Normally people don't focus their eyes on the sun; its always in the sky but noone usually has a reason to look at it so you never focus on it, and most know not to. During an eclipse thats exactly where you want to look and focus on, and focussing on it makes it even more dangerous (guessing due to time spent with same area of your retina receiving the focussed energy from it rather than moving around your view and unfocussed spot).

During totality (darkest part of 100% eclipse) your eyes start to adjust for the dark sky and open up, allowing even more light in; as sun comes out of totality it gets very very bright again quite quickly, and your eyes might not be ready for it - use protection as you see the diamond ring start to get bigger/brighter (or only look at the diamond ring on the way into totality for safest view, when its getting darker).

The same effect as totality is almost had by using eclipse glasses or other dark viewers that cover your head/eyes, and switching from that view back to unprotected view of the sun could be equally as dangerous for the same reason.

(I'm not an expert).


What do falsifiability and verifiability actually mean as a precondition of a scientific hypothesis?

[I am hesitant to post this answer, as I am unclear that this is a question about science rather than philosophy as I said in a comment.  But I've written it now, so.]
First of all, all scientific theories rest on a mass of assumptions not all of which are testable.  For example if I'm doing astronomy I'm assuming that I'm not just living in a simulation, or that, in fact, anything outside my own mind exists at all.  I can't test those ideas, but I'm going to assume them or I can't do any astronomy because there's no point.
Without assumptions like that all you can do is maths: Fermat's last theorem is true whether or not Fermat or Andrew Wiles, exist, say.
But there are assumptions which are made at a level above these kind of things which also matter in astronomy.  For measuring the spatial curvature of the universe one is very important: the cosmological principle.

The cosmological principle.  Viewed on a sufficiently large scale, the properties of the universe are the same for all observers.

(From William C. Keel (2007) The Road to Galaxy Formation (2nd ed.), Springer-Praxis, via Wikipedia.)
Here an 'observer' doesn't mean 'someone on Earth', it means 'someone anywhere you like'.
The cosmological principle implies two things: that on large enough scales the universe is homogeneous (it looks the same everywhere) and isotropic (it looks the same in all directions).
The cosmological principle is not fully testable I think.  We can check that the universe looks isotropic here, and people do this.  If it's not then the cosmological principle fails.  But if it does look isotropic then the principle still may be false: perhaps we are in some special location where the universe looks isotropic.  If we could get far away (very far away) and test that the universe looks isotropic from at least two other places as well, then we could conclude that the part of the universe we've observed is also homogeneous I think.
But it could still be fooling us somehow: the cosmological principle is really just something we assume is true.  It's a very reasonable assumption, I think, because what it's really saying is that where we are is not special: the universe is not, in fact, built around us.  The cosmological principle is the inverse of the Ptolemaic model, where everything goes around the Earth, really, and even more it is an inverse of the 'nothing but me exists' model which I said I was going to rule out at the start.
Well, if we assume that the cosmological principle is true, then the bit of the universe we can observe is just the same as all the other bits of it on large enough scales.  So, if we assume that general relativity is a good description of the universe on large scales (that's a lot more testable) and probably some other things which are testable then if we can measure the spatial curvature (or equivalently, the matter density) of the part of the universe we can see, we can conclude that this holds for all of it.
That is the sense, I think, that claims about the testability of measurements of spatial curvature are true.


Why are helium resonance lines called "resonance lines"?

It appears to be a conventional label that is applied to transitions between the ground state and another energy level (some definitions specify the first excited level) of an atom and is used in all the physical sciences, not just astrophysics.
e.g. He I (58.4 nm) is a transition from $^1$P to the $^1$S ground state.
In fact all atomic/ionic transitions can be considered resonant phenomena, but the term "resonance line" is applied only to these particular types of transition, perhaps because they are usually the strongest lines in the spectrum from that species, since in most cases, the ground state is the most populated. So there is nothing "particularly resonant" about them
A "resonance line" is:

A spectral line caused by an electron jumping between the ground state and the first energy level in an atom or ion. It is the longest-wavelength line produced by a jump to or from the ground state. Because the majority of electrons are in the ground state in many astrophysical environments, and because the energy required to reach the first level is the least needed for any transition, resonance lines are the strongest lines in the spectrum for any given atom or ion.

https://www.oxfordreference.com/view/10.1093/oi/authority.20110803100415843
And here it is used in that context in astrophysics: (Morton 2013)

The tabulation emphasizes resonance lines, i.e., lines whose lower level is the ground state.

Or in Chemistry: From https://goldbook.iupac.org/terms/view/R05341

The radiative decay of an excitation level may proceed to the neutral ground state and would thus occur at the same energy as the corresponding line in the absorption spectrum. Such a line is called a resonance line and the process is called resonance emission.

Or from the Basic Atomic Spectroscopic Handbook:(https://www.nist.gov/pml/basic-atomic-spectroscopic-data-handbook)

The strongest persistent lines usually include one or more resonance lines, i.e., transitions to the ground level or term.


In terms of mean angular velocity, the distribution of rotation rates among main sequence stars is well known. Allen (1963) compiled data on mass, radius, and equatorial velocity, which was then expanded upon by McNally (1965), who focused on angular velocity and angular momentum. It became clear that angular velocity increases from low rates for spectral types of G and below before rising to a peak around type A stars and then slowly decreasing.
Equatorial velocity continues increasing to mid-B type stars, before slowly decreasing, but because of the increased radii of O and B type main sequence stars, the peak in angular velocity occurs before this. As part of Jean-Louis Tassoul's Stellar Rotation notes, many O type stars have rotational periods similar to that of the G-type stars like the Sun!
The distribution is not smooth and uniform (McNally noticed a strange discontinuity in angular momentum per unit mass right for A0 and A5 stars; see his Figure 2); Barnes (2003) observed two distinct populations in open clusters, consisting of slower rotators (the I sequence) and faster rotators (the C sequence). Stars may migrate from one sequence to another as they evolve. Interestingly enough, stars on the I sequence lose angular momentum $J$ faster than stars on the C sequence:
$$\frac{\mathrm{d}J}{\mathrm{d}t}\propto-\omega^n,\quad\text{where}\begin{cases}
    n=3\text{ on the I sequence}\\
    n=1\text{ on the C sequence}\\
  \end{cases}$$
Here, of course, $\omega$ is angular velocity. These results obey Skumanich's law.
Oblateness can be determined from mass, radius, and angular velocity as
$$f=\frac{5\omega^2R^3}{4GM}$$
Using this and McNally's data, some quick calculations get me the following table:
|Spectral type|$f/f(O5)$|
|--|-------|
|O5 | 1    |
|B0 | 1.28 |
|B5 | 1.84 |
|A0 | 1.67 |
|A5 | 1.35 |
|F0 | 0.482|
|F5 | 0.0387|
|G0 | 0.000314|


The term "stellar component" refers to a part ("component") of a system that is in the form of one or more stars.

For example, the Alpha Centauri system has three stellar components: Rigil Kentaurus, Toliman and Proxima Centauri. The planet Proxima Centauri b is also a component of the Alpha Centauri system but it is not a stellar component because it isn't a star.

Another example: the stellar component of a galaxy refers to the fraction of the galaxy that is in the form of stars, as opposed to dust, gas, dark matter, etc.

In the case of your question, it is asking what the inclination would be if the companion were a star (i.e. a stellar component of a binary star system).


Can quantum entanglement cause nucleosynthesis to happen in stars?

You are referring to pyconuclear reactions - these are reactions that are initiated even when the temperature is effectively zero. They are caused by zeropoint oscillations of particles that are trapped in a deep potential well and hence are purely a quantum mechanical effect.

This is far from just of theoretical interest. It may be that pyconuclear reactions occur in the condensed matter of white dwarfs; possibly even initiating the carbon burning that can destroy the star in a type Ia supernova.


The short answer is that waves that are "in the apparatus" are indeed stretched. However the "fresh waves" being produced by the laser are not. So long as the "new" waves spend much less time in the interferometer than it takes to expand them (which takes roughly 1/gravitational wave frequency), then the effect you are talking about can be neglected.

Details:

There is an apparent paradox: you can think about the detection in two ways. On the one hand you can imagine that the lengths of the detector arms change and that the round-trip travel time of a light beam is subsequently changed and so the difference in the time-of-arrival of wavecrests translates into a phase difference that is detected in the interferometer. On the other hand you have the analogy to the expansion of the universe - if the arm length is changed, then isn't the wavelength of the light changed by exactly the same factor and so there can be no change in the phase difference? I guess this latter is your question.

Well clearly, the detector works so there must be a problem with the second interpretation. There is an excellent discussion of this by Saulson 1997, from which I give a summary.

Interpretation 1:

If the two arms are in the $x$ and $y$ directions and the incoming wave the $z$ direction, then the metric due to the wave can be written
$$ds^2 = -c^2 dt^2 + (1+ h(t))dx^2 + (1-h(t))dy^2,$$
where $h(t)$ is the strain of the gravitational wave.

For light travelling on geodesic paths the metric interval $ds^2=0$, this means that (considering only the arm aligned along the x-axis for a moment)
$$c dt = \sqrt{(1 + h(t))}dx \simeq (1 + \frac{1}{2}h(t))dx$$
The time taken to travel the path is therefore increased to
$$\tau_+ = \int dt = \frac{1}{c}\int (1 + \frac{1}{2}h(t))dx$$

If the original arm is of length $L$ and the perturbed arm length is $L(1+h/2)$, then the time difference for a photon to make the round trip along each arm is
$$ \Delta \tau = \tau_+ - \tau_- \simeq  \frac{2L}{c}h$$
leading to a phase difference in the signals of
$$\Delta \phi = \frac{4\pi L}{\lambda} h$$
This assumes that $h(t)$ is treated as a constant for the time that the laser light is in the apparatus.

Interpretation 2:

In analogy with the expansion of the universe, the gravitational wave does change the wavelength of light in each arm of the experiment. However, only the waves that are in the apparatus as the gravitational wave passes through can be affected.

Suppose that $h(t)$ is a step function so that the arm changes length from $L$ to $L+h(0)/2$ instantaneously. The waves that are just arriving back at the detector will be unaffected by this change, but subsequent wavecrests will have had successively further to travel and so there is a phase lag that builds up gradually to the value defined above in interpretation 1. The time taken for the phase lag to build up will be $2L/c$.

But then what about the waves that enter the apparatus later? For those, the laser frequency is unchanged and as the speed of light is constant, then the wavelength is unchanged. These waves travel in a lengthened arm and therefore experience a phase lag exactly equivalent to interpretation 1.

