Is there a significant temperature difference on the moon from apogee to perigee or vice-versa?

Roughly speaking, the moon's orbit has a semi-major axis of about 400,000 km. The semi-major axis of the Earth's orbit is 149.6 million km. The intensity of sunlight received by the Moon depends on the inverse of the square of the distance, so the difference amounts to $((1.496e8 + 4e5)/(1.496e8 - 4e5))^2$ = 1.011, or about 1%. So, not very much.

The variation in Earth's distance from the Sun is about five million km, so that's going to have more of an effect than the lunar orbit itself (as Chappo pointed out in their comment). That works out to a variation of about 6.9% in insolation from perihelion to aphelion. The maximum possible variation -- full moon at aphelion versus new moon at perihelion -- would be $((1.521e8 + 4e5)/(1.471e8 - 4e5))^2$, which works out to about 8.1%.

The difference in insolation from new moon to full moon at perihelion is about 1.09%, versus a difference of 1.06% at aphelion.


Wher can I find a list of blueshifted regions in the Milky Way?

I don't think there is a catalogue containing information about blue/red shift per region of the sky. You probably need a catalogue containing blue/red shifts per object and then calculate the average blue/red shift per region.
Most catalogues with blue/red shift are however targeted towards galaxies. They may also contain stars but you will have to filter the stars from the background galaxies. If I remember correctly, the Sloan Digital Sky Survey (SDSS) contains a flag specifying whether an object is a galaxy or a star. So you might use the SDSS data to get a large set of stars with blue/red shift. This flag has been set automatically using pattern recognition. I do not know how good the classification was.

You can access the data at the SDSS SkyServer. There are a lot of objects in SDSS so you will need time to acquire the data and knowledge about SQL to query the database.

Even then, you still don't have information about the spiral arm in which the object is located although you might use a method such as was used by Belokurov et al. (2006) to detect the field of streams to separate the different spiral arms.


Why do scientists think life requires water?

When you lose your keys on a dark night you look where the probability of finding them is highest (which is a combination of being likely to see them with their probability of being there), which is not necessarily where they are most likely to be. In the case of the search for extra-terrestrial life we have a fairly good idea what the signature of carbon based life using water as its primary solvent (conventional life) looks like. We also know that such life is possible. Other unconventional forms of life could exist but we have a very poor idea of what signature to look for, so we devote most of our effort to looking for the signature we will recognise.

We do not need to prove that unconventional life is impossible in order to spend most effort on searching for the conventional. It is sufficient to suspect that the probability of finding such life is higher than unconventional just because the probability of detection is so much higher. If we find nothing looking for conventional life, then a larger proportion of the effort will be devoted to searching for the unconventional.


Any studies of the detection of dark matter in or near the solar system?

We don't know what dark matter is, and we have never directly detected the exotic dark matter that is believed to make up much of the mass of the universe.

We expect dark matter to be in the form of a particle that interacts only with gravity and the weak nuclear interaction. Such a particle would be invisible (as light wouldn't interact with it) and it could pass through normal matter.

Since such dark matter doesn't interact much, except gravitationally, it would be spread out in a large elliptical cloud around the galaxy. There would be dark matter everywhere in the galaxy, including in the solar system, but there are no clumps of dark matter. There are no dark matter planets or stars.

In fact, we would expect about 1 proton mass of dark matter in every 3 cubic cm. That is a very small amount of mass. In the solar system, there is a billion-billion times more normal matter than exotic dark matter. And as I've noted, nobody has ever directly observed a dark matter particle, we don't know what they are.


Can a black hole be torn apart by SMBHs?

@peterh's answer contains the essential point, but I think lacks enough background. Let me try to go a bit deeper.

First of all, let's be clear that we're talking about the classical Black Holes (BHs) of General Relativity (GR), and not the true BHs of the as-yet unknown theory of quantum gravity (TQG). To answer your question, TQG does not have a role, because your question is about a regime which is far from where TQG effects would be relevant.

It's contrary to intuition, but the vicinity of a SMBH is a less exotic and less stressed environment than the vicinity of an ordinary BH. If you fell toward a solar-mass BH, the tidal forces would have torn you apart well before you got to the event horizon, while in a SMBH, it happens well after you pass the event horizon. Likewise the Hawking radiation (which can be taken as a measure of the stress in spacetime) is greater outside a solar mass BH than it is outside a SMBH. Wikipedia has a decent article on this which, while it contains at least one howler, on the whole seems to be reliable.

Bottom line: the tidal forces which tear things apart are weaker outside a SMBH than outside a smaller BH. But we've already observed several smaller BHs collide using LIGO, and they don't shred each other, but merge. So we can confidently say that a SMBH will not shred a smaller BH.

Another way to understand why a BH does not shred is to think about what's actually there. The answer is nothing. If you could survive the tidal forces (perhaps by falling into a SMBH) and could avoid the matter which generally forms an accretion disk around a BH (maybe by finding one in intergalactic space where there's no matter to pick up) and fall through the event horizon (EH), you'd notice nothing at all. It's just ordinary empty space outside the hole, at the EH, and inside the EH. Locally (i.e., it the cabin of your (now doomed) spaceship), there's nothing to see and nothing to detect, other than the ever-increasing tidal forces which will eventually tear you apart.  The BH is nothing but curved spacetime and there's nothing else there. What's to tear apart?

I'm using words. The mathematics of GR, which is very difficult but is also well-understood, describes exactly what happens when to BHs approach one another and Hawking's Area Theorem does prove that under GR, merger is the inevitable result.

What about the Theory of Quantum Gravity? Obviously we don't know much about it, since it hasn't been discovered yet, but one of the few things we're sure of is that it will give results exactly* the same as GR at "large" distances. And "large" is this case is probably anything bigger than an atom. Even if the limit is "bigger than a bread box", the two BHs are far too large for their collision to have any TQG effects. The TQG effects all happen down at the singularity the GR predicts at the center of the BH -- but whatever happens there is lost to the outside world.


What (actually) is the " deprojected half-light radius" of this almost-all-dark-matter Galaxy?

The half light radius is the radius from within which half the luminosity emerges.

"Deprojected" means that the authors must have fitted some model to the 2D distribution of light, which can then be mathematically deprojected to give them a 3D model for luminosity as a function of radius, that they can then integrate to give a number for the half light radius.

In section 3, the authors explain that they have done this by fitting a "Sersic profile" to the surface brightness distribution https://en.m.wikipedia.org/wiki/Sersic_profile  The Sersic profile actually has the 2D half light radius as one of its parameters. But if you imagine looking through a ball of stars, this 2D measurement of the half light radius is an underestimate of the true 3D half light radius, because the surface brightness profile is more sharply peaked than the 3D stellar density distribution that produces it.

The authors appear to approximately correct (deproject) this by multiplying the half light radius by 4/3. They also make a small correction for the non-sphericity of the galaxy.

The deprojection factor depends (slightly) on the $n$ index of the Sersic profile and needs to be found by doing a numerical integral. The details can be found in the appendices of Wolf et al. (2010. http://arxiv.org/abs/0908.2995 ), who also provide expressions to directly estimate mass from the projected half light radius and the line of sight velocity dispersion.


Considering our methods of exploration, how likely is it that there are unfound planets (not dwarf planets) in our solar system?

There are no undiscovered planets between the sun and Neptune.
Objects closer (to Sun) than Neptune that are large enough to be considered planets (and not dwarf planets) can't remain 'hidden'. If it's there, the light from the sun will bounce off it and we will see it. As it moves in its orbit, we will notice the position in the sky change, so we will know it isn't a star.
I would like to give a more broad answer to this question though:
What is the maximum size of an undiscovered solar system object and how does it change as you get further from the sun?
The further out a solar system object is, the harder it is to detect. The rate at which it gets harder is severe; the light we receive from an object scales roughly as $1/r^4 $.
($1/r^2$ for the light travelling from the Sun to the object, and again, $1/r^2$ for the light travelling from the object to us on earth).
We are able to detect some very small earth-crossing asteroids. Some as small as ~50 metres across. At a guess (and this is just based on my intuition, not any calculations), there are probably no undiscovered objects larger than 1 km close to earth.
As you travel to the outer solar system (Jupiter to Neptune), the number of bodies increases dramatically. There are currently ~700,000 known solar system bodies and most of them occur in this area. It is believed that all asteroids larger than 10 km have been found.
In the area immediately beyond Neptune (30 AU to 100 AU), we have been finding many pluto-like objects in the past two decades, objects with diameters of 500 km or more. For reference, Pluto is about 2200 km across. It is entirely plausible that there are some, if not many, similar objects at this distance range that have not yet been found. Some of these would classify as dwarf planets, in that they are round, orbit the sun, but have not cleared 95% of their orbit of other matter.
And finally - and this is where it gets exciting - where it gets really far out, there may be an undiscovered large planet, out at ~700 AU. Referred to as Planet Nine, this is a hypothetical object that some (noteworthy) astronomers believe exist because of patterns they see in the orbits of other distant dwarf planets. From calculations, they estimate that it would be as heavy as 10 earths and travel between 200 and 1200 AU. However, these distances are so large and the sunlight out there is so dim, that even with the best telescopes and two years of looking, they haven't found it.
Finally, I'd like to share two graphs. The first one is of distance (of closest approach, or perihelion) vs diameter of various outer solar system bodies. In the bottom right corner there is a noticeable lack of dots, indicating roughly where we don't yet have the capability of seeing objects that small and distant.
The second is an expanded but slightly less accurate diagram of the same.

Image 1 from Johnston's trans-Neptunian objects page. Image 2 from xkcd.


How do supernova explosions reconcile with light speed constraints?

Interestingly enough, the limit on the time needed for such a change to propagate through the star isn't from the speed of light, but from something called the dynamical timescale, $t_{\text{dyn}}$:
$$t_{\text{dyn}}\sim\frac{R}{v}\sim\sqrt{\frac{R^3}{GM}}\simeq\frac{1}{\sqrt{G\bar{\rho}}}$$
where we've assumed that the speed of a pressure wave, $v$ is on the order of the escape velocity of the star. $R$ is the star's radius, $M$ is its mass, $G$ is the gravitational constant, and $\bar{\rho}$ is its mean density (density is not constant throughout a star).

