Is the Earth Really Spinning? (honest question)

You are already starting to get it.

  That would make sense at the poles

What about one meter from the poles?
Or a kilometre?

As long as you can see the celestial pole in the sky, you can see the stars revolve around it at night.

Let us see if you are able to see the celestial pole.

Texas was about 30 degrees north last time I checked:

That explains the circular movement of the stars, the Sun and the Moon.

This is true for all locations on the Earth, except for the equator:

Is the Earth spinning? That depends, you can always choose a frame of reference that suits you. However, only one of them are non-rotating, the Inertial frame. In all the others we have fictitious forces acting, like centrifugal or Coriolis forces.

We can test if the Earth rotates by watching a pendulum throughout a day. The pendulum would then seem to slowly rotate during this period of time, meaning some fictitious "force" is acting on it. That means that we are located in a rotating frame of reference, and thus the Earth rotates.


Q: "Is there a cosmic, rather than technological, upper limit to what a telescope can resolve?

  Space radio interferometers could have a baseline of millions of kilometers, but is there a point where a larger baseline doesn't improve the resolution anymore because the photons observed are distorted before they arrive?'

Deviation of the wavefront from a blackbody spectrum and the processes that create them is reasonably well understood, the phases of the incoming wavefronts can be measured to enhance the directionality and reduce the distortion. Correlation of the different frequencies allows detection of even small amounts of distortion.

See: "The evolution of CMB spectral distortions in the early Universe" (29 Sept 2011), by J. Chluba, R.A. Sunyaev and the Video: "Spectral distortions of the CMB and what we might learn about the early universe", by Jens Chluba, and "Science with CMB spectral distortions" (27 May 2014), by Jens Chluba.

The method is explained in the paper "Cosmic Microwave Background constraint on residual annihilations of relic particles" (9 Aug 2000), by McDonald, Scherrer, and Walker, page 2:

  "In this paper we compute the energy injected into the CMB by annihilating particles as a function of their mass and annihilation rate (i.e., the product of cross section and abundance squared). We derive constraints on the particle properties by comparison with the observed limits on chemical potential ($\mu$) distortions, and Compton-$y$ distortions (§II). We compare these constraints to similar constraints obtained from the production of deuterium by photodissociation of primordial helium (§III.A), and from the diffuse photon background produced after recombination by extragalactic annihilations (§III.B), and annihilations in the Milky Way halo (§III.C).

  II. DISTORTIONS OF THE CMB ENERGY SPECTRUM

  We consider first the effect of annihilation products on the CMB energy spectrum. The distortion of the spectrum takes place in two steps: first the high energy annihilation products rapidly dissipate their energy into the background photons and electrons, and then the low energy background evolves more slowly in an effort to restore the Planck spectrum. The permanence of distortions produced after $z ≃ 10^6$ is simple to understand in the following way: A Planck spectrum with a given photon number density must have a specific energy density. For $z ≃ 10^6$, photon nonconserving processes (double Compton scattering and bremsstrahlung) are inefficient
  in the background plasma. Therefore, if energy is injected into the CMB but not the the correct number of photons, a Planck spectrum cannot be restored. We now discuss in more detail the form of the distortions pro-
  duced in different redshift intervals. ...'.

See: "New Technology Offers to Broaden Vision for Radio Astronomy". Even a tiny antenna is very capable, this tiny antenna can see in seven directions simultaneously:

[Click image to zoom]

Adaptive optics are used with some optical telescopes, no reason not to apply the same algorithm to VLBI.

"Adaptive Optics: ESO's Very Large Telescope sees four times first (laser) light" (07/11/2016):

  "... even at the best optical sites on Earth such as the 2600 m summit of Cerro Paranal or the 4205 m summit of Mauna Kea in Hawaii, refractive-index variations in the atmosphere lead to a seeing-limited resolution of 0.4 arcsec in contrast to the theoretical diffraction-limited resolution of less than 0.02 arcsec for the 8.2 m VLT telescopes.

"PKS 1954-388: RadioAstron Detection on 80,000 km Baselines and Multiwavelength Observations" (5 May 2017), by Edwards, Kovalev, Ojha, An, Bignall, Et al.:

  "1 INTRODUCTION

  A major challenge in astronomy is the struggle to observe objects with an angular resolution sucient to probe the underlying physical mechanisms. The longer wavelengths of radio-astronomy initially made the quest for high angular resolution more difficult, but the relative ease of preserving phase information enabled the technique of Very Long Baseline Interferometry (VLBI). Intercontinental VLBI routinely achieves milliarcsecond-scale angular resolutions, and extending the baselines between telescopes into space, with satellitebased telescopes, currently yields the highest angular resolution achieved in astronomy.".

There's a newer paper than the one you linked to, using a longer baseline, with better results. See: "PSR B0329+54: Substructure in the scatter-broadened image discovered with RadioAstron on baselines up to 330,000 km" (13 Sept 2016), by Popov, Bartel, Gwinn, Johnson, Andrianov, Fadeev, Et al., the conclusion on page 8:

  "7 SUMMARY AND CONCLUSIONS

  Here we summarize our observations and results and give our conclusions.

  (i) We made VLBI observations of PSR B0329+54 with RadioAstron at 324 MHz on projected baselines up to 330,000 km or 350 M$\lambda$. Our goal was to investigate scattering properties of the ISM which affect radio observations of all celestial sources. While the results of such observations are in general influenced by the convolution of source structure with the scattering processes, pulsars are virtually
  point-like sources and signatures in the observational results can be directly related to the ISM scattering properties.

  (ii) Visibility function at short ground-ground baselines manifests a single bright spike in delay-rate space that vanishes on long space-ground baselines. Thus, the scattering disk of PSR B0329+54 was completely resolved on ground-space baselines of 15,000 to 30,000 km. The FWHM of the
  angular diameter is 4.8 ± 0.8 mas at 324 Hz.

  (iii) The diffractive length scale or size of the diffraction spot near Earth is 17, 000 ± 3, 000 km.

  (iv) With the assumption of turbulent and large-scale irregularities in the plasma, the effective scattering screen is located at d/D = 0.6 ± 0.1 or somewhat more than half of the distance from Earth to the pulsar.

  (v) At longer projected baselines, up to 330,000 km, significant visibility amplitudes were detected, although none were expected from the scattering disk. They are scattered around a mean which stays approximately constant up to the longest baselines. This result indicates that substructure was discovered in the scatter-broadened image of PSR B0329+54.".

Your last question:

  Q: "I'm instead asking about cosmic limitations due to for example interstellar and extragalactic gas which scatters light."

Because you are looking at so many frequencies from so many angles for an extended period of time it's possible to use a supercomputer to see through (around) atoms, fog, dust.

Here are some layperson's explanations:

Ted Talks video: "See invisible motion, hear silent sounds"

CNN YouTube: "New technology makes troops invisible"

Ascendent Technology YouTube "Thermal Infrared & Visible CCTV PTZ Camera Fire Monitoring & Detecting See Through Smoke & Haze"


What is the difference between ICRS and ICRS2?

Not much has changed. Each update is consistent with its predecessor.

The original ICRF is defined by 212 extragalactic sources. The ICRF2 is defined by 295 sources, 97 of which are shared by ICRF1

The reason for the change was greater accuracy. The 295 sources in ICRF2 allow for locations accurate to less than 40 micro-arcseconds (µas) compared with 250 µas for ICRF1. The ICRF2 is consistent with ICRF1, to within the latter's precision.

Both reference frames agree with the J2000.0 frame, to within the precision of the J2000.0 frame. The ICRF is several orders of magnitude more precise than can be achieved by optical astronomy.


Is this photocomposition of Andromeda's galaxy and the Moon accurate?

The Andromeda Galaxy's angular size is about 178x63 arc-minutes and the Moon is about 31 arc-minutes wide. Relatively speaking, the Andromeda Galaxy's apparent size is nearly 6x2 Moons.

So from the photo, their relative sizes in the sky appears to be very roughly accurate. The galaxy appears to have been superimposed from another image just to show what the galaxy might look like in the sky if it were brighter.


Like many bodies in the outer solar system, Enceladus is covered in water ice. It is also strongly believed to have liquid water underneath the ice, as we have observed (with the Cassini probe) plumes of water from its polar regions.

The water is salty and alkali due to the dissolved salts and the reactions of water with the underlying rocks. The plumes contain sodium chloride.

So there is a salty ocean under the ice of Enceladus.  There is also likely to be water under the ice of other moons, such as Europa.


Yes, nebulae can often have very distinct colours. What produces those colours can depend on what elements are in the nebula material and what the temperature and density are.

Generally speaking, green colours in a nebula are due to forbidden transitions in ionised Oxygen, though can feature the hydrogen $\beta$ Balmer line. Red colours can be due to Hydrogen $\alpha$, but there are also lines due to ionised sulphur and nitrogen that often make contributions.

If you take a spectrum of a nebula it usually completely obvious what elements are responsible, because the spectra consist of very sharp emission lines at the wavelengths that correspond to the well-known electronic transitions in particular species. An example is shown below, which is the spectrum of a typical planetary nebula.
Here, the labels mean the following: OIII means oxygen with its outer two electrons removed; NII is nitrogen with its outer electron removed etc., H$\alpha$, $\beta$, $\gamma$ are the $n=3 \rightarrow 2$, $n=4 \rightarrow 2$ and $n=5 \rightarrow 2$ transition of the hydrogen Balmer series. The wavelength scale is in angstroms where 10 angstroms is 1 nm. So 5000 angstroms is the green part of the spectrum, 6500 angstroms is red and 4000 angstroms is blue.


How can light reach us from 14 billion light years away?

If a star 300,000 km away from you explodes, it'll normally take the radiation from that explosion one second to start vaporizing you.
If the space between you and that exploding star inflates to 2.6X10^23km (28 billion ly) one half second after the explosion, the radiation will take 14 billion years, plus a half second, to reach you, and won't vaporize you.
Of course the expansion need not be instantaneous. You can plug in about any function you want there. The important thing is not the distance between you and the star at the time of explosion, but the total distance the photon must travel on its path to you. That distance must take into account the expansion of space between the emitted radiation and you, the observer.


Why is the detection of gravitational waves such a "Big" deal?

The way I see it, there are three reasons it's such a big deal.

