Does the Reflection Method for Looking at the Sun Work For Refractors?

Actually, your Newtonian may run the risk of suffering from the projection method. The secondary, planar mirror in the Newtonian will be subject to heating. Although it is not in the prime focal point, the light rays are already pretty much bundled up, and this may lead to heating and ultimately to failure of the secondary mirror!

With refractors, you don't have anything in the optical path up until the eyepiece, so in a sense it is more safe.

However: I strongly suggest against using simple projection methods. One alternative is to use a Herschel-wedge. However, these are again not that useful on Newtonians, because they filter the light at the eyepiece end of the telescope. It is probably better to buy a solar filter which is rather inexpensive and can be fitted to any kind of telescope. Without wanting to promote any specific brand, the company Baader has a good explanation on the topic on their website. The important thing is that the filter is securely fastened to the optical tube assembly, and that there are no scratches or cuts in the filter foil.

Another much more expensive method is to use a specialized solar telescope. These usually use very narrowband filters and coatings on the telescope lenses. Often they are specialized to one of the hydrogen or calcium spectral lines, showing great amount of structure in the suns atmosphere.


Why are larger hex mirrors round and the smaller aren't?

Start by  drawing the largest inscribable circle (circle completely contained within the mirror assembly).  Then  draw, e.g., the smallest superscribing circle (completely contains all the mirrors), and calculate how many more hexagons are needed to fill the additional area covered.

Finally, draw the circle showing the design entrance aperture of the telescope to see how much of the aperture is lost vs. how much of the outer hexagons' area is unused in each case.

Then do a cost vs.  quality tradeoff.  Take into consideration (for the smaller telescopes) the delta cost of using more, smaller hexagons.


How are cardinal points on horizon defined?

Geographic Meridian
Every point on Earth is on some meridian:

A (geographical) meridian (or line of longitude) is the half of an imaginary great circle on the Earth's surface, terminated by the North Pole and the South Pole, connecting points of equal longitude, as measured in angular degrees east or west of the Prime Meridian.

From where you are standing, North is the direction on the horizon along this meridian or "line" of longitude that extends to the Earth's geographic North pole, and South is the direction on the horizon 180 degrees opposite, towards the South pole.
East and west are the perpendicular directions, which are also found extending lines of latitude from where you are standing towards the horizon.
Locally you can think of the region under your feet as flat, and so North, South, East and West all form right angles even far from the equator.
Close to the poles, none of that changes mathematically, and exactly on the poles those terms suddenly no longer have any meaning.
Celestial Meridian
meridian:

In astronomy, the meridian is the great circle passing through the celestial poles, as well as the zenith and nadir of an observer's location.
Consequently, it contains also the north and south points on the horizon, and it is perpendicular to the celestial equator and horizon. A celestial meridian is coplanar with the analogous terrestrial meridian projected onto the celestial sphere. Hence, the number of celestial meridians is also infinite.

Source


Why would astronomers want to eclipse the Sun?

As Conrad Turner notes, that talk is about a device for blocking the light from another star (one you think might have planets orbiting around it). It's not for blocking the light of the Sun!

If you try to look at a planet orbiting around another star, the glare from the star makes it very hard to see the planet. By placing a specially shaped device in front of the telescope (in this case, thousands of kilometers in front), between the telescope and the star you're interested in, you block most of the light from the star, but not the light from any planets near that star.

https://en.wikipedia.org/wiki/New_Worlds_Mission


Determining needed exposure time given SNR

Here's a simple start, and I suggest you must include your sensor's noise -- not flat-fielding, but just the accumulated readout noise in a pixel.
You can use the Poisson SNR rule for the incoming light, so optical noise $N_o = \sqrt{S_o}$.  Just add the electronics noise, again as a simple model, $N_e = C_(readout) + k\tau$ where $\tau$ is the integration time and $k$ represents leakage current into the pixel well.   $S_o$ also equals some constant times $\tau$ , so just combine the two noise terms in an RSS manner, i.e. $ N_{tot} = \sqrt{N_o^2 + N_e^2}$ .


Does gravity bend light, and how much time does it take for light to cross gravity of a Black Hole?

You are quite right: Einstein's theory says the curvature of space is locally deformed. The essence of this is captured in the spacetime 'metric', a mathematical tool that tells us what space looks like and, derived from this, what is meant by 'a straight path', which photons take. If there is no source of gravitation present, the path of a photon will be what you know intuitively as a straight line. However, for some mass concentration (e.g. a black hole, as you say), this path will be bended such that the mass concentration acts as a lens. This is immediately clear from this image from the CFHTLenS survey:

Crucially to your question I think, you must remember that photons do not experience time and their speed is equal to $c$, the speed of light. Photons are not unaffected by their movement through a gravitational field, mind you: but this shows up as a gravitational redshift (for a time-varying potential), rather than a time delay (other than the slightly elongated path, perhaps).

Besides that, the age of the Universe is typically not quite measured as you say, but rather through parameter estimation in e.g. the cosmic microwave background. The effect of gravitational lensing needs to be taken into account for that, but not in the way you presume.

Also, remember that an extreme gravitational field such as that of a black hole is relatively rare, and even if it did delay the photon for a 1000 years that is still a tiny fraction of the age we would, according to your way of thinking, infer; the photon would have to encounter a LOT of such black holes for this 'effect' (which does not occur) to have a huge impact.

As a general point, you seem to have some (Interstellar-induced?) misconceptions about the universe, but I think that's not for me to address here.


Is night sky getting brighter day by day?

The rate of star formation in the universe peaked about 3.5 billion years after the Big Bang, or 2.7 billion years after it, according to a different study—in any case, for most of the universe's lifetime fewer and fewer stars have been forming each year. Though I don't have precise data on the death rate, on the whole the number of stars is decreasing as more old stars die than new ones form.

Meanwhile, the light from old stars is being redshifted out of existence by the expansion of the universe, which is itself accelerating.

I'm sure there's a mathematical model that can quantify this properly, but I think those two facts, combined, must be the answer to why the sky isn't getting brighter.


The most comprehensive catalogue of strong gravitational lenses that I am aware of is the CASTLeS database with about 120 lens systems. Several of them have extended features, arcs, etc., while others are point sources. They provide the redshifts of lens and source where available, as well as time delays between multiple images and the velocity dispersion (a proxy for the mass) of the lens for a few.

You would have to create a comparison dataset without strong lensing yourself, but that should not be difficult: Go to Hubble's Ultra Deep Field web page, download the images with the same filters as used in CASTLeS and cut out likely looking galaxies. The UDF contains mostly very high-redshift galaxies, so the chance of accidentally picking a strong-lens system is very slim.

(Disclaimer: I've only ever used ACS UDF images, not the NICMOS ones; if it's too hard, just skip the IR data.)


If Alpha Centauri A's solar system exactly mirrored our own, what would we be able to detect?

This is a broad question and too broad for me to answer comprehensively. It should be broken down into doppler methods, transits and direct imaging; and that's before we get to questions of detecting Kuiper belts, radio emission etc.
I'll stick for the moment with what I know about detection of planets using the doppler wobble technique.
Doppler Technique
The reflex radial velocity semi-amplitude of a star for the case of a planet of mass $m_2$ orbiting a star of mass $m_1$, in an elliptical orbit with eccentricity $e$, and orbital period $P$ and with an orbital axis inclined at $i$ to the line of sight from Earth is:
$$  \left( \frac{2\pi G}{P}\right)^{1/3}\frac{m_2 \sin i}{m_1^{2/3}} (1-e^2)^{-1/2}.      $$
A (very) detailed derivation is given by Clubb (2008).
So I built myself a little spreadsheet and assumed that all the planets were seen optimally at $i=90^{\circ}$ (they could not all be seen optimally, but the smallest inclination would be about $i=83^{\circ}$ for Mercury, so it doesn't make too much difference) I'll also assume the mass of Alpha Cen A is about $M \simeq 1.1M_{\odot}$.
The results are

Planet
RV semi-amplitude (m/s)

Mercury
$8.3\times 10^{-3}$

Venus
$8.1\times 10^{-2}$

Earth
$8.4\times 10^{-2}$

Mars
$7.5\times 10^{-3}$

Jupiter
$11.7$

Saturn
$2.6$

Uranus
$0.28$

Neptune
$0.26$

The limits of what are possible are well illustrated by a planet around Alpha Cen B, claimed to be in a 3 day orbit and with a mass similar to the Earth (Dumusque et al. 2012, and see exoplanets.org). The radial velocity semi-amplitude detected here was $0.51\pm 0.04$ m/s, and some spectrographs, notably the HARPS instruments, are routinely delivering sub 1 m/s precision. Thus Jupiter and Saturn would be detectable, Uranus and Neptune are right on the edge of detectability (remember you can average over many RV observations), but the terrestrial planets would not be found (Earth detections would require precisions below 10 cm/s. Remember also that the weaker signals would have to be dug out from the larger signals due to the Jupiter- and Saturn-like planets.
However, there is a second limitation: to find a planet using the doppler method you need to observe for at least a significant fraction of the orbital period. Given that current m/s precisions have been available for only $\sim 5$ years, it is unlikely that Saturn would yet have been detected.
A picture that illustrates the situation can be obtained from the exoplanets.org website, to which I have added lines that approximate where RV semi-amplitudes would be for 10 m/s and 1 m/s precision (assuming the Alpha Cen A mass and circular orbits). I've marked on the Earth, Jupiter and Saturn. Note that few objects have been discovered below the 1 m/s line. Also note the lack of planets between the 1 and 10m/s lines with periods longer than a couple of years - the recent increase in sensitivity has yet to feed through to lower mass, longer period exoplanet discoveries.

