860^\circ/hr}$$
From pairs of actual observations reported to MPC,
ZTF Palomar (I41) saw a sky motion of  566°/hr at 10:23 UT,
and ATLAS Mauna Loa (T08) recorded 439°/hr around 11:46 UT.


Many hot stars are born in multiple star systems because the cores of these stars tend to split (see Jeans instability). With lower mass stars this still can happen. However, there are other ways. For example, in a young star cluster, close encounters with other stars can cause a star to be captured by another one.


How much starlight do black holes absorb?

tldr; extremely little. Space is big, even biggest black holes are ridiculously tiny in physical size

The amount of starlight absorbed by black holes is really minuscule. A black hole is not going to be sucking light in: it can be fairly well approximated to be an absolutely black object around the size of its event horizon (or photon sphere, does not really matter for the purpose for this thought experiment, since photon sphere is 1.5x EH).

Now, let's think of occlusions by more familiar objects: the Moon and the Sun. Now let's think our Moon would be a black hole. Retaining its current size, it would need to have mass in range of about 500 solar masses - pretty big for a BH. It would be also orbiting really close to our Sun; so close, it would probably get eaten by the Moon-BH relatively soon. Also, we would be observing this system really close.

But even in this case, we would still see the Sun most of the time unobstructed. Obstruction means absorption of light, thus there's really not much of this happening.

The amount of absorption of light from a star can be calculated by figuring out the ratio of black hole surface area to the surface area of a sphere with the radius equal to distance between the star and the black hole. In most cases you will be using square kilometers for the BH area, and square light years for the sphere. Which is almost exactly the same as comparing single atom mass to grams.

For the second question, star shine does not really contribute much to the temperature of space, so CMB will dominate this until very very far in the future, when all the stars are already gone


Why does Titan's atmosphere not start to burn?

Titan is one of Saturn's moons. Titan has a dense atmosphere, at about 1.5 bars. It also seems to have lakes of liquid methane.

For a conventional combustion, you would need a good Methane-Oxygen mix. Every combustion is basically an oxidization. Apart from very energetic events, such as SL9 hitting Jupiter, you will need quite a bit of oxygen to get a nice, explosive fireball in Titan's atmosphere.

According to Wikipedia, you need two Oxygen molecules for every Methane molecule you want to burn.


How do we know the radius of the universe is 47 billion light years?

The observable universe's edge, by definition, is the farthest part of the Universe that light could have reached us from. Let's say light from the observable universe's edge has just reached us. The longest it could have possibly travelled for would be 13.8 billion years, since the Universe is only 13.8 billion years old.

As such, one might assume that the light came from 13.8 billion light years away, and hence the observable universe's radius should be 13.8 billion light years. The problem is that this overlooks a key phenomenon: the expansion of the universe.

The location from where the light had been emitted has grown increasingly distant from us. In other words, by the time the photon has reached us and we are able to see the object emitting it, the object has moved much farther than when it emitted the photon. This is why we are able to see objects of distances far beyond 13.8 billion light years.

So to work out the actual numbers, let's suppose an object emitted a photon at the beginning of the universe $t=0$, and it just reached us today ($t_0 = 13.8 \ Gyr$). Evidently, this object lies at the edge of the observable universe, so finding its distance allows us to find the observable universe's radius.

Let's consider a given time $t$ in the past. Suppose our photon travelled for a small interval of time $dt$; it follows that the distance it just travelled away from its source would be $dx = c \ dt$. Due to the expansion of the universe, that $dx$ will expand over time; compared to today, it was a factor $a(t) = \frac{dx(t)}{dx(t_0)}$ smaller (this is called the scale factor at time $t$). Hence, the distance it gained from its source grew to be $dx(t_0) = \frac{dx(t)}{a(t)} = \frac{c \ dt}{a(t)}$ by today.

At every time $t$, it increased by an amount $dx$. So if the photon reached us at time $t_0 = 13.8 Gyr$ after it was emitted, then the total distance it gained from its source is the sum of these changes:
$$r=\int_0^{t_0} dx(t_0) = \int_0^{t_0} \frac{c}{a(t)} dt \approx 46 \ Gly$$
Which is therefore the distance of the object from us.


Does the Earth ever cross the Moon's path with respect to the CMB?

Not within plus/minus a few hundred years at least! The Moon's orbital plane around the Earth (viewed in an inertial frame) rocks back and forth a bit (Lunar Nodal Precession with a period of about 18.6 years), but that's not going to help here.

You can play with the numbers in this program.

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader

degs = 180./np.pi

load = Loader('~/Documents/yournamehere/SkyData')
data = load('de405.bsp')
ts   = load.timescale()

timez = []
dayz  = np.arange(1, 31, 0.1)
dayz  = np.arange(1, 366, 0.1)

#for year in range(1700, 2200):
for year in range(2017, 2028):
#for year in range(2000, 2030):

    #time    = ts.utc(year, 7,  dayz)  # July
    time    = ts.utc(year, 1, dayz)   # January
    timez.append(time)

earth = data['earth']
moon  = data['moon']

RAdegs, Decdegs = [], []

for time in timez:

    RA, Dec, dist = earth.at(time).observe(moon).radec()
    RAdeg, Decdeg = degs*RA.radians, degs*Dec.radians

    # lazy way to block the wrapparound
    dRAdeg = RAdeg[1:] - RAdeg[:-1]
    dropit = np.abs(dRAdeg) > 100
    RAdeg[:-1][dropit]  = np.nan
    Decdeg[:-1][dropit] = np.nan

    RAdegs.append(RAdeg)
    Decdegs.append(Decdeg)

RAdegpts  = [157.5, 157.5+180]  # coordinates from the question
Decdegpts = [-24.0, +24.0]

if 1 == 1:
    plt.figure()
    for RAdeg, Decdeg in zip(RAdegs, Decdegs): #[::5]:  # plot every 5th year
        plt.plot(RAdeg, Decdeg)
    plt.scatter(RAdegpts, Decdegpts)
    plt.title("2017 throuth 2027 continuously", fontsize=20)
    plt.show()


Aligning the equinoxes to the cardinal points on a circular calendar

As Mike G said, the Sun's ecliptic longitude is ideal for this application.
I see from your profile that you're a coder, with some knowledge of Python. Here's a short Python script that prints solar ecliptic longitudes for each day for any given year. (It actually prints 367 days, I figured an extra day or two may be useful). The code mostly avoids features peculiar to Python, so it should be fairly easy to translate into other languages.
My code uses the algorithm that Mike G has mentioned from Wikipedia. (I also tried another algorithm from that page, but it's not sufficiently accurate).
"""
 Approx Sun ecliptic longitude

 Prints daily positions for a year.
 For dates between 1950 and 2050,
 it should be accurate to within
 0.01 degrees. For details, see
 https://en.wikipedia.org/wiki/Position_of_the_Sun#Ecliptic_coordinates

 Written by PM 2Ring 2020.12.6
"""

from math import sin, radians
from datetime import timedelta, date

# Approximate solar ecliptic longitude
# Uses day zero = Noon UTC, 1 Jan 2000
# so we subtract half a day
def eclon_approx(n):
    n -= 0.5
    L = 280.46 + 0.9856474 * n
    g = 357.528 + 0.9856003 * n
    gr = radians(g)
    el = L + 1.915 * sin(gr) + 0.020 * sin(2 * gr)
    return el % 360

oneday = timedelta(days=1)
day_zero = date(year=2000, month=1, day=1)

