It's a pretty common tendency to fix problems by buying more eyepieces, but that's rarely how it works in practice. There are a few steps you need to take before buying more glass.

Collimation

Check collimation first. This should be your first item on the list, full stop. The performance of the scope decreases rapidly when it's out of collimation (for solar system objects). Check the documentation that came with your scope. Also check these links:

http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/

http://garyseronik.com/a-beginners-guide-to-collimation/

If the scope is not in good collimation, a new eyepiece is a waste of money. All the eyepiece does is magnify the real image generated by the scope in the primary focal plane - so if that image is out of whack, then no eyepiece will give you the performance you're looking for.

Check collimation every time you use the scope. With practice, it will end up taking only a couple minutes.

Seeing

It varies with time and location. Sometimes it's better, other times it's bad. Go on the CDS site, search for your location and use the Seeing forecast; good seeing is represented by dark blue, bad seeing by light blue or white.

http://www.cleardarksky.com/csk/

Plan your observations according to the forecast.

Seeing only affects fine-detail objects like the Moon or the planets. It does not affect faint DSOs (galaxies, nebulae, etc) because your eye does not resolve fine detail on DSOs anyway.

Thermal equilibrium

If the primary mirror is not at the same temperature as the air, "local seeing" conditions (vortices of air on the primary) will blur the image. Take the scope outside 1 hour before you start observing and let it acclimate.

For a scope 8" or bigger it's often a good idea to install a fan on the back of the primary to accelerate the process - but get everything else done first.

http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/

Light pollution

Only affects DSOs (galaxies, nebulae, clusters). It does not affect solar system objects such as planets. The only remedy is to load the scope up in a car and drive away from the city.

However, all Messier objects are visible in most amateur telescopes even from very light-polluted places like major cities. They just don't look as spectacular as from a dark sky place.

Finding focus

Make sure the scope is actually focused properly. This might be more tricky than it seems. Use a star nearby and adjust until the star is as sharp as possible.

You could also make or buy a Bahtinov mask, which will give you proof positive of perfect focus (adjust focus with the mask on, until all 3 lines intersect each other in the center).

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

https://www.astrozap.com/scripts/prodList.asp?idCategory=74

With all of the above taken care of you should definitely see Jupiter's equatorial belts. They become visible even in 2" of aperture, and a slightly larger instrument should show them pretty clearly, even in a modest eyepiece. Under good seeing, with a perfectly collimated instrument, you can even start noticing the curls and the vortices in the belts even with an aperture as low as 4" ... 6".

Jupiter getting more and more blurry as you increase magnification is normal. It's not a sign that the eyepiece is bad. What's not normal is not seeing the belts in a 10" scope. That indicates that either collimation is bad, or seeing is exceptionally bad, or both; typically it's collimation.

Your eyepieces should actually give decent performance in the center of the field of view.

You also need to let your brain-eye system adapt to the image. You don't have to observe from a very dark spot, Jupiter is bright enough to not require perfect dark adaptation. But your brain and your eye need a bit of time until they can extract all details from the image. Let it drift through the field, stay relaxed, follow the image and just take it in slowly. You will start seeing details after 3 or 4 minutes that were not visible at first (assuming good collimation, seeing, etc).

Now on to the eyepiece. Your instrument is a f/4.7 newtonian, 1200 mm focal length. A 10 mm eyepiece will give you 120x magnification. This should not be too much even under bad seeing - provided the scope is collimated.

I've observed Jupiter a lot at 1200 mm f.l. in a smaller instrument. I've found that a 8.8 mm eyepiece (136x) was a reliable workhorse, never needed anything with less magnification. When seeing got better, I could push it up to 6.7 mm (180x) or even 4.7mm (255x). But speaking in general, Jupiter is low contrast and does not benefit from pushing the magnification too hard.

Mars OTOH does very well under very high magnification. But that assumes your collimation is top-notch and seeing is very good.

Provided that you take care of collimation, etc first, you could try and get better eyepieces. The ones that came with your scope are not that bad actually, they do fine in the center, it's the edge performance that's weaker. But if you do get new eyepieces, you might as well get something better. I've found that the Explore Scientific 82 degree series provides a good bang for the buck - not too far from the performance of top glass, but much cheaper than top prices.

https://explorescientificusa.com/collections/82-series-eyepiece

But the options here are very numerous, so do your own research.

You might also benefit from having a barlow, maybe a 2x or a 2.5x. However, the much higher magnification generated with the barlow will not work unless collimation, seeing, etc are all taken care of!

It's a little difficult when you juggle collimation and seeing at the same time, one you can control (collimation), the other you can't (seeing), but realize that this will take some time to figure out, so go one step at a time.

Also understand that this is not the kind of problem that is solved by simply throwing money at it.


What is the difference between wind, jet and outflow?

I'm not sure there are very clear official definitions, but here is my take on accepted usage in the star formation community:

Outflows - general term describing both winds and jets.

Jets - collimated outflows along magnetic poles.

Winds - outflow from the stellar surface in all directions, not just along the poles.

Stellar winds are what comes from individual bright stars. Jets can be protostellar jets from young stars, or on a larger scale from galactic nuclei.


Gravitational lensing is the effect that large amounts of gravity have on the path of light. Here's an example of the effect:

The animation shows a black hole passing in front of a galaxy (simulated). In theory any object with a surface mass density larger than the critical surface mass density would be able to produce similar such effects.

What's happening is that light goes in a straight line along the spacetime fabric. When a very massive object is in the fabric, it bends the fabric - since the light follows a straight line on the fabric, the line is also bent. This means that the light goes in a not-straight line, and thus a lensing effect.


How does a planet-sized robot standing on a planet affect its gravity?

Of course it would have had some effects! Gravity obeys the mathematical function:
$F = {G\ m_1\ m_2\over D^2 }$
where G is the constant of gravity (approx. $6.67430 \times 10^{-11} N m^2 kg^{-2}$), $m_1$ and $m_2$ are the masses of each body, and $D$ is the distance between each body (center-of-mass to center-of-mass).
Supposing the robot’s mass is about as much as the planet’s mass, it would definitely have thrown it off-course as it approached the planet—but, then again, the planet would also have affected the robot’s course.
It would not, however, have “affected the planet’s gravitational forces,” as the gravitational force exerted by a body depends only on its mass.
I would presume, though, that there are,  on planet Cybertron (though we can’t see them on this image) bodies of water, or simply loose objects. These would have been affected by Unicron’s presence/approach, and would have been somewhat displaced by his presence/approach, as there was a force (basically) as strong as Cybertron’s gravity pulling them in the direction where Unicron was at that moment. Such objects on the opposite side of Cybertron may have been crushed (if they were not made very resistant), for example, and such objects on the side of Cybertron facing Unicron would have “lost their weight” (at least partly) and—no, not float—being easier to displace…
A lot is not know in the context of your question. Maybe reading the comix where this comes from would help know Unicron’s trajectory, etc., but I’ll leave that for others.
Be well!


How did LIGO verify that the gravitational waves originated 1.3 billion years ago from two specific black hole collision?

The original paper states the methodology used to determine the location of the gravity waves. The bottom line is that they were detected in two different locations. Using the difference in timing, one can estimate the location in the sky to an area of about 600 degrees squared where the origin of the wave could have been. The two signals had an offset of 7 ms. Pure speed of light differences would have been 10 ms. This can be used to effectively identify a circle in the sky where the merger might have happened. Of course, this doesn't reveal the distance.

The distance was figured by a number of things. First of all, the frequency pattern was used to estimate the mass of each black hole. Using these masses, one can find the estimated total power of the event. Using this estimate, and seeing the estimate of the amplitude from the signal, one can use the inverse squared law to estimate the distance to said event.


Does the milky way have relativistic mass against galaxies which are moving away from it at high speeds?

What the mass increase from special relativity tells you is that if you see a physical system (say, a rocket with a physics laboratory on board for convenience) flying past your location at a certain relative velocity and you measure the mass of an object in that laboratory, for instance by measuring its gravitational effect on another mass in the same laboratory (using lasers to measure distances and velocities and so on), you will get an answer which is higher than the one an observer in that laboratory would get.

In and of itself this doesn't even tell you much about what you'd measure about a far distant laboratory, especially one far enough away that the expansion of the universe significantly effects your measurements.

Furthermore, since these two galaxies on opposite sides of the observable universe cannot actually  exchange messages and observe one another, the mass increase is not really a consideration -- you can't measure the mass of an object in the other laboratory, because you can't get a signal there and back.


Data Analysis: Image Sharpening through Posterisation

1) your original image is blurred and possibly overexposed. I can't tell if the blurring is due to poor focus or inaccurate tracking. I think you really need much better, sharper images that show the true boundary of the planet before you can start to measure diameters.

2) with a sharp image, there should be no need for posterising the image.

3) "Is there a reason why the middle posterised object gives the most accurate value" - well we could claim that it marks the "average" estimated diameter, but in my opinion the answer is: coincidence.  If you posterised it to give, say, four bands instead of three, which would you take as your "true" diameter? And without knowing the real diameter of Jupiter in advance, how would you ever know?