In conclusion: only Jupiter would have been so far found by the doppler technique.
Transit techniques
I'll also add a few comments about the transit technique. Transit detection will only work if the exoplanets orbit such that they cross in front of the star. So high inclinations are mandatory. Someone who is better at spherical trigonometry should use the published data for the solar system to work out how many (and which) planets transit in some highly optimal orientation. Given that the planets have orbital inclinations with a scatter of a few degrees, then some straightforward trigonometry and a comparison with the solar radius, tells you that these orbits will generally not all transit for any particular viewing angle. Indeed a number of the Kepler-discovered multiple transit systems are much "flatter" than the solar system.
The Kepler satellite is/was capable of detecting very small transiting planets thanks to its very high photometric precision (the dip in flux is proportional to the square root of the exoplanet radius). The picture below, presented by the NASA Kepler team (slightly out of date now), shows that planetary candidates have been discovered that are down to the size of Mars. However these tend to be in short period orbits because a transit signal needs to be seen a number of times, and Kepler studies this patch of sky for about 2.5 years (when this plot was produced).
So from this point of view, possibly Venus would have been seen, but none of the other planets could be confirmed.
However, there is a wrinkle. Alpha Cen A is way too bright for these kinds of studies and way brighter than the Kepler stars. You would have to build a special instrument or telescope to look for transits around very bright stars. Some of this work has been done by ground-based surveys (mainly finding hot Jupiters). A new satellite called TESS (Transiting Exoplanet Survey Satellite, launched in April 2018) is a two year mission, focused on finding small planets (Earth-sized and bigger) around bright stars. However, most of its targets (including Alpha Cen) are only observed for a 1-2 months, so only the inner parts of their planetary systems will be probed.


If the moon wasn't receding from Earth, what would be the impact on the weather and tides?

Tidal amplitudes roughly scale inversely with distance to the fourth power (this comes from the tidal force scaling as $r^{-3}$ and the Earth's potential scaling as $r^{-1}$). At its formation, the Moon was approximately seven Earth radii away from Earth, or about ten times closer than it is today. Thus, the amplitude of the tide on Earth would have been approximately $10^4$ times larger than it is today. As the typical ocean tide is 1 m, this means ocean tides would have been 10 km high on Earth! Of course, the collision that formed the moon turned the surface of the Earth molten, so really we're talking about a 10 km tall molten tide, which is far less scary...(kidding).

Imagine all the noise the ocean makes when it crashes on a beach. That noise requires energy to produce, and is partially driven by the action of raising/lowering the ocean via tides. Now imagine how loud a 10 km wave crashing would be. This is why the Moon receded quite quickly from the Earth shortly after forming, the tides enable the dissipation of a lot of orbital energy.

As for the hypothetical effect of this supertide on the present-day Earth's weather, the Earth's troposphere (which contains pretty much all of the Earth's weather phenomena) extends 10–20 km from the ground. Considering the height of the ground is changing by a similar amount due to the tide, I would imagine the effects on the weather would be quite dramatic.


Is the Object in this Photo Mars?

Almost certainly yes if it didn't move like a plane or hover and make noise like a quadcopter.
in-the-sky.org' skymap shows Mars at 117 degrees East and 25 degrees altitude at 21:43 local time in Woerden in central Netherlands yesterday. Clicking on Mars yields more information including a charts showing that it's close to opposition.
I took a photo from about 25 degrees north almost the same day I think :-) It was higher in the sky for me but just as bright and similarly cloudy. The big photo is intermediate focus. The two below it are best and bad focus. Even when defocused to 200 pixels in area it show brightly, when in focus it is saturated and so just looks like a few white pixels.


Calculate apsides without knowing eccentricity

You mention this is a system in which a vessel is orbiting a planet. In that case it's fair to assume the mass of the vessel is negligible compared with the planet mass and
the planet will sit in one of the focal points of the ellipse.

You can then use Kepler's second law to find the semi-minor axis as
$$ \frac{1}{2} r v = \frac{\pi a b}{P} $$
where $r$ the current distance and $v$ the current velocity, $P$ is the period and $a$ and $b$ give the respective semi-major and semi-minor axes.

From $b$ you find the eccentricity as
$$ b=a \sqrt{1-e^2} $$

And the apsides follow as
$$ \mathrm{periapsis} = a ( 1 - e ) $$
$$ \mathrm{apoapsis}  = a ( 1 + e ) $$


I didn't read the SciAm article, but: Any cosmological theory compatible with "the" mainstream big bang theory doesn't assume material to have been contained in "Our Universe" (needs to be defined) from the very beginning.
Instead a very hot, unified state, is assumed, which then underwent several symmetry breakings (https://en.wikipedia.org/wiki/Symmetry_breaking, https://en.wikipedia.org/wiki/Higgs_mechanism), before "material" came into existence, as we observe it by now.

So, the very hypothetical, presumed four-dimensional star just needs to trigger an initial state causing the big bang. Whether "Our Universe" is a brane universe embedded in a higher-dimensional universe, is another mere hypothesis. Whatever "star" or "black hole" in a higher-dimensional universe would mean, the physics would be very different from the physics we can observe. For instance, inverse square laws wouldn't hold due to the additional dimension(s), hence orbital dynamics, electromagnetism needed for atoms, etc., all would either fail, or work rather different. Presuming the existence of stars in such a hypothetical universe is very daring, already.


Was time different before the great inflation?

Food for thought on density and time dilation.  Density isn't gravity.  We tend to think it is, but it's not.  Variation in density gives us gravity.

If you have an infinite universe, very young, hot and dense, but it's uniform, then you have no time dilation.  One way to think about this is to ask if light passing through the universe is red shifted or blue shifted.  If the universe is uniform and infinite, it's the same in all directions around the photon, so the photon is neither red nor blue shifted, so, contrary to what our intuition about relativity tells us, there might not have been a great deal of time dilation in the very young universe.

Of-course the ray of light passing through expanding space is enormously red-shifted, but that doesn't affect time dilation.  The gravitational red or blue shift in the very young universe might still be close to zero meaning very little time dilation anywhere.

As the early universe grew less uniform then you had time dilation variations in denser parts of the early Universe but with FTL expansion and gravity limited by the speed of light, it's unclear if there ever was massive time dilation in the young universe (at least, unclear to me, if somebody here knows 100%, feel free to post and/or correct me).

Now if we assume the universe is finite and flat and the big bang happened at a point, which, generally isn't the model most people like these days, then you could run into enormous time dilation inside the tiny young and dense universe, but it still expanded FTL, so the calculations get weird and too hard for me.   If the expansion of space happens faster than gravity can keep up, gravitational time dilation might not apply, at least not until after the rate of expansion slowed down enough for gravity to permeate the local regions.  I couldn't even begin to run the math on this and you'd have different answers for different models too, I'd think.   I agree with Andy though, it would need to be accounted for or the models wouldn't be much good.


Is Polar Lights observation possible at the latitude of 56° N?

Yes. The Aurora form in a ring, roughly centred on the Magnetic pole. The pole does wander slowly, but it has been located in the Canadian Arctic for some time.

As the pole is on the American side of the Arctic ocean, aurora are seen at lower latitudes in America than in Russia, nevertheless, when a powerful solar storm occurs, aurora can be seen at surprisingly low latitudes, even into the subtropics, as Wayfaring stranger notes in her comment.

Aurora in Moscow is a notable, but not exceptional occurrence, as this news report indicates http://www.themoscowtimes.com/article/507096.html


Projection is a safe way to view the partial phase of the eclipse.

A pinhole produces a small image of the sun, but doesn't require any special equipment. A larger image can be obtained by projecting through a telescope:

Point the scope at the sun, but never look down a scope that is pointing anywhere near the sun, neither look along the tube. Use its shadow to line it up with the sun.  Hold a white card half a metre or so behind the eyepiece, and focus the telescope to make a sharp image of the sun. You will want to provide some shade for the image.

Image from Bakersfield College

This works well with binoculars, you need some way to mount, or prop up the binoculars, and you will need to cover up one of the lenses (or you will get two overlapping images.

If you have a larger telescope, you will want to cut down the aperture, to prevent overheating.

You will be able to see sunspots on the disc of the sun and the growing bite of the sun hidden behind the moon (which will be otherwise unseen). You can practice on any sunny day, You should be able to see some sunspots, and perhaps faculae on any day.

Other alternatives are specially designed solar viewing glasses. They must be unscratched and meet ISO 12312-2:2015 standard. Welders glass is also possible if rated "14" (the darkest possible).

It is not possible to use any form of glasses with a telescope. It is possible to get an aperture filter, which may be useful if you have the right telescope, but the projection method is better if you want more than one person to view at the same time.

Any kind of improvised solar filter is unsafe. This includes: stacks of sunglasses, pop-tarts wrappers, exposed film and viewing reflections in water.

During totality, the sun is hidden by the moon, and so it is not dangerous to look at the solar corona during totality with the naked eye. If you had been wearing solar viewing glasses, be ready to replace them. You MUST put them back on before totality ends.

Photography is hard. The images of the solar eclipse are made with professional telephoto lenses and good deal of skill. Instead you can photograph the scene. A timelapse of the eclipse can be a very effective way to record the event, especially if you have some interesting foreground and background. The eclipse itself will be a tiny dot you can still document how the scenery and shadows change during the event.


How would one calculate the synodic period between 3 planets?

Normally three planets would not line up along a simple line like that.  Planets have orbits with different inclinations, so at best they'd be in the same plane.

Working that out requires a detailed calculation of their motions.  The simplest approximation is to actually assume everything has the same inclination and treat the orbits as circles (which they're not precisely).

For this simple model, a planet's angular position $\theta$ at any time $t$ is given by :

$$\theta(t) = \theta_0+t\frac {2\pi} T$$

Where $T$ is the orbital period and $\theta_0$ is the initial angle.

What you want is the value of $t$ satisfying :

$$\theta_1(t) = \theta_2(t) + 2m\pi = \theta_3(t)+2n\pi$$

Where the $n$ and $m$ values are integers.

Now doing this for two planets is easy, we get :

$$t_{12}=\left(\frac {\theta_{10}-\theta_{20}}{2\pi} - m\right)\frac{T_1T_2}{T_2-T_1}$$

$$t_{13}=\left(\frac {\theta_{10}-\theta_{30}}{2\pi} - n\right)\frac{T_1T_3}{T_3-T_1}$$

Remember we have to work out those $m$ and $n$ values to get a result !

But the problem is that we need to get integer values and all the other values are real numbers.  This means that there may be no (exact) solution that produces integer values and hence no one time when they all line up.

Now in the real world their are some other issues :

The orbital inclinations mean we need to work in three dimensions
The orbits will be ellipses, not circles, and if we want even more accuracy they're not even convenient shapes like that (look up the three body problem to get an idea why).
The speed of light is finite.  So what "lined up" means is not as simple as it seems.  There might be only one observer who would say they were lined up and every other observer would consider them not lined up.
Planets have non-zero sizes, so we can allow them to be lined up within a range of values, not a simple set of numbers, but a whole set of ranges of values.
Again with time in general relativity, whose time are using as a reference.  That's a really complicated one in general relativity (more than it might seem), but fortunately to a good approximation we can ignore relativity a lot of the time or use small corrective factors.
All our measurements for the parameters of the orbit will have a finite accuracy and some uncertainty range.  So, particularly over very long periods of time, our calculations can become inaccurate.