@interact
def main(year=2000):
    day = date(year=year, month=1, day=1)
    d = (day - day_zero).days
    for i in range(367):
        angle = eclon_approx(d)
        print('{:03d}, "{:%b-%d}", {:6.2f}'.format(i, day, angle))
        day += oneday
        d += 1

Here's the output for 2021:
000, "Jan-01", 280.79
001, "Jan-02", 281.80
002, "Jan-03", 282.82
003, "Jan-04", 283.84
004, "Jan-05", 284.86
005, "Jan-06", 285.88
006, "Jan-07", 286.90
007, "Jan-08", 287.92
008, "Jan-09", 288.94
009, "Jan-10", 289.96
010, "Jan-11", 290.98
011, "Jan-12", 292.00
012, "Jan-13", 293.02
013, "Jan-14", 294.03
014, "Jan-15", 295.05
015, "Jan-16", 296.07
016, "Jan-17", 297.09
017, "Jan-18", 298.11
018, "Jan-19", 299.13
019, "Jan-20", 300.14
020, "Jan-21", 301.16
021, "Jan-22", 302.18
022, "Jan-23", 303.20
023, "Jan-24", 304.21
024, "Jan-25", 305.23
025, "Jan-26", 306.25
026, "Jan-27", 307.26
027, "Jan-28", 308.28
028, "Jan-29", 309.30
029, "Jan-30", 310.31
030, "Jan-31", 311.33
031, "Feb-01", 312.34
032, "Feb-02", 313.36
033, "Feb-03", 314.37
034, "Feb-04", 315.39
035, "Feb-05", 316.40
036, "Feb-06", 317.41
037, "Feb-07", 318.43
038, "Feb-08", 319.44
039, "Feb-09", 320.45
040, "Feb-10", 321.47
041, "Feb-11", 322.48
042, "Feb-12", 323.49
043, "Feb-13", 324.50
044, "Feb-14", 325.51
045, "Feb-15", 326.52
046, "Feb-16", 327.53
047, "Feb-17", 328.54
048, "Feb-18", 329.55
049, "Feb-19", 330.56
050, "Feb-20", 331.57
051, "Feb-21", 332.58
052, "Feb-22", 333.58
053, "Feb-23", 334.59
054, "Feb-24", 335.60
055, "Feb-25", 336.60
056, "Feb-26", 337.61
057, "Feb-27", 338.61
058, "Feb-28", 339.62
059, "Mar-01", 340.62
060, "Mar-02", 341.63
061, "Mar-03", 342.63
062, "Mar-04", 343.63
063, "Mar-05", 344.63
064, "Mar-06", 345.64
065, "Mar-07", 346.64
066, "Mar-08", 347.64
067, "Mar-09", 348.64
068, "Mar-10", 349.64
069, "Mar-11", 350.64
070, "Mar-12", 351.64
071, "Mar-13", 352.63
072, "Mar-14", 353.63
073, "Mar-15", 354.63
074, "Mar-16", 355.62
075, "Mar-17", 356.62
076, "Mar-18", 357.62
077, "Mar-19", 358.61
078, "Mar-20", 359.60
079, "Mar-21",   0.60
080, "Mar-22",   1.59
081, "Mar-23",   2.58
082, "Mar-24",   3.58
083, "Mar-25",   4.57
084, "Mar-26",   5.56
085, "Mar-27",   6.55
086, "Mar-28",   7.54
087, "Mar-29",   8.53
088, "Mar-30",   9.51
089, "Mar-31",  10.50
090, "Apr-01",  11.49
091, "Apr-02",  12.48
092, "Apr-03",  13.46
093, "Apr-04",  14.45
094, "Apr-05",  15.43
095, "Apr-06",  16.42
096, "Apr-07",  17.40
097, "Apr-08",  18.39
098, "Apr-09",  19.37
099, "Apr-10",  20.35
100, "Apr-11",  21.33
101, "Apr-12",  22.31
102, "Apr-13",  23.29
103, "Apr-14",  24.27
104, "Apr-15",  25.25
105, "Apr-16",  26.23
106, "Apr-17",  27.21
107, "Apr-18",  28.19
108, "Apr-19",  29.17
109, "Apr-20",  30.14
110, "Apr-21",  31.12
111, "Apr-22",  32.10
112, "Apr-23",  33.07
113, "Apr-24",  34.05
114, "Apr-25",  35.02
115, "Apr-26",  35.99
116, "Apr-27",  36.97
117, "Apr-28",  37.94
118, "Apr-29",  38.91
119, "Apr-30",  39.88
120, "May-01",  40.85
121, "May-02",  41.82
122, "May-03",  42.79
123, "May-04",  43.76
124, "May-05",  44.73
125, "May-06",  45.70
126, "May-07",  46.67
127, "May-08",  47.64
128, "May-09",  48.61
129, "May-10",  49.57
130, "May-11",  50.54
131, "May-12",  51.50
132, "May-13",  52.47
133, "May-14",  53.44
134, "May-15",  54.40
135, "May-16",  55.36
136, "May-17",  56.33
137, "May-18",  57.29
138, "May-19",  58.25
139, "May-20",  59.22
140, "May-21",  60.18
141, "May-22",  61.14
142, "May-23",  62.10
143, "May-24",  63.06
144, "May-25",  64.03
145, "May-26",  64.99
146, "May-27",  65.95
147, "May-28",  66.91
148, "May-29",  67.87
149, "May-30",  68.82
150, "May-31",  69.78
151, "Jun-01",  70.74
152, "Jun-02",  71.70
153, "Jun-03",  72.66
154, "Jun-04",  73.62
155, "Jun-05",  74.57
156, "Jun-06",  75.53
157, "Jun-07",  76.49
158, "Jun-08",  77.45
159, "Jun-09",  78.40
160, "Jun-10",  79.36
161, "Jun-11",  80.31
162, "Jun-12",  81.27
163, "Jun-13",  82.23
164, "Jun-14",  83.18
165, "Jun-15",  84.14
166, "Jun-16",  85.09
167, "Jun-17",  86.05
168, "Jun-18",  87.00
169, "Jun-19",  87.96
170, "Jun-20",  88.91
171, "Jun-21",  89.86
172, "Jun-22",  90.82
173, "Jun-23",  91.77
174, "Jun-24",  92.73
175, "Jun-25",  93.68
176, "Jun-26",  94.63
177, "Jun-27",  95.59
178, "Jun-28",  96.54
179, "Jun-29",  97.50
180, "Jun-30",  98.45
181, "Jul-01",  99.40
182, "Jul-02", 100.36
183, "Jul-03", 101.31
184, "Jul-04", 102.26
185, "Jul-05", 103.22
186, "Jul-06", 104.17
187, "Jul-07", 105.12
188, "Jul-08", 106.08
189, "Jul-09", 107.03
190, "Jul-10", 107.98
191, "Jul-11", 108.94
192, "Jul-12", 109.89
193, "Jul-13", 110.84
194, "Jul-14", 111.80
195, "Jul-15", 112.75
196, "Jul-16", 113.71
197, "Jul-17", 114.66
198, "Jul-18", 115.61
199, "Jul-19", 116.57
200, "Jul-20", 117.52
201, "Jul-21", 118.48
202, "Jul-22", 119.43
203, "Jul-23", 120.39
204, "Jul-24", 121.34
205, "Jul-25", 122.30
206, "Jul-26", 123.25
207, "Jul-27", 124.21
208, "Jul-28", 125.16
209, "Jul-29", 126.12
210, "Jul-30", 127.07
211, "Jul-31", 128.03
212, "Aug-01", 128.99
213, "Aug-02", 129.94
214, "Aug-03", 130.90
215, "Aug-04", 131.86
216, "Aug-05", 132.81
217, "Aug-06", 133.77
218, "Aug-07", 134.73
219, "Aug-08", 135.69
220, "Aug-09", 136.65
221, "Aug-10", 137.61
222, "Aug-11", 138.57
223, "Aug-12", 139.52
224, "Aug-13", 140.48
225, "Aug-14", 141.44
226, "Aug-15", 142.41
227, "Aug-16", 143.37
228, "Aug-17", 144.33
229, "Aug-18", 145.29
230, "Aug-19", 146.25
231, "Aug-20", 147.21
232, "Aug-21", 148.18
233, "Aug-22", 149.14
234, "Aug-23", 150.10
235, "Aug-24", 151.07
236, "Aug-25", 152.03
237, "Aug-26", 152.99
238, "Aug-27", 153.96
239, "Aug-28", 154.93
240, "Aug-29", 155.89
241, "Aug-30", 156.86
242, "Aug-31", 157.82
243, "Sep-01", 158.79
244, "Sep-02", 159.76
245, "Sep-03", 160.73
246, "Sep-04", 161.70
247, "Sep-05", 162.67
248, "Sep-06", 163.63
249, "Sep-07", 164.60
250, "Sep-08", 165.58
251, "Sep-09", 166.55
252, "Sep-10", 167.52
253, "Sep-11", 168.49
254, "Sep-12", 169.46
255, "Sep-13", 170.44
256, "Sep-14", 171.41
257, "Sep-15", 172.38
258, "Sep-16", 173.36
259, "Sep-17", 174.33
260, "Sep-18", 175.31
261, "Sep-19", 176.28
262, "Sep-20", 177.26
263, "Sep-21", 178.24
264, "Sep-22", 179.22
265, "Sep-23", 180.19
266, "Sep-24", 181.17
267, "Sep-25", 182.15
268, "Sep-26", 183.13
269, "Sep-27", 184.11
270, "Sep-28", 185.09
271, "Sep-29", 186.08
272, "Sep-30", 187.06
273, "Oct-01", 188.04
274, "Oct-02", 189.02
275, "Oct-03", 190.01
276, "Oct-04", 190.99
277, "Oct-05", 191.98
278, "Oct-06", 192.96
279, "Oct-07", 193.95
280, "Oct-08", 194.94
281, "Oct-09", 195.92
282, "Oct-10", 196.91
283, "Oct-11", 197.90
284, "Oct-12", 198.89
285, "Oct-13", 199.88
286, "Oct-14", 200.87
287, "Oct-15", 201.86
288, "Oct-16", 202.85
289, "Oct-17", 203.84
290, "Oct-18", 204.83
291, "Oct-19", 205.83
292, "Oct-20", 206.82
293, "Oct-21", 207.81
294, "Oct-22", 208.81
295, "Oct-23", 209.80
296, "Oct-24", 210.80
297, "Oct-25", 211.80
298, "Oct-26", 212.79
299, "Oct-27", 213.79
300, "Oct-28", 214.79
301, "Oct-29", 215.79
302, "Oct-30", 216.79
303, "Oct-31", 217.79
304, "Nov-01", 218.79
305, "Nov-02", 219.79
306, "Nov-03", 220.79
307, "Nov-04", 221.79
308, "Nov-05", 222.79
309, "Nov-06", 223.79
310, "Nov-07", 224.80
311, "Nov-08", 225.80
312, "Nov-09", 226.80
313, "Nov-10", 227.81
314, "Nov-11", 228.81
315, "Nov-12", 229.82
316, "Nov-13", 230.83
317, "Nov-14", 231.83
318, "Nov-15", 232.84
319, "Nov-16", 233.85
320, "Nov-17", 234.85
321, "Nov-18", 235.86
322, "Nov-19", 236.87
323, "Nov-20", 237.88
324, "Nov-21", 238.89
325, "Nov-22", 239.90
326, "Nov-23", 240.91
327, "Nov-24", 241.92
328, "Nov-25", 242.93
329, "Nov-26", 243.94
330, "Nov-27", 244.96
331, "Nov-28", 245.97
332, "Nov-29", 246.98
333, "Nov-30", 247.99
334, "Dec-01", 249.01
335, "Dec-02", 250.02
336, "Dec-03", 251.03
337, "Dec-04", 252.05
338, "Dec-05", 253.06
339, "Dec-06", 254.08
340, "Dec-07", 255.09
341, "Dec-08", 256.11
342, "Dec-09", 257.12
343, "Dec-10", 258.14
344, "Dec-11", 259.16
345, "Dec-12", 260.17
346, "Dec-13", 261.19
347, "Dec-14", 262.21
348, "Dec-15", 263.22
349, "Dec-16", 264.24
350, "Dec-17", 265.26
351, "Dec-18", 266.28
352, "Dec-19", 267.29
353, "Dec-20", 268.31
354, "Dec-21", 269.33
355, "Dec-22", 270.35
356, "Dec-23", 271.37
357, "Dec-24", 272.39
358, "Dec-25", 273.40
359, "Dec-26", 274.42
360, "Dec-27", 275.44
361, "Dec-28", 276.46
362, "Dec-29", 277.48
363, "Dec-30", 278.50
364, "Dec-31", 279.52
365, "Jan-01", 280.54
366, "Jan-02", 281.56