In practice, the "buildup time" for the phase lag is short compared with the reciprocal of the frequency of the gravitational waves. For example the LIGO path length is about 1,000 km, so the "build up time" would be 0.003 s compared with the reciprocal of the $\sim 100$ Hz signal of 0.01 s and so is relatively unimportant when interpreting the signal (the detection sensitivity of the interferometer is indeed compromised at higher frequencies because of this effect).


Are red dwarfs (M dwarfs) known to be strong infrared sources?

Yes, red dwarfs are strong in the infra-red. This is because they are much cooler than our sun, and just like a red hot poker which has cooled from being yellow hot, their emissions are redder. It takes less energy to emit red light than to emit longer wavelengths, and still less energy to emit in the infra-red. Being much smaller than the sun and with less fusion taking place, red dwarfs have less energy.They are known to emit flares, sometimes quite large ones.


How official is TNO 486958, 2014 MU69's new name Arrokoth?

The International Astronomical Union describes this minor planet naming process:

Minor Planet Center assigns provisional designation
MPC assigns permanent designation
Discoverer suggests name
IAU working group reviews name
MPC publishes name

  Accepted names become official when they are published, along with their accompanying citations, in the Minor Planet Circulars, issued monthly by the Minor Planet Center.

The last page of the Minor Planet Circular published 2019-11-08 says:

  (486958) Arrokoth = 2014 MU69
      Discovered 2014 by M. W. Buie and the New Horizons Search Team.
      Arrokoth is the word for "sky" from the Powhatan language of native people from the Chesapeake Bay region.
  Institutions in this region played a prominent role in facilitating the discovery and exploration of this ancient and distant object.

so it's as official as it gets. As with these near-Earth asteroids:

(99942) Apophis = 2004 MN4
(101955) Bennu = 1999 RQ36
(162173) Ryugu = 1999 JU3

we can expect (486958) Arrokoth to displace 2014 MU69 as the prevailing identifier in new publications.


Can we learn anything from observing galaxies edge-on, like the Spindle Galaxy?

Yes, we can certainly learn things from edge-on galaxies. The easiest things to learn, as you suggest, have to to with the vertical structure of the stars (and gas). For example, our knowledge of how thick galaxy stellar and gas disks are -- and whether this thickness varies with distance from the galaxy center -- comes almost exclusively from studying edge-on galaxies. It is also in many cases easier to study faint stellar and gaseous halos in edge-on galaxies, because in face-on galaxies, these things would be superimposed on -- and largely lost in the glare of -- the face-on disk.

Traditionally, it was also easier to study the radial structure of disks in edge-on systems, because the fact that the line of sight through a disk intersects more stars when the galaxy is edge-on than when it is face-on meant you could make observations at higher signal/noise at large radii. Thus, the detection that many disk are "truncated" (that is, that there is a radius beyond which their brightness falls off more rapidly than it does inside that radius) was first made using photographic data of edge-on spirals.

Some elements of bulge structure have been traditionally easier to study in edge-on galaxies. The realization that some galaxies had "boxy" or "peanut-shaped" bulges -- along with the existence of "cylindrical" off-plane stellar rotation in such bulges and their likely nature as the vertically thickened inner parts of bars -- came about from observations of edge-on galaxies. (Although we now know how to identify such structures in moderately inclined and even face-on galaxies as well.)

(Note that our view of the Milky Way is essentially that of an edge-on galaxy, albeit one seen from the inside.)


Is it possible to calculate the time of day given latitude, longitude and solar elevation angles (zenith, azimuth, elevation)?

Of course it is! How do you think sundials work? That said, there are better, more precise ways to keep time. I don't know how to do it for an image; that is really a computational problem rather than an astronomy question. You might be better off asking it at Stack Overflow.


How do black holes form from stars that go supernova

Stars are not torn apart by supernovae, or at least their cores are not$^*$. At the centre of the explosion is a proto-neutron star. Indeed it is the formation of this proto neutron star that drives the explosion.
If the core is massive enough, then the neutron star will be too massive to be supported by any source of pressure and would quickly collapse to a black hole. This might be very likely if the supernova is insufficiently energetic to expel the entire envelope and some of it falls back onto the proto neutron star, causing it to implode.
It also seems likely that many black holes could be formed without a supernova explosion at all. A direct collapse black hole is probably required to produce $>10M_\odot$ black holes.
These possibilities and the observational evidence for them are discussed by Mirabel (2016).
$*$ The exceptions being pair instability supernovae at very high masses, which may set an upper limit to the mass of black hole that can be produced in this way.


What would happen if we released water into deep space?

Water can't exist in a vacuum.  Water in space (atmosphere, no gravity) tends to form into a ball, but take away the atmosphere and it would rapidly fly apart.  It actually wouldn't look like boiling.  You get boiling on Earth because water is held in a pot by gravity.  In empty space the water would rapidly and somewhat explosively disburse

This article says it would look something like what happens when you throw hot water into very very cold air and it rapidly turns to snow.  That's probably a pretty good visual image if you imagine no gravity and everything would just continue to fly apart, not slow by air resistance or fall to the earth.

In cold space the water would freeze, but some of the liquid would also effectively also boil away, even in very cold temperature because liquid isn't stable in a vacuum.  As I said, some of it would boil, but it wouldn't look like boiling, It would fly apart faster than it would freeze but you'd also see it freeze fairly rapidly.   If the water was enclosed, freezing would take some time, but in a vacuum and spreading out, it would freeze into snow and ice crystals rapidly.

A percentage of the water would become gas and that would help carry some heat away from the remaining water, speeding up the freezing process.  That gas would fly away at high velocity and not be visible.   If you collected and weighed the ice-crystals after the water was tossed into space you'd probably find that a percentage of the water escaped as gas.

You can get an idea of water turning into ice by heat transfer and escape from forming into gas molecules in a video that show what happens to water in a vacuum chamber.   A vacuum chamber can't replicate the rapid spreading out due zero gravity but it demonstrates the water turning to ice, even at room temperature.   Here's one video of that.


Are the stars in constellation located in a plane or they are in different distances

Constellations and asterisms are generally not proximate in space, but rather happen to be nearby only when viewed from Earth. From Wikipedia's article on asterisms:

  Like constellations, asterisms are in most cases composed of stars which, while they are visible in the same general direction, are not physically related, often being at significantly different distances from Earth.

A recognizable example of this would be the brightest star in Taurus, Aldebaran.  Aldebaran appears in the V-shaped bull's head, mostly formed by stars in the Hyades star cluster. However, while those stars are actually close to one another, Aldebaran is less than half as far away.

  Since the star is located (by chance) in the line of sight between the Earth and the Hyades, it has the appearance of being the brightest member of the more scattered Hyades open star cluster that makes up the bull's-head-shaped asterism; however, the star cluster is actually more than twice as far away, at about 150 light years.

Similarly, the stars of Orion's belt are actually hugely separated in space.

In addition, it's worth noting that the stars have different velocities across the sky, so the constellations will look increasingly different over long periods of time. Here is one prediction for 50,000 years time, and the effect will get proportionately larger as time goes on.


For the fist question it is theoretically possible, and it is explained in the popular Twin Paradox example.

For the second one I'm not sure if there are some advances but as far as I know there is an important problem about relativistic speeds: the mass of the object traveling at such speed increases dramatically to the point it would weight so much it would collapse(It wouldn't collapse actually).


two black holes orbiting each other within their event horizon

Inside the event horizon is a weird place to be. Inside a black-hole space-time is flowing towards the singularity faster than the speed of light (as measured from outside the black-hole. The effect is that any thing inside a black hole will, in a finite (and usually short) amount of time end up at the singularity.

Models of merging black holes exist such as video from nasa. It requires super-computers to solve Einstein's equations numerically.

As I noted in a comment, no stable orbit can exist within 1.5 times the radius of the event horizon. The orbital velocity is c at that distance (1.5 Schwarzschild); it's also called the "photon sphere". Within that radius all orbits are unstable. And the radiation of gravitation waves means that as two black holes orbit each other, they will lose energy, and so their orbits will decay. As the black holes approach, the event horizons are distorted, and merge into an droplet shape


Buying a telescope for the first time. How important is the focal ratio and is a computerised telescope worth it?

We can compare these scopes on a few grounds:

Celestron 114 LCM

Greater focal length
Alt-Az mount (intuitive)
Computerized (easy)

Orion 9851 SpaceProbe 130 EQ

Cheaper
Larger aperture
Greater highest useful magnification
Non-computerized (the experience of learning the sky through one's own methods, alone with the telescope, is one which you should not deny yourself, in my humble opinion)
EQ mount (allows for simple star tracking)

These are just some items which stand out to me as being important. The question you have asked is largely one of opinion—there is no "right" answer here. It depends entirely on what you want. You say you want to focus on planetary observations, a field where extreme magnifications are common and useful. For deep space, larger apertures and shorter focal lengths will aid.

The focal ratio is a quick way to judge the "speed" of the telescope's optics. It is found by taking the ratio between the focal length and aperture. The Orion scope in this instance has an f-ratio of f/6.9, while that of the Celestron is f/8.8. This implies that the view through the Orion scope will be slightly brighter, broader, and less-magnified than that through the Celestron, for any given eyepiece. However, these ratios are quite close and are both within a range commonly used both for deep-sky and planetary observations.

My final piece of advice is that for a beginner, the subtle differences between these scopes in these regards simply does not matter. You are unlikely to experience any significant difference between these two scopes. Because of that, I would personally recommend going for the Orion telescope—it's cheaper, will give you greater magnifications for planetary observations, and its lack of computerization will force you to truly learn the sky. Should you desire easier set-up, less time navigating through the sky, and so on, then the computerized Celestron may be preferable.

Telescopes are beautiful instruments. Whichever you choose will serve you well.

Clear skies!


What’s the object between the Earth and The Sun currently showing in Google maps?

That's the Moon alright, and it's definitely real and definitely there. If you go outside and look at the Sun right now, the Moon will be almost but not quite on top of it, though it's impossible to make out due to the Sun's glare. If it were any closer we would have had a solar eclipse around the time of the New Moon that took place 20 minutes ago. Two weeks from now at Full Moon, it'll be in even better alignment, though on the opposite side of the Earth from the Sun this time, which will result in a total lunar eclipse. Mark your calendar!


This answer applies to the simplest case: a static binary and a point source.

In the binary case, $u(t)$ describes how close each point in the source trajectory comes to (what you have defined as) the origin of the binary axis.

