Why is it strange that outer stars in a galaxy are travelling with the same speed as inner while the total mass is also increasing?

This is accounted for in real models of the rotation curves of galaxies. The rotation curve of a galaxy is more complicated than the motion of planets in the solar system, for the reasons that you describe.

A trivial model would consider the galaxy as spherically symmetric and would use the shell theorem to estimate the centripetal acceleration of a star at radius $r$. Thus
$$ m r \omega^2 = G\frac{m M(r)}{r^2},$$
where $m$ is the mass of the star, $\omega$ is the angular velocity and $M(r)$ would be the mass of all gravitating matter at radii $<r$. Note that this only applies to a spherically symmetric distribution of mass.

To make further progress demands that you know the density distribution of the gravitating matter. Let's just assume that density $\rho$ is constant for the moment. Then
$$ r^3 \omega^2 = G \int \rho\ 4\pi r^2\ dr = \frac{4\pi}{3} G \rho r^3$$
$$ \omega = \sqrt{\frac{4\pi G \rho}{3}}$$

Thus the angular velocity would be constant with radius and the rotation speed, $v = \omega r$ would increase with radius. This I think is the situation that your question supposes and so yes, if there was a constant (or perhaps slowly declining) density of material in the Galaxy then this would produce a rotation curve that increased (or was flat).

The trouble is that the density of material in our Galaxy inferred from the matter that we can see is not constant with radius. It declines rapidly and exponentially, such that there is very little visible matter beyond a radius of about 15 kpc. If we take a situation where we go to radii beyond the gravitating matter, then a similar treatment to the above suggests that
 $$ m r \omega^{2} = G\frac{mM}{r^2},$$
where $M$ is now the total mass of the (visible) Galaxy. In this case
$$ \omega = \sqrt{\frac{GM}{r^{3}}}$$
and the rotation velocity $v = \omega r$ should decline as $1/\sqrt{r}$ (like it does in the solar system).

It is the fact that the rotation velocity of stars and gas at large radii ($>15$ kpc) continues to be flat or even increase that leads to the conclusion that the mass that we can see is not all that there is. i.e That we need "dark matter" to explain the high rotation speeds at radii where there is very little visible matter.

Realistic models of the galaxy do not make the assumption that the visible matter is spherically symmetric (it isn't). But the conclusions I qualitatively set out above hold in the same way.


Solar Telescope Projector Screen? [& intro telescope for solar eclipse]

For your purposes, I'd really recommend just purchasing the EclipSmart. The telescope itself is of reasonable quality. If you have a camera tripod, bring it with you to view the eclipse$-$the stock tripod is so flimsy as to be useless. If you dont have one, buy one. Additionally, you should invest in different eyepieces. The telescope comes with a 20-mm EP, which will provide relatively low magnification. I'd recommend getting a 6-mm EP, and possible an intermediate one as well.

Clear skies!

P.S. I would recommend against getting the XX14i as your first telescope. I think instead you should get a smaller EQ, in the 4-6-in bracket. Having to manually find your target will teach you much about the skies, and having to do so with a smaller aperture will humble you. I worked a 4" Alt-Az for a decade as a kid before moving to a 4.5" EQ for a year, and then finally an XT12i as a highschool sophomore. Having to find my targets in my relatively bright skies made the sights I see through my 12" that much more incredible. I think allowing yourself the same experience will be a decision you won't regret. Don't fall for aperture fever!


Lack of objects between heliopause and Oort cloud?

There may be Sednoids there.

Sednoids are a hypothetical class of "inner Oort Cloud objects" named after their prototype, Sedna. Sedna's aphelion is ~936 AU, bringing it close to the inner boundary of the Oort Cloud. Sednoids may have aphelions ranging from about 100 AU to 1,000 AU.

The problem is, only two Sednoids have beet detected to date, 90377 Sedna and 2012 VP113. Brown et al. 2004) suggested that ~500 may be detectable; surveys simply haven't tracked objects in that area.

Why are Sendnoids where they are? Three ideas have but put forth:

A planet at ~70 AU scattered these objects into elliptical orbits.
A close pass by a nearby star.
Interactions with other stars in the Sun's original cluster.

These objects would fill in the space between the Kuiper Belt/scattered disc and the Oort Cloud itself.


What is main radiative mechanism in Gunn-Peterson effect?

The Gunn-Peterson trough is a feature in the spectrum of a (background) source near the Lyman $\alpha$ line, caused by a high density of intervening, neutral hydrogen (HI).

The background source may be a quasar, but could also be e.g. a gamma-ray burst or just a regular galaxy with a strong Ly$\alpha$ line.

Absorption vs. scattering

The cross section of hydrogen for Ly$\alpha$ photons is quite large, so the mean free path of the photons is very small. Because Ly$\alpha$ excites hydrogen to the first excited level, the electron can only fall back to the ground state again, emitted another Ly$\alpha$ photon, in another direction. Hence, hydrogen doens't absorb Ly$\alpha$ photons, but rather scatters them.

Inside the (host) galaxy, the HI density is so large that Ly$\alpha$ is all the time scattered both into and out of the LOS. However, when the photons escape the galaxy and enters the circum- and intergalactic medium (IGM), the density decreases. Along the LOS to the galaxy, a photon hitting a hydrogen atom is "lost", and no new photons are added to the observed flux, so effectively, it becomes an absorption process rather than a scattering process.

The figure below (from Laursen 2010) illustrates this phenomenon: the three photons are initially emitted in random directions, but end up traveling toward the observer on the right. On their way through the IGM, two of them are scattered out of the line of sight (LOS), and the probability of a photon from another galaxy being scattered into the LOS is negligible.

Reionization

In the early days of the (post-recombination) Universe, hydrogen was mainly neutral, but as star formation emerged, these stars started to reionize the IGM. This process started already at $z\simeq15$–$20$, when the Universe was ~200 million years old , but until $z\simeq6$, when the Universe was ~a billion years old, there was still so much HI left that Ly$\alpha$ photons scatter all the time in the IGM.

Blue wavelengths are eventually redshifted to Lyman $\alpha$

Now when light leaves a galaxy, it gets redshifted due to the expansion of the Universe. That means that after a while, light blueward of the Ly$\alpha$ line is redshifted to become Ly$\alpha$, and may hence scatter. As the spectrum travels through the IGM, it redshifts continuously, so that every wavelength of the "original" Ly$\alpha$ line eventually becomes Ly$\alpha$.

If the HI density in the IGM is high enough, that means that everything blueward of the Ly$\alpha$ line is wiped out. This is the Gunn-Peterson trough. If the density is low, but with some neutral clouds dispersed, then an absorption line occurs exactly at the wavelength that, at the location of the cloud, happens to have been redshifted to Ly$\alpha$. When there are many such clouds around, you'll gets lots of absorption lines, forming the so-called Lyman $\alpha$ forest. The fewer the amount of such clouds, the more sparse the forest will be.

The Gunn-Peterson trough doesn't have to start at Ly$\alpha$, but since this is the strongest transition, it is most distinct here. But you will often see a corresponding absorption at Ly$\beta$ (and even Ly$\delta$ and Ly$\gamma$, as well as the stronger metal lines like MgII and CIV). This is visible in the figure below (from Becker et al. 2015)

You see that immediately blueward of Ly$\alpha$, everything is wiped. As you go along the spectrum toward the blue, that part corresponds to later and later epochs, because is takes longer time to redshift that part into Ly$\alpha$. Hence, eventually the Universe becomes less and less neutral, and the absorption is no longer complete.


Dark Flow: statistical limits on existence

The dipole in the microwave background indicates motion of the Milky Way and thus of the whole Local Group, at least, at about 600 km/s in a certain direction.  The straightforward explanation is that the density irregularities nearby from superclusters and voids result in a net gravitational acceleration that, over the age of the universe, resulted in this velocity.  A number of superclusters (Hydra-Centaurus, Shapley, and Norma) happen to line up along this direction and reasonably fit the picture.  Also there are some big voids in the opposite direction.
There are two basic ways to quantitatively understand what is the cause.  One can make measurement of local flows around these superclusters and voids to determine their masses and thus their net effect on the MW velocity.  Or one can try to measure the "dipole motion" on larger scales, that is at what size scale does this motion drop to 300 km/s and what scale does it drop to near 0?  A number of groups have tried to do this and the results are in disagreement.  In particular, one group found no drop off in motion out to fairly large distances.  Therefore, an alternate explanation was proposed that the dipole motion is primordial and due to objects beyond the horizon that interacted with us before inflation.

Will we ever settle this?  I am sure that we will eventually.  There is much work to do measuring the distances of galaxies out to 100 Mpc or so with greater and greater accuracy using new methods (SN Type 1a, surface brightness fluctuations, Tip of the Red Giant Branch) and larger telescopes with better instrumentation and then modeling the peculiar flow on all scales.  If the local distribution of matter cannot explain fully the dipole moment in the MWB, then perhaps Dark Flow from before inflation will be necessary to explain it.


Who is Seyfert, and what is "the so-called Seyfert flare"?

As has now been corrected, BBC has misunderstood term "Seyfert flare", instead calling it a "Sifter flare".

A "Seyfert flare", is not really a common term, but the authors refer to an energetic outburst from the type of active galaxies called Seyfert galaxies (after Carl Seyfert).

Like a quasar, a Seyfert galaxy is powered by gas accretion onto a central, supermassive black hole, although they're less luminous by roughly two orders of magnitude.

In this case, there is evidence for a highly energetic ($10^{56\text{-}57}\,\mathrm{erg}$) explosion occurring only a few million years ago, resulting a an ionizing, bipolar cone extending outward from the Milky Way's central black hole, Sgr A*.

This powerful flare resulted in huge, 10 kpc-scale, X-ray/gamma-ray-emitting bubbles. The line ratios of $\mathrm{C\,IV}/\mathrm{C\,II}$ and $\mathrm{Si\,IV}/\mathrm{Si\,II}$ points toward ionizing radiation energies of at least 50 eV.

The orientation of these cones is seen in this figure from the paper, which will be on the arXiv tomorrow. The first-author, Joss Bland-Hawthorn, explains in a video here.


Eclipses of Jupiter's Moon during Retrograde motion

The fact that eclipses are either "early", "on time" or "late" only depend on the distance between Jupiter and the Earth. Since the speed of light is not infinite, a longer distance means a longer delay between the actual eclipse and the time it is observed from Earth.