$t_{\text{dyn}}$ can vary greatly among different stars, depending on their characteristics. It can be on the order of seconds to minutes to hours to days, and it can determine the time it takes for parts of the supernova to take place. It's a lower limit, though; large dynamical changes can take more time than $t_{\text{dyn}}$, just not less. For more information, see Section II.B of these notes.

We have to ask ourselves, what precisely is $v$? I waved the specifics away as a "pressure wave", but is there more to it than that? Unfortunately, we still don't have great models of what goes on inside a supernova - I'm sure you've heard that before - but we do have inklings of the process, which I touched on a bit here.

Essentially, as the core collapses, matter at the inner regions begins to fall in, reaching extremely high densities. Some strange things happen here, including the production of neutrinos. At first, degeneracy pressure slows the collapse of the inner star, and a shock wave forms. The so-called "supernova bounce" happens, as this wave travels through the outer layers of the star.

The shock wave is further energized by the release of neutrinos, created via electron capture (note that some have already escaped the outer layers):
$$e^-+p\to n+\nu_e$$
This adds to the energy of the shockwave, which eventually blows away the outer layers of the star. Therefore, we can see how the dynamic timescale relates to the shock wave and the time of the resulting explosion.

The exact processes causing the bounce and shock wave are not yet fully understood; this is just a basic picture.


Wavelengths for observed objects -- emitted or observed?

If the redshift is $z$, then the wavelength of light you observe is at a wavelength that is $1+z$ times longer than the wavelength emitted in the frame of reference of the galaxy.

In this case $z=0.373$, so when you observe light centered on say the 3.4 micron band, then the light was emitted at a wavelength of 3.4/1.373 microns.


If an astronomical object appears to be very bright, and does not appear to 'twinkle' very much, then it is probably a planet, rather than a star. Jupiter is almost on the opposite side of the Earth to the Sun right now, so it will rise in the East when the Sun is setting in the West. Since Jupiter does look like a very bright star to the naked eye, and since it is in the part of the sky that you describe, then the object that you are seeing is almost certainly Jupiter.

As for recognising actual stars, you really need to learn the layout of the constellations to some extent. This will allow you to find your way around the night sky. A useful tool for this is a program like Stellarium. This will show you the positions of stars and other astronomical objects, including the planets.


What is the temperature 55 km beneath the surface of Mars?

If we look at Mars' possible geothermal gradient (see Earth's) which is about 25 °C per km.   Using the low estimate of Mars's gradient to be 1/4 that of Earth's Source, that's a bit over 6° C per km. so 55 km, 330° C.  Added that to Mars' average surface temperature of -55 C, you're talking 275° C or 527° F at 55 km underground, and that's a low estimate.

Estimates of the thickness of Mars' crust are often less than 55 km and if you're talking about digging into mantle or even, half way to the mantle, that's both an unsustainable dig and unlivable temperatures.

Source

  The average thickness of the planet's crust is about 50 km (31 mi),
  with a maximum thickness of 125 km (78 mi)

and

Source

Also, small point, but the 55km estimate assumes uniform temperature.  The simpler pressure calculations don't take into account temperature fluctuation.   With warmer temps and warmer lighter air at the bottom of the dig, which would be inevitable, you'd probably need 60 km, maybe 65 km of depth to achieve 1 atm.  There's also circulation issues when you have cold heavier atmosphere above warm lighter atmosphere and there's drainage/flooding problems for the lowest part of a valley, so to avoid flooding, you might want to keep the surface temperature below freezing, to avoid runoff and silt build-up, or dig even deeper so there's a lake of sorts that catches the melting ground-ice around the dig and the living area is above the lake.

This doesn't mean that deep valley living on mars will never be possible, but it's likely impossible with it's current atmosphere.  With extensive terraforming, tapping gases from underground and/or from it's ice-caps or by crashing comets into Mars (that would take a lot of comets), over time, enough of an atmosphere could be created where deep valley living might work.    A simulated atmospheric pressure can also be created with a bubble-dome of sorts, or one of my favorites, a stretch-wrap around the entire planet, but for now that's way beyond our means.

For the near term, living underground on Mars is probably the way to go.   Dig 50 or 100 feet underground and you'd have the means to create a near vacuum seal, which on Mars, you'd want to do, you could create earth like air pressure and you'd be protected from solar and cosmic rays and access to plenty of underground water.   underground on Mars is likely the way to go, at least, in the near future.   There's no shortages of articles on this.  Here's one 6.

Humans can also live at less than 1 atm.  People have settled as high as 16,700 feet, which is somewhere in the neighborhood of 55% of 1 atm.


Can we know the orbital planes of extraterrestrial planetary bodies?

I presume what you mean is how does the plane of the orbit compare to the equatorial rotation plane of the star?

The answer is, you can sort of estimate this, by using something called the Rossiter-McLaughlin effect (see also Rossiter 1924; McLaughlin 1924).

You can find plenty of information on the web - I'll add a couple of links when I have a moment - but to summarise:

The rotation of a star broadens its spectral absorption lines. The hemisphere coming towards us emits light that is is blue shifted, the hemisphere receding is redshifted. If we now take a transiting planet, during the eclipse it crosses the disc of the star and obscures regions that are blue or red shifted by various amounts.

Now what you do is measure the line-of-sight velocity of the star. During the transit you would not expect this to vary due to the "Doppler wobble" caused by the exoplanet, except that if the planet obscures a blue shifted portion of the stellar disc, the net spectral absorption line that remains shifts to the red, and vice-versa. The pattern of red, then blue shift (or vice versa) as the transit progresses is known as the Rossiter-Mclaughlin effect.

A schematic showing how the Rossiter-McLaughlin effect works and how a different transit geometry with respect to the rotation axis of the star leads to a different line-of-sight-velocity signature in the spectral lines of the star. (Image credit: Subaru Telescope, National Observatories of Japan.)

If the planet orbits in the same plane and in the same direction as the stellar rotation (as the planets in our solar system nearly do), then the blue shifted limb of the parent star is obscured first, followed by an equal amount of redshift as the planets moves to obscure the receding stellar limb (see image above, left). Thus the stellar absorption lines show a redshift followed by a symmetric blueshift. If the planet was retrograde it would be symmetric but occur in the opposite order. A Polar orbit would show no RM effect. Inclined orbits would have an asymmetric RM effect - ie perhaps more blueshift than redshift (see image above, right).

The RM effect cannot give an exact orientation, it gives the projected angle between orbital and rotation axes on the plane of the sky. Nevertheless, that is sufficient for us to know that a lot of the transiting exoplanets have highly misaligned orbital and stellar rotation axes.


Does the light from a torch reach to the moon?

Visible light from a torch has a 50% (blue light) to 90% (red light) chance of making it through the Earth's atmosphere, depending on the amount of dust and aerosols in the atmosphere and how close the Moon is to the horizon.

Light travels in straight lines. The Moon has an angular diameter of only 0.5 degrees as seen from the Earth. Most ordinary torch beams diverge by much more than this, so only a small fraction of the photons from a torch would pass through a circle of angular diameter 0.5 degrees. Hence only a small fraction of the emitted photons would hit the Moon.

Finally, you said that the torch was pointed "towards the Moon". Actually, you pointed it to where the Moon was 1.5 seconds ago, and the torch light reaches the distance of the Moon a further 1.5 seconds later. Thus the Moon has travelled 3 seconds in its orbit. However, since the orbital period (360 degrees) is a month, then it only covers an angle of 1.5 arcseconds in 3 seconds, so this shift is not relevant for your experiment (though it is for those bouncing lasers off reflectors on the Moon, which proves that light does in fact reach the Moon).


What is the difference between a moon and a random chunk in the rings

Your question:

  There also are satellites in between the rings. What distinguishes a
  moon from any other chunk in orbit around a planet?

Wikipedia should be taken with a grain of salt, but they define moons or moonlets within a ring system as creating a gap or partial gap, sometimes described as looking like a propeller.

  In 2006, four tiny moonlets were found in Cassini images of the A
  Ring.[44] Before this discovery only two larger moons had been known
  within gaps in the A Ring: Pan and Daphnis. These are large enough to
  clear continuous gaps in the ring.[44] In contrast, a moonlet is only
  massive enough to clear two small—about 10 km across—partial gaps in
  the immediate vicinity of the moonlet itself creating a structure
  shaped like an airplane propeller.[45] The moonlets themselves are
  tiny, ranging from about 40 to 500 meters in diameter, and are too
  small to be seen directly.[9]

A moon, I would think, would need to clear out an entire gap within a ring.   Clearing out a partial gap or "propeller" as noted above can be as small as 40 meters.  Often the moonlets themselves aren't visible but the propellers they create in the ring system is visible.

Some more details and images here and here


What are the best candidates for solar lensing of exoplanets, and what would the results be?

I've seen this question for the past few days and have shied away from it because it seemed too difficult, but here goes:

550AU is the focal point of the sun, so theoretically anything on the opposite side would be focused into a perfectly coherent image right? Unfortunately due the density of the sun (Mass to radius ratio specifically) only radio waves that just miss the sun would be focused to that point, all other shorter wavelengths don't diffract enough and simply 'collide' with the sun or form a lense ring at a greater distance. So radio-observation only. Explained at this article https://physics.stackexchange.com/a/25501

A star such as a neutron star however would make an excellent candidate for stellar lensing because it has a high density and low radius, meaning much shorter wavelength radiation is able to go around the star.

This wouldn't be particularly effective at discovering new exoplanets, only observing them in more detail.


When will the number of stars be a maximum?

TL; DR Somewhere between now and a few hundred billion years time. (For a co-moving volume) Now read on.

If stellar remnants are included, then the answer is very far in the future indeed, if and when the constituents of baryons begin to decay. So let's assume  that "stars" means those things that are undergoing nuclear fusion reactions to power their luminosity. Let's further assume that the stellar mass function, ($N(m)$ is the number of stars per unit mass) we see in the neighbourhood of the Sun is representative of populations in all galaxies at all times (difficult to make a start, without assuming this).

