Why are gravitational perturbations stronger at larger semi-major axes?

"Why is it that the secondary feels more perturbations from the gravitational influence of the sun the further away it is from the primary?"

In brief, it is because the net perturbing acceleration on the secondary is just the (vector) difference between (a) the accelerative attraction towards the perturbing body experienced by the secondary, and
(b) the accelerative attraction towards the perturbing body experienced by the primary.

Thus, the closer the secondary is to the primary, the more nearly equal in size and direction are those two attractions towards the perturber, and the closer to zero is their vector difference.  Another result is that the the more similar are the changes in velocity in size and direction produced by the perturbing accelerations in the primary and the secondary, and the closer to zero is the resulting perturbation in their motions relative to each other.

This has been known for a long time as a consequence of Newton's 6th corollary to the laws of motion: "If bodies, anyhow moved among themselves, are urged in the direction of parallel lines by equal accelerative forces, they will all continue to move among themselves after the same manner as if they had been urged by no such forces."

Because there can be immense variety in the possible trajectories of the primary and secondary apart from the perturbations, any detailed illustrations can speedily develop massive and intricate trigonometrical expressions.

But in all cases, including that of the Kozai-Lidov effect, the scale of the effect, however intricate its form, depends on the size of the net perturbing accelerations.

A highly simplified configuration can at least show by example how a net perturbing force, affecting the relative motion of primary and secondary, is almost directly proportional to the first power of the distance between primary and secondary -- though it also depends of course on further factors due to changes in angular configuration.

The diagram below indicates some highly simplified configurations.

Suppose first that the primary (E) and secondary (M) are for one instant in line with the perturbing body (S), with M at M1 between E and S.  Let s stand for distance ES, and d for distance EM (with d << s).  Suppose also that the mass of E and M are negligibly small relative to the mass of S (though not negligible relative to each other).

With these approximations, the accelerative attractions of S on M and S on E are respectively $ k/(s-d)^2 $ and $ k/s^2 ,$ and the net perturbing acceleration on M is the difference $ ( k/(s-d)^2 - k/s^2 ) .$

Putting $ s(1-d/s) $ for (s-d), and using the binomial expansion of $ 1/(1-d/s)^2 $ , one sees that the terms in $ k/s^2 $ cancel, leaving the net perturbing force as $ k/s^2 * (2d/s) $ , plus terms in higher powers of d/s, i.e in $ k d^2/s^4 $ and so on.

Where d is very much smaller than s, the higher-power terms in d/s can be neglected, and then the net perturbing acceleration on M at M1 in the chosen example-configuration is closely approximated by  $ +2 k d / s^3 $ , away from E and towards S.

If instead the configuration has M at M2 so that E is in line between M and S, then the net perturbing acceleration on M clearly becomes $ ~ -2 k d / s^3 $ , i.e. away from E and away from S.

If instead M is at M3, with the line EM3 at right-angles to ES, and if also the angle ESM3 can be treated as small enough that its cosine can be approximated as 1 and its sine as d/s , then it is easily found that the net perturbing acceleration on M at M3 is approximately $ k d/s^3 $ towards E, again neglecting higher powers of d/s.

If M is at an intermediate position M4, and D represents the angle ESM4, it can be seen by using a little more trigonometry that the net perturbing acceleration on M at M4, under the assumptions already made, has a component parallel to line ES of approximately $ +2 k d cos D / s^3 $, and a component perpendicular to line ES (acting always towards line ES) of approximately $ k d sin D / s^3 $.

All the components are proportional to the E-M separation d, to the extent that the omitted series of terms in higher powers of d/s can be treated as negligible as has been done here.


500+ kps.  There's a pretty good answer over at https://space.stackexchange.com/questions/3948/escape-velocity-for-our-galaxy, but the question appears to fit here as well.

For a mass all located inside a circular orbit, escape velocity is a simple function of orbital velocity, but that ignores most of the mass outside that circular orbit (all of it if its distribution is spherically symmetric).  The Milky Way has a lot of mass outside the Sun's orbit, though.


Isn't it absurdly unlikely that our Moon would constantly be "facing" us?

The key concept is called Tidal locking. Earth's gravity forced the Moon to rotate in that way during the first few tens of millions of years after the Moon formed and then kept it that way.
The wikipedia page describes the mechanism:

Consider a pair of co-orbiting objects, A and B. The change in rotation rate necessary to tidally lock body B to the larger body A is caused by the torque applied by A's gravity on bulges it has induced on B by tidal forces.[5]

The gravitational force from object A upon B will vary with distance, being greatest at the nearest surface to A and least at the most distant. This creates a gravitational gradient across object B that will distort its equilibrium shape slightly. The body of object B will become elongated along the axis oriented toward A, and conversely, slightly reduced in dimension in directions orthogonal to this axis. The elongated distortions are known as tidal bulges. (For the solid Earth, these bulges can reach displacements of up to around 0.4 metres (1.3 ft).[6]) When B is not yet tidally locked, the bulges travel over its surface due to orbital motions, with one of the two "high" tidal bulges traveling close to the point where body A is overhead. For large astronomical bodies that are nearly spherical due to self-gravitation, the tidal distortion produces a slightly prolate spheroid, i.e. an axially symmetric ellipsoid that is elongated along its major axis. Smaller bodies also experience distortion, but this distortion is less regular.

The material of B exerts resistance to this periodic reshaping caused by the tidal force. In effect, some time is required to reshape B to the gravitational equilibrium shape, by which time the forming bulges have already been carried some distance away from the A–B axis by B's rotation. Seen from a vantage point in space, the points of maximum bulge extension are displaced from the axis oriented toward A. If B's rotation period is shorter than its orbital period, the bulges are carried forward of the axis oriented toward A in the direction of rotation, whereas if B's rotation period is longer, the bulges instead lag behind.

Because the bulges are now displaced from the A–B axis, A's gravitational pull on the mass in them exerts a torque on B. The torque on the A-facing bulge acts to bring B's rotation in line with its orbital period, whereas the "back" bulge, which faces away from A, acts in the opposite sense. However, the bulge on the A-facing side is closer to A than the back bulge by a distance of approximately B's diameter, and so experiences a slightly stronger gravitational force and torque. The net resulting torque from both bulges, then, is always in the direction that acts to synchronize B's rotation with its orbital period, leading eventually to tidal locking.

The same process is, very slowly, slowing the Earth's rotation to synchronize with the Moon, although I'm not sure that process will ever complete.


What is the fastest spinning rotation of a Neutron star?

As it turns out, the fastest spinning neutron star found yet is a pulsar 18000 light years away in the constelation of Sagittarius which scientist catalogued as PSR J1748-2446ad. Pulsars are neutron stars that rotate, are highly magnetic and emit a strong perpendicular beam of electromagnetic radiation.

This pulsar's speed is such that:

  At its equator it is spinning at approximately 24% of the speed of light, or over 70,000 km per second.

PSR J1748-2446ad rotates a little over 700 times a second, and scientists have this to say on the theoretical limits of the rotation speed of a pulsar (from here):

  Current theories of neutron star structure and evolution predict that pulsars would break apart if they spun at a rate of ~1500 rotations per second or more, and that at a rate of above about 1000 rotations per second they would lose energy by gravitational radiation faster than the accretion process would speed them up.


How can an amateur astronomer verify the position of near Earth objects?

Is there any publicly available, NEO-related database out there? Or is
  there a specific institution a hobby-astronomer can/should turn to to
  be able to learn more about individual NEOs?

Yes, there is NASA's Near Earth Object Program that catalogues all  detected NEO's and had advanced reporting and seearch capabilities (a bit overwhelming number of options actually).

There is also ESA's own (actually DLR's, the German space agency) database under the EARN (European Asteroid Research Node) project within the SSA-Programme (Space Situational Awareness) called The Near-Earth Asteroids Data Base:

  An on-line data-base of Physical and Dynamical properties NEOs, with
  the corresponding bibliographical references is available at the EARN
  site. This data-base, compiled and maintained at DLR by Gerhard Hahn,
  contains information on all known near-Earth objects
  (Atens-Apollo-Amor Asteroids), and is updated on a regular basis. It
  is also linked to, and reachable from the NEO Dynamics Site NEODyS at
  the University of Pisa, which provides extensive information on all
  dynamical aspects of NEOs, including close approach predictions and
  collision probabilities.

Honestly though, I find the NEO Program one from NASA a lot more convenient to use, even if a bit overwhelming. For extensive research though, I'd suggest referring to both of these databases, with the latter probably more conveniently used through the already linked to NEODyS Project.


After what time interval do the closest approaches of Mercury to the Earth repeat?

@Glorfindel's explanation is very clear. I wondered about how big of an effect "everything else" would have, but I couldn't figure out how to add an image to a comment. Here is a graph from 2018 to 2023 of the distance between Mercury and Earth. You can see every 3rd minimum is a bit deeper than the others just like Glorfindel said. The details of each local minimum change in a somewhat irregular way (about 10% of the closest distance variation), but the broad pattern fits the simple explanation well.

Here is python code to generate such a graph yourself for a different time period. It uses skyfield to access JPL's data on the planetary positions and matplotlib to graph it.
from skyfield.api import load
import numpy as np
import matplotlib.pyplot as plt

planets = load("de421.bsp")
ts = load.timescale()

times = ts.utc(2018,1,np.linspace(0,365*5,10000))

distances = (planets["Mercury"] - planets["Earth"]).at(times).distance()

plt.plot(times.utc_datetime(), distances.km);
plt.title("Distance from Mercury to Earth in km versus time");

plt.savefig("Mercury-Earth-2018-2023.png");
plt.close()


Could the universe have evolved differently?

Given chaos theory, things could have been different. Given the laws of physics, there are limits to how much difference there could have been.

To put it bluntly, a subatomic particle could have zigged instead of zagging, causing a fusion event that led to the formation of a star not to take place in one spot but another.
But it would not prevent stars from forming.


Obtaining TESS input catalog (TIC) IDs for Gaia `source_ids`?