How big would the supernova when a neutron star collides with another neutron star?

When two neutron stars collide, the resulting supernova is called a "kilonova," producing much more energy than a regular supernova. Additionally, heavier elements are formed, such as gold, platinum, etc. Other effects of this collision include a gamma-ray burst and/or gravitational waves. The object formed after the event can either be a heavier neutron star, or a black hole (if the mass of the merger exceeds the maximum mass of a neutron star, about 3 solar masses).
As of right now, we have not detected a neutron star merger that is visible to the naked eye; almost all of them occurred in different galaxies, billions of light-years away. I hope this helps.


Is it possible neutron stars are actual elements?

To be considered an element, they would have to have a positively charged nucleus. They don't. The neutron star is largely neutral.

They would have to have a cloud of surrounding electrons. These electrons would have to share "orbitals" with other electrons around other, nearby neutron stars. That doesn't happen.

Finally, at these scales, gravitational interaction predominates. Even if neutron stars did have some positive charge, and did have electrons orbiting them, interaction with other neutron stars would still depend a lot on gravity.

Finally, even if a neutron star was charged, it's not clear whether electrons around them would actually fall into some kind of orbitals governed by quantum mechanics. QM tends to not happen at such a scale.

In conclusion, no, they are completely unlike atoms and their nuclei.


How to calculate solar coordinates. (Right Ascension, Declination, and ultimately the Altitude and Azimuth of the Sun)

For the Sun at JD 2458599.125, RA = 32.8° and dec = 13.2° are correct regardless of geographic location.
In the alt/az calculation,

LHA should be 323.6° and converted to radians, and
the expression for alt needs a call to Math.Asin.

Trigonometric expressions are easier to verify if they are free of inline unit conversions.
Store angles in radians where possible; otherwise make it obvious with a name suffix such as _deg.

When computing the local hour angle, you can approximate GAST with GMST.
The difference eqeq is smaller than the approximation error of the Sun position algorithm you're using, and the alternative GMST formula is valid for D within a few centuries of J2000.0.


The Neupert Effect describes the commonly observed phenomenon that hard X-rays occur predominantly during the rise phase of soft X-rays during a solar flare, as illustrated by the graph below:

In the graph, the first peak is that of the hard X-rays, which you can see occurs during the rise of the other line which represents the soft X-rays.

Source:

The Neupert Effect - Solar Physics at Montana State University


Can someone explain a Geostationary Orbit in a simpler way?

The centrifugal force increases with the radius of the orbit for a fixed angular speed. The gravitational pull decreases with the radius of the orbit around Earth. The geostationary orbit is, where the two forces annihilate for the geosynchronous angular speed. The gravitational pull takes the role of the centripetal force to keep the satellite on the circular orbit.

That's the principle, which keeps satellites, moons, planets on their respective orbit. If there is some additional force, they leave their former orbit.


Why did the Event Horizon Telescope take so long to take a photo of a black hole?

A single telescope does not have the right angle size to picture the black hole that far away. The scientists had to use 8 telescopes to take a picture of the same point in the sky and merge the data from all sites using super computer. The data retrieval must point to the same object at the same time. Given the diverse condition of the 8 locations, there are only 5 days in a year when all of them have clear enough skies to point to the same location in a year.

Event Horizon Telescope


Are there more images with different filters available for the April 6 Hubble image of Jupiter?

As those observations are from only a few days ago, I think the data will still be embargoed, usually one year, until it's publicly available.

To confirm this, I've looked at the Hubble Legacy archive, where Hubble data is made publicly available. When you look there, the last public data for Jupiter is from February 2016.
Although I couldn't use the retrieved data from 2016, as it's not the FITS format, but it downloaded something.

So either you'll have to wait it seems, or you contact the project PI and ask nicely ;)

In the HST proposal page from the MAST archive (that links to the legacy archive) you see the photos they've taken listed, but the data is not available, I guess due to embargo.

Edit:
However I'm not sure if the legacy archive has all the data, as there's one dataset from 2011 with which I've worked that should be there, and I can't find it. Weird.


Is there a way to tell what the surface of a planet is like?

The answer to your question for that specific planet is No.

There is no way at this point in time to tell if Kepler-442b is a big barren rock or a water world or like earth for that matter. They only know it is there because it passes in front of the star it is orbiting and that dims the light. That is the only reason they know it is there.

This planet is not visible.


How is the Gould Belt younger than the sun?

Everything is in motion in our Galaxy. The Sun has executed some 20 laps of the Galaxy since it was born and may have migrated inwards or outwards to some extent. The Sun's location has nothing to do with the Gould belt or vice versa.

The Gould belt stars formed just 30 million years or so ago. The position of the Sun relative to the Gould belt is a coincidence and it will not be in the Gould belt in another 30 million years or so, because it has motion relative to it.


How does the Moon's gravity affect Earth's oceans despite Earth's stronger gravitational pull?

Everything in the universe has a gravitational influence on everything else in the universe. It isn't a question of the strongest gravitational pull winning out and all the others doing nothing.

The Earth is the strongest pull on the oceans, but the Moon and the Sun both have easily measurable effect in addition to the Earth's. Other bodies (Venus, Jupiter,  a small asteroid in another galaxy,....) all have much smaller effects which will be hard or impossible to detect amidst the noise due to waves and so on.


Detect Passage of Massive Relativistic Object

Probably not.

I believe that your question can be stated "Can we detect a lingering effect on the local environment in interstellar space a decade or two after the passage of a relativistic starship?"

The first step is to ask just how relativistic? If it's just "really fast" (say, 0.99c) there would be one set if effects, but if it was much fast -- 0.99999c, for instance -- some new effects would come into play.

The second is to ask about the density of gas in the vicinity. If it's like normal interstellar space (typically a million atoms per cubic meter) its passage will be much harder to detect than if we're in something like a Giant Molecular Cloud where densities can be 100 to a million times greater.

First off, you won't see any lingering electromagnetic radiation: That will necessarily be light years off by a decade later and you specified local effects.  That leaves two things that I can see: (a) Effects on the pre-existing gas, and (b) bits of ship that remain behind.

The interstellar medium is nearly entirely atomic hydrogen, either neutral or ionized. The temperature is typically 100-1000 K. The mean free path of these atoms is very large: 105 miles up to a million times that. The mean speed of the atoms is about 1 kps.

That means that any effect the passage of the ship has on the local interstellar medium will be entirely erased by the diffusion of the interstellar medium (IM) in a short time.  Imagine that the ship blows a tube of vacuum in the IM a million miles across. Neutral hydrogen atoms from outside of this region (just a few mean free paths away) will move in at 1 kps. In a few million seconds, the void created by the ship's passage will be refilled. (A  million seconds is about 10 days, so a single year is plenty of time for the IM to diffuse back to equilibrium.)

If the IM is much denser, the effect may linger as long as a decade, but we're not talking about the IM anywhere near us.

The second potential source is bits of the ship left behind. Hitting the IM at .99c is precisely the same as the IM hitting the ship at .99c, and that's quite hard radiation -- the IM's protons are hitting at 5 GeV. Unless the ship has force shields of some sort, this is likely to cause some spallation -- atoms of the ship's front end eroded into space. This might be detectable, but the problem I see is that those bits are unlikely to be dumped into space with zero transverse velocity and will thus quickly leave the local area and be undetectable a decade later.

The ship it may dump enough energy into the IM to heat it -- creating a long tube of hotter than normal IM along its path. The problem -- again -- is that the mean free path is so long that the hot atoms would immediately decamp for distant places and the colder atoms outside the ships's range would diffuse in, erasing the tracks.

If the speed were insanely high (as if .99c wasn't insane enough), possibly you might see some sort of effect where a bow wave of plasma builds up big enough to clear a really wide tube along the path, long enough that it can't fill in in a decade, or enough energy is dumped into the IM that it's still excited a decade later.  The problem is that that energy -- which is large -- is coming from the ship's kinetic energy through friction with the IM. (a) This slows the ship and (b) it probably does very bad things to the ship's structure.

The last effect is that we might see some side effect of the ship's drive or shields. But since they're basically magic (Clarke's Third Law) I don't see how anyone can give you a scientific answer.


So far, opinion is divided, as evidence has not been found to prove either way. Back in 2015, a study published in Nature, and linked via the Verge, here suggested that no, brown dwarfs do not exhibit star-like behaviours, but are in fact much closer to planets (due to them being below critical mass for hydrogen fusion in their cores)

All planets that have magnetic fields within our Solar System have aurorae, so Hallinan and other researchers believed that if brown dwarfs were more like planets, they too would exhibit aurorae.
...each time LSRJ 1835+3259 rotated — every 2.84 hours — it pulsed a radio emission and an aurora appeared at a location on the brown dwarf's surface.
..."In the case of the brown dwarf, the aurora is so much more powerful than anything we've seen in our Solar System," says Hallinan. "Jupiter's aurorae are 1,000 times more powerful than the Earth's aurorae. Well this brown dwarf has aurorae that are at least 10,000 times more powerful than Jupiter's aurora." The aurorae's colors are also a bit more diverse, emitting red along with blue and green.
...it strengthens the argument that brown dwarfs are more closely related to planets, kind of like "beefed-up Jupiters." "Typically, people tend to think of these things as failed stars," says Nichols. "My preferred explanation is they’re more like very successful planets, and this study demonstrates more planet behavior than stellar behavior."

However this 2017 article in the Astrophysical Journal which looked at the same data from LSRJ1835+3259 suggests otherwise, identifying the magentic data not as aurorae, but as effects more like that of our own sun. A nice summary of the article on Howstuffworks.com:

As the object rotated, a powerful magnetic region came into view, more powerful than the magnetic fields associated with sunspots that we observe on the sun. Sunspots are magnetically active regions that can trigger coronal mass ejections, solar flares and produce powerful streams of solar wind — all of which can generate powerful geomagnetic storms on Earth.
In an interview with New Scientist, the researchers point out that LSR J1835+3259 is very young (approximately 20 million years old) and the powerful magnetic field could be interacting with the object's protoplanetary disk (if it has one). But if this active magnetic region is long-lasting and representative of its global magnetic field, LSR J1835+3259 is way more "star-like" than we give brown dwarfs credit for.
So rather than calling brown dwarfs "failed stars," perhaps we should call them overachieving planets or magneto-dwarfs.