That @interact decorator before the main function isn't standard Python, it's a feature of Sage. You can play with an interactive version of this script on a SageMath server here.


Can I leave a planet without achieving escape velocity?

The basic concept here, that you instead of relying on having a large enough velocity that the Earth can not pull you back in time just have a low and constant velocity, and keeps thrusting to counteract gravity instead, is not flawed by default, given your assumptions

  Assuming that my fuel supply could last long enough, and I wasn't
  concerned about travel time, could this method allow my rocket to
  "leave" without achieving escape velocity?

However escape velocity decreases by distance, so even though you are travelling at only 100kph, that is eventually going to be above escape velocity, as the escape velocity goes towards zero when the distance goes towards infinity. You can not really escape a gravitational field, so the only meaningful metric is if you have enough kinetic energy to not be pulled back. That is not possible by your constraint of not having escape velocity.

In a sense though, you can escape the Earth's gravitational field, using the common approximation of sphere of influence. It defines a region for Earth's gravitational influence, and ignores it outside of it. The border of the SOI can be reached without ever going at escape velocity.


What is "g-98" in plots of stellar spectra?

It looks like the floating "g-98" is just an extension of the header of the plot (made by IRAF), which has wrapped around the image for lack of room on the right-hand side. Looking at the N0 plot, we see that the header should read

NOAO/IRAF V2.10.4EXPORT perry@sparky.sao.arizona.edu Mon 06:04:51 03-Aug-98

which just contains information about the software, user and creation date behind the plot.

(Hat tip to Pela for confirming my guess!)