You can query TIC by Gaia ID. Using the MAST Portal, MAST Catalogs, TESS Input v7, Advanced Search form, I specified:

GAIA ID: 4052922352453886976

and got a single result:

TIC ID (ID):            50559830
RA (ra):                275.945292 (18:23:46.870)
Dec (dec):              -25.672353 (-25:40:20.47)
TESS Mag. (Tmag):       13.579
Version (version):      20180610
UCAC ID (UCAC):         322-168973
2MASS ID (TWOMASS):     18234686-2540204
GAIA ID (GAIA):         4052922352453886976
GAIA G Mag. (GAIAmag):  14.5068

Since the MAST portal currently provides TICv7, the Gaia ID is probably from Gaia DR1.
The TIC release notes say Gaia DR2 data are new in TICv8.

  9  GAIA  The matched Gaia DR2 identifier has been updated in TICv8.
  Gaia IDs in TICv8 are included for stars that are found in the Gaia-provided Gaia-2MASS look-up table.
  These identifiers have been updated by the Gaia team from DR1 to DR2, and may not be consistent between TICv7 and TICv8.


Can concentration of gas in exoplanet atmosphere be found out from wavelength and absorbing radius from spectral data?

Sure, but it is not straightforward. There's a lot of degeneracies involved, such as many molecules sharing similar absorption bands, the presence of clouds, scattering, absorption happening at many different pressure ranges, solving the chemical equilibirum equations of the species and the temperature-profile modifying the overall spectrum as well.
There's a lot of "Forward" models available online, which "solve" for the atmosphere (Give you the theoretical spectrum) of a planet given for some initial conditions, such as metallicty, Carbon to oyxgen ratio, or concentration of molecules.  What you want in your case is to do the inverse, i.e. start from your data and go to the mixing ratios of the different species. You can do that with "Retrieval" models, which you should again find information about online.
Assuming that you are working with transit data (which is what I understood from your post), you can often assume a constant temperature for the entire atmosphere at the equilibrium temperature of the planet, since transmission spectrum are usually less affected by the temperature profile, as they act more like passive absorbers.


How do I prevent or reduce shake when observing with binoculars?

Don't hold the binoculars in your hands. Humans are made of meat. wobbly wobbly meat. There's apparently devices that let you mount binoculars on tripods - (this google search would be a start). Those and a tripod would probably be helpful in decreasing shake.

I suppose it would affect mobility a little but that's a tradeoff.


How does one determine the effective temperature of a star from its spectrum?

Temperature ($T_{eff}$) can be quite tricky to determine accurately as it interrelates to a number of other fundamental measurements.
Firstly, remember that the spectrum we observe from stars are pin-point, they give us the entire overall result and not a specific location or part of the star. We need to dissect the various parts to arrive at the fundamental parameters. We arrive at our results by iterating the values of the fundamental parameters until a model spectrum matches the true spectrum we observe. The issue is, like you say, the existence of a whole lot of uncertainties.
The first of these (although it doesn't have a large effect) is the Uncertainty Principle itself. This creates natural line broadening due to the emitted photon having a range of frequencies. The width of the line is determined by;
$$\Delta E \approx \frac{h}{T_{\text{decay}}}$$
where $\Delta E$ is the uncertainty in the energy,
$h$ is the Planck constant, and
$T_{\text{decay}}$ is the amount of time the electron stays in a high energy state before decaying.
Fundamental parameters
The rotation of the star causes a Doppler shift effect on the line spectra making it broaden. The faster the rotation, the broader (yet smaller) the line. Like the Uncertainty Principle, this is natural broadening as it doesn't impact the abundance of any particular element in the star.
Measuring the rotational velocity ($V_{\text{proj}}$) depends on both its axis of rotation and our line of sight to the star. Therefore, we use a combination of both velocity about the equator ($v_e$) and the star’s polar inclination ($i$) to determine the projected radial velocity;
$$V_{\text{proj}} = v_e \sin i$$
Temperature ($T_{eff}$)impacts wavelength in such a way that higher temperatures impart higher random motions on the atoms. When these photons collide with an atom, they can cause the atom to become ionised, i.e. lose an electron. Different energy levels (and therefore temperature) will create different abundances at the various ionisation stages of atoms.
The temperature of the stellar photosphere decreases as we move away from the core. Therefore the line profile represents a range of temperatures. The wings of the line arise from deeper, hotter gas which displays a larger range of wavelengths due to increased motion. The higher the temperature, the broader the wings of the line profile ([Robinson 2007, pg 58][1]).
Here you can see the effect of various temperature values on the synthetic spectral line of FE I 6593 A. Red: $T_{eff}$ = 4000K; Black: $T_{eff}$ = 5217K; Blue: $T_{eff}$ = 6000K;
$T_{eff}$ on spectral lines" />
Microturbulence ($v_{\text{mic}}$) is the non-thermal localised random motion of the stellar atmosphere. It works in a similar way to temperature - an increase in the motion of atoms creates a wider range of wavelengths observed and therefore broader line profiles.
In strong lines, saturation can occur when there aren’t any more photons to be absorbed. As microturbulence in these areas increases, it presents more opportunities for photons to be absorbed. This broadens the wings of the line profile increasing the overall strength of the line. We can use this fact to determine $v_{\text{mic}}$, by ensuring that the strength of the lines (equivalent width) have no correlation with their abundances.
Finally, surface gravity which is a function of the star’s mass and size:
$$\log g = \log M - 2 \log R + 4.437$$
with $M, R$ being in solar units and $g$ in cgs.
A star with a higher mass but smaller radius will invariably be denser and under greater pressure. By definition, denser gas has a higher number of atoms per unit of area (abundance), leading to stronger spectral lines.
A gas under pressure provides more opportunities for free electrons to recombine with ionised atoms. For a given temperature, ionisation is expected to decrease with an increase of surface gravity, in turn increasing the abundance of atoms in the neutral or low ionisation states.
The measuring of $T_{eff}$
As we've seen, there are a number of ways in which a spectrum of a star can be altered. The one you're interested in is temperature. As temperature is interrelated to all the other fundamental parameters, we need to treat them together as a whole and tease out the value of $T_{eff}$.
We begin with a synthetic spectrum and modify its properties iteratively until it matches the shape of the star’s spectrum. Adjustments of one parameter will invariably affect the others. The spectra will match when the temperature, surface gravity, and microturbulence values (amongst others) are correct. This is obviously very time consuming although programs exist to help.
Atmospheric properties can also be determined by other less time consuming means. Photometric colours can be used as a proxy for temperature, and absolute magnitudes for surface gravity. However, these determinations can suffer from inaccuracies due to interstellar extinction and are at best a close approximation.
[1] Robinson, K. 2007, Spectroscopy: The Key to the Stars (Springer)


Suggestions for a beginner-friendly telescope

Cheap telescopes are just fine
I won't recommend a specific telescope, but I can tell you the following: I live in the middle of Copenhagen, and with my random 15 year old 4" telescope, which at that time cost \$300-$400, I can see the Moon's craters and mountain ridges, Jupiter's belts and moons, Saturn's ring, Venus' phases, and some nebulae. My telescope is manual, i.e. has no motor to track the sky, but I'd guess that today for the same price, you could get a telescope that does have a motor and a small computer that can find celestial objects, which makes things much easier.
Stay home
In my experience, going outside the city doesn't really add much to the experience (for amateur telescopes). To really see dark skies, you need to be 50-100 km away from the nearest large city, I'd say. Maybe that's not far from you, but when buying a telescope, you should also take into account that larger telescopes are more difficult to transport and set up, so if it's too large, you might end up never pulling yourself together to take it out.
Stay inside
If you take your telescope outside, it takes some time before it has reached the same temperature as the air, and during this time, there can be quite a lot of turbulence, which makes the image flicker. If I quickly want to show my 7 yo something, I simply use it from inside my living room, looking through the window. Two minutes, and he's impressed (until something else catches his attention).
What you see is not spectacular
Also, you should be prepared that there's no way that you can see anything like the colorful pictures that you can find on the internet. You will be able to see that Jupiter and Mars have slightly red colors, but apart from that, most things will look black/white. The reason is that the eye needs much more light than can be collected through a telescope in order to see colors. The largest telescope that I've looked through with the eye was a 31" telescope, and even that didn't reveal much colors.
To see colors, you need to take a picture with a long exposure. Many telescopes have the possibility of attaching a camera to it, but for this you definitely need a telescope with a motor such that it moves along with the sky, or else everything will just leave long white trail across the image.
You should also be prepared that planets and the like will look quite small. You could see something like this (very approximately to scale):

Know what you see
The beauty of what you see in an amateur telescope, even if it's small and dim, lies in knowing what it is that you see. Knowing that some nebula is 1000 lightyears away, and that its ever so slightly reddish color is hydrogen being illuminated by young, massive stars makes it much more fantastic to see.
My telescope is of medium quality. What you get for an expensive one is probably more clear images, but the most important factor is the light-collecting area, i.e. a 5" will show you more than a 4". I'm sure users like Florin Andrei can tell you more here.


If Jupiter were to absorb all other planets in the solar system, would it reach degeneracy pressure?

No, hydrogen burning doesn't start until 82 $m_{\rm Jup}$ and deuterium burning starts at 13 $m_{\rm Jup}$, so evidently 1.4 $m_{\rm Jup}$ is not enough for this experiment.

Your misconception is that

  Jupiter's mass is about the maximum a planet can be before it starts to fuse hydrogen in its core and undergo a massive transformation.

which is simply not true.


If earth changed its orbit, what would happen to the moon?

Aside from the fact that it's 81 times easier to move the Moon than the Earth, so, in such a scenario, the Moon could be moved too with an extra 1.23% of time/money/energy, so, it really would make sense to move them both, not just move the Earth and risk losing the moon or worse, crashing into it.

But (lets credit the need to save 1.23% to a congressional budget committee) and just move the Earth, as you propose.   Would the Earth lose the Moon?  . . . It depends, but not necessarily.

The first, idea to get out of the way is that there's no gravitational string connecting the Earth and Moon, so if the Earth was accelerated fast enough (say, launching millions of very high velocity rockets off the same side of the earth at the same time) there's no attachment that would pull the Moon with the Earth.   This is one of those scenarios where the rubber sheet analogy works quite well.  Accelerate the Earth and the corresponding gravity well moves away from the Moon and the Moon moves corresponding to the decrease (or increase) in the slope of Earth's gravity well, so the Moon's velocity carries it relative to the new slope.  But that's the argument for a single acceleration.   For multiple gravity assists, you have to account for the moon being in different positions so the relative movement wouldn't all be in the same direction.