Seeming conflict between most distant objects and age of universe (both estimated)

No conflict. The universe is expanding. The furthest/oldest object we can see emitted its photons approximately 13.4 billion years ago, but the universe has expanded since then and that object is now around 32 billion light-years away. See this answer or search the astronomy stack exchange for "age of universe expansion".


Universe’s Glow Shows There Are Fewer/More Galaxies Than We Thought, Say Scientists

This is a tricky science result to explain and, as often turns out to be the case, not quite as secure or solid as the press release recyclers report.
The original paper is by Lauer et al. (2021) and measures the cosmic optical background (COB) using images taken with New Horizons, which is way beyond the orbit of Pluto. The advantage over things like HST is that the images are not contaminated by light scattering from zodiacal dust. Most of the paper deals with how they remove various instrumental signatures, get a figure for the dark sky brightness (about 27 mag per square arcsec in the V-band) and then attempt to explain this by examining the "light budget" from various known sources.
There are 4 main known sources. Faint stars below the detection limit of New Horizons (which is only about $V\sim 19$); ditto for galaxies; there is scattered light from bright stars out side the field of view; and there is a diffuse scattered light from dust in the Galaxy. The genuine COB is their measured sky brightness minus the contribution from stars and scattered light from our Galaxy.
They make an estimate of these 4 components and the sum comes to less than their dark sky brightness, but only by around 2 error bars; so a weakly significant result.
Now it gets confusing. In the text they say that the integrated light from faint galaxies (IGL) accounts for 50% of the COB, and may be consistent with it at the 2 error bar level. See Fig 18 in the paper.
Previous studies, based on HST data, suggested that if you extrapolated galaxy counts down to galaxies of about $10^6 M_\odot$, then there would be ten times as many unseen galaxies as you could count in the Hubble deep fields, mainly at $V>30$. However, if you work out how much IGL these would contribute it turns out to be larger than the COB that Lauer et al. have measured. Hence the headline that there may be fewer galaxies in the universe than previously claimed. I say may, because this claim is not one of the main conclusions of the paper, and in the text they say there may not be a significant discrepancy once the uncertainties are considered. This aspect of the work has been highly exaggerated in the press releases.
Lauer et al.'s estimate of the IGL is based on adding up the galaxies that have been observed down to $V\sim 30$. So you do need some extra faint galaxies to boost the IGL enough to fully explain the COB, but just hiding these at magnitudes $V>30$ may not be sufficient. The suggestion is that galaxies with $V<30$ are being undercounted. This is the origin of the second part of your report. It is that there may be more galaxies at brightnesses that ought to have been seen in the deep HST images. The implications of that are not explored.
Personally, I would like to take a look at how they have treated faint Galactic stars. Extrapolating from local luminosity functions may not be appropriate for estimating the contribution of stars down to $V=30$ in the halo of our Galaxy. If there are more lower mass objects in the halo than locally then this might have a bearing on the (2 sigma) discrepancy between the (apparent) COB and the IGL.


I kinda found a solution, unfortunately it's a few hundredths of a degree off from those specified by Horizons.  And it requires you to define a custom kernel for the observer's reference frame, which isn't bad as long as your doing a lot of observations from one location.  The code below is in Python and requires spiceypy ("pip install spiceypy").

import spiceypy
import math
#By Greg Miller (gmiller@gregmiller.net) 2019
#Released as public domain

spiceypy.furnsh('erotat.tm')

#Lat,Lon position of observer e.g. Louisville, KY
#Needs to match definition of "MYTOPO" in custom kernel mytopo.tf
lat=math.radians(38.2464000)
lon=math.radians(274.236400)

#convert lat,lon to xyz
obspos = spiceypy.georec(lon, lat, 0.0, 6378.1366, 1.0/298.25642)

#time of observation in UT
et = spiceypy.str2et('2007 JAN 1 00:00:00')

#Get the xyz coordinates of the moon relative to observer in MYTOPO frame
moon, lt = spiceypy.spkcpo('MOON', et, 'MYTOPO', 'OBSERVER', 'LT+S', obspos, 'EARTH', 'ITRF93')

#reclat function only wants a length 3 array
temp=moon[0:3]

#convert to polar coordinates
[r, lon, lat] = spiceypy.reclat( temp )

#Convert polar coords to Az, El
az = -lon

if  az < 0.0:
    az += spiceypy.twopi()

el = lat

print( 'Moon Az (deg):        {0:15.6f}\n'
       'Moon El (deg):        {1:15.6f}\n'.format(
           az * spiceypy.dpr(),
           el * spiceypy.dpr() )  )

spiceypy.unload('erotat.tm')

This is the contents of the erotate.tm kernel spec file.  You must also download all of the kernel files individually, you can generally find them by searching for their names.  Note that the last one "mytopo.tf" is one you must supply, an example is below.

\begindata

KERNELS_TO_LOAD = ( 'naif0008.tls'
                    'de414.bsp'
                    'earthstns_itrf93_050714.bsp'
                    'earth_topo_050714.tf'
                    'pck00008.tpc'
                    'earth_000101_070725_070503.bpc',
                    'mytopo.tf' )

\begintext

Below is an example mytopo.tm file.  The lon,lat go in the TKFRAME_1234567_ANGLES and is the only field you'll need to edit.  Note the odd format for lat/lon compared to how it's specified in the Python code above.  The lon is just the negative, the lat is the negative of 90-lat.

\begindata

  FRAME_MYTOPO             = 1234567
  FRAME_1234567_NAME       = 'MYTOPO'
  FRAME_1234567_CLASS      = 4
  FRAME_1234567_CLASS_ID   = 1234567
  FRAME_1234567_CENTER     = 399

  TKFRAME_1234567_SPEC     = 'ANGLES'
  TKFRAME_1234567_RELATIVE = 'IAU_EARTH'
  TKFRAME_1234567_ANGLES   = ( -274.236400, -51.7536, 180 )
  TKFRAME_1234567_AXES     = (       3,        2,   3 )
  TKFRAME_1234567_UNITS    = 'DEGREES'

\begintext


Are asteroid complexes and types the same?

Yes they are basically the same. The original asteroid taxonomies were defined from photometry (brightness) in a few filters such as the Eight Color Asteroid Survey of Tholen (1984). This resulted in asteroids with similar slopes being grouped together into the main complexes or classes. Once we got more detailed spectra of asteroids subtle differences between sub-members of e.g. the S-complex showed up and so the large complex was broken into more sub-classes such as Sc, Sq etc.


Is it odd that our Sun has so many planets?

From an exoplanet-finding point of view, the Sun has between one and three planets.

The major exoplanet-finding techniques in current use involve watching for either periodic Doppler shifts as the planet's gravitational pull causes the star to wobble, or periodic brightness shifts as the planet transits the star.  Both require that the planet is large enough and close enough to generate a measurable signal and that the orbital period is short enough to let astronomers distinguish periodic variations from one-off variations; the transit method additionally requires that the planet's orbit crosses the star from the point of view of Earth (which favors close-in orbits).  Looking at the Solar System with these techniques:

Mercury: too small
Venus: Maybe visible
Earth: Maybe visible
Mars: too small
Jupiter: Highly visible
Saturn: Orbital period too long
Uranus: Orbital period too long
Neptune: Orbital period too long
"Planet 9": Orbital period too long

If you look at this graph of exoplanet discoveries, Jupiter is solidly in the cluster of blue Doppler discoveries, Saturn is just past the "we've been watching for one full orbit" right-hand edge of that cluster, Earth and Venus are somewhat below the sloped minimum-period-mass line, and everything else is nowhere near the detection range.

The reason the Sun has far more known planets than any other star is simply because we've got a better look at it.


Is it possible for a person to not see the new moon at different places on earth?

+taupunkt basically answered this: the thin lunar crescent after New Moon is only visible for a short while after sunset.  While the sun is up, the moon is invisible and it becomes visible when, after sunset, the sky turns dark enough.  However, the moon will set typically 1-2 hours after sunset, and will become invisible some time before that due to atmospheric extinction.  Whether or not, and if so for how long, you can actually see the moon's crescent depends on the local cirumstances: the distance between the sun and moon (which correlates with the lunar phase and depends on your geographical longitude), the angle between the ecliptic and the horizon (depends on your latitude), whether you're in a mountainous area (and if so, whether you're on a mountain top or in a valley), atmospheric humidity, the weather (a few clouds at the western horizon is all it takes), whether you use your naked eye, binoculars or a telescope, your eyesight and experience, etc.

Because of this, it is difficult to predict when you can see the first/last moon crescent.  I use this paper to do that for the Netherlands (Google translation, should be ok since it's a table) and for the naked eye.  One thing that strikes me is that of the 25 cases of a first/last crescent sighting this year, the lunar phase ranges from 1.2-9.2%, with a mean of 4.0%, so 0.8% seems to be rather optimistic.  You may allow the use a telescope though.

The table doesn't list how long the moon is visible for (the prediction of whether the moon is visible or not is sufficiently uncertain by itself), but it will definitely not be more than the time difference between the rise/set times of the sun and moon (~1-2h).  If we use half of the average of those numbers, we get 45 minutes, which may well be optimistic but is not too unrealistic either (for a ~4% crescent).  The other ingredient is how long the lunar phase is at 0.8%.  Assuming that this means between 0.75% (below which it would be rounded down to 0.7%) and 0.85% (above which it would be 0.9%) this was the case for ~97 minutes for the last time this happened (July 27).  Together with the 45-minute observation window, this gives a ~142-minute window of opportunity, or about 10% of a day, indicating that the crescent would be visible from ~10% of the earth.

Remember that we assumed a 4% crescent here, not 0.8%, so this 10% is an overestimation.  My guess is that it is a huge overestimation, since in most places you will not be able to see the moon with the unaided eye at all.  Even if the moon were somehow visible for an instant everywhere, the fraction of the earth the 0.8% crescent is visible from would drop to ~6.7%.  If you want a more precise definition of 0.8%, e.g. 0.80%, or between 0.795% and 0.805%, the moon spends about 10 minutes at the desired phase and this would be visible from 0.7% of the earth.