For superior planets, retrograde motion occurs around opposition, at a time when the distance between the planet and the Earth is minimal. So you should expect eclipses to appear early during retrograde motion. But note that the early eclipses are not a consequence of retrograde motion: they are both consequences of keplerian motion.


If two stars collide, what is the probability that they merge to form a single star?

Fairly good.
Two stars of mass $M$ falling from infinity straight towards each other until they merge at distance $2R$ will get kinetic energy $GM^2/R$. This is a lot, for two suns it is $1.8978\times 10^{41}$ J. However, compared to the binding energy of even a single star, $\approx 3GM^2/5R$ this is less(the sun has binding energy $2.2774\times 10^{41}$ J, and a double mass $2^{1/3}R$ radius same density merged star $7.2302\times 10^{41}$ J, 3.17 times more). So there is not enough energy released to blow up a star, but is is about a quarter of it: a lot of matter is going to get ejected or end up in orbits through a heated envelope that will take a while to simmer down.
The key issue is whether the cores get slowed down enough by the encounter to remain bound, becoming a common envelope binary. A direct hit clearly would work, but glancing collisions may allow the cores to miss each other: now the question is whether the envelope can absorb enough kinetic energy. A rough estimate may be that there is significant slowing if the mass scooped up/pushed aside $\pi r_{core}^2 \rho_{envelope} R$ becomes comparable to $m_{core}$. For two sun-like stars with $r_{core}=0.2R_\odot$ this seem to happen, but much hydrodynamics may occur complicating things.
Glebbek's dissertation on stellar mergers estimates a rough condition for the orbital angular momentum to exceed the maximum spin angular momentum of the merged star as $$\frac{r_p}{R_1+R_2} > k^4\frac{(1+q)^{\xi+4}}{2q^2}$$ where $k^2\approx 0.05$, $\xi\approx 0.6$, $r_p$ the periastron distance, and $q=M_2/M_1$. This typically is exceeded: there is a lot of angular momentum that needs to be shed (for example by blowing off a lot of heated gas). For example, two sun-like stars having $r_p=R_\odot/2$ has a LHS of 1/4 and a RHS of 0.0303.
That dissertation also contains numerical simulations of various merger cases.


Are there examples of the effect of shock waves on chemical evolution in of interstellar medium or circumstellar medium?

Shock waves in the interstellar medium (ISM) can be caused by a spiral density wave, supernova shock, or turbulent flow collision. These waves sweep, compress and heat the ISM, thus modifying which chemical reactions can occur. These physical conditions can harbor the chemical reactions to create complex molecules such as NH$_3$, CH$_3$OH, H$_2$O and more.

Based on a chemical/dynamical model, Bergin et al. (2004) investigates the formation of molecular clouds behind shock waves. In the $10^6$ years after the shock wave has passed, the temperature of the ISM is rather high (10 000 K) and molecules are photodissociated. As the ISM cools down, H$_2$ can start to form, followed by CO (which needs to be shielded from photodissociation by the H$_2$). Within 20 million years after the passage of a shock wave, atomic gas has been turned into molecular material.

The chemistry that occurs due to a supernova shock could even explain where some molecules found in comets come from. During a shock, water vapor is formed through reactions with hydrogen and oxygen. Bergin et al. (1998) shows that H$_2$O and CO$_2$ formed in the shock can deposit on grains.


Are black holes solely responsible for hyper velocity stars?

No, black holes are not the only cause of HVSs, although it is thought to be the most common mechanism.

Hyper velocity stars are believed to be caused

when binary stars come close enough to a supermassive black hole for one of the pair to be captured while the other star is ejected at high velocity. This appears to the main mechanism for HVSs. See for example this Harvard Smithsonian Center for Astrophysics page, and this University of Utah page.
when one star of a binary goes super nova, the second star may be ejected at high velocity. See US 708: Hypervelocity Star Ejected by Supernova Breaks Galactic Speed Record and How do Hypervelocity Stars End up Breaking The Speed Limit?.

There may be other causes, since not all HVSs seem to be explained by these mechanisms. Perhaps some of them are not from this galaxy at all - they may be just passing through.


Would blocking sunlight on Venus help make it habitable?

The fundamental problem is the heat capacity of the planet. If you keep it out of sunlight long enough it will eventually cool to the temperature of the background ratiation, bit it will take a long time. If we just treat it as a blackbody surrounded by a 0 K universe the cooling time will roughly be $t_{cool}\approx 2.5\cdot 10^{11}$ seconds if we want to reach 300 K. That is about 7900 years. Wait for about 29,000 years and you can start scooping up frozen carbon dioxide. And this assumes good heat conduction and that the interior is not hotter - I think there will be bad geological issues when trying to change the crust temperature.

Another issue is of course the total lack of sunlight. Martyn Fogg describes a neat solution in his book Terraforming: orbit a reflector in a polar orbit, angled at 45 degrees so it reflects sunlight to the planet. This would for the right orbit give a daily cycle despite Venus rotation issues and the shade shield.

Whether this is a sensible start for terraforming or just evidence that the technology does not help very much depends on your level of optimism and timescales. For exploration you may want a method that does not transform your exploration target as much.


Did interstellar asteroid Oumuamua pass near other stars?

A recently released paper on arXiv (Bailer-Jones et al., AJ) has made use of the recently released GAIA-DR2 data set, and specifically the 7 million stars that have full 6D information, to try and find plausible home stars for 'Oumuamua. They found 4 potential systems of which the closest encounter was at 0.6 parsecs with the M2.5 dwarf HIP 3757 at a relative velocity of 24.7 km/s, 1 million years ago.  However all of the encounter stars have quite high encounter velocities.  A giant planet or a binary system could provide the needed velocity but none of the 4 systems are known to be binaries or to harbor planets. However 7 million stars with full information is still a pretty small fraction of the total, a situation which will improve with later data releases.


Would two Earth-like planets on opposite sides of a Sol-like star at about 1 AU be a stable configuration?

It would not be a stable configuration, but it wouldn't be too far from the stability.

The are 5 points where a third body could have a semi-stable orbit in a 2-body system:

The Earth is the blue point. The "anti-Earth" on the other side would be in the third Lagrange point ($\rm{L}_3$).

However, this is not a stable orbit. Any small deviation, including the effect of possible fourth bodies, would change the system so, that it results yet more deviation. Only $\rm{L}_4$ and $\rm{L}_5$ are stable, and even this has some requirements (the third body can be at most $\approx$ 3.5% of the second).

Corresponding this, the space on the other side of our Sun was examined by some space probes already, and there is nothing.


There is a very nice project called d3-celestial by Olaf Frohn on github. In contains a data file describing the Milky Way as polygons, see here. A demo showing this Milky Way can be found here. And even better, the source for this data is cited, pointing to the Milky Way Outline Catalog by Jose R. Vieira.

Depending on your project, the json format from d3-celestial might be easier to read than the one from Jose R. Vieira.
Note that you don't have to worry about these "contours" moving on a time scale of hundred years, but this is another question.


apparent and absolute magnitude of meteors and space debris

The formula appears to compare the squares of distances, not taking into account the difference in extinction (which is small for small zenith distances, and for different altitudes provided they're high enough), this means $M - m = 5log(\frac{h}{cos(Z)})-5log(h_1)=-5log(cos z) - 5log(\frac{h_1}{100})$. Also, I believe the formula should say -5 cos z ... the meteor from the picture provided is farther away from the one at zenith and at the same altitude and should have higher (numerical value) magnitude.


Most of my information is sourced from here:
https://apod.nasa.gov/apod/ap990511.html

To summarize the source, Bok globules are extremely cold because of their composition -- they're generally just clouds of pre-stellar material which block out all light. This makes the interior of Bok globules "shielded" from interstellar radiation that would otherwise heat them up.

Also, you claim that the globules should have retained some heat from gravitational collapse. However, since the globules don't encounter much friction in space, and there aren't many major collisions that produce changes in kinetic energy, there isn't much energy to dissipate. After all, even Bok globules aren't dense objects by any stretch of the imagination.


Does New Moon Affect Stargazing After MoonSet?

For most people the Moon does not significantly impact on stargazing, even when it is full, unless they are trying to view something in a direction close to the Moon.

For people doing astrophotography, the Moon has a bigger impact since the sky is noticeably brighter and results in fainter objects not being able to be photographed.

And for astronomical observatories, some operations are impossible at worst and severely restricted at best since the telescopes are usually larger than "home" varieties and cameras much more sensitive but most noticeably a few days either side of the full moon.

So, depending on what you are actually doing, a ~38% Moon will not be a problem and if you are looking more than about 30° away from the Moon you will not even notice it.


How did matter first come into existence?

Actually, I also believe that before anything should have come into existence nothing should have been in existence. So everything should have got created out of nothing as we get +ve numbers and -ve numbers from zero as follows.
0 => + 4 - 4 or 0 => +5 - 3 - 2
Wherein, the +ve numbers represent the matter and the -ve numbers represent antimatter. Of course though splitting of zero can be explained but how we explain it in physics is an unresolved mystery.


Is it possible that Titan is a kuiper object captured by Saturn?

Titan is roughly ten times more massive than Pluto or Eris, the most massive known Kuiper Belt objects (KBOs). (Titan is in fact more massive than any other moon in the Solar System except Ganymede.) It would be a rather strange coincidence if the object that was by far the most massive of the KBOs was in orbit around Saturn, well interior to the Kuiper Belt.

Titan's orbit is prograde (it orbits in the same direction that Saturn rotates) and barely tilted with respect to Saturn's equator (less than half a degree), the rings, and most of the other moons. That strongly suggests it formed out of an accretion disk around Saturn, as the Galilean moons formed around Jupiter.

Titan also has an extremely dense atmosphere, unlike any of the KBOs.

So it's rather unlikely that Titan is a captured KBO.

(Neptun's moon Triton, by the way, is thought to be a captured KBO. But it is only slightly more massive than Eris or Pluto, and it has a peculiar retrograde orbit that is very difficult to explain if it formed around Neptune, but easier to explain if it was captured. And since Neptune is the furthest out of the (known) major planets, it's not that surprising that it could have captured a KBO.)