The solution is non-trivial. It involves finding the roots of a 5th order complex polynomial. You can look for example at how Etienne Bachelet's pyLIMA package codes it in Python. If you want to do it yourself, you first need to identify the location of the caustics and critical curves for the given arrangement. For that you need your two masses (or just the mass ratio) and their separation in units of the Einstein radius. The coefficients of the polynomial (A, B, C) are described in Schneider & Weiss 1986 (eqn 9b).

Then you need to define the trajectory of your source relative to the binary axis. You also have to choose your origin (some people use the midpoint between the masses, others use the largest of the two masses or the barycenter).

For each point in the trajectory (instantaneous source position) you have to evaluate the corresponding magnification, that is, find the roots of the polynomial for that position. You will then get five solutions, which correspond to the images of the source. However, not all 5 solutions correspond to real images. To find out which are the real ones, you have to substitute them in the inverted equation and see if they map back to the given source position. You will always have three (outside the caustics) or five (inside the caustics) real images.

You can see an example below.

You can see how different source trajectories generate different light curves at this link.


It looks to me like you have two problems.

You need to adjust the focus. All telescopes come with a knob that alters the focus. This is pretty easy.
the shadow you see is from the secondary mirror, and it is off centre. I don't know what the word for this is (off-collimation?), but I had it the other night and I fixed it by wiggling the secondary mirror and making sure it was flush with the hole it's supposed to fit in.

Try using the focus and if it's still bad, take another shot and post it here and maybe someone with more experience than me can help.


Parking a telescope at a Lagrange point: is this a good idea from a debris point of view?

Hmmm no, it wouldn't be cluttered with debris, and yes, it's a good idea to park the JWST (James Webb Space Telescope) at the Sun-Earth L2 point.

The five Lagrange points are unstable, for one because of the gravitational anomalies of the two massive bodies of the Lagrange system, eccentric orbits, and there are many other factors to their instability. At the same time, they are least gravitationally attractive points around two massive bodies.

Think of L-points as parking your car on a flat space at the top of the hill. You'll have to approach with some control and then try to balance at your parking spot, if you don't plan using your handbrake and also remain there stationary:

    Visualisation of the relationship between two massive bodies and their five Lagrange points (Source: Wikipedia)

There wouldn't be any debris clutter there, or any other matter like smaller particles, at least not any more likely than elsewhere around them, only transient in nature and possibly even less likely than elsewhere since all the other mass particles would gravitate towards the more massive bodies of two Lagrange system centers in their vicinity.

No body with rest mass would stay there on its own accord, not unless it has active attitude control to position itself there and constantly adjust for changes in gravitational attraction vector as the two mass bodies rotate around their axes, change distance while orbiting each other, or L points being influenced by other mass bodies of the same planetary system.

At the same time, JWST will be shielded by the Earth from any Solar activities and also Sun's interference with JWST's sensitive equipment as it starts observing the Universe in the infrared spectrum. Most of the orbital debris from our own space exploration missions is cluttered in the LEO (Low Earth Orbit) belt at roughly an altitude of 500-1500 km above the Earth's surface:

             Source: Active Debris Removal: EDDE, the ElectroDynamic Debris Eliminator, Jerome Pearson et al. (PDF)

Now, the JWST will be deployed to a very large 800,000 kilometres (500,000 mi) radius halo orbit around the Sun-Earth L2 point, that is 1,500,000 kilometers (930,000 mi) from the Earth, around 4 times farther than the distance between the Earth to the Moon. So not exactly in Earth's shadow and it will still use large deployable sunshield, but that's a long way away from where we're littering our LEO or even GEO/GSO (Geostationary/Geosynchronous orbits) with GEO being the farthest of these frequently used orbits, at 35,786 kilometres (22,236 mi) above the Earth's equator (and its graveyard orbit a bit farther than that). JWST will be placed in halo orbit nearly 40 times as distant from the Earth as you'd expect any space debris cluttering our planet's orbits at.

Yes, it is a very good idea to place JWST around the Sun-Earth L2 point.


Do all orbiting objects have barycenters?

The barycenter of a system consisting of two objects of mass m1 and m2 with their centers separated by a distance d is always along the line connecting the two centers and m1/(m1+m2) of the way along the line from the center of m1 to the center of m2.

So for the Sun and the Earth, where the Sun weighs in at about 333,000 Earth masses, the barycenter will be about 1/333001 of the way from the center of the Sun to the center of the Earth, or about 280 miles from the center of the Sun in the direction of Earth.

(So it might as well be fixed in place for all you or I could tell without high-quality instruments, but it does wobble in a 270-mile orbit once each year.)

The presence of the rest of the planets makes this all a lot more complicated, of course and the Sun actually traverses a much more complicated path than it would if the Solar System was a simple two-body system. In this case, the barycenter is possibly better thought of as the center of mass.

OTOH, since Jupiter is so much heavier than everything else, to a decent first approximation the Solar System is a binary system. The barycenter for the Sun-Jupiter pair is actually outside the Sun, at about 1.09 Solar radii.

See my answer at What point does Earth actually orbit? for more.


The process of determining photometric redshift includes, e.g.,:
1. objects of interest
2. types of data (e.g., one epoch multiple filters, one filter multiple epochs)
3. method of analysis (including model, and algorithm)

From what you mentioned, those are just algorithms. So, you might take a look around on other aspects as listed.


How tall are the "fractured" linear structures on Europa's surface?

Apparently they have been inferred to be "220 meters". See the references in:

S. D. Kadel, S. A. Fagents, "TROUGH-BOUNDING RIDGE PAIRS ON EUROPA", , Lunar and Planetary Science XXIX. (PDF)


How much larger would a star have to be to cause thermonuclear reactions if it was made out of mostly rock like Earth, instead of gases?

It really depends what you mean by "rock". At the temperatures and pressures at the cores of stars (and at which nuclear fusion reactions are possible), "rocks" as I suspect you are thinking of, do not exist.

Thermonuclear reactions do not occur because the gas is "flammable", they occur because the kinetic energies of the nuclei in the gas (at these temperatures, atoms are fully ionised) are sufficient to get them close enough together for nuclear fusion to take place.

"Rocks" are made of atoms of silicon and oxygen (for example) in the form if silicates. But these are dissociated at fairly low temperatures compared to the centres of stars. Oxygen and Silicon thermonuclear fusion ignition requires temperatures in excess of $10^9$ K, and these temperatures are only reached late in the lives of stars of mass $>8 M_{\odot}$.

It is possible to have stars with solid cores. This is thought to be the fate of white dwarf stars as they cool. Most white dwarfs are made of a mixture of carbon and oxygen and this "crystallises" once the core of the white dwarf cools below about a few million degrees.

Ordinarily, the cores of white dwarfs are inert as far as nuclear reactions are concerned, because the temperatures are too cool. However, if the white dwarf is massive enough (or mass is added to it), then the central densities climb, and for white dwarfs of mass $\sim 1.38M_{\odot}$ it is thought that the densities become high enough ($\sim 10^{13}$ kg/m$^3$) to start nuclear fusion of carbon via the zero point energy oscillations in the crystalline material (so-called pyconuclear reactions). Such reactions in degenerate matter are highly explosive and might result in the detonation of the whole star in a type Ia supernova.


Can I build a telescope with a lens from a CD-ROM as eyepiece?

This article gives a construction guide for how to make an eyepiece using a CD ROM player lens: How to make an eyepiece for a telescope.

As to what the focal length of your  eyepiece will be, you can refer to this paper, in particular Figure 1 from which I derived the following relation:

$f=0.73r$

where $f$ is the focal length of your lens, and $r$ its radius (this is a rough approximation, of course). So given the fact that a typical CD drive lens is about 5mm in radius, its focal length will be only a few millimeters.

Now, assuming the focal length of your telescope is 1000mm, the magnification will be a few hundred times. What you will be able to see with that depends on the width of your objective lens and how well you will have built your eyepiece, but you should probably be able to see the Moon with your custom eyepiece. As for the planets, it will be hard to get something good given how much you are magnifying—not impossible, but be aware that magnifying several hundred times requires a good weather, a good telescope and a good eyepiece.


What types of signals is SETI (and other such efforts) looking for?

What signals do SETI receive?
SETI uses radio telescopes to look at large portions of the sky across a range of wavelengths. They believe that we are most likely to find alien civilisation buy looking for microwaves.

How would we know if they were from aliens?
There are radio signals from many different sources bouncing around the universe, many of which we don't quite understand yet. However, natural sources of energy tend to produce a range of wavelengths. SETI are hoping to find "narrow-band signals", which  are strong signals at a single wavelength. These (as far as we know) only come from devices made to produce them, so only a technologically advanced species would emit these signals.

Is there a maximum distance for detecting these?
Yes, these signals would decrease with distance. It is not so much that they become indistinguishable form natural sources, as they are a different type of signal, but as the waves travel they spread out, like dropping a pebble in a pond, so the signal is much weaker as you get further away. Also, space is not empty. If the waves are emitted from a planet in a galaxy, there will probably be lots of dust and gas (interstellar medium) in the galaxy which will absorb the radio waves, so it might be like looking for car headlights on a foggy day - we will see something vague and fuzzy but it will be much weaker if there is more matter between us and them.

If you want to find out more about SETI they have a website with FAQs and news. If you want to get involved there is a project called SETI@home which uses spare processing power on your home computer to analyse the data they are receiving and search for these narrow-band signals.


How close are we to be able to detect and measure gravitational lensing inside the milky way?

Well, there are microlensing experiments, the results of which have largely removed MACHOs as a significant source of Dark Matter.  The background field of stars is usually a nearby galaxy (a Magellanic cloud one or Andromeda, say), or the galactic bulge, and the experiment looks for lensing objects between us and there.  In particular, they're pretty much always looking for the microlensing effect within our own Milky Way or its halo.  These experiments have detected a number of microlensing events (and many, many more candidates that were due to other things), so, yes, this is not only possible, it is already done.

The wiki page I started with linking has this to say, in particular:

The first two microlensing events in the direction of the Large Magellanic Cloud that might be caused by dark matter were reported in back to back Nature papers by MACHO and EROS in 1993, and in the following years, events continued to be detected. The MACHO collaboration ended in 1999. Their data refuted the hypothesis that 100% of the dark halo comprises MACHOs, but they found a significant unexplained excess of roughly 20% of the halo mass, which might be due to MACHOs or to lenses within the Large Magellanic Cloud itself. EROS subsequently published even stronger upper limits on MACHOs, and it is currently uncertain as to whether there is any halo microlensing excess that could be due to dark matter at all. The SuperMACHO project currently underway seeks to locate the lenses responsible for MACHO's results.