The first, as you say, is the (further) confirmation of Einstein's theory of gravity. Newtonian gravity doesn't have gravitational waves. Their existence was already quite established by Hulse and Taylor's discovery and analysis of the binary pulsar PSR B1913+16.  The system's orbit is decaying almost exactly as predicted by General Relativity.  (Note that in the classic figure, the curve is not a fit: that is the prediction!) The direct detection of gravitational waves, however, better confirms that the waves behave as we expect they would, in the sense that the observations match the so-called "chirp" that we expected.

The second reason is because this is now a new way of measuring things in the Universe. Note that the announcement told us (roughly) the masses of the two black holes.  It's actually very difficult to weigh black holes!  For example, this 2010 paper discusses constraints on 23 stellar mass black holes, all from the fact that they're in binary systems, and the motion of the partner can be observed.  Even then those are mostly mass functions that still depend on the inclination of the orbits, which is often unknown.  So the gravitational waves give us a nice measurement, not just confirmation that the phenomenon exists.

With the method proven, there are some fascinating potential applications.  One of my favourites is that LIGO (or similar) could potentially detect the formation of a black hole in a core-collapse supernova.  Just like the detection of neutrinos from SN1987A confirmed some of our basic ideas about these events, detecting the gravitational waves from the birth of a black hole would inform our understanding further still.  And it might even tell us, for example, how massive the black hole is at birth.  There's no end to the novel ways that our theories could be tested!

The final reason is simply the astonishing technical achievement.  They measured motions of at most a few thousandths of a femtometer over the 4km arms.  That's mind-bogglingly precise.  Hopefully this will give more impetus to other gravitational wave detectors.  Here's hoping that eLISA will fly!

The parallel with the Higgs boson is quite apt.  In both cases, we built a big machine to find something that we were pretty sure should be there. (Though the LHC was a tad more expensive...)  Now it's a case of using the new tool to observe the Universe, and seeing if the waves behave as we think they should. And who knows, maybe some completely unexpected signal will appear to confound us completely.


Do the planets, asteroid belt, kuiper belt, and scattered disc lie on the same plane?

They are not exactly in the same plane, but roughly are.

For example this graph representing the centaurs, the Kuipier belt and scattered disc objects show that most objects are around the same plane, with few objects with more than 30 degrees of inclination:

This is the inclination of the asteroids:

Again, few have more than 30 degrees of inclination.

The planets are also all roughly in the same plane. Lets take the ecliptic as a reference (which is almost the same thing as Earth orbital's plane):

The Sun's equator has an inclination of 7.25 degrees.
Mercury has an inclination of 7.00 degrees.
Venus has an inclination of 3.39 degrees.
Mars has an inclination of 1.85 degrees.
Vesta has an inclination of 7.13 degrees.
Juno has an inclination of 12.98 degrees.
Ceres has an inclination of 10.59 degrees.
Pallas has an inclination of 34.84 degrees.
Interamnia has an inclination of 17.29 degrees.
Hygiea has an inclination of 3.84 degrees.
Jupiter has an inclination of 1.30 degrees.
Saturn has an inclination of 2.48 degrees.
Uranus has an inclination of 0.77 degrees.
Neptune has an inclination of 1.76 degrees.
Orcus has an inclination of 20.57 degrees.
Pluto has an inclination of 17.15 degrees.
Ixion has an inclination of 19.58 degrees.
Salacia has an inlination of 23.94 degrees.
Varuna has an inclination of 17.2 degrees.
Haumea has an inclination of 28.19 degrees.
Quaoar has an inlination of 7.99 degrees.
1992 QB1 has an inclination of 2.19 degrees.
Makemake has an inclination of 29.00 degrees.
Eris has an inclination of 44.04 degrees.
Sedna has an inclination of 11.92 degrees.

Most of those bodies have a low inclination to the ecliptic, but Pallas is an interesting exception. I don't know why Pallas has such a high inclination but I would bet on a Kozai mechanism due to Jupiter's influence.

Plutinos and Kuipier belt objects are influenced by Neptune, and part by the Kozai mechanism.

Eris is a special and curious case, it is too far from Neptune, although there is still some weak gravitational Neptune's influence there, but I don't know why its inclination is so high. This contrasts with Sedna, which have a more reasonably lower inclination.

Further, the Oort cloud is expected to be full of bodies of essentially unrestricted inclinations and possibly very excentric orbits, as a result of gravitational interactions with passing stars on the past. By the way, a passing star in a distant past might be an explanation for Eris orbit and inclination.

Note: All of the data here, as well as both of the images, I found on wikipedia.


Why does a right-handed circular polarized wave get lagged when going through ionized plasma?

What is the mathematical difference between their refraction indices?

The conventions on the signs of the direction of magnetic field vectors and of charges.

The tutorial specifies several things that can have a sign;

direction of the light propagation
direction of the magnetic field
sign of the particle's charge (electrons have negative charge)

Let's look at WIkipedia's Faraday effect; Interstellar medium where an equation for $RM$ or rotation measure is given:
$$RM = \frac{e^3}{2 \pi m^2 c^4} \int_0^d n_e(s) B_{\parallel}(s) ds$$
This expression includes $e^3$ and $B_{\parallel}$. Notice that the power of $e$ is odd. The assumption is that the interstellar medium is negative electrons, so they've replaced $q^3$ with $e^3$ but if it were positrons they'd have to write
$$q^3=-e^3$$.
Also remember that $B_{\parallel}$ can be positive or negative since the field could point either way relative to the direction of propagation.
The tutorial shows some of the intermediate steps where we get the idea that both the sign of the charge and the directions of the field will be important.

Why right handed circular polarization gets lagged when going through ionized plasma?

It could be either the RCP or LCP that is delayed more than the other, depending on the direction of the magnetic field relative to the direction of propagation, and on the sign of the lightest charged particles in the plasma that can respond most quickly to the electromagnetic wave.
In this example, the field just happened to be drawn parallel to the propagation, and normal plasma with negative electrons. In this case RCP is delated more than LHP. Had either one of those been opposite in the example, the result would have been oposite.


Can the range of the observable universe be extended through an intermediary?

The simple answer is no, but it might be easier to understand if we look at the difference between the different cosmological horizons.

Since no information can travel faster than light, and nothing can have travelled for longer than the age of the Universe, then the limit of the observable universe is calculated by multiplying the speed of light by the conformal time since the Big Bang (the age of the Universe, adjusted for the scale factor representing the expansion of the Universe). This limit is also known as the particle horizon.

The current limit is a sphere of ~46.5 billion light-years in radius. If we could detect any signals emitted from matter that far away, we'd be seeing the Big Bang itself! However, the most distant light (EMR) signal we will ever be able to detect is the cosmic microwave background radiation (CMBR), which represents the earliest light after photon decoupling, emitted from a shell ~45.7 billion light-years away.

The CMBR was emitted about 380,000 years after the Big Bang, while the earliest stars didn't form for at least another 100 million years, and the earliest galaxy we've observed so far is GN-z11 which dates back to about 400 million years after the Big Bang. Nonetheless, there's been enough time since then for the light from those earliest structures – albeit highly redshifted and extremely weak – to reach us directly, so there's no need for an "intermediary".

While the particle horizon defines the current limit of the observable universe, what about future observations? In a Universe with accelerating expansion, the outer parts of the observable universe are now receding from us faster than the speed of light, and information from those areas may never reach us at all. The point at which light emitted now can never reach us is called the cosmic event horizon. The current cosmic event horizon is around 16 billion light-years away, which is a lot less than the ~46b ly to the edge of the observable universe.

To make this a bit clearer, let's look at the path the information would take via the intermediary. To avoid confusion, let's call your distant object a supernova rather than just a nova, and let's assume the intermediary is at the midpoint of a straight line between us and the supernova.

The information – whether light waves or gravitational waves – leaves the supernova and travels at the speed of light to the intermediary, where (let's say) the information is immediately relayed in a boosted signal – also travelling at the speed of light – towards Earth. Apart from the momentary delay in relaying the signal at the midpoint, the information will be following exactly the same path as if it had left the supernova and travelled directly to Earth. In other words, if we can receive the information about the distant supernova from the intermediary, we must also be able to receive the information straight from the supernova itself.

Now, in the case of a supernova that's just beyond our cosmic event horizon, the intermediary can receive the light from the supernova, but the space between us and the intermediary will have expanded in the billions of years it takes for the supernova's light to get to them. By that time, the Earth is receding from the intermediary so fast that their relayed signal can never overcome the accelerating expansion of space on the second leg of its journey. We never receive their signal, as they too have receded over the cosmic event horizon.


Would time go by infinitely fast when crossing the event horizon of a black hole?

(I will assume a Schwarzschild black hole for simplicity, but much of the following is morally the same for other black holes.)

  If you were to fall into a black hole, my understanding is that from your reference point, time would speed up (looking out to the rest of the universe), approaching infinity when approaching the event horizon.

In Schwarzschild coordinates,
$$\mathrm{d}\tau^2 = \left(1-\frac{2m}{r}\right)\mathrm{d}t^2 - \left(1-\frac{2m}{r}\right)^{-1}\mathrm{d}r^2 - r^2\,\mathrm{d}\Omega^2\text{,}$$
the gravitational redshift $\sqrt{1-\frac{2m}{r}}$ describes the time dilation of a stationary observer at a given Schwarzschild radial coordinate $r$, compared to a stationary observer at infinity. You can check this easily: plug in $\mathrm{d}r = \mathrm{d}\Omega = 0$, the condition that neither the radial nor the angular coordinates are changing (i.e. stationary observer), and solve for $\mathrm{d}\tau/\mathrm{d}t$.

The conclusion is that if you have the rocket power to hover arbitrarily close to the horizon, you will be able to see arbitrarily far into the history of the universe over your lifetime. However, that doesn't actually cover what happens to an observer that crosses the horizon. In that case, $\mathrm{d}r\not=0$, and the coefficient of $\mathrm{d}r^2$ above becomes undefined at the horizon: as in the other question, the Schwarzschild coordinate chart simply fails to cover the horizon and so is ill-suited for talking about situations cross the horizon.

But that's a fault of the coordinate chart, not of spacetime. There are other coordinate charts that are better adapted to questions like that. For example, the two Eddington-Finkelstein charts are best suited for incoming and outgoing light rays, respectively, and the Gullstrand-Painlevé chart is adapted to a freely falling observer starting from rest at infinity.