Comparing different magnitudes. Is this statement correct?

Textbooks typically express the flux vs. magnitude relation something like this:

$$m_2 - m_1 = -2.5 \log_{10} \frac{f_2}{f_1}$$

which we can transform into this:

$$\frac{f_2}{f_1} = 10^{(m_1 - m_2) / 2.5}$$

so the original version of your formula was correct.

If m2 - m1 = 0.5, then f2 / f1 = 0.63.
Some readers find comparisons like "x% fainter" confusing, so I would say method 2 is more sensitive than method 1 by a factor of 1.58.


At what distance from Earth would our Sun be the same apparent magnitude as the next brightest star in the sky?

One way to answer would be to consider the brightest star in our sky (other than the Sun), which is Sirius.  Then determine how far you would have to be from our Sun for it to be as bright as Sirius is from here.

That turns out to be 1.8 light years.  That's not even halfway to the nearest star, so if you're in any other star system, then our Sun is just another star.  If you're anywhere in our solar system, even way out in the Oort cloud, then our Sun is way brighter than anything else.


Would stars have formed in the Universe if atomic hydrogen couldn't make molecular hydrogen?

The role of H$_2$ is to allow primordial gas to cool down sufficiently to allow the collapse to start and then to hold the gas as a relatively low temperature as it gets much more dense. The formation of H$_2$ is essential because atomic hydrogen simply has no way of cooling itself below temperatures that are capable of exciting the $n=2$ level and then emitting Lyman alpha photons. This means that without H$_2$ the gas can't cool effectively once it gets below about $10^4$ K and it will never reach the densities required to form stars.
The ongoing formation of H$_2$ in primordial gas phase reactions (no dust required) allows a collapsing cloud to cool down to around $\sim 200$ K.
Without molecular H$_2$ you get stuck with large ($\sim 10^4 -10^6M_{\odot}$) pressure-supported "protogalaxies" that won't collapse any further. These are formed at redshifts of around 20-40 from growing primordial density perturbations and have initial temperatures of a few thousand K. The ongoing formation of H$_2$ in primordial gas phase reactions (no dust required) allows a collapsing cloud to cool down to around $\sim 200$ K.
The importance of H$_2$ cooling is outlined in reviews by Glover 2005 and reiterated more recently by Klessen 2018. A telling quote is from the review by Bromm 2013, who says

If the gas were unable to cool, there would be no further collapse, and consequently no
gas fragmentation and star formation. The gas would simply persist in hydrostatic
equilibrium, roughly tracing the density profile of the Dark Matter. Early on, it was realized
that cooling in the low-temperature primordial gas had to rely on molecular hydrogen
(H2) instead (Saslaw and Zipoy 1967).

Conversely, in gas that is enriched with heavier elements, H$_2$ cooling is totally unimportant because there are more efficient routes for a collapsing cloud to radiate away energy (e.g. Glover & Clark 2012).


Stellar life cycle flow chart with mass conditions and time scales

My slapdash (and perhaps messy because I rendered it in PowerPoint) attempt at addressing stellar evolution. Open the flowchart in a new tab in fullscreen to see a better rendering of it.


Is there any way for a planet orbiting a red dwarf in the habitable zone to not be tidally locked?

Leconte et al. (2015) suggested that the presence of an atmosphere could prevent or at least slow tidal locking. The star should exert two separate torques: one on the atmosphere and one on the solid body of the planet:
$$T_a=-\frac{3}{2}K_ab_a(2\omega-2n),\quad T_g=-\frac{3}{2}K_gb_g(2\omega-2n)$$
where
$$K_a\equiv\frac{3M_*R_p^3}{5\bar{\rho}a^3},\quad K_g\equiv\frac{GM_*R_p^5}{a^6}$$
for stellar mass $M_*$, planetary radius $R_p$, mean density $\rho$, semi-major axis $a$, mean motion $n$, rotation rate $\omega$, and response to torques $b_a$ and $b_g$. The two torques could be equal, and assuming that the atmosphere transfers some angular momentum to the surface of the planet, this could prevent tidal locking. There are several equilibria at which this could occur:


What is this planet or satellite in this picture?

As in this question about an image of a partial solar eclipse, the object in question is:

round and about half the apparent size of the Moon
fainter and slightly bluish
directly opposite the Moon relative to the center of the image

I believe it is a reflection of the Moon from coated surfaces in the camera lens.


Could some bacteria theoretically evolve to survive in space?

Currently there is no life that we know of that evolved in, or can survive for prolonged time in a vacuum.

To say 'Life in space is possible because there is anaerobic life' is a completely wrong point, that ignores the effects of vacuum on life.

Sure, there are extremophiles that can survive for a limited time in space (see also this list) when going into hibernation modes. However all life as we know it needs liquids and/or gases to survive. Those would evaporate once the cell/water bear would become active again. Activity is necessary to sustain and multiply life, so no, nothing that we know of can stay active and alive in space.

  Cosmic rays and high energy particles could damage DNA, but there might be someway life could protect itself against them

Until life comes up with such mechanisms, we just don't know. The problems with liquids and gases evaporating would need some tougher solutions even.

Bacteria are also pretty heavy 'particles' and therefore cannot float anywhere near the 'edge of space'. There are reports of bacteria in the stratosphere, but that requires already strong upwinds to put them there.


You cannot see the event horizon. That being said:

A non-rotating black hole, free of external influences, has perfect spherical symmetry. All its properties are exactly the same in any direction, period. This is the Schwarzschild metric.

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

Even if it is electrically charged, if it's non-rotating, and free of external influences, it is still perfectly spherically symmetrical - this is the Reissner–Nordström metric.

https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric

As soon as it begins rotating, the black hole is no longer in perfect spherical symmetry. It acquires an ergosphere, which is shaped like a flattened ball. This is the Kerr metric.

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

Significant external factors can distort the shape of the event horizon. Two black holes merging will go through a process where spherical symmetry is lost temporarily:

https://www.youtube.com/watch?v=JOQMvyLYmd4

Technically, anything near a BH would slightly distort the metric of spacetime, but in practice this would not be easily measurable unless the external factor is very large (another very massive object). So for most practical purposes, the metrics described above would apply.


What variables are needed to calculate simple horseshoe orbit times?

That depends on the accuracy you want to work with. To zeroth order, as outlined in Murray & Dermott, "Solar System dynamics", Chapt 3., you can do the following:

The zero-velocity contours that are plotted in your image will not be coinciding with particle orbits to infinite precision, but they're a good zeroth order approximation for objects with low eccentricity w.r.t the star ($v_{r}/v_{\theta} \ll 1$)
A particle on those orbits is on a regular Keplerian orbit of radius $r_{\rm H}$, out of reach of the gravitational influence of the planet at $r_{\rm P}$. Thus, to get the largest "chunk" of traveltime of one part of the horseshoe orbit, either inner or outer part, you can work out with relative velocities and the assumption of keplerian velocities.
Caution is required as to which orbital time you're interested in: If $a$ is Earth's distance to the sun, and the horseshoe is $d$ away from a perfectly circular orbit, thus alternating between distances $a\pm d$, then the orbital time in the rest-frame of the sun will be $v_{-}=\sqrt{\frac{GM}{a-d}}$, for a large number of orbits until close encounter, and after that $v_{+}=\sqrt{\frac{GM}{a+d}}$.
Therefore, the simplified case that the satellite is on its own Keplerian orbit is true for most of the time. Armed with this knowledge, we can approximate the Planetocentric recurrence time as $t_{\rm rec}=\frac{2\pi a}{v_{K} - v_{-}}$ as a simple catch-up time between objects on different orbits.
The relative velocity $v_{\rm rel} = v_{K} - v_{-} = \sqrt{\frac{GM}{a}}-\sqrt{\frac{GM}{a-d}}$ can be expanded in the limit of $d/a\ll1$ into $v_{\rm rel} \approx \frac{1}{2} \sqrt\frac{GM}{a} \frac{d}{a}$ and thus I derive from this $t_{rec}\approx \frac{4\pi}{\sqrt{GM}} \frac{a^{5/2}}{d}$. As expected, the cycle time diverges for $d=0$, as in the co-orbiting case with Earth this time must be infinite. An upper limit on $d$ cannot be derived from this, one would need to turn for the full solution for that.

Just out of curiosity, I've plugged in a few values into this formula and written something quick in python:

import numpy as np

#Basic physics quantities
G      = 6.678e-8 #cgs units
pi     = 3.141592
navo   = 6e23 # particles per mole
sigma  = 5.67e-5   #erg cm-2 s-1 K-4
kb     = 1.38e-16  #erg/K
km     = 1e5 #kilometers in cm
mearth = 5.98e27  #g
msun   = 2.0e33   #g
au     = 1.49e13  #cm
yr   = 365*24*3600
rearth   = 6370e5
rjupiter = 74000*km

#
# Returns the approximate horseshoe-cycle time in the reduced 3body problem
# Masses of bodies: m0>>m1>>m2
# Semimajor axis distance is from m0 to m1, radial distance is a(m0->m1)-a(m0->m2)
#
def hs_cycle(mcentral, semimajor_axis, radial_distance):
    return 4*pi/np.sqrt(G*mcentral)*semimajor_axis**(5./2.)/radial_distance/yr

#
# https://en.wikipedia.org/wiki/(419624)_2010_SO16 around the Sun
#
# Quoted cycle time ~350 years, with d=0.004 AU
#
print("Predicted 2010_SO16 cycletime [years] = ", hs_cycle(msun, 1*au,0.004*au), " predicted = 350 yrs")

#
# Janus/Epimetheus around Saturn
#
# a = 151410 km, d = 25 km, as stated in https://en.wikipedia.org/wiki/Epimetheus_(moon)
# Quoted cycle time = 8 years (from comments)
#
print("Predicted Janus/Epimetheus cycletime [years] = ", hs_cycle(95*mearth, 151410*km,50*km), " predicted = 4 yrs")

#
# 3753 Cruithne
#
# a = 1 AU and semimajor axis difference from https://en.wikipedia.org/wiki/3753_Cruithne
# Quoted cycle time = 770 years
#
print("Predicted 3753 Cruithne cycletime [years] = ",hs_cycle(msun, 1*au, (1.0-0.99774)*au), " predicted = 770 yrs")

and the results I get are

Predicted 2010_SO16 cycletime [years] =  495.7747141830971  predicted = 350 yrs
Predicted Janus/Epimetheus cycletime [years] =  11.542076781209305  predicted = 8 yrs
Predicted 3753 Cruithne cycletime [years] =  877.4773702355546  predicted = 770 yrs

So the formula may be off by up to a factor of ~2. This is for sure simply because reality is more complex than a simple circular-orbit approximation, but also due to the quality of the values used.
Wikipedia isn't well known for researching particular values well. I took those that I found there. For SO16 that was particularly confusing to select, so I took those two that were mentioned in the same line of text, hoping they would come from the same source.