If I have understood correctly, stars form in big clouds of gas and dust that are pulled together by their gravity. And the stars are often ignited when something disturbs the cloud, such as a passing massive body, or a wave like the once which create spirals in galaxies.

You have the sequence backwards. Protostars ignite by themselves once their cores reach a critical temperature and pressure. The gravitational collapse that eventually results in protostars (and then a stars) oftentimes needs a little help from an external disturbance such as a passing star or a supernova shock.

  Are there any phenomena on earth that behave like these clouds, which could be used to illustrate and explain star formation?

Not really. The interstellar gas clouds that eventually collapse to form stars are very large, very massive, very cold, and very tenuous. The best way to understand them is to understand them directly rather than looking for a poor analogy. There's nothing on Earth like them.


How bright would the sun appear from the hypothetical Planet Nine proposed by Caltech?

Between $1/40,000$ and $1/8,000,000$ of the brightness as seen from Earth, depending on what the actual orbit would turn out to be, and where the planet is in its $15,000$ year orbit period.

Brightness drops as $\dfrac{1}{r^2}$ with distance from the light source. Earth is at $1~\textrm{AU}.$ The theoretical planet is at $200~\textrm{AU}$ when it's closest to the sun, and up to $2800~\textrm{AU}$ at the point furthest from the sun at the upper end of the estimated orbit.

So e.g. $1/200^2 = 1/40,000$ of the brightness (luminance) as seen from Earth.

For how a human would experience it, we can convert to exposure value as used in photography: The difference in exposure value (photographic 'stops') is $\log 2$ of the luminance ratio, so we would have $15$ to $23$ stops less light than on Earth.
Sunny noon on Earth is $15~\textrm{EV}.$

So the brightness at noon on the planet surface would be:

At $200~\textrm{AU},$ planet orbit is closest to the sun: About $0~\textrm{EV},$ roughly the same as a dimly lit interior
$400~\textrm{AU},$ lower bound on semi-major axis: $-2 ~\textrm{EV},$ similar to a landscape lit by the full moon
$1500~\textrm{AU},$ upper bound on semi-major axis: $-6~\textrm{EV},$ similar to landscape lit by a quarter moon
$2800~\textrm{AU},$ upper bound on aphelion (the point on orbit most distant from the sun): $-8~\textrm{EV}.$ This would be dark, but you would probably still see enough to avoid running into things.


What are the stars/constellations a beginner/enthusiast can easily identify?

In the Fall (early morning) and the Winter (early-late evening), Orion is a large an easily identifiable constellation.

In the Spring (early morning) and the Summer (early-late evening), Cygnus and Pegasus are two large and easily identifiable constellations.

Check out an application called Stellarium.  It's free and available for every current platform and has a large database.


How have we measured the radius of the Earth (old and new)?

This is a pretty broad question so I'll limit myself to describing one instance of how the radius of the Earth is measured. Specifically, I'll talk about how it's measured today, with current technology.

As technology has progressed, our ability to precisely measure facets of the Earth has drastically improved. At this point, we can measure the exact shape and size of the Earth to such precision that asking about the "radius of the Earth" is no longer meaningful. Primarily because we know the shape of the Earth to such precision that we know it is not a perfect sphere with a fixed radius.

The current standard used to describe the exact shape of the Earth is the World Geodetic System, the latest edition of which is referred to as WGS84. In effect, this is a coordinate system centered on the Earth that precisely describes the shape of the Earth. This is done by using sensitive satellite measurements to fit the spherical harmonics to the Earth's geoid. The current, most updated version of WGS84 has more than 4.6 million harmonic terms with an accuracy in "radius" at any given point of 10 km. That is, you could pick any precise location on the Earth and use WGS84 to calculate the radius of the Earth at that exact point to an accuracy of 10 km. This radius would of course vary as you moved about the Earth.

If you really want to break this down into a simple number, the WGS model, at the crudest level, considers the Earth to be an oblate spheroid with an equitorial radius of $R_{eq} = 6\:378\:137\:\mathrm{m}$ and a polar radius of $R_{p} = 6\:356\:752.3\:\mathrm{m}$.


Why is gravity only an attractive force?

Because mass is positive

To expand your quote concerning the gravitational force into an equation:

$$F_G = -\frac{Gm_1m_2}{r^2}$$

The force of gravity, $F_G$ is proportional to the product of the masses and inversely proportional to the distance, $r$, squared. Let's break this down and see what might cause $F_G$ to be positive.

In this equation, $r$ cannot be negative because it's a distance between two locations. Two locations cannot be a negative distance apart. And even if they somehow were, the squared would take care of that anyway.

$G$ is the universal constant and always positive. You might argue that it could possibly be negative, but that's not possible. $G$ actually doesn't really exist. It doesn't describe anything fundamental to the physics of the universe. $G$ is simply a bookkeeping constant that allows us to get the right answer for the force based on any choice of units for mass and distance. Technically, if one uses the "correct" units for mass and distance (e.g., the Planck units), then $G=1$ and effectively doesn't exist. Since $G$ is just a scaling factor that depends on the choice of units, it will only be a positive number.

That leaves us with the masses. These are the only things which could possibly be negative. Of course, to get a positive, repulsive force, one mass would have to be positive and the other negative. But what exactly is a negative mass? Mass is the metric which describes "how much" of something there is. How can you have less than nothing of something?

Why can mass not be negative?

If you want to look at this another way, you can show that if mass could be negative, you'd get nonsensical results! Assuming of course, all other aspects of physics were the same. Recall from Newton's second law that

$$F = ma$$

Let's say there are two blocks sitting on a table. One block has a mass $m_1>0$ which is positive and the other has a mass $m_2<0$ which is negative. Ignore all other forces on these two blocks for the moment.

I go up to $m_1$ and I apply a force to push this mass forward. The acceleration that is induced is: $a = F/m_1$. Necessarily, the direction in which $m_1$ moves is the same direction in which I'm pushing. That's all well and good.

Now I go over to $m_2$ and I apply the same force, attempting to push it forward on the table. The acceleration induced on $m_2$ will be: $a = -F/|m_2|$. Note I made $m_2$ positive and pulled out the negative sign. You can see that if my force is forward, the direction the mass moves will be backwards! But here's the problem, my hand is in the way because it's trying to push to mass. As the mass tries to move backwards into my hand, it will be applying a force back on my hand, which by Newton's third law, necessarily mean's my hand is applying more force on the block, which then applies more force on my hand, ... and suddenly infinite forces are being applied or equivalently, these objects are infinitely accelerating. This is described by the concept of Runaway Motion.

If this seems strange to you, that's because it is. If negative masses existed, we'd live in a very weird universe. Fortunately, we live in a universe where physics makes sense, mass is positive, and by extension gravity is always attractive.


Are planets more likely to have dense materials at their cores?

Essentially, a planet can be considered as a self-gravitating fluid. As was already known to Archimedes, under gravity low-density fluids are buoyant and rise, while high-density fluids sink. So, with some local exceptions, planets are always well structured from the densest material in the core to lower-density material further up.


How would an exoplanet be found from earth if our view of its star system is "from the top"?

First of all, it is not that you can always find the planet (if it is there).