Then again, you may allow the use of a telecope, so that you can see the moon even during day time.  If you also don't worry about the weather, the window of observation would be 12 hours (since the moon is up for observers at about 50% of the earth's surface) and with the 97-minute time window for which the crescent is between 0.75% and 0.85%, the 0.8% crescent would be visible from about 57% of our planet's surface.  For 0.795%-0.805%, this would be 51% of the earth.

The conclusion is that a 0.8% crescent of the moon is visible from beteen 0% and 60% of the surface of the earth, depending on your exact definitions and your choice of accuracy and optical aid.


Do planetary radio frequency change upon alignments?

If you're talking about the electromagnetic radiation (in the radio spectrum) produced by the planets themselves, then the answer is no.

BTW, not all planets generate a lot of radio EM. Jupiter does generate some amount, but the other planets are quite a bit more quiet. Still, a very small amount is probably generated by most planets out there.

In any case, this is a strictly local phenomenon. Local conditions within and on that planet predominate in driving the production of radio EM.

By comparison, distances between planets are HUGE. Whether they are "aligned" or not is of no importance to what happens on the surface. Anyway, they appeared "aligned" only to us, on Earth, and even then the "alignment" is always imperfect (they are never exactly on a line).

Shortly: no.


How can many stars be formed from the remains of one supernova?

Stars don't "come from" a supernova.
Stars come from the interstellar gas in the galaxy, particularly where it is more concentrated into nebulae.
This gas is mostly hydrogen and helium, but it is "enriched" with heavier elements from old stars, including from stars that have exploded in supernovae. Over the billions of years since the galaxy formed, there have been many supernove which have gradually increased the amount of heavier elements in the interstellar gas.
The shockwave from a supernova can cause the gas to become squeezed and more dense, and this can trigger the gravitational collapse that leads to star formation, but the gas for those stars doesn't come from one particular supernova.
As a nebula collapses into many stars, some (the largest) will be large enough to supernovae (which can trigger subsequent periods of star formation).


The solution to your mystery about why habitable planets are found around nearby low-mass stars, but distant G/K stars is all to do with observational selection effects and biases. There are many G/K stars within 100 light years of the Sun, but almost none have been examined in the detail required to reveal small, habitable planets around them.

Almost all the small "Earthlike" planets in habitable zones have been found by the transit technique. It is much easier to find such planets around small stars for two reasons:

The transit technique yields a signal that depends on $(R_p/R_{*})^2$. Thus small planet transits are easier to see around small stars.

The probability of a transit depends on the ratio of the star size to orbital radius. This sounds like it works against finding planets around small stars; and it does at a fixed orbital radius. But, because the habitable zone is much closer-in to a smaller, less luminous star, then the habitable planets around such stars are more likely to transit. In addition, their orbital periods will be tens of days rather than the months to a year around G/K stars and this makes it much easier to observe repeated transits.

In fact, the combination of these two selection effects makes it possible to find habitable zone planets around small stars all over the sky from ground-based observatories.

The downside is that small stars are low luminosity, so the planets that can be found are found around nearby examples at tens of light years.

In contrast, habitable zone planets around G/K stars are mostly found in months-year orbits by Kepler, which stared at a single small field for 4 years. The majority of the stars targeted by Kepler for exoplanet searches were 10th-15th magnitude G/K stars (there are few brighter examples in such a limited patch of sky), which are at distances of hundreds to thousands of light years.


If the Oort cloud has billions of objects larger than 20 km, then why do we have so few comets?

Most Oort cloud objects stay in the Oort cloud, and we never see them.

Comets don't last long. A comet is made of a mixture of ices and dust. A comet that enters the inner solar system has one of a few outcomes:

it could pass out of the solar system, and either never be seen again, or enter such a long orbit that it isn't seen again for an extremely long time (like Hale Bopp)
it could fall into the sun (or travel so close that it is completely broken up) (like comet ISON)
it could get caught into a short period and over time have all its ices boiled away by the heat of the sun. (asteroid 2015 TB145. Halley's comet is in the process of losing its ice with each orbit it grows smaller)
it could hit a planet. (Shoemaker-Levy 9)

In all cases, the comet is lost, destroyed or deactivated.

As comets have a short lifespan, they are rare.


Can Hawking Radiation ever be detected and does it exist?

How do we know Hawking radiation exists?
Quantum Physics (Quantum Field Theory) and General Relativity are considered as the two main theories in physics, which can explain a broad range of phenomena at the scale of atoms and subatomic particles to the large scales of the universe, beyond the ability/validity of classical, nonrelativistic physics. They also offer a number of extraordinary predictions, many of which have been experimentally confirmed. So, we are sure about the validity of these theories (and we also know their limitations). If we accept the fundamental principles of these theories, the existence of Hawking radiation would be unquestionable, no matter how we interpret this phenomenon.
Hawking, traditionally, obtained it in 1974 by use of the study (quantization) of quantum fields near the event horizon of the Schwarzschild black hole background for different observers in different locations (this subject is called quantum field theory in curved background). Also, using the Heisenberg's uncertainty principle and applying the basic/elementary relations of quantum mechanics and general relativity, one can qualitatively obtain the Hawking temperature without requiring the knowledge of QFT and field quantization (see also Yakov Zeldovich, who first predicted radiation in rotating black holes). Hawking radiation is also predicted by a number of different approaches in physics, so at present there is no doubt about its existence, but the main problem is how to detect it. The are a number different methods for deriving the Hawking radiation such as evaluating thermal partition function of black holes (in zero-loop approximation of quantum gravity path integral formalism), unitary decay of near extremal D-brane states (in string theory) etc, which (I think) are beyond the scoop of this question. By the way, for these reasons, we expect that any quantum gravity theory which combines both quantum physics and general relativity (if exist any) must predict the Hawking radiation as well, undoubtedly.
Is there any confirmed observational evidence for Hawking radiation?
Not yet. The existence of black holes is theoretically possible and now it has been experimentally verified, no doubt (see Reinhard Genzel's works). But, It seems we have serious difficulties for observing Hawking radiation. A straightforward calculation for the Hawking radiation of static (Schwarzschild) black holes results
$${T_{Hawking}} = \frac{{\hbar {c^3}}}{{8\pi G{M_ \odot }{k_B}}}\left( {\frac{{{M_ \odot }}}{M}} \right) = 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{{\rm{K}}^\circ},$$
where ${M_ \odot }$ is the mass of son and $M$ represents the black hole mass (${k_B}$= the Boltzmann constant, $G$ = Newton constant). The formation of black holes is possible if the initial mass of collapsing object be greater than the so-called Chandrasekhar limit, $1.4$ ${M_ \odot }$ (i.e., $2.765 \times 10^{30}$ kg). This means that a solar-mass black hole ($M \sim {M_ \odot }$) has a temperature of about $60$ ${\rm{nK}}^\circ$, and larger black holes ($M > {M_ \odot }$) have even lower temperatures. Now, compare this extremely tiny value with the cosmic microwave background radiation (CMB, which is faint cosmic background radiation filling all space), i.e. ${T_{CMB}} \approx 2.73\,{{\rm{K}}^\circ}$. In conclusion, the temperature (Hawking radiation) of realistic black holes is significantly lower than the background temperature of the universe. Qualitatively, this argument is also valid for rotating black holes. This is the main reason why it is generally believed that the Hawking radiation cannot be observed, at least with the current technology of our detectors.
Physicists do not stop looking for Hawking radiation
But this is not the end of the story. Until writing this answer, since any direct observational evidence is potentially ruled out, we have to seek indirect experimental evidence, e.g., e.g., by means of simulations or analog systems. The subject of analog gravity and its different models have been invented for this purpose. Analogue gravity is a research programme which investigates analogues of phenomena in general relativity (such as black holes) within other physical systems, typically (but not exclusively) condensed matter systems, with the aim of gaining new insights into their corresponding problems. Naively, the mathematics governing both the black hole and analog systems are similar. So, there exist a number of analog gravity models which share the same essential properties with black holes such as event horizon and Hawking radiation. For example:

In this paper (published in Nature Physics), the author created a narrow, low density, very low temperature atomic Bose-Einstein condensate, containing an analog black hole horizon and an inner horizon, as in a charged black hole, and in this analog black hole model, he's observed the Hawking radiation emitted by the black hole.

In this paper (published in Phys. Rev. Lett.), the authors utilized the analogy between the propagation of fields around black holes and surface waves on moving water and they experimentally (omitting details) confirmed the Hawking process.

In this paper (published in Phys. Rev. Lett.), an (famous) analog of a black hole, known as  sonic black hole, in a Bose-Einstein condensate has been created in which sound waves, rather than light waves, cannot escape the event horizon.

This is a nice link in Physics Stack Exchange about analog gravity which is a water flow analogy with a black hole. It is a discussion about a good (home) experiment with a black hole analog! Also, this one maybe useful for you.

If you search in web about these analog black holes, you quickly find many interesting papers for confirmation of Hawking radiation, especially in Nature Astronomy Journal.
Is there any astrophysical candidate for detecting Hawking radiation?
Well, yes, but we are not sure about the existence of them. As proposed by Hawking, primordial black holes are well-motivated candidates for detecting black hole (Hawking) radiation. These are hypothetical types of black holes that formed soon after the Big Bang (as a result of fluctuations in the early Universe). Studies speculate that, depending on the model, primordial black holes could have initial masses ranging from $10^{-8}$ kg to more than thousands of solar masses. This is an interesting, remarkable result and I explain why.
In fact, a black hole that initially has mass $M$ will eventually evaporate to nothing. According to the Hawking radiation formula of a Schwarzschild black hole, i.e.,
$${T_{Hawking}} = \frac{{\hbar {c^3}}}{{8\pi G{M_ \odot }{k_B}}}\left( {\frac{{{M_ \odot }}}{M}} \right),$$
Hawking radiation further decreases the black hole mass and black holes with very small mass would experience runaway evaporation, creating a burst of radiation at the final phase (The evaporation would actually be pretty catastrophic, as it would release about as much energy as a large thermonuclear bomb during this time). But, the problem is that if we can observe these explosions during our time. Straightforward calculations by implementing the Stefan-Boltzmann law show that the life time of a black hole is as
$${\tau _{{\rm{life}}}} = \frac{{256{\pi ^3}k_B^4}}{{3G\sigma {\hbar ^4}}}{(GM)^3} = (2.095 \times {10^{67}}\,{\rm{year}})\,{\left( {\frac{M}{{{M_ \odot }}}} \right)^3},$$
where $\sigma$ is the well-known Stefan-Boltzmann constant. For the solar-mass black holes this is much greater than the age of the universe (i.e., $13.8$ billion years), so they are not good candidates. On the other hand, by use of the above formula one can show that primordial black hole with mass lower than $10^{11}$ kg would not have survived to the present due to Hawking radiation. But, interestingly, for primordial black hole (if they exist) with mass lower than the solar mass ($2 \times 10^{30}$ kg) and also greater than $10^{11}$ kg, it is possible to detect/observe the final stage of their life, which is a big explosion equivalent to a hydrogen bomb yielding millions of megatons of explosive force! Along these lines, there exist a number investigations, e.g., see these [1], [2], [3] and [4].
Of course, as usual, there exist some technical/fundamental difficulties to experimentally detect Hawking radiation in primordial black holes. This is the beginning of another story (For a good discussion see this link).