Anyone finding more consistent values, is free to comment.


Where is the Solar system's barycenter located?

The solar system barycenter (SSB) is sometimes inside the Sun and sometimes outside.
As an observer outside the solar system could detect with Doppler spectroscopy, the Sun is what's wobbling around.
The Sun's offset from the SSB is a vector sum of roughly:

0.00496 au ±5% away from Jupiter
0.00272 au ±6% away from Saturn
0.00083 au ±5% away from Uranus
0.00155 au ±1% away from Neptune

The other planets contribute much smaller amounts to the total.
Each planet's contribution is proportional to the product of its mass and its orbital distance.
When these components add constructively as in 2020-2023, the center of the Sun can be as far as 2 R☉ away from the SSB.
When they cancel as in 2029-2030, the center of the Sun is within 0.5 R☉ of the SSB.
The solar radius R☉ is 0.00465 au, shown here with a dashed line.


How Does Earth Move In Relation To The Stars

The orientation of Earth wrt the stars repeats each sidereal day (thus ~23:56h). The visible night sky only changes each (synodic 24h) day due to the different direction the sun is at when considering the star's rest frame.

If you have the star's position from like now, you simply rotate Earth for each our by 360/(23.933)° = 15.04°.
If you want to do that for an arbitrary date, take the stars' position: they come with a datum like J2000 which means their position on January 1st 2000 UTC. C calculate the sidereal days passed since then, accounting for the incomplete rotation of the current day which gives the rotation around the Earth's axis as above.

If you want to go into more detail and longer time, and account for centuries and longer, you will need to take into account the nutation and presession of the Earth axis.


Do the neutrons in neutron stars emit the radio waves?

The electromagnetic radiation comes from accelerated charged particles, mainly electrons and positrons.
The surface of a neutron star is not made of neutrons. It is a totally ionised gas of nuclei plus electrons, with a composition that could range from iron-peak elements to hydrogen and helium accreted from the interstellar medium.
As a result of their rotation and strong dipolar magnetic fields then there are also very strong electric fields at the surface of the neutron star. These fields  rip electrons and protons from the surface and accelerate them away -  electrons from the magnetic poles and protons at lower magnetic latitudes. The classic work on this is by Julian & Goldreich (1969).
The acceleration due to these electric fields is easily sufficient to overcome gravity. The particles are forced to spiral along the magnetic field lines by the Lorentz force, and produce synchrotron or curvature radiation directly. The photons from these processes are then capable of creating secondary electron-positron pairs that are accelerated in opposite directions, releasing more photons in a cascade.
No neutrons are harmed in the making of this radiation, other than the ones in nuclei at the surface of the neutron star that get bombarded by high energy accelerated electrons and positrons.


What are the masses of the two stars (given the information provided)?

You need to use the generalized (a la Newton) form of Kepler's third law; see here.


Is the Oort Cloud actually opaque, or is the "cloud" used euphemistically because minor planets inside of it are too dim to be seen?

Well, given that we can see the stars and that the Oort "cloud" is closer than the nearest star, then the answer to your main question is obviously that the "cloud" is not opaque.

I think it is called a cloud because it consists of many individual, small "particles" that don't interact with each other. Also, the term cloud avoids giving the impression that the Oort objects are arranged in any disk-like configuration - they should be much more spherically symmetric. The word cloud also avoids giving the impression that the objects fall between two relatively tight orbital radius limits. i.e. The Oort cloud is not like the Kuiper belt.


Is a year really 365.24 days, or is it 365.2564 days like I remember?

Both are correct.

The Sidereal Year is the length of time it takes Earth to complete an orbit around the sun, relative to the fixed stars. It is 365.2564 days.

The Tropical Year is the length of time it takes for the Sun to complete a cycle around the Ecliptic and return to the position in the cycle of seasons; e.g. from Vernal Equinox to Vernal Equinox.  It's about 365.24217 days, about 20 minutes shorter than the Sidereal year because of the precession of the equinoxes. This is the year that the Gregorian Calendar is attempting to emulate.


In case of light losses in your optical system you obviously have to correct for this. Anything attached in front of your detector could potentially reduce your photon count, so you may have to re-calibrate this to interpret your count rates correctly. You have to examine your setup and make either some theoretical or, if possible, experimental analysis of the effect.


What is meant by *topocentric* right ascension and declination?

You are correct that topocentric coordinates are for the position of "close" objects, corrected for observing from the Earth's surface instead of the theoretical center.

The topocentric and geocentric coordinates would use the same system (J2000, true, etc.) There is no convention to use one in favor of another (but there may be preferences in some cases).


Can the gravitational potential in a spiral galaxy be positive?

I think your recreation is essentially correct. For example, if you look at Figure 1 in that paper, you can see that the potential goes from positive to negative as a function of the azimuthal angle ("phase").

What you're missing is that the potential and density functions they define are perturbations, which are intended to be added to an axisymmetric galaxy model. The idea is that the axisymmetric disk is modulated by their perturbation, so that the total density is less than average (but not less than zero!) where their perturbation is $< 0$ and greater than average where their perturbation is $> 0$. (Similarly, the total potential is $< 0$ everywhere; in the regions where their perturbation is positive, the total potential becomes less negative, but never $> 0$.)

As they point out (pp.4-5), "The impressions invoked by the density distributions of Figures 3 and 5 can be somewhat misleading. These densities must be regarded as perturbations to an azimuthally uniform stellar disk with the same vertical scale height."

And: "In Figure 10, a disk component with the same radial dropoff and scale height as the perturbation, and just sufficient amplitude to make the net density everywhere positive, has been added to the perturbation density." and "In Figures 12 and 13, various amounts of perturbation density are shown added to a representative full stellar disk."


Could the expansion of universe be a side effect of gravity?

Well, everything is possible I suppose. And we do see plenty of material that talks about the expansion of the universe as a kind of repulsive gravity. But I would say no, the expansion isn't due to gravity, and it isn't some side effect of gravity. It's something different.

  I mean maybe gravity "attracts things" at a "small" distances and repels them at "larger" distances?

It doesn't. Gravity attracts things at all distances. It follows the inverse square rule. Some people like Mordehai Milgrom say it doesn't quite follow the inverse square law, see MOND. But he isn't saying it repels at large distance.

  Do we have any theory/hypothesis/paper on this topic? Is it possible to prove or disprove?

We have general relativity. In general relativity gravity sucks. It doesn't blow.

  Could gravity have a repulsive term that dominates at cosmological distances and could it be such a term that causes (accelerated) expansion?

Like I said, everything is possible. But I will say no, the universe is expanding because of dark energy, not because of gravity. I rather like Erwin Schrödinger’s cosmic pressure myself. See the 2014 Cambridge companion to Einstein by Michel Janssen and Christoph Lehner: “Schrödinger [1918] had pointed out another way of treating the cosmological constant: moving it from the left-hand side of Equation [7], where it represents a contribution to space-time curvature, to the right-hand side, where it represents a contribution to the energy-matter distribution. Then it would correspond physically to a kind of cosmic pressure”. Unfortunately Einstein didn’t like it. You can read more in Alex Harvey’s 2012 paper How Einstein Discovered Dark Energy. He gives a translation of Schrödinger’s paper and refers to Einstein’s response. If you were to ask a modern cosmologist, I think the answer would be that Schrödinger was pretty close to the mark. See page 252 of Relativity, Gravitation and Cosmology by Robert J Lambourne:


Where is the center point for the Supergalactic coordinate system?

According to Wikipedia, the origin of the system is:

The zero point (SGB=0°, SGL=0°) lies at (l=137.37°, b=0°). In J2000 equatorial coordinates, this is approximately (2.82 h, +59.5°).

In the same article there is a recent referred paper on this topic, which explains [The zero point] is one of the two regions where the SGP is crossed by the Galactic plane.

Perhaps you already know this website, which shows many good pictures. I was trying to find a better figure, but I guess you have to push your imagination according to the definition.

EDIT: According to this, SG coordinates are similar to the galactic ones, so that the North Supergalactic Pole is defined in Galactic coordinates, which means that they have the same origin (i.e., the Sun).


Would humans hear gravity waves from a binary BH fusion nearby?

As your ear is extremely small compared to the wave length of the gravity waves, remember they travel at the speed of light not the speed of sound. The structures in the ear would not respond, also the ear would change in size (by a  ridiculously small amount) itself, thus nullifying the effect anyway. LIGO works by comparing the light travel time of two laser beams at precise right angles, your ear is not directional like that.


GW190521 black hole merger total mass calculation and missing mass, how does this happen?

I remember the first time I'd read about a gravitational wave detection and the resulting mass I also stopped when I realized the numbers didn't add up. At first it's shocking. While we might be aware that very tiny changes in mass are associated with huge amounts of energy release, it's absolutely amazing to think of a mass the size of our Sun radiated as energy!
But that's exactly what happens in the mergers of black holes. The fraction of mass converted to energy is much smaller when two normal stars merge, gravitational waves are produced but are small in comparison. But when the objects are extremely dense so that they can orbit very close before the merger, as in the case of neutron stars and black holes, those gravitational waves are powerful and a significant fraction of that mass will radiate away as ripples in space itself.
Answer(s) to Fraction of initial mass lost (radiated) by neutron star mergers compared to black hole mergers? indicate that this fraction can be up to several percent in these cases.
We can convert an absurdly tiny amount of those distortion waves back into other forms of energy as a thought experiment, though not practical. For more on that, see answers to

Does a gravitational wave loses energy over distance?
Transfer of energy from gravity back to other “more familiar” forms of energy?
Does shortening the path length of an excited etalon do work? What about LIGO?