Despite not solving the dark matter problem, microlensing has been shown to be a useful tool for many applications. Hundreds of microlensing events are detected per year toward the Galactic bulge, where the microlensing optical depth (due to stars in the Galactic disk) is about 20 times greater than through the Galactic halo. In 2007, the OGLE project identified 611 event candidates, and the MOA project (a Japan-New Zealand collaboration) identified 488 (although not all candidates turn out to be microlensing events, and there is a significant overlap between the two projects). In addition to these surveys, follow-up projects are underway to study in detail potentially interesting events in progress, primarily with the aim of detecting extrasolar planets. These include MiNDSTEp, RoboNet, MicroFUN and PLANET.


Unable to locate Big Dipper through Polar Axis Finder

My guess is that the drawing is a "cartoon" and not to scale.

The declination of Polaris is +89° 15′ 50.8″ so it is only 0.746° away from the pole, (where the "+" is in the center of your drawing).

Since the inner circle's 143 pixel diameter corresponds to 1.472°, the outer circle's 824 pixel diameter is only 8.48°, or 85.76 North declination. The big dipper and Caseopia are at 60° to 50° North declination, way way outside the view of your scope.

They are shown only to provide a general direction to help decide how to rotate your finder.

 Source


Calculating the Angular Momentum of a planet

The big essential fact about momentum is that it is conserved (in the presence of a central force, as is the case here). So the Angular momentum of Pluto today is the same as it was yesterday, and the same as last year and (excepting perturbations) the same as it has ever been.
It is easiest to calculate for a body that is moving perpendicular to its position vector. This is always true for circular motion, about the centre of the circle.  It is not true for elliptical motion, except at apoapsis and periapsis.
At these times $L = mvr$.  For Pluto the periapsis speed is $v = 6.10\, km/s$ the distance is 4.44 billion km and the mass is $1.31 × 10^{22}$ kg.  To get the angular momentum you've got to multiply them together. If you want SI units, convert those km to m first.
The angular momentum today is the same.
Alternately you can use the relationship
$L= \sqrt{\mu \ell}$ where $\mu = GM =1.33×10^{20}$ and $\ell$ is the semi latus rectum or $\ell=a(1+e^2)$, and you have to plug in the semimajor axis for Pluto and eccentricity of its orbit.
Or you can look it up, since it is constant $L=3.6\times 10^{38}$


Mathematically, the motion of the Pluto-Charon system can be decomposed into two parts: The motion of Pluto-Charon about the Sun, and the motion of Pluto and Charon about one another. If one sets the reference point to be the center of Pluto, the path the Pluto-Charon system would appear to follow about the Sun would be an epicycle, which is a far more complicated means of describing the pair's trajectory through space. By instead setting this reference point to be the barycenter, the trajectory followed by the Pluto-Charon system is conveniently a Keplerian ellipse.

Charon is 11.6% the mass of Pluto, and is (on average) 19,571 km away from Pluto (Source 1, 2). The barycenter is the point at which the masses of the two bodies "balance", and for the Pluto-Charon system it lies at a distance of $$\frac{m_{\text{Charon}}}{m_{\text{Pluto}}+m_{\text{Charon}}}\times\text{distance}=\frac{0.116}{1.116}\times\text{distance}=0.104\times 19,571\,\text{km}=2,034\,\text{km}$$ from the center of Pluto. As Pluto is only 1,153 km in radius, the barycenter lies ~900 km above its surface. The Pluto-Charon system is the only known (minor or major) planet for which this is the case, although the Earth-Moon system will likely satisfy this criteria billions of years in the future (see Is the Moon a Planet? on Physics SE). There are some instances of binary asteroids, however, for which the barycenter lies outside the surface of both bodies.


Do we observe Mars Eclipse or Jupiter Eclipse?

No.

There are two types of eclipse: A lunar eclipse is when the moon goes into the Earth's shadow. We can see a lunar eclipse because the moon is so close to the Earth. If you were standing on the moon, The Earth would look bigger in the sky than the sun. If the Sun, Earth and moon are aligned, the Earth will block out the light from the sun.

Mars and Jupiter are further away. If you were standing on Mars, The Earth would look like a bright star, it would be too far away to block all the light from the sun, even if the Sun, Earth and Mars were perfectly aligned.  So there can be no eclipse.

The second type of eclipse is a solar eclipse, when the moon blocks of light from the sun. Since Mars and Jupiter orbit further from the sun than the Earth, they can never block of its light. Venus and Mercury sometimes do cross the face of the sun. This is called a "Transit" of Venus. But all the planets are too far away to block of all the light, so there is no eclipse.


Is the light we see from stars extremely old?

Yes, the speed of light in vacuum (or c) is 299,792,458 m/s and one light-year is the distance the light travels in one Julian year (365.25 days), which comes out as 9.4605284 × 1015 meters. Since c is the maximum speed at which all energy, matter, and information in the Universe can travel, it is the universal physical constant on which the light-year (ly) as one of the astronomical units of length is based.

That means that visible light as an electromagnetic radiation cannot travel faster than c and in one Julian year it can traverse a maximum distance of

  d = t * c

d is distance in meters

t time in seconds

c the speed of light in vacuum in meters per second

If we calculate this distance for a 4.243 ly distant object, that comes out as 4.243 * 365.25 * 86,400 s * 299,792,458 m * sˉ¹ or exactly 40,141,879,395,160,334.4 meters (roughly 40 trillion kilometers or 25 trillion miles).

That is the distance the light traveled since it was last reflected of (or in our case emitted from, since Proxima Centauri is a red dwarf star) the surface of a celestial object to be 4.243 Julian years later visible at our observation point, in this case our planet Earth from where the distance to Proxima Centauri you quoted was measured.

The more powerful the telescope, the further into the past we can see because the light is much older! This goes the same regardless of the distance of the object you're observing, but astronomy is particularly neat in this regard and we can observe objects that are so distant that we see them from the time when they were still forming.

For further reading on other units used to measure far away objects you might be interested in reading this question on the parsec.


So far there have not been any published measurements of the albedos of the TRAPPIST-1 planets. While albedos have been measured for some exoplanets, the planets in question are irradiated far more severely than even the innermost TRAPPIST-1 planets.
Using the formulae in my answer to this question about computing planetary apparent magnitudes, we can make a back-of-the-envelope check. The definition of absolute magnitude becomes the apparent magnitude of a planet as the magnitude it would have at opposition if it were located 1 au from TRAPPIST-1, and the observer located 1 au from the planet.
Defining the absolute magnitude of the TRAPPIST-1 planets as the apparent magnitude if the planet were 1 AU from TRAPPIST-1, and the observer were 1 AU from the planet, the uniform disc model gives the following formula for the absolute magnitude $H$:
$$H = 5 \log_{10} \left( \frac{671\,000\ \rm km}{D_{\rm p} \sqrt{A_{\rm g}}} \right)$$
Where $D_{\rm p}$ is the diameter of the planet and $A_{\rm g}$ is the geometric albedo. The 834,000 km value is the equivalent of the 1,329 km value used for asteroid diameter estimation in our Solar System, adjusted for the fainter absolute magnitude of TRAPPIST-1 (using the V-magnitude 18.798 and parallax 80.4512 mas from SIMBAD, I calculate an absolute magnitude of 18.33).
The least favourable case for visibility of an outer planet from an inner one would be for planet h viewed from planet b: TRAPPIST-1h is both the smallest and outermost of the TRAPPIST-1 planets. To pick an unfavourable case for visibility, let's use a geometric albedo of 0.05 similar to the P-type asteroid 46 Hestia. Using a diameter of 0.752 times the Earth (NASA Exoplanet Archive), the absolute magnitude works out as roughly 12.5. The apparent magnitude is then given by
$$m_{\rm p} = H + 5 \log_{10} \left( \frac{d_{\rm p\ast} d_{\rm po}}{1 \mathrm{\ au}^2} \right) - 2.5 \log_{10} q(\alpha)$$
where $d_{\rm p\ast}$ is the star–planet distance, and $d_{\rm po}$ is the planet–observer distance, and $q(\alpha)$ is the phase integral at phase angle $\alpha$.
Again using the uniform disc model which gives $q(0) = 1$, the apparent magnitude of h from b, when b and h are located on the same side of TRAPPIST-1 is about -0.2 for the Hestia-like case, which is similar to Saturn viewed from Earth. On the other hand, the angular diameter in this configuration is about 3.7′, so unlike Saturn it would be visible as a disc rather than a point.


What does "forward facing" mean for space telescopes?

This can best be described by two slides from this presentation.

"Forward-facing" implies looking towards Earth in the spacecraft's orbit, in the direction of the spacecraft's velocity vector:

"Backward-facing" implies looking in the opposite direction, away from Earth and in the opposite direction to that of the spacecraft's velocity vector:

These are from the C9 campaign, which also used a forward-facing orientation.


Are the planets Mercury, Saturn, Uranus and Neptune losing mass?

Anything that's working, spending energy is losing mass,  Einstein Mass-energy equivalence of SR along with radiant energy(electromagnetic and gravitational radiation) would be effecting all the planets, but at a different rate for each one.  https://en.m.wikipedia.org/wiki/Radiant_energy
  And.   https://en.m.wikipedia.org/wiki/Mass_in_special_relativity
  These should help I think


Data on the composition and orbits of asteroids

The JPL Small-Body Database Search Engine is a good place to start.
If diameter < 20 m, then probably absolute magnitude H > 25.
Fortunately we know of over 4000 asteroids in that size range, mostly after close approaches to Earth.
Unfortunately the orbit and magnitude are all we know about most of those, due to limited observational data.
JPL SBDB lists only about 40 asteroids with known Tholen spectral type M, all with perihelion q > 1.5 AU and diameter > 30 km.
The picture is a little brighter for type S, with over 300 known, a few with diameter < 2 km and minimum orbit intersection distance < 50 LD.


There is no connection between the date of the solstice and the perihelion. It is merely a minor coincidence that perihelion occurs close to the solstice.

The relationship isn't fixed. Precession in the Earth's orbit (caused by gravitational perturbations of Jupiter and other planets) will change the relative time of perihelion and solstice over a period of thousands of years.


Why aren't secondary mirrors offset to get rid of diffraction spikes due to the support vanes?