The number of stars that have been born is equal to the sum over time (the integral) and over mass of $N(m)$ multiplied by the rate at which mass is turned into stars in a comoving volume of the universe $\Phi(t)$.

We then need to subtract a sum over time and mass of the rate of stellar death in the same comoving volume. The rate of stellar death is the rate of stellar birth at a time $t-\tau(m)$, where $\tau(m)$ is the mass-dependent stellar lifetime. We ignore mass transfer in binary systems and assume that multiples can be treated as independent stellar components.

Thus the number of stars at time $t$ is approximately
$$N_*(t) = \int_0^{t} \int_m N(m) \Phi(t') - N(m)\Phi(t'-\tau(m))\ dm\ dt'\ .$$
To find where this is a maximum, we differentiate with respect to time and then equate to zero. i.e. We look for the time when the stellar birth and death rates are the same.

I was going to (and possibly still will) attempt some sort of analytic approximation, but Madau & Dickinson (2014) have done it better and taken into account the metallicity dependence of stellar lifetimes and the chemical evolution of galaxies. The star formation rate peaked about 10 billion years ago, is more than an order of magnitude lower now and is exponentially decreasing with a time constant of 3.9 billion years.

The integrated stellar mass is shown in their Fig 11 (shown below). It is still increasing today, but at a very low rate and has not passed through a maximum. The reason for this is that most stars have masses of 0.2-0.3 solar masses and lifetimes much longer than the age of the universe. Even if these stars are added at a very slow rate, their death rate is zero at present.

If star formation did continue at a low-level then the number of stars would only begin to significantly diminish once the stars near the peak of the stellar mass function, that were born at the earliest times, start to die. The lifetime of a 0.25 solar mass star is around a trillion years (Laughlin et al. 1997).

On the other hand if star formation ceased now then the number of stars would immediately begin to diminish.

Perhaps we could argue that the current exponential decline will continue and the peak
will come in another few billion years when stars of 0.8-0.9 solar masses begin dying off. However, that is futurology given that we have no first principles theory that explains the time-dependence of star formation, so I believe the best answer that can be given is somewhere between now and a few hundred billion years time.

Note that this answer assumes a co-moving volume. If the question asked is phrased in terms of the observable universe then because the number of stars has nearly reached a plateau, then the answer becomes close to whatever age the volume of the observable universe is maximised. I say "close to" because you have to factor in that the observable universe includes stars in distance slices at all cosmic epochs. I am unwilling to undertake this horrendous calculation, but note that the current concordance cosmological model has our observable universe slowly increasing from around a radius of 45 billion light years now, to about 60 billion light years in the far future Davis & Lineweaver 2005, and this may compensate for a slow decline in the number of stars in a co-moving volume.


Scientifically important discoveries with the help of amateurs

I guess it depends on what you consider scientifically important, but users at Planet Hunters have helped with the discovery of several exoplanets by analyzing data from the Kepler Space Telescope, as well as the famous Tabby's star (KIC 8462852) which has led to a paper by Tabetha Boyajian co-written by the users who participated in the analysis of the light curve of the star.

The star is currently still the subject of numerous studies and publications.


What is the acceleration of the stars' speed in the Universe? Positive or negative?

This is the classic misconception about the expansion of the Universe: there is no center of the Universe. The whole universe expands as one, not from a single point nowhere.

Knowing this it makes no sense speaking of the direction of an object due to the Universe. You could compare the directions of two objects in it though.

Therefore star's direction depend on where and how they where created and also the interaction with other stars and their home galaxy.


Time after sunset until star can be seen

It depends on the star's brightness. Astronomical darkness (full night, essentially) begins when the sun is 18 degrees below the horizon. Then, you can see all of the ones you're going to be able to see.

You will see stars before that point; which ones and how many will depend on how bright they are relative to the brightness of the twilight sky.


Solar wind, Earth wind and planetary winds?

From going through the literature the paper by Wang et al. 2021 is citing, I am nearly certain that the term "Earth wind" must be a recent invention, perhaps by those authors themselves.
It is however correct to call the solar wind a 'wind'. This is because a wind is a pressure-driven bulk motion of a collectively coupled gas. The solar wind, at its base is driven by the enormous pressure gradients via the 1 million Kelvin hot solar corona.
While Earth's primary, hydrogen-rich atmosphere might have experienced such a phenomenon, it is incorrect in naming the current atmospheric escape as 'wind'.
But now vocabulary aside, current atmospheric escape rates at the terrestrial planets are governed by various ionic escape processes, the most important of them being polar cusp escape. So the data you are actually looking for is data on polar cusp escape from various orbiters. Fortunately, Gunell et al., (2018) have given a recent compendium on this (see their table A.1), and attempted a simple modelling of the data. There is no data available on polar escape for Mars and Venus, as this concept doesn't apply to those planets. The other escape processes are on average smaller for hydrogen, but larger for oxygen ions.
Similarly, I think there is no data available on the escape of gases from the gas giants. The magnetospheres of Jupiter and Saturn have been studied, and their particle populations characterized, but I think it is unclear how much of the magnetispheric content gets ultimately lost. There is no reliable data available on polar escape for Uranus and Neptune, as those planets were only ever visited by one singular flyby (that of Voyager 2).


"Black holes evaporate with time" means they lose mass and energy with time. This phenomenon is termed as Hawking radiation (named after Prof. Stephen Hawking who gave the theory) or simply black hole evaporation. According to him when a particle or antiparticle pop into existence at the event horizon instead of annihilating each other as they are supposed to, one of the the two will fall in the black hole and other will escape. Some of the mass inside the black hole is converted into energy which later converts into particles ( at the event horizon in this case)
The black hole loses energy/mass in form of  particle ( mass and energy are equivalent)


Would I actually be able to see Ceres without shining a really bright torch onto it?

Ceres is 2.76 AU away from the sun, so the sunlight is seven times weaker. That is comparable to the levels of illumination 30 minutes before sunset on Earth, so you are not going to have any problem seeing the surface.

The albedo of Ceres is quite low, comparable to worn asphalt. But it is still easy to see it.


Observing Comet 45P - reliable sources of position data

The best way to get ephemeris of solar system bodies is to use JPL HORIZONS:

http://ssd.jpl.nasa.gov/horizons.cgi

You put "Target Body" as 45P, your "Location", "Time", and click "Generate Ephemeris". This will give the RA/DEC information for time by default.

If you want some advanced data other than the listed, e.g., phase angle, V-magnitude, RA/DEC change rate, etc, you can tune "Table Settings". I usually skip daytime and airmass < 2.

There can be some mismatch between the ephemeris and real observation. This is not because JPL HORIZONS gives wrong calculation, but because the reported observation data is not perfect. It calculates the orbit of the target from previously reported observational data to IAU MPC. IAU MPC provides the orbital elements inferred from that reported data, and these elements are used to generate ephemeris from JPL HORIZONS. So if the report is slightly imperfect and thus the inferred orbital element differs from the true value by a slight amount, it may cause the mismatch at a time far from the original observation. This is the reason we need more and more data continuously and update the catalog.


Dust mass-loss rate from a massive star given a set of parameters?

Roughly: the mass loss rate for a spherically symmetric wind is given by
$$\dot{M} = 4\pi r^2 \rho v_w,$$
where $\rho$ is the density in the wind, $r$ is the radius at which that density is measured and $v_w$ is the wind velocity.

Here you have $v_w$ and have the mass of material within two concentric radii - so you can work out the average density at some average radius in the shell.

The age of the shell is not relevant to the mass-loss rate, but assuming the wind parameters are unchanged over that time, they do tell you ho much mass has been lost in total $\simeq \dot{M} \tau$.


What's the difference between semi-major axis and apoapsis?

If the semi-major axis of an ellipse is a and the eccentricity is e,
then the distance from the center C to a focus F1 or F2 is c = ea.
Apoapsis is a + c and periapsis is a - c.

Ellipse with e ≈ 0.78:

For the Earth's orbit around the Sun, a = 149.6 million km and e = 0.0167, so

c = ea = 2.5 million km
aphelion = a + c = 152.1 million km
perihelion = a - c = 147.1 million km


What are the equations governing stellar evolution (Luminosity, Mass, Temperature, Radius)

What you want is hydrodynamic stellar evolution. The following starts with a brief review of hydrostatic case to give the fundamentals. The last paragraph discusses the hydrodynamic case.

For hydrostatic case, see this, or this. Simplified stellar evolution assumes hydrostatic equilibrium (i.e., time-independent fashion). The system of equations compose of i) mass conservation, ii) mass transportation (i.e., hydrostatic equilibrium), iii) energy conservation, and iv) energy transportation. The system also requires i) the equation of state, ii) opacity, and iii) nuclear energy generation for being solved in a close form.
The close-form solution is typically in the form of polytrophs (i.e., Lane-Emden equation). More complicated model, which is still hydrostatic equilibrium, will include other factors such as composition, rotation, and binary interaction. Since it is static and equilibrium case, the time-dependent terms are zero.

For hydrodynamic case (i.e., time dependent) as you wanted, there is no close-form solution, therefore only hydrodynamic simulation is the method for the study. The simulation model uses the set of equations mentioned above without dropping the time-dependent terms. See, e.g., MESA.


Is the Sun really burning and getting smaller?

I won't bother answering how the Sun "burns" its fuel — that is already answered in this question. Your second question is whether the Sun is getting smaller; the answer depends on what you mean by "smaller". We'll analyze this in two ways:

Radius

If we define the radius of the Sun by distance between the core and photosphere, then the Sun is not getting smaller. In fact, it is growing over time, as demonstrated in this graph:

Image courtesy of Ribas (2009), colored by user RJHall on Wikipedia under the Creative Commons Attribution-Share Alike 3.0 Unported license

The height of the photosphere is increasing ever so slightly, even in the Sun's main sequence stage; specifically, Table 3 of Sackmann et al. (1993) states that at the Sun's zero-age main sequence, it was $0.9R_\odot$ in radius, and at its terminal age main sequence it will be $1.6R_\odot$. That means that, over the course of about 10.9 billion years, the Sun's radius will have increased by $0.7R_\odot$ — about 70% its current radius.