If the axis of the Earth was through Greenland, what would the north celestial pole be?

If you assume that the shift was due to natural process, it would still point more-or-less towards Polaris.  (If the shift is due to magic or advanced technology, all bets are off.)

I think your question may be based on a false premise, that the celestial sphere is fixed relative to the Earth.  It isn't at all.  Over the timescales we're dealing with it's basically just fixed and unchanging.

The rotating Earth acts like a gyroscope and its axis points in some direction.  To a good approximation, that direction is fixed in space due to conservation of angular momentum.  So in the absence of external torques, the Earth just sits there and spins like a top.

(To a better approximation, there are some external torques and the gravitational effects of the Sun and Moon causes the Earth's axis to slowly precess. It wobbles (a bit like a top slowing down) once every 26,000 years, actually chaning where it points to in space.  I think this is probably an irrelevant detail though, for your question.)

So, if the Earth's axis passed through Greenland, what would happen?  (Other than the climate changes due to the polar and equatorial regions moving around, of course.)

We have in fact seen something like this in the Earth's past, and it's called Polar Wandering. We observe that the Earth's north pole has moved relative to the Earth's surface and actually has moved quite a lot. But what happened was that the Earth's axis stayed fixed relative to the stars and the Earth's bulk shifted relative to its axis of rotation.

The True Polar Wander page referred to above has an excellent diagram:

So, the answer is that if the shift occurred because of changes to the Earth's moment of inertia, the axis of rotation would still point towards Polaris.  If it happened due to outside influences (a giant impact, aliens, magic) it might wind up pointing most anywhere.


Where did the hydrogen come from in a type II supernova?

Good question - the answer is that stars are not in general well mixed - or rather, the nuclear-burning core is not well-mixed with the rest of the star.
That means that a star will finish hydrogen burning even when about 80% of the hydrogen in the star is still available, but is situated outside the core and cannot be mixed into the core (or at least not very efficiently).
By the time a type II supernova explodes, a large fraction of its envelope (the region outside the core) is still hydrogen.
The Sun's hydrogen comes from the big-bang. Hydrogen is not manufactured in stars, it is a raw material that was created in the first minutes of a hot big bang 13.7 billion years ago. The same is mostly true for helium, but the heavier elements in the Sun are the products of nucleosynthesis of the billion or so stars that lived and died in our Galaxy before it was born. The material that they eject through stellar winds and supernovae of all types is thoroughly mixed into the interstellar medium.


How much more life could the Sun acquire via star lifting?

TL;DR: the main sequence lifespan of the sun can be increased by a factor of 12.2.

Perhaps the most complete astrophysical analysis of stellar engineering for extending Earth habitability is Martin Beech's book Rejuvenating the Sun and Avoiding Other Global Catastrophes (2008).

In order to maintain the biosphere the sun's interior need to be mixed (in order to avoid the red giant phase) and material needs to be removed (in order to avoid a too great luminosity).

Material removal

Removing material is tricky because the potential energy at the Sun's surface is $U_\odot = GM_\odot/R_\odot=$ 1.9e11 J/kg. While just 2e-6 of the mass-energy embodied in the matter and 0.3% of the potential fusion energy it is still cumbersome. If a Dyson sphere was used at perfect efficiency the maximal mass extraction rate would be \begin{equation}
\dot{m}=\frac{L_\odot}{U_\odot} = \frac{L_\odot R_\odot}{GM_\odot} \approx 2.0061\cdot 10^{15} \text{kg/s},
\end{equation}
a typical mass for a minor asteroid. The timescale to disassemble the sun using just solar luminosity is \begin{equation}\tau_{disassemble}=\frac{3GM_\odot^2}{5R_\odot L_\odot} = 18.852\text{Myr}.\end{equation} Long before this time changes in pressure would reduce luminosity, making this estimate an underestimate by about an order of magnitude. I get 556.93 Myr, a bit more than Criswell's 300 Myr.

Another idealized model is that the lifting process fuses hydrogen (using the $p-p$ process) to power further lifting. In this case each kilogram of lifted mass produces $6.45\cdot 10^{14}$ J, allowing lifting of 3380 kg more fuel. This could of course speed up arbitrarily if there was technology for it.

I am leaving out how to do it. Beside Criswell's classical proposals (speeding up rotation, sucking off plasma using magnetic fields etc.) Beech suggests a stream of ramscoops or a stellar wrap could do it. This is not really key to the question.

Luminosity decrease

Luminosity can also be decreased by increasing non-thermal core pressure, perhaps through strong magnetic fields or rapid internal rotation. This would extend the main sequence lifetime as \begin{equation}\tau_{NTPS}\sim \tau_{MS} \left(1-\frac{p_{NT}}{p_T}\right )^{-7.5}.\end{equation} A 10% increase over current pressure would double the sun's main sequence lifetime and about halve the luminosity at hydrogen exhaustion. Achieving this is nontrivial given the current solar structure (the magnetic fields are generated in the convective outer envelope), but Beech notes that were the mass reduced below about half the current mass the modified sun would be full convective and magnetic fields would also produce a higher $p_{NT}$ in the core.

Another way of reducing luminosity is increasing the stellar opacity $\kappa$. In order to reduce the luminosity by a factor of 100 (and the surface temperature by a factor of 10) the opacity needs to be increased by a factor of 100. However, Beech notes that merely adding heavy elements will not help since they increase mass and hence luminosity. While one can envision "dialysis" where helium is removed and replaced by larger $Z$ elements this appears complex, and there are no local readily available sources of enough high $Z$ material in the solar system.

Mixing

Engineering enhanced mixing before the mass is removed is nontrivial. Impactors and lasers simply do not penetrate deep enough to matter. Beech suggests employing a sub-stellar mass black hole oscillating through the sun to mix it. If the hole is small, the time until absorption is long.

Result

With mass-loss and complete mixing Beech finds the main sequence lifetime increased up to a factor of 12.2, with reasonable factors between 4 and 6.

Beech's scenario has the problem that the late stage sun while having an acceptable luminosity at Earth's orbit will be far hotter, releasing significant UV radiation. The end result is in many ways similar to a blue straggler star.


Which Lowell Observatory telescope is this? (used to draw Moon maps for the Apollo program)

That's the historic 24-inch Alvan Clark refractor.
It was installed in 1896 and restored in 2014-15.
Besides the moon mapping project, Percival Lowell used it to observe Mars, and Vesto Slipher used it for some of the first measurements of galaxy redshifts.
Its current mission is education; in mild weather the observatory lets visitors look through it.

Roger Vine's Scope Views has more photos and a detailed description.


Is there any planet out there which is half gaseous and half rocky ? is this possible?

Probably.   Relatively little is known about exoplanets because they're very hard to get a good look at, but there's no reason why a rocky world couldn't accumulate enough ices and/or gas to also resemble a gas giant.   Now half rocky half gas giant (hydrogen/helium) might be rare.   Half rocky half "ices" is very possible and those have likely already been observed.  It helps to understand planet formation and elemental abundance.

There's likely a size limit to rocky planets with thin atmospheres because mass tends to accumulate and retain atmosphere and with hydrogen and helium being the most abundant gases in the universe, once a planet gets massive enough, it should collect and retain hydrogen and helium and begin to resemble Jupiter or Saturn, even if it started out as a rocky world.

Rocky worlds, need elements like Magnesium, Silicon, and Iron, often bound with oxygen. close to 90% of the mass of rocky worlds in our solar system comes from those 4 elements and 90% isn't a bad estimate for rocky exoplanets.

Based on our solar-system:

Iron makes up about 0.117%, Silicon 0.065% and and Magnesium 0.051%, for about .234% of the mass of the solarsystem.  Oxygen bound to those elements, using Earth's composition as an estimate adds about another 50%, up to about 0.35%.   Hydrogen and Helium make up about 97.6%, which leaves about 2% of the mass of the solar system in the form of ices and heavier gases, not hydrogen and helium.  Those are primarily water, ammonia, CO2 and Methane, with smaller amounts of other gases/ices.

These numbers are rough but good enough for an estimate.   In our solarsystem there's about 6 times as much ice and heavier gases by mass than there is rocky material and there's close to 40 times as much hydrogen and helium as the other elements.

Planets need to be quite large to hold hydrogen and helium.   Jupiter and Saturn are the only planets in our solar-system that are hydrogen and helium abundant.    Even the smaller gas giants, Uranus and Neptune have comparatively little hydrogen and helium.

So there's probably no such thing as a small gas giant.   Gas giants need to be large or they don't exist at all.  (I should make a footnote that as a gas giant loses atmosphere, you can get a small gas giant, but that would be in a transition phase).

I don't like the term "Ice giants", though it's often used to define Neptune and Pluto because they are primarily ices (water, methane, CO2, NH3), not primarily hydrogen/helium.     I don't like calling them "ice giants" when they can also be hot, so I prefer the term Neptune like planets or Neptunes.

Uranus (a Neptune type planet) is mostly made up of ices and heavier gases with an estimate of just 3%-10% hydrogen and helium.  Estimates of about 70%-90% of ice/not hydrogen/helium gas and the remaining, about 7%-20% more dense material.   That would be some other elements, perhaps some solidified carbon (diamonds), higher sulfur and lower Iron by percentage than Earth as well as some Earth like silicates. magnesium-oxides, Iron and Iron-oxides, though the high internal temperature might not leave much in the way of chemical bonds towards the center of the planet where temperatures reach 9,000 degrees C.

Neptune is similar to Uranus but more massive and more dense.  It's thought to have more water and perhaps a larger internal mantle, though like Uranus, the majority of it's mass is ices and heavy gases (not hydrogen/helium).

Neptune and Uranus have hydrogen in the form of Methane, Ammonia and Water, but I'm counting that hydrogen as part of their "heavy gases/ices".   They both have a comparatively low percentage of free hydrogen (3%-10%) making a clear distinction between Neptune like planets and gas giants like Jupiter and Saturn with which the majority of their mass is hydrogen.

It helps to think of how planets form.   Smaller planets have too little gravity to retain hydrogen and helium gas and solar-systems are mostly too warm for hydrogen to freeze, so planets are made up of solid material for the most part, either rocky or ices that clump and stick together during formation.   When they have enough gravity, then they can begin to hold onto an atmosphere and further from the star makes this easier.