  If this is correct, would you see the whole universe's future "life" flash before your eyes as you fall in, assuming you could somehow withstand the tremendous forces, and assuming black holes don't evaporate?

No. I think this is best seen from the Penrose diagram of Schwarzschild spacetime:

Light rays run diagonally. In blue is an example infalling trajectory, not necessarily freely falling. Note the two events where it crosses the horizon and where it reaches the singularity. Shown in red are inward light rays that intersect those events. Thus, the events that the infalling observer can see of the external universe consist of the region between those light rays and the horizon. The events occurring after that won't be seen because the the observer will have already reached the singularity by then.

Now suppose the observer tries a different trajectory after crossing the horizon, accelerating outward as much as possible in order to see more of the future history of the external universe. This will only work up to a point: the best the observer can do is hug the outgoing light ray (diagonally from lower-left to upper-right) as much as possible... but since the observer is not actually allowed to go at the speed of light, seeing all of the future of history will be impossible. The best the observer can do is to meet the singularity a bit more on the right of the diagram.

By the way, since the light ray worldlines have zero proper time, trying to do that will actually shorten the the observer's lifespan. If you're in a Schwarzschild black hole, you would live longer if you don't struggle to get out.

The above is for an eternal, non-evaporating black hole, as that's what you're asking about here. (The 'antihorizon' is there because the full Schwarzschild spacetime is actually an eternal black hole and its mirror image, a white hole in a mirror 'anti-verse', which not shown on this diagram. That's unphysical, but not relevant to the situation we're considering here.)

  If it is correct that black holes evaporate due to Hawking radiation, would you be "transported" forward in time to where the black hole fully evaporates?

An evaporating black hole is morally the same as above: only an ideal light ray can reach the point when the black hole fully evaporates; everyone else gets the singularity. (Since this ideal light ray right along the horizon would be infinitely redshifted, arguably not even that.) You can repeat the above reasoning on its Penrose diagram yourself:

Addendum:

  I have thought a bit about this, and does this solution take into account the relativistic time effects near the horizon of the black hole (e.g. is my understanding correct that the observer would observe time in the universe passing approaching infinitely fast when approaching the event horizon)?

How much time dilation happens depends entirely on what coordinates we're talking about (more generally, which frame field). What a given observer will actually see, however, is completely independent of choice of coordinates. In particular, Penrose diagrams illustrate the light cone structure of the given spacetime, and what an observer can in principle see depends entirely on what light rays intersect the observer's wordline. So yes, it's taken into account by default.

If you're actually falling in it, no, your understanding is mistaken, for reasons explained above. For additional motivation, flip the question around: what does the very distant stationary observer see of the infalling object? On the above Penrose diagram, outwardly directed light rays are diagonal, from lower-left to upper-right. Draw some some outward light rays from the blue infalling worldline. You will see that no matter how far into the far future (up on the diagram) you pick an event outside the black hole to be, you can connect that event with an outward light ray originating from the blue infalling worldline before it crosses the horizon. The conclusion would be that an observer that stays outside the black hole would be able to see the infalling object arbitrarily far into the future. No matter how much time passes for someone who stays out of the black hole, the image of the infalling object would still be visible as it was before it crossed the horizon. (In principle at least; in practice it will get too faint to see after a while.)

Thus, the usual result of "infinite gravitational time dilation makes the image of the infalling object hover forever near the horizon" is also straightforwardly deducible from the diagram, and so is completely consistent with the infalling object being able to see a finite part into the future of the external universe. Perhaps it is best to emphasize that the situation is not actually symmetric: what the external observer sees of the infalling object is not some straightforward flip-around of what the infalling object sees of the external universe. The black hole itself breaks that symmetry.


How do we know that the mass ratio of hydrogen and helium in the sun is approximately 3:1?

You're asking two questions here but they are related, so that may be OK.

The mass ratio of the Sun is probably not 3:1 though, as the Sun has been turning hydrogen into Helium for nearly 5 billion years.   The interior of the Sun has a considerably higher helium ratio as a result, but the outer edges of the sun - which is what we see and read from spectroscopy, is close to the 3:1 mass ratio.

If this is accurate the Sun overall is closer to about 2.6:1.

Scientists can measure the ratio by observing spectral lines.  Stars make this easy because of how hot they are, but that information is limited to their surface.   We can't see the core of the sun.  Still more than half of the elements have been observed in the solar surface.

See on solar spectroscopy

Or a related question from Physics Stack

The 25% helium by mass is confirmed both by spectroscopy and by quantum physics models based on the modeled expansion as the early Universe goes from hot to cold.   Helium is more stable than hydrogen, but, it requires an intermediate step in it's formation because deuterium has to form first.  Also, though I think this plays a smaller role, free neutrons decay and become protons which become Hydrogen.

More massive elements need hotter temperatures to form but the early universe went from hot (Quark soup), to less hot (formation of hadrons), to static - 92% Hydrogen, 8% helium by element or 75-25 by mass with tiny trace amounts of Lithium.   It stayed that way until the first stars and the formation of heavier elements.

That's a very simplified version.  You can visit the stack exchange answer below for more details or this link here.

Both spectroscopy and quantum physics agree on the ratio, which is always a plus when the same answer is reached two different ways.   There is a lingering lithium ratio problem, but even so, the 75%-25% ratio is pretty good.


What are the current observational constraints on the existence of Nemesis?

Few important points about WISE:

it was able to detect anything with a temperature above 70-100 K, whereas the coolest brown dwarfs are in the 500-600 K range (the coolest was discovered by WISE itself, see Mainzner et al., 2011);
it was able to detect objects larger than 1km up to 3 AU from the Sun, or objects of 2-3 Jupiter masses in a distance up to 7-10 light-years (see here and here);
the closest brown dwarf detected by WISE is a brown dwarfs binary system, located at 6.6 light-years from the Sun.

So I would say we can be pretty confident that Nemesis existence has been ruled out.


What decided how the Kepler space telescope was pointed?

Changing fields breaks the timeline of the observations.  If you have variation with a period similar to when you switch, you might completely miss it.  Even if the periods are different, it degrades your observations and makes it harder to draw conclusions.  The more uninterrupted viewing you get, the better.

This mission document covers some of the reasons for the choice of the field.

Continuously viewable during the mission (you can't pick something where the earth, sun or moon constrains your viewing).
Rich in stars.  Observe as many as possible.
The size of the sunshade limits the field to be more than 55 degrees from the ecliptic.

Although it can image lots of stars, the actual field isn't all that large.  Attempting to pick a single field with some fraction nearby would be difficult, and was not Kepler's mission.  In fact the resource of follow-up observations from ground telescopes was mentioned as one reason a northern field was chosen over a southern one.


If Earth had no atmosphere, how many people would be killed by meteors every year?

Three millions.

For this answer to be any fun, let's first assume that before removing the atmosphere, we have equipped all humans with oxygen supplies. In return, they are not allowed to go indoors or otherwise protect themselves.

The size of meteors — as well as most other things in the Universe — has a distribution that is well described by a power-law. The larger the meteor, the rarer they are.
Small meteors can be lethal only by a direct hit, while large meteors in the real world can be lethal due to the effect they have on the environment; dust can cover the sky for years and prevent the Sun from keeping us warm. However, since we have removed the atmosphere, the debris will fall down immediately, and the only way to kill us is by the explosion. But, as will be shown below, the dangerous ones are the small ones because the rate at which large ones falls off so quickly.

The number $N$ of meteors entering Earth's atmosphere every year with a diameter $D$ (measured in meters) or larger is given by (Brown et al. 2002)
$$
\frac{N(>D)}{\mathrm{year}} \simeq 37 \left( \frac{D}{\mathrm{m}} \right)^{-2.7}.
$$

Minimum size needed to be lethal

What makes a meteor lethal is really outside astronomy, and someone may be able to come up with a better answer, but for the sake of the argument, let's assume that the energy needed to kill a human by a direct hit is equal to the energy of a bullet fired from the world's smallest firearm, the 2mm Kolibri, which has a muzzle energy of $E_\mathrm{lethal} \simeq 4\,\mathrm{J}$.

Earth orbits the Sun at 30 km/s. Depending on the direction from where the meteor arrives, it will have different speeds, but it will at least enter with the escape velocity of Earth, i.e. 11 km/s, and a typical speed is $v_\mathrm{impact} = 17\,\mathrm{km/s}$ (Marcus et al. 2010). Thus, the minimum mass to kill by a direct hit is as small as
$$
m_\mathrm{lethal} = 2 \, E_\mathrm{lethal} \, / \, v_\mathrm{impact}^2 \simeq 30 \,\mu\mathrm{g}.
$$
At an average density of 4 g/cm$^3$, this means that a small grain of a diameter of 0.2 mm will kill you. According to the formula above, the number of such grains (or larger) that hit Earth every year is $2\times10^{11}$.
Since the surface area of Earth is 510 million $\mathrm{km}^2$, that's $n_\mathrm{hits}\sim5\times10^{-4}\,\mathrm{hits}\,\mathrm{m}^{-2}\,\mathrm{year}^{-1}$.

Some 7 billion people live on Earth. They each occupy roughly $1\,\mathrm{m}^2$ (and weren't allowed to protect themselves, remember), so in total they occupy an area of $A_\mathrm{folks,tot} = 7 \times 10^9\, \mathrm{m}^2$.

Thus, the number of people being killed by direct hit meteors every year is
$$
N_\mathrm{kill} = n_\mathrm{hits} \, \times \, A_\mathrm{folks,tot}
 = \mathbf{3\,million\,year^{-1}},
$$
or one person every 10 seconds.

Death by explosion

How about non-direct hits? Let's assume that the minimum size to kill you would have an energy equal to that of a hand grenade holding 100 g explosives, and that you're killed if it hits within a few meters of you, so that you "occupy" 10 m$^2$. Similar to above, you can work out that this corresponds to a meteor of ~1 cm being lethal, and that the rate at which such meteors is only $2\times10^4$ per year, so only 2-3 people would be killed in this way. Even larger explosions are even more rare. Thus, the probability of killed by a explosion is negligible compared to direct hits, so the final answer would be the above, i.e. 3 million per year.

Conclusion: Let's not remove the atmosphere.


Does time slow down because the universe is expanding at an accelerating rate?

Yes, time does run slower for far-away objects, as observed from our point of view; this is a prediction of general relativity. And yes, because expansion accelerates, this time dilation slowly, very slowly, becomes more pronounced (this would happen even if the expansion didn't accelerate, but just continued at the same rate).