No, it's not. The radiation field in the interior of the Sun is very close to a blackbody spectrum.
If you look in any particular direction the brightness (power per unit area) you see is $\sigma T^4$, where $\sigma$ is Stefan's constant. Even at any particular wavelength it is always the case that a blackbody of higher temperature is brighter than a blackbody at lower temperature.
Given that the interior temperature might be $10^7\ \mathrm K$, then the surface brightness is $5.7 \times 10^{20}\ \mathrm{W/m^2}$, compared with the $1400\ \mathrm{W/m^2}$ you would get by looking directly at the Sun (please don't do this). Note that most of this power comes out at X-ray wavelengths, but because of the properties of a blackbody, the brightness at visible wavelengths will still be plenty brighter than that of the solar photosphere (see below).
A possible source of confusion is this term "opacity". When things are in thermal equilibrium, which the interior of the Sun is, then they emit the same amount of radiation as they absorb. So high opacity also means high emissivity.
Details for interest:
The opacity, $\kappa$ in the solar interior ranges from 1 cm$^2$ g at the centre to about $10^5$ cm$^2$ g just below the photosphere. To estimate the mean free path of photons we need to multiply this by the density $\rho$ and take the reciprocal:
$$ \bar{l} = \frac{1}{\kappa \rho}\ .$$
The density varies from 160 g/cm$^3$ at the centre to about 0.001 g/cm$^3$ just below the photosphere. Thus the mean free path is about  6 micrometres at the center and is actually quite similar just below the photosphere (it peaks at around 2 mm about three quarters of the way out towards the surface).
Thus your "view" of the stellar interior is of a foggy sphere with radius of no more than a few times $\bar{l}$. The fog however is tremendously bright - as outlined above.
The brightness at particular wavelengths is proportional to the Planck function
$$B_\lambda = \frac{2hc^2}{\lambda^5} \left(\frac{1}{\exp(hc/\lambda k_B T) -1}\right).$$
Thus at $\lambda=500$ nm (visible light), the ratio of brightness for blackbodies at $10^7$ K (solar interior) to 6000 K (solar photosphere) is $4.2\times 10^{4}$. i.e. Even just considered at visible wavelengths, the interior of the Sun is about 40,000 times brighter than the photosphere.


Would all neutron stars inevitably collapse into black holes due to quantum tunneling?

The materials inside a neutron star are densely packed (extremely close to each other) and as time goes by say about an eon later, most of the neutrons would eventually tunnel into each other

Based on what ?

The material's collapse was halted from an astonishingly high speed collapse by the neutron degeneracy pressure that exists precisely because the universe doesn't want to let two neutrons fuse.  The enormous forces involved in the collapse and the enormous forces keeping the mass compact are not high enough to overcome this.  That's why it exists in the first place.

Over time "tunnelling" works in many, many ways. If two neutrons were to fuse in this manner the list of possible by products is enormous, and a black hole is most unlikely, IMO.  We've a lot of experience of whacking particles into other particles in extreme conditions ( think CERN ) and we haven't seen any sign of a black hole yet ( and believe me, they're looking for them ).  So I think the odds of a fuse into a black hole are extremely low.  Over time it's more likely the neutron star would "evaporate" into other particles.  But over those time scales you're looking at the cold death of the universe and protons decaying and so on.

And if, for argument sake, a mini black hole formed from two neutron's "fusing" it's life expectancy would be minuscule as tunnelling effects ( predicted by Hawking ) suggest it would "evaporate" in the blink of an eye.  So isolated fusing events, even f they happened, would be unlikely to produce a black hole that grew, but instead one that just blew up as fast as it was created.

  and voila! the neutron star becomes black hole? sound suspicious to me but can it happen?

No evidence I am aware of suggests neutrons will fuse in the manner you're describing.

"Neutron Star" is probably something of a misnomer.  Apart from neutrons, current models ( "educated guesses" ) suggest that the core of such a body might be a lump of something called quark-gluon plasma, which is essentially mush made from neutrons.  This is what you get when you fuse neutrons together -  a form of mush made up of the stuff that neutrons and protons are made from - quarks.

So theory and evidence suggests that you don't get a black hole from just waiting for a neutron star to change into one.


Generate East and North unit vectors tangent to a point on the celestial sphere, using vectors?

If $\mathbf{\hat{n}}$ points to the north celestial pole, then the eastward tangent vector is

$$\mathbf{\hat{u} = \frac{\hat{n} \times \hat{w}}{\|\hat{n} \times \hat{w}\|}}$$

and the northward tangent vector is

$$\mathbf{\hat{v} = \frac{\hat{w} \times \hat{u}}{\|\hat{w} \times \hat{u}\|}}$$

$\mathbf{\|\hat{w} \times \hat{u}\|}$ analytically should be 1 but can be slightly different numerically.


Planetary tails: How many have we observed yet?

Yes, apart from WASP-69b, there has been reports of other exoplanets emitting some form of planetary tails. Let's start from the Wikipedia:

KIC 12557548 b is a small rocky planet, very close to its star, that is evaporating and leaving a trailing tail of cloud and dust like a comet. The dust could be ash erupting from volcanos and escaping due to the small planet's low surface-gravity, or it could be from metals that are vaporized by the high temperatures of being so close to the star with the metal vapor then condensing into dust.
In June 2015, scientists reported that the atmosphere of GJ 436 b was
evaporating, resulting in a giant cloud around the planet and, due to
radiation from the host star, a long trailing tail 14 million km (9
million mi) long.

More on the dust particles that is trailing from the exoplanets KIC 12557548 b as well as KOI-2700b:

The observed tail lengths are consistent with dust grains composed of
corundum (Al2O3) or iron-rich silicate minerals (e.g., fayalite,
Fe2SiO4). Pure iron and carbonaceous compositions are not favored. In
addition, we estimate dust mass loss rates of 1.7 ± 0.5 M⊕ Gyr-1 for
KIC 12557548b, and > 0.007 M⊕ Gyr-1 (1σ lower limit) for KOI-2700b.
Source: Dusty tails of evaporating exoplanets, I. Constraints on the dust composition, R. van Lieshout1, M. Min and C. Dominik, A&A, Volume 572, December 2014, DOI: 10.1051/0004-6361/201424876

Astronomers using NASA's Hubble Space Telescope have confirmed the
existence of a baked object that could be called a "cometary planet."
The gas giant planet, named HD 209458b, is orbiting so close to
its star that its heated atmosphere is escaping into space.
Source: https://www.nasa.gov/mission_pages/hubble/science/planet-tail.html

Astronomers have confirmed that the planet Gliese 436b seems to be
trailing a gigantic, comet-like cloud of hydrogen.

Source: https://skyandtelescope.org/astronomy-news/exoplanet-with-a-comet-tail-06302015234/

Disintegrating planets allow for the unique opportunity to study the
composition of the interiors of small, hot, rocky exoplanets because
the interior is evaporating and that material is condensing into dust,
which is being blown away and then transiting the star. Their transit
signal is dominated by dusty effluents forming a comet-like tail
trailing the host planet K2-22b, making these good candidates for
transmission spectroscopy.
Source: Eva H. L. Bodman et al 2018 AJ 156 173, DOI: 10.3847/1538-3881/aadc60


The role of 'the interference of waves' in VLBIs'

You can think of the object being photographed as a collection of light-emitting points. The light emitted from each point is spherically symmetric, at least over the small angle that we see.
If you sample one of those point sources at two points, and the source lies exactly on the perpendicular bisector of the line segment between the sampling points, then light of the same amplitude and phase will reach both sampling points at the same time. If the source is slightly off that plane, then the matching light will reach the sampling points at slightly different times. This allows you to disentangle the light coming from different directions.
I don't know anything about real-world VLBI, but in principle the way you extract an image from the raw data is by simulating a camera. You introduce electromagnetic waves with the strengths you measured at the spacetime locations where you measured them, and simulate them reflecting off a planet-sized mirror (or refracting in a planet-sized lens) and hitting a detector. You can show that light from any point in your source object will be focused to a point on the detector by your mirror/lens, i.e. it constructively interferes there and destructively interferes elsewhere on the detector. By linearity, the sum of the light from all the source points forms an image of the source.


What causes a neutron star to become a pulsar?

It is thought that all neutron stars are pulsars when they are first born, as all spin extremely rapidly and have strong magnetic fields shortly after birth, arising from their dramatic collapse from a few Earth-sized massive star core into a city-sized neutron star. Conservation of angular momentum means that an initially slow-rotating core becomes a rapidly spinning neutron star, which also drives a dynamo action that generates a strong magnetic field.

The emission from a neutron star comes as a result of pair-production in its magnetosphere, which has an approximately dipolar shape and thus the strongest field strength at its poles. This field is rarely aligned with the spin axis, thus giving the "lighthouse" effect where the beam sweeps over an observer at twice the spin frequency. These pair-produced particles radiate energy via synchrotron emission (primarily in the radio), sapping energy from the magnetic field. As the neutron star ages, the magnetic field eventually decays below a critical point where electrons/positrons cannot be pair-produced. This is called the "death line," illustrated here and denoted by the word "graveyard" on a P-P_dot diagram:

After crossing this threshold, the pulsar loses its ability to emit radio waves. Pulsars only remain pulsars for the first ~10^7 years of their lives, and thus about 99% of neutron stars are no longer pulsars.