Edit: by the way, I noted your previous question on this subject. Did you read and use those answers? How sure your magnified image is really the size you think? You could try to verify this by photographing a double star of known separation...


Is Cosmos seriously using that exact number? Egads... if they are, don't take it too seriously, but otherwise they're probably conceptually correct.

  How do we know it was dark energy?

In cosmology, the ΛCDM model fits numerous observations, most notably those observed by the WMAP satellite, but also others, to the Friedmann-Robertson-Walker family of solutions of general relativity, which describe a homogeneous and isotropic universe. The $\Lambda$ refers to the cosmological constant, which is the simplest model of dark energy, and is equivalent to an intrinsic energy density of the vacuum--hence, it must be Lorentz-invariant, and since time and space mix in a Lorentz transformation, the only way for that to be so is for its pressure to be equal to exactly the negative of its energy density. With $\Lambda>0$, the pressure is negative, and this is what's responsible for the outward gravitational acceleration of the universe.

  Given we can't define or detect dark energy, why do we believe this is the cause?

The cosmological constant was defined eight decades before the discovery of cosmic acceleration, and we believe it to be the cause because it's theoretically very simple and it fits observations. There are other models for dark energy, though.

  How do we even know this new acceleration happened?

Cosmological evolution is described by the Friedmann equations, which has the principal contribution of radiation, matter, and dark energy, in addition to spatial curvature of the universe. As the universe expands, both radiation and matter get diluted, but dark energy does not. Therefore, if the universe keeps expanding, the proportion of dark energy becomes dominant over both radiation and matter.

Most likely, the article is simply an exaggerated reference to the transition between the matter-dominated era and the dark-energy-dominated era. The particular number $6,771,500,000$ years seems comically precise for much of anything in cosmology, but since the nine-year best-fit WMAP+eCMB+BAO+H₀ (see above WMAP link) for the age of the universe is $13.772\pm0.059\,\mathrm{Ga}$, I propose the following explanation:

The author hears that the age of the universe is $13.772$ billion years, and completely ignores any error bounds on that estimate.
The author hears from somewhere else that the dark-energy-dominated era began when the universe was around $7$ billion years old, and again treats this number as exact.
The author then subtracts the two.

This accounts for all but the last $0.5\,\mathrm{Ma}$, which about one-ten-thousandth. I conclude that my theory of the origin of the number agrees rather well with the empirical evidence of the author's words. Perhaps the $0.5$ as the next digit was rationalized as some sort of midway-point in the author's mind.

As to where $7$ billion years comes from, I have two guesses.

In cosmology, the Hubble parameter is sometimes taken to be unitless, by scaling to $100\,\mathrm{km/s/Mpc}$. For example, we sometimes say that $h = 0.704$ when we mean that $H_0 = 70.4\,\mathrm{km/s/Mpc}$. Thus, $100\,\mathrm{km/s/Mpc}$ is kind of a de facto standard of comparison. It might be a coincidence, but it's rather suspicious that if one ignores the contribution of radiation (which is justifiable, since it's negligible except in the very early universe) and pretends that $H_0 = 100\,\mathrm{km/s/Mpc}$, then the age of the universe when the matter and dark energy densities are equal would be
$$\frac{2\ln(1+\sqrt{2})}{3\sqrt{1-\Omega_\text{M}}}\frac{1}{H_0}\bigg\vert_{\Omega_\text{M} = 0.2865} \approx 0.70H_0^{-1}\bigg\vert_{h = 1.0} \approx 6.8\,\mathrm{Ga}$$
I'm using the Ryden reference given in wiki for matter-dominated era above. Thus, I think someone did a guesstimate using a nice, round number for the Hubble parameter and rounded to $7$ billion years.
The transition between matter-dominated and dark-energy-dominated is a somewhat arbitrary matter of convention. We could take it mean when matter and dark energy densities are equal, as above, but on the cosmological scale, pressure is three times as important as energy density (cf. Friedmann equations). Thus, we could take it instead as the time when the matter density was twice the dark energy density. This would be earlier than the $9.8\,\mathrm{Ga}$ given in the Ryden reference, but arguably more appropriate because the Friedmann equations would predict zero acceleration, i.e., it would be the transition between inward and outward acceleration. I haven't calculated when that would be, but this sort of conventional fuzziness is might be responsible for the discrepancy between the wiki's matter-dominated era and the dark-energy-dominated era pages. Although the latter claims $5\,\mathrm{Ga}$ instead, rather than $7\,\mathrm{Ga}$ (but it doesn't explain the source).

So Cosmos may or may not be right about dark-energy-dominated era beginning when the universe about $7$ billion years old, but concluding that it happened $6.7715\,\mathrm{Ga}$ ago is certainly taking some numbers as too exact to be justifiable.

  Something in the red shift?

I think at this point you're better off making a separate question on specific issues with the ΛCDM model, if you have any. For the purposes of this question, I'm taking ΛCDM as a given, as almost certainly Cosmos is doing as well.

  Yes that is a lot of questions, but the main question is how do we know an accelerated expansion happened at that time?

The standard cosmological model is that dark energy is a cosmological constant. Thus, its energy density does not change. It then becomes straightforward that at some point, cosmological expansion makes matter too dilute to compete with dark energy, which is when dark energy starts dominating. It was there all along, but before then, matter was dense enough to be dominant instead.


Would dark energy save the earth for a while as the sun heats up?

I don't think this is a bad question, and I think Pela is essentially correct, the effect is very small.

Dark energy doesn't expand objects, for example, cause those are bound together, but dark energy should expand space between the sun and the earth though the effect is tiny.

If we take this number, 68 KM per second per 1 million light years (link below), that's 245,000 KM per hour per million light years or about 1 KM every 4 hours per lightyear.   With 63,000 Astronomical units per light year, between the sun and the earth, we can expect the earth (assuming it's in a perfectly stable orbit around the sun - which it's not, but lets say it is), a velocity of 1KM/30 years.   And that's over the age of the universe.   The earth/sun is only 1/3rd the age of the universe, so dark energy has accelerated the earth maybe 1 KM/90 years velocity since the formation of the solar system.   That's a pretty small push.  A velocity change of a bit a meter per year over the age of the solar-system.   The effect of the solar wind and the Jupiter and Venus orbital tugging are probably greater than that.

Source:  http://www.space.com/26279-universe-expansion-measurement-quasars-boss.html

Now, over a few billion years, a small velocity change of 1 meter per year would add up - so I think dark energy is part of the solar system's long term orbital calculation.  I might be wrong on that, but I think it has a measurable effect over billions of years even in distances as small as planetary orbits.

But would it save the earth - I doubt it.  The sun's heating up accelerates over time.   It will at some point in the future, move too fast and get too hot for the earth, so, we'll have to save ourselves, either by moving the earth or terraforming mars, or building space colonies or some combination of all 3.   The good news, we have tens of millions, maybe hundreds of millions of years to figure that stuff out.

Dark energy has been useful though.   It's probably kept us from crashing into Andromeda and/or the great attractor - however far away that is.   Certainly delayed those collisions (We're never going to crash into the great attractor, I gather, thanks to dark energy), and crashing into Andromeda, when it happens, might harmless anyway, but I'm just kind of thinking out-loud.

I do think it's a good question as there probably is some small expansion due to dark energy between the earth and the sun.   If my numbers are wrong, I welcome correction.


Identifying common galaxy spectral lines

You need to compare it with the spectrum of a similar galaxy at a known redshift, that would probably enable you to identify features with known rest wavelengths.

If you can find such a template, then the best way of estimating a redshift for a galaxy spectrum like this, consisting of mostly weak and blended absorption features, is to cross-correlate your template galaxy with this one. The cross-correlation "lag" gives you the difference in redshift.


Estimating the abundance of type-Ia supernovae using magnitude limited survey

The rudiments of a solution would be take your apparent magnitude limit and work out how far away you can see type Ia supernovae and how far away you can see core collapse supernovae.

Within the latter distance, 30% of the supernovae would be type Ia. Beyond that distance, all of the observed supernovae would be type Ia.

You then need to be assuming something about the density of supernova progenitors and thinking about any spread in their absolute magnitudes. And you didn't mention any observational uncertainties.


Calculating the Diameter of Jupiter through Image

Your barlow is marked "2x" but I'll bet it isn't.

How is your camera mounted on the barlow? More importantly, what distance is between the camera's focal plane and the barlow?

An experiment to try (or at least to think about).  Get an eyepiece extension tube (or improvise something) and put it between the camera and barlow, increasing the spacing. You will have to refocus a bit and you will get a different (I guess bigger) image size, even though the barlow is still "2x". To get an exact "2x" barlow you might need a particular spacing, and eyepieces are normally closer to the eyepiece than cameras because of the extra bulk of the camera.  (I think this is the most likely cause of your discrepancy; your camera is currently mounted further back than an eyepiece would be so projection is making the image larger.)

A second experiment to try: without the barlow, photograph the moon and see how close you get to its expected size.  If it's close, then our assumptions about the barlow are in question.

Third thing to try: are you absolutely sure about the distance from Earth to Jupiter? (It's probably OK but I'm always suspicious about online calculators.)


