How to do logarithmic wavelength binning (in Python) in order to find redshift?

I think you could just take the natural log of your wavelength scale, then re-bin this onto a new log of wavelength scale generate using np.linspace().
Your new scale would be generated using the maximum and minimum values of your raw log of wavelength scale then it is up to you to define the amount of bin in the new homogenous log scale.
To re-bin the raw log of wavelength scale onto the homogenous log of wavelength scale you would assign fractions of the old flux values into the new bins where the fractions are determined by the fraction of the old bin (that the flux value was in) that overlaps the new bin.
In the end you will need this to be on kilometres per second or a unit of velocity. I expect you would simply multiply the scale by the speed of light although I have experienced slight loss of precision when I do this.


Orion Skyquest xt8i Intelliscope Dobsonian Telescope Lens

The telescope you have is quite well regarded, and the eyepieces that come with it are of reasonable quality.  If you can see the moon clearly, including craters and other features, then you should be able to see that the planets, particularly Jupiter and Saturn,  are not like stars.  At the very least they will look like disks, not points of light, and saturn's rings will be unmistakable. The 10mm lens will give magnification of 120, which is more than adequate.  My guess is that you are not looking at the planets, but stars close to them.
Make sure sure the finder scope is properly aligned - use the moon for that - and then try again.


Good textbooks for a descriptive astronomy course

Kaufman & Freedman's Universe is a very basic astronomy book aimed at non-astronomy science students or astronomy 1st year students, although for the latter it is perhaps a bit too basic. It covers introductions to all the topics you mention, except perhaps exoplanets (but my edition is from the late 90's; my guess is they have it covered too by now). It even gives suggestions for observing projects at the end of each chapter, along with exercises and extra literature. It is also very rich in pictures and explanatory diagrams.


Is there an alternative (potentially easier to understand) Euler diagram of the new solar system body classification?

Per the IAU (Resolution B5, adopted by the 2006 IAU General Assembly), there are five supposedly mutually exclusive categories of things in the solar system:

The Sun (only one), which is distinguished by the ability to fuse hydrogen into helium in its core.
The planets (eight, might possibly become nine if Michael Brown is correct), which are distinguished by having the ability to "clear their orbit" (a well-defined concept with numerical meaning) before the Sun turns into a red giant.
The dwarf planets (unknown number, most likely well over 100), which are distinguished as being bodies that don't qualify as the sun or as planets, but are sufficiently massive to more or less have pulled themselves into a roundish shape.
Small solar system bodies (unknown number, well into the millions) which are distinguished by not being distinguishable as any of the other categories.
The satellites of planets, dwarf planets, and small solar system bodies (unknown number, the number of which change over time as satellites are captured or rejected) which are distinguished by being in orbit about a planet / dwarf planet / SSSB rather than the Sun.

Contrary to that resolution, these boundaries are fuzzy rather than sharp, particularly when one looks outside of our solar system. (Note well: The IAU definitions pertain to our solar system only. There are currently eight planets in the entire universe, with the possibility of growing to nine if the conjectures regarding a largish planet beyond Pluto turn out to be correct.)

Consider the first boundary, that which separates planets from stars. This becomes fuzzy with the smallest red dwarf and the largest super-Jupiter. There's no clear-cut boundary here. In fact, there's a good amount of debate over this very issue. This boundary is quite sharp in our solar system, but it almost certainly is anything but sharp in star systems beyond our own.

The second boundary, the one between planets and dwarf planets (which, despite the name, are not planets) also has the potential for being fuzzy. In our solar system, that boundary is once again rather sharp. Any time you can come up with a boundary that is separated by five orders of magnitude you have a nice sharp classification boundary. However, given the hugeness of the universe, there is almost certainly some star system out there, somewhere, where this boundary is as fuzzy as can be.

The third boundary, the one between dwarf planets and small solar system bodies, is fuzzy in our own solar system. Objects less than 100 meters in diameter are, as a rule, anything but "round". Objects over a kilometer in diameter are, as a rule, quite "round" (when the shape is resolvable). Astronomers have found lots and lots of objects between 200 and 600 meters in diameter. Squint hard enough at the smaller ones and you might be able to convince yourself that they are "roundish". Focus intensely on the larger ones and you might be able to convince yourself that they are not quite round. There's no sharp boundary here, even in our own solar system.

The fifth category, satellites vs non-satellites, opens yet another can of worms. Planets (particularly the giant planets) can temporarily capture objects as satellites, only to lose them as satellites after a short time (tens or even hundreds of thousands of years is a "short time" in this context).