Anyway, if this is the case, when we observe the system not totally face-on, we resort to radial velocity measurements. The principle is simple, the actual much less.
The orbiting of the planet around the star causes the star to "oscillate" around the center of mass.
Here you can see a good example:

However, when the system is exactly face-on, the Doppler shift is useless, but still the gravitational effect of the planet onto the star is useful and used.
Here is what happens in the latter case:

It is clear that the star undergoes a change in position.
Then usually, the radial measurements go along with astrometrical measurements, to identify the position of the star along the planet's orbit.

Of course, you need some conditions are verified to allow this measurement.
First of all, you planet must be massive enough to bring a sensible change in the position of the star. Then your instrument must be sensible enough to resolve the tiny effect originated by this change.
Possibly, also other conditions play a role, but I am not an expert (you can take a look at here).

For a general comprehension, also a wiki for Methods of detecting exoplanets.


Can a telescope ever increase the apparent luminance of an extended object?

Can there be no telescope design... that makes extended objects appear "brighter"... than with the naked eye, by somehow overcoming the limitation of the observer's pupil?

I've left out all of the parenthetical qualifications and I'll answer this, let's see if it gets at the heart of the question.
tl;dr: No, because etendue; the same reason that a wall doesn't get brighter when we walk towards it and why we can't go outside with a magnifying glass and concentrate blue sky.

I thought I'd leave the other answer in place for contrast and as a teaching moment, but the down voters had a go at it, so now you can't see it unless you have 10k reputation.
I'd claimed that a pair of 7x50 binoculars would make the object 49 times larger in solid angle but collect $(50/6)^2$ or about 69 times more light.

If my fully dark-adapted pupil is 6mm in diameter then the aperture is 8.3 times larger in diameter than my pupil, but the image is only 7 times larger. We square the ratio to get the ratio of surface brightnesses, so it will appear to be
$$\left( \frac{50/6}{7/1} \right)^2 \approx 1.42$$

However the OP pointed out in a comment that this would produce an exit pupil larger than the 6 mm entrance pupil of the eye.

Thanks for the answer. I'm not sure about the conclusion though. With 7x50 binoculars, the exit pupil is 7.14 mm in diameter, meaning not all of the incident light falls on the retina since the eye pupil is smaller. Specifically, the fraction of of light that does enter the eye is $(6/7.14)^2=0.705$, which is precisely the reciprocal of the factor 1.42 that you calculated. So in the absence of transmission losses I think the luminance remains the same as that seen with the naked eye.

That was an Aha! moment, nature is smart, or at least I'm not.
I replied:

omg I think that I have I failed to recognize something as fundamental as conservation of etendue. Now it looks like my answer is wrong. :-( ...

This failed for the same reason that a wall doesn't get brighter when we walk towards it and why we can't go outside with a magnifying glass and concentrate blue sky on a sheet of paper. In classical mechanics the analogy is conservation of phase space and Liouville's theorem


Is it possible to know if a planet is located in the Habitable Zone knowing it's properties?

The habitable zone is, roughly, defined as the region around a star in which a planet could have liquid surface water. So if the surface temperature of the planet is between 0 and 100*C then the planet is in the habitable zone. The converse, however, is not true.

The reason for this definition is that the sort of chemical reactions that are needed for life work best when there are complex molecules dissolved in some kind of solvent, and the best polar solvent is H2O.

If you know the distance a planet is from its star, the temperature can be estimated from the properties of the star and the atmosphere of the planet. Not every planet in the habitable zone actually has liquid water. Venus, Earth and Mars are all in the Habitable zone of the sun, since they could all have liquid water if they had the right atmosphere. But of course only Earth actually does have liquid water in any substantial amounts.

So just being in the habitable zone, doesn't make a planet certainly suitable for life, and even if water does exist on the surface, it does not mean that humans could inhabit the planet.


Why does the exoplanetary system TOI-178 challenges current theories of planet formation?

The original research paper is published is found on A&A and available via arxiv, too.
One of the mysteries to understand in planet formation is the distinction between terrestrial planets and gasous planets - there is no continuous distribution curve. That means that different processes must be involved. Understanding how the involved processes interact, what conditions are required to trigger the individual processes, and how they shape a final system is important for the overall understanding. And this is easiest, the more complete information you have about a system and the better you can deduce its past evolution from the current state. The laplace chaining of the MMR of the planets helps in this endevour.
Thus using this somewhat surprising stability of the system, we can tackle further issues with the formation of the individual planets itself and possibly also the planetesimal-disk interaction. In their summary they state:

In this respect, the different planets of the TOI-178 system lie on
both sides of the radius valley (Fulton et al. 2017). Therefore,
reconstructing the past orbital and atmospheric history of this planet
may provide clues regarding the origin of the valley.

The conjected 'decreasing density' of planets from center outward directly comes from the density of available solid material in the protoplanetary disk: further inward at higher temperatures only the heavier elements can condense. Barring any migration of planets and swapping radial order, this defines the radial density distribution. You see the very same density variation inside our own solar system. The equation of state of matter at high pressures in the planets centres somewhat slightly distorts this, but the general trend applies.


What is the "lost light" in this unusual Hubble Deep Sky image?

Let me see if I can explain the main aim and accomplishment of this work.

First off: the picture you're puzzling over is a "luminance RGB" image, in which the bright regions are represented by color (a sort of pseudo-true color using near-infrared images), with the second faintest parts represented with black and the very faintest parts with white. The latter aren't qite "garbage", as Hobbes suggests in their answer, but they are relatively speaking the noisiest parts of the image, so there's little real information be found there.

This paper (Borlaff et al.; see the link in Hobbes's answer) is about re-processing near-infrared HST images originally taken about ten years ago as part of the Ultra Deep Field. Previous processing of these images (e.g., Koekemoer et al. 2013 ["HUDF12"] and Illingworth et al. 2013 ["XDF"]) was focused on getting information about the smallest, faintest galaxies, which are mostly the really distant, high-redshift galaxies. Because of this, the crucial stage of sky subtraction had some biases: in particular, it tended to treat the faint outer regions of the large, nearer galaxies as part of the sky to be subtracted. This is actually fine for the analysis of the small, distant galaxies, but means that if you do want to analyze the outer regions (outer disks, faint stellar halos, remnants of merger structures, etc.) of the larger, closer galaxies, you have the problem that their outer regions are over-subtracted (hence the "missing light") and thus unmeasurable.

(The "sky" being subtracted is a combination of emission from certain atoms in the tenuous outer atmosphere above HST, sunlight scattered from dust grains in the inner solar system, and the so-called "extragalactic background" = the combined light from unresolved distant galaxies.)

The abstract mentions four improvements the new study implemented when they re-processed the HST images: "1) creation of new absolute sky flat fields, 2) extended persistence models, 3) dedicated sky background subtraction and 4) robust co-adding."

I would suggest that the third item is perhaps the most important: they implement a method which does not subtract off the faint outer regions of the larger galaxies, and thus the resulting images still have information about the outer parts of these galaxies.

The plot below (extracted from Figure 20 of the paper) illustrates the kind of improvement they were after. It shows the surface brightness (in the F105W near-infrared filter) of one of the largest galaxies (a giant elliptical -- I think it's the big, round, yellow galaxy in the lower middle of the color image) as a function of radius (measured in elliptical annuli). The red triangles were measured using the XDF-processed image, the blue squares used the HUDF12-processed image, and the black points use the newly re-processed image produced as part of this paper [ABYSS]. You can see that the XDF points drop off at a radius of about 55 kpc, the HUDF12 points fall off at around 90 kpc -- but the light from this galaxy can be traced out to 140 kpc in the ABYSS-reprocessed image. This also means you can potentially analyze structure in the outer part of the galaxy out to ~ 150 kpc, which wouldn't be possible with the previous versions of the image.