Asteroid Gravity Assist - how close to the Earth?

A passing asteroid would affect both the Moon and the Earth, but for optimal Earth movement, you'd want it pass within, say, a few thousand KM of Earth's surface, outside of the atmosphere but much closer to the Earth than the Moon, where it's gravitational tug on the Earth would be over 1,000 times the tug on the Moon at closest pass.  The entire path wouldn't be a 1000 to 1 ratio, but with a close enough pass we can effectively say the Asteroid is giving the Earth a much larger gravitational tug, where the tug on the Moon is somewhat negligible.  In effect, moving the Earth's orbit but not moving the moon's orbit nearly as much.

Asteroids passing that close would be inside the Roche limit and probably would begin to break apart, but that wouldn't change the gravity assist or be a problem for the Earth if the velocity was sufficient to pass through and not remain in a near-Earth orbit.

If it was a single asteroid of high mass with a near-Earth pass that effectively moved the Earth but not the Moon, Then any of your scenarios are possible.  The Earth could be moved sufficiently where it could be pulled away from the Moon (a change in velocity of about 1,500 km/hour in the right direction would do that) or the Earth could be pulled into the Moon (a change in velocity of about 3,600 km/hour), or the orbit could be sufficiently jiggled where one of those two scenarios would happen eventually if not right away.

To change the Earth's velocity that much, with a single asteroid pass, you'd need that asteroid to be, (if my math is right), a few times the mass of the Moon, and an object with that mass passing within a few thousand KM over the Earth's surface, lets just say that would be a bad idea for a number of reasons. - so, it's better to do it the way you suggested.  Many asteroids, over a long time.

Many asteroids - what would happen to the Earth-Moon system

Each asteroid, passing within a few thousand km of Earth would move the Earth slightly, but as the Moon moves around the Earth, many of those movements would cancel out.   They would move the Earth towards the Moon when the Moon was on the same side of the Earth as the asteroid, and away from it when the Moon was on the opposite side.  Similarly, eccentricity could be increased or decreased.   There's no one answer as to whether the Earth would lose the moon without adding up the position of the moon and averaging it out.     Moving the Earth could be timed, where, in effect the Moon would be carried along with it.  This works both ways too.   We could move the Earth's orbit by properly timed thrusters or explosions on the surface of the Moon, sometimes moving the Moon away from the Earth, sometimes closer, but gravity assists are probably easier with less energy involved, as Jupiter could be used to accelerate the asteroids, which would reduce the energy input needed.

The net effect is that you can move the Moon with the Earth, or the Earth with the Moon, but pushing on just one of them.   That's possible with gravitationally bound objects.

So, depending on the timing of the asteroid gravity assists, you could create any of your four scenarios, but without specific timing, there's at least a pretty good chance the Moon stay with the Earth.  The 2nd scenario, the Moon getting tidally torn apart, that would only happen if it got very close to the Earth.


Can Drizzle Integration increase resolution beyond theoretical resolution of telescope?

Drizzling can't actually do any better than the theoretical resolving power of the combination of your telescope and the atmosphere. What it can do is at least partly compensate for having pixels that are too large to properly sample the resolution of your telescope + the atmosphere.

The Nyquist Sampling Theorem basically says that you should have at least 2 pixels per resolution element in order to properly sample the resolution. Or, conversely, if you want to achieve a resolution of X arcsec, you want pixels that are no more than X/2 arcsec in size. So if the theoretical combination of your telescope diameter and the atmosphere had a resolution of 0.5 arcsec (so we're talking about really good atmospheric conditions), you want pixels that are 0.25 arcsec or smaller; otherwise, the pixelization overwhelms the intrinsic resolution. But note that you don't really gain anything from going to even smaller pixels: you'll just have a nice blurry image without visible pixelization. Having pixels that are 0.1 arcsec in size won't make the image any sharper than pixels that are 0.2 arcsec in size, if the resolution of your system (telescope + atmosphere) is ~ 0.5 arcsec.

In your hypothetical case, 0.52-arcsec pixels are good enough for resolutions down to about 1 arcsec. Ideal drizzling would allow you to effectively have pixels that were about 0.25 arcsec in size, which would accommodate a resolution of about 0.5 arcsec. But you can never do better than the intrinsic resolution of your telescope + the atmosphere.


How many judges sat on Galileo's trial over heliocentrism and who were they exactly?

The seven who convicted were
Felice Centini, Desiderio Scaglia, Antonio Barberini, Berlinghiero Gessi, Fabrizio Verospi, Guido Bentivoglio and Marzio Ginetti. And as you note Francesco Barberini, Laudivio Zacchia and Gaspare Borgia did not. These three didn't vote to acquit, they were just absent from the signing of the condemnation - we don't know why

All information from The Galileo Case: Trial, Science, Truth page 142
By Mario D'Addio. Gracewing Publishing, 2004 ISBN 9780852446652


Z is the atomic number, A is the atomic mass.

A white dwarf is typically made of carbon (A=12, Z=6) or Oxygen (A=16, Z=8). So Z/A = 0.5.


Assume that a star has a uniform mass density. Then how does the temperature scale as a function of the distance to the center?

I am not sure what you mean by looking for an answer without derivations? Of course you have to solve the maths to get a mathematical answer.

The handwaving answer is that the star has to be hotter in the middle, because at constant density the pressure depends on temperature and you need a negative pressure gradient to support the weight of the gas above.

The answer is:

The equation of hydrostatic equilibrium is
$$\frac{dP}{dr} = -\rho g,$$
where $\rho$ is the mass density and $g$ is the gravity at radius $r$, which by Newton's shell theorem is due to the total mass interior to $r$.

This equation is easily derived by considering the force equilibrium on a slab of gas in the star (your error above was in not recognising that it is the difference in pressure across the slab that provides the upward force).

If the density (and composition and ionisation state) are constant, then this equation can be written as
$$ \frac{\rho k_B }{m} \frac{dT}{dr} = -\rho \frac{4\pi G r^3 \rho}{3r^2},$$
where $m$ is the mass per particle and a perfect gas has been assumed.

Integrating with respect to radius and assuming $T(r=0)=T_0$, we obtain
$$T = T_0 - \frac{2\pi G\rho m}{3 k_B}r^2$$

Finally, if the star has radius $R$, then this will be radius at $T=0$, since nothing could be supported beyond that. In which case
$$T = \frac{2\pi G\rho m}{3 k_B}(R^2-r^2)$$

Or in terms of mass $M$
$$T = \frac{GM m}{2 k_B R^3}(R^2-r^2)$$

Constant density is of course a poor approximation. The above treatment gives a central temperature of 7 million Kelvin for the Sun, assuming $m\simeq 10^{-27}$ kg.


Where can I find a list of retrograde Milky Way stars?

The Milky Way's outer halo has many globular clusters with a retrograde orbit (about 40% of all clusters in Milky Way). One of the more prominent example include Kapteyn's star which is highly retrograde due to it being ripped from a dwarf galaxy and merging with the Milky Way.
However, the structure of the halo is a topic of an ongoing debate. Several studies have claimed that the halo consists of two distinct components. The "inner" halo consists of more metal-rich, prograde stars and "outer" halo consisting of metal-poor, retrograde stars. These findings have been challenged many times due to the argument on the topic of "duality of motion".
More readings here:

Carollo et al. 2007
Kravtsov 2001
Kordopatis et al. 2020


It is impossible to measure anything that happened in the Planck Era, so physicists are unsure whether the universe actually began from a singularity or whether the singularity in the Big Bang theory represents an inability to describe the universe at that time. In fact, "Most physicists believe that this is a mathematical artefact and does not describe what actually happened."
It would be more accurate to say that the Big Bang is the beginning of space-time, than that it "created" space-time.

The Big Bang is a description of how the universe evolved in its earliest stages, not exactly a resolution to why the universe came to be. The latter is really a question for philosophy, not something that can be addressed by science per se. Science informs our philosophy, but does not make philosophical claims.


What else could it be?  Very simply, it could all be a coincidence...

What they have done is noted that objects in the outer solar system have their orbits grouped in interesting ways. Someone suggested this might indicate there was a heavy planet (or core of a giant protoplanet) that influenced their orbits in this way; the recent announcement is the result of examining this idea in detail.  It is likely to be caused by gravitational influence of such a body.

Of course the TV and newspapers have picked this up and said a planet HAS been discovered. But there might be nothing at all; the loose clustering could all be a coincidence - though that seems, statistically, unlikely at the moment.

Here is the paper for anyone who wants to check.  (Beware, lots of mathematical terminology.)


Warm and Hot dark matter density profiles

There are indeed models for halos made of warm dark matter. The profiles will depend on exactly what the dark matter is and what its temperature is. I am no expert, but a starting point could be a paper by Vinas et al. (2012), who claim that the profiles are much flatter in the centre than those of CDM halos.

http://arxiv.org/abs/1202.2860

Hot dark matter is defined as such during the early universe and will be nearly uniformly distributed at the epochs of early structure formation. However at later epochs, hot dark matter with rest mass can cool down. For example, the cosmic neutrino background would have been highly uniform when it decoupled and when  most galaxies formed. But because of the non-zero neutrino rest mass and the expansion of the universe, these neutrinos are now non-relativistic and will be affected by cosmic structures and become "clumpy". The same will be true for any dark matter particle with a rest mass greater than about 0.1 eV.


Has the Lyman-alpha forest ever been used to test Arp's theory?

All possible tests of the Arp model have been failed. High redshift quasars have Lyman alpha forests, low redshift galaxies do not.

The Arp model suggests that the Lyman alpha forests are somehow intrinsic to the quasar. There are many problems with this idea.


What will "the Earth being consumed by the Sun" look like?