What shape are black holes theorized to be?

The event horizon would be spherical. What is usually depicted in drawings is the accretion disk, where all the debris gather before entering the event horizon.


What is the highest recorded temperature in space

Well this infograhic puts gas heated by a supernova at about 55 million Celsius.

And that's not even the highest temperature we can just "find" out there, even if we discard supernovae and stellar cores.  The gas in a cluster of galaxies (the ICM) can have temperatures on the orders of 10 million to 100 million Kelvin.  Such a high temperature gas emits high energy photons such as X-rays via bremsstrahlung radiation, and these are observable (and have been observed) by X-ray telescopes.

There's this review article from 2003 that covers the basic methodologies and measurements made on ICMs, including (but decidedly not limited to) temperature.


Is there online data on asteroid axial tilts?

From this article, http://hubblesite.org/pubinfo/pdf/2010/33/pdf.pdf, here's the rotation axis for 4 Vesta: RA=307.5°±3.1°, Dec=43.1°±1.2°. Update, per TildalWave's comment: RA=305.8°±3.1°, Dec=41.4°±1.5°

Rotational motion of Ceres.

An old article (1986): Magnusson, P., Distribution of spin axes and senses of rotation for 20 large asteroids, Icarus 68:1 1-39 http://www.sciencedirect.com/science/article/pii/0019103586900722


Does the sun blow a bubble into the interstellar medium?

At the moment, the Solar System is inside the Local Interstellar Cloud (which is itself within the Local Bubble). The LIC has a neutral hydrogen number density of $\sim0.1\text{ cm}^{-3}$, which is a bit lower than the mean galactic neutral hydrogen number density. The Local Bubble itself is less dense than the LIC, by about one or two orders of magnitude. On the whole, this region of space is less dense than the ISM, on average.

So, if we take atomic hydrogen density to be a good benchmark, then the LIC and surrounding regions are in fact less dense than the Solar System on average.


Shouldn't all moon orbits be inherently unstable?

The answer to your simple question is that while a circular orbit could remain in the absence of any other gravitational fields, in reality you'll have an ellipse and it can be stable from the beginning, remember stable has a wide range of values.

If you allow drag through tidal forces etc then the orbit will change, eg our Moon is slowly getting further away from us, but we still consider this stable over realistic timeframes.

Additionally, the actual orbit will track out a sort of helix as the planet travels round the sun.


What causes celestial bodies to move like they do from Earth's surface?

The apparent motion of planets is complex but predictable.

The apparent motion is due to the combination of three different motions:

The rotation of the Earth,
the motion of the Earth around the Sun,
and the motion of the planet around the Sun.

The Planets move in ellipses, slightly perturbed by other planets.

These motions are predictable, and their combination is straightforward trigonometry. The resultant motion is complex, the planets appear to loop relative to the stars, but entirely predictable. We know exactly where the planets will be in the sky for any time for many thousands of years in the future or the past.


Is the curvature of Mars visible from the top of Olympus Mons?

The impression I get from the Wikipedia page on Olympus Mons is that from the top of the volcano, you can't even see parts of Mars that aren't Olympus Mons. Quoted from above:

Because of the size of Olympus Mons and its shallow slopes, an observer standing on the Martian surface would be unable to view the entire profile of the volcano, even from a great distance. The curvature of the planet and the volcano itself would obscure such a synoptic view. Similarly, an observer near the summit would be unaware of standing on a very high mountain, as the slope of the volcano would extend far beyond the horizon, a mere 3 kilometers away.

Emphasis mine.
Wikipedia sources the bold line to this source in the Way Back Machine, which works out the math. Near the bottom of the page, there is a section starting with the phrase:

Years ago someone asked me if a person could see the horizon of Mars from the top of Olympus Mons. That is not as simple as one may think. If Mars was a prefect sphere and the slopes of Olympus Mons were smooth then the distance to the horizon could be calculated fairly easy. However, the surface of Mars has hills and valleys, and Tharsis bulges out from the mean spherical surface of Mars several miles, so one must consider that as well.

That's pretty much your question. The conclusion is:

Second, if this same [6 foot tall] person stood at the top of Olympus Mons (88704 ft) we would find that he or she could see a lot further to the horizon, some 265 miles away, if nothing was in the way:
$\cos A = (11,121,762.53 / (11,121,762.53 + 88704 + 6)$
$A = 7.2^\circ = 0.126\ \textrm{radians}  \leftarrow (0.126 \times 11,121,762.53 ft) / 5280 = 265 \textrm{miles}$
At those angles a person 6 feet high would be looking down at an angle of 7.2 degrees in order to see the horizon of Mars, so they would surely look down into the hump positioned at 50 miles away or at the 2.5-degree grade of the slope instead of over it to the horizon.

So it would seem that the horizon, as viewed from the top of Olympus Mons, would still be part of Olympus Mons. Indeed, you wouldn't even be able to see most of Olympus Mons from the top.


How do I get from Star Atlas to real life?

The Right Ascension of a star is fixed (ignoring precession of the equinoxes). What you need or want to know is the "sidereal time" - that is to say the "time" based on which RA is at your local meridian at a given time. The sidereal day is slightly shorter than the mean solar day (ie the 24 hour day) and so sidereal and solar times drift apart - only being equivalent at the vernal equinox.


What is the gravitational force between us and Andromeda?

The masses of the Milky Way and Andromeda (depending on exactly whay you consider "part of" each) are estimated as about 850 Billion solar masses and 1.7 Trillion solar masses respectively. Their mean distance is about 2.54 million light years, so it's easy to calculate the force using $$F = GMM'/R^2$$ We get a force of $$6.6\times 10^{29} N$$.


Does interstellar cloud obstruct or reduce the visibility?

Yes it is possible that a cloud reduce the visibility because the matter in it will absorb some of the light which is going through.

Yes, we cannot see all the universe and so we can see interstellar clouds up to a certain distance. Indeed, light takes time to reach us.

If there is a star behind a cloud you will have difficulties to determine its true luminosity. In general, we define the luminosity of stars thanks to a reference star.

I'm not sure to really understand the fourth question ...


Is there a upper limit to the number of planets orbiting a star?

There do exist somewhat trivial configurations, which are stable in the long term and which include arbitrarily many bodies. Consider, for example, a set of $N$ circularly moving bodies of the same mass $m$, which obeys the constraint $mN\ll M$, where $M$ is the mass of the star. So long as $mN\ll M$, the bodies move dominantly in the gravitational field of the star and are hence moving stably over long term period. However, as $N$ is arbitrary, one concludes that there is no upper limit on the number of planets, provided that their total mass is small.

A more physical example would be a protoplanetary disc, or an accretion disc, which is a limit $N\rightarrow \infty$ of an arbitrary planetary system (not not necessarily circular) of a given mass. A yet more physical example is an asteroid belt, consisting of a large number of bodies on, roughly, stable orbits. Finally, during planet formation process the star goes through stages, when it is surrounded by sets of pebbles and asteroids, which keep their structure constant over a large number of orbits (roughly, of order $10^5$). And these all are real physical examples of planetary-like systems.

The answer to your question would start to alter, though, if you start imposing additional conditions apart from $N\rightarrow \infty$. For example, if you require that bodies do not collide in the long term, some of the above named systems would not work (for example, accretion disc model), but some other would (sets of concentric particles). If you additionally require that the object should obey the definition of a planet, that is have some range of masses, then interesting things will start happening when the total mass of the planets will start being comparable to the mass of the star. So the limit would certainly exist. Finally, you might be more strict about what do you really mean by stability here, and that could also have a bearing on the answer.

To summarize, unless you impose any constraints, there do exist N-body systems orbiting a star in a stable fashion and having arbitrarily large $N$.


The north celestial pole doesn't have a right ascension. It's like the north terrestrial pole not having a longitude. Asking for it creates an exception. In any case, the RA of a point, other than the poles, isn't dependent on time or the observer's location. Your simulator has an error in its error handling.


The short answer is no, with some caveats to the effect of sort of, depending on how loose an analogy you want to make.

Sound propagation in a fluid is limited by the speed of sound, which can be used to define a "sound cone" structure analogous to the causal light cone structure in spacetime. This is a described by an acoustric metric, which could have an acoustic horizon when the speed of the fluid exceeds the speed of sound, and even an analogue of Hawking radiation.

A general acoustic metric for a perfect fluid has the form
$$g_{\mu\nu}=\alpha^2\begin{bmatrix}-(c^2-v^2)&-\vec{v}\\-\vec{v}&\mathbf{1}\end{bmatrix}\text{,}$$
where $\alpha$ is a conformal factor. The Schwarzschild black hole can be put in this form, as in the Gullstrand–Painlevé chart is not just spatially conformally flat, but exactly Euclidean at every instant of time. One can imagine that a Schwarzschild black hole is like a drain sucking space down to the singularity at the local escape velocity.

The rotating Kerr black hole cannot be put in this form, although its equatorial slice can be. See Visser and Winfurtner (2005) for details, as well for a discussion as to why interpreting the metric as corresponding to a physical fluid is problematic even in the Schwarzschild case due to the the conformal factor.