There is a 1.6 meter off-axis telescope for solar observation at Big Bear Observatory called the New Solar Telescope (NST).  Off-axis is particularly helpful in solar observations because it reduces stray light and eliminates a rotating diffraction pattern which can be severe when the sun is the target.  However, for most astronomical work the additional costs are just not worth it, yet. The extra mechanical costs may be greater than the extra costs for the optics.  The total weight of optics and supporting structures of professional telescopes can be many tons.  There is additional strain required to rotate and support an asymmetric structure like this, i.e. bigger motors are needed.  Galaxies are not bright enough to produce diffraction patterns and the issue of stray light from nearby bright objects is far less of a problem than the foreground stars, zodiacal light, and night sky light actually in the line of sight.  If a nearby star is producing a bothersome diffraction spike, one can usually rotate the field to avoid this. I am not saying that stray light or diffraction effects are never a problem, just that this is not a top problem and, if we need to trade down in size at all for an off-axis design, most astronomers would not accept it. Perhaps if relative costs for these telescopes come down in the future, they will be more common.


At what depth below the Sun's surface does the density reach that of water?

The sun's density is 1 gm/cc at approximately 50% of the way down towards the core. if radius of Sun is R then at R/2 the density will be that of water.

@astrosnapper's comment links to this answer in Physics SE. A reverse image search of that unsourced graphic leads to

Thesis: Multi-spacecraft analysisof the solar coronal plasma
NASA: The Solar Interior
Whatever this is shown below
the link in that image http://www.sns.ias.edu/~jnb/SNdata/Export/BP2004/bp2004stdmodel.dat provides some raw data
the reference in that image is standard solar model BP2004 J. N. Bahcall and M. H. Pinsonneault, Phys. Rev. Lett. 92 (2004) 121301 What do we (not) know theoretically about solar neutrino fluxes? (also in ArXiv)


How can gravity lead to the Big Crunch scenario?

The amount of matter in the universe is directly related to the curvature of space itself.  We can look at the Friedmann Equations to see how this works:
$$
H(t)^2 = \frac{R'(t)^2}{R(t)^2} = \frac{8\pi G}{3}(\rho_{m} + \rho_{r}) + \frac{1}{3}\Lambda - \frac{c^2}{R^2\mathscr{R}^2}
$$
This is the equation for determining the scale factor for finding distances in the universe at any given time.  Here, $H$ is the Hubble Constant, $R$ is the scale factor, $G$ is the gravitational constant, $\rho_m$ is the density of matter (both dark matter and baryonic matter), $\rho_r$ is the density of radiation (photons), $\Lambda$ is the cosmological constant due to dark energy, and $\mathscr{R}$ is the radius of curvature of space.

Next, we define a quantity we call the "critical density" ($\rho_c$).  This is the density of matter needed to make the universe go from an "open", hyperbolic geometry to a "closed", spherical geometry.  $\rho_c=\frac{3H^2}{8\pi G}$. We create a value $\Omega=\frac{\rho}{\rho_c}=\frac{8\pi G \rho}{3H^2}$ to take into account this critical density, making our equation:

$$
H^2 = \Omega_m + \Omega_r + \Omega_\Lambda - \frac{c^2}{R^2\mathscr{R}^2}
$$

From here, we input our present-day values for everything.  We define $R$ to be 1 and $H$ to be $H_0$ at the present day.  We also choose to neglect the density of radiation and dark energy.  Simplifying the equation a little gives us:

$$
\frac{c^2}{\mathscr{R}^2} = H_0^2(\Omega_{m,0} - 1)
$$

From here, we can see that the curvature of the universe is dependent on $\Omega_{m,0}$, which is directly related to the density of matter in the universe.  Specifically, for $\Omega_{m,0}\gt1$, $\mathscr{R}$ will be positive, meaning this will be a spherical, closed universe.  If $\Omega_{m,0}\lt1$, $\mathscr{R}$ will be negative, meaning the universe will expand forever, having hyperbolic curvature.  If $\Omega_{m,0}=1$, then $\mathscr{R}=\infty$, which is a flat universe.

So, we can see that having a dense enough universe, gravity will curve space and make it spherical.  You can also use the Friedmann Equation to calculate the deceleration parameter of space - how quickly the expansion is slowing or speeding up:

$$
q_0=\frac{\Omega_{m,0}}{2}
$$

Here, we can see that in a positively curved universe, $q_0>0$, meaning that the universe's expansion will be decelerating.  Eventually, the expansion rate will be negative, and then it will begin collapsing in on itself.

One thing to keep in mind when thinking about this is that gravity can, in a way, be considered a 'fictitious force'. Gravity is the force objects feel as they attempt to travel in straight lines through curved spacetime.  Matter is what is responsible for the curving of spacetime, so by increasing the density of matter in the universe, you increase the curvature, and thus make objects want to move closer together, which will increase the density, which increases the curvature.  Thus you have a feedback cycle, where you will ultimately collect all matter at a single point, and $\mathscr{R}=0$, so spacetime will have a 0 curvature radius, meaning spacetime has collapsed as well.


How is the heat production inside moons compliant with Conservation of Energy?

The heat production inside moons is due to tidal heating. Due to this, the orbit of the moon would change.
The orbital and rotational energy of the moon is dissipated as heat either in the surface ocean or in the interior of the moon. This is especially seen in the moons of large planets, for example Io.
Io, being Jupiter's innermost moon, is pulled by Jupiter on one side and the other moons on the opposite side. Due to this, Io's orbit is highly elliptic. During this orbit, the shape of the Io changes continuously (the surface flexes by ~100m). This causes a large amount of friction in the moon.

"Tidal heating on Io" by Lsuanli - Own work. Licensed under CC BY-SA 3.0 via Commons.
Usually, the tidal heating leads to tidal dissipation and circularizes the orbit. However, it hasn't happened in Io because it is in orbital resonance with other moons of Jupiter.
If the orbital period of the moon is less than the rotational period of the planet, the moon is decelerated if the moons are out of the synchronous orbit, the opposite thing happens, conserving energy and angular momentum.
On the other hand, Europa, another of Jupiter's moons is tidally locked (it has same orbital period as the rotational period of the larger body) to the Jupiter, which causes tidal flexing on the surface.


Has the most luminous object in the universe been found?

As it stands I think the answer is no. The most luminous object in the (observable) universe is probably hidden behind many magnitudes of dust extinction. There are lots of faint, infrared-detected objects that may be extremely luminous galaxies in the early universe (and indeed some have turned out to be) but it is difficult to get their redshifts, so it is difficult to be sure we have a complete luminosity function. We now know that objects with central black holes $>10^{9} M_{\odot}$ already exist in the universe after a billion years or so. To get to that size there must have been an awful lot of accretion and an awful lot of accretion luminosity.

In addition, the most luminous galaxy/quasar is going to be a rare object and only ridiculously small areas of the sky are covered by things like the HST ultra-deep field. i.e. There is almost bound to be a more luminous one in some other patch of sky that has not (yet) been surveyed. The logistical problems of getting a UDF over the whole sky (or even a significant fraction of the sky) means that isn't going to happen any time soon.

Finally, the brightest quasar is certainly not the most luminous. 3C273 is quite a modest object, in terms of luminosity, but it is very close to us as quasars go. But perhaps you didn't mean bright in that sense? Brightness is normally reserved to refer to the apparent brightness. Luminosity is the extinction-free, distance-corrected power.


Gravitational attraction of Sun on distant object

The full equation for the time for an object to drop is
$$t =  \frac{ \arccos \Big( \sqrt{ \frac{x}{r} }\Big)  + \sqrt{ \frac{x}{r} \ ( 1 - \frac{x}{r} ) } }{ \sqrt{ 2 \mu }  }  \, r^{3/2},$$
where $x$ is the radius of the sun, $r$ is distance of the object, and $\mu=GM=1.327\times10^{20}$. (in SI units, so you will need to convert your distance to metres, and get a time in seconds.)

As $x<<r$ you can assume $\frac{x}{r}=0$ so the formula simplifies to
$$\frac{\pi r^{3/2}}{2\sqrt{2\mu}}$$

For an object released at 0.1,0.5 or 1 light year that is 3000000 years, thirty million years or a hundred million years.

For the speed on impact, you can assume that the body will reach very close to the escape velocity of the sun, which is about 620 km/s. It matters little whether the body is released from 0.1, 0.5 or 1.0 lightyears, because the body will pick up speed only when it approaches the sun.

The last part of you question is unclear. If the body is at rest relative to the sun, and the various gravitational affects of other stars are ignored, then it will fall towards the sun. If on the otherhand the body is at rest relative to the centre of the galaxy, then it is moving extremely fast relative to the sun, much faster than the sun's escape velocity. It won't fall anywhere near to the sun, instead the sun will deflect it slightly as it falls towards the centre of the galaxy.


I might venture into some math which is slightly above what you need, but I wanted to show you the reason behind the correct equation to use. To start with, the pressure due to radiation, in the most general sense, is given by

$$P_{\mathrm{absorption}} = \frac{\langle S\rangle}{c} \cos(\theta)$$
$$P_{\mathrm{reflection}} = 2\frac{\langle S\rangle}{c} \cos^2(\theta)$$

The top equation represents the case where some light hits your object and the light gets absorbed, and the bottom equation represents the case where light hits and bounces off your object (really it's re-emitted). Your case is the bottom equation so I'll work with that from now on and drop the reflection subscript. Without needing to explain exactly what it is, just accept that $\langle S\rangle$ is the average energy of the photon being absorbed or emitted. Now, that's for a single photon. If we have a lot of photons, generally the energy distribution of all the photons together is said to be described by some function, such as $I_\lambda$ (the subscript refers to wavelength). We can define the full pressure then by not only integrating over all wavelengths of all the photons, but importantly all the directions that the photons may hit or be emitted.

$$P = \frac{2}{c}\int_0^\infty\int_0^{2\pi}\int_0^{\pi/2} I_\lambda(\lambda) \cos^2(\theta)\sin(\theta)\:d\theta\:d\phi\:d\lambda$$

Note now that we've replaced $\langle S\rangle$ with $I_\lambda(\lambda)$. Also, since we're assuming this is pressure on a flat surface, we're not integrating over every angle, but only half of them $-$ any photon, in order to hit the solar sail, has to come from the direction the sail is facing, and can only reflect in the direction it came from, i.e., we're discluding photons "behind" the sail. For that reason, the $d\theta$ integral only goes to $\pi/2$.