The exact fate of Earth as the sun becomes a red giant is somewhat uncertain because of two factors.
The first is that as the sun expands, its surface gravity $GM_\odot/R_\odot$ will decrease and gas can more easily escape as solar wind. This will reduce the mass of the sun, making Earth's orbit spiral outwards. But how much expansion depends on details of the far future solar wind we do not know yet, and this makes it slightly uncertain whether Earth can get into an orbit beyond the maximal solar radius. (The gas will also produce some drag, but it is likely negligible in this case)
The second factor is tidal interactions. As the sun expands it rotates more slowly, and in particular will rotate slower than Earth's orbital speed. That means that the tidal bulge induced by Earth will always be trailing the planet, creating a backward pull slowing it and dragging it inwards. This effect is small when the surface is far away, but grows rapidly as it approaches. Hence the sun can gobble up Earth even if the orbit is larger than the solar radius - but much depends on somewhat weakly understood tidal interactions.
Once the planet is inside the envelope it will rapidly get dragged in. To get a timescale, consider the time to push aside an Earth mass of gas: $\tau = M_\oplus /  \pi R_\oplus^2 \rho v$. For $\rho=$ 0.1kg/m$^3$ I get $\tau=0.4972
$ years. Maybe a bit longer if the orbit got wider and slower. Since the density of Earth is much higher than the gas, the planet will not be tidally disrupted but instead leave a hot plasma wake (that paper has more detailed calculations of infall timescales etc.).
The shocked gas will form a plasma sheath that erodes the planet at some rate, leaving a trail of metal-enriched material. Eventually the gravitational potential energy is released as heat, but this is mixed up with the rest of the solar luminosity over maybe a million years or so.
If Earth survives the giant phase it will orbit the white dwarf core (at a distance set by mass-loss expansion of the orbit minus gas drag inspiral; about twice the current orbit if drag is small). At this point it will become an airless (due to luminosity-induced atmosphere loss earlier), frozen rock over a few hundred million years or so as it cools from earlier heating.


If Mars orbited the Earth how distant would it have to be to cause the same tides?

The Tidal acceleration between 2 bodies is calculated with this formula:

$$
a_T = \frac{2GM_MR_1}{Dm^3}
$$

Where $M_M$ is Mar's mass, $R_1$ Earth Radius and $D_m$ the distance to the Moon. If you equal this to Moon's Tidal acceleration you will get $D_M$ as distance to Mars to get the same Tidal acceleration having $\frac{M_M}{M_m}=8.73328184501$:

$$
D_M = \sqrt[3]{8.73}D_m\simeq2.06Dm
$$


Can the electromagnetic counterpart of a gravitational be seen by 'small' telescopes?

Unlikely.
The Neutron star merger GW170817 had an optical counterpart (SSS 17a) The optical magnitude peaked at about +18, this is beyond the abilities of most amateur set ups.
Black hole mergers may be intrinsically more energetic, but less bright in the electromagnetic spectrum, as the black holes don't have a surface to interact electromagnetically.
However each gravitational wave detection is examined to see if there is a corresponding gamma ray burst, or optical transient.  Combining results of gravitational, neutrino and optical observations has been called "multi messenger astronomy".


Evidence for homogeneity of the universe

In the standard big bang model, homogeneity is an assumption - therefore it cannot predict homogeneity. Homogeneity is not a requirement of the model and indeed you can build "inhomogeneous big bang models" - though they must still be consistent with the observational evidence of a high degree of homogeneity on large scales within the observable universe.
The obvious observational evidence for homogeneity on large scales (and of course, in the observable universe) is the smoothness of the cosmic microwave background (CMB). The CMB is very, very uniform - with perturbations (bar the large-scale dipole) at the level of 1 part in $10^5$.
We can say very little about the unobservable universe, since it is... unobservable. And because it is unobservable, it has no influence on us and cannot affect the utility of the models we use to explain the observable universe.
NB: the big bang is just a model of how the universe works, with adjustable parameters including dark matter, vacuum energy density and homogeneity, not a theory.


Terrestrial Exoplanet Skies – I've Built a Visual Sky Chart. Is it Accurate?

I think the star,its halo and cloud can't be darker than the sky color even early M type still look blindingly bright orange-red on the sky not dim orange-red.
The sky color, if heavier i think it will look more "desaturate"(brighter too but not bright to the point it's brighter than the star) and thecolor shift toward the red side(you are correct) that dense sky will be affected by the color of starlight very intensely to the point it's almost have the same color of the starlight
example: 10 bars sky on a planet around K5V sun would look light orange-ish grey also the cloud would look bright orange, same color for the star's halo.
example2: 10 bars sky on a planet around A5V sun would look bright pale teal/cyan with yellowish white cloud same color for the star's halo.
example3: 5 bars sky on a planet around A5V sun would look powder blue with pure white cloud, same color for the star and its halo.


Nothing can be stationary inside the event horizon and everything, even light must travel inwards. Therefore there is no way that your analogy can work and is why using the idea of escape velocity is a really bad idea in General Relativity.
i.e. You cannot be at a fixed $r$ coordinate to lift a weight. You must be falling inwards, as must any weight you lift, as indeed must a light beam you direct radially outward.


Are there other pictures of Titan surface from Huygens?

One second of googling reveals the whole archive:

http://esamultimedia.esa.int/docs/titanraw/index.htm
(note you can click onto the strips to inspect them!)
The archive depicts the whole decent of the Huygens lander onto Titan, with the first picture being on the ground, and some of them seem to be taken from higher up altitudes, pages 17-27 you can see the flow deltas)
The three pictures per strip show the views of the three cameras into different directions.

However the location on the planet doesn't change, just the viewing angle around the spacecraft upon descent. This is because Huygens was a lander without any additional propulsion, as it was already considered to be a massive feat if the lander would make it to the surface at all.
Also a propulsion system would have been more expensive, and more or less useless, as the expected lifetime of the lander in Titan's atmosphere was a few hours, which happened to be true.

So summa summarum, from the final position in which the lander was fixed, yes this is the only picture of Titan's surface.


How much mass does the Sun lose as light, neutrinos, and solar wind?

The solar neutrino luminosity is about 2.3% of its electromagnetic luminosity (i.e. light). So the extra mass lost in the form of neutrino energy is 2.3% of your original calculation.

The average mass loss in the form of a wind and coronal mass ejections is about $4\times 10^{16}$ kg/year, but varies with the solar cycle (and from cycle to cycle) (Mishra et al. 2019).

4.5 billion years ago? It depends how exact you want to be. The Sun is thought to be 4.57 billion years old, so 4.5 billion years ago it would have been 70 million years old.

A 70 million year old Sun would have been on the hydrogen burning main sequence and about 20% less luminous than it is now, so you can scale your luminosity-mass and neutrino mass loss rates by about 0.8.

However, the solar wind was probably much stronger than it is now. The observational constraints on this are weak, but theoretical models suggest the mass loss rate in the wind scales as rotation rate $\Omega^{1.33}$ (Johnstone et al. 2015). Unfortunately, we still don't know how fast the Sun rotated in its infancy; it could have been anything from about 10 to 100 times it's rotation rate now. That means the mass loss rate in the wind would have been 20-500 times what it is now. Thus mass loss from a wind would dominate.

But maybe you meant $\sim 4.5$ billion years ago, in the sense that you wanted an answer for before the Sun became a star. i.e. Before hydrogen fusion began at a few million years after the Sun's birth. In that case, the wind losses might have been as per the 70 million year old case (with similar uncertainties), but there would be no neutrino losses (no nuclear reactions) and the luminosity of the Sun could have been a factor of 10 higher as a contracting pre main sequence star. In that case, mass loss from a wind would probably still be the biggest contributor.


The period (at least in physics) is defined the time an oscilating system needs to get back to its starting point (for a sinus curve its 2*pi).

Now when they say an eclipsing binary has a period of T years, it means that its brightness varies as seen from earth when they are eclipsing (thats how we detect eclipsing stars and planets by the way), and that the curve of its brightness is like an oscilation that gets back to its starting point every T years.

The meaning of this is, that this system consisting of 2 stars will be exactly as you see it now again in T years.


It's usually a good idea to use the main page and follow links there instead of guessing URL; a search for "kepler telescope data" and "tess telescope data" quickly revealed these links:

Kepler: https://keplerscience.arc.nasa.gov/data-products.html which links for some data to https://exoplanetarchive.ipac.caltech.edu/index.html

TESS: https://heasarc.gsfc.nasa.gov/docs/tess/data-access.html

There no general access to derived data needed for a particular publication (though it's good practise and meanwhile becoming more common to offer that in supplementary online data).


Sirius is the most likely identification.  It is visible in the South East at 6:30 from the USA, it is the brightest star in the sky and it's brightness makes the twinkles caused by the atmosphere more noticable. Similarly its brightness makes the colours caused by twinkles more apparent.
By shaking the tripod a picture of the apparent colours of Sirius can be captured
https://epod.usra.edu/blog/2012/01/sirius-twinkling.html
There are several other bright stars in that region of the sky: Betelgeuse, Rigel and Procyon, any of which could have been your star, but Sirius is the most likely simply as it is the brightest.  The stars will be back in (almost) the same positions tonight.  If you can find Orion and Orion's belt, and follow the direction down you will get to Sirius, which should confirm your identification.


Do the axes of rotation of most stars in the Milky Way align reasonably closely with the axis of galactic rotation?