However, Hamilton and Lisle (2008) found that the Doran chart of Kerr spacetime can be interpreted as a six-dimensional "Lorentz river" characterized not only by a velocity but also a twist. This really quite different from the acoustic fluid analogy, as the 'river' does not spiral inwards, but rather has an intrinsic twist that rotates infalling objects. Still, it is interesting in its own right.

Image by Andrew Hamilton. (Twist not shown.)

References:

Visser, M., Weinfurtner, S. "Vortex analogue for the equatorial geometry of the Kerr black hole", Class. Quant. Grav. 22:2493-2510 (2005) [arXiv:gr-qc/0409014]
Hamilton, A. J. S., Lisle, J. P., "The river model of black holes", Am. J. Phys. 76:519-532 (2008) [arXiv:gr-qc/0411060]


What happens if two galaxies collide very quickly?

Have 2 people stand in opposite corners of a large empty room each with a jar of marbles.  Have them roll these at each other all at once as fast as they can, and you will get a pretty good answer to your question.

97% (or thereabouts) of a given galaxy is empty space.  Most of one is going to pass harmlessly through the other, though you are likely to get a few collisions.  At or near the speed of light, gravity will have very little effect as well, so a few orbits will be perturbed, but any given star will hardly notice.

I suspect (though I haven't the expertise to say definitively) that the radiation increase during the ~100,000 years or more that this collision would take (assuming the galaxies were of approximately the size of the Milky Way and that they hit exactly edge on) would be significant to any creatures living in your galaxies.

If by the off chance (I'm talking about lightning-striking-a-shark-currently-eating-a-recent-lottery-winner kind of chance) the supermassive black holes at the center of these two galaxies collided, the speed would be no obstacle, and they would combine.  The angular momentum would certainly rip the new black hole from one of, or likely both, of the colliding galaxies.  This would be bad for them.  I propose sending this as a "What If?" question to XKCD.  He seems to have the time to do the math on questions of this nature and follow as far as it leads.


How high were the tides back when the Moon was much closer to Earth?

To expand on Keith's comment, the equation for the tidal force is based upon the difference between the acceleration due to the orbiting object, and the gravity of the body itself, i.e. the difference between how hard the Moon pulls on you compared to how hard the Earth pulls.  This results in an overall equation of:
$$|a_{tidal}| = 2\Delta r G \frac{M}{R^3} $$
where $\Delta r$ is the radius of the Earth (6371 km), G is the gravitational constant (6.674 x 10$^{-11} m^3kg^{-1}s^{-2}$, M is the mass of the Moon (7.342 x 10$^{22}$kg), and R is the distance between the two (here either 22,500km or 384,400km). To check our equation we can calculate the current tidal acceleration: using 384,400km, we arrive at a value of 1.099 μm/s$^2$, which is close to the accepted value of 1.10 μm/s$^2$ (2).  Calculating the tidal force using the distance of 22,500km, we arrive at a value of 0.0055 m/s$^2$, or about 5,000 times the current tidal force.  That could make for some massive tides, however, were there oceans?
Current research suggests that the Moon formed as the result of a major impact, around 4.5 billion years ago (3), while the first oceans formed around 4.4 billion years ago (4).  This leaves a gap of around 100 million years between that 22,500km distance and the formation of the oceans, but it is close enough that the tidal force is likely not that different.  However, translating the height of the tides is difficult.  As the average tide is about 0.6m (5), a tide 5,000 times higher would be almost 3km high, which seems impossible.  Additionally, the height of the tide appears to be more strongly affected by geography than by the tidal force, as the current range of tides is from 0-15m in height.  The earliest evidence we have of tides is about 2.5 billion years ago (6), but unfortunately, the geologic record only records the pattern of the tides, not the height.  We can tell that the Earth was spinning faster then, with a day 4.5 billion years ago being about 4-5 hours long, but it may have been too cold to melt the oceans.
So, to sum up:  The tidal force with the Moon at that distance would be 5,000 times greater than it is today, but there may not have been oceans, and the tide height is affected more by the shape of the ocean than the tidal force itself.  However, something like the water planet in Interstellar, with massive, 3,000km-high tides sweeping over the Earth every couple of hours is possible, although a boring ice planet is possible too.


finding Dec and R.A of a star by having Dec and R.A of another star

You can measure the position in the sky of any star; this is its altitude and azimuth. The star's position in the sky depends on its actual position in space, your position on Earth and the rotation of the Earth.

Measuring the position requires some kind of angle-measurer. Such as a quadrant or sextant. You also need knowledge of which direction is North. This can be determined, even if the pole star is not visible, from the location of other stars. A telescope can increase accuracy but the same measurements can be done either with or without a telescope.

You can use your position, and the position of Gamma Cephei in the sky (by altitude and azimuth) to determine the local sidereal time. You can then use the position of Gam Del and the sidereal time to determine the ra and dec of Gam Del.

Other questions discuss the conversion between alt-az to ra-dec Software to convert RA and DEC into ALT and AZ


Which elements are an indication of habitable exoplanets?

While not strictly relevant to your question, I'm very much looking forward to what the James Webb Telescope might tell us about exo-planet atmospheres.   That's probably the thing I'm most looking forward to in astronomy discoveries over the next 5-10 years.

The one thing I've heard that they are looking for specifically is a combination of Oxygen (which implies Photosynthesis) and CH4 which, in combination with oxygen implies bacterial digestion of plant matter (aka cow farts).   Those 2 things together are of particular interest because CH4 and O2 tend to interact with each other, so, if not regularly produced they aren't likely to both be present, at least, in measurable amounts.   Other factors like plate tectonics, Methane escaping from the ground or under the oceans, comet impacts and perhaps even lightning might make such readings a little bit unreliable, but as I understand it, that combo, CH4 and O2 is the big one to look for.

Other interesting things might be a CO2/O2 ratio, also implying photosynthesis and perhaps telling us something about plate tectonics or O2 respiration.   Variation in H20 content which could tell us something about weather, if the planet has freezing/thawing periods like Mars or vast deserts and few oceans and little water or occasional hurricanes.   Dust particles could tell us something about the soil or volcanic activity.  I would think NH3 content in abundance might be an indicator of little or no life, but I'm not positive there.  Water with ammonia oceans might be possible and still have life, though we might find that not a good planet to live on.

Finally habitable is a term that needs a bit of exploration.  Lets use the Earth as an example some 2 billion years ago, there was Oxygen in the atmosphere, there was CO2 in the atmosphere, there were oceans, there was a magnetic field, there was even an ozone layer, so it had everything Knu8 listed, but, there were no trees, there wasn't even any soil.  There might have been a thin microbial layer living on the rocky surface or, if too hot, perhaps in caves.  There were no fish, there wasn't even any good plants or krill like things growing in the oceans for fish to eat.  It was mostly one celled life mostly just in the oceans and perhaps lakes.  Earth 2 billion years ago might not be our first choice for habitation.  It was also prone to photosynthesis driven snowball earth periods as well as having no soil and no trees.

There's also the question as to whether we'd want to inhabit an "earth-like" planet.  Bacteria can be pretty hostile and if there's life on the planet, it might not be safe to settle, and there would also be ethical questions.   If there's no life, then the Ethics isn't an issue, but there's likely a lot more terraforming.   It's all relatively moot given the difficulty of that kind of space travel, but "habitable" is a more difficult question than it first appears.   Our technology might be able to make a dead planet habitable over time, where as a planet with life on it might appear more habitable but could pose unforeseen problems.  I'm not sure which would be more ideal.

I'm not an expert in this, but what to look for in exoplanet atmospheres for habitability is a little complicated and I've only touched on it briefly.   Fun thing to think about though.


Why doesn't the black hole in the center of the Milky Way glow similarly to the famous M87 image?

M87 is an active galaxy with an accreting, rotating central black hole of mass $M\simeq6.5\times10^9M_{\odot}$ (Akiyama et al. 2019). It is the inner disk surrounding this black hole that produces, among radiation, the 1.3 mm synchrotron emission observed by the Event Horizon Telescope, as well as the relativistic jets emitted perpendicular to the disk's plane.
The Milky Way is not currently an active galaxy, and while there is some evidence for a cool accretion disk around Sgr A* (Murchikova et al. 2019), it is sparse. The strongest constraints on Sgr A*'s accretion rate are about five orders of magnitude lower than M87*'s $\sim0.1M_{\odot}\;\text{yr}^{-1}$ (Di Matteo et al. 2003). The very low accretion rate onto Sgr A*, a black hole three orders of magnitude less massive, combine to indicate that any emission of the same sort from the Galactic center should be much less intense than that seen by the EHT from M87*.


How would Earth's magnetic field change if the outer core was solid?