This time dilation  is a well-known effect, and is always taken into account when doing observations. For instance, when observing distant supernovae, one is often interested in how their luminosities decrease as a function of time. This is called their lightcurve. In order to compare lightcurves at different redshifts, they are usually converted to their restframe, i.e. how they would look if you were "standing next to the supernova" (e.g. Goldhaber et al 2001).

However, time dilation does not work exactly as you seem to think. At galaxy at a redshift of $z$ has its time dilated by a factor of $1+z$, so time runs twice as slow for a galaxy at, say, $z=3$ than for a galaxy at $z=1$. Galaxies with redshifts larger than $z\sim1.5$ recede faster than the speed of light, and time does not at all stop here. Only for $z\rightarrow\infty$, i.e. at the beginning of time at Big Bang, does the time dilation approach infinity.


Antimatter is present on Earth and is being naturally created all the time (by Beta decay) as well as being created as product of cosmic ray collisions and in particle accelerators.

However the universe appears to be principally made of matter and so the antimatter thus created is annihilated: and although the production of antimatter through nuclear decay is essentially continuous, it is not of a high volume so the energy created in this way is not significant in the normal course of events. The real question is why is matter so much more abundant than antimatter: which is an open question though various theories have been proposed about why we might live in such a universe.


Why are gas giants colored the way they are?

The simple answer is that the composition in terms of elements doesn't matter for the colour, since that is determined by the molecular composition as well as state (pressure and temperature) of the visual layers (which contain a small but significant fraction of other elements than H and He). Given the differences in the giant planets temperature and pressure, this molecular composition is most likely different.


What does an Einstein-De Sitter universe look like?

Bendiness = space-time curvature.

And yes, it is flat (k=0) for the EdS model. Essentially it is the epoch at which only matter is thought to dominate (no contributions from radiative pressure or dark energy, $\Lambda$).

It is only a model. The EdS universe is thought of as a static Universe which is expanding indefinitely. It just makes solving certain equations in Cosmology a little easier, but it was perhaps due to Einstein's stubbornness in accepting a dynamic Universe.


Is it possible to overcome the problem of blind spot(s) of current gravitational wave detectors?

The blind spots are caused by the way the detectors work. They are sensitive to a gravitational wave (GW) changing the relative path length along interferometer arms at right angles to each other.

Gravitational waves come in two polarisations (plus and cross). These polarisations cause alternate perpendicular expansions and contractions in space, but are rotated by 45 degrees with respect to each other (plot from Kalmus 2009).

A GW source would normally be a mixture of both. The sensitivity to each polarisation depends on the orientation of the interferometer arms with respect to the source direction. For example, a source that is directly "overhead" will only be detected in the polarisation state that lines up with the arms and not in the other, because it would make the arm lengths change by the same amount. e.g. Imagine your detector is lined up with the x and y axis indicators in the picture above, then only the plus ($+$) polarised waves would be detected.

However, if the waves come from a source in the plane of the interferometer, then neither polarisation causes a relative difference in the arm lengths if the source lies along the bisector of the two arms or along an equivalent line at right angles to this. Thus there are 4 blind spots on the sky. Note that if a source is perfectly polarised (e.g. an edge-on merging binary system), then there will be addditional blind spots.

There are several solutions: make your interferometer 3 dimensional (i.e. add a third arm at right angles to the other two); make a triangular interferometer (this is what is planned for space interferometers); build another interferometer with a different plane/orientation to the first.

As an example, the plots on the left in the picture below show the sensitivity of LIGO (Washington and Louisiana) to unpolarized GWs at the top. These two interferometers are nearly lined up and nearly in the same plane, so you see the 4 blind spots clearly. The middle plot on the left shows how things improve when you add VIRGO (in Italy) - there are essentially no blind spots in the combined instrument, and then in a planned 6-detector network at the bottom. (Plot from Andersson et al. 2013).


There is a class of meteorites called Lunar Meteorites and several of those have been found on earth.

Of course, it doesn't necessarily indicate whether we have actually witnessed a certain lunar meteor shower. Considering that nearly 100 metric tons of space debris falls on earth each day, one would imagine that it'll be colossally impractical to attempt mapping a meteorite to the moon, using a reference to a recorded meteor shower.

However, it might be worthwhile to take a look at this link which addresses rather interesting questions:

 How did lunar meteorites get here?

  Any rock on the lunar surface that is accelerated by the impact of a meteoroid to lunar escape velocity of $2.38 km / s$ or greater will leave the Moon's gravitational influence. Most rocks ejected from the Moon become captured by the gravitational field of either the Earth or the Sun and go into orbit around these bodies. Over a period of a few years to tens of thousands of years, those orbiting the Earth eventually fall to Earth, whereas those orbiting the Sun may also eventually strike the Earth in a few tens of millions of years after they were launched from the Moon.

 How do we know that they come from the Moon?

  Chemical compositions, isotope ratios, mineralogy, and textures of the lunar meteorites are all similar to those of samples collected on the Moon during the Apollo missions. These meteorites are rich in a mineral called anorthite and have high concentrations of aluminum and calcium. Because of some unique aspects about how the Moon formed, the lunar highlands are composed predominantly of anorthite. To the best of our knowledge, anorthite is much less common on asteroids, and on the surfaces of any other planets or planetary satellites.

  Quotes derived and paraphrased from the linked article.

An up to date catalog of all lunar meteorites found so far indicates that the heaviest lunar meteorite is a rock (pair) named Kalahari 008 & 009, weighing ~$14kg$.


If it disappeared towards east or so, as I  think because the other "star" should be Venus, that it can be a satellite - or even more an airplane - as seen from "its rear". Basically at first it went far along the line of your line of sight looking still, until finally you could see the curvature of its path. From crude geometric feeling it was most likely an airplane, a lower altitude make it easy the seeing "from the rear" (think about what you see observing the rear light of a car moving on an horizontal road towards a down hill. It looks more or less two lights in the dark then it disappears. Or a ship approaching horizon). About absence of blinking: if it indeed was an airplane mirroring the sunlight, not way to see the less intense "navigation" lights.


How does the Gamma Ray Burst that occurred when 2 black holes merged compare to other GRB's?

The interpretation you suggest in the second paragraph is incorrect. It is understandable, since there is a debate in the literature - different papers come to potentially contradicting conclusions.

"Excluding a possibility that the event is associated with substantial gamma-ray radiation, directed towards the observer" simply means that no observable GRB is found, hinting that may be the original detection was caused by instrumental background: each detector has it's own instrumental backgrounds, only real events should be seen by all sufficiently sensitive instruments.

In principle, it is always possible that the merger generated beamed emission, directed somewhere else. No instrument should have seen it directly, and there is no simple way to know if it happened.

This upper limit is derived from the observation of another satellite (INTEGRAL/SPI-ACS, Savchenko et al 2016) than the original detection (Fermi/GBM, Connaughton et al 2016).
Also an alternative analysis of the Fermi/GBM data (Greiner et al 2016) suggested that no event can be found in the GBM data - their opinion is that it was a background fluctuation of some kind.

Right now, the teams who reported these conflicting results are working together, trying to come up with a consistent picture, which might be, in principle, a GRB with this or that properties associated with the GW150914, an unrelated GRB with some properties, or no detection whatsoever. This work is centered on cross-calibrating and comparing instruments, and is also useful to avoid these kind of uncertainties in the future.

One could try to characterize the spectral properties of this event, following the approach of the original Fermi/GBM team. But unfortunately, the measurement appeared to be in very unlucky conditions for Fermi/GBM (in bad direction). Which is why the signal was very weak (below what would be usually reported for a real GRB, though recently the attempts were made to decrease these thresholds, see Goldstein et al 2017 ), and spectral characterization is loose. You can look for some details to Veres et al 2016.
With these large uncertainties the spectrum is compatible to that of known short GRBs.

The luminosity estimate depends on the spectrum, but is seems to be at the lower end of the short GRB sample (see e.g. Wanderman et al 2015)
But because the uncertainties are large, the event, if real, might be unusual as well.

The INTEGRAL observation, non-detection, would imply much softer (perhaps, unusual for a short GRB) or/and a weaker burst, possibly incompatible even with the highly uncertain Fermi/GBM data.

Duration of this possible GRB possibly associated with GW event is the easy part and is about 1 s long, typical for a short GRB (Kouveliotou  et al 1993).


The chemical make-up of the comet is indeed relevant. From Wikipedia (Exocomet):

  The exocomets can be detected by spectroscopy as they transit their host stars. The transits of exocomets, like the transits of exoplanets, produce variations in the light received from the star. Changes are observed in the absorption lines of the stellar spectrum: the occultation of the star by the gas cloud coming from the exocomet produces additional absorption features beyond those normally seen in that star, like those observed in the ionized calcium lines. As the comet comes close enough to the star, cometary gas is evolved from the evaporation of volatile ices and dust with it.

This is not to say that you can determine anything else (besides their existence and chemical composition) about the comets.


Difference in observing hot stars that have been reddened and stars that are cool and red

Reddening does not change the details of the spectrum of a star. The differences in absorption lines that are present in a hot star vs a cool star are still there after reddening. So if you have a spectrum then distinguishing the two scenarios is easy.

If you don't have a spectrum but have two colours then it is still possible to distinguish the scenarios by tracing the star back along an extinction vector in a colour 1 vs colour 2 diagram until it intercepts an intrinsic locus for (e.g.) main sequence stars. This can be definitive, but can be ambiguous depending on the types of star and particular colours.

If you just have a colour, you can't tell.


Assume you have a spherical blackbody.

The solar flux at the radius of the Earth is given to a good approximation by $L/4\pi d^2$, where $d = 1$ au. This is $f=1367.5$ W/m$^2$ (though note the distance between the Earth and the Sun has an average of 1 au).

If it is a blackbody sphere it absorbs all radiation incident upon it. Assuming this is just the radiation from the Sun (starlight being negligible), then an easy bit of integration in spherical polar coordinates tells us that the body absorbs $\pi r^2 f$ W, where $r$ is its radius.