(I should point out that I'm friends with, and have co-authored papers with, a couple of the authors, so I may be a bit biased -- but I think this is really impressive work!)


How do I calculate the positions of objects in orbit?

You must solve the Kepler's equation:

$M = E - e \sin E$

(where $M$ is the mean anomaly, $e$ the eccentricity of the orbit, $E$ the eccentric anomaly).
As you can see, this is a transcendental function, so you will probably need to iterate through a calculator, in order to solve it.
Many of useful formulas to understand the parameters dependence are reported on the Wikipedia page as well.

Here is another page with some examples. Just use the formulas and try to play with numbers, at some point you will realize at least orders of magnitude of the binary system elements.


Can the Universe be spatially closed and expand forever?

Yes, if the Universe is matter-dominated and spatially closed, then a cosmological constant:

$\Lambda > \frac{4\pi G \rho}{c^2}$

Where $\rho$ is the critical energy density (of matter) of the Universe, will cause it to expand forever.

NB Where $\Lambda$ is equal to this value you get the (unstable) Einstein static Universe.


The calculations seem reasonable. The conclusion is wrong.

The Earths antumbra starts 1.3 Gm from Earth. At less than 1.3 Gm there is umbra, penumbra and non-shadow. The moon is 0.4 Gm from Earth.

$$0.4 < 1.3$$

Hence it is impossible for the moon to be in the Earth's antumbra.

However, the Moon's antumbra forms at almost exactly one lunar distance. We notice that the moon and the sun are almost exactly the same size in the sky. When the moon is further from the Earth, an annular eclipse can form, as a small part of the Earth will then be in the moon's antumbra.

I've made a scale model of the sun-earth-moon shadows, at https://ggbm.at/CbbvmBFq


The de Sitter spacetime is maximally symmetric, and so having a full set of Killing vector fields, it must have a static form produced by a timelike Killing vector field.
One way to get without worrying about coordinate transformations is to take the $M\to 0$ limit of the general spherically symmetric lambdavacuum solution, which is the the Schwarzschild–de Sitter spacetime:
$$\mathrm{d}s^2 = -\left(1-\frac{1}{3}\Lambda r^2\right)\mathrm{d}t^2 + \left(1-\frac{1}{3}\Lambda r^2\right)^{-1}\mathrm{d}r^2 + r^2\,\mathrm{d}\Omega^2\text{,}$$
where $\mathrm{d}\Omega^2 = \mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2$ is the usual unit $2$-sphere. Except for minor notational differences, this is also identical to the static de Sitter slicing presented here in terms of different slices of a hyperboloid in Minwkoski $\mathrm{E}^{1,4}$, with $\alpha\equiv\sqrt{3/\Lambda}$.

In these coordinates, the timelike Killing field is simply $\partial_t$, which corresponds to a family of static observers with time dilation relative to the static observer at the origin given by
$$\dot{t}\equiv\frac{\mathrm{d}t}{\mathrm{d}\tau} = \frac{1}{1-r^2/\alpha^2}\text{,}$$
which diverges as $r\to\alpha$.

For geodesics, the Killing field also generates a conserved specific energy $\epsilon = \left(1-r^2/\alpha^2\right)\dot{t}$, which makes solving for radial geodesics not difficult. What's interesting about them is they take an infinite amount of coordinate time $t$ to reach $r=\alpha$, but only a finite amount of proper time $\tau$, and they have no trouble at all continuing beyond it. Additionally, time dilation for static observers becomes infinite there. This behavior is analogous to the event horizon of a black hole, except that the location of the de Sitter horizon is dependent on the choice of origin.

  Can Hubble red shift be interpreted as time dilation?

Yes. If we wish, we can think of the redshift due to a particle (e.g., galaxy) at some radial coordinate as a combination of two things: the special-relativistic time dilation of a galaxy with respect to a local static observer and the gravitational time dilation between static observers given above. In fact, the $\epsilon$ parameter above is exactly the (per-mass) energy of the particle as measured by a local static observer, i.e., the relative Lorentz gamma.


Why do massive rapid-rotating stars move bluewards off the zero-age main sequence (ZAMS)?

The blueward evolution is a simple consequence of the homology relations for main sequence evolution.
If the star is rotating very fast, then this induces strong mixing of the interior, such that the star can be treated as chemically homogeneous. As main sequence hydrogen burning proceeds, the entire star is gradually turned into helium, with a consequent increase in the mean molecular weight $\mu$.
Therefore, the evolution can be thought of as a sequence of models with similar mass, but increasing $\mu$.
The luminosity evolution is
$$L \propto \mu^{7.5} M^{5.5} R^{-0.5}\ .$$
The strong $\mu$ dependence comes about through the strong dependence of opacity on $\mu$ for stars in radiative equilibrium.
On the other hand, the radius is only weakly dependent on $\mu$. That is because the central temperature depends on $\mu M/R$ (from hydrostatic equilibrium), but the strong temperature dependence of the CNO cycle, means the central temperature hardly increases as burning continues. As a result
$$ R \propto \mu^{2/3} M^{4/5}\ .$$
If we assume a blackbody, then $T_{\rm eff}\propto L^{1/4}R^{-1/2}$ and thus
$$ T_{\rm eff}\propto \mu^{1.54} M^{0.98} R^{-0.12}\ .$$
If we start off with 10% He by number, then $\mu=0.56$. But if we end with 80% He, then $\mu=1.2$.
On the face of it, this would appear to more than double the surface temperature during the main sequence evolution. The effect will be mitigated by mass loss and the disproportionate effect on opacity contributed by whatever metals are present (these are not zero metallicity stars).


Reference: First spectrographic observations by Fraunhofer

With further research I found an Answer in "The Cosmic Century" by Longair (Cambridge).
When Longair says

  In 1823, Fraunhofer made further observations of the spectra of the planets and the brightest stars, anticipating by about 40 years the next serious attempts to measure the spectra of the stars (Fraunhofer, 1823)

the article referenced is Kurzer Bericht von den Resultaten neuerer Versuche über die Gesetze des Lichtes, und die Theorie derselben (A short account of the results of recent experiments upone the laws of light and its theory.) which was actually fetched by my research in the NASA ADS (https://ui.adsabs.harvard.edu/abs/1823AnP....74..337F/abstract). But the title did not mention any planet or star and I could not access the full text, so I thought it was not the right reference.


How many celestial bodies can I see with a 90x telescope?

The key issue for a telescope is not magnification but light gathering capability - which is (crudely) the size of the main aperture. Thus a 90x magification on a very large (wide) telescope would let you see a very large number of things (if you are in an area where the sky is dark), but 90x on a small telescope would let you see a number of interesting things (the Moon, planets, some nebulae and star clusters) but not relatively faint objects.

Small, cheaper, telescopes are still worth buying if you want to dabble, seeing the Moon through a telescope is always amazing and a small telescope can show lots, for instance.

Also, for a small telescope don't go for big magnifications - 90x would be fine for a low cost 50mm refractor for instance.