The Moon Titan, as an example, is small to have an atmosphere, but it's quite far from the sun and largely made up of ices on it's surface, so as the ice on it's surface melts, it basically replenishes it's atmosphere.  Titan is still out-gassing, and it losing atmosphere, but it loses atmosphere slowly enough that it retains an atmosphere, but in a sense, it's in a transition, where it's atmosphere is outgassed.  When it runs low on surface ice that readily thaws, it should begin to resemble the icy moons of Jupiter.

So, back to your original question.   Many combinations are possible, some are not.   Baby Jupiters (the mass of Neptune) are probably unlikely, though a gas giant close to it's star that's lost a lot of it's atmosphere could resemble a baby Jupiter, but like Titan, that would be transitional.

Rocky Jupiters are unlikely because hydrogen outnumbers rocky material by over so much, over 200 to 1 in most of the milky-way.   Now, some solarsystems are likely more "metalic" than others, so that ratio will have some variation, but it's once there's enough mass for a gas giant to form, hydrogen is likely to be the abundant element and Helium #2 and any rocky core would be dwarfed by the hydrogen and helium.

A rocky Neptune however, no reason why not.   A planet with 8 earth masses of Earth like material and 8 Earth masses of ices (and 5%-10% hydrogen) would basically be a rocky Neptune.   It would likely be a little smaller and certainly denser, but not all that different in outward appearance.

Water worlds is a common term that might qualify as a "rocky-neptune".   We don't know exactly what they are made of, but density estimates suggest a high percentage of water (and presumably CO2, CH4, NH3), which is the majority of Uranus and Neptune.

Here's a chart of exoplanets of less than 20 Earth masses.   I suspect there's a considerable margin of error in these estimates, but it more or less agrees with the rocky-neptune argument.   Planets with 2 or 3 grams per cc would be in that range.

See chart:  www.hpcf.upr.edu/~abel/phl/hec_plots/exoplanet_df.png

Source.

Baby Neptunes of just 1 or 2 Earth masses might be possible too, but they'd probably need to be quite cold and far from their star or they'd be in danger of losing their atmosphere.

Iceworlds like Pluto can get quite small, but Pluto has very little atmosphere.  Planets generally need to be fairly large to retain their atmosphere.   If they are too cold, that atmosphere freezes.   Titan, as I mentioned above, is in what could be called a slow transition where it's atmosphere comes from it's surface and it loses it slowly.   Ceres, based on it's density, is an icy moon like object too, though it's lost nearly all it's surface ice, but it probably has a lot of water/"ices" below it's surface.


How much is usually deposited in a Mars regional dust storm?

Philip R. Christensen wrote an article back in 1986 called Regional dust deposits on Mars: Physical properties, age, and history. The abstract says

Dust deposited during global storms is subsequently removed only from dark regions, resulting in a net accumulation in the low‐inertia, bright regions. The thickness of these current dust deposits is 0.1–2 m.

This rather rough approximate for global storms seems to be a upper boundary, because e.g. Bruce Cantor states in the abstract of the article MOC observations of the 2001 Mars planet-encircling dust storm, the dust storms observed by the two rovers do deposit significantly less material (but can still cover solar cells):

The storm caused substantial regional albedo changes (darkening and brightening) as a result of the redistribution (removal and deposition) of a thin veneer of surface dust at least 0.1–11.1 μm thick.

However, the abstract also says that local storm events might significantly redistribute material:

The redistribution of dust by large annual regional storms might help explain the long period (∼30 years) between the largest planet-encircling dust storms events.

Both articles are paywalled, so I could not dig deeper here, so let us explore a bit a very local phenomenon, namely the dust devil. Dust devils have been frequently observed on Mars.
S.M. Metzger, Matt R. Balme et al. give an estimate of mass transport per dust devil in their article In-Situ Measurements of Particle Load and Transport in Dust Devils:

We find that dust devils [...]  can  loft  about  1  kg  of  PM10  material  to  heights  of  several  hundred  meters over their lifetime. Furthermore, applying the 5-15% TSP/PM10 ratio to the TSP data  from  the  five  large  dust  devils  sampled  in  1996  suggests  that  larger  examples  can  deliver  up  to  20  kg  of  fine  particulates  to  great  height.

References

Hall and Lemmon: Mass eject from Martian dust devils as functionof height and devil morphology


Different times of sunset at the same location due to elevation

The page you refer to is about sunrise/sunset throughout the year, but what you really care about is differences in the geometry on one day. This is sometimes expressed as different observers seeing the horizon at different distances, or sometimes as an angle, a “dip” of the horizon when seen from higher altitude.  Thinking about that geometry should get you started on working out the answer.


Less stars in the night, compared to 15-20 years back

In my own case, if I see fewer stars than I did when I was younger, it's partly because my eyes have deteriorated with age. I don't know whether that applies to you.

The most likely explanation is increased light pollution. If you can manage to get to an area far enough away from city lights, on a cloudless and moonless night, you should see just as many stars as you did when you were younger.

  Stars are dead?

No.  All the stars you can see in the night sky are within a few hundred light-years of Earth. That means that you're seeing them as they were no more than a few hundred years ago. Most stars live for billions of years; the Sun, for example, is about 5 billion years old and is expected to live for another 5 billion or so years in essentially its current form. Some of the brighter stars have shorter lifespans (because they're larger and "burn" their nuclear fuel more quickly), but even they have lifespans of at least millions of years.

No naked-eye visible stars have "died" in the last several centuries. We do sometimes see stars explode (as novas or, more rarely, as supernovas), but I don't believe any of the stars that have done this were naked-eye visible before they exploded.

  Current position of the earth in galaxy

No.  The stars do move relative to each other, but not quickly enough for the motion to be visible over a human lifetime. Barnard's Star has the fastest proper motion of any star in the sky, but it only moves at about 10 arcseconds per year; it would take it nearly 2 centuries to move the width of a full Moon. And Barnard's star isn't even naked-eye visible. The stars you see in the night sky are in very nearly the same apparent positions as they were thousands of years ago.


What does it mean for a planet to be just 300 kms across?

The reports that Goblin is a planet are incorrect. It is a dwarf planet, at best. Dwarf planet and planet are mutually exclusive terms. Planets are objects whose masses are so large that they have collided with, ejected, or captured the vast majority of the smaller stuff that share the same orbit as the planet.

Moreover, Goblin might well not even be a dwarf planet. The distinguishing characteristics of a dwarf planet is an object that is large enough to have pulled itself into a more or less spherical shape via self gravitation but is not anywhere near large enough to have the potential to clear the neighborhood of its orbit about the Sun.

The threshold that distinguishes dwarf planets from small solar system bodies is the potato radius. Objects somewhat smaller than the potato radius look like lumpy potatoes, while objects larger than that have sufficient mass to more or less pull themselves into hydrostatic equilibrium. The potato radius depends a bit on composition, and probably also depends on distance from the Sun. (Water ice is quite rocky at trans-Neptunian distances from the Sun.) The potato radius for rocky objects is about 300 km (diameter = 600 km), while for icy objects it's about 200 km (diameter = 400 km).

With an estimated diameter of 300 km, Goblin is probably too small to qualify as a dwarf planet, let alone as a planet.


How does energy (from fusion reactions still inside the sun) still have gravitational attraction?

I think it is best not to think of energy having a gravitational attraction, rather that energy, like mass, (since they are equivalent keeping in mind c) will curve spacetime.  Electric charges, magnets, and electromagnetic fields will curve spacetime, Scientific American. So whether it is mass or energy, they both will affect the curvature of spacetime, and the sun's total impact on spacetime will not change.


How well can we in principle determine $T_{\textrm{eff}}$ of a star?

The question is compromised by saying that you allow arbitrarily perfect measurements.

If we have a bolometer that can measure the amount of flux from a star, at a distance that is known to arbitrary accuracy, with arbitrarily good spatial resolution, then what we do is measure the bolometric luminosity from a 1 m$^2$ area at the centre of the stellar disk. This flux is $\sigma T_{eff}^4$.

Now of course, stars do not have homogeneous atmospheres (spots, granulation, meridional flows, non-sphericity due to rotation...), so the result you would get would depend on exactly what 1 m$^2$ bit of atmosphere you were looking at. So, with my arbitrarily accurate instruments I would have to measure the luminosity from every 1 m$^2$ patch over the entire surface of the star. Each one would give me another estimate of $T_{eff}^4$; each would be somewhat different. That would be difficult, but the form of your question allows me to ignore those problems.

At this level of precision, the utility of a single $T_{eff}$ for the whole star is questionable, but if you wanted one then it would be the flux-weighted mean of all the above measurements, and as far as I can see one can instantaneously determine it to whatever accuracy you desire. Of course it will then vary if you have a variable star, and it will vary from point to point with time due to granulation; so the accuracy of the $T_{eff}$ could depend on how quickly and by how much it varies compared with how long it takes you to do your arbitrarily accurate measurements.

I think to get a better answer, you do need to specify some realistic observational constraints - such as (a) you cannot resolve the star at all, or (b) that you can resolve it, but observations can only take place from an earth-bound observatory (thus not allowing you to take flux measurements from the whole surface at once).

One thing occurs, is that in unresolved observations, even with an absolutely accurately measured luminosity (assuming isotropic radiation) there is still the issue of what radius to use. The radius at which the radiation escapes the star (at optical depth $\sim 2/3$) is ill-defined and wavelength dependent. An error bar of maybe tens of km is appropriate here, since atmospheres are 100-200 km "thick". For a solar-type star this would limit $T_{eff}$ accuracy to $\sim 0.1 K$ !


Are epicycles and mean Earth/Sun actually useful astrophysical concepts?

I'll say yes here, but only if the epicycles are used as approximations. Here's how they would be useful:

The eccentricity of the celestial body's orbit is low, i.e. $e<<0$. This means that it would be nearly circular, and epicycles could therefore be reasonably accurate. For example, it would not be a good idea to use epicycles to model Mercury's motion, because it has a high eccentricity. Comets would also be poor choices. Triton, on the other hand, has low eccentricity.
Only a vague approximation is needed. If you aren't looking for calculations with a high degree of accuracy, you could use epicycles. It would be a bad idea to use epicycles to put a spacecraft into a precision orbit, but a better idea to use them in a toy simulator. NASA, however, has higher standards.
You don't need to use general relativity. One of general relativity's most important predictions was the precession of Mercury's orbit. Over long periods of time, general relativity triumphs over classical theories when it comes to this accuracy. It would be hard to use epicycles to model this - although potentially possible.