Additionally, we can only see a pulsar if its beam crosses over the Earth while the pulsar rotates. As the beam angle is usually tens of degrees, the total fraction of pulsars whose beams cross the Earth is approximately 10% the total that exist.

In the lower-left of the diagram you can see very short period pulsars, these are millisecond pulsars and have been spun up by accreting mass from a companion. These pulsars have very long lifetimes, but only represent a small subset of the neutron star population.


I agree with your supervisor: If your paper is going to a peer-reviewed journal, you should cite peer-reviewed material. I also agree with you that the NASA version is clearer than the original figure, but if you're only going to refer to the image — i.e. not show a reproduction — I'm sure the reader can figure it out.

Alternatively, you could create your own (color) version with the annotations that are relevant for your work, and then write something like "…based on the schematic diagram of Urry & Padovani (1995)".


The first one is alpha (α); the second one is delta (δ). You can study the shapes of the letters here:

https://en.wikipedia.org/wiki/Greek_alphabet

Here's what they mean (information from a comment to your question):

  Orientation models use three Euler angles to describe body
  orientation. The first two angles are the right ascension and
  declination of the north pole of a body as a function of time. The
  third angle is the prime meridian location (represented by "W"), which
  is expressed as a rotation about the north pole, and is also a
  function of time.


By convention, when observing an exoplanet, is the observer's line of sight (on Earth) perpendicular to the plane of sky?

Yes, the reference plane for exoplanet orbital inclination is our sky plane.
From the NASA Exoplanet Archive documentation:

  The Observed Inclination is the orbital inclination with respect to the plane of the sky....
  0 degrees correspond to an orbit in the plane of the sky, face on with respect to our line of sight.
  90 degrees corresponds to an orbit that is edge-on with respect to our line of sight, perpendicular to the plane of the sky.

If you query the archive for exoplanets with discovery method "Transit," you get inclinations mostly between 85° and 90°.

If the only available observation is a fluctuation in the star's radial velocity, then the planet's mass m and orbital inclination i are unknown, but a minimum mass m sin i can be computed.
The actual mass is greater unless i = 90°.

The orbit diagram from Wikipedia can apply to extrasolar systems with one change: the reference direction is celestial north instead of ♈.


Why does the Moon sometimes look Yellow?

Because light with higher frequency, that is bluer colors, scatters more in Earth's atmosphere and celestial objects thus look redder (yellow is towards red) while the atmosphere looks blue. See Rayleigh scattering.
Here is video with a lengthy and somewhat entertaining explanation:
https://www.youtube.com/watch?v=SRh75B5iotI


Has any moon achieved "retrograde equatorial orbit"?

This is directly related to another question: Why are asteroids with zero orbital inclination rare?
If captured, irregular moons are randomly oriented in space then there is very little chance of them having either inclination angles near zero or near $180^\circ$. This is because, if they are uniformly distributed in space, the fraction of orbits within a given inclination range $i_1 < i < i_2$ is given by
$$ f = \frac{\int^{i_2}_{i_1} \sin i\ di}{\int_{0}^{180^\circ} \sin i\ di}$$
$$f = \frac{1}{2}\left(\cos i_1 - \cos i_2\right)$$
Thus for $i_1 =177^\circ$ and $i_2=180^\circ$, $f= 0.00068$ (i.e. about 0.07%).
Given that there are only $\sim 100$ irregular satellites, it is not that surprising that none are found with $i>177^\circ$.
Note there are physical reasons why the regular satellites have low inclinations - they are not uniformly distributed and therefore they can have very low inclinations.


Can circumstances exist such that as seen from some solar system all stars are obscured?

A planet with no stars?
Our solar system could certainly pass through a Bok Globule, which is a dense cloud of opaque interstellar dust and gas.  They are typically about a light year across, so we could be stuck in one for quite some time.  While this would obscure the other stars while we passed through, I don't think Earth would be a very good place to live during that time.  Gas and dust falling into the Sun would probably strip or change our atmosphere and end life on Earth.
When stars form in a Bok Globule, the gas in the star's sphere of influence gets pulled into the interstellar disk and then either absorbed by the star and planets or ejected out of the stellar system.  I suspect that a star couldn't form without leaving big holes in the globule.
But suppose life emerged on a tidally locked planet. If only the sun-facing side of the planet was warm enough to support life, but the dark facing side was fatally cold (even near the day/night transition zone), then the inhabitants might never see the other stars due to light pollution from their sun.
Another possibility could be that a planet was in a binary star system.  If the primary star was much more massive than the other star, and the planet was in an L1 orbit with the less massive star around the more massive star, then even if the planet was turning, there would be no night.  Here is a picture from wikipedia to illustrate the orbit.

Unfortunately the L1 orbit is unstable, so a planet in such an orbit wouldn't stay there for long.
On a planet in constant daylight, the other planets would also be difficult or impossible to see.  Highly reflective gas giants in nearby orbits would certainly be visible, though.
Orbital calculations without a fixed celestial sphere?
It would be much harder to deduce a model for the motion of the planets without fixed stars in the back ground.
But say one has a single observatory, and the only things one measures are the times and azimuth angles for the high magnitude solar system bodies (the Sun, Moon, Mercury, Venus, Mars, Jupiter, and Saturn) as they ascend into the sky and descend out of it (for a fixed elevation angle).
If one gathers these for a few years, they will have enough data to have an overdetermined system to calculate the latitude of my observatory, the Earth's axial tilt, the Earth's rotation rate, and the orbital elements of all the bodies they can observe.
One could do it as Kepler did, by matching geometric shapes to the observations.  Better yet, one could do it as Newton did, with a theory of gravity.


Could the KIC 8462852 (Alien megastructure star) be explained by orbiting gas clouds?

No.

The light curves don't look at all like anything transiting. The ingress and egress profiles and the timing are way beyond that kind of explanation. Not planets, not comets, not clouds, not alien super structures.
No infrared light from dust or gas has been detected, as they should've been if the starlight had heated small particles. And since this is an F star with about 50% larger diameter than the Sun, which contains 99% of the mass in our Solar System, you can imagine how huge any comet/dust/gas cloud has to be to cover 22% of it.
No brightening of the starlight has been detected, as this allegedly huge"cloud" orbits behind the star and should reflect light to us. Shouldn't those huge icy comet clouds reflect some starlight?
Jupiter is about as large as a planet can get, even with 10+ times more mass, its volume would not increase by much. Gas giants differ much more in mass than in volume. An object 20 or 200 times more massive than Jupiter would be a roughly Jupiter sized red dwarf star, and that is not what has been observed at "Tabby's star".

The anomalies correlate strictly with the telescope's orbital period and orientation and are obviously a man made artefact because of some unforeseen malfunction of the instrument, nothing astrophysical is involved.


How small would you have to crush an object for it to become a black hole?

Well, I wasn't going to answer but the other two answers are wrong, or at least incomplete. If you wish to make a black hole from a stellar-sized object, then there is no need to compress it to as small as the Schwarzschild radius (though that would certainly work and would certainly be the answer for smaller objects with negligible self-gravity). Instead, you just need to compress it to a size at which it cannot be possibly supported by any plausible equation of state against further gravitational collapse. It turns out that this is somewhat bigger than the Schwarzschild radius and hence the density required is considerably lower. More details below.

There is a radius, larger than the Schwarzschild radius at which a neutron star, quark matter, whatever its equation of state, cannot be supported against collapse.

There are limits imposed by causality and General Relativity on the structure of compact stars. In "Black Holes, White Dwarfs and Neutron Stars" by Shapiro & Teukolsky, pp.260-261, it is shown, approximately, that even if the equation of state hardens to the point where the speed of sound equals the speed of light, that $(GM/Rc^2)<0.405$.

The Schwarzschild radius is $R_s=2GM/c^2$ and therefore $R > 1.23 R_s$ for stability. This limit is reached for a neutron star with $M \simeq 3.5 M_{\odot}$. A more accurate treatment in Lattimer (2013) suggests that a maximally compact neutron star has $R\geq 1.41R_s$.

If the equation of state is softer, then collapse will occur at smaller masses, and higher densities but at a similar multiple of $R_s$.

Thus it is not necessary to compress matter within $R_s$ to form a black hole.

The picture below (from Demorest et al. 2010) shows the mass-radius relations for a wide variety of equations of state. The limits in the top-left of the diagram indicate the limits imposed by (most stringently) the speed of sound being the speed of light (labelled "causality" and which gives radii slightly larger than Shapiro & Teukolsky's approximate result) and then in the very top left, the border marked by "GR" coincides with the Schwarzschild radius. Neutron stars become unstable where their mass-radius curves peak, so stable neutron stars are always significantly larger than $R_s$ at all masses.


I don't think this particularly useful way to think about it. The equator is the equator, Cancer is 23.4 North, and Capricorn is 23.4 South. The equator is a great circle on the sphere of the sky, but the Tropic of Cancer is not a great circle.  These are fixed.

Now (in Northern Hemisphere summer) the sun is directly above a point on (or near) the tropic of cancer.  This doesn't make the tropic of cancer a "virtual equator". It just makes it the tropic of cancer in summer.

It makes a difference because the path of the sun seen from the equator on midsummer is not the same as the path of the sun seen from the tropic of Capricorn on the equinox.  On the equinox, the sun traces a great circle in the sky (from wherever you view it) On midsummer's day the sun doesn't trace a great circle (anywhere).


What do the fusing 'onion layers' of a pre-supernova star look like to scale?

Pre- supernova models often characterise the compactness of the core using a "compactness parameter" defined as
$$ \upsilon = \frac{(M/M_{\odot})}{R(M)/1000\ {\rm km}},$$
where $M$ is usually chosen to be $2.5M_{\odot}$ and $R(M)$ is the radius within which $M$ is contained.

Pre supernova models by Farmer et al. (2016) show that the central $2.5M_{\odot}$ of a massive star included the Carbon-burning core in a $15M_{\odot}$ (initial mass) star, but only contained the oxygen-burning core in a more massive star (models are presented up to 30$M_{\odot}$. i.e. Your answer is going to be mass and composition dependent (these models are for a solar metallicity initial composition).

In the $15M_{\odot}$ model (with mass-loss), $\upsilon \sim 0.08$ at core collapse, which means the carbon burning core would have been contained within 31,000 km. The $30M_{\odot}$ model is more compact with $\upsilon= 0.58$, indicating that the oxygen-burning core would have been contained within 4,300 km.