There is very likely to be a random scatter.
Unlike planets orbiting the Sun in the Solar System, most of the stars in the Galaxy did not form at the same time as the Galaxy itself. There is therefore no strong reason to suspect that the angular momentum vectors would be aligned for similar reasons. On the other hand, the Galactic gravitational potential does depart from spherical symmetry in its inner regions, because the visible matter, which becomes dominant in the inner regions, is concentrated into a disc - so presumably this, or perhaps the tidal forces exerted by this on molecular clouds, could imprint some angular momentum preference.
The evidence is sketchy but suggests random orientations, at least in the solar neighbourhood. I refer you to Detection of exo-planets , where I discuss this in the context of detecting transiting exoplanets.
In a series of papers, myself and colleagues have investigated the distribution of spin-axes in clusters of stars. The idea here, which is not far-fetched, is that large clouds from which clusters form will have some angular momentum. The question is how much of that angular momentum is inherited by the stars it forms, or to what degree can turbulence in the collapsing gas essentially randomise the spin vectors of collapsing fragments. Our technique was to combine rotation periods (latterly from Kepler observations) with careful measurements of projected equatorial velocities ($v \sin i$, where $i$ is the spin inclination to the line of sight) to get projected radii ($R \sin i$) and then to model the distribution of $R \sin i$ with various assumptions about the spin-axis distribution. In all three of the clusters we have studied (Pleiades, Alpha Per, Praesepe), the distribution was consistent with a random distribution, with quite strong limits on the amount of alignment that was possible (Jackson & Jeffries 2011; Jackson, Deliyannis & Jeffries 2018; Jackson et al. 2019). The technique has been replicated in a fourth cluster, NGC 2516, by Healy & McCullough (2020), with the same conclusion.
Other authors have claimed alignments in some cases. Notably, using Kepler asteroseismology of red giants in two clusters in the Kepler main field, Corsaro et al. (2017) claimed a quite tight alignment of spin axes, pointing almost towards us in each case. Since the Kepler field is not far from the Galactic plane and these were distant clusters, then the spin axes would be almost in the Galactic plane (a bit like Uranus and the Sun). However, the likelihood of finding such a result if individual clusters had random average angular momentum vectors raised question marks - the probability of seeing that vector pointing towards you is very low. Work by Kamiaka et al. (2018) shows that the asteroseismological estimates may be systematically biased towards low inclinations.
A further piece of evidence for some alignment was in the orientations of bi-polar planetary nebulae towards the Galactic bulge. Rees & Zijlstra (2013) found a non-random distribution that suggested that the orbital angular momenta of binary systems, responsible for the bipolar shape of the nebulae, were oriented aligned with the Galactic plane (again, like Uranus around the Sun). The result is highly statistically significant but as far as I know has not been followed up despite its obvious implications for estimations of transit yields from exoplanetary surveys.
I think the biggest argument that there is no significant effect for average stars in the field of the Galaxy, is that the exoplanet people working on the TESS survey (which covers the whole sky), would have found a drastic spatial dependence on their yield of transiting planets as a function of Galactic latitude. The majority of transiting planets (or at least hot Jupiters) have orbital axes coincident with the spin axis of the star (like planets in the Solar System). If these orbital axes were aligned with Galactic north (or any other direction) it would mean you would see far fewer transiting planets when looking towards those directions. I have heard no reports of such a spatial dependence.


What are the concrete technical arguments supporting the idea that the wave function of the universe can be written as partition function?

Similar approaches have been generalized to more general spacetimes.

Technical details for dS space and further references in this paper about the holographic principle, p.43 ff., may provide some idea.


Is the gravitational field of the sun uniform?

The Sun rotates, taking about 24.5 days to complete a revolution at the equator, a bit longer at the poles. This makes the Sun have a bit of an equatorial bulge, which in turn means that the Sun's gravity field is not quite uniform. The Sun's J2 is very small and the effect is well known.
Very large stars are asymmetric due to being close to the Eddington limit. The Sun is too small of a star to have the kind of gravitational anomalies that very large stars and rocky planets exhibit.


Trans-Neptunian Dwarf Planet near Heliopause

Heliopause is believed to be at a distance of about 123 AU away from the Sun. Now, if you look at the list of farthest body in the solar system, you will see two bodies located approximately at that distance: 2018 VG18 and predicted FarFarOut (124.12 AU and ~140 AU from the sun as of June 2020 respectively). They are considered the farthest TNOs in the solar system.
For comparison, Voyager 1 and Voyager 2 are the two space probes which left the solar system is now 149.36 AU and 124 AU from the sun as of June 2020. Also you can see some of the near-parabolic comets originating from that distance.


Is it ever possible that the graviation from the mass of kinetic rotational energy will overcome centrifugal force?

In some sense, the answer is trivially 'yes':
Due to spacetime curvature, there exist no circular orbits below $\frac{3}{2}r_s$, with $r_s$ the Schwarzschild radius.
So whenever you get an object sufficiently dense, that its radius is below its $\frac{3}{2}r_s$, the mass equivalent to any nonzero rotational energy overcomes the centrifugal force of an orbit at $\frac{3}{2}r_s$.
Below $\frac{3}{2}r_s$ even zero rotational energy overcomes the centrifugal force, since the latter is then pointing inward.

This doesn't require a Black Hole.


Is there a way to calculate the mass of of a stellar remnant given the initial mass of the star?

For massive stars and their remnants, one can only be guided by theoretical measurements at present, since we do not have any masses for isolated neutron stars and black holes and no measurements of the masses of the progenitors of any neutron stars or black holes (only limits). Here are two plots taken from the extensive review by Limongi (2017). These show the relationships you want (the red lines) between the final remnant mass (on the y-axis) and the initial progenitor mass on the x-axis. The calculations are highly dependent on metallicity, so plots are given here (there are more in the paper) for solar metallicity ([Fe/H]=0) and a thousandth of solar metallicity ([Fe/H]=-3). There is also a dependence on rotation, which may be important.

You have to appreciate that these red curves are highly uncertain and model-dependent.
The situation is much clearer at lower masses. The masses of white dwarfs, even isolated white dwarfs, can be estimated quite well and their presence in binary systems and clusters of stars gives the means to estimate their ages. Combining this with their cooling ages gives a good estimate of the time that their progenitor spent as a "normal" star and then stellar evolution models give an estimate of the progenitor mass. Both the modelling and measurements are much more secure and so the initial-final mass relation for white dwarfs is in quite good shape.
The plot below is from a summary by Kalirai (2013). It is likely to be superseded in the next year or two by the wealth of data coming in on white dwarfs from the Gaia satellite. The names on the plot are the clusters in which the calibrating white dwarfs were found. You will see on the plot that Kalirai gives a simple formula that represents this relationship and might be what you are looking for. I believe the dashed line is from some stellar evolution model.


I don't think you'll find a single agreed shape for a rotating neutron star, not least because we don't have an agreed single model for the equation of state of the material in a neutron star (which is more complex than the name suggests).

I found one openly available paper (I'm sure there are more) which will give you a rough flavor for the complexity of modeling the shape of neutron stars.  As you'll see the difficulty of no single model for an equation of state (EOS is the shorthand typically used) is just one issue.

I think "ellipsoid" should be considered as an approximation, although it's not something I'd consider written in stone.

Remember that to be useful a paper has to provide not just a model for what the shape might be, but also someone has to provide a way to measure this, which is challenging.  I think one of the hopes for the new era of gravitational wave astronomy is to be able (eventually) to make more useful and measurements that help us investigate the interior of neutron stars.

So this is an open question, I think.

@Rob-Jeffries asked a question in comment about typical numbers for the deformation, and I answered in comment but comments can be removed by the system, so I'm adding that information as an edit :

In the first section of the paper I linked to they do quote fractional deformations as being typically $10^{−5}$, perhaps $10^{−4}$ in special cases and in extreme cases up to $10^{−3}$. However another paper gives an analysis based on crustal rigidity and a very small deformation for a particular neutron star. The paper I initially liked to describes an upper limit based on gravitational wave considerations, I think, rather than a general analysis.


Orbital velocity of a planet - why is my calculation off by about 10%?

Well done you. I double checked the calculations and couldn't fault what you had done. So I contacted the lead author of the paper about it and here is the response:

"After checking the numbers in our paper, I found an error: we actually used a mass of 1.0 MSun for J1407 in our simulations, instead of the 0.9 MSun as stated. This accounts for the difference in pericentric velocities (as well as the different semi-major axes, which would be smaller in the 0.9MSun case).
We will attempt to correct this in the published version, and send a correction to arXiv."


What is the gravitational force felt on Earth from the other planets in our solar system?

Because of the inverse square law for Newtonian gravity we have the acceleration due the gravity $g_b$ at the surface of the Earth due to a body of mass $m_b$ at a distance $d_b \gg r_e$ (where $r_e\approx 6371 \mbox{km}$ denotes the radius of the Earth, note all distances will need be in $\mbox{km}$ in what follows) is:
$$
g_b=g\times \frac{m_b}{m_e}\times \left(\frac{r_e}{d_b}\right)^2
$$
where $g$ is the usual accelleration due to gravity (from the Earth at the Earth's surface $\approx 10 \mbox{m/s}^2$, and $m_e\approx 6.0 \times 10^{24} \mbox{kg}$. We get the maximum acceleration due to a body when that body is at its closest to the Earth, which is what we do from now on (except for the Sun and Moon where the mean distance is used).

Now for the Moon $r_b\approx 0.384 \times 10^6 \mbox{km}$, and $m_b\approx 7.3 \times 10^{22} \mbox{kg}$, so the accelleration at the Earth's surface due to the Moon $g_b\approx 3.3 \times 10^{-5} \mbox{m/s}^2$

Then putting this relation and Solar-System data into a spread sheet we get:


Proxima Centauri can't explode as supernova as it is about 2 orders of magnitude too small and light.

But if a supernova were to happen from that distance? How bright would it be?

You can use the formula:
$$\text{Apparent magnitude} = \text{Absolute magnitude}+5log_{10}(distance)-5$$
where the distance is in parsecs.

Proxima centauri is 1.3 parsecs distant.

A supernova typically has an absolute magnitude between -15 and -19 (although they can go brighter or dimmer in the case of hypernovas etc). So Let's use -15 as the low end and -19 as the high end.

Plugging into the equation, we get an apparent magnitude of between -18 and -23.

For comparison, the sun as seen from earth has an apparent magnitude of about -26, so the supernova would be between 1500 and 15 times dimmer than the sun from earth.

Perusing the list in the Wikipedia article on apparent magnitude, we find the brightness of the supernova would be between the brightness of the Sun viewed from Jupiter and the Sun viewed from Pluto. In any case, that is plenty of light to read by, and plenty of light to see colour by.


What is the brightest star (relative magnitude) in M31?

The Variable stars that Hubble studied in M31 when he showed that it was a galaxy are among the most luminous. They include Var-A1, a luminous blue variable, and magnitude about 16.5. Var A-1 is one of the most luminous stars known.

Nasa has a catalog of stars in the m31 field, The brightest star in this catalog is an 11.4 magnitude star. But I think that database needs careful interpretation.


How much mass does our sun need for exploding like supernova?

The smallest stars that could explode in supernova have a mass of at least about 8 times the mass of the sun. So it is impossible for the sun to become a pulsar at the end of its life. It will form a red giant which will then lose its outer layers and form a white dwarf.


Why isn't most hydrogen in the universe molecular (diatomic), instead of atomic (monoatomic)?