Quick answer for now; the Earth's field has been more and more carefully measured by precise tracking of satellites over time. It stared with very careful visual tracking/timing and interferometrically tracking and recording the Doppler shift of the radio signals from late 1950's satellites like Sputnik and Vanguard and then many many others over the years, then on to pairs of satellites that track each others' distance GRACE and GRACE-FO in order to build an Earth Gravitational Model.

It's expressed as a bunch of coefficients of spherical harmonics. The monopole term is $GM$ the standard gravitational parameter, there is no dipole moment since the center is defined as the Earth's center of mass, and the quadrupole moment is expressed using $J_2$ and $J_{22}$, the first of which represents Earth's oblateness which is roughly a 0.1 % effect in low Earth orbit and is what makes Sun-synchronous and Molniya orbits possible.

The spherical harmonics of a given published model only work outside of (essentially) all of Earth's mass. Very near the Earth's surface (as opposed to in orbit) you would have to correct them for the shell of the mass of the atmosphere above you which is no longer affecting you as well as nearby mountain tops that rise above you.


Claim that 30-m class telescopes will have resolution far superior to Hubble: true?

Adaptive optics, as it says on the GMT website:

  One of the most sophisticated engineering aspects of the telescope is
  what is known as “adaptive optics.” The telescope’s secondary mirrors
  are actually flexible. Under each secondary mirror surface, there are
  hundreds of actuators that will constantly adjust the mirrors to
  counteract atmospheric turbulence. These actuators, controlled by
  advanced computers, will transform twinkling stars into clear steady
  points of light. It is in this way that the GMT will offer images that
  are 10 times sharper than the Hubble Space Telescope.

and on the TMT web site:

  In addition to providing nine times the collecting area of the current
  largest optical/infrared telescopes (the 10-meter Keck Telescopes),
  TMT will be used with adaptive optics systems to allow
  diffraction-limited performance, i.e., the best that the optics of the
  system can theoretically provide. This will provide unparalleled
  high-sensitivity spatial resolution more than 12 times sharper than
  what is achieved by the Hubble Space Telescope. For many applications,
  diffraction-limited observations give gains in sensitivity that scale
  like the diameter of the mirror to the fourth power, so this increase
  in size has major implications.


A neutron star without a flat surface and an atmosphere?

There is a number of issues with the question, but let me sketch out some kind of answer, so you get something out of it.

The atmosphere of a neutron star is a topic that's a bit speculative. Estimates vary a lot. Regardless, neutron stars can have an atmosphere - sure, gravity is huge, but they are also extremely hot. Some molecules are bound to jump up a bit.

X-ray measurements have shown the atmosphere of a certain neutron star to be 10 cm thick. That's not 10 meters, but it's better than nothing. Other neutron stars surely have different kinds of atmospheres, some thicker, others thinner. It looks like this is a matter for further research.

https://spacemath.gsfc.nasa.gov/weekly/6Page78.pdf

You say the relief on the surface could be up to 5 mm height differential. Earth has mountains 8.8 km tall, and ocean holes 11 km deep - that's a 20 km differential, more or less. That's 4 milion times greater than 5 mm. Keep in mind, the surface material on a neutron star also has different properties from the normal stuff on Earth, different composition, and it's under extreme compression and extreme magnetic fields, which give it different mechanical properties. So 5 mm is quite believable. However, that is an extreme, and most surface should be smooth as glass. Any surface irregularity should be unstable in the long run.

Depending on the model, in some cases, extremely hot neutron stars might have surfaces that are more like a fluid than a solid. Emphasis on "might".

None of the above is set in stone. Neutron star physics is an area of active and intense research.

I recommend the book called Dragon's Egg by physicist Robert L. Forward. It's sci-fi, but the author was a bona fide scientist with extensive work in the field of gravitational wave detection. In his own words, Dragon's Egg is "a textbook on neutron star physics disguised as a novel". It's fun to read, too, one of my favorite hard sci-fi novels.


Why do we have the cosmological constant?

Reason 1:

Let's look at the Friedmann equations without the cosmological constant.

$$  \frac{\dot{a}^2 }{a^2} = \frac{8 \pi G \rho}{3}-\frac{kc^2}{a^2}$$

The term on the LHS is just the Hubble constant squared $H^2$ which can be measured the direct measurement of recession velocity of galaxies

The density term can be said to be a combination of $\rho_{matter}+\rho_{dark- matter}$ both of which can be measured directly;$p_{matter}$ by observation of matter in our galaxy and other galaxies while $\rho_{dark- matter}$ by rotation curves of galaxies.

The curvature constant $k$ can be estimated today by the anisotropy measurements in the CMBR.

As it turns out the parameters don't fit and we need more mass-energy in the universe(almost 2-3 times of that we had estimated).

So comes along Dark energy or basically the cosmological constant. Cosmological constant or the dark energy are just two ways of looking at the equation,either as just a constant or a form of mass-energy(though we have solid reasons to believe the latter).

And this is our picture of the universe today:

Reason 2:

Now historically the cosmological constant was necessary for an altogether different reason.

The second Friedmann equation without the cosmological constant looks:

$$   \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) $$

Now this predicts for normal type of matter,the universe must decelerate.($\ddot{a}<0$)

Now,people measured the redshift of the type-1a supernovae and found out the quite paradoxical result that the universe was being accelerated in its expansion.

Since normal matter can't explain this type or behaviour,we again have to look at Dark Energy(or the cosmological constant).And so with the cosmological constant the equation becomes:

$$ \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} $$

Thus $\ddot{a}>0$ is possible.

Therefore the cosmological constant is necessary to both explain the current rate of expansion and the accelerated expansion.

So finally the accelerated expansion can be explained and today we have the $ΛCDM$ model of the universe.

References:

1:http://en.wikipedia.org/wiki/Friedmann_equations

2:http://hyperphysics.phy-astr.gsu.edu/hbase/astro/univacc.html

3:http://en.wikipedia.org/wiki/Lambda-CDM_model


The inner planets are not resonance orbits with each other,  this means that the ratio Earth-year:Martian-year is irrational. And the exact postions of the planets don't recur.
The positions do approximately recur, but the time between approximate recurrences depends on how accurate you want the recurrence to be.  "accurate to 90 degrees?"  "15 degrees?"  "1 degree?"  Are you only counting right-ascension, or are you also looking at the declension too (the planets don't orbit in exactly the same plane)  Are you willing to make Keplarian approximations, or do you want to include perturbations.
The motions of the planets are simple enough physics, but not simple ratios.  So predicting a "recurrence to a particular accuracy" needs physics to forecast.


Would the dynamic pressure of the Sun’s solar wind be strong enough to trap plasma ions within an open container?

So you basically said that we could use the Sun to propel a rocket. We already have rockets like this equipped with solar sails. These solar sails have a "sail" that catches the stuff thrown out of the Sun. I'm no expert on this way of travel, but that is the basic description of how it works.


Why the chosen orientation for the M87 black hole image?

North is to the top and East is to the left. I believe it is a standard image in celestial coordinates.

It is important to standardise so that the image can be compared with large scale features like the M87 jet.


Why is the speed of light 299,792,458 meters/sec?

Speed of light being finite is one of the fundamentals of our Universe.

If it were infinite, this would have a major implication in causality.

Besides, in non-quantum physics, light is just an electromagnetic wave. Eelectromagnetic field is described by Maxwell's equations, which predict that the speed $c$ of electromagnetic waves propagating through the vacuum depends on the dielectric permittivity $ε_0$ and the magnetic permeability $μ_0$ by the equation $c = {1\over\sqrt{ε_{0}μ_{0}}}$ so you can not have an infinite speed of light unless electric permittivity or magnetic permeability were zero, which in turn would cause all sorts of odd things to electromagnetical attraction (and thus, to matter existence beyond elemental particles).


if the universe is infinite shouldn't it have experienced an infinite number of universe ending events?

The universe has a finite age, about 13.7 billion years. So there has not been time enough for infinite events in our past light-cone.
There are rather few events in mainstream physics that are universe ending. Closed universes eventually implode, and if certain forms of dark energy exist they can cause a Big Rip where the universe expands to infinity in a finite time. But obviously observers cannot look back on them in their past. This is also true for other local disasters like false vacuum decay.


Do all moons orbit their planets on the ecliptic?

Of course not.

See Neptune's moons. Triton's orbit has a tilt of 157º from Neptune's rotation plane, which is in itself tilted 30º from Neptune's orbit, just 2º away from the Ecliptic.

Or look at http://en.wikipedia.org/wiki/Moons_of_Uranus


Is it possible to use a laser to propel a spaceship?

Yes, it might be possible to send laser-propelled spaceships to the stars.
There are a whole series of technical challenges in the laser technology and the materials that such a spacecraft might be made of.
The basic idea is to accelerate mini-spaceships with very low mass, but a large, reflective surface area, up to relativistic speeds. The project that I know of is called "Breakthrough Starshot".
Contrary to what you say in your question, the laser would not be aimed at the target star, it would be aimed at the spaceships! These would have to be set off on a carefully calculated trajectory, because during their decades-long journey, the relative positions of the Sun and target star would change significantly. The impulse phase would take place within the Solar System and the craft would cruise most of the way.
The intensity of the lasers required would be immense compared to the light from a star. The baseline plans involve something like $10^9$ W/m$^2$ delivered to the light sail for 10 minutes. This is to be compared with about $10^3$ W/m$^2$ from the Sun at the orbit of the Earth.


Yes, all of those contribute to the total irradiance, which is the amount of sun power falling on a particular area, measured in Watts per square meter. You can imagine a 1m$^2$ "window" perpendicular to the sun's rays - excluding atmospheric and weather effects, the amount of sunlight passing through that window never changes. But when the sun is directly overhead, that window illuminates a 1m$^2$ patch of ground, while later in the day that same sun power is spread over a larger area. Only the angle of the sun affects how big that area is, and that is determined, as you point out, by time of day, season, and latitude.