At collapse the iron core might have a mass of 1.4-1.8$M_{\odot}$, will be supported by electron degeneracy pressure and should have a size a bit smaller than a typical carbon white dwarf (radius of a few thousand km).

You can compare these sizes with the size of the entire red supergiant star, which might have a radius of a few au (e.g. Betelgeuse).

The shell-burning regions would be found out to slightly larger radii than this I think, but these numbers are a reasonable estimate. Looking in detail at the models I don't think that the final stages of pre- supernova evolution look anything like the strictly stratified onion-shell picture that is seen all over the internet.


Is there a prediction when our solar system would fall apart?

The process whereby the Sun will become a red giant and then turn into a white dwarf is neither short nor simple. It's absolutely not the case that it will expand, blow away a chunk of its mass, contract again, and that will be it.

According to Wikipedia (emphasis mine):

  For the Sun, four thermal pulses are predicted before it completely loses its outer envelope and starts to make a planetary nebula. By the end of that phase – lasting approximately 500,000 years – the Sun will only have about half of its current mass.

There's a good lengthy coverage of the last days of the Sun here. It cites a study and then concludes that

  The bottom line is that the orbits are unpredictable and where the
  planets find themselves once the star dies is dependent on when, how
  much and how quickly the sun shed half its mass. This makes a major
  difference in how the solar system will look in the end and how easily
  the planets may be ejected from their orbits.

So the simple answer to your question is: as the Sun loses mass, the orbits of the outer planets will become larger. When they become wide enough, they will escape into interstellar space. If the hypothesized Planet Nine exists, it could help kick out Uranus and Neptune. Jupiter (and Mars, I guess) may or may not remain. Earth will in all likelihood be dragged into the Sun during the red giant phase as the Sun's outer layers expand to fill its orbit.


How is it that all planets (and moons) in our solar system are in equilibrium orbits?

I think at our current stage of solar system evolution, because of the fact that we have been able to evolve to our current level of sophistication, it could be considered to be very stable and in a very calm period in its evolutionary history. Unstable objects usually will be flung out very early on in the formation of such a system, hence why we do not see them today. However, just look at the Earth-Moon evolutionary history. It is theorised that our current Moon was captured due to a collision between Earth and perhaps a Mars (or similar size) object. This is anything but stable.

The reason why, is of course, related to gravity. Or more correctly, the gravitational potential of the system. All systems want to thermalise (in the same sense that a room of air molecules want to reach a thermal equilbirium). Although our solar system is anything but thermalised, it is constantly working to achieve this. Hence, why, at our present state of the solar system's evolution, we seem like we are in a fairly calm state. At this state, to go into more depth, we can employ Bertrand's theorem which tells us that for a central potential with an $r^{-1}$ dependance on radial distance the orbits will be stable. The stability of orbits in three spatial dimensions is due to the fact that the gravitational potential decreases with distance $r$ as $r^{-1}$.


What is the shape of the Sun's orbit around the Earth taking into account elliptical orbits?

The Earth moves in an elliptic orbit around the sun (or around the barycenter). If, in helocentric coordinates the Earth is at position (x,y), then in Geocentric coordinates the position of the sun is in position (-x,-y)

So the locus of the Sun in Geocentric coordinates exactly matches the locus of the Earth in Heliocentric. The path of the sun in geocentric coordinates is an Ellipse.


What exactly is this VLA radio image of Jupiter showing?

A uniform disk has been subtracted to better show the fine banded structure on the planet.

This means the surface of the planet is far brighter (in the raw data) than the surrounding magnetosphere. The planet itself has been darkened considerably in the image, otherwise it would be a pale circle and show little or no detail.

The press release from the VLA's site itself doesn't even show the wide field image however; it concentrates on the image showing detail of the spot itself. It mentions:

  an innovative data-reduction technique to "unsmear" the data from many hours of observing

There is an earlier 2006 paper by some of the same researchers showing earlier attempts, with a much more coarse radio map of Jupiter. This goes into more detail about the earlier image processing techniques though (requires good knowledge of Fourier analysis etc.)

(Note: I can't read the current Science paper in full, it's behind a paywall.)


We don't know (yet). It could have started with the explosion of a very large star, or a primordial black hole formed very soon after the Big Bang. See Wikipedia:

The origin of supermassive black holes remains an open field of research. Astrophysicists agree that once a black hole is in place in the center of a galaxy, it can grow by accretion of matter and by merging with other black holes. There are, however, several hypotheses for the formation mechanisms and initial masses of the progenitors, or "seeds", of supermassive black holes.
One hypothesis is that the seeds are black holes of tens or perhaps hundreds of solar masses that are left behind by the explosions of massive stars and grow by accretion of matter. Another model hypothesizes that before the first stars, large gas clouds could collapse into a "quasi-star", which would in turn collapse into a black hole of around 20 M☉. These stars may have also been formed by dark matter halos drawing in enormous amounts of gas by gravity, which would then produce supermassive stars with tens of thousands of solar masses. The "quasi-star" becomes unstable to radial perturbations because of electron-positron pair production in its core and could collapse directly into a black hole without a supernova explosion (which would eject most of its mass, preventing the black hole from growing as fast). Given sufficient mass nearby, the black hole could accrete to become an intermediate-mass black hole and possibly a SMBH if the accretion rate persists.
Another model involves a dense stellar cluster undergoing core-collapse as the negative heat capacity of the system drives the velocity dispersion in the core to relativistic speeds. Finally, primordial black holes could have been produced directly from external pressure in the first moments after the Big Bang. These primordial black holes would then have more time than any of the above models to accrete, allowing them sufficient time to reach supermassive sizes. Formation of black holes from the deaths of the first stars has been extensively studied and corroborated by observations. The other models for black hole formation listed above are theoretical.

(emphasis mine)


Tidally locked Venus, is it possible and consequences?

It may not be possible for Venus to become tidal locked
I don't think we know if it's possible for Venus to become tidally locked. Correia et al. 2008

expect the equilibrium rotation to differ from the synchronous motion

for planets like Venus with thick atmospheres relatively close to the Sun.
This might be best illustrated with a graphic from Auclair-Desrotour et al. 2016:

The gravitational torque is driving towards a tidal lock, but the thermal atmospheric derived torque is in opposition.
Observations may show that Venus is slowing down, according to this esa article. However, according to the article,

the planet could have weather cycles stretching over decades, which
could lead to equally long-term changes in the rotation period

A tidal-locked Venus might not have big day/night temperature variation
This article: The Gale Winds of Venus Suggest How Locked Exoplanets Could Escape a Fate of Extreme Heat and Brutal Cold suggests that the super-fast winds of Venus could occur even if Venus was tidally locked.  These winds exceed 100 m/s, which, according to the article is 60 times the surface rotation speed.  Even though Venus has an incredibly slow rotation rate currently (~243 days) the difference between night and day temperatures is very small.  We wouldn't expect much more deviation if Venus became tidally locked, as the heat transport is primarily driven through various other transport mechanisms than the Sun.  A relatively small amount of the Sun's energy makes it to the surface of Venus.
Edit:
In the comments, uhoh asks about the super-fast winds of Venus:

Is this mechanism keeping Venus' back side warm discussed anywhere in
a peer-reviewed state[...]?

As a quote from Horinouchi et al. (the "SR" in the quote refers to Super-Rotation, which is the super-fast winds):

SR may also occur in tidally locked exoplanets—those that always face
the same way toward their host star and are therefore heated only on
one side. Zonal (east-west) flow around the rotation axis, including
SR, can transport heat from the dayside to the nightside of those
exoplanets

Showman and Polvani go into greater detail in their paper: Equatorial superrotation on tidally locked exoplanets.


Predicting/simulating magnetospheres of exoplanets?

The magnetosphere is usually not generated by the atmosphere, but by the interior ferromagnetic flows of a planet. I say usually not, because Venus is an exception for the terrestrial planets in our solar system.
But in general rocky planets will exhibit a wide range of interior flow conditions, depending on their individual composition and state of cooling.
Gaseous planets can be a bit easier in that case, if one accepts uncertainties about the properties of metallic hydrogen.

If you then simply go an assume you know all those complicated properties, then sure, you can simulate something. But it's not a scientifically testable prediction only playing around with parameters, as long as it's not tied to a specific object.


Latest sunrise/earliest sunset out of phase with solstice?

The point here is that sunrise and sunset does not only move towards or away from noon, but noon itself moves around 12:00. There is a difference between "civil time" (the one counted by wall clocks) and "solar time" (the one marked by sun clocks). This difference is called the "Equation of Time" and its geometrical shape is called the Analema.

So in order to have the earliest sunset you need a combination of both a short day and a noon moved before 12:00. Same reasoning for latest sunrise.

You can see more about the Equation of Time on Wikipedia: http://en.wikipedia.org/wiki/Equation_of_time


How to make a telescope for viewing planets, moon and DSOs using a convex lens of aperture 100 mm and focal length 200 mm and other lenses at home?

You would be better off using lens 5 and 3. Though don't expect much, a simple single lens for the objective and for the eyepiece will give a lot of chromatic aberration (false colour). Refracting telescopes usually use at least two lens of different glass types (crown & flint often in a doublet) and eyepieces usually have 4+ lens.

You should be able to see craters on the Moon, the 4 Galilean Satellites of Jupiter, a couple of belts on Jupiter and Saturn's rings. You wont be able to reduce chromatic aberration as simply as that, the type of glass (it's refractive index) and curvature of the lens all need to be matched. You would be much better off buying a ready built objective and eyepieces if you want to make a refractor yourself, though it would probably be cheaper to buy a ready made telescope.


Relation between age and spectral type of star

Yes, the spectral type changes with age. The spectral type is a function of temperature, gravity and chemical composition at the photosphere. All of these can change during a star's life.
A star spends most of its life on the main sequence and changes in temperature and gravity are relatively slow. But thereafter there are comparatively rapid changes. For example, the Sun will become a K-type giant when it is older.
Second question. Spectral type is based on the appearance of the spectrum. However it turns out that when you put the spectral types in the correct order, they are also a sequence of temperatures and gravity.
We say O stars are hotter than M stars because that is a physically meaningful thing to say. The appearance of the spectrum and categorisation is just taxonomy.