Note well: I'm not saying Pluto should once again be promoted to the status of "planethood". That very sharp boundary (at least in our solar system) says that it most definitely is not a "planet".

But what about asteroids versus comets, which is what started this series of questions? That boundary is ridiculously fuzzy. Fuzzy logic (a real and serious concept) doesn't come close to addressing how "fuzzy" this is. Centaurs and formerly active comets are the boundary cases. There is no boundary. There instead is a soft melding from one to another.


How astronomers distinguishes between natural and artificial signals coming from outer space?

We could begin to conclude if a signal is from some Extra Terrestrial Intelligence (ETI) if the signals are very difficult or impossible to explain by non-life origins or terrestrial origins. One commonly cited example is with a very narrowband radio signal. In natural radio emission, the frequencies are spread out due to processes like Doppler shift which raise and lower the frequency or radio emission. The Wow signal was so narrowband that it behaved just like one would expect for ETI, but having only one detection that disappeared makes it hard to be sure it wasn't interference or a problem with that particular antenna.

I would suggest that ETI detection should have some of these properties, but feel free to add more:

No natural abiotic terrestrial phenomenon that can produce it
Verifiability - multiple detection methods, instruments and teams that can confirm the signal is not due to errors in processing, analysis or local interference
Evidence for non-random patterns such as prime numbers

KIC 8462852 currently has a feasible solution that dust escaping from a comet or multiple comets can obscure light from its host star. This natural explanation has ways of being confirmed (such as wavelength dependence to the stellar blockage), which will help with the first criterion. Hopefully, the flux dips observed in Kepler's days ~800 and ~1500 will repeat in the future but if they don't, it fails the second criterion. As with definition life, it is hard to come up with a rule book. However, if we see a lot of the properties we associate with ETI, we can start the classification process. As more evidence comes in, like polarization, multi-wavelength and Doppler shift data for KIC 8462852, we can evaluate which criteria are satisfied.


Orbital elements of solar system using Pyephem

There is a website: http://orbitsimulator.com/formulas/OrbitalElements.html that has a javascript program for converting state vectors to orbital elements and back.

The source of that website is possible to convert to python: Here is a Body class that has a method for calculating orbital elements based on that website. The method takes one argument, a body the is the Principal (ie the Sun)

G = 0.0002946        # in units of seconds, AU and solar masses.
class Body:
    """A body has attributes r and v, which are its position and
velocity in cartesian coordinates and a mass. implied units are
solar masses, AU and seconds."""
    def __init__(self,r,v,mass):
        self.r = np.array(r,dtype="float")
        self.v = np.array(v,dtype="float")
        self.mass = mass
        self.GM = self.mass*G

    def orbital_elements(self,principal):
        '''view-source:http://orbitsimulator.com/formulas/OrbitalElements.html'''
        mu = G*(principal.mass+self.mass)
        # calculate relative position,velocity
        r = self.r - principal.r
        v = self.v - principal.v
        try: #catch division by zero
            R = np.linalg.norm(r)
            V = np.linalg.norm(v)
            a = 1/(2/R - V**2/mu)  # semi major axis

            h = np.cross(r,v)
            H = np.linalg.norm(h)

            P = H**2/mu
            Q = np.dot(r,v)

            E= np.sqrt(1-P/a)  #eccentricity

            e = [1-R/a,Q/np.sqrt(a*mu)]
            i = np.arccos(h[2]/H)
            Omega = 0
            if i!=0:
                Omega = np.arctan2(h[0],-h[1]) #Longitude of acending node

            ta = [H**2/(R*mu) -1,H*Q/(R*mu)]
            TA = np.arctan2(ta[1],ta[0])
            Cw = (r[0]*np.cos(Omega)+r[1]*np.sin(Omega))/R

            if i==0 or i==np.pi:
                Sw = (r[1]*np.cos(Omega) - r[0]*np.sin(Omega))/R
            else:
                Sw = r[2]/(R*np.sin(i))

            omega = np.arctan2(Sw,Cw) - TA  #argument of periapsis
            if omega<0: omega += 2*np.pi

            u = np.arctan2(e[1],e[0])
            M = u - E*np.sin(u) # mean anomaly

            return(a,E,omega,Omega,i,M)
        except ZeroDivisionError:
            #meaningless, but avoids crash
            return(0,0,0,0,0,0)


How can I estimate the M/L ratio from a fitted SED on a galaxy?