Because of this, the length of your shadow is a good tool to estimate the angle of the sun and your risk of sunburn. I tend to be a bit sun-sensitive, but if my shadow is longer than I am tall (i.e. the sun is below 45 degrees), I'm usually fine - that's most of the day in winter, but excludes the hours around noon in the summer. That angle will of course have to be adjusted based on your personal sensitivity.

As for your experience, I would imagine that people tend to expose more skin in the hottest month of the year (August), increasing their sunburn risk over a period with higher irradiance but less exposure (June). Sunburns also will also often show up in the afternoon because they're a result of accumulated damage throughout the day. If an individual takes X hours of exposure to burn, they're more likely to hit that mark later in the day, rather than during the peak of irradiance.


Turn celestial coordinates into light years

Rectangular celestial coordinates are used regularly. The most common examples are the JPL ephemerides for solar system objects. Wikipedia's page on geocentric equatorial coordinates includes the equations required to convert right ascension and declination to xyz.

The most commonly used reference system for astrometry is the ICRS coordinates that are closely aligned with the J2000 equatorial system. Implementaions of ICRS were used for the Hipparcos and Gaia astrometry data.


Extragalactic Magnetic Fields - General Information That I'm Missing

On this question, the Widrow (2002) review is still pretty up-to-date. You can also have a look on the introduction of Beck (2013), a recent PhD thesis on the subject with a straightfoward introduction to the subject.
The main questions remain the same for decades now:

What is the origin of the magnetic seed fields?
How to amplify magnetic field up to observed values in the local Universe?
How to get large scale magnetic field, in particular if the magnetic seed fields are produced locally?

Origin of the magnetic seed fields
Where do we start? That is no easy question, with a diversity of answers. One of the most standard ones is the Biermann battery you mentioned (one significant problem with Biermann batteries is the timescale needed to amplify such small fields; see below) in the early Universe. Another one is that the magnetic field can be generated at small scales by first generations of AGN and supernovæ and then distributed at larger scales (dynamo can help but distribution at larger scale can still be an issue; see below). Lastly, you can have some exotic processes to generate seed fields in the first instants after the Big Bang, typically linked to inflation processes and such.
Magnetic field amplification
If there is an initial magnetic field — which is not so easy to demonstrate — there are few mechanisms that could amplify magnetic fields efficently enough to explain the observed magnetic field. The most standard one is the $\alpha-\omega$ dynamo, combining a twist of the magnetic field lines in a differentially rotating disk with turbulent motions in the ISM that allows a regeneration of the magnetic fields (supernovæ explosions, stellar winds and other instabilities could do the work). Another possible mechanism is the magneto-rotational instability (MRI), that can easily generate a significant, large-scale field from almost any given seed field. There are other candidates such as Parker instability, or other flavors of dynamos.
A central question then is of timescale: if these mechanisms can generate magnetic fields, the needed timescale to amplify a seed field to the current value may be far larger than the age of the Universe.
Distribution
Magnetic field is everywhere, not only in the ISM and at larger scale in galaxies, but also in galaxy  clusters, in the IGM and so on. If the origin of the magnetic seed fields is of cosmic origin (exotic processes in the early Universe), you can explain the observed distribution; if not, things get tricky again. Turbulence, galaxy collisions and mergers may be of help, but the question is still largely open.


How fast does a spacecraft have to be to enter a primordial black hole without being torn apart?

Let's try a crude back of the envelope calculation to get orders of magntitude:

The tidal force near the event horizon is of the order of $G M/r^3$ which is something like $10^{18} g/m$. So a probe with a mass of 1 gram and diameter of 1cm would experience about $10^{12} N$ of force trying to "spaghettify" it by accelerating ends of it at about $10^{16} g$ relative to the centre. If it experienced that force for a time $t$ the ends would likely move about $10^{17} t^2 m$ relative to the centre, assuming $t$ is short enough that relavity does not complicate things, so if we want to limit that to say 1mm we need $t < 10^{-10}$. So basically we want the probe to go from a few Schwarzchild radii down to 1 in about 100ps in its internal time frame, so that the tidal forces do not have time to tear it apart. Using the Newtonian approximation this is a few times faster than the speed of light, so what we learn is that the probe must be going at relativistic velocities, to have a chance of surviving. From its perspective, that flattens out the gravitational field around the black hole in the direction of travel, so that the tidal force seems to even briefer in duration.


How many observations does it take to determine the orbit of a TNO

The main problem with determining an object's orbit is we only know the position with certainty in two dimensions.  The distance to the object is largely unknown.  This accounts for the large uncertainty in the period of newly discovered TNOs.  Many possible orbits could fit the early observations, and therefore the uncertainty is large.

As time goes on, and more observations are made, more and more of these possible orbits are excluded and the orbit fit becomes more exact.

So to answer your question, it really depends on how exact you want the answer, and how good your observations are.  The more observations you take, and the better they are, the closer your orbital fit becomes.  And more importantly, the bigger the arc of the orbit your observations cover, the more accurate the orbit becomes. It is impossible to give a specific number.

To give a practical example, Pluto takes 248 years to orbit the Sun.  It was discovered in 1930, so we've only observed it for 85 years, or about 1/3rd of one orbit.  One of the problems the New Horizons probe sent to Pluto had to overcome was that we didn't know, at launch, exactly how far away Pluto was, so therefore we didn't know exactly when New Horizons would pass the dwarf planet. That made it difficult to tell the probe where to point to take pictures at closest approach.


We are in Green Lake, WI, 8/12/18, seeing bright orange “planet”/“star” almost due South, perhaps 20 degrees above horizon, what is it?

Not all sky apps get DST right or compute planetary positions accurately.
Stellarium at your location shows Mars 17 degrees above the SSE horizon at 22:15 CDT, or 20 degrees above due south at 23:45.
Skymaps.com northern edition also shows Mars above the horizon before 10 PM in mid-August.
As Mars is unusually close to Earth this summer, it appears brighter than usual--and orange as usual.


Information about known objects in the Universe

Here's an API: Documentation for Catalog/Image webservices provided by CASG/MAST.

You can also find more search resources in Github, e.g. HackTheUniverse.


Existence of planets larger than their host star?

Since the smallest stars are still the size of gas giant planets, the question ends up coming down to whether gas giants exist around stars at the bottom of the main sequence. Close-in gas giant planets are rare around low-mass stars, though there do seem to be long-period ones. This means the largest planetary radii for the systems in question are going to be similar to Jupiter, rather than inflated hot Jupiters. An exception would be the case of very young systems before the planets cool and shrink but in that case the star would also still be contracting so you probably don't win there.

A problem is that these stars are extremely faint, so the radial velocity method is tricky - this may change a bit once more RV instruments that operate in the infrared (e.g. the Habitable Zone Planet Finder) come online. The long orbital periods for giant planets around these stars would also require longer observation times to make a detection. Unfortunately the long orbital periods would make transits unlikely, so most probably we wouldn't be able to determine the planet's radius and would not know for certain that the planet is larger than the star.

Direct imaging has spotted a few objects of a few Jupiter masses at fairly wide separations from objects near the hydrogen-burning limit, e.g. 2MASS J02192210-3925225 with an object at the deuterium-burning limit is located about 150 AU from a 0.1 solar mass star. It's not entirely clear what to call these objects and they may be very low-mass brown dwarfs rather than planets. Furthermore these systems are so young that the stars have not yet contracted to their main sequence radii. For low-mass stars this can take several billion years, by which time the planets will have cooled off and become much fainter (and less detectable). These kind of wide-separation systems may also end up being disrupted by stellar encounters.

The other approach that works for detecting these kind of systems is gravitational microlensing, which tends to find objects near the system snowline, i.e. on scales more similar to our planetary system. An example of the kind of system that might have a planet larger than its star is KMT-2016-BLG-1107Lb, where the parameters suggest a ~3.3 Jupiter mass planet orbiting a ~0.087 solar mass star at ~0.34 AU. Unfortunately the uncertainties in the parameters are typically large because the lens systems are usually invisible. This means we also don't have radius information, so we can't say for sure that this system definitely has a planet larger than its star.

So it looks like systems do exist where a planet can be larger than the main-sequence star it orbits, though so far there is no confirmed case due to the difficulty of making the necessary observations.


Central temperature and pressure for constant density vs linear density model of stars

Because the total mass of the star has to be conserved.

If you consider a linear density model, for a star of fixed mass and radius,
you have to make the central density higher than the average density in a constant density model.

Furthermore, because the mass of each shell goes as $\rho r^2$, there needs to more shells with a. density higher than average. This means the average density of a mass shell is higher than in the constant density model and more of the mass is at smaller radii.

Hydrostatic equilibrium demands that the pressure gradient is proportional to $-\rho M(r)/r^2$, where $M(r)$ is the mass inside radius $r$. At any radius, this will be larger on average in the linear density model because more of the mass is at smaller radii.

If the pressure gradient is higher, then the central pressure will be higher. The value of the central temperature then just comes from gas laws.


How can I find out which known stars lie within or close to M8

Have a look at the following references - found relatively easily using the following query in the NASA ADS site.

UBVRI and Hα Photometry of the Young Open Cluster NGC 6530
This paper by Sung & Bessell (2000) contains an extensive photometric catalogue. The main table can be queried interactively.

The Lagoon Nebula and its Vicinity
This is a review article by Tothill et al. (2008), from which you can get pointers to most of the significant literature (pre-2008) on the region.


Luminosity Schechter function for galaxies

Total Luminosity with $\alpha=-1$:
$$
\begin{aligned}
L_{tot}&=\int_0^{\infty}L \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\
&=\int_0^{\infty}\Phi_0 e^{-\frac{L}{L^*}}dL
\end{aligned}
$$
Now put $L'=L/L^*$ and you get:
$$
L_{tot}=\Phi_0L^* \int_0^{\infty}e^{-L'}dL'=\Phi_0L^*
$$
Total number:
$$
\begin{aligned}
N_{tot}&=\int_0^{\infty} \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\
&=\int_0^{\infty} \Phi_0 \frac{1}{L}e^{-\frac{L}{L^*}}dL
\end{aligned}
$$
which diverges at the lower limit.

Observe with $\alpha=-1$ we have no gamma functions.

Also IIRC with $\alpha \in (-1,0)$ when computing $L_{tot}/N_{tot}$ the gamma functions cancel leaving $(1+\alpha)L^*$