Yes, the atomic hydrogen is probably mostly left over from the Big Bang. [Edited to add: Not sure how much that is true and how much present-day atomic hydrogen is the result of recombination.] And, yes, ${\rm H}_{2}$ does get dissociated by high-energy photons -- and also by cosmic rays, which can penetrate dense, dusty clouds that block most of the high-energy photons.
The real issue, as I understand it, is that's actually very difficult to make ${\rm H}_{2}$ by colliding two H atoms in the gas phase. This is because when two atoms collide, the resulting (temporary) molecule almost always has too much energy to remain stable and will very rapidly dissociate back into the two atoms -- unless you can somehow get rid of the excess energy before it does so.
Now, you can do this if the reaction actually involved producing extra components (like a more typical molecular reaction -- e.g., $AB + C \rightarrow AC + B$), since then the excess energy can be put into the kinetic energy of the products (while still preserving momentum). But in a reaction like ${\rm H} + {\rm H} \rightarrow {\rm H}_{2}$, there's no second body to allow that.
You could also do this if the "extra component" is some third body that happens to hit the other two at the same time (without becoming part of the molecule), since it can rebound with excess kinetic energy. But while this can work in dense gasses (e.g., in a laboratory on Earth), it will happen much too rarely in interstellar or intergalactic space.
You can also do this if the temporary molecule can emit a photon carrying away the excess energy. However, such transitions are strongly forbidden, especially in the case of a symmetric molecule like ${\rm H}_{2}$, which means that on average they will take too long to occur.
Finally, you can get rid of the energy if the reaction actually takes place on a surface that acts as a catalyst by absorbing the excess energy. This is thought to be the primary way ${\rm H}_2$ is formed in the interstellar medium: two H atoms that are adsorbed onto a dust grain combine on the grain's surface.
I mean, if the loose protons and electrons shot out by stars can (usually) find each other to form atoms, which is why neutral atomic hydrogen is more common between the stars and planets than ionized hydrogen, why can't the lonely atoms find each other to form $H_{2}$?
In addition to the fundamental problem outlined above, it's also easier (as Alchemista pointed out) for protons and electrons to "find each other" because they are attracted by their opposite electrical charges, something which doesn't happen in the case of neutral atoms.


In stellar astrophysics, "burning" means nuclear fusion, not chemical combustion. So a star burns hydrogen to helium. (Incidentally, normal chemical burning of hydrogen in air produces water).
This might seem to be confusing terminology, but it's not an issue in practice because the regions in stars where fusion takes place (the core, and shells surrounding the core, in big old stars) are far too hot for molecules to exist, so chemical processes simply cannot occur there.

Hydrogen fusion in a star core is not a single nuclear reaction. There are two (known) sets of reactions: the proton–proton chain, and the catalytic CNO (carbon-nitrogen-oxygen) cycle. From Wikipedia:

The proton–proton chain, also commonly referred to as the p-p chain, is one of two known sets of nuclear fusion reactions by which stars convert hydrogen to helium.
It dominates in stars with masses less than or equal to that of the Sun, whereas the CNO cycle, the other known reaction, is suggested by theoretical models to dominate in stars with masses greater than about 1.3 times that of the Sun.

Hydrogen burning is the main way that young stars produce heat and light. (Older stars can burn heavier elements, if they have sufficient mass). But hydrogen burning is also important because it's the main way that neutrons are produced: the burning of heavier elements generally does not change the number of neutrons. (Neutrons are also created when a heavy star collapses into a neutron star).
In a Sun-like star, hydrogen burning is a very slow process. It takes a lot of heat and pressure to overcome the electrical repulsion between two protons and fuse them into a diproton. However, the diproton is extremely unstable, with a half-life of much less than a nanosecond. Usually, a diproton just splits up into two protons, but sometimes one of its protons transforms into a neutron, converting the diproton to a deuteron (a deuterium nucleus). The probability of diproton to deuterium conversion is very small, in the order of $10^{-26}$. Various p-p chain reactions convert deuterons to helium nuclei.
The slowness of deuterium production is a good thing: it means our Sun will continue to burn hydrogen for billions of years. OTOH, it also means plain hydrogen will never be a practical fuel for an artificial fusion reactor.
The density in the Sun's core is ~150 g/cm³, which is much greater than any terrestrial material (eg, the density of lead is only 11.3 g/cm³). So a cubic metre of solar core matter has a mass around 150 metric tons. But the rate of hydrogen fusion is so slow that it only produces around 276.5 watts of power, which is similar to the heat production of a cubic metre of compost, as mentioned on Wikipedia. Of course, a 1 m³ compost heap won't keep pumping out heat for billions of years. ;)

Wikipedia has a good collection of articles on the various stellar nuclear fusion processes, including:

The triple-alpha process, which burns helium to carbon.
The carbon-burning process, which converts carbon to oxygen, neon, sodium, and magnesium.
The neon-burning process
The oxygen-burning process
The alpha ladder
The silicon-burning process


What happens when two black holes collide?

There are many hypothesis saying that there will a collision between
  the milky way and Andromeda galaxy...so what happens when two black
  holes will collide??

There's a few answers in the links in my comment above copied, here and here, but it's such a fun question I thought I'd answer it anyway.

With Andromeda and the Milky way's super-massive black holes in their respective centers, those objects are so massive, they are likely to be largely unaffected by any stars in their path and they will just fly towards each other at whatever rate and direction gravity dictates and by the time they get fairly close to each other, their respective gravity should have them moving quite fast towards one another.   I don't know if anyone knows if they will orbit around each other for an extended period of time or spiral in fairly quickly.   That will depend on the angle of approach.  It's possible they could pass by each other, miss and each could get sent far out away from the other, part of an enormous orbit around each other, taking perhaps billions of years to lead to an eventual collision.

There's a few simulated videos out there on what happens when 2 super massive black holes spiral into each other.   Here's one.

As Super-Massive black holes approach, each will pass through roughly half of the other galaxy, disrupting the orbits of any stars they pass hear-by, though they'd likely need to pass within less than a light year to have significant effects on the star's direction, which wouldn't happen that often, but it would happen.

Sagittarius A is huge for a black hole, but quite small compared to the space between stars.  It's estimated to be about 44 million KM in diameter, which means it would fit (Just barely) inside the orbit of Mercury and our Sun.   Mostly it will fly past stars.   Andromeda's super massive black hole may be several times larger, but still fairly small compared to the distance between stars it's likely to pass on the way towards their mutual collision.

It's possible that one or both stars will pass through each other's galaxy relatively collision free but it's possible that one or both of them will get close enough to a star to generate a large accretion disk.   While black holes don't have charge, their accretion disks do, and if that happens, that could be an interesting and not very well understood interaction between the 2 super-massive objects and their accretion disks.   Also, as they spiral in towards each other, several of their near-by stars will be cast any-which way, some of them, inevitably inside towards one of the 2 black holes.   It could be a very impressive show.

Finally, as they merge, which, might take quite a bit of time if they end up orbiting each other, perhaps millions, even billions of years.   If/when they do merge, there could be some curious and not very well understood gravitational wave effects. bending and stretching of space like ripples.  (Gravity already bends space, but not in measurable ripples.   Our observation's of gravity warped space is a smooth curve.

I'll re-post this article here from comments that says it's possible for 2 black holes to repel each other if they bend space in opposite spin-directions and approach each other on a level plane.   Less close and they could still easily disrupt outer planet orbits, casting some planets every which way.

  And how will it effect the other objects revolving around them??

Imagine jellybeans in a salad spinner going as fast as you can spin it, and you remove the top.  That's basically what will happen to any near-by stars.   Both galaxy centers are quite crowded with stars (and perhaps several black holes) orbiting their centers.   The gravity assists from 2 super-massive objects moving towards each other will be significant and, basically stars will be flying all over the place.   Stars have tiny masses compared to those objects and they could be sent in any direction.


What is the ratio of "real" stars in the sky?

It's almost 100% stars.

In good conditions, you can see perhaps 2000 stars. (There are about 6000 naked-eye visible stars; of these, 3000 are above the horizon at any time, and about 1000 are hidden because they're too close to the horizon and blocked by the atmosphere.)

The number of non-star objects you can see without assistance is tiny in comparison:

The Sun (obviously not at night, and of course it's a star)
The Moon
The International Space Station, but only a tiny fraction of the time
The visible planets: Mercury, Venus, Mars, Jupiter, Saturn (and maybe Uranus if you have excellent conditions and better eyes than mine)

A handful of other artificial satellites might be visible, but only rarely. Comets can be very visible, but again that's rare. Visible asteroids are even rarer. Vesta, a large asteroid, may be barely visible, but I've never seen it.

Note that some of these objects are quite obviously not distant stars just based on their appearance.

Some galaxies (Andromeda, and the two Magellanic Clouds if you're far enough south) are visible, but they don't look like stars. A few of the brighter nebulas and globular clusters might be visible; the latter are groupings of stars, so I don't know how they'd count.

Meteors and aircraft can be visible, but they're within the atmosphere, and probably not covered by your question.


Sample bias' contribution to "Planet 9" hypothesis, why was this not thoroughly addressed before?

I think the underlying premise of the question -- e.g., "Why was this sample bias not thoroughly addressed before?" -- is somewhat incorrect. Previous papers, including papers by those making the "Planet 9" claim, have attempted to address sample biases; a secondary issue is that the new paper uses data unavailable to previous studies (and also ignores some of the previous data).
To begin with, the original paper by Batygin & Brown (2016) does include a brief discussion of the possible effects of selection biases (in Section 2). This was, admittedly, pretty cursory, but they have followed it up with more extensive analyses (Brown 2017; Brown & Batygin 2019). From the abstract of the latter paper: "To determine if observational bias can be the cause of these apparent clusterings, we develop a rigorous method of quantifying the observational biases in the observations of longitude of perihelion and orbital pole position. From this now more complete understanding of the biases, we calculate that the probability that these distant KBOs would be clustered as strongly as observed in both longitude of perihelion and in orbital pole position is only 0.2%."
So Napier et al. are not, as you suppose, the first to consider selection biases (as they indeed acknowledge). What, then, is the difference? Part of it is that Napier et al. argue for performing an analysis using detailed simulation taking full account of individual survey characteristics if they are known. In order to do this, they focus on objects found in three recent surveys. This means that they deliberately ignore the 6 objects used for the original claim of Batygin & Brown: "The six ETNOs considered in the Batygin & Brown (2016) (BB16) analysis were discovered in an assortment of surveys with unknown or unpublished selection functions, making it difficult to establish that the observed angular clustering was indeed of physical origin." Instead, they use 14 objects "detected by three independent surveys with characterized selection functions, all published since BB16."
(Note that the Batygin & Brown 2019 bias analysis also used 14 objects, but not the same 14 -- they included the original 6 objects that Napier et al. exclude, but did not have access to objects detected after their study, which Napier et al. do. We are in a situation where there are very few data points, and there is the potential for divergent results based on the small number statistics of different studies.)
Eric Jensen has already posted links to some Twitter discussions by Batygin on this topic; I can point you to a Twitter thread by the other original author (Mike Brown), as well as a recent blog post by him which tries to understand why they and Napier et al. get such different results.