The luminosity L isn't estimated, it's observed. You make models of stellar populations with many different parameters and those give you a set of SEDs. For each set of parameters, you will have an SED covering a range of wavelengths (a continuous line).

From your observations, you will have an SED of a few points depending on how many wavelengths you observed. You use $\chi^{2}$ fitting or some other technique to compare all your model lines with those few points and see which fits best. This gives you the best fit value for the parameters in your model, one of which is M/L.

You have the actual luminosity of the galaxy from the observations (at some wavelength) so you multiply the estimated M/L by the measured L to get an estimate of M.


Calculating the apparent magnitude of a satellite

This is for satellites with unknown size and orientation but known standard magnitude (Standard magnitude can be found on the satellite info page of heavens above, the number is called intrinsic magnitude)
The proper formula is

            double distanceToSatellite = 485; //This is in KM
            double phaseAngleDegrees = 113.1; //Angle from sun->satellite->observer
            double pa = phaseAngleDegrees * 0.0174533; //Convert the phase angle to radians
            double intrinsicMagnitude = -1.8; //-1.8 is std. mag for iss

            double term_1 = intrinsicMagnitude;
            double term_2 = 5.0 * Math.Log10(distanceToSatellite / 1000.0);

            double arg = Math.Sin(pa) + (Math.PI - pa) * Math.Cos(pa);
            double term_3 = -2.5 * Math.Log10(arg);

            double apparentMagnitude = term_1 + term_2 + term_3;

This will give the apparent magnitude of the satellite.
Note: I gave the formula in C#


Has the velocity of photons coming from black holes already been measured?

No.

Photons can't come from the singularity of a black hole, or from beyond the event horizon.
While some very good candidates for black holes exist, none have been certainly observed.
We have never observed Hawking radiation: it remains theoretical.
We have observed radiation from the accretion disk of likely black holes, however the matter in this disc is orbiting in 'normal' spacetime.
Direct measurement of the speed of individual photons is pretty much out of the question. We can measure the speed of light by creating photons at one point and then timing when those photons arrive at another. You can't measure a photon's passing, as to measure it you would have to interact with it. If you interact with it, it is no longer the same photon. This leads to...
When we speak of a photon travelling from one point to another, remember that it is a quantum process: the photon interacts with matter and with itself. Photons are not classical particles. Everything about them needs to be considered through quantum electrodynamics.
The flashes from pulsars are coming from a region of intense gravity. They do propagate at the speed of light: pulsars measured from different locations flash at slightly different times due to light travel time. There is nothing special about the space-time outside of a black hole as compared to that around a neutron-star, only a difference of degree. Gravity is still pretty intense around a neutron star, $g\approx10^{12}\mathrm{ms^-2}$
Gravitational redshift from white dwarfs is well known. Gravitational redshift is predicted by relativity, which assumes a constant speed of light.
Constant speed of light is pretty fundamental to relativity. If a theory breaks this, a lot of other physics is broken (exceptional claims require exceptional evidence)


How fast would an object at cruising altitude(around 10,600 m) need to travel to stay on the opposite side of the Earth to the Moon?

Well, as a first approximation,  you know the Earth rotates once every 24 hours while the moon moves "only a little" in that time.  This means your object must appear to an observer on the Earth' surface to rise and fall 12 hours opposite to the moon.  First thing, then, is you better be moving East to West, just as the moon appears to do.

Using the Earth's radius as approximately 6400 km,  your object is essentially at the Earth's surface to 2 significant digits.  The orbital circumference is thus $2*\pi*6400$ km.  You want to cover that distance in  24 hours, or 1676 km/hr .

If you want to get very accurate results, you'll need to take into account the moon's eccentricity with respect to the equator (so that you are always on the opposite end of a diameter thru the Earth's center from the moon) as well as the fact that it moves approximately 1/28 of a revolution per day.


Why did Earth need material from comets to assemble the raw materials for life?

The earth in its early stages was far too hot for water to be in liquid form and the water vapor that would have been around at the time would have been blasted away via solar wind (http://en.wikipedia.org/wiki/Solar_wind). Anyone claiming to know exactly how water came to be on Earth is likely daft, but comets and asteroids seem to be the most common theory for the source. MinuteEarth, a YouTube creator has a great video about it but as a new user on this particular sub-site I cannot link multiple things at once.

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


What phenomena was this video from SOHO actually showing?