If you don't understand the math at this point, that's fine but here's the important takeaway: If we only integrated over all wavelengths, you'd get the equation without the three that you listed. What that means is that the equation without the three doesn't take into account all the possible angles and orientations of the incoming photons. It simply assumes all your photons hit your surface perpendicularly, and then bounce off perpendicularly. This is of course not physically accurate as the photons can come in from a variety of angles.

If you do the angle integrals out, you find that you get the following:

$$P = \frac{2}{c}\int_0^{2\pi}\:d\phi \int_0^{\pi/2} \cos^2(\theta)\sin(\theta)\:d\theta\int_0^\infty I_\lambda(\lambda)\:d\lambda$$

$$P = \frac{2}{c}\frac{2\pi}{3}\int_0^\infty I_\lambda(\lambda)\:d\lambda$$

You can see there in the last line that the three popped out of those integrals over all the possible angles (specifically the $d\theta$ integral). Now, you get the rest of this by assuming some sort of function for $I_\lambda(\lambda)$ which describes the energy distribution of the photons as a function of wavelength. Since your light comes from a star, you know that's just going to be the Planck distribution (specifically the equation for $B_\lambda(\lambda,T)$). This integral is non-trivial, but if you again run through the math, you find

$$P = \frac{2}{c}\frac{2\pi}{3}\int_0^\infty \frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda k_BT}-1}\:d\lambda = \frac{2}{c}\frac{2\pi}{3}\frac{\sigma T^4}{\pi}$$

$$\boxed{P = \frac{4\sigma T^4}{3c}}$$

This is the true equation for radiation pressure on the solar sail, assuming reflection is occurring. This includes the three because we need to account for the photons hitting and bouncing off at all possible angles (for the solar sail). Note one important differences here is that my equation already has the extra "2" due to reflection factored in. Your last equation differs from mine in that it throws in that extra "2" for reflection. So either it erroneously included the factor of "2" twice, or (more likely) its account for reflection at every angle, not just all possible angles. Remember above, we only integrated the $d\theta$ angle up to $\pi/2$ for the reason explained there. If we integrated up to $\pi$ to include every angle, then we'd get an extra "2" as well and my equation would match your last one. Your solar sails though, don't receive or re-emit photons at every angle, and so my equation differs by that factor of "2".


Are 3D coordinate data from Sloan DSS-III available & easily accessible to non-pros?

SDSS DR12 Catalog Data
looks like a good starting point,
apparently pretty open to those willing and able to figure it out.
Their
SciServer Compute
site hosts Jupyter notebooks to query CasJobs in SQL.

The Large Scale Structure galaxy catalog under
BOSS value added catalogs
may also be relevant.


Formula for sun's atmospheric refraction below the horizon

Without taking temperature and pressure into account, from my old code I found this approximate formula: (angles in degrees)

$\Delta \alpha_{reflect} = tan[(90°-\alpha_{technical}) \times 0.9679] \times 0.023626$

If this calculated reflection angle exceeds $\frac{34.9}{60}$ degrees, use that.

Observed elevation would be $\alpha_{observed}=\alpha_{technical}+\Delta\alpha_{reflect}$

Example

a point technically (without atmosphere) 34.9' (minutes of arc) below horizon, appears on the horizon.
a point technically on the horizon, appears 28.1' above the horizon.


An answer to your question is contained within What is the largest hydrogen-burning star? The hottest observed main sequence stars are of type O3V, with photospheric temperatures of about 50,000 K.

However, it is indeed possible that hotter main sequence stars may exist in the present-day universe, but have simply evolved into Wolf-Rayet stars (and lost a lot of mass). Indeed, Crowther et al. (2010) claim to see evidence for such objects in the cluster R136 in the Large Magellanic Cloud.

How hot could such objects be? A theoretical study by Bromm et al. (2001) suggests that very massive, metal-rich present-day stars might reach 65,000 K at a few hundred solar masses (see their Fig.1).

The hottest zero age main sequence stars ever would probably be very massive ($\sim 1000M_{\odot}$) ultra-metal poor, or even metal-free population III stars. These actually begin life as He burning stars until they produce enough carbon to commence the CNO H-burning cycle. Equation 6 and Fig.1 in Bromm et al. (2001), who provide a theoretical study of such objects, suggests effective temperatures of about $1.1\times 10^{5}$ K for such stars, with very little detailed dependence on mass and metallicity.

I guess such a star might be classified as O1, but according to Bromm et al. they are basically blackbodies once the mass exceeds about 300 solar masses.


The mass of the Sun is determined from Kepler's laws:

$$\frac{4\pi^2\times(1\,\mathrm{AU})^3}{G\times(1\,\mathrm{year})^2}$$

Each term in this component contributes to both the value of the solar mass and our uncertainty. First, we know to very good precision that the (sidereal) year is 365.256363004 days. We have also defined the astronomical unit (AU) to be 149597870700 m. Strictly speaking, the semi-major axis of the Earth's orbit is slightly different, but by very little in the grand scheme of things (see below).

At this point, we can solve for the product $GM$, known as the gravitational parameter, sometimes denoted $\mu$. For the Sun,
$$\mu_\odot=132712440018\pm9\,\mathrm{km}^3\cdot\mathrm{s}^{-2}$$

So solve for $M_\odot$, we need the gravitational constant $G$, which, as it turns out, is by far the largest contributor to the uncertainty in the solar mass. The current CODATA value is $6.67384\pm0.00080\times10^{-11}\,\mathrm{N\cdot m}^2\cdot\mathrm{kg}^{-2}$, which all combines to give
$$M_\odot=1.98855\pm0.00024\times10^{30}\,\mathrm{kg}$$
where my uncertainty is purely from the gravitational constant.

The value $1.9891\times10^{30}\,\mathrm{kg}$ (and nearby values) probably come from an older value of the gravitational constant of $6.672\times10^{-11}\,\mathrm{N\cdot m}^2\cdot\mathrm{kg}^{-2}$, which is still supported by some measurements of $G$.


What happens when a huge star is being consumed by a black hole

A large star near a black hole may lose some of its atmosphere by having it spilling over its Roche lobe and into an accretion disk around the black hole. That disk will also radiate significant amounts of UV and X-rays that can blow away more atmosphere. That could in principle lighten the star, lowering the pressure and temperature in the core, and extend its lifespan a bit. Usually the amount of mass lost is just a few percent, so this is not a big difference. However, as discussed in the first link, for some heavy stars significant amount of matter may be lost (especially since the star itself may radiate so much that it loses its outer envelope). It does not prevent the star from eventually going supernova, but can change what kind of object it turns into (neutron star or black hole).


How did the Milky Way and Andromeda Galaxy separate before they began their collision course?

I thought it was a fun question, I gave it some thought.  Layman's version, so corrections are welcome.

The very early universe, the first few seconds or so, it was quite dense.  This is fairly well publicized and the specifics and details don't really matter for this question, but the energy and stuff in the early universe that later formed the Milky-way and the energy and stuff that formed Andromeda were cosmic neighbors and in the first seconds of the universe, quite close to each other.  The basic Hubble rule is that things that are close (in cosmological terms) were always close, like the dots on a balloon that gets blown up.  Everything moves away from everything else, but things that were next to each other were always next to each other.  On smaller scales, like stars within galaxies or planets, this isn't true because of gravitational assists and different orbital paths, but its true for galaxies.   (there might be some small variation on this with "orbiting" dwarf galaxies, but it's still by and large, true).

Another important point to make about the early universe is that it was much more uniform, so, because of the uniformity, gravity was pulling nearly equally in all directions.   The nearby somewhat more massive clouds of matter and energy didn't dominate gravitationally because there was similar mass in all directions.    You might think that a denser universe means stronger gravity in the direction of nearby matter, but that doesn't happen if the matter is spread evenly enough in all directions.

So, generally speaking, in the very young universe, the stuff that became the Milky way and the stuff that became Andromeda started out much closer, but they didn't pull each other much early on because of uniformity.

Over time, local gravity overcame uniformity.  I couldn't tell you when that happened (maybe someone here can), but it took some time before the makings of galactic groups and clusters began to form.   We know from the background radiation map that the Universe was still quite uniform as late as 380,000 years after the big bang.   See picture:

(Edit due to correction).  And at 380,000 years, the known Universe is thought to be about 1/1100th it's current radius.   We know from the cosmic background radiation that at that time, the Universe was still largely uniform, though non-uniformity did lead to the formation of galaxies, clusters and super-clusters, so it was important, but it still took some time for local groups and galactic clusters to form.

See here.

  On this map, the hot regions, shown in red, are 0.0002 Kelvin hotter
  than the cold regions, shown in blue.

and

  These cosmic microwave temperature fluctuations are believed to trace
  fluctuations in the density of matter in the early universe, as they
  were imprinted shortly after the Big Bang. This being the case, they
  reveal a great deal about the early universe and the origin of
  galaxies and large scale structure in the universe.

But once local gravity overtook uniformity, then galaxies can start drawing each other together more efficiently.   The spiderweb pattern of galaxies shows that there is a tendency for this to happen, as galaxies draw other galaxies towards them, even as space expands, so you get lines of more dense regions with more galaxies and pockets of empty space, mostly free of galaxies.

Source of picture for further reading.

So, what basically happened with Andromeda and the Milky-way is that expansion combination with uniformity set them fairly far apart even as they were still forming, but they were still next door neighbors as far as galaxies go.

Both the Milky-way and Andromeda have absorbed smaller, closer dwarf galaxies, perhaps many times, so they weren't the closest 2, but they were the closest 2 non dwarf galaxies (at least, that we know of).

You might think that galaxies could orbit other galaxies, but the nature of Hubble expansion doesn't give galaxies much tangential velocity towards other nearby galaxies, so basically, 2 galaxies are either moving towards each other due to mutual gravity or away from each other due to expanding space.   Now in a 3 or more galaxy system with similar masses, they can get some tangential velocity, but generally speaking, the direction between any 2 large galaxies is mostly either towards or away, not an orbit.

Another thing to keep in mind, once the uniformity effect is overtaken by local gravitation of local groups or clusters, is that massive objects at these distances don't exactly pull each other quickly.   It takes a long time for gradual acceleration to turn into relevant velocity.

At our current distance from Andromeda (about 2.5 million light years), and it's current mass (about 1.5 trillion solar masses), it's G force on us, using the formula below (Math can be added if needed):

comes to about 1/2.8 trillionths of a meter per second squared.  That's less than 1/20,000th the gravitational pull Pluto has on Earth.   A pull that small is going to accelerate things towards each other extremely slowly, so (I think) for the first billion or few billion years, the expansion of space between the Milky way and Andromeda likely exceeded any gravitationally caused velocity between them.