As for where the planets are - Venus, Mars, Jupiter, Saturn are generally easy to spot if you look up their positions online and know your way round the constellations. Because the planets are not, as stars effectively are, points at infinite distance, but small disks they (on most nights) do not twinkle as the stars do.


A great explanation; all credit goes to: https://groups.google.com/forum/#!original/sci.physics.research/PZh0yawL2nA/YKbv-EpWJPsJ

  The explaination of a black hole in terms of escape velocity is not a
  good way to think of a black hole. The escape velocity is merely the
  intial speed at which a projectile must be fired from a large body in
  order so that, given that only gravitational forces act on it (which
  is a scenario very far from reality) it will never return to that
  body.

  However, anything that passes through the event horizon of a black
  hole will never be on the other side. Even if it were to constantly
  accelerate away from the black hole, it would still be going inward.
  This is not true of Newtonian gravity. In fact, in Newtonian gravity,
  there are no black holes, because there is no universal speed limit.

  The best way to think of an event horizon is where spacetime is so
  bent that all paths 'curve inwards', IE, the lightcones all point
  inward, so there is no chance of escape.

  BTW, in GR, it's not good to explain the effects of gravity as a
  FORCE. It is as much a force as the Centrifugal Force or the Coriolis
  Force. There is an amount of force needed to stay the same distance
  from an object, IE, to overcome gravity, however. And as you approach
  the black hole, the amount of force required approaches infinity. But
  to say that there is an infinite amount of force PULLING on someone is
  not very accurate.


Why are only the poles of Mars frozen, despite too cold temperatures on lower latitudes?

Just like a cold winters night on Earth does not guarantee snow, so does a cold Mars not guarantee ice everywhere.
In order for ice or snow to form, the moisture that is later solidifying has to be available at the cold location. Moisture in this case is $CO_2$ with traces of water, leading to an ice cap made of $CO_2$-ice, with a ring of $H_2O$-ice 'trailing' around its outer radius.
The atmosphere is in fact only cold and dense enough for $CO_2$ to solidify at the poles, so no mystery there. Otherwise the entire atmosphere would condense out into a thin layer of $CO_2$-ice. To understand the formation of the $H_2O$ ice, we have to take into account how the water is transported inside the martian atmosphere.
Mars possesses only one global circulation cell (see early modelling on this by Haberle et al., (1993), and a recent review on the martian climate for the interested reader by Read et al., (2015)).

((Figure from Haberle et al.,(2015))
The reason for the single cell is a combination of Mars' small size and rotation period, as outlined in this answer. The effect of the single cell is that the condensable moisture is only transported essentially pole-to-pole directly, with only small variations of ~10-15 degrees latitude during the martian year.
Hence, no other location on Mars than the poles can accumulate a significant enough vapour pressure in order to form ice.


Are exoplanets which are above escape velocity of particular interest?

The problem is that planet ejection is something that takes place over a very short timescale compared to the ages of the systems in question, so we're unlikely to see it actually take place. The main prospect of detection would be imaging of the ejected planets, for which you'd want a young giant planet as these retain enough heat from their formation to be sufficiently luminous in the infrared for detection even when not illuminated by a nearby star.

There has actually been a claim for the detection of a (proto)planet in the process of being ejected from a binary star system: the case of TMR-1C as reported by Terebey et al. (1998) "A Candidate Protoplanet in the Taurus Star-forming Region". Unfortunately the nature of the object is not confirmed, see Terebey et al. (2000) "The spectrum of TMR-1C is consistent with a background star". There has been some back-and-forth on this: Riaz & Martín (2011) "Large-amplitude photometric variability of the candidate protoplanet TMR-1C
" reject the background star hypothesis, while Riaz et al. (2013) "A near-infrared variability campaign of TMR-1: New light on the nature of the candidate protoplanet TMR-1C" regard it as a strong candidate for being a young stellar object (YSO).

Another relevant case might be the circumbinary planet HD 106906 (AB) b, which may have been saved from ejection by a stellar flyby, see De Rosa & Kalas (2019) "A Near-coplanar Stellar Flyby of the Planet Host Star HD 106906".

Ejected planets are of interest due to their contribution to the population of low-mass objects in star clusters and the field (this population may well contain both ejected planets and low-mass brown dwarfs), and the impacts on the orbital architecture of the surviving planets after episodes of planet-planet scattering. One example of the latter I found is Carrera et al. (2019) "Planet-planet scattering as the source of the highest eccentricity exoplanets", there are many other papers on this topic.


What does it mean that the Sun`s rotation axis is: RA 286.13° , Dekl: 63.87°?

You can think of the position given in your title as the coordinates of the star that is directly over the Sun's north pole. (Similar to how Polaris is almost directly over the Earth's north pole.) Likewise, the ecliptic plane has a north pole which can be located by right ascension and declination.

7.25° to 7.26° appears to be the correct calculation, based on the following:

North pole of sun: 286.13° RA, 63.87° declination
North pole of ecliptic: 270° RA, 66.5738° declination (per Wikipedia)

$$cos(angle)=sin(dec1)sin(dec2)+cos(dec1)cos(dec2)cos(RA1-RA2)$$
$$cos(angle)=sin(63.87)sin(66.5738)+cos(63.87)cos(66.5738)cos(286.13-270)$$
$$angle=7.26$$


Are binary neutron star mergers needed to explain the abundance of gold?

The creation of some very heavy neutron-rich elements, like gold and platinum, requires the rapid capture of neutrons. This will only occur in dense, explosive conditions where the density of free neutrons is large. For a long time, the competing theories and sites for the r-process have been inside core-collapse supernovae and during the merger of neutron stars.

My understanding is that it has become increasingly difficult for supernovae to produce (in theoretical models) sufficient r-process elements to match both the quantity and detailed abundance ratios of r-process elements in the solar system (see for example Wanajo et al. 2011; Arcones & Thielmann 2012). The conditions required, particularly a very neutron-rich environment in the neutrino-driven winds, are just not present without the fine tuning of parameters (see below).

Instead, the models invoking neutron star mergers are much more robust to theoretical uncertainties and successfully produce r-process elements. The question mark appears to be only over their frequency at various times in the evolution of a galaxy and exactly how much enriched material is ejected.

The announcement of GW170817 makes this all more plausible. A neutron star merger has been seen. The behaviour of the optical and infrared emission after the event matches the expectations of merging neutron star models (e.g. Pian et al. 2017; Tanvir et al. 2017). Of particular note is the developing  opacity and fading in the blue and  visible, with the spectrum
becoming dominated by the  infrared with broad spectral features. This is the expectation for an expanding cloud of material that is heavily polluted by the presence of lanthanides and other r-process elements (Chornock et al. 2017). The reasonable agreement between the observations and models suggests that indeed a large quantity of r-process elements were produced in this explosion.

To go from there to the claim that the origin of gold is solved (as claimed in the press conference) is a step too far. The amount of r-process material produced has large uncertainties and is model-dependent.
The rate of mergers is only constrained to about an order of magnitude in the local universe and is not measured/known in the early universe. What could be said is that this channel for r-process production has been directly observed and so must be taken into account.

On the other hand, r-process production by the supernova channel is not yet ruled out. Some simulations at least, that involve rotation and magnetic fields appear to be still "in the game" (e.g. Nishimura et al. 2016). It could be that the presence of significant r-process material in very old metal-poor stars requires a supernova channel, since the merger of neutron stars takes some considerable time to occur (e.g. Cescutti et al. 2015; Cote et al. 2017).