If we do the math $\left (0.7R_\odot \over 10.9 \ \text{Gyr} \right )$, we could find that the Sun's average increase in radius (from beginning to end of main sequence) is about 4.5 centimeters each year (1.77 inches). This calculation shouldn't be taken too seriously, since the growth rate is not linear, but it demonstrates how slowly the Sun has been growing.

The Sun will continue to grow at a very slow rate until it leaves its main sequence stage. Once it enters the red giant branch, it will grow many times its current size, to approximately 1 AU.

Mass

The Sun is losing mass for two reasons: the nuclear fusion in its core converts some mass into energy, and high energy protons and electrons are constantly ejected from the Sun's corona as solar winds. Wood et al. (2002) shows that the Sun's mass-loss rate $\dot{M}$ is (approximately) inversely proportional to the square of its age $t$:

$$\dot{M} \propto t^{-2.00 \pm 0.52}$$

As of now, $\dot{M} = 10^{-14}\frac{M_\odot}{\mathrm{yr}}$. After the Sun leaves the main sequence, it will lose its mass at a much faster rate. As stated in Sackermann et al., the rate at which it a post-main sequence star loses mass can be generalized with Reimers' law:

$$\dot{M} = -\eta (4 \times 10^{-13}) \frac{LR}{M}$$

where $\eta$ is a value depending on the star (derived to be ~0.6 for the Sun), $M$, $L$ and $R$ are in solar units, and $\dot{M}$ is in solar masses per year. This is simplified, but gives us an idea of how the Sun's phase affects its mass-loss rate.

By the time the Sun enters the red giant branch, it will have only 72.5% its current mass, and its luminosity and radius will reach $2300L_\odot$ and $170R_\odot$, respectively. Plugging those values in, we find that the rate of mass loss will be $1.3 \times 10^{-7} \frac{M_\odot}{\mathrm{yr}}$. It will lose mass even faster in the asymptotic giant branch, when its mass will become ~54.1% of what it is now, meaning $\dot{M}$ will be $2.5 \times 10^{-7} \frac{M_\odot}{\mathrm{yr}}$.


Relation between intensity of Sun and the light that is emitted from surface of sun

We take a unit area dA of sun,, which is emitting light directly to our line of sight

You are, I think, imagining the light as if it was directed out from a point, with directions of photons outward from the center.

What actually happens is that the light is emitted from every point on the surface and what's more every point on the surface can send light in any and all directions.

So the Sun is a diffuse source of light, not a point source of light.

  But there are also numerous other unit areas throughout sun's surface whose light does not reach our eyes directly.

So because of what I have described - light being emitted in all possible directions from all over the Sun's surface - we do see light from all parts of the surface.  And that light does reach our eyes directly.


Why does the author believe that the central mass that gas cloud HCN-0.009-0.044 orbits is smaller than our solar system?

The authors were able to successfully model the motion of the gas streams as Keplerian orbits around an object of $\sim30000M_{\odot}$. In doing so, they derived some key quantities, such as the pericentric distances of these two gas components (the "Balloon" and the "Stream"). One has its closest approach at $\sim0.21$ pc; the other has its closest approach at $\sim0.07$ pc ($14000\text{ AU}$). This places the pericenter well inside the Oort Cloud, and therefore inside the Solar System.

Therefore, given that it's highly unlikely that the Balloon skims the surface of the central object, we can say that the object's radius is certainly less than that of the Oort Cloud, and by extension, the Solar System. This is a key part of the argument that the object is, in fact, an intermediate mass black hole, as they note on Page 5. It's basically the same logic used to constrain the size of Sagittarius A* based on those infrared flares observed last year.


On the numbers

Rogue planets are thought to be more abundant than stars in the milky-way and depending on where you make the cut-off, rogue dwarf-planets for example, are expected to enormously outnumber the number of stars.   older article quoting the MOA survey.

This article - OGLE survey, or, Portland's Optical Gravitational Lensing Experiment estimates:

  Based on a statistical analysis of more than 2,600 microlensing
  events, drawn from six years of observations on about 50 million
  stars, the OGLE team estimates that there is perhaps one Jupiter-mass
  rogue planet for every four stars in the galaxy.

It's not an experiment it's a survey, but I guess they couldn't resist the name OGLE, so they went with that, so, props on the name, but I digress.

There's still some disagreement between MOA and OGLE, though I think the consensus is that OGLE gives a slightly better estimate of 1 Jupiter-mass rogue for every 4 stars.

Due to the nature solar-system formation, we expect that a number of planets get cast out during the early period and larger orbiting objects are good at casting out smaller ones when they pass to close, but it's almost impossible for smaller objects to cast out a larger one.    We expect that the number of smaller rogue planets should increase significantly, perhaps even exponentially.  But the OGLE survey had difficulty locating smaller planets, 1-10 Earth masses.   That may be because they are more difficult to see, or they might be less common than expected.

Binary star systems, for example, with two or more much more massive than Jupiter objects could cast out larger objects like Jupiter-mass planets fairly easily, so there are significant unknowns on the number of smaller than Jupiter rogue planets.  Estimates are high, observations are low.   NASA's WFIRST might be powerful enough to resolve this after it's launch in 2020.

This article a study by the Kavli institute, suggests an aggressive estimate of 100,000 rogue-planets (or Nomad planets) for every star in the Milky-way, based on the 2011 MOA estimate of two Jupiter-Mass objects for every star and extrapolating based on how common Pluto sized objects are thought to be relative to Jupiter sized (about 50,000 to 1).    If we adjust for the 1 to 4 instead of 2 to 1, that estimate is shaved down to about 12,000 pluto or larger rogue planets per star.

Whatever the real numbers are, if we count all the way down to dwarf planet size, Rogue planets should significantly outnumber stars and an estimate of 3-4 magnitudes isn't unreasonable.   For planet sized, say Mercury to Neptune, better observations would help with this estimate a great deal, but a bad estimate of Mercury and up Rogue planets outnumbering stars 10 or 20 to 1 isn't crazy.

I'm fussing on the numbers because it's relevant to how many pass through the solar system.   Using this question/answer as an approximation and because the escape velocity from a solar system is usually tiny compared to the escape velocity from a galaxy or the orbital speed around the galaxy, generally speaking, rogue planets should have star-like orbits around the Milky-way.    If a star passes within one light-year of our sun about once every million years, even if you set a relatively high estimate of 100,000 rogue planets (pluto size and up), that's still one every 10 years, and that's within 1 light year.   If you want to estimate within the Kuiper Belt - where we might expect to observe a pluto sized rogue planet passing through.   The target area of the Kuiper belt is about a million times smaller than the 1 lightyear radius, so a rogue planet passing through the Kuiper belt - by this math, once every 10 million years.

The reason why it's never been observed is that they are usually very far away and too small to be noticed.  Even with today's best equipment, it's nearly impossible to see a Pluto sized rogue planet passing through a lightyear or half a light year away and if they only pass through every 10 or 100 or more years, well, that's why we've never seen one.

It's possible, in our lifetime, someone might see one or two rogue planets pass within 1 light year of us, or inside the Oort cloud just past 1 light year, but we'll almost certainly need an upgrade in telescopes first.

on capture

In a two body system, gravitational capture is impossible and a rogue planet to our solar system is very much like a two body system.   The problem is the rogue object accelerates towards the larger object (the star) and it adds velocity.   That added velocity means it flies away just as fast.    In a two body system, objects are either already in orbit or they fly past.

In a loose sense, all the stars in the galaxy create an N-body system where capture becomes possible, but the problem is that the stars are so far apart that any secondary body gravitational effects from other stars is close to negligible.  The chance of Alpha Centauri (for example) slowing down an object so our Sun can catch it, is close to zero.   If it happened at all, it would be a very distant and very tenuous capture.   That's not to say it never happens.  Over 4 billion years of our solar system, capture may have happened several times, but I wanted to point out why capture is difficult and likely rare.  Most Rogue planets that got close enough would likely just pass through.

Capture becomes statistically more likely when a massive 3rd body is close.

If a Pluto-sized rogue planet passed close enough to one of our 4 large planets, or, the theoretical planet X, then it's direction and velocity could be significantly changed, and the odds of capture way up.   Needless to say, those 4 planets and the theoretical "Planet Nine" make very tiny targets in the vastness of space, but that's one way capture could happen and over the 4 plus billion years the solar system's been around, capture probably has happened from time to time.   The trick is getting a close enough look at distant orbiting objects to see if their chemistry indicates they aren't from our solar-system.

In addition to Rogue planet capture, when two stars pass close enough to each other, two stars exchanging Oort cloud objects probably happens too and might be the more common method of object exchange.  Oort cloud objects already have a relatively slow orbital speed around their star which makes striping them away from one star and being captured by another, somewhat more likely.   It would also depend on the relative velocity of the two stars also and how close they passed.   But this method of exo-object capture might be quite a bit more common than rogue object capture.

If an object like Sedna is eventually identified as not being from our solar-system, it would be hard to know if it had been a rogue planet or a "stolen" object from the Oort or Kuiper belt from a close passing star from long ago.


I really love this question. Were we to establish a colony on our moon, we would probably want a calendar/time system which actually corresponds to physical patterns and cycles on the moon, but also something which is easy to track and keep in sync with Earth, since the two civilizations would certainly be in close contact.

In a way, we already do something this here on Earth with time zones, though time zones are just offsets within the same second/minute/hour/day system, so people throughout the world are in different times, but they stay in sync. But beyond that, our entire calendar system (we currently use the Gregorian Calendar), is mostly based on our own planet's positions relative to the sun. It just wouldn't make sense on the moon. Earth days on the moon would just be arbitrary 24-hour blocks of time; they wouldn't correlate at all to whether it's light or dark outside.

So, if we really wanted to keep some logical synchronization between earth and moon calendars, we would probably want to adjust here on earth to a lunar-based calendar. Most older civilizations did this, using a full lunar cycle to indicate a "month." But here's the (cool) thing: the moon is a little lopsided and way gravity works, the exact same side is always facing us. We never see the back of the moon. So, it's rotating at exactly the same speed as it's revolving around us, meaning a "day" on the moon lasts one full lunar cycle: just over 29 earth days. So even the idea of a day, how long a day is, and what you could accomplish in a day is so far from Earth's that it's hard to even compare the two.