We have an excellent reference of what would happen to the Earth's magnetic field if it had a solid core: we can look at Venus' magnetosphere. Its internal magnetic field is dominated by the remanent crustal magnetization and its external magnetic field by the induced currents and their magnetic field in the ionosphere. The magnetic field mostly misses any dynamo part due to Venus' very slow rotation (even though it also has a similar internal structure like Earth).
Thus as a result, the Earth's magnetosphere would shrink considerably had it as little rotation and/or no liquid outer core. With only a solid core remaining, it would also only be able to produce some small remanent magnetization which is usually MUCH less than any self-feedback dynamo like Earth, Jupiter, Sun,... are able to produce.
Erosion of the upper atmosphere would be much stronger and we could enjoy polar lights about continuously and even at low lattitudes as we'd loose considerable magnetic field strength, especially of the dipole moment.


How bright is it on Pluto in the middle of the day?

Back when New Horizons was preparing for its flyby back int 2015, NASA's website set up a tool to allow you to experience the brightness of light at High Noon on the subsolar point on Pluto.
From Space.com: NASA's 'Pluto Time' Shows You How Bright It Is on Dwarf Planet

To an observer on Pluto's surface, the sun would be about 1,000 times
dimmer than it is here on Earth, NASA officials said. At noon, the
sunlight would be strong enough for you to read a book, they added.
(Though it might be hard to keep your eyes from straying upward to
take in the exotic Pluto sky, with the big moon Charon looming just
overhead.)

The NASA Pluto Time website can be found here:
https://solarsystem.nasa.gov/planets/dwarf-planets/pluto/plutotime/
In general, the time it will return for you will be a little bit after sunset (or before sunrise), but still bright enough for pretty good vision. For my current location, at about 41° N latitude , on October 31, 2020, it's giving me a local time of five minutes after sunset.


What proportion of planetary systems have been found with 'Hot Jupiters'

In 2011, about 20% of the exoplanets found were hot Jupiters. That is a lot, but it is strongly biased simply because there are the easiest planets to detect. You can detect exoplanets by transit (that means that the planet passes in front of its host star which decreases the luminosity of the observed star during the transit), hot Jupiters being closer to their host star and bigger than smaller planets will therefore be easier to detect. You can detect exoplanets by radial velocity, which means that you detect variation of the position of the host star due to planet's gravity. The closer and the heavier the planet is, the larger the variation will be, and therefore the easier to detect.

That's why it is important to say "In 2011"; as our detection techniques (and telescopes) improve, the proportion of hot Jupiters should decreases.

Edit:
To put that in perspective, you can read this article from Jason Wright in which he tries to estimate the "real" ratio of hot Jupiters orbiting around "normal" stars (F, G and K type stars). This ratio is actually much smaller than the currently observed ratio, about 1.2%.


Compute Planet's Apparent Visual Magnitude

Lets assume we're dealing with a superior planet, which is to say, the planet is orbiting at a greater distance than the Earth. This effectively ensures that there is no planetary phase to deal with.
Now, the Sun has a luminosity of $L_{sun}$ such that the Solar flux as seen by the planet is
$$
 f_{planet} = \frac{ L_{sun} }{ 4 \pi d_s^2 }.
$$
The cross section of the planet is about $A = \pi r_p^2$, so for a given albedo $a_p$ the reflective luminosity of the planet will be
\begin{aligned}
 L_{planet} &= a_p f_{planet} A \\
            &= a_p \pi r_p^2 \frac{ L_{sun} }{ 4 \pi d_s^2 }.
\end{aligned}
Given that we know the absolute magnitude of the Sun, the absolute magnitude of the planet follows as
\begin{aligned}
 V_{planet} &= -2.5 \log_{10}\left[ \frac{ L_{planet} }{ L_{sun} } \right] - V_{sun} \\
            &= -2.5 \log_{10}\left[ a_p \frac{ r_p^2 }{ 4 d_s^2 } \right] - V_{sun}
\end{aligned}
The apparent magnitude of the planet as seen from Earth can then be calculated as
$$
 m_{planet} = V_{planet} + 5 \log_{10}\left[ d_{e-p} \right] - 5
$$
where the distance between Earth and the planet, $d_{e-p}$, should be in parsecs and of course depends on the orbital phase of the two respective planets.
So the one additional parameter required was the Solar luminosity. There are probably a bunch of subtle effects with the albedo and the geometry of the system that are not taken into account here, but it should be a fair approximation.


What exactly are the "ν6 secular Sun-Jupiter-Saturn" and the "1:4 Sun-Jupiter" resonances?

The 1:4 (or 4:1) Jupiter resonance is a mean-motion resonance: an asteroid there takes 1/4 as long to orbit the Sun as Jupiter does.
Perturbations by Jupiter at recurring ecliptic longitudes alter the asteroid's orbital period, so asteroids do not remain long in this state.
The Kirkwood gaps in the asteroid belt include this and other mean-motion resonances with Jupiter.

The $\nu_6$ secular resonance (also called $g = g_6$) is more complex: in certain combinations of orbital period and inclination, an asteroid experiences perihelion precession at the same rate as Saturn (the 6th planet) does, resulting in a gradual increase of the asteroid's orbital eccentricity.


How to get the solar zenith from the longitude and latitude?

Direct Normal Irradiance (DNI) is the the amount of solar radiation from the direction of the Sun.

Diffuse Horizontal Irradiance (DHI) is the amount of solar radiation that does not arrive on a direct path from the Sun.

Global Horizontal Irradiance (GHI) is the total amount of solar radiation. It is the sum of Direct Normal Irradiance (DNI), Diffuse Horizontal Irradiance (DHI) and ground-reflected radiation. Ground-reflected radiation is usually insignificant (compared to DNI and DHI) so for all practical purposes GHI is the sum of DNI and DHI.

Thus $GHI = DHI + DNI \times cos (Z)$ where $Z$ is the solar zenith angle.

Solar zenith angle is derived from latitude.

Solar zenith occurs at local solar noon, which is derived from longitude.


How does the solar wind destroy the atmosphere of a planet lacking a magnetic field?

What determines how long a planet will keep its atmosphere is the strength of the planet's gravity versus the random thermal motion of the molecules in its atmosphere.

The atmosphere is a gas, and all gasses have molecules that move around with velocities that are dependent on the temperature.  High-temperature gasses move faster, and low-temperature gasses move more slowly.  Each molecule of gas will have a velocity $v_{\text{gas}, \text{average}}=\sqrt{\frac{3k_BT}{m_{\text{gas}}}}$.

In addition, each individual molecule is pulled to the Earth by gravity.  Heavier gasses, such as water, carbon dioxide, and oxygen, are pulled more than lighter gasses, such as hydrogen and helium.  The escape velocity of the planet is given by $v_{\text{escape}}=\sqrt{\frac{2GM_{\text{planet}}}{R_{\text{planet}}}}$.

If a gas molecule has a high enough thermal energy, then if it is at the top of the atmosphere, it could fly off into space, the thermal motion overcoming gravity's pull.  The lighter it is, the less thermal energy it needs to have.  Of course, this all depends on probability and statistics, but, in general, if $v_{\text{gas}}\ge0.2 \times v_{\text{escape}}$, that species of gas will be reduced to 1/2 its initial amount after 1 billion years.

Solar wind consists of charged particles and high energy photons, which are deflected by a planet's magnetic field.  However, in the case of a planet like Mars, without a magnetic field, the particles interact with the atmosphere.  They tend to split molecules apart, resulting in lighter gasses (a process called "photodissociation").  These gasses, being bound less strongly by gravity, are more free to fly off into space under thermal motion.  As more and more gas leaves the planet, the atmosphere becomes thinner and thinner until, ultimately, only the heavy gasses are left.  In addition, when a molecule is split, the energy imparted to it by the impact may be enough to eject it from the atmosphere (a process called "sputtering").

Planets (and moons) that are closer to the sun are subject to more intense solar radiation, and also have higher atmospheric temperatures.  This is why distant moons, such as Titan, have thick atmospheres of hydrocarbons.  Planets close to the sun, such as Venus, can retain their atmospheres for a variety of reasons.  Firstly, Venus's atmosphere is primarily carbon dioxide, which is much heavier than Hydrogen, and more difficult to split via photodissociation than water.  Planets like Earth also have hydrogeological cycles which can recycle gas from rocks and water back into the atmosphere.

For more information on this, read up on Jeans Escape.  A good set of notes for beginners can be found here.


Saint-Exupery describes visit to plateau where he easily finds several meteorites. Is this realistic?

This is only a partial answer; meteor falls are pretty common and so their density will be high in flat areas without a lot of flowing water, they're hard to identify without a good eye or at least a sensitive metal detector.
Hopefully a more quantitative source for the areal density of meteor finds can be found. The numbers are there, and if all rocks in the area that Saint-Exupéry describes are light in color, then picking up and checking every dark rock and checking to see if it has indications of burning or melting, this might be possible. Of course if it were this easy to collect them there then it seems others would have gone and collected a lot more. But those people have no incentive to document it since they can (often) be sold for cash apparently.
Note that the meteors shown in the image from the PRI article linked below are not dark but the dearth of dark meteorites found in Antarctica is a different story (1, 2) though there's been a lot of progress since 1986 when the document you link to was published.
According to Arizona State University's Meteorites page Meteorite Locations:

Since there is an estimated one meteorite fall per square kilometer per year, geologically stable desert regions can show significant accumulations of meteorites. Some desert regions have dozens of different meteorites per square kilometer, though they can be difficult to distinguish from normal terrestrial rocks.