Why can space telescopes see through a planetary nebula?

Let's say we have a spherical shell of some material emitting light, much bigger than the star it surrounds. If we look right down the center of the shell, our line of sight takes us through only a small amount of gas - not enough for there to be significant emission. On the other hand, if we look at the nebula near the outer edge, our line of sight takes us through substantially more material:

Inspired in part by an image in this piece by Phil Plait.
If you look through more light-emitting material, you'll see stronger emission than if you look through less material, so in this toy model, we would perceive the edges to have strong emission. So a spherical nebula wouldn't lead to uniform emission throughout our cross-sectional view of it.
This simple spherical model holds for some cases, but certainly not all, particularly because many nebulae aren't spherical (Wikipedia makes the claim that only ~20% are truly spherical, which I haven't been able to verify.) For example, we know that the Ring Nebula isn't actually a sphere because the central surface brightness is much lower than the projection effect I described above would predict. Something else clearly must be going on. In this case, the discrepancy is solved by the fact that the Ring Nebula is a bipolar nebula, with its axis along our line of sight, with a dense ring of gas around its equator. Therefore, there's more material near its edges than its center from our perspective, and so we see more emission.


How can a planet not rotate at all and yet have 1 year that is day and 1 year that is night?

Your premise is incorrect. We used to think Mercury was tidally locked, but since 1965 we now know it is in a 3:2 spin-orbit resonance, which gives it (long) days.

Anything you read that says it is tidally locked is old, or itself using old reference material.

Here are some quick sketches to show the orbit, relative orientation of the planet, and the sidereal and solar days. If you like it as an answer I'll redraw with a graphics tool instead of my iPad Pen app...


How do I plot orbit planets and satellites?

One keyword to search for is "spice kernel".
SPICE stands for Spacecraft Planet Instrument C-matrix Event. That is a type of data format in which orbital data are commonly provided by ESA, NASA and other institutions dealing with detailed data; it is the recommended format by IAU to use. Often the so-called kernels for objects and missions are also readily available for download and digestion of tools (e.g. here or here).
Quite a few poplular solutions are python modules like HelioPy, SpicePy or PoliAstro and possibly this tutorial might do what you want. They usually have good documentation with extensive examples for simply copy&paste on how to use them. That said, python is not the only option, there are solutions in other languages as well.
It's also worth to note that some missions come with their own dedicated tools like Cassini's JCSN or Cassini orbit plotting tool.
For natural bodies AstroPy, possibly along with JPLEphem might suffice and prove quite versatile.


How are image credits expressed in Astronomy presentations?

There is no conventional format, really. For publications purposes if one uses data from a telescope or image or code created by someone else or similar, the source usually requests being cited or acknowledged in a form specific to their own taste. Some examples include what you quoted or “This paper makes use of the following ALMA data: ADS/JAO.ALMA#******. ALMA is a partnership of ... “.
Answering your question.

For ESA and NASA there is probably a difference between “/“ and “&”.
For a publication in a journal you must follow the exact wording.
If you just give a presentation no one minds if you use your favorite formatting while acknowledging all parties involved.


What constraints are there on solid material in the ice giants?

NASA conducted a study on the Ice Giants in 2016 after recognition as a high-priority flagship mission target in the 2013-2022 Planetary Decadal Survey (overview of the Decadal Survey from The Planetary Society), a community document which helps set priorities and goals. The full report and summary are available from LPI's Ice Giants Mission Planning page and is intended to review the science and mission options leading into planning for the next decadal survey report. In the Science section starting on page 29, the introduction has a table of the major components of the different planet types (terrestrial, gas giant and ice giant) where ice giants are listed as ~10% gas, ~65% ice, ~25% rock with a total mass of ~15 times that of the Earth. This is in comparison to terrestrial planets which are almost 100% rock and gas giants which are ~85% gas. This is illustrated in the figure below taken from the report (their Figure 3-1):

In the lead-in to the subsections on the various science questions and subsections, the report goes on to say:

  We  therefore  consider  the  objectives  of determining  the  interior  structure  of an ice  giant  (Section  3.1.1)  and  the  bulk  composition (including noble gases and key isotopic ratios, Section 3.1.2) as the highest-priority science to be done.

Section 3.1.1 on interior structure states that there is a presumption that these "ice giants" are dominated by water but there is very little direct evidence. It then explains that ice giants are normally modelled with three layers: an outer Hydrogen/Helium envelope, below that a thick fluid ionic water layer (or water mixed with other ices like ammonia and methane), and at the center, a small rocky core. However the limited amount of gravity field data from the Voyager 2 flybys showed that both planets are not as centrally condensed (less mass in the center) as expected in the simple 3 layer structure and also:

  "there is no strong motivation, beyond computational convenience, for a 3-layer structure (Helled et al. 2010)." Link to Helled et al. 2010 journal article (free access)

Both Uranus and Neptune have bulk densities a little under compressed water, implying that there has to be Hydrogen/Helium (H/He) envelope and which is indeed seen in spectra of the visible surface. Below this envelope though little is known apparently. The low density and the limited gravity data would imply a mostly water interior with other gasses such as ammonia or more H/He to keep the density lower than water and a very small amount of rock in the core Nettelmann et al. 2013 (arXiv paper link). This model however predicts an ice:rock ratio of more than 10:1 which is not seen in any of the other icy moons and the model does not explain well the heat flux coming from the interior. This presentation by Nettelmann at a 2014 ice giants meetings talks more about these models, the constraints from the gravity data from Voyager 2 and the heat flux problems and shows possible internal structure diagrams.

The interior structure section of the Ice Giants report ends by calling for measurements of the higher order gravitational moments J2, J4 & J6 to higher precision, particularly for Neptune. However this is limited to the bulk structure and doesn't constrain the layering which would need a Doppler imager on a spacecraft to measure planetary oscillations analogous to solar helioseismology and/or an entry probe. The Doppler imager is discussed more in Sections 3.3.1 and 3.3.2 of the report which note that these new instruments have the potential to provide great information on the interior structure, but that Doppler imagers are technologically immature and that even with the largest telescopes on Earth or near-Earth spacecraft it has not been possible to see if the oscillations exist and the oscillation amplitude may be too small to detect. The report also notes such an instrument is also large, power hungry and produces a large data volume, all risks on a constrained mission to the outer planets but the benefits outweigh the risks. (Using up most of your mass and power budget on an instrument that may fail, or not produce a measurable result, makes people very nervous...)

In summary, there are some limited constraints on the bulk components of the ice giants but this is still a poorly constrained and understood problem which is a subject of ongoing research but is need of additional constraints and data from a dedicated mission. Such a mission would need to be flagship class which makes it expensive and would need broad support from the majority of the planetary community and would likely need to be the top-ranked mission from the next decadal survey covering the 2023-2032 timeframe to stand a chance of being funded and flying.


From a physics point of view, there's no obvious mechanism that could provide significant, consistent increases in the earth's rotation speed.

The moon's gravitation appears to be the main driver of changes to the earth's rotation.  And since it revolves more slowly than the earth's rotation, it tends to pull energy from the rotation, lengthening the day.  This is completely consistent with the evidence that points to ancient years on earth having more (and therefore shorter) days than is true in the modern era.   This is also consistent with the observed gradual recess of the moon away from the earth.

It is also possible to directly measure the length of a day and see that days on earth are significantly longer than they were in 1900.  Again, consistent with a slowing rotation.

The implications of this for the evolution of sleep cycles would indeed be better in another location (perhaps Biology SE).  But I'd suggest that this is a very slow process.

Remember that this is something that increases the length of the average day.  But there are lots of other variations between days during a year.  An organism might for instance synch to something like sunrise.  But in spring, sunrises come more than a minute earlier than 24 hours apart each day, and in fall they come more than a minute later than average each day.

Meanwhile, it was probably more than a million years ago that the average day was shorter by a full minute.  So the increase in day length over time is definitely happening, but at a rate that is almost irrelevant to biology.


Is there a limit to the distance of detectable gravitational wave events?

Several factors influence whether a source of gravitational waves at a certain distance is observable by a certain instrument. One way to compute the limit to the distance is explained in Abadie et al 2010 and is as follows:

Distance. The amplitude of gravitational waves decreases roughly with the inverse of the luminosity distance $\propto D^{-1}$. So sources that are further away will be more difficult to observe and at some point they will be less/not visible.
Power of the source. The stronger the source the more easy it's signal can be detected. For binary systems with larger total mass $M$ and larger reduced mass $\mu$ you will observe higher amplitude waves. The amplitude of the signal $\vert \tilde{h}(f) \vert$ can be expressed as (the expression is from Abadie et al 2010 see Finn and Thorne 2010 for a derivation)

$$ \vert \tilde{h}(f)\vert = \frac{2c}{D} \left(\frac{5 G \mu}{96 c^3} \right)^{1/2} \left( \frac{GM}{\pi^2c^3} \right)^{1/3} f^{-7/6}$$
Sensitivity. The detector can be more or less sensitive. The more sensitive the detector the lower the luminosity or distance of the objects that it can observe.

The sensitivity can be expressed by the noise power density $S_n(f)$ (as function of frequency $f$) which is specific for the instrument (and you will be able to see graphs of this in many publications). A signal can be observed if it is stronger than the noise. Abadie et al 2010 use as limit a (conservative) signal-to-noise of $\rho$ =8, which means that the signal must be 8 or more times stronger than the background noise in order to be detected.