Maybe one idea is to have weeks or months here on Earth which would line up with days on the moon. Earth would still have its 24-hour days and the Moon would have its 709-hour days, but at least we could predict Moon time using our own calendars. One proposal is the The Hermetic Lunar Week Calendar which continually adjusts the length of a week to keep the months in sync with the moon, so that a full month is always one lunar cycle (or lunar day). It would be weird but in a future where humans are legitimately split between Earth and Moon, it would probably make sense for both to keep a calendar like that, which takes into account the length of a day on both bodies.


Will a Ball placed close to Sun fall into it?

Short answer is yes. The only way radiation pressure would be a factor was if the ball was in orbit around the Sun, this would mean that the slight pressure it feels would cause it to drift away slowly (assuming its made of the most heat resistant material on Earth!). But in your question you said "placed" near the sun not in orbit around it, so the strength of the sun's gravity would overcome all other forces and pull it in.


What is the nature of "rock vapor" in this description of the formation of the Moon?

Planetary Scientist Sarah Stewart's research is on the formation of the moon, not, as far as I can tell, as much on the chemical composition and precise temperature of the atmosphere after impact, so I don't know if plasma is all that relevant to her work, but I think she'd have to model and account for total energy and temperature, similar to what you did in your question.

If I understand you correctly, you want to know what Earth's atmosphere was like, lets say a weeks or a few weeks or maybe a year after the giant impact.   Dr. Stewart's team has a word for this type of planet, a Synestia

Plasma temperature is tricky for 2 reasons.  One, there's not a specific temperature where gas becomes plasma.  Unlike melting points of boiling points, which happen at specific temperatures (and specific pressure for boiling points), the plasma phase of matter is closer to a dimmer switch that turns on gradually than a specific plasma point.  Similar to the temperature where fusion happens, individual electrons are unpredictable, so heating a gas, it will turn into a plasma gradually.

It's also possible for a rock vapor, take SiO2 as a baseline, to retain it's double bonds as a gas and at the same time, be a low level plasma, emitting some electrons, so it can be both rock vapor and a plasma.   That's not possible for water, for example, because those bonds are too weak.  Water splits into Hydrogen and Oxygen a couple thousand degrees lower than when the individual molecules begin to enter the plasma state.

Another problem is pressure.  The center of the Earth is plasma temperature (low level Plasma but it's in that temperature range), but people generally don't call that state of matter a plasma.

I think your 40 km/s estimate is too high, because Theia was thought to be a Trojan object before it collided with Earth, so the collision rate should be not much more than escape velocity, maybe 12 or 13 km/s because they shared the same orbit.

40 km/s meteor collisions on Earth happen because they approach at a different inclination, where the orbital directions are much less lined up, that's how you get 11 to 70 km/s for meteors, but Theia was probably on the low side of that, perhaps 14 or 15 km/s tops depending on it's eccentricity - if I can make a bad guess.

I'm sure that plasma temperature happens during giant impacts.  But the temperature is highest where the two objects collide, so initially, the highest temperature corresponds with the highest pressure.   After the impact you have the explosive rebound, because collisions of this magnitude are more like large explosions than anything else and after that you can model where the temperature goes as the Earth begins to settle.

Models would have to account for how the heat moves around and through the planet, how much heat ends up burred vs goes into rock vapor, heat of vaporization, heat lost due to expansion of rock vapor, how quickly heat radiates away (I would think it would be highly opaque, so radiation would be somewhat slow),

Bigger factors would be how much is lost in rebound and ejected material and how much is transferred to angular momentum.  There's also the uncertainty on how massive Theia was.  I think later estimates put it at 1/2 to 1/3rd the mass of Mars.

Another way to look at this question is, after formation, the synestia would have layers, similar to any gas giant, though gas giant layers aren't well understood, we could use the sun as an example.  There might be convection, conduction and condensation layers, and perhaps lapse rate could be applied, and obviously gravity would be lower with the material more spread out.   You might also have layers where the pressure was sufficient that different types of matter would form, like, hot enough to be liquid but enough pressure to be a solid, similar to Earth's core.

All that said, trying to calculate the lower atmospheric temperature of this theoretical, recently formed synestia is a little bit more math than I want to do, and I'd probably get it wrong anyway even if I did the math.  But it seems entirely reasonable that the lower atmosphere was at plasma temperatures if much of the upper atmosphere was at rock-vapor temperature.  But if you can get a temperature model, that would be a step in the direction of a plasma model.   I'd guess the low level plasma temperature for rock vapor would begin somewhere in the 5,000 or 6,000 C range, but it's a hard thing to look up as different compounds have different plasma temperatures.  There's even some cold plasma, like florescent bulbs work on that property, but they require an electric field.

I don't know if my long "I don't know if there was plasma" counts as an answer, but it's a fun question and I thought I'd give it a shot.   I like her idea a lot and I've read a few articles that indicated problems with the more traditional giant impact models, so she may end up being right.


Why is the cloud around Eta Carinae shaped the way it is?

It seems likely due to rotation.  The disklike "skirt" could be due to material that is orbiting the star that gets thrown out by the eruption, and the bipolar nature of the eruption can be because material is preferentially ejected from the poles of the star.  It may seem odd that the poles, where there is less angular momentum per gram, would be better at ejecting material, but this might be due to a greater role for radiation pressure in causing the eruption.  Radiation pressure produces forces that are proportional to the radiative flux, which can come out preferentially in the polar directions because this becomes the path of least resistance when rotation makes the shape of the star highly oblate.


What can a regular person use to view the stars?

Binoculars.

This isn't going to attach to a smartphone, not everything in life is best seen through a screen. The camera on a smartphone is remarkable, but your eyes are much more sensitive in low light. Binoculars will make a lot more stars visible, even in heavily light polluted places.

A pair of binoculars is cheap enough for students. They are categorised as, for example 10x50, which means 10x magnification, and 50mm lenses. You want the second number to be big, which means the binoculars gather more light. 10x50 is often thought to be a good compromise between light gathering and weight. You won't want "zoom" or "night vision" or other such features.


How did our ancestors discover the Solar System?

1. Ancient cultures observed the sky

Night skies are naturally dark and there was no light-pollution in ancient times. So if weather permits, you can easily see a lot of stars. No need to tell about the Sun and the Moon.

Ancient people had good reasons to study the night skies. In many cultures and civilizations, stars (and also the Sun and the Moon) where perceived to have religious, legendary, premonitory or magical significance (astrology), so a lot of people were interested in them. It did not took long to someone (in reality a lot of different people independently in many parts of the world) to see some useful patterns in the stars that would be useful to navigation, localization, counting hours, counting days and relating days to seasons, etc. And of course, those patterns in the stars were also related to the Sun and the Moon.

So, surely all ancient cultures had people who dedicated many nights of their lifes to study the stars in detail right from the stone age. They would also perceive meteorites (falling stars) and eclipses. And sometimes a very rare and spetacular comet.

Then there are the planets Mercury, Venus, Mars, Jupiter and Saturn. They are quite easily to notice to be distinct from the stars because all the stars seems to be fixed in the celestial sphere, but the planets don't. They are very easily to notice to be wandering around in the sky with the passage of the days, specially for Venus, which is the brightest "star" in the sky and is also a formidable wanderer. Given all of that, the ancient people surely become very aware of those five planets.

About Mercury, initially the Greeks thought that Mercury were two bodies, one that showed up only in the morning a few hours before the sunrise and another only a few hours after the sunset. However, soon they figured out that in fact it was only one body, because either one or the other (or neither) could be seen in a given day and the computed position of the unseen body always matched the position of the seen body.

2. The Earth seems to be round

Now, out of the stone age, already into ancient times, navigators and merchants who travelled great distances perceived that the Sun rising and setting points could variate not only due to the seasonal variation, but also accordingly to the location. Also, the distance from the polar star to the horizon line also variates accordingly with the location. This fact denounces the existence of the concept nowadays known as latitude, and this was perceived by ancient astronomers in places like Greece, Egypt, Mesopotamia and China.

Astronomers and people who dependend on astronomy (like navigators) would wonder why the distance from the polar star to the horizon varied, and one possibility was that it is because the Earth would be round. Also, registering different Sun angles in different locations of the world on a given same day and at a given same hour, also gives a hint that the Earth is round. The shadow on the Moon during a lunar eclipse also gives a hint that the Earth is round. However, this by itself is not a proof that the Earth is round, so most people would bet on some other simpler thing, or simply don't care about this phenomenon.

Most cultures in ancient times presumed that the world was flat. However the idea of the world being round exists since the ancient Greece. Contrary to the popular modern misconception, in the Middle Ages, almost no educated person on the western world thought that the world was flat.

About the Earth's size, by observing different Sun positions and shadows angles in different parts of the world, Erasthotenes in ancient Greece calculated the size of Earth and the distance between the Earth and the Sun correctly for the first time as back as the third century B.C. However, due to the confusion about all the different and inconsistent unit measures existent back then and the difficulty to precisely estimate long land and sea distances, confusion and imprecision persisted until the modern times.

Ancient cultures also figured out that the shiny part of the Moon was illuminated by the Sun. Since the Full Moon is easily seen even at midnight, this implies that the Earth is not infinite. The fact that the Moon enters in a rounded shadow when exactly in the opposite side of the sky as the Sun also implies that it is the Earth's shadow on the Moon. This also implies that Earth is significantly larger than Moon.

3. Geocentrism

So, people observed the Sun, Moon, Mercury, Venus, Mars, Jupiter, Saturn and the fixed sphere of stars all revolving around the sky. They naturally thought that the Earth would be the center of the universe and that all of those bodies revolved around the Earth. This culminated with the work of the phylosopher Claudius Ptolemaeus about geocentrism.

Altough we now know that the ptolomaic geocentric model is fundamentally wrong, it could be used to compute the position of the planets, the Sun, the Moon and the celestial sphere of stars, with a somewhat acceptable precision at the time. It accounted to include the observation of planets velocity variations, retrograde motions and also for coupling Mercury and Venus to the Sun, so they would never go very far from it. Further, based on the velocity of the motion of those bodies in the sky, then the universe should be something like:

Earth at the center.
Moon orbiting the Earth.
Mercury orbiting the Earth farther than the Moon.
Venus orbiting the Earth farther than Mercury.
Sun orbiting the Earth farther than Venus.
Mars orbiting the Earth farther than Sun.
Jupiter orbiting the Earth farther than Mars.
Saturn orbiting the Earth farther than Jupiter.
The celestial sphere of stars rotating around the Earth, being the outermost sphere.