If it is then able to reach thermal equilibrium and it entire surface is at the same temperature, then it will re-radiate all this absorbed power. Hence
$$\pi r^2 f = 4 \pi r^2 \sigma T^4,$$
where $T$ is the "blackbody equilibrium temperature". Hence
$$ T = \left( \frac{f}{4\sigma}\right)^{1/4} = 278.6\ K$$


Often for molecules to form in interstellar space, dust is used as a catalyst. The reason is that in typical interstellar environments, densities are so immensely low that even for just two atoms to meet, the probability is so small that formation time scales are very long. For 3+ atoms, the chance decreases rapidly. Instead, an atom can stick to a dust grain and wait for ages until other atoms stick. The atoms slowly "crawl" around on the surface of the dust grain, eventually meet and make bonds. If the formation of a bond releases energy (is exothermic), the molecule can be ejected from the grain surface. This process is call adsorption.

The dust also helps shielding the molecules from stellar radiation which would otherwise easily destroy them. This is why molecular clouds are also very dusty. But actually UV irradiation helps with the formation of very complex molecules by ionizing less complex molecules that subsequently can make bonds with other atoms and molecules.

If the environments are extremely dense, as when a star dies and ejects its gas either as a supernova or a planetary nebula, molecules can also form. This is probably also how the dust itself is formed, although it might also be formed later on (this is currently debated; the problem is that supernovae are so powerful that their shock waves tend to destroy dust shortly after it forms, and planetary nebulae are created from stars that live so long that they can't explain the abundance of dust in the very early Universe where they wouldn't have had the time to live their lives).

As Stan Liou and LocalFluff says, methane is actually an "easy" molecule to form, both because it's rather simple, and because its constituents, hydrogen and carbon, are the most and the fourth most abundant elements in the interstellar medium, respectively.

In fact, far more complex molecules are regularly found in interstellar space, as can be seen on this list.


What does the ']' in the spectral line "CIII] 1909 Å" mean?

Under conditions of very high temperature and very low pressure forbidden conditions can occur, they shift the color of the excitation radiation from the ultraviolet into the visible spectrum giving nebula their beautiful colors. These are represented using brackets.

See Wikipedia's Selection Rules, section Angular Momentum:

  "Semi-forbidden transitions (resulting in so-called intercombination lines) are electric dipole (E1) transitions for which the selection rule that the spin does not change is violated. This is a result of the failure of LS coupling.".

That notation would read:

Element.
Ionization state.
Forbiddenness - superallowed, allowed, first forbidden, second forbidden, third forbidden, fourth forbidden, etc. Think of "forbidden" as meaning "likelihood". Represented with one or two brackets.

See note below: "In many-electrons systems transitions that violate rule 4 change the total spin (semi-forbidden transitions) and are called intercombination lines (e.g. CIII]). Transitions which violate the propensity rule 5 and/or 6 are strictly forbidden, and are labelled by two square brackets (e.g. [CIII]). Table 1 gives examples of the 3 types of transitions.

  Do not confuse the use of two brackets around an element with their usage around abbreviations for massive hot blue emission stars. With massive Wolf-Rayet stars, their spectra groups are enclosed in square brackets: e.g. [WC]. Most of these show [WC] spectra, some [WO], and very rarely [WN].

The "double Lambda", in the example you provided, is a 'half doublet'. A doublet for that element would be OIII]λλ1661,1666 another example would be SiIII]λλ1882,1892. One paper where this is discussed is: "CIII] Emission in Star-forming Galaxies at z ~ 1" (Mar 4 2017), by Xinnan Du, Alice E. Shapley, Crystal L. Martin, and Alison L. Coil. A doublet is two distinct spectral lines formed independently but which are so close that previously inaccurate instruments and poorer understanding of these mechanisms led many to believe a single line at a particular location. See: "Performance Analysis of Standard Fourier-Transform Spectrometers" 2007 by Douglas Cohen.
The Å ($\unicode{x212B}$$\unicode{x212B}$ or $\unicode{x00C5}$$\unicode{x00C5}$, a Latin capital letter "A" with a ring above it) is a Swedish letter, the symbol for the ångström, a unit of length equal to 10$^{−10}$ m (one ten-billionth of a metre) or 0.1 nanometre.

The Wikipedia webpage: "Spectroscopic notation" should be helpful:

  Ionization states

  "Spectroscopists customarily refer to the spectrum arising from a given ionization state of a given element by the element's symbol followed by a Roman numeral. The numeral I is used for spectral lines associated with the neutral element, II for those from the first ionization state, III for those from the second ionization state, and so on.10 For example, 'He I' denotes lines of neutral helium, and 'C IV' denotes lines arising from the third ionization state, C3+, of carbon.

See also: Wikipedia's Ionization energy, section Values and Trends.
More about doublets:

"The Grism Lens-Amplified Survey from Space (GLASS). XI. Detection of C IV in Multiple Images of the z = 6.11 Lyα Emitter behind RXC J2248.7-4431" (2017), by Schmidt, KB; Huang, KH; Treu, T; Hoag, A; Bradač, M; Henry, AL; Jones, TA, et al.

"The UV Properties of the Narrow Line Quasar I Zwicky 1" (Jun 26 1997), by Ari Laor (Technion), Buell T. Jannuzi, Richard F. Green, Todd A. Boroson (NOAO).

Note:

From "Spectroscopy - IPAG Grenoble" (.PDF), page 2:

  "2.3 Selection rules for Hydrogenic atoms

  In general, there is always a non-zero probability for any transition between two states to occur. However, in some instances, the probability is exceedingly small. Under some approximations (e.g. L-S coupling, dipole approximation, ...) the matrix element might be strictly zero. However, higher-order terms generally lead to non-zero, yet small, probability. The cases in which probabilities vanish, under given approximations, are called selection rules.

  The dipole transitions described above are called dipolar electric transitions. The detailed computation of the key term $\scriptsize \left| \overrightarrow \epsilon \cdot \overrightarrow \mu_{if} \right|^2$ allows to determine the cases for which a
  transition is allowed:

  $\scriptsize{\overrightarrow \mu}\!_{i f}$ does not vanish: dipolar electric transition between states $i$ and $f$ is allowed;
  $\scriptsize{\overrightarrow \mu}\!_{i f}$ vanishes but there exists non-zero higher-order terms in the development of $e^{i \overrightarrow k \cdot \overrightarrow r}\!\!$: dipolar electric transition is said to be semi-forbidden, but e.g. electric quadrupolar transition are allowed, though with much smaller rates;
  $\scriptsize{\overrightarrow \mu}\!_{i f}$ and all higher-order terms of the development also vanish: the transition between $i$ and $f$ is forbidden. Other higher-order terms may still allow the emission/absorption of photons.

  The selection rules that apply to one-electron systems are:

  energy level: ∆n any
  orbital angular momentum: ∆l = ±1
  parity must change between i and f
  magnetic quantum number: ∆m = 0,±1
  spin does not change: ∆s = 0 (always true for the H atom)
  total angular momentum: ∆j = 0,±1

  In a multi-electron atom, these rules apply to the jumping electron. These rules completely determine the spectra of one-electron atoms, such as HI and HeII, and also the alkali metals. Diagrams which show the allowed transitions are called Grotrian diagrams.

  2.4 Selection rules for many-electrons systems

  We have thus far discussed the interaction of single-electron atoms with a radiation field. The results can be generalized to many-electron systems, by considering the total electric dipole moment, $\scriptsize\overrightarrow D = \sum{_i} −e\overrightarrow {r_i}$. For multi-electron atoms, the selection rules for dipolar electric transitions are:

  total angular momentum: ∆J = 0,±1 but J = 0 → 0 is strictly forbidden
  magnetic quantum number: ∆M$_J$ = 0,±1
  parity must change between i and f
  spin does not change: ∆S = 0
  one electron jumps, ∆n any, ∆l = ±1
  orbital angular momentum: ∆L = 0,±1, 0 − 0 forbidden

  The first three rules are rigorous and must be satisfied by all dipolar electric transitions. The last three are not necessarily satisfied in complex atoms, and are called propensity rules. Transitions which fulfill rules 1–3 are allowed. Transitions that violate 4 change the total spin and are called intercombination lines (e.g.CIII]). Transitions which violate the propensity rule 5 and/or 6 are strictly forbidden, and are labelled by square brackets (e.g. [CIII]). Table 1 gives examples of the 3 types of transitions.

  $$\begin{array}{ccccccrrr} \\
\text{Species} & f \qquad \leftarrow \qquad i & λ (\unicode{x00C5}) & A_{ul}(s^{−1}) & \Delta J & \text{Parity}^\dagger & \Delta S & \Delta l & \Delta L \\
\hline
\,\text{NII} & 2{p ^{2}} \; {^{3}\!P} ^e_0 \leftarrow 2p3s \; {^{3}\!D} ^o_1
& 1084.0 & 2.18×10^8 & -1 & o \rightarrow e & 0 & -1 & -1 \\
\;\text{CIII]} & 2{s ^{2}} \; {^{1}\!S} ^e_0 \leftarrow 2s2p \; {^{3}\!P} ^o_1
& 1908.7 & 114 & +1 & o \rightarrow e & -1 & -1 & -1 \\
\text{[CIII]} & 2{s ^{2}} \; {^{1}\!S} ^e_0 \leftarrow 2s2p \; {^{3}\!P} ^o_2
& 1906.7 & 0.0052 & +2 & o \rightarrow e & -1 & -1 & -1 \\
\end{array}
$$
  $$\text{Table 1: Examples of allowed, semi-forbidden, and forbidden transitions."}$$


How can we tell that the sun is moving with theories such as the theory of relativity?

We don't know. All that we know is that the Earth moves relative to the Sun and relative to other things in the wider universe. That is what we mean by relative motion.

A measurement of absolute motion requires you to define an absolute reference frame. We don't have one of those - you can choose your reference frame to suit your purposes. The beauty of relativity is that one of its axioms is that the laws of physics remain unchanged whichever reference frame you pick.

Conrad suggests using the cosmic microwave background as an absolute frame of reference. Whilst I agree that this is a useful frame of reference - it demonstrates with what velocity we (or any other object) are moving with respect to the local co-moving volume; there is no sense in which this is absolute or has any special properties beyond its convenience. The laws of physics remain unchanged whatever your velocity with respect to the cosmic microwave background.


Does the luminosity of a star on its main sequence increase or remain approximately constant?

The luminosity of a star remains approximately constant during its time on the main sequence. There is a slight increase over time, which is, as you guessed, related to the changing composition of the star due to fusion.