The overall picture is still uncertain. A review by Siegel (2019) concludes that the best fit to the available evidence is that some rare type(s) of core collapse supernovae (known as "collapsars") are still the best bet to explain the Milky Way r-process elements. The primary evidence for this is the presence of Europium (an r-process element) enhancements in some very old halo stars and the general trend of decreasing Eu/Fe with increasing Fe, suggesting a more alpha-element-like production site for the r-process - i.e. supernovae.


Might Oort cloud comets be exchanged between solar systems?

You can exclude the 'considering the distance' piece - of course Oort Cloud objects could transfer between different gravitational fields.

However what is it you think will make this transfer? Without some sort of gravitational impetus why would one of these objects leave the solar system? And if you do manage to slingshot one out of the solar system at a high enough speed to exit the Sun's gravity, remember that most directions end up very far away from any other solar systems.

Tl;Dr sure, but not very likely


Do bigger telescopes equal better results?

Magnification is not the main purpose of a telescope. A very small telescope will have a lower resolution than a larger one, but for larger ones, the atmosphere limits the amount of detail visible.

Large telescopes capture more light than small telescopes. More light means that dimmer objects are visible, and there is more light to be analysed. A very dim object, such as a planet orbiting another star, needs a very large mirror to collect enough light to analyse spectroscopically. And this is the reason why larger telescopes are made. Of course larger telescopes cost a lot, and cost is the reason that telescopes like the "Overwhelmingly large telescope" was not built.

Much of the development in telescopes in the last 25 years has been not in ever larger mirrors, but in adaptive optics, or using two telescopes together as an interferometric telescope.


How did Mars come to have a 24 hour 39 minute day?

"It's believed that the Earth was rotating about once every 5 hours
  before the theorized collision with a Mars sized coorbiting object
  referred to as Theia."

Almost. Theia did not have to be co-orbiting, just an intersecting orbit. We have no idea what the Earth's spin was before the collision, but it is theorized that the Earth rotation had a 5 hour period after the collision with Theia, at the time of the Moon's formation from the debris.

The fact that Mars and Earth have such a similar period is a coincidence, perhaps you are asking why Mars is spinning so fast? Well actually Mars is not the odd man out, Mercury and Venus are. Most planets spin fast. exactly which spin orientation is somewhat arbitrarily determined by the vagaries of the ways the planetesimals collided to form them. The fact that Venus and Uranus have unusual spin orientations is just the way things turned out.

Both Mercury and Venus used to spin much faster. Mercury's spin was tidally slowed down by the Sun and Mercury's orbit was (and still is being) driven further away by the Sun (just like the Moon and Earth: Why is the Moon receding from the Earth due to tides? Is this typical for other moons?). Eventually Mercury was held in that 2:3 resonance. Which, by the way had a certain amount of luck involved (see: Mercury’s capture into the 3:2 spin-orbit resonance as a result of its chaotic dynamics ). Venus, we are not so sure of.

The tidal force from the sun is much much less for Venus than for Mercury, but much more than for Earth. However Venus has a dense hot massive atmosphere, which can be forced into both gravitational bi-modal (two peaks) tides and thermal uni-modal (one peak) tides. The bulge lags behind the tidal forcing peak, which creates a torque by the sun to slow it down. This is fiendishly complex (See: Long term evolution of the spin of Venus - I. Theory )

P.S. Actually Phobos, and probably Deimos, are thought to be constructed fairly recently (millions of years) from debris from a collision of Mars with a large asteroid. There is no way to capture a whole asteroid into orbits that close.


Looking for planetary position calculation

If you're only interested in the positions, you'll probably be best off just using a program instead of diving into celestial mechanics. Depending on what you exactly want with "planetary position", here are three different approaches to finding a planet's position at a given time:

Stellarium is a free program to visually solve for the position of a planet, and it gives some precise data for the positions of the planets too, but it needs to be installed and run on your computer so it won't be the quickest way to find the position.
If you're looking to get a planet's coordinates on ecliptic and/or equatorial planes, you can let this web app do the calculations for you for a given time.
Also, a quick Google search came up with this web app meant for working with zodiacs.


How much mass does an object in space need to keep a human on its surface?

Any object with mass (even you) has gravity. The mutual attractive force between two objects is given by the formula
$$ F = G \frac{M_1 M_2}{R_{12}^2}, $$
where the two mass are $M_1$ and $M_2$ and $R_{12}$ is the separation of their centres of mass.

So to answer your question we need to define some sort of parameter that specifies what you mean by "meaningful gravity".

One way of doing this would be to demand that the forces due to tides from the Sun are greater than the gravitational force holding you to the body in question. Even here though we need to make an assumption about how far away the object is from the Sun - let it be $r$. Now let your mass be $m$, the mass of the body be $M$, the radius of the body be $R$, and the mass of the Sun be $M_{\odot}$.

To stay attached to the object when you are at the "subsolar point" - i.e. the body rotates so that you are closest to the Sun, then we require
$$G \frac{M m}{R^2} > G \frac{m M_{\odot}}{(r-R)^2} - G\frac{m M_{\odot}}{r^2}$$

We can assume that $r \gg R$, so cancelling out $Gm$ and performing a binomial expansion on the middle term, keeping only the first two terms of the expansion:
$$\frac{M}{R^{2}} > \frac{M_{\odot}}{r^{2}}(1 + \frac{2R}{r}) - \frac{M_{\odot}}{r^{2}}$$
$$ \frac{M}{R^{2}} >  2M_{\odot}\frac{R}{r^3}$$
Thus this sets a limit on the density of the object in question
$$ \frac{3M}{4\pi R^3} = \rho > \frac{3}{2\pi} \frac{M_{\odot}}{r^3}$$

At $r=1\ au$, this means the density only needs to exceed $3 \times 10^{-4}$ kg/m$^3$, which will be satisfied by any solid body. NB: This limit would be a limit where you would actually be pulled from the surface by the Sun's tides.

A more stringent requirement might be to ensure that you can't jump off the surface. On Earth, an average person might be able to jump vertically about 50cm. Using the equations of uniform acceleration (SUVAT), we know $u^2 = 2g s$, where $g$ is the gravitational acceleration, $s$ is the height jumped and $u$ is the initial upward velocity. This tells us that you can jump upwards at about 3m/s. Assuming this is the same on any other body (its hard to say how well you could jump in much lower gravity) you could equate this to the escape velocity of the object, given by $v_{esc} = (2GM/R)^{1/2}$. Thus this gives a constraint of $M/R > u^2/2G$.

If we set a realistic density of $\rho = 5000$ kg/m$^{3}$ for an asteroid, we can replace $M$ with $4\pi R^{3} \rho/3$, to say that you would be able to jump off an asteroid if it were smaller than:
$$ R < u \left(\frac{3}{8\pi G \rho}\right)^{1/2}$$
For the numbers discussed this means $R< 1.8$ km and $M<1.2\times 10^{14}$ kg.

Many more details at
https://physics.stackexchange.com/questions/46318/is-there-a-small-enough-planet-or-asteroid-you-can-orbit-by-jumping

Incidentally, as you will have noticed, the result doesn't depend on your mass.