In fact, the ptolomaic model is a very complicated model, way more complicated than the copernic, keplerian and newtonian models. Particularly, this could be compared to softwares that are based on severely flawed concepts but that are still working due to a lot of complex, tangled and unexplainable hacks and kludges that are there just for the sake of making the thing work.

4. The discovery of the Americas

Marco Polo, in the last years of the 1200's, was the first European to travel to China and back and leave a detailed chronicle of his experience. So, he could bring a lot of knowledge about what existed in the central Asia, the East of Asia, the Indies, China, Mongolia and even Japan to the Europeans. Before Marco Polo, very few was known to the Europeans about what existed there. This greatly inspired European cartographers, philosophers, politicians and navigators in the years to come.

Portugal and Spain fough a centuries-long war against the invading Moors on the Iberian Peninsula. The Moors were finally expelled in 1492. The two states, were looking for something profitable after so many years of war. Since Portugal ended its part of the war first, it had a head start and went to explore the seas first. Both Portugal and Spain were trying to find a navigation route to reach the Indias and the China in order to trade highly profitable spices and silk. Those couldn't be traded by land efficiently anymore due to the fact that the lands on West Asia and North Africa were dominated by Muslim cultures unfriendly to Christian Europeans, a situation that were just made worse after the fall of Constantinople in 1453.

Portugal, were colonizing the Atlantic borders of Africa and eventually they managed to reach the Cape of Good Hope in 1488 (with Bartolomeu Dias).

A Genovese navigator called Cristoforo Colombo believed that if he sailed west from the Europe, he could eventually reach the Indies from the east side. Inspired by Marco Polo and subestimating the size of Earth, he estimated that the distance between the Canary Islands and the Japan to be 3700 km (in fact it is 12500 km). Most navigators would not venture in such voyage because they (rightly) tought that Earth was larger than that.

Colombo tried to convice the king of Portugal to finance his journey in 1485, but after submitting the proposal to experts, the king rejected it because the estimated journey distance was too low. Spain, however, after finally expelling the Moors in 1492, were convinced by him. Colombo's idea was far-fetched, but, after centuries of wars with the Muslims, if that worked, then Spain could profit quickly. So, the Spanish king approved the idea. And just a few months after expelling the Moors, Spain sent Colombo to sail west towards the Atlantic and then he reach the Hispaniola island in Central America. After coming back, the news about the discovery of lands in the other side of the Atlantic spread quickly.

Portugal and Spain then divided the world by the Treaty of Tordesillas in 1494. In 1497, Amerigo Vespucci reached the mainland America.

Portugal would not be left behind, they managed to navigate around Africa to reach the Indies in 1498 (with Vasco da Gama). And they sent Pedro Álvares Cabral, who reached the Brazil in 1500 before crossing the Atlantic back in order to go for the Indies.

After that, Portugal and Spain quickly started to explore the Americas and eventually colonize them. France, England and Netherlands also came to the Americas some time later.

5. The Earth IS round

After, the Spanish discovered and settled into the Americas (and Colombo's plan in fact didn't worked). The question that if it was possible to sail around the globe to reach the Indies from the east side remained open and the Spanish were still interested on it. They eventually discovered the Pacific Ocean after crossing the Panama Ishtums by land in 1513.

Eager to find a maritime route around the globe, the Spanish crown funded an expedition leadered by the Portuguese Fernão de Magalhães (or Magellan as his name was translated to English) to try to circle the globe. Magellan was an experienced navigator, and had reached what is present day Malaysia traveling through the Indian Ocean before. They departed from Spain in September 20th, 1519. It was a long and though journey that costed the lives of most of the crew. Magellan himself did not survived, having died in a battle in the Phillipines on 1521. At least, he lived enough to be aware that they in fact reached East Asia by traveling around the globe to the west, which also proves that the Earth is round.

The journey was eventually completed by the leadership of Juan Sebatián Elcano, one of the crewmen of Magellan. They reached Spain back through the Indian and Atlantic Oceans on September 6th, 1522 after traveling for almost three years a distance of 81449 km.

6. Heliocentrism

There were some heliocentric or hybrid geo-heliocentric theories in ancient times. Notably by the Greek philosopher Philolaus in the 5th century BC. By Martianus Capella around the years 410 to 420. And by Aristarchus of Samos around 370 BC. Those models tried to explain the motion of the stars as rotation of the Earth and the position of the planets, specially Mercury and Venus as translation around the Sun. However those early models were too imprecise and flawed to work appropriately, and the ptolomaic model still was the model with the better prediction of the positions of the heavenly bodies.

The idea that the Earth rotates was much less revolutionary than heliocentrism, but was already more-or-less accepted with reluctancy in the middle ages. This happens because if the stars rotated around Earth, they would need do so at an astonishing velocity, dragging the Sun, the Moon and the planets with it, so it would be easier if Earth itself rotated. People were uncomfortable with this idea, but they still accepted it, and this became easier to be accepted after the Earth sphericity was an established concept.

In the first years of the 1500's, while the Portuguese and Spanish were sailing around the globe, a polish and very skilled matemathical and astronomer called Nikolaus Kopernikus took some years thinking about the mechanics of the heavenly bodies. After some years making calculations and observations, he created a model of circular orbits of the planets around the Sun and perceived that his model were much more simpler than the ptolomaic geocentric model and was at least as precise. His model also features a rotating Earth and fixed stars. Further, his model implied that the Sun was much larger than the Earth, something that was already strongly suspected at the time due to calculations and measurements and also implied that Jupiter and Saturn were several times larger than Earth, so Earth would definitively be a planet just like the other five then-known planets were. This could be seen as the borning of the model today know as Solar System.

Fearing persecution and harsh criticism, he avoided to publish many of his works, sending manuscripts to only his closest acquaintances, however his works eventually leaked out and he was convinced to allow its full publication anyway. Legend says that he was presented to his finally fully published work in the very day that he died in 1543, so he could die in peace.

There was a heated debate between supporters and oppositors of Copernic's heliocentric theory in the middle 1500's. One argument for the opposition was that star parallaxes could not be observed, which implied that either the heliocentric model was wrong or that the stars were very very far and many of them would be even larger than the Sun, which seemed to be a crazy idea at the time.

Tycho Brache, which did not accepted heliocentrism, in the latest years of the 1500's tried to save geocentrism with an hybrid geo-heliocentric model that featured the five heavenly planets orbiting the Sun while the Sun and the Moon orbited Earth. However, he also published a theory which better predicted the position of the Moon. Also, by this time, the observation of some supernovas showed that the celestial sphere of the stars were not exactly immutable.

In 1600, the astronomer William Gilbert provided strong argument for the rotation of Earth, by studing magnets and compasses, he could demonstrate that the Earth was magnetic, which could be explained by the presence of enourmous quantities of iron in its core.

7. With telescopes

All of what I wrote above happened without telescopes, only by using naked eye observations and measurements around the globe. Now, add even some small telescopes and things change quickly.

The earliest telescopes were invented in 1608. In 1609, the astronomer Galieu Galilei heard about that, and constructed his own telescope. In January of 1610, Galieu Galilei, using a small telescope, observed four small bodies orbiting Jupiter at different distances, figuring out that they were Jupiter's "moons", he also could predict and calculate its positions along their orbits. Some months later, he also observed that Venus had phases as seen from the Earth. He also observed Saturn's rings, but his telescope was not powerful enough to resolve them as rings, and he tought that they were two moons. These observations were incompatible with the geocentric model.

A contemporary of Galilei, Johannes Kepler, working on Copernicus' heliocentric model and making a lot of calculations, in order to explain the differing orbital velocities, created an heliocentric model where the planets orbits the Sun in elliptic orbits with one of the ellipse's focus in the Sun. His works were published in 1609 and 1619. He also suggested that tides were caused by the motion of the Moon, though Galilei was skeptical to that. His laws predicted a transit of Mercury in 1631 and of Venus in 1639, and such transit were in fact observed. However, a predicted transit of Venus in 1631 could not be seen due to imprecision in calculations and the fact that it was not visible in much of the Europe.

In 1650 the first double star were observed. Further in the 1600's, the Saturn rings were resolved by the use of better telescopes by Robert Hooke, who also observed a double star in 1664 and developed microscopes to observe cellular structures. From them on, many stars were discovered to be double. In 1655, Titan were discovered orbiting Saturn, putting more confidence on the heliocentric model. More four Saturnian moons were discovered between 1671 and 1684.

8. Gravitation

Heliocentrism was reasonably well-accepted in the middle 1600's, but people was not confortable with it. Why the planets orbits the Sun afterall? Why the Moon orbits Earth? Why Jupiter and Saturn had moons? Although Keplerian mechanics could predict their moviment, it was still unclear what was the reason that makes them move that way.

In 1687, Isaac Newton who was one of the most brilliant physic and mathematic that ever lived (although he was also an implacable persecutor of his opponents), provided the gravitational theory (based on prior work by Robert Hooke). Ideas for the gravitation theory and the inverse square law already were developed in the 1670's, but he could publish a very simple and clear theory for gravitation, very well-fundamented in physics and mathematics and it explained the motions of the celestial bodies with a great precision, including comets. It also explained why the planets, the Moon and the Sun are spherical, explained tides and it also served to explain why things falls to the ground. This made heliocentrism to be definitely widely accepted.

Also, Newton gravitational law predicted that Earth rotation would make it not exactly spherical, but a bit ellipsoidal by a factor of 1:230. Something that agreed with measures done using pendulums in 1673.

9. What are the stars and the Solar System afterall?

In the early 1700's, Edmund Halley, already knowing about newtonian laws (he was a contemporary of Newton) perceived that comets who passed near Earth would eventually return, and he found that there was a particular instance of sightings every 76 years, so he could note that those comets in reality were all the same comet, which is called after him.