The so-called Vogt-Russell theorem tells us that a star's mass and composition are the only two factors that affect its structure and, by extension, its luminosity.$^{\dagger}$ Most main sequence stars have negligible mass-loss rates (at least, compared to evolved stars like AGB stars), and so therefore compositional changes should be the only thing dictating luminosity evolution.

It turns out that the luminosity is related to the mean molecular mass $\mu$ by (approximately)
$$L\propto\mu^4$$
which is not insignificant. Fusion turns hydrogen to helium in the star's core, and hence there should be an increase in $\mu$ and thus in luminosity (roughly a factor of two). The change is small compared to the orders of magnitude by which two main sequence stars may differ from one another, but it isn't zero.

As an aside, there is indeed a change in the central temperature (though I would note that your use of the Stefan-Boltzmann law requires a change in the surface temperature) to accommodate the increase in luminosity, but the energy generation rate is very sensitive to changes in temperature, particularly for stars using the CNO cycle.

$^{\dagger}$ This is, strictly speaking, not true. Rotation, as Rob Jeffries mentioned, is also important for certain stars, and it can have a role on the evolution of a star's luminosity.


Where a database of different z values of redshift can be found?

There are some great databases with a lot of information in them, when you found the galaxy you looked for, somewhere on the page there should be redshift information if it is has been measured at some point.

The NED/ipac database is quite nice for galaxies (https://ned.ipac.caltech.edu/).
Or SIMBAD is also great (http://simbad.u-strasbg.fr/simbad/).


There is no chance of a collision in the short term.  2020 CD3 is in a rather chaotic orbit of the Earth that extends well beyond the moon and doesn't come very close to Earth Each orbit is different, but no orbit brings it closer than the moon. The orbit is irregular because it is perturbed both by the moon and by the gravity of the sun. At its furthest point from the Earth. The gravity field that it moves in is not a simple "inverse square law" because of the combination of Earth, Moon and Sun.

It is likely to escape Earth in the next few months, and it will go back to be a sun orbiting asteroid, we know this because we understand how gravity works, so we can predict its future path. Its future path involves it leaving the neigbourhood of the Earth.  However as its orbit does bring it close to the Earth's orbit there is a chance of it colliding with the Earth at some point in the future. It would make a bright fireball, but would not be dangerous, as it is only a couple of metres across. Objects like this collide with the Earth frequently.

Its orbit is not yet well enough known to determine exactly how it will return to a solar orbit, and so we can't yet predict whether it will be temporarily captured or even collide with Earth on its next close approach. But the probability of a collision is low.


Could the big bang have created super massive black holes?

Arxiv prodives Haiman 2012 paper The Formation of the First Massive Black Holes. The leading theory, based on WMAP, is that the first perturbations resulted in dark matter halos, which were the first gravitational concentrations of matter.

  the first nonlinear objects in the universe were born inside ∼ 10^5 M⊙
  dark matter halos at redshifts of z ∼ 20−30

The non-linear objects do not refer precisely to black holes. The key part for us:

  [...] question of how the  first SMBHs  were  assembled. It is  worth
  emphasizing that this is an unsolved problem – indeed, it is not
  entirely clear even whether the first nonlinear objects in the
  universe were stars or black holes, and whether galaxies or their
  central black holes formed first.

The article proceeds to tell a very wide range of theories, but the summary is - we are quite clear about simulating the early dark matter halos but we really don't know how to clearly show whether/how it led to SMBH formation.

I'd like to emphasize what is (maybe) obvious: SMBH requires a very concentrated gravity. We assume that a very early universe despite being dense didn't have enough gravity. All theories are generally about how density dropped far down and in what ways could it start to grow again, but now only inside the concentrations (i.e. when the gravity started to affect universe, the density was much lower that inside stars or black holes).


What could a cloud of mini radio dishes see?

Partial answer:

I imagine this array would have very high resolution due to the large number of large and small baselines

and

I expect time synchronization would be an issue, but perhaps they could find a clever way to synchronize.

The 1 meter dishes are small and so for the dish to have any relevance to the project the wavelength has to be a lot smaller.
To do interferometry your synchronization will have to be extremely good; a 3 cm wavelength would be 10 GHz and the period is 0.1 nanosecond. You'll need to have all kinds of expensive and sophisticated electronics to maintain a stable and drift-free timebase at that level of accuracy.
You might be able to leverage your long baselines but fairly wide fields of view for these small dishes by using timing somehow. If you had ten distant dishes each pointed in 50 different directions and there was a sudden event, you might be able to localize it by arrival time differences, a bit like how lightning detectors work.
I don't know anything about it, but there is also something called intensity interferometry but it's not clear if that's actually helpful at all. See Hanbury Brown and Twiss effect and (paywalled) R. Hanbury Brown, R.Q. Twiss (1954) LXXIV. A new type of interferometer for use in radio astronomy

A new type of interferometer for measuring the diameter of discrete radio sources is described and its mathematical theory is given. The principle of the instrument is based upon the correlation between the rectified outputs of two independent receivers at each end of a baseline, and it is shown that the cross-correlation coefficient between these outputs is proportional to the square of the amplitude of the Fourier transform of the intensity distribution across the source. The analysis shows that it should be possible to operate the new instrument with extremely long baselines and that it should be almost unaffected by ionospheric irregularities.

See Boffin : a personal story of the early days of radar, radio astronomy and quantum optics
See also Dainis Dravens' 2010 BOSON INTERFEROMETRY From astronomy to particle physics, and back especially the radio part.
Only somewhat related, with some discussion of the effect before going on to discuss the optical implementation (but using electronic correlation): Intensity interferometry: Optical imaging with kilometer baselines
I wrote a related answer here: https://astronomy.stackexchange.com/a/42131/7982


How much volume of gases could be added to Mars atmosphere that the planet gravity can retain?

There are hard limits on what types of gas that Mars can retain based on its temperature and mass
(Graph of what gasses an astronomical body can retain).

Volume wise, it's not clear. Currently Mars is still losing its atmosphere, so it can't even retain that amount of atmosphere. But if you continually added gas to Mars, there isn't an end point where the planet could no longer hold the atmosphere being added. It just increases the pressure of the atmosphere at the surface and only slightly extends the height of the atmosphere. And after it stopped being added, Mars would slowly lose the atmosphere again.


There isn't such a great difference between Earth and Mars (at about 25 hours) and the gas/ice giants (between 10 and 20 hours). Earth had a day of about 6 hours in the early Solar system, but has been slowed by lunar tides. Small planets will not accrete gas in their early formation, and so will not pick up angular momentum from accretion discs, which can explain why Jupiter and Saturn have rapid rotation rates.

Mercury and Venus are slowed by Solar tides. In the case of Mercury, it is in a resonance between its orbit and the solar tide. In the case of Venus, there are significant atmospheric tides, that slow it down (It actually spins backwards, and the full explanation is incomplete.)

So the answer is three-fold: Distant gas giants pick up lots of momentum in their formation, they are not much affected by Solar tides, and the differences between them and the rocky planets are not actually so great anyway.


What is the organic material in Saturn's rings?

It is indeed a much more interesting cocktail of chemicals.
Before plunging into Saturn, Cassini collected some invaluable data about composition of its atmosphere and how the complex ring system works.
Cassini experienced some "heavy ring rains" in Saturn's innermost D ring, which turns out to be dust grains. The probe tried to use the Ion and Neutral Mass Spectrometer (INMS) to analyse the chemical content of this rain even though it was designed to analyse gases but the rain hit Cassini at such tremendous velocity that INMS instantly vaporized. It was able to collect some data about its composition which include molecular hydrogen, water, butane and propane and trace amount of  methane, ammonia, carbon monoxide, nitrogen and carbon dioxide. A second study was done when Cassini used Cosmic Dust Analyzer(CDA) and inspected 2,700 grains of dust and found that, while most of the material (around 95 percent) was water ice, there was also trace amount of silicates.
The material flowed from the ring downpour was huge. Cassini measured 22,000 pounds (10000 kg) of material per second falling from rings. Most of the chemical slurry is being purged from the D-ring and in turn the C-ring is somehow replenishing the D ring (which is playing a role in the composition of the planet's ionosphere and atmosphere).
Cassini also predicted the composition of E-rings by inspecting the ice grains ejecting from Enceladus. New kinds of organic compounds have been detected in the plumes bursting from Saturn's moon Enceladus. Powerful hydrothermal vents eject material from Enceladus' core, which mixes with water from the moon's massive subsurface ocean before it is released into space as water vapor and ice grains. The newly discovered molecules, condensed onto the ice grains, were determined to be nitrogen and oxygen-bearing compounds. The organic material found were categorized in three types: (i) Amine (ii) Carbonyl  (iii) Aromatics
Additional information of E-ring came from the compositional profile of ice particles from Saturn’s moon Rhea, from a series of spectra recorded on Cassini’s Rhea flyby (R4) in 2013. No striking change in composition is observed but the ice grains had a varying size distribution. Sodium salts and organic compounds were more frequent in relatively large ice grains, whereas pure water ice particles become more abundant in smaller E ring grains.
Source:

https://www.nasa.gov/feature/jpl/nasas-cassini-reveals-new-sculpting-in-saturn-rings
https://www.smithsonianmag.com/smart-news/saturns-rings-rain-organic-compounds-its-atmosphere-180970475/
https://www.sciencealert.com/saturn-ring-rain-heavy-chemical-composition-organic-compounds-cassini
https://www.nasa.gov/feature/jpl/new-organic-compounds-found-in-enceladus-ice-grains
https://www.space.com/saturn-moon-enceladus-organic-compounds-in-plumes.html
Khawaja, Nozair. (2017). Organic compounds in Saturn's E-ring and its compositional profile in the vicinity of Rhea. (link)


I don't think any of those would give a supernova, because the first two don't cross an important mass boundary, and the last one might make a black hole from a neutron star, but that's not going to be much of a supernova because it's already very compact and there's not going to be much to have a core bounce off of, nor is the white dwarf massive enough for a type Ia.  They should all produce some significant emission, depending on how fast the mass is introduced.  Perhaps somewhere between what we would call a supernova, and something more along the lines of a classical nova (for the first one) and quasiperiodic X-ray flaring (for the next two).   They could get pretty bright, and I don't know if the nomenclature has individual names for each one of those, but they don't quite sound like supernovae.


Minimum telescope needed to see the rings of the Saturn

Large binoculars are already enough. If you have some at home, try them. At our recent public outreach event, we had a 20x magnification on them.