PRI's Want to find a meteorite? Antarctica might be the best place to look says:

This winter, Lanza has been a rookie member of the ANSMET (the Antarctic Search for Meteorites) field team. For 40 years, the project has sent teams of scientists to the bottom of the globe to recover meteorites from all over the solar system, including chunks of the moon, comets, even Mars.
On this trip? Lanza and her colleagues recovered a total of 569 meteorites.
“They are just waiting for us to come and find them,” Lanza says. “You can imagine that blue ice ... and then you just look and there's some dark spots on them. And those are rocks and frequently those rocks are from space.”
Meteorites fall to Earth all the time, but Antarctica is a particularly good place to find them.
“We're lucky with Antarctica,” Lanza says. “It's this big white sheet covered in glaciers. And so these meteorites just become embedded in the ice and start flowing with the glacier. And in some places the glacier will run into something — like the mountain range that we were in — the Miller range. It will slow down and then the winds ... will start to remove some of that ice and that acts to concentrate the meteorites in these locations. So we can actually go there and find many more meteorites than you might imagine would fall in a single location.”

Nina Lanza knows space rocks. In her day job as a staff scientist at Los Alamos National Laboratory, she operates the Curiosity Rover’s ChemCam, using a rock-vaporizing laser to analyze the Martian surface.


Why does squinting make hard-to-see objects clearer?

Squinting works the same way as a pinhole camera.

Ideally, light from a single point source entering your eye anywhere on your pupil will be focused on a single spot on your retina. But this works perfectly only if you have perfect vision; otherwise light entering near the top of your pupil may be directed to a slightly different spot on your retina than light entering near the bottom.

By squinting, you block out some of the light from the edges, effectively making your pupil narrower, creating a sharper but dimmer image on your retina. (You may find that it improves the vertical resolution more than the horizontal resolution.)

If you happen to be nearsighted (as I am), you can see a similar effect by looking through a small pinhole, or through a small aperture made with your fingers. If the light is bright enough, you'll see a dimmer but sharper image.

The irregular shapes may be interference from your eyelashes.


Kepler space telescope undetected planets

That's right. The inclination of the orbital plane around stars is considered to be random throughout the galaxy, thus the planets we can detect by the transit method is just a tiny fraction of the planets that we should expect in our stellar neighbourhood.

The transit method allows for planetary detection only when the line of sight from Earth to the system is contained, or almost contained, in the orbital plane of the planet. This means that only a tiny range of orbital inclinations on each star are good for detection.

Why did I say almost? Because there is some range of inclinations that still would yield a transit. This range is not fixed, and it depends on the distance of the planet to its host star. As you can see in this diagram:

Planet A is closer to the star and thus creates a wider shadow. If an observer is located in that shadowed region far away it can detect planet A. Planet B instead is farther from the star and thus its shadow is narrower. It is interesting to note, that even if both planets here share the exact same orbital plane there are places from where you would only detect planet A and never detect planet B (see the green arrows). This is the reason we have a bias towards planets orbiting closer to their star.

This effect is in fact quite strong: consider our Solar System from an exoplanetary perspective. If you were located in a random star in the sky, what are the chances you would spot an Earth transit? Well, it turns out that it is way more probable to detect a Mercury transit, even if Mercury is the smallest planet, just because of its vicinity to the Sun. A recent paper showed this diagram of the regions of the sky where some alien inhabitants would spot a transit for each of our planets:

As you can see Mercury has the wider strip. Also it's interesting to note that due to these differences in the size of the orbits (let's use the semi-major axis, $a$, as a reference) and due to small differences in orbital inclinations there is no place in the entire sky from which an alien could detect simultaneously more than four of our planets by the transit method. No place in the universe where all the Solar System's planets would be detectable.

The detection method also depends on the relative sizes of the star, $R_s$, and the planet $R_p$: A larger star has a larger disk (as viewed from Earth) that can be easily photobombed by a planet and a larger planet can photobomb more easily if it is larger.

The result is that the probability of detecting a planet increases as we increase both/either $R_p$ and $R_s$ and increases as we decrease the distance to the host star $a$. The relation is then of this form:

$P \sim (R_s+R_p)/a$

This relation imposes several observational biases. We can see exoplanets that are large and closer to their star, but we can't see planets that are small and farther. That is the reason the first detected exoplanets are the so-called hot Jupiters: giant planets much closer to their stars than Mercury is to the Sun. This diagram shows all the exoplanet detections plotted on size vs. orbital distance:

As you can see, small planets are only detectable if they have very small orbits around their stars. We have yet to find a planet the size of Earth (quite small) and with a 365 day orbital period (1 AU distance) using the transit method. There is no reason to think that this is representative of the overall population of planets. The black region of the plot is probably filled with dots, but our instruments can't scout that region yet.

The Kepler telescope had a camera with a field of view on which it could detect more than half a million stars, but the actual number of stars monitored during the mission was around 150,000 stars (these stars had good signals and were perfect targets for the mission). For these 150,000 stars Kepler found 2,345 exoplanets distributed in 1,205 stars. So we can say that for each star targeted by Kepler, the average probability of finding some planets there is around $0.8\;\%$. That should give you an estimate of the occurrence of orbital inclinations that result in transits.

The truth is that this number is too small, because Kepler has several more biases. For example, Kepler only confirmed planets after three transits were detected. Since the Kepler mission lasted for four years and four months we can say that in the best case scenario Kepler was able to detect a planet with an orbital period as long as two years and two months, but this is not even the case since for that to happen a transit should have been detected just at the beginning of the mission, halfway, and at the exact end of it, and this coincidence didn't happen. Thus Kepler had no chance to discover any planet with periods longer than two years (enough for Earth, but not enough for our Jupiter for example), even if the orbital inclination matched perfectly for the transit. So you might expect more possible transits than those actually portrayed by the Kepler telescope.

In fact, for planets close to their stars, it has been estimated that the probability of a random alignment to allow a transit gets up to $10\;\%$. For the case of stars as big as our Sun and planets at the same distance as Earth, the probability of this random occurrence drops to $0.47 \;\%$. So with all the diversity of planets (in terms of sizes and distances to their host star) it is reasonable to expect a $0.8\;\%$ detection rate for Kepler (if we also add the time restriction to observe three transits).

A $0.47\;\%$ is an amazing number! It means that for every Earth-like planet we detect by the transit method we should expect another 213 Earth-like planets orbiting other stars that are undetectable by the transit method.

This kind of reasoning has been expanded. We have many difficulties to detect them, but if you mathematically model that difficulty and the corresponding biases associated with the known instruments and you assume random configurations, you can see that each discovery yields statistical significance to the amount of possible planets that are really out there. There are so many detections now that we can finally establish with statistical confidence that there are more planets than stars in our galaxy (even if we have probed an infinitesimal fraction of the entire population), even if this was something that could be expected we have now strong evidence for that thanks to Kepler. This means that there could be around a trillion or more plants just in the Milky Way. Now we are also able to establish some statistical constraints on the occurrence of Earth-like planets (orbiting in the habitable zone of their sun-like star) thanks to Kepler. There are probably around 11 billion planets in our galaxy with these specifications.

TL;DR

There are many more planets than the ones we can detect by the transit method, between 10 and 100 times more depending on the size and orbital period of the planet you are searching for.


How can we tell if the asteroid 2013 TV135 will hit Earth in 2032?

The answer to the first question is: more observations.

  That is, what process and techniques are used to calculate and increase the accuracy of our collision threat estimation?

You just have to know the orbit.

But the orbit of asteroids is very complicated to predict, because they are influenced, sometimes dramatically, by the planets and by smaller bodies, like other asteroids, that we might not be able to see.

On top of this we cannot measure directly the orbital speed of an asteroid. We can only observe changes in the angular position. If we know the distance (which might not be easily measurable) we can at least put a lower limit to the speed.

The basic steps to study the orbit of an asteroids are:

find it and try to get a first guess about the position and velocity (this requires multiple observations far enough in time to be able to measure an angular movement)
plug the position and velocity info (and likely their errors) into a computer simulation that simulate the solar system dynamics. Fast forward the simulation to predict the future orbit.
observe again the asteroid and compare its position with the simulation. Re-estimate position and velocity. Go to point 2.

As the orbit can change significantly a continuous interplay between 2. and 3. is necessary.

Once you have an orbit you can estimate the probability for the asteroid to pass near enough to the Moon or Earth to be deflected and hit us.

But thankfully the solar system is huge, we are tiny and moving between the much much larger Sun and Jupiter :D


Limiting magnitudes for different telescopes

Your calculated estimate may be about correct for the limiting magnitude of stars, but lots of what you might want to see through a telescope consists of extended objects-- galaxies, nebulae, and unresolved clusters.  The magnitude estimates of these objects tend to consider them as point sources, but as extended objects they have much less of a contrast difference with the sky.  In my experience in an equally light-polluted location with a similar-size telescope, the sky brightness makes the limiting magnitude of extended objects between 7 and 8, even with a light-pollution reduction filter in use.
That isn't to say a telescope is worthless in the city; you can still observe the Moon and planets, resolved clusters are pretty, and you can observer the brighter extended objects.  But bringing the telescope to a dark-sky location makes an incredible difference with many more objects visible.