The only remaining problem with the heliocentric model was the lack of observation of parallax to the stars. And nobody knew for sure what the stars were. However, if they in fact are very distant bodies, most of them would be much larger than the Sun. At the first half of the 1700's, trying to observe parallax, James Bradley perceived phenomena like the aberration of light and the Earth's nutation, and those phenomena also provides a way to calculate the speed of light. But the observation of parallax remained a challenge during the 1700's.

In 1781, Uranus were discovered orbiting the Sun beyond Saturn. Although barely visible to the naked eye in the darkest skies, it was so dim that it escaped observation from astronomers until then, and so were discovered with a telescope. The first asteroids were also discovered in the early 1800's. Investigation on pertubations on Uranus' orbit due to the predicted newtonian and keplerian movement eventually leaded to the discovery of Neptune in 1846.

In 1838, the astronomer Friedrich Wilhelm Bessel who measured the position of more than 50000 stars with the greatest precision as possible, could finally measure the parallax of the star 61 Cygni successfully, which proved that stars were in fact very distant bodies and that many of them were in fact larger than the Sun. This also demonstrates that the Sun is a star. Vega and Alpha Centauri also had their parallaxes measured successfully in 1838. Further, those measurements permitted to estimate the distance between those stars and the Solar System to be on the order of many trillions of kilometers, or several light-years.


Moving from mathematics to astronomy/astrophysics

There are fewer placements than qualified applicants so competition is strong.  Moreover, suppose this was the other way round:  A person doing a PhD in (say) planetary nebulae comes to the maths department saying that they are really interested in maths, they're good at equations and calculus and really like all that topology stuff (you know, rubber sheets and so on) so can they start doing cohomology of differential graded categories?
It's not impossible, and a strong foundation in maths is a prerequisite for progress in Astronomy but the question you should ask is why should a university hire you, when there are 5 others with PhDs in astrophysics?
Talk to your supervisor and careers advisor before burning any bridges.


How do you determine which star, between two, is more luminous?

One way to determine the luminosity of a star is to first determine its distance, d.  Then one can use the formula, Luminosity = Flux * 4 * Pi * d$^2$.  Flux is related to the apparent brightness that is measured with a telescope (and CCD or photometer).  mag = -2.5 * Log(Flux) + c.

If the star is relatively nearby, one can use parallax to get the distance.  If the star is in a binary system, one can solve for the distance from the period, orbital velocities and apparent size of the orbit.  If you know the velocities throughout one orbit, then one knows how far it has physically traveled.  Comparing this with the apparent size gives an approximate distance.  However, since one usually does not know the inclination, this causes uncertainty.  If it is an eclipsing binary, then you know that it is not very inclined and this works well.  On the other hand, if it is eclipsing, one can usually determine the radii of the binary pair and then get distances from Stefan Boltzmann Law.

Another method for estimating luminosity is to locate the star in the HR Diagram (Luminosity vs Color).   The color (difference between the magnitudes in two different filters) locates it horizontally, and stellar class locates it vertically in the diagram.  One needs a spectrum of the star to determine the class.

And we have one more trick up our sleeves.  If the star is a periodic variable such as a Cepheid or an RR Lyra, then there are relations between period of variability and luminosity.


Orbital period for a 1 light year object?

To answer this, let's visit Kepler's third law which tells you the orbital period of a body, if you know how far away it is.

$$\frac{P^2}{a^3} = \mathrm{constant}$$

As it turns out, if you use the correct units, for the orbital period, $P$, and the semi-major axis, $a$, namely years and astronomical units (AU) respectively, then that constant is equal to 1! You have our moon orbiting the sun at one light year, so this needs to be converted to AU. Google tells me that $1\:\mathrm{light\:year}$ is $63241.1\:\mathrm{AU}$. For this specific case, you get that

$$P = a^{3/2} = 63241.1^{3/2} = 15,903,734\:\mathrm{years}$$

Notice that this formula didn't depend on the mass of the moon! (Technically the mass of the moon is in there, but it's pretty negligible and so doesn't matter much.)

You may be scratching your head over the units $-$ how can a timescale squared equal a distance cubed?. Technically there's some trickery going on because that constant isn't unitless and I've implicitly assumed those units are hanging around, without explicitly saying so. Rest assured, the units do make sense for this equation.


If a person were to die on the Moon or Mars, would the body decompose?

Space, as Randall notes, is really dry. Mars, (recent discoveries notwithstanding) is not much moister.

In these conditions, bodies mummify.

The microbes that live in you, wouldn't survive the freezing, dessication and radiation. There is no real upper limit on how long a mummified body could exist in space. There would be nothing to cause the body to change, and so it would remain. There would be a slow breaking down of surface proteins, due to UV light, and eventually micrometeorites would erode the body, but these processes would take many millions of years.


Would space-based telescopes need to remain in the shadow of Earth to avoid light interference from the sun?

Its a nice idea, but it wouldn't work.

To stay in the Earth's shadow a telescope would need to orbit the Earth in the same time the earth orbited the Sun so that it always stayed on the opposite side. For its orbit to be this long it would need to be a long way out, remember geosynchronous orbit (one orbit in 23h 56min 4sec) is at a distance of 42,164 km. For a 1 year orbit the telescope would be so far away that its would not be in orbit around the Earth but the Sun, as it would have a much larger gravitational influence. It would also be too far away to be in the Earth's shadow cone.


How did Batygin estimate when Planet Nine was ejected?

Page 12 of Batygin & Brown (2016) says that a speculative formation scenario can be drawn from recent solar system formation simulations by Bromley & Kenyon and by Izidoro et al. These suggest that the core of a nascent ice giant may have been ejected very early in the solar system's history in order to explain the properties of the observed planets; the formation of Uranus and Neptune was probably accompanied by at least one other ice giant.

The source of the claim you mention is actually an article by Eric Hand in Science Magazine.  He points out that possibly, to explain why this planet was ejected yet still remains part of the solar system in a much wider orbit, then you need it to have been slowed down by residual gas in the protosolar disk. So I assume the 10 million year is an (uncertain) upper limit on the dispersal of the disk and I would guess the 3 million year lower limit is just how long it takes to form a 10 Earth mass ice giant core at the distance of Neptune.


Will Gaia detect inactive neutron stars?

I will throw out a suggestion that could hardly be considered an "answer", but it would be interesting if someone could put flesh on the bones of it.

When I first read your question I thought you were asking whether gravitational perturbations from an unseen neutron star would cause a measurable effect on the positions of other stars. I think the answer to this is certainly no unless the neutron star was so close to another star that they were (or could be practically considered as) a binary system.

Binary systems featuring neutron stars could definitely exist (they won't be common though) and Gaia should see them if they are close enough and the companion star is bright enough. A variant of this would be a neutron star that just happens to drift (probably at high speed) close to another star, producing a notable change in the photocentre of the visible star over the 5-year Gaia mission. This is unlikely because close encounters between unrelated stars are very unlikely.

But then I started thinking about lensing. Even the presence of planets in our solar system causes a gravitational lensing signature - the positions of stars are distorted by $\sim 1-10$ microarcseconds close to the positions of planets and this is detectable by Gaia. The distortion can be spotted because of course the planets move with respect to the background stars.

In GR, the lensing angular deflection is
$$\alpha = \frac{4GM}{c^2r},$$
where $r$ is the closest approach of the light to the lensing object of mass $M$. For a lensing object at distance $d$ and a background star at an angular separation of $\theta$ from it, then
$$\alpha = 0.04 \left(\frac{M}{M_{\odot}}\right)\left(\frac{d}{\rm 1\ pc}\right)^{-1} \left(\frac{\theta}{\rm 1\ arcsec}\right)^{-1}\ \ \mu{\rm \ arcsec}$$

Neutron stars are of order 1-2$M_{\odot}$; the closest they might come is a few pc. The astrometric precision of Gaia will be about 20 microarcsec for a star with $V =15$, falling to about a milli-arcsec at $V\sim 20$.
The signature of a neutron star moving with high proper motion might be that it passes very close in the foreground to a bright star, causing its position to wobble backwards and then forwards by an amplitude given by the above formulae.

To get a 20 microarcsecond signature that might be detectable, a neutron star at one pc would have to pass within 0.003 arcseconds of a $V=15$ star.  The surface density of such stars varies according to Galactic latitude but is around 1000 per square degree for stars of $V\leq 15$ (Bahcall & Soneira 1980) at intermediate Galactic latitudes. If the neutron star has a proper motion like Barnard's star of around 10 arcseconds/year, then it traverses 50 arcseconds during the Gaia mission. If background stars are evenly spaced, the chance of coming within 0.003 arcseconds of one is about $10^{-5}$ (if I've done my sums right).

So, almost no chance of seeing this.


If $r_p$ is the perihelion distance (the closest distance of the planet in its elliptical orbit from the Sun), and if $r_a$ is the aphelion distance (the farthest distance from the Sun), the major axis is
$$
2a = r_a+r_p
$$
where $a$ is the semi-major axis (half of the major axis).
The eccentricity of the orbit can be calculated from the perihelion and aphelion distances using
$$
e = \frac{r_a-r_p}{r_a+r_p}
$$
Once the eccentricity is known, the semi-minor axis can be calculated from
$$
b = a\sqrt{1-e^2}
$$
The minor axis, of course, is twice the semi-minor axis, or $2b$.


What is the solar system orientation to its direction of travel?

This is Venus, high in the morning sky, amid the faint pillar of light called the Zodiacal Light. The glow is sunlight reflected off cometary dust in the inner solar system.
  Above is the centre of the Galaxy area of Sagittarius.  Alan Dyer

Since the  Zodiacal Light lies in the ecliptic plane, the orientation of the Solar system in the Galaxy is something like this:

The Sun moves through the local interstellar gas cloud approximately into the direction of Scorpio with a speed of about 25 km/sec.

Since Scorpio is fairly close to the galactic center (Sagittarius), the Solar System is moving approximately edge-on through the gas cloud as in the NASA diagram.