The good news is, Andromeda and the Milky-way have had a long time to pull on each other.  About 13 billion years or so.   1/2.8 trillion m/s^2 over a billion years is works out to about 11 km/s.  Over 13 billion years, 143 km/s.   And, the actual velocity that the Milky-way is moving towards Andromeda isn't too far away from that, about 110 km/s.  That right there gives us a ballpark Netwonian estimate for how far apart they were when they started.   (Now, footnote, for objects of somewhat similar mass, you have to take the mass of both of them, not just the larger one, so it's closer to 15 or 16 km/s per billion years), but that doesn't change the overall numbers too much.

Dark Energy expansion (which currently, between the Milky-way and Andromeda is about 60 km/s), and that makes the calculations a bit more tricky, especially since dark energy expansion might not have been constant throughout the last 13 billion years.  Related article here.

At the current rate they are moving together, about 110 km/s or, 250,000 mph or 1/2,680 c, In one billion years at current velocity Andromeda will be 375,000 light years closer (and, in reality, a bit more as it will continue to accelerate by mutual gravity and dark energy expansion will be reduced as they get closer).

And if we work backwards, a billion years ago it was a bit less than 375,000 light-years further away than it's current 2.5 million light year distance, with diminishing distances further away over previous billion year intervals.   The farther back you go, and difficulties with knowing how close they were when local gravity became a key factor over uniformity makes estimates pretty difficult, but they were probably never too much further apart than they are now.  Perhaps double, as a very rough guess.   Because gravitational acceleration drops off at the square of the distance, I find it hard to believe that Andromeda was ever much more than twice as far as it is now.

I would imagine, based on my assumptions above, that by the time the galaxies were recognizable as galaxies, say, when the universe was a billion or 2 billion years old (give or take), Andromeda and the Milky way were probably moving away from each other initially, but over time, the gravitational attraction was able to overcome overcome the expansion of space.


How many supernovae events are on their way from within the Milky Way?

The supernova rate is only estimated/known to a factor of 2-3, so high levels of accuracy are not possible or warranted.
If we assume high mass stars are born uniformly in a disc of radius  $r$  light years, then you can work out a "surface density rate" of supernovae of $1/100 \pi r^2$ per square light year, per year (adopting your rate of one per century over the Galaxy).
Then, consider a thin annulus of radius $x$ and width $dx$ around the Earth. The number of supernovae that explode per year will be
$$dN = \frac{2\pi x}{100 \pi r^2} dx$$
and the number that have exploded in the last $x$ years, who's light is yet to reach is just this multiplied by another factor of $x$.
If the Sun was at the centre of the Galaxy the calculation is then simple
$$N = \frac{1}{50r^2}\int_{0}^{r} x^2\ dx = \frac{r}{150},$$
with $r$ measured in light years. If $r\sim 30000$ light years (I think 50,000 is a bit big), then the number is 200.
Unfortunately, that isn't the geometry. Instead of being a circular annulus around the Sun, you have to work with an annulus that is truncated where it reaches the "edge" of the Galactic disc. I may add to this answer later, but I doubt this will change the number above by very much.


What is the lower limit of a coronal mass ejection?

No, as ‘coronal mass ejection’ is defined ambiguously and refers to an event that is not physically bound to some exact limit and therefore cannot have an undisputable lowest limit value, in the mathematical sense, in any parameter.
For example, the Oxford dictionary states:
“
coronal mass ejection
Noun - Astronomy
An event in which a large cloud of energetic and highly magnetized plasma erupts from the solar corona into space, causing radio and magnetic disturbances on the earth.
“
Clearly, any limit, if fixed to some exact value of mass, flux, duration, etc., would have to be fixed arbitrarily.
As a thought exercise, consider if someone came along and presented you such a ‘lower limit’.
You, or a third party, can respond with: “well what about the situation where an event occurs just one electron (or one Planck second, etc.) below such a limit? Would that be considered a CME or not?”


Oh sorry, I just understand what is going on. Being the far side of the moon, did not mean that it's never exposed to the sun.
So the Expression "Dark Side Of The Moon" is not so true.

Has  written in Wikidepia :

Tidal forces from Earth have slowed down the Moon's rotation to the point where the same side is always facing the Earth—a phenomenon called tidal locking. The other face, most of which is never visible from the Earth, is therefore called the "far side of the Moon". Over time, some parts of the far side can be seen due to libration.[7] In total, 59 percent of the Moon's surface is visible from Earth at one time or another. Useful observation of the parts of the far side of the Moon occasionally visible from Earth is difficult because of the low viewing angle from Earth (they cannot be observed "full on").

The phrase "dark side of the Moon" does not refer to "dark" as in the absence of light, but rather "dark" as in unknown: until humans were able to send spacecraft around the Moon, this area had never been seen.[1][2][3][4] While many misconstrue this to think that the "dark side" receives little to no sunlight, in reality, both the near and far sides receive (on average) almost equal amounts of light directly from the Sun. However, the near side also receives sunlight reflected from the Earth, known as earthshine. Earthshine does not reach the area of the far side that cannot be seen from Earth. Only during a full Moon (as viewed from Earth) is the whole far side of the Moon dark. The word "dark" has expanded to refer also to the fact that communication with spacecraft can be blocked while the spacecraft is on the far side of the Moon, during Apollo space missions for example.


Where do we define the "surface" of a gas planet?

There are two common definitions in use for the surface of gas planets:

The 1-bar surface: As pressure increases, the deeper in we go into the gas planet, we will hit a pressure of 1 bar at some altitude. Gas at this altitudes will usually sit deep enough in the gravitational well and be of a near-uniform density and temperature, as to not be influenced by exterior parameters, for example the solar wind. Therefore, the altitude of the 1-bar level will remain essentially constant, for short astronomical times.
The $\tau=2/3$-surface: This is the altitude, from which photons can escape freely into space. This happens at an average optical depth $\tau$ of 2/3. It is essentially what you see in your image as the limit of the black background. For the sun one end of the photosphere is the average $\tau=2/3$-surface, and for transiting exoplanets this is identical to the measured transit radius at that wavelength.

There is no hard relation between those two surfaces, but in general their altitude will not be different by more than a scale height, as at around 0.1-1 bar the gaseous atomic and molecular bands become enormously pressure broadened, which makes the atmosphere quickly opaque at most wavelengths, for the usual gas giant components.


Does supermoon's white light splits into seven colours

The answer by @comprehensible nicely states that whatever the effect might be, it is related to your camera. Of course the moon's light has a spectrum all the time, just like the sun, but we don't see it until we look through some dispersive material or object.

I thought I would just add a little back-up information as well.

I took your photo and rotated, cropped, and flipped it left-right on my laptop, then made 1D plots of the three color channels (R, G, B) normalized to local (R+G+B) by projecting down to the new x-axis. Note that this is very unscientifically done, but it's still interesting.

If the dispersion were diffraction, then moving away from the moon you'd expect to see the order B, G, R, B... since diffraction peaks are at larger angles for longer wavelengths. It would repeat if there were multiple orders. Of course this is messy geometry and the source (the moon) is very wide, so this is only guessing.

Surprisingly, this is the order that appears! So it's possible that there is some periodic pattern somewhere on the lens or in the camera. Since the characteristic size of one cycle is of the order of the diameter of the moon or about 0.01 radian, if there scratches or streaks on the lens the period owould be of the order of 100 wavelengths. There is an obvious periodicity in the camera's CCD (assuming it's not a film camera) but then the ray tracing would be more complicated (multiple reflections) and this simple analysis wouldn't work.

This is just speculation, and thin film interferece can also produce color effects also, and non-uniform thickness (like a soap bubble) can make color gradients as well.

Also note, our perception of color is a complex subject, and while our eye and brain together try to reconstruct a rich color behavior, the simple up and down wiggles of the intensity in the three color channels of your camera's sensor is all that is actually in the photo.

The color modulation is quite strong, so I am guessing that whatever is producing the 1 directional flair is also the thing that has the periodicity.

Why don't you try a similar experiment and point at strongly overexposed LEDs or maybe street lights (Sodium or Mercury vapor) and see if you get a clearer effect. This would be just for fun. This is just an artifact of your camera and has nothing to do with the moon.

Here's the Python script I used to plot the image after I rotated/cropped/flipped it on my laptop:

import numpy as np
import matplotlib.pyplot as plt

fname = "Moon Spectrum.png"

img = plt.imread(fname)

img1D = img.sum(axis=0).T[:3]

img1Dnorm = img1D/img1D.sum(axis=0)

plt.figure()

plt.subplot(2,1,1)
plt.imshow(img)

plt.subplot(2,1,2)
for thing, color in zip(img1Dnorm, ['red', 'green', 'blue']):
    plt.plot(thing, color=color)

plt.xlim(0, img1Dnorm.shape[1])

for lab, x, y in [('B',  58, 0.35), ('G',  80, 0.39), ('R',  97, 0.31),
                  ('B', 126, 0.36), ('G', 141, 0.46), ('R', 166, 0.40) ]:        plt.text(x, y, lab, fontsize=14)

plt.show()


General relativity can be used to discuss, theoretically, what happens to objects as they approach the event horizon of a black hole; it all has to do with your point of view. Outside viewers, those not affected by the gravity of the black hole, would view an object approach the event horizon slowly and pretty much never enter past the horizon. For the observers point of view time stops for the object. From the point of view of the object, its own time runs as normal, and it is the time outside of the event horizon that runs faster than normal. For more general information at a really easy to follow format you might check out What is Relativity by Jeffery Bennett.


How is the Lithium Depletion Boundary used to determine the age of a stellar cluster?

When low mass stars are very young, they are termed pre main sequence (PMS) stars. These PMS stars have larger radii than main sequence stars of the same mass, and energy transport in their interiors occurs primarily through convection. The convection ensures that the star is thoroughly mixed and chemically uniform.

As the PMS star radiates away its gravitational potential energy, it contracts. The virial theorem tells us that as it does so, its interior becomes hotter. Roughly speaking, the core temperature is proportional to $M/R$, where $M$ is the mass and $R$ the radius.

Nuclear fusion of hydrogen will not commence until the core temperature reaches more than 10 million K, however there are other fusion reactions that become possible at lower temperatures - namely deuterium burning at around 1 million K and then lithium burning at around 3 million K. The latter reaction is not energetically important in the star's life because there is not much lithium in the star to begin with (about 1 part in a billion), however this lithium can be observed in the photosphere of the PMS star (via the 670.8 nm Li I resonant absorption line) and the convective mixing means whatever we see at the photosphere also represents the abundance at the core.