What are the Earth-like features of Titan?

To expand a bit on Undo's answer, there are some geomorphological features on the surface of Titan that resemble similar features on Earth, suggesting similar processes.

For example, according to this NASA image comparison, some lake features have features that are similar to groundwater based salt pans on Earth. Image below:

The feature on the left is Ontario Lacus on Titan compared to the Etosha Pan of Earth on the right.  It should be noted, that the Titan feature is considerably larger than the Earth analogue.


What are the natural life sustaining resources available in the Jovian moon system?

The moons Europa, Ganymede, and Callisto all have substantial amounts of water ice on their surfaces, and for the latter two, also silicate rocks. There is also some CO2 in the ice as well as ammonia and methane. These give potential sources of carbon and nitrogen, and with enough energy you could make glucose and amino acids - Food.

Io is a crazy place, with lava flows, and sulphur covering the surface. Its not an easy place to mine

One potential source of energy is ${}^3\mathrm{He}$ This is a potentially useful fuel for a fusion reactor. It is found (in very low concentrations) in shaded parts of the moon. It may also be found on Jupiter's moons.

The minor moon Carme may have originated in the Kuiper belt (long ago) and have an interesting combination of carbon based substances. Amalthea has an interesting red colour - maybe carbon compounds again. There are quite a few moons that began as captured asteroids, but none seem to have been metallic.

There are lots of problems with life on Jupiter's moons: Cold - No air - radiation - etc


Could asteroid 'Oumuamua actually be round?

The hypothesis that the object has an extreme aspect ratio comes from both the max/min ratio of the light curve and the shape of the light curve, along with some assumptions of the uniformity of the albedo as you mentioned.

If you read the BBC article Bizarre shape of interstellar asteroid and click from that page on the link that says which is published in Nature journal. it will take you to an open access draft of the Nature Letter. In it you can see the shape of the light curve. I've chopped up some sections to show below.

The "dip" is quite sharp. Remember magnitudes are a log scale, and so the dip would be even sharper on a linear scale. Since 5 magnitudes are a factor of ~100, a 2.5 magnitude dip is a factor of ~10.

This is what you would get when a high-aspect ratio objects so that is viewed end-on. A black spot would darken the curve much more gradually, being viewed for a longer fraction of the cycle.

This is not proof of the high-aspect ratio hypothesis, and the authors don't claim this. However, it's the best way to fit the curve with the fewest assumptions, as far as I can tell.

Screen shot of the on-line Accelerated Preview of the Nature Letter A brief visit from a red and extremely elongated interstellar asteroid K. Meech et al. doi:10.1038/nature25020


Conversion of magnitudes to Jansky and MAGPHYS?

Caveat this with the fact that I am not familiar with this code.

The input asks you for the flux through the filter, but a filter has a bandpass, so if you have a 10 nm bandpass filter and you measure $1 W/m^2$, then you measured $0.1 W/m^2/nm$ (sorry, I'm a physicist so I never got used to measuring wavelengths in Hz).

As for your second question, the factor of 23 is because in cgs units a Jansky is $10^{23} erg/s/Hz/cm^{-1}$ and that TOPCAT equation is in cgs (the equation is $m_\text{AB} = -\frac{5}{2}\,log_{10}\,f_v - 48.6$) and if you are normalizing it to Janskys, you need to divide by $10^{-23}$.


Official degrees of earth’s rotation per day

First, we need to decide which definition of "day" to employ. There are several types of days:

Apparent solar day: the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer;
Mean solar day: a more uniform, averaged solar day without seasonal variations;
Stellar/Sidereal day: the time needed for the Earth to rotate once relative to the stars;
SI day: a unit of time containing exactly 86,400 SI seconds defined by caesium atoms.

Since we generally refer to the traditional day/night cycle when we say "day", this means a form of solar time. To get a more averaged value, let's use the mean solar day.

The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition :

$$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) \text{ radians}$$

Where $Tu$ is the Julian UT1 Date - 2451545.0

So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about 360.98561°. However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you 361.0°, a figure that will likely remain true for at least several millenia.

If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h UTC every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and VLBI observing distant objects in the universe. Official reports of its orientation, the Earth Orientation Parameters are published on the IERS Website.

As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year:

As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation.


Could human life thrive on a planet in a pulsar star system?

Only five pulsar planets have been confirmed or have garnered enough evidence to make a strong case for their existence. None of them are like the terrestrial planets in the Solar System insofar as the way they formed and their orbital movements.

PSR B1620-26 b. This planet is in a circumbinary orbit around a pulsar and a white dwarf, with a semi-major axis of ~23 AU. It is theorized that it formed around a Sun-like star, which became a binary star system after encountering a neutron star (another partner must have been ejected from the system, as is the case with triple-star encounters). The Sun-like star promptly became a white dwarf, and the planet's orbit would have widened after the encounter with the neutron star, which is a pulsar.

Basically, this planet did not form around the pulsar. The system was captured so to speak, and so neither the Sun-like star/white dwarf or planet would have been affected by the supernova. Additionally, the planet is fairly far away from the two stars in the system, meaning that effects from the pulsar are even less negligible.

The point of this is that life on this planet is possible - especially given the planet's old age - but it would be hard for it to survive. If life formed before the encounter with the original binary system, it would have been disrupted a bit by the orbital changes, reducing the amount of starlight it received. Today, it would receive little light, as the white dwarf has a low luminosity. There wouldn't be enough energy to sustain life - as is the case for most pulsar planets.

Also, as Rob Jeffries pointed out, this planet is a gas giant, not a terrestrial planet, meaning that human-like life could not develop on it and any arriving humans bent on colonization would not be able to live on it normally.
The planets of PSR B1257+12. Three planets orbit this pulsar, at distances of about 0.19, 0.36, and 0.48 AU - fairly close to the pulsar and within range of a decent amount of harmful radiation, relative to PSR B1629-26 b. The major reason that these planets are not a good place for life is that they are thought to have formed after the supernova that led to the evolution of the star to a pulsar, from remnants of debris from the supernova. The star itself is quite young - ~3 billion years - and the supernova would have happened relatively soon thereafter. The planets themselves are less than 1 billion years old, meaning that life has not had time to develop.

My issue with the age of the planets isn't based on the short time period for life to develop so much as on the idea that conditions might be hellish, comparable to those experienced in the Hadean eon. Geologically, these planets would have had different starts than Earth did, and they most likely would not have encountered the problems of the Late Heavy Bombardment, but life might not have a good chance at starting at all for another 500 million years.
PSR J1719-1438 b. This planet is unusual because it is thought to be the remnant of the companion star - severely damaged by the supernova. It is also composed of exotic materials, for a planet - crystallized carbon. The conditions here are not conducive to life.

Looking at this list we see three distinct types of pulsar planets emerge:

Captured planets
Newly formed planets
Former companion stars

The latter two are not good places for life, in part because they may not be composed or compounds life can use. The first group may be the best chance, as their formation could have been relatively normal. The problem lies in the fact that they might have been rogue planets, floating along without good sources of energy from stars.

All of this isn't even taking into account the radiation from the pulsar itself. The only escape is to have an orbit far from the pulsar, as with PSR B1620-26b, but then there's no outside source of energy. The best chance would be to have the planet in a circumbinary orbit around the pulsar and a companion star, as in this planet's case. At least then there would be some helpful light.


Why does Sirius proper motion "wiggle" with a 1 year period, when Sirius A and Bs binary orbit is a 50 year period?

Your vantage point changes throughout the year as the Earth moves around the solar-system barycentre (close to the centre of the Sun). That means the positions of nearby stars "wiggle" with respect to more distant background stars, with a period of one year; the amplitude of this wiggle is how you determine the parallax distance to the object.

In the case of Sirius AB, this wiggle is superimposed on a motion caused by the binary orbit of the two components in the Sirius AB system and by the motion of the Sirius system with respect to the Sun projected onto the plane of the sky - the proper motion.


No, this is not possible.

During the stellar formation, some angular momentum will always be present. And any "braking" effects (magnetic, relativistic, tidal etc.) will become weaker as the rotation slows down. So the rotation will never completely stop, because any forces slowing down the rotation will weaken as well.


Proper Terminology For Satellite Going Into Planets Shadow

SUMMARY: The word is 'eclipse'

If you visit https://www.skyandtelescope.com/observing/a-jupiter-almanac/ and click on "Phenomena of Jupiter's Moons in 2018" you get to https://s22380.pcdn.co/wp-content/uploads/WebJphenTab2018.pdf . The latter document refers to such events as "an eclipse by Jupiter's shadow".

Since Sky and Telescope is a fairly reliable source, I'd say this is a good answer, though an answer from NASA or IAU would be better.

Note that it is also a lunar eclipse as viewed from Jupiter and a solar eclipse as viewed from the moon being shadowed.