(Image: NASA, ESA, Z. Levay (STScI) and A. Fujii)


This phenomenon is the equation of time, the annually varying difference between apparent solar time (what a sundial indicates) and mean solar time (related to civil time).
It has two components:

The eccentricity of the Earth's orbit around the Sun causes a cycle with 7.5-minute amplitude and 12-month period, with zeroes at perihelion (Earth moving fastest and closest to the Sun) and aphelion (Earth moving slowest and farthest from the Sun).
The oblique angle between the Earth's equator (rotation plane) and ecliptic (orbital plane) causes a cycle with 10-minute amplitude and 6-month period, with zeroes at equinoxes (Sun crossing Earth's equator) and solstices (Sun farthest north or south of the equator). This answer explains how that works.

These combine so that apparent solar time is 17 minutes ahead of mean solar time in early November and 14 minutes behind in mid February.
Between those times of year, the apparent Sun falls several seconds behind the mean Sun each day, affecting both sunset and sunrise times as you observed.

For further explanation, see the US Naval Observatory or Wikipedia.


How Would a Neutron Star Actually Appear?

Your question is too general, you need to get to specific examples.

First, very few neutron stars are pulsars. Pulsars are either a brief phase during a pulsar's spin-down at the start of a neutron star's life, or they are the product of the spin-up of a neutron star in a binary system. Most neutron stars fall in neither of these categories.

A standard neutron star will look like any other star at a similar temperature. Most of them will be very hot indeed - 100,000 K or more, though the cooling histories of neutron stars are still uncertain and depend on some exotic physics. Such an object is "white hot" - it emits black body radiation at all frequencies visible to the eye (as well as lots more at UV wavelengths).

How close would you have to get for it's apparent luminosity/magnitude to match the Sun? Well that depends on the size and temperature of the neutron star. Most are thought to have a diameter of 20 km. The way you would do the calculation is equate the blackbody radiative flux per unit area at a given distance to the solar radiation constant of about 1300 W per square metre. However, there are two wrinkles for a neutron star: First, the radiation is gravitationally redshifted, so the temperature we measure is lower than the temperature at the surface. Second, General Relativity tells us that we can see more than just a hemisphere of the neutron star - i.e. we can see around the back - and this increases the flux we observe. These are roughly factor of two effects, so just to get an order of magnitude estimate, ignore GR and assume a 10 km radius NS with $T=10^{5}$ K.

Using Stefan's law for a blackbody, then at a distance $d$, we have that
$$\frac{4 \pi r^2}{4\pi d^2} \sigma T^4 = 1300\ W\ m^{-2},$$
where $\sigma$ is the Stefan-Boltzmann constant.

For $r=10$ km, then $d=7 \times 10^{8}$ m, which is coincidentally about a solar radius. Of course this distance depends on the square of the temperature, so a younger NS with $T=10^6$ K, then $d \sim 1$ au.

These are the distances where the total flux at all wavelengths would be similar to that from the Sun. To do the calculation just for the visible range we need to account for the bolometric correction, which converts a visual magnitude to a bolometric magnitude. The bolometric correction for the Sun is $\sim 0$, whereas the bolometric correction for a very hot star could be -5 mag. This means that only 1% as much flux from the hot neutron star emerges in the visible band compared with sunlight. This means that the distances calculated above, if we require the visual brightness of the neutron star be similar to the Sun, must be reduced by a factor of 10.

To turn to pulsars. Note that the pulsed radiation does have an optical component and pulsed optical radiation has been seen from a number of pulsars. Optical synchrotron emission would just appear to be a periodic, intense brightening of the pulsar, as the beam sweeps across the line of sight. If you were not in the line of sight, then you would not see the pulsed optical emission. If you could observe the beam passing through nebulosity or some other medium around the pulsar then yes there may well be some effects you could see in terms of ionisation or scattered light coming from along the beam path.

Lastly, the gravitational lensing effect. Yes, this should be strong close to a neutron star. The deflection angle (in radians) is given by
$$ \alpha  = \frac{4GM}{c^2 b},$$
where $b$ is how close the light passes to the neutron star and $M$ is the neutron star mass. Expressing $b$ in terms of the 10km radius of the neutron star:
$$ \alpha \simeq 0.83 \left(\frac{M}{1.4M_{\odot}}\right) \left(\frac{b}{10 km}\right)^{-1},$$
where strictly speaking this formula is only valid for $\alpha \ll 1$.

So consider a planet directly behind the neutron star at a distance of 1 au. The light from this would only need to be bent through an angle of $\sim 2 \times 10\ km/1\ au \sim 10^{-7}$ radians in order to be seen from a planet diametrically opposite at a distance of 1 au. So this is easily possible. However, the image would likely be highly distorted, especially if the neutron star was spinning. It would not look dissimilar to this simulated black hole image, but with a bright neutron star in the middle rather than a black disc.


Which side of earth will face toward sun when earth and sun will be get closer and be bonded by Gravitational Force?

We all know that one day the distance between the Earth and the sun will be zero.

This may be true, as the sun will get much bigger and engulf the Earth.  But this is irrelevant to the rest of the question, which seems to be about tidal locking.

  They will be bonded with gravity.

The Earth is already gravitationally bound to the sun. But it isn't tidally locked. The Earth+moon has too much angular momentum to ever become tidally locked to the sun. It will be engulfed by the sun before it becomes tidally locked.

  which side of the earth will face toward the sun

As noted above, the Earth will continue to rotate and will not become tidally locked.

Even if it would become tidally locked it would be impossible to calculate. It is like putting a battery into a clock and asking exactly what time the battery will run out.  You can't measure the present nor predict the future with enough accuracy to forecast such things

In other words. This won't happen, and if it does we can't calculate it.  It's impossible.


The GREAT receiver attached to SOFIA to detect far-infrared radiation incorporates elements reminiscent of both radio and optical detectors, but since it works by mixing a signal from a tunable oscillator with the oscillating field from the incoming radiation rather than by the photoelectric effect as in most optical detectors, you might consider it more akin to a radio telescope receiver.


Are hot stars like O-type stars entirely composed of helium?

The lines that appear in a stars spectrum mainly reflects its temperature not its composition, see here

O-type stars start out with the same sort of composition as other stars, that is they are mainly H and He (approximately 75% and 25% by mass) with traces heavier elements.


$$
    K_\text{min}=2\pi\sqrt{a^2+3(ma^2)^{2/3}}
\tag{1}
$$
where the quantities $m$ and $a$, both of which have units of length, are defined by
$$
    m=\frac{GM}{c^2}
\hskip2cm
    a=\frac{J}{Mc}.
\tag{2}
$$
The maximum spin is a=m (an extremal Kerr black hole), the borderline case between a black hole and a naked singularity.
More details from Chiral Anomaly here:
https://physics.stackexchange.com/a/469282/


Will James Webb see Population III stars?

According to Rydberg et al. (2013) "Detection of isolated Population III stars with the James Webb Space Telescope", the answer for non-lensed population III stars appears to be no. For lensed stars, the detection would be difficult but the expected number of detections is very small:

  To detect even one 60 M⊙ Population III star when pointing JWST through the galaxy cluster MACS J0717.5+3745, the lensing cluster with the largest Einstein radius detected so far, the cosmic star formation rate of Population III stars would need to be approximately an order of magnitude higher than predicted by the most optimistic current models.

Prospects may be somewhat better for detecting population III galaxies. These would likely be small (dwarf) galaxies in low-density regions, as larger galaxies likely became rapidly polluted by supernova ejecta. From Zackrisson et al. (2012) "Detecting gravitationally lensed Population III galaxies with the Hubble Space Telescope and the James Webb Space Telescope":

  We find that the ongoing HST survey Cluster Lensing And Supernova survey with Hubble (CLASH), targeting a total of 25 galaxy clusters including J0717, potentially could detect a small number of pop III galaxies if ∼1 per cent of the baryons in these systems have been converted into pop III stars. Using JWST exposures of J0717, this limit can be pushed to ∼0.1 per cent of the baryons. Ultradeep JWST observations of unlensed fields are predicted to do somewhat worse, but will be able to probe pop III galaxies with luminosities intermediate between those detectable in HST/CLASH and in JWST observations of J0717.

(The designation J0717 is an abbreviated form of MACS J0717.5+3745 referred to in the Rydberg et al. paper.)


How is the cost of JWST distributed on different parts of development and operations?

So most of the costs should end up as salaries to engineers.

Even in a project that is solely people-based, most of the costs do not end up as salaries to engineers. Salary is typically a bit less than a half of the cost. Those engineers get benefits. Health care in the US isn't cheap. They get holidays, vacation, and sick leave. They receive matching funds on their 401K contributions. In the US, employers and employees both contribute to the cesspool lockbox sometimes known as "Social Security". The employer share is part of the cost of having an employee (and companies charge for it). The employer also pays for unemployment insurance and other items on behalf of their employees. Those engineers sit at desks in an office or a cubicle, and have a nice computer on which they do their work. They use the employer's servers, printers, copy machines, conference rooms, break rooms, and coffee machines. All of that costs money. These overhead costs are the first of several indirect costs.

The accountants who write those engineers their paychecks aren't charging to any project. They're still paid by the company, as are the executives of the company. The executives also typically don't charge their time to a project, and those executives can be paid quite handsomely. This leads to another category that adds to quantity that companies charge the government, "general and accounting". Finally, for-profit companies need to keep their shareholders / stakeholders happy. They need to earn a profit. This leads to a third category that companies charge the government, "profit".

When you add up all of those indirect costs, they come close to and oftentimes exceed direct costs (i.e., chargeable salary). Note that this is for a project where the only costs are the people working on it.

Building a spacecraft notches things up a lot. Cleanrooms are expensive. Control rooms are expensive. The flight computers are expensive. The bolts and fasteners are expensive. The custom, one-off sensors are expensive. The spacecraft's thermal control system is expensive. The spacecraft's structure is expensive. Launch is expensive. Every piece of hardware, every fabrication facility is expensive.

Finally, cost overruns, requirements changes, and delays can represent a significant portion of the cost of a poorly managed project. JWST has experienced numerous cost overruns, requirements changes, and delays. Big contractors have learned that the only way to win big government contracts is to underbid. They would lose their shirts if the government kept to that initial contract. But it never does. The government instead changes the requirements, changes the payments, and changes the schedule. Those changes are beyond expensive.