The Li burning reaction is extremely temperature dependent (like $T^{20}$ or thereabouts), so it turns on like a switch once the core reaches the appropriate temperature (e.g. see Bildsten et al. 1997). The time it takes for a PMS star to reach this core temperature basically depends on its mass. More massive, and hence more luminous stars contract faster and reach the Li burning temperature quicker. Once they do so then the Li in the star is rapidly and thoroughly consumed by fusion. The relationship between the age at Li destruction and the luminosity of the PMS star at that time is the age-LDB luminosity relationship that you refer to.

The result is that if you look at a bunch of stars in a cluster (assuming they all have the same age), then the more massive, more luminous PMS stars will have destroyed their Li, whilst the lower mass, lower luminosity stars will still contain their original Li content. The luminosity at the reasonably sharp transition between these two regimes is known as the LDB.

LDB ages are arguably the most accurate way to find the ages of stars in clusters. All age determinations are to some extent dependent on what physical ingredients are in stellar evolution models, but the sensitivity of the age-LDB luminosity relationship to various uncertainties is quite weak (e.g. Burke et al. 2004) - we basically understand the physics of a contracting, fully convective ball of gas quite well. The LDB ages can also be precise, because the very sharp turn-on of Li burning and its rapidity should lead to a sharp transition between stars with low luminosities that have Li and stars with only slightly higher luminosities that don't.

Interestingly, in the last couple of years it has become apparent to us (and others) that there are some model ingredients that are not completely understood, namely the effects of dynamo generated magnetic fields and dark starspots. Both of these may lead to the suppression of heat transport, either throughout the star or just at the surface, slowing the PMS contraction so that at a given age, the PMS star has a cooler core temperature. This might delay the onset of Li burning and mean that the currently determined LDB ages are underestimates by 10-20% or so (e.g. Jackson & Jeffries 2014; Somers & Pinsonneault 2015 ).


Why didn't the Big Bang produce heavier elements?

I think that your thought process is flawed in that you assume that by drastically increasing the temperature you are guaranteed to get heavy elements. As odd as this may sound, this isn't the case (especially during the Big Bang Nucleosynthesis (BBN)) for a few reasons. In fact, if you took a hydrogen-only star and made it go supernova, you wouldn't get heavy elements like you see in current stars going supernova.

BBN Timescale

One major point to consider is that the BBN era is calculated to be only ~20 min long. That isn't really much time to form elements. Sure, supernovae happen in an instantaneous flash, but there are other things going on there, which I'll get to in a second. The main point here is that fusing takes time and 20 min isn't that much time to form heavy elements.

Deuterium

To get heavy elements, you need to build up to them. You can't just smash together 50 protons and 50 neutrons and get tin. So the first step is to smash together a proton and neutron to get deuterium, but here you already run into a problem known as the deuterium bottleneck. As it turns out, the huge temperatures actually (and somewhat counterintuitively) impede the creation of deuterium. This is mainly because the deuteron will end up having so much energy that it will be able to overcome the binding energy (and deuterium has pretty low binding energy being that its only two nucleons) and will likely break apart again. Of course, given the density and temperature you can still get a good amount of deuterium simply by force of will, but not as much and not at the rate you'd expect otherwise. Another point that makes deuterium form less frequently that you'd naïvely expect is that the proton to neutron ratio before BBN was about 7:1 due to the proton being more favorable to be created since it has a slightly lower mass. So 6 out of 7 protons didn't have a corresponding neutron to combine with and had to wait for deuterium to form first before it could combine with anything.

Tritium, Helium, Lithium, Oh My!

Deuterium is then the catalyst for forming all the next stages of particles in your soup. From here you can throw them together with various other things to get $^3\mathrm{He}$, $^3\mathrm{H}$, and $^4\mathrm{He}$. Once you've got a good amount of deuterium, tritium, and helium isotopes floating around, you can start making lithium and if you're lucky a bit of beryllium.

To Boron and Beyond

But now, once again you run into a bottleneck, and one more severe than the deuterium bottleneck. You can't easily jump to heavier elements with what you have on hand. The next fusion chain, and the way stars do it, is the triple-alpha process which helps to form carbon but to perform this chain and build up enough carbon you need a lot of time. And we only have 20 minutes! There just isn't time to form the carbon we need to progress along the fusion cycle. As I hinted at the beginning, pure hydrogen stars also wouldn't produce heavy elements upon supernova for this reason. They're able to produce heavy elements now because they've had billions of years before their SN event to build up a base amount of carbon, nitrogen, oxygen, etc. that can aid in the heavy element fusion processes.

So you don't have the time to follow the triple alpha process and make carbon — what about other processes? Surely the temperatures are high enough that you can do different fusion methods not seen in stars. Well... no. You can't even smash together lots of $\mathrm{He}$ or $\mathrm{Li}$ to get really heavy elements because of the fact that heavy nuclei are only stable if they have way more neutrons than protons. And we already said there was a big neutron deficiency early on so the chance you have enough neutrons hanging around to smash together to get, say $^{112}\mathrm{Sn}$ (that's tin with 62 neutrons), is pretty small. What's more, you can't even try to either skip carbon by making something slightly heavier or form something intermediary between lithium and carbon. Again, this is because of stability issues. So with no other options, you have to shoot for carbon after lithium, and as stated above, you just don't have time for that.

TL;DR

Overall, BBN is limited to getting only to lithium because of limited time, proton to neutron abundance ratios, and fusion bottlenecks that slow things down. All these come together to produce ~75% $^1\mathrm{H}$, ~25% $^4\mathrm{He}$, ~0.01% $^2\mathrm{H}$ and $^3\mathrm{He}$, and trace amounts of $\mathrm{Li}$.


Do all objects get larger in an "expanding universe"?

It would, if there were no interactions.

However, everyday matter is held together by other kinds of forces (electromagnetic, mainly). These are quite strong, and any "stretching" of space will be counteracted by these forces "pulling" the atom/molecule/object back to its original size.

So the expansion of the universe only matters on the larger, cosmic scale, between objects like galaxies which are separated by a large enough distance that they don't affect each other too much, gravitationally speaking.


Modern telescopes, even those that can be purchased from stores for, say, under $10000 (which is already a lot), were produced with computer-controlled equipment to polish the mirror and make the lenses and mirrors, and the lenses usually have nano-coating to reduce reflections and increase light transmission.

Overall you have a century of progress & experience building them since then. Mounts are better as well, mechanically and electronically (you have auto-guiding whereas it was manual a century ago, with an astronomer watching a star through a reticle). In other words, modern telescopes, even amateur ones, are vastly improved.

You can compare photos (from various books such as https://www.amazon.com/Day-We-Found-Universe/dp/0307276600) to modern astrophotography done by amateurs (eg https://www.astrobin.com/explore/top-picks/). The modern ones (also benefiting from sensors more sensitive than film) are much better.


When did the Moon become tidally locked to Earth?

"Protection" isn't the only effect of Earth. Here is a different POV: Earth may have accelerated impactors by gravity assist.

A different approch is the thinner crust, as suggested for the near side, which may have allowed asteroids to penetrate Moon's crust, such that lava could flow into the basins, or which may have favoured volcanism on the near side (see "Lunar interior" on this site).

A third approach is the protective property of Earth preventing the near side to be covered with many new craters, hence leave the maria visible.

According to Wikipedia the time to lock tidally is about
$$t_{\mbox{lock}}=\frac{wa^6IQ}{3G{m_p}^2k_2R^5},$$
with $$I=0.4m_sR^2.$$
For Moon $k_2/Q = 0.0011$, hence
$$t_{\mbox{lock,Moon}}=121\frac{wa^6m_s}{G{m_p}^2R^3}.$$
With Earth's mass $m_p=5.97219\cdot 10^{24}\mbox{ kg}$, Moon's mass $m_s=7.3477\cdot 10^{22}\mbox{ kg}$, Moon's mean radius of $R=1737.10\mbox{ km}$, $G=6.672\cdot 10^{-11}\frac{\mbox{Nm}^2}{\mbox{kg}^2}$we get
$$t_{\mbox{lock,Moon}}=121\frac{wa^67.3477\cdot 10^{22}\mbox{ kg}}{6.672\cdot 10^{-11}\frac{\mbox{Nm}^2}{\mbox{kg}^2}\cdot{(5.97219\cdot 10^{24}\mbox{ kg})}^2(1737.10\mbox{ km})^3},$$
or
$$t_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}wa^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}.$$
Parameters are $w$ the spin rate in radians per second, and $a$ the semi-major axis of the moon orbit.

If we take the the current simi-major axis of the moon orbit of 384399 km
and a maximum possible spin rate of
$$w=v/(2\pi R)=\frac{2.38 \mbox{ km}/\mbox{s}}{2\pi\cdot 1737.10\mbox{ km}}=\frac{1}{4586 \mbox{ s}},$$
with $v=2.38 \mbox{ km}/\mbox{s}$, Moon's escape velocity, 1737.1 km Moon's radius,
we get
$$t_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}\cdot \frac{1}{4586 \mbox{ s}}\cdot (384399\mbox{ km})^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}\\
=501416\mbox{ s}^{-1}\cdot \mbox{ km}^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}=
5.01416\cdot 10^{14} \mbox{ s}.$$
That's about 16 million years, as an upper bound.

If we assume a higher Love number for the early moon, or slower initial rotation, the time may have been shorter.

The time for getting locked is very sensitive to the distance Earth-Moon (6th power). Hence if tidal locking occurred closer to Earth, the time will have been shorter, too. That's likely, because Moon is spiraling away from Earth.


Galaxies are far away but very big. Planets are nearer but very small.

Even for planets that are not completely swamped by the much brighter light of their host star, their angular size is much, much smaller than even quite distant galaxies and certainly smaller than the best achievable angular resolution by any telescopes on Earth or in space.

A quick example will suffice.

The angular resolution of the Hubble Space Telescope (HST) is about 0.1 arcseconds. That is, unless something subtends an angle of much more than 0.1 arcseconds then you are never going to get a good resolved image of it.

A big spiral galaxy is about 50 kpc in diameter. Such a galaxy will still have an angular diameter of 0.1 arcseconds when it is at a distance of $10^{11}$ pc. i.e. We can get images of such things (if they exist and if they are bright enough) right across the observable universe.

A planet the size of Jupiter subtends an angle of 0.1 arcseconds at a distance of 0.01 pc. i.e. We could not resolve the surface of Jupiter even at only a hundredth of the distance to the nearest star.