You need a relatively heavy mount though, or at least have an attachable weight to a tripod. This is needed in order to damp shaking of your mount, which can severely impact the viewing experience. It would be not unusual for the mount to be more expensive than the binoculars.
Also the connector mount-to-binoculars should be some kind of smart design, because you might want to scan around the sky, i.e. change the binoculars orientation by a few degrees every so often during gobservations.

However if you don't want to use binoculars, then telescopes with ~ x100 magnification are a safe bet for bright objects. If you want to observe faint objects, you have to think also about the telescope diameter, field of view,...
For telescopes there is also the issue that small children often cannot properly view through the one-eyed telescope pieces, or are unable to process the instructions like "keep your eyes two fingers from the viewing thingy", and then report seeing nothing. This is considerably easier for kids with regular binoculars.


Viewing a solar eclipse through a leafy tree

It seems that the station has, in fact, given out extremely dangerous advice - simply because it was poorly phrased.

Viewing a solar eclipse is incredibly dangerous - mostly under circumstances where you're looking towards the Sun. As Wikipedia warns,

  Looking directly at the photosphere of the Sun (the bright disk of the Sun itself), even for just a few seconds, can cause permanent damage to the retina of the eye, because of the intense visible and invisible radiation that the photosphere emits. This damage can result in impairment of vision, up to and including blindness.

You couldn't pay me enough to so much as face toward the Sun and look at my feet and attempt to view a solar eclipse out of the corner of my eye!

There are safe ways to view an eclipse, the most common being using a pinhole camera to indirectly view the eclipse. Not only are you not looking directly at the Sun, but you can actually turn away and look at the projection.

But here's where I think the station messed up. According to CNN,

  Or let a tree do the work for you. "Overlapping leaves create a myriad of natural little pinhole cameras, each one casting an image of the crescent-sun onto the ground beneath the canopy," NASA says.

So NASA actually advocates using the leaves as a pinhole and using indirect viewing methods - i.e. looking anywhere but the Sun - as opposed to watching the eclipse through the leaves.


How would I search within an angular distance of an object corresponding to a distance in parsecs?

Ken G's answer is essentially correct, but there is one important thing to keep in mind: The distance $r$ between the galaxies that is interesting for you is the distance they had at the time they emitted the light you see (you don't care about what has happened to them since that time and how far apart they are today; you care about the physical properties then). But the angle you observe them to span is not angle they spanned when they emitted the light (since they were closer to you in the past, so the light they emitted in your direction at that time would "miss" you by the time it reached that place).

This means that, in your calculation, the distance from you to the galaxies should not be the "physical" distance $d$ (i.e. the actual distance right now to the galaxies, which also corresponds to the comoving distance), but the so-called angular diameter distance, $d_A$. These are related (in a flat universe) by
$$
d_A = \frac{d}{1+z},
$$
where $z$ is the redshift of the galaxies (I assume they have almost the same redshift; if they are too different, then the distance from each other is not simply given by the projected distance). Note that in a non-flat universe, the formula is slightly more complicated; see the link above.

The comoving distance, in turn, is obtained by integration:
$$
d(z) = \frac{c}{H_0} \int_0^z \frac{dz'}{\left( \Omega_\mathrm{M}(1+z)^3 + \Omega_\Lambda \right)^{1/2}},
$$
where $c$ is the speed of light, and $\Omega_\mathrm{M}$, $\Omega_\Lambda$, and $H_0$ are the present matter and dark energy density parameters and Hubble constant, respectively. Here, I've again assumed a flat universe, and neglected radiation since when that was important, there were no galaxies anyway).

Obviously, this effect is not big for small redshifts, but $d_A$ has the interesting property that it increases with physical distance only out to a certain point (roughly $z=1.6$), after which it decreases, because the galaxies were closer to us in the past.

So the physical distance $r$ between two galaxies spanning an angle $\theta$ is
$$
r = \frac{\theta}{d_A}
$$
or
$$
\boxed{
\frac{r}{\mathrm{kpc}} = \frac{\theta / \mathrm{arcsec}}{d / \mathrm{Mpc}} (1+z)
 \, \frac{10^{-3} \times 180\times3600}{\pi},
}
$$
where the last factor accounts for converting Mpc to kpc and radians to arcseconds.

In Python, arbitrarily for $z=2.2$ and $\theta=12.3''$,  it's as easy as

from astropy.cosmology import Planck15
from astropy import units as u
z     = 2.2                               # redshift of galaxies
theta = 12.3 * u.arcsec                   # angle
r_ang = Planck15.kpc_proper_per_arcmin(z) # phys. dist. per angle
r     = r_ang * theta                     # physical distance
print('Distance between galaxies: ', r.to(u.kpc))

which will print Distance between galaxies:  104.22540165050975 kpc.


How can a supermassive black hole be 13 billion years old?

The answer to this is unknown at the present time.
The issue is that an accreting "seed black hole" can only accrete at a limited rate. The limitation is provided by radiation pressure from the material it is accreting. This provides negative feedback and defines a maximum accretion rate for spherical accretion known as the Eddington limit, which defines a maximum luminosity and hence a maximum accretion rate (since accretion provides the luminosity).
The net result is that the black hole can grow exponentially, with a growth timescale of about 30-40 million years. That is, it can roughly double in mass in that timescale.
Now the first, primordial stars may have been much more massive than big stars in the universe today, as a result of their pure hydrogen/helium composition. Even if they were able to form up to a few hundred solar mass black holes, it would still take almost a billion years to build a $10^9 M_\odot$ black hole, even if it accreted at the maximum Eddington rate for all of that time.
This is a bit of a stretch so alternative solutions are being sought. These include

Maybe you can form even bigger black holes from early collapsing gas clouds. There are some ideas that if the primordial gas gets ionised by the very first massive stars then it doesn;t fragment into "stellar" objects. This can enable you to form very, very massive "stars" (sometimes called "quasi-stars"), which have black holes at their centres rather than nuclear-fusing cores. These might grow rapidly to be of order $10^5 M_{\odot}$ This can shave of many exponential growth timescales by starting off with a more massive seed.

Maybe you can merge black holes. We now know from gravitational wave detections that mergers can proceed between black holes with tens of solar masses to form $\sim 100M_\odot$ black holes. Perhaps in the early universe there may have been clusters of high-mass black holes that managed to merge with each other on short timescales. These would then form a massive seed black hole that could grow relatively quickly.

There are some ideas about how the Eddington limit can be circumvented and mass accretion can proceed at higher rates, perhaps by suppressing the radiative efficiency of the accreting black hole (e.g. Li et al. 2012).


Gaia's original science and technology report (see page 221, see also the summary) gives an analysis of the Lissajous orbit. From what I understand Gaia will be placed in a small amplitude Lissajous orbit, giving it an orbital radius of $\sim400000$ km away from $\sim100000$ km along the Sun-Earth axis.

In addition to the fact that this orbit is intrinsically unstable, variable radiation pressure from the sun causes stochastic destabilization of the orbit. The prediction is that once in orbit, a small velocity correction of $\sim1$ m/s is needed about once a month in order to maintain the orbit.

From a summary on Planck's orbit it seems that this applies to Planck as well.

That the SEL2 point is being 'crowded' doesn't seem like such a big problem to me. Once the satellites are turned off (both Planck and Herschel are no longer active) their orbit will rapidly destabilize, effectively removing then from the region. But more importantly, the orbital radius is comparable to the distance to the moon, this really is a tremendously large amount of space. Since these satellites are only a few metres across, the probability of a collision is likely negligible.


The expansion is on a scale larger than galaxies, so within a galaxy, two black holes or neutron stars might pass close to each other from time to time as they move under the influence of the galaxy's gravity. If a third object gets involved (eg a star) then they could end up being captured into orbit . Much more likely, however, is both components of a binary star (which are common) evolve inbto black holes or neutron stars so they are orbiting one another "from birth".
This is a difference between general relativity and Newtonian gravity. In GR no orbit is ultimately stable -- energy is slowly lost as gravitational waves and the objects spiral in. If the gravitational fields are not very large, however, this effect is much too small to detect, and overwhelmed by other effects, so it seems like planets around the sun are orbiting stably. Only when you have objects with a very strong gravitational field in a very close orbit is it noticeable. Even then we're only detecting a few such events per year within a billion light years of Earth, so it's not common.


The presentation linked to seems to primarily  discuss General Circulation Models, solutions to the equations of fluid dynamics which predict the behavior of a planet's atmosphere, a component of more complete Global Climate Models (confusingly, both are GCMs).

Spectral data on exoplanets is extremely limited to date and poorly constrains the composition, but a lot can be inferred from a planet's orbit and eccentricity (how far from circular the orbit is). At the simplest level, planet temperature depends on how much light it receives from its parent star, and this will vary as a planet goes around a non-circular orbit. Starting from there and making reasonable assumptions about the composition of the planet (usually starting with a solar system analogy) you can test out a GCM model in the various extreme cases we observe (hot Jupiters etc.), to be subjected to future observational constraints.

One exciting new development is looking at a planet's phase curve (the observed brightness of the planet as it orbits its parent star) and using that to infer properties of the atmosphere. One team recently found that Kepler 7b's phase curve  may best be explained by the presence of clouds:

Demory, et al. 2013. “Inference of Inhomogeneous Clouds in an Exoplanet Atmosphere.” arXiv:1309.7894 [astro-Ph] (September 30). http://arxiv.org/abs/1309.7894.


This is a common problem with Newtonian type telescopes and DSLRs, the sensor in the camera is too far back to reach focus. As mentioned in the comments a low profile focuser may help or move the main mirror up the tube, I've seen this done on some Skywatcher Newtonians to solve similar problems.


Calculating distance of comet C/2019 Y4 (ATLAS)

It appears that you are doing a simple parallax calculation, using a baseline obtained from the Earth's rotation over an hour. You appear to be neglecting the most significant change, the motion of the comet.

If you plot the position of C/2019 Y4 in SkySafari, you will see that around this date the comet's location against the star background shifts about 37 arc minutes per day, or a bit over 1.5 arc minutes per hour. The location is based on the orbital elements of the comet (MPEC 2020-C76), but also includes the effect of the Earth moving in its orbit, as mentioned in the comments.

There is no simple way of determining the distance of a solar system object from two position measurements alone. A rough estimate of the orbital elements requires at least three position and time measurements; the MPC orbital elements referenced above are based on many observations over several months.