It's a polar crown cavity or coronal prominence cavity. Here is another Youtube video explaining it. It references A new look at a polar crown cavity as observed by SDO/AIA

  Conclusions.
  We propose to define a polar crown cavity as a density depletion sitting above denser polar crown filament plasma drained down the cavity due to gravity. As part of the polar crown filament,
  plasma at different temperatures (ranging from 50000K to 0.6 MK) is observed at
  the same location on the cavity dips and sustained by a competition between the gravity and the curvature of magnetic field
  lines. The eruption of the polar crown cavity as a solid body can be decomposed into two phases: a slow rise at a speed of 0.6 km/s,
  and an acceleration phase at a mean speed of 25 km/s

The video on this page shows one forming nicely.


Can we visually perceive exoplanet transits with amateur telescopes?

Amateur equipment is good enough. But you cannot detect it with a naked eye.
The change in flux for a passing exoplanet in transit is roughly 1%...2% at most for the larger exoplanets - and it is a gradual change. That's a change you do not notice with the naked eye, but it needs photographic equipment to create a sequence of images which allow analysis of brightness of all seen stars in order to detect the variation in the one interesting one.
With an appropriate camera and stable mount your 70mm telescope will suffice to detect these light changes in a sequence of moderately long-exposure images over the course of a few hours (exposure time a few minutes each).
Many of the professional ground-based exoplanets surveys don't use or didn't use much larger telescopes (e.g. see the HATnet programme). The difference often is more in the grade of automation, the sensitivity of the photographic equipment and automation of data processing pipeline.


Super-pressurized ice probably wouldn't be static.  It would (almost certainly) move around and shift with the seasons and planetary climate changes and all that good stuff, so I'm extremely doubtful that anything would stay "littered" on the floor in this scenario.

Other factors would be whether the planet had life, which can drop shells or scales and other stuff, which falls to the ocean floor, and volcanic activity which likely happens on most planets and would eject some matter to cover whatever was on the surface over time.

Plate tectonics, obviously another factor though I'm not sure how plate tectonics would affect a thick layer of ice.

This planet is too hot for life (probably) and likely too large for plate tectonics, but I'm sure the hot ice still shifts and moves around and the interior heat has to escape from time to time, either cracks in the surface or volcanoes.   I think you'd need a nearly dead surface like the moon or mars for things to "collect" like that.   I don't think an exotic ice planet would cooperate in that way.

update,

  what I am trying to do is come up with a good scenario where life "as we know it" could arise and flourish on an ocean planet. To me that meant needing some sort of ocean floor that wasn't just some exotic form of ice. I've seen standard models of ocean planet interiors, but hadn't really considered that there will be a wide variety (ie. the interior of Ganymede is probably quite different than Europa). So, in the right conditions I suppose volcanism and plate tectonics may be the answer (super-earths may be more tectonically active than Earth)

Lots to cover here, I'll give it a shot briefly.

Life may have arisen on earth near the ocean floor but not on the ocean floor.  it was the hot vents of gas that Extremophiles lived in.   I don't think they require the ocean surface, just the volcanic jets with methane and other gasses that they could use.  (I think they use Methane, not 100% sure).   I think they require liquid water, but not ground.

Since the deep below the surface of planets tends to be hot, exotic ice would usually be quite hot, I would think, just solid under very high pressure, I suspect that only Extremophiles could live near exotic ice in those high temperatures.   High pressure is less of an obstacle but high temperature might limit the options for life to the very primitive/tiny (I think).    The good news is, nearer the surface, you might get lower temperatures and that might open the door to more advanced life.   I think what's on the surface is generally more interesting than what's on the floor.    Ganymede and Europa both have frozen surfaces with occasional breaches from tidal forces.   A larger planet might have exotic ice deep down but more earth like oceans towards the surface.   I think that would be a lot more interesting to us.

I don't know how many varieties of water worlds there are.   Neptune is a water world of sorts though it's usually categorized as an ice giant and it probably doesn't have what we would consider oceans but it has a lot of water.

On, Plate Tectonics, it might not be essential to life on water worlds.   Volcanism can exist without plate tectonics and Volcanism is probably essential to life on water worlds.   Don't get me wrong.  Nothing beats plate tectonics for circulation and recycling of materials and preventing unwanted buildup of waste, so plate tectonics is really cool and better than just volcanism, but I think active Volcanism would be sufficient for life in oceans.   It's really hard to say anything definite about this subject though, as it's based on theory.

It's probably possible to have volcanic eruptions even though dozens or maybe even hundreds of miles of exotic ice because the heat has to go somewhere, eventually, assing it's likely to build up over time, so either by circulation of eruption, the heat has push through at some point.  This even happens on so called "dead" planets like Mars or even the Moon.   Mars still has the occasional volcanic eruption, just not very often.