I realize you're talking about Earth's artificial satellites, and this answer involves Jupiter's natural satellites, but I think the same terminology applies.

I happen to be working on Jovian lunar eclipse calculations and ran across this resource incidentally.


What fraction of starlight, seen from Earth, is actually reflected light?

A truly negligible amount. You only need compare the brightness of the planets as viewed from the Earth with the brightness of the sun. A very rough calculation (considering the relative magnitudes of the planets and the Sun) suggests that light emitted directly by the sun is 100 million times brighter than light reflected off planets.

You note that the sun is not a perfect black body, and so does reflect some light. However it only reflects light reflected on to it from the stars, which are a very weak light source. And most starlight that does fall on the sun is absorbed. I haven't tried to estimate how much of the sun's light is reflections of starlight, but it is a very very small proportion. I'd guess much less than a billionth.

There is one situation where reflected light is significant, and that is when very bright stars illuminate dust and gas, to form a reflection nebula. These are common and at least part of the orion nebula is reflected starlight. Other reflection nebulae are too dim to see.

While a I don't have an exact figure, it is clear that nearly all starlight is directly emitted from the star, and it takes special equipment to see any reflected light at all.


Is there a way to know if there are/are not solar-system-like systems of objects without a star?

Brown dwarfs (ie sub-stellar objects that are too small to support hydrogen fusion) do exist and some brown dwarfs have been found to host planets.  An example of this is the brown dwarf 2M1207.  It is a very dim object, about 25 Jupiter masses. It is still quite hot, so not completely "dark", and it hosts a planet with a mass of 3-10 Jupiters.
There are doubtless other similar bodies, and smaller rogue planets with moons, though such bodies would be very hard to observe. Light would rarely pass through such a system (the gaps between systems are just too great) but if it did, the gravity of the system could cause light to be focussed.  This is called "microlensing" and is an observed phenomenon.
The idea that such bodies could explain dark matter is a good one, and it has even been given a name "MACHO" for "Massive Compact Halo Objects".  The trouble with the MACHO hypothesis is that it doesn't fit with our understanding of how matter forms after the big bang. If matter behaves how we observe to behave, then there is no way that the big bang could have created enough normal matter to explain the observed amount of mass, this means that the excess mass must be in some other form, perhaps as a "WIMP" (Weakly Interacting Massive Particle) And yes the names were chosen to mirror each other.
We seem to live in a universe dominated by WIMPs and not MACHOs.


Calculation to Convert from Equatorial to Heliocentric coordinate systems

No, this is not possible. Equatorial coordinates only specify the direction of a certain point in space (as seen from Earth), not the distance. During a solar eclipse, the Moon and the Sun have (approximately) the same equatorial coordinates, but the distance to the Sun is 0 or almost 1 AU.
It is possible to define equatorial or ecliptic coordinates with the Sun as center, but even then there is no general formula. You'll need the full position (i.e. three coordinates) of the Earth (relative to the Sun) or vice versa, and the full position of the object under observation to be able to convert the coordinates.


Can an event horizon form with the observer already inside?

The event horizon is the boundary in spacetime between the black hole interior and exterior. If you exist before the black hole forms and you end up at the singularity, then you necessarily passed through it at some point. You can't get inside without crossing the boundary. From the perspective of some foliation of the spacetime, the event horizon encloses no area when it first appears, then it expands outward at the speed of light or greater, eventually passing by your location.
Nothing physically happens at the event horizon; you have to know global properties of the spacetime to work out where it is. There would be no sign that you'd crossed it.
The experience from the inside would be like a big crunch on a smaller scale (so a small crunch, I guess). The singularity is in the future, not the spatial center.


I think it is referring to the speed and Lorentz factor $(\beta = v/c$ and $\gamma = [1-\beta^2]^{-1/2})$ of the gas as a whole. Within the gas, there could be particles moving with a variety of velocities.

So if you pick up a ball of gas at 10,000 K (ouch) and throw it at 100 m/s then the bulk speed is 100 m/s, but obviously the particles in the gas have their own individual velocities.


How accurately can you tell the time using a photo of the stars?

To make a clock you need something that changes predictably with time, and stars are nearly constant, making them poor clocks.

The stars do move across the sky each night, but since the morning stars of winter become the evening stars of  spring, and the sky looks different in different locations you need to know the date of the image, the location of the photo, and the exact direction of the photo. That could give sub 10 min accuracy. Unfortunately while date and location may often be recorded with a photo,  exact direction is not, unless the horizon is visible in the image.

Without that information, and it is possible to 'get lucky'. For example, if there are two or more planets in the shot, their locations could be measured to give at least the date of the photo. Similarly if a rare event like a nova is visible that can offer dating information. If the shot happened to catch a bright meteor, and the date was known well enough, you could get almost 1 second accuracy. Variable stars might also give the chance of at least dating the image.

With no planets you can use the very slow proper motion of the stars. With a good enough telescopic shot of the night sky you can be quite accurate, but with a wide view of a camera you wont be able to get enough accuracy to give any useful date.

The exposure time of a typical camera is to short to capture any stars. If you just point and shoot at tye sky, you just get black. To capture images of stars  you would need a timed exposure.  A telescope would help too.

So if you photographer wanted to create an image which could be timed accurately, and they had time to prepare for it, or it was a lucky shot this might be possible, otherwise not.


Succinct explanation of black hole mass, diameter, shape?

I should start off by saying that black holes are enigmas. In describing our universe, we have the General Theory of Relativity (GR) to help us describe gravity and objects which are very massive. We also have quantum mechanics (QM) and quantum field theory to help us describe small objects such as particles. For the most part, these two realms are completely separate and these two important theories don't mix. After all, you don't need QM to describe the physics of orbiting the Sun, and you don't need GR to describe electrons orbiting an atomic nucleus.

A problem arises though, when you begin talking about objects which require both QM and GR, and black holes fall very nicely into this regime. Often we find that when we mix the two theories, we wind up with nonsensical answers. It is highly probable that our understanding of the mechanics of a black hole is incomplete. What we really need is a quantum gravity theory to talk about black holes and as yet we don't have one.

That being said, let me lay out the basic structure of a black hole based on what current theory suggests to be true. Fortunately, black holes are relatively simple in structure, as dictated by the No-Hair Theorem (which basically says you only need a few parameters to completely describe the state of a black hole, as opposed to what you'd need to fully describe the physics of a person or even an atom). The (somewhat simplified) anatomy of a black hole is shown below.

In this diagram there are a few important parts. The first being the singularity. When you put a bunch of mass together, gravity acts to squeeze that mass into a ball. That is what drives the fusion in our Sun and makes our planet round. Normally there is some resistive force that stops the matter from collapsing completely, such as the structural integrity of rocks in the Earth or fusion in the Sun. At some point though, if you pack on enough mass, that resistive force can't hold out and gravity wins. In this case, all the matter collapses down as far as it can go. It becomes a point of matter known as a singularity. This singularity has no size, it is literally (and mathematically) a point. All the mass of the black hole exists here. So when they say a black hole is 20 billion solar masses, they mean that the singularity contains the matter equivalent to 20 billion Suns, all confined to that single point.

The other important concept for a black hole is the event horizon. I should stress that the event horizon is not a physical object in space. It is a mathematical boundary. Without getting into the nitty-gritty and in basic terms, the event horizon is where our current physics "stops working". We cannot describe inside the event horizon with our current physics because we are in that region where both QM and GR are important and they don't play well together. The event horizon is also the boundary where, if crossed, you're trapped forever in the black hole. There's a lot more physics involved with the event horizon, but that's the gist. Usually when someone refers to the size of a black hole, they're talking about the event horizon. Thus if they say the black hole is 118 billion kilometers across, they mean that the event horizon has a diameter of 118 billion kilometers. Note, from the image above, that the radius of the event horizon is easily calculated from only the mass of the singularity (and a few physical constants).

There are further components to a black hole such as the ergosphere or photon sphere, and often black holes have accretion disks of material around them. I'll leave those concepts to another question though.


If jupiter was the densest planet on the solar system with low volume, could it be a star without multiplying its mass 80 times?

To achieve thermonuclear fusion you need high temperatures and high density.  If you were able (by magic) to compress and heat the core of Jupiter then fusion would occur.
But there is no mechanism that can cause Jupiter to increase it's temperature and density like this. In the core of a proto-star, gravity will provide the force needed to increase the pressure and density and the release of gravitational energy will provide the temperature to allow fusion to start.
Moreover, when fusion does start, the release of energy causes the core to stop contracting and prevents the density from continuing to increase.  The release of fusion energy stabalises the collapse of the ball of gas, and allows stars to shine for perhaps billions of years.  If you started fusion in the core of Jupiter (for example by setting off a nuclear bomb) it would cause that part of core to expand, lowering density and ending the fusion.  The stable state for a body the size of Jupiter is to be a planet, not a star.
Increasing density by changing the composition to one with larger nuclei actually makes it harder to start fusion.  Larger nuclei have a larger positive charge. This makes it harder to get them to fuse. So you need much higher temperatures.
So the only way for Jupiter to become a star is to increase it's core temperature and density. But that can't happen except by greatly increasing its size.