You're confusing error and uncertainty.

In school labs or other situations where the true value of a quantity is known, the difference between a measured value and that true value is sometimes referred to as the "error" in the measurement. A lower error is taken as an indication that the results are successful.

When making observations, the true value of a quantity is seldom known. Indeed, that's usually the purpose of making observations: not as a pedagogical tool but as a way to determine something about the world. In this case, we're concerned with the uncertainty in the measurement - that is, essentially how confident we are in the value we determine. More quantitatively, we might give something like a $3\sigma$ uncertainty that explains how likely it is that the true value falls within 3 standard deviations of the measurement. That's a way of showing how robust the results are.

It's also helpful for comparing sets of measurements - if one group says that (and I'm making up numbers here) that their measurements have determined the Hubble constant to be $H_0=67.0\pm1.2\text{ km}\text{ s}^{-1}\text{ Mpc}^{-1}$ and another group says that their methods have found it to instead be $H_0=74.00\pm0.75\text{ km}\text{ s}^{-1}\text{ Mpc}^{-1}$, then we can see from the uncertainties that the two results are in fairly strong tension with one another.

If you're determining something like redshift, you presumably have some raw data measuring other quantities. Perhaps you've identified H$\alpha$ emission from a source and found it to be at $\lambda=8540\pm30$ angstroms, while the rest wavelength is $\lambda_0=6562.8$ angstroms. Then you can use standard uncertainty propagation methods to determine the uncertainty in the redshift of the source. In this case, the redshift is
$$z=\frac{\lambda}{\lambda_0}-1$$
The uncertainty in $z$ is then
$$\sigma_z=\sqrt{\left(\frac{\partial z}{\partial\lambda}\right)^2\sigma_{\lambda}^2+\left(\frac{\partial z}{\partial\lambda_0}\right)^2\sigma_{\lambda_0}^2}$$
where I've included $\sigma_{\lambda_0}$, though realistically it should be much smaller than $\sigma_{\lambda}$. Keep in mind also that $\sigma_{\lambda}$ has itself presumably been determined through error propagation from the raw data collected.

(Side note: As pela pointed out, this assumes that the redshift is entirely due to the motion of the object you're looking at, and that's not always the case. I originally picked Lyman $\alpha$ has my line, which turns out to not be particularly useful. Additionally, the toy redshift ended up being quite low and therefore even more sensitive to other not-recessional contributions, like stellar winds. H$\alpha$ is, I think, a better example, especially with this higher redshift.)


How is it possible that we haven't discovered anything in the Oort cloud yet?

It's not very surprising that no Oort Cloud objects have been detected via occultations. They're extremely rare, even for our most advanced space-based observatories.
According to Ofek & Nakar 2010, published about one year after the launch of Kepler, the telescope would be able to detect between $0$ and $\sim100$ occultation events of Oort Cloud objects, assuming that it monitored $\sim10^5$ stars for a time $\tau=3$ years - both of which ended up being underestimates.$^{\dagger}$ Part of the reason for the uncertainty is that the parameters of the Oort Cloud are not tightly constrained. Varying the inner radius $r_{\text{min}}$ from 1000 to 5000 AU, or varying the index $\alpha $ of the number density distribution $n\propto r^{\alpha}$ from $-4$ to $-3$, can produce changes on the expected occultations by a couple orders of magnitude.
The problem becomes apparent when you consider that 1) occultations are intrinsically short and infrequent and 2) they're one-off events. It's not the same as detecting a transiting exoplanet, or observing an expected occultation from a minor planet whose orbit is already known. Therefore, it's hard to tell whether an event is a true occultation or simply a glitch of some sort. The second point of the paper was to present methods for validating possible events.
Ground-based surveys of Kuiper Belt Objects (KBOs) have been performed, but they suffer from atmospheric effects, as you might expect. Given that occultations would last on the order of 1 second, atmospheric scintillation becomes a problem for the high-cadence searches required (Alcock et al.). Given that some ground-based surveys just for KBOs a few kilometers across, like the Taiwanese-American Occultation Survey, have returned no detections (Zhang et al. 2008), it might not be surprising that ground-based Oort Cloud searches are also extremely difficult.

${\dagger}$ By comparison, the CoRoT mission, assuming it monitored $\sim10^4$ stars over the same period, was expected to detect essentially no occultations, according to the same analysis.


Recommendation for learning about stellar astrophysics

Best book I have used for this topic is: An Introduction to Stellar Astrophysics by Le Blanc.

Excellent book, very good introductory start into this topic.


What is the axial tilt of a planet measured relative to?

If you look at the Astronomical Almanac for the year 2011 as an example, in the table at the top of page E3, you find two measures of axial tilt. The third column is the declination of the planet's north pole for the mean equinox and equator of date 2011 January 0, 0 hours terrestrial time. So this is measured with respect to the Earth's orbit. The last column is the inclination of the planet's equator to the planet's orbit.

If you look at the right ascension column in the same table, you see that on that particular date the right ascensions are all different.


Milky Way as seen by the human eye: where exactly do those photons come from?

I believe that the answer is mainly the first (directly from stars), although there will certainly be some of each of the others. The diffuseness comes from the fact that there are many more relatively dim stars than rods in the retina of the human eye.  My argument for this is that magnified photographs of most of the Milky Way at least, break down into individiual stars, as seen in this image from astropixels.com


Is the angular resolution of a telescope irrespective of used eye-piece?

The resolution of a telescope is the resolution of the image created by the primary mirror at the focal plane.  It provides the minimum separation between two equal brightness stars that appear separate in that image.  Often, astronomers put a photographic plate or a CCD at the focal plane, create long exposure photos and these can then be examined at arbitrary resolution (with a magnifying glass, microscope, or zoom software on a screen).  But, no matter how much you magnify the image, stars separated by less than the resolution will appear merged.

It is true that if you use, say, a wide-angle lens when observing directly with a telescope the resolution that you get can be limited by the resolution of your eye.  What is happening is the telescope is forming an image on your retina that has the full resolution, but your eye cannot make use of the finest details.  On the other hand, a hawk looking through the same eyepiece may get full resolution benefits since hawks' eyes have better resolution than humans'.


How does the stellar evolution for low and intermediate mass stars differ?

Where does the evolution actually stop? Is the Helium Core Flash the last thing a Sun-like stars experiences or does it in fact follow the rest of the tractory?

Stars with masses below about 2 solar masses undergo a helium flash but their evolution does continue afterwards, as shown in the second diagram (Fig. 13.4). So no, the core flash is not the last thing a Sun-like star experiences.
The tracks are sometimes separated like this for the technical and practical reason that many programs that create stellar models struggle to model the helium flash. To get around this, the models sometimes stop at the tip of the red giant branch and are then restarted at stable core helium burning by "engineering" a model to the correct state, without going through the flash.


Will the Earth ever be tidally locked to the Moon?

As the moon orbits Earth, tidal forces slow down the Earth's rotation by 2 milliseconds per century. Eventually, in tens of billions of years, the Earth and Moon would achieve a double tidal lock, where both are stuck with one side facing the other as they orbit the Earth-Moon barycenter. In 7.5 billion years, the Sun will expand past the Earth's current orbit, but the Earth may drift out further, preventing it from being vaporized.

However, this is beside the point, because in about one billion years, all of Earth's water will have boiled away, meaning that there would be no more ocean tides, and thus the Earth-Moon system will likely never achieve a double tidal lock.

References:

Will the Moon ever leave Earth’s orbit? - Giles Sparrow, Space Answers
Harvesting Lunar Eccentricity? - Terry R. McConnell, Syracuse University
When Will Earth Lock to the Moon? - Fraser Cain, Universe Today
The Sun Will Eventually Engulf Earth--Maybe - David Appell, Scientific American
Distant future of the Sun and Earth revisited - K.-P. Schröder and Robert Connon Smith, MNRAS
When will Earth lose its oceans? - CNRS, Science Daily


How does the evolution of a solar system not break the second law of thermodynamics?

The total entropy actually increases, as the molecular cloud shrinks under gravity.

It may seem that as the molecules are getting closer, they are more ordered, which means less entropy. That is however only one part of the process. The second (important) part is: when the molecules are closer, they also have higher kinetic energy (since they descended into lower gravitational potential). So the gas is getting hotter, as it shrinks.

The increase in temperature of the gas is increasing its entropy, because the molecules occupy more momentum space. This increase in entropy via temperature is bigger than the decrease in entropy via the shrinking itself.

Later on, the hot condensed gas (or hot planet) radiates the heat into the space and cools down. You end up with a cold planet which indeed has lower entropy than the original gas cloud, because it is no longer hot. But the entropy increase was carried away by the radiated photons. So in total - entropy of the universe increased (the radiated photons are out there somewhere).

You can find some more detailed discussion of this topic on the excellent web page by John Baez or here.


How to calculate the movement of the object passing near other object in space?

Since you're interested in the equations of motion, I would solve this problem by using Lagrangian mechanics. Essentially, find the kinetic and potential energies for these two bodies, A and B.

Construct the Lagrangian:

$$L = T - V$$

where T is the kinetic energy, and V is the potential energy. Then use the Euler-Lagrange Equation to achieve the equations of motion (I would add it here, but I'm not sure of the specifics of your problem).

Two-body motion can always be constrained to a plane, so you may have to throw a constraint in there through the use of a Lagrange multiplier.


Laser Collimator producing oval not circle on the Primary

First of all, what do stars look like through the scope, particular just inside and outside focus?  If they are oval, then read on.

I'm not familiar with this telescope, but I suspect the most likely sources of the problem are
1. The corrector lens you mention (which is a relatively unusual thing to have) is distorted.  If you can't remove this, then you are stuck with it.
2. The adjustment screws on the secondary mirror may be too tight, thereby distorting the shape of the .  Try loosening each screw by a small amount, say 1/8th of a turn, and see if this improves things.  Make sure the telescope tube is horizontal when you do this in case the secondary slips off its mount.


Which is more rare: Lunar eclipse or Solar eclipse?

Solar and lunar eclipses occur about equally often - between two and four times per year.

However, when you do not intentionally travel around the world chasing solar eclipses, you are more likely to observe more lunar eclipses. The reason is that solar eclipses can only be observed from a comparatively small area while lunar eclipses can be observed from anywhere were the moon is visible when they occur, which is half the planet.