This signal-to-noise ratio is determined by an integral of the ratio of
the frequency-domain waveform amplitude $\vert \tilde{h}(f) \vert$ and the noise power density $S_n(f)$.

$$\rho = \sqrt{4 \int_0^{f_{ISCO}} \frac{\vert \tilde{h}(f) \vert^2}{S_n(f)} \text{d}f }$$
where $f_{ISCO}$ is the frequency of the innermost stable circular orbit of the binary system

In that article (Abadie et al 2010), the limit of the distance for the detection of wave events from binary blackholes with mass $10 M_{\odot}$ was estimated at 2187 Mpc which is pretty close to the distance of 2840 Mpc estimated for GW170729 (which is heavier).

Note that the limits for binary neutron stars are more often reported and are easier to find. For instance in Moore et al 2015 you can read in more detail about the increase of the limit for LIGO from 80 to 100 Mpc in the recent years. The first image shows plots of $S_n(f)$ as function of $f$ and of $D$ as function of time (during the experiments improvements have been made and the distance was changing).

Rate of occurances. When certain events have a higher probability to occur then it may also be more likely to observe them at a far away certain distance. The computation for these rates include the aspects of distance. Also as stated by Abadie et al

  The real detection range of the network is a function of the data quality and the detection pipeline, and can only be obtained empirically.

In Abbott et al 2016 a computation is performed to determine the probability to observe a particular event at a certain distance. The distance, for $40-40 M_{\odot}$, ranges up to a roughly $z=0.6$ (or using $d \approx z c / H_0 \approx 0.6 \times 3 \times 10^5 / 74.2 \approx 2.5 Gpc$), which is plotted in the last figure of that reference.

Conclusion: the observation of GW170729 at about 3Gpc is about the limit of the current instruments

References

Predictions for the rates of compact binary coalescences observable by ground-based gravitational-wave detectors J Abadie et al 2010 Class. Quantum Grav. 27, 173001
Gravitational waves from a compact star in a circular, inspiral orbit, in the equatorial plane of a massive, spinning black hole, as observed by LISA Lee Samuel Finn and Kip S. Thorne 2000 Phys. Rev. D 62, 124021
Gravitational-wave sensitivity curves C J Moore et al 2015 Class. Quantum Grav. 32 015014
Supplement: The Rate of Binary Black Hole Mergers Inferred from Advanced LIGO Observations Surrounding GW150914 B. P. Abbott et al 2016 ApJS 227 14


I'm not sure what the current status of neutron starquake theory is, but last time I looked there were a couple of ideas.

The outer parts of a neutron star consist of a solid crust of neutron-rich nuclei (plus electrons, and a little deeper, free neutrons). Neutron stars are born rotating extremely fast, but they spin down due to the loss of rotational kinetic energy through magnetic dipole radiation. As a result, they are born oblate, but as the spin rate decreases this is not the lowest energy configuration. It used to be thought that starquakes (pulsar glitches) were caused by cracking of the crust and rearrangement towards a more spherical configuration.

The second idea is that it could be due to changes in the very intense magnetic field of the neutron star. The magnetic field is tied very tightly to the crustal material, so if the magnetic field shifts, possibly associated with the neutron star spin down, then it will put the crust under stress, which can be relieved by a starquake.

White dwarf stars also contain solid, crystalline material, however they do not spin so fast and do not have similar ultra-strong magnetic fields. I have not heard that they experience starquakes.

"Normal stars" do not have a crust or any other solid layer, so there can be no similar starquake phenomenon.


Wikipedia reports the age of HD 140283 as the figure you cited, 14.46±0.8 billion years. Now look at the reference it refers to. This is the paper by Bond et al that presents measurements of some characteristics of the star and the subsequent calculations, as well as a lengthy section on where the error came from (800 million years is, after all, quite a lot - even in Astronomy!).

The paper describes an approach to determining ages of stars without all that tedious mucking around with globular clusters. It suggests determining "ages of extreme Population II subgiants in the solar neighborhood based on direct trigonometric parallaxes, combined with state-of-the-art theoretical isochrones appropriate to the  detailed composition of each star." In short, the idea is to test stars to see if there ages are consistent with the calculated age of the universe (13.77±0.06 billion years - note the much smaller uncertainty).

The paper cites some previous estimates of the star's age - ~14 billion years and 13.5±1.5 billion years. It's worth noting this because the authors appear to pride themselves on how they tried to reduce the error of their measurements. The team next measured the star's parallax, also attempting to reduce error. With that in hand, as well as measurements of temperature and visual magnitude, they used stellar evolution models to determine its age. They also compensated for extinction from their measurements. The authors list the error from parallax, photometry, and extinction as 310 million years. The measurements then (page 11) list the age "implied" by the models and measurements to be 14.46 billion years.

Where is the extra 490-million year-uncertainty? Take a look at Figure 1, near the end. It is due to a variety of factors, the largest of which is the uncertainty of the "initial oxygen abundance" - although more properly the uncertainty in the ratio of oxygen to hydrogen.

It does come down to what David H said - the error measurements due comply with a universal age of 13.77 billion years. I would note, though, that the uncertainty in the measurement of the age of HD 140283 is much greater (800 million years) than the uncertainty of the age of the universe (37 million years). If the uncertainty due to the oxygen/hydrogen ratio could be eliminated, a more accurate age could be calculated, but until this error is addressed (preferably by another team), I think it's safe to say that it's more likely that the measurements for HD 140283 are off than that the measurements for the universe, and so what could have been a paradox is somewhat resolved.

TL;DR - There's a heck of a lot of error in the age of HD 140283, which fits reasonably with current models of the age of the universe, and I would trust the measurements of the age of the universe much more than the measurements of the age of this star, especially because the method used here is not widely used.


Star visibility above horizon map likewise day & light map or moonlight world map

As long as we can live with a few tens of kilometers accuracy, we can treat the Earth's surface as a sphere.
The subpoint of the object (the point on Earth at which the object appears at the zenith, overhead, or the central point on Earth seen from the view of the object if they could see Earth clearly) is simply given by
$$\text{lat} = \text{Dec}$$
$$\text{lon} = \text{R.A.} - \text{GST}$$
where GST is Greenwich Sidereal time
To make your plot we draw a great circle around the subpoint, then make a simple Equirectangular projection or lat/lon projection.
In Python for example, calculate the latitude and longitude of the subpoint,
latsub = np.radians(dec)
lonsub = np.radians(RA - GST) # where these have already been converted to degrees

Pick a bunch of points to make a circle
theta = np.linspace(0, 2*np.pi, 10001)

Draw a unit circle around Earth's equator, assuming the subpoint is at the north pole (dec = +90 degrees).
x, y, z = [f(theta) for f in (np.cos, np.sin, np.zeros_like)]

Tilt it by 90 - dec
tilt = np.pi/2 - np.radians(dec)
x, z = [x * f(tilt) for f in (np.cos, np.sin)]

calculate the latitude and longitude of the tilted circle by shifting the longitude (subtracting the longitude of the subpoint)
loncirc = np.arctan2(y, x) - lonsub
latcirc = np.arctan2(z, np.sqrt(x**2 + y**2))

Put it all together and convert units properly, and you get something like this. I did it for the Sun but you can do this for any object for which you have the R.A. and Dec.

import numpy as np
import matplotlib.pyplot as plt

# for right now, from Google:
RA = (3 + 29/60 + 38/3600) * 360 / 24 # deg for the Sun
dec = +19. # deg for the Sun
GST = (16 + 15/60) * 360 / 24 # deg

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]

latsub = np.radians(dec)
lonsub = np.radians(RA - GST) # where these have already been converted to degrees

theta = np.linspace(0, twopi, 10001)
x, y, z = [f(theta) for f in (np.cos, np.sin, np.zeros_like)]

tilt = halfpi - np.radians(dec)
x, z = [x * f(tilt) for f in (np.cos, np.sin)]

loncirc = np.arctan2(y, x) - lonsub
loncirc = np.mod(loncirc + pi, twopi) - pi
latcirc = np.arctan2(z, np.sqrt(x**2 + y**2))

dloncirc = np.abs(loncirc[1:] - loncirc[:-1])
jump = np.argmax(dloncirc)

latcirc, loncirc = [thing[:-1] for thing in (latcirc, loncirc)]
loncirc[jump] = np.nan
latcirc[jump] = np.nan

locd, lacd = [np.degrees(thing) for thing in (loncirc, latcirc)]
losd, lasd = [np.degrees(thing) for thing in (lonsub, latsub)]
losd = np.mod(losd + 180, 360) - 180 # put it within the maps -180 to +180 range

plt.plot(locd, lacd)
plt.xlim(-182, 182)
plt.ylim(-90, 90)
x, y = 121.5, 25 # Taipei
plt.plot([x], [y], 'or')
plt.text(x, y, ' taipei')
plt.plot([-losd], [lasd], 'oy')
plt.text(-losd, lasd, ' sun')

plt.title('Sunlit Earth 08:40 local time TPE, sunrise was 5:09')
plt.gca().set_aspect('equal')
plt.show()


Can an astronomical telescope view objects on Earth?

Yes, but many (not all) telescopes made for astronomy will invert the image. Presenting the image upright is important for terrestrial instruments, but it doesn't matter for astronomy. Also, it usually requires fewer optical components to make an instrument that inverts the image.

Also, terrestrial instruments are made to operate in an environment where there's plenty of ambient light (daylight), whereas many telescope designs (such as truss dobsonians) are built to operate in a dark environment and are usually hampered by excess ambient light.

So, in a nutshell, yes, you could turn a telescope towards objects here on Earth, and it would work, but for best results and easiest operation just get some good binoculars.

Finally, "power" is a much misunderstood characteristic of optical instruments. It is technically true that a large dobsonian could in theory outperform any pair of binoculars you could care to try, but in practice a large optical stack is difficult to swing around and is fickle with regard to collimation, thermal equilibrium, etc. If you're willing and able to put up with the operating difficulties, with the necessity to flip the image around, etc, then I guess that indeed a large dobsonian could surpass anyone's little binoculars in a terrestrial context.

But very large magnification always reduces the field of view. Go ahead and crank magnification all the way up, and it's like looking through a straw. It's hard to locate anything. Astronomers have specific methods to deal with this difficulty, but for a terrestrial observation the narrow field of view is a liability, not an advantage.

There's a reason why various optical instruments have evolved into different shapes and designs to operate in different environments. "Horses for courses", as they say.