But water worlds certainly can have plate tectonics.  There's nothing in the water that would prevent it from happening.  Plate Tectonics is, as I understand it, primarily a factor of the size of the planet.   Gas planets - different story, but planets with a hard surface, Earth sized, a tiny bit smaller to a fair bit but not much bigger are good candidates for plate tectonics (I think).  There's some debate on how large, I think, still going on.   But I remember reading that ocean/water worlds might even be more likely to have plate tectonics.   Plate tectonics is definitely something we'd look for if we ever get a close enough look at other planets in different solar-systems (exoplanets).

Just my thoughts on this.  Not meant to be complete or definitive.


It depends on the kind of star you want to make.

Basically, what you really need is hydrodgen; as soon as enough hydrodgen is gathered somewhere and collapse, you end up with a star.

But, if you just have hydrodgen, you will get a very massive star, and with a history different than the stars observed in the local Universe. Why so?

hydrodgen is a poor coolant, therefore the Jeans' mass (the minimum mass for a density structure to gravitationally collapse), which strongly depends on temperature, will be much higher if the temperature is higher. In practice, it means that the formed stars will be much more massive.
you need at least a small fraction of carbon to launch the CNO cycle in stellar cores (CNO cycle is one of the two ways to burn hydrodgen and to turn it into helium); stellar cores are then hotter and denser, and stars become hotter and more luminous.

So if you want to make a star as we typically observe in our galaxy, you will need also some molecules (hopefully there are plenty of them) and also some carbon. Else, you'll get big, massive, hot and luminous stars. Like the first generation of stars (called Population III stars, you can see here if you want more dirty details) in the early Universe.


The Hubble constant effectively has a unit of inverse time only: ${\rm s}^{-1}$:
It is usually given as ${\rm km}\cdot{\rm s}^{-1}{\rm Mpc}^{-1}$. Yet looking at it, km and Mpc are both units of length, so they cancel eachother out when converted to the same unit. Thus add the appropriate unit conversion from km to Mpc and it remains an inverse time only:
$$t = \frac{1}{H_0} = \frac{\rm Mpc}{60 {\rm km}\cdot{\rm s}^{-1}} = \frac{3\cdot 10^{22} {\rm m}{\rm s}}{60\cdot 1000 {\rm m}} = 5\cdot 10^{17}{\rm s} = 15{\rm Ga}$$
Add dimensionless factors to the equation to account for different cosmological models and inflation theories. This is anyway only a ballpark number - but an easy one to get for the age of the universe.


At what distance would it be safe for humans to stare at the sun?

If the sun is visible as a disc, then it is certainly can cause damage. As you move further from the sun the disc appears smaller, but the surface brightness remains the same. The eye can resolve a disc of about 1 arcminute, and the sun is about 30 arcminutes when viewed from Earth. So you would need to be at least 30 times further from the Sun than the Earth is, which puts you almost exactly at Neptune.

From that distance, the sun would be a magnitude -19 star. If you stared at it, it could probably still damage your eyes, however, your eyes are constantly jumping (saccades) which might move the image of the sun around the retina enough to prevent damage. There is, of course, no atmosphere to protect you from UV, so let's hope that's not a problem.

Beyond Neptune the sun gets dimmer, as it can't appear to get any smaller, and there is probably little risk by the time you reach the main part of the Kuiper belt.


How do focal length, angular magnification and field of view relate?

Magnification and Field of View are directly opposite. A low magnification will give a large field of view, and a high magnification will give a small field of view. (Field of view indicates how "wide" is the portion of sky that is viewable in the eyepiece.)

Think of looking at the Moon with low power. Let's say the Moon fills 1/4 of the view, meaning that you could get 4 Moons lined-up side by side across the view. When you increase the magnification, the Moon looks bigger, so it fills more of the field of view. And therefore, you can get fewer Moons lined-up side by side. If you double the magnification, the Moon fills 1/2 of the view, meaning you can now only get 2 Moons side-by-side in the view.

Note that the above example is unusual for a telescope, depending on the design of course. Most astronomical telescopes used by amateurs will show an area of sky slightly larger than the Moon when using "low" power. Being able to fit 4 Moons is a bit of an exaggeration, for easy of explanation.


Upgrading telescope from 5" to 8" or 10"?

The first thing I would recommend is to join an astronomy club or society.  There's most likely one near you, and membership is generally very affordable (in the US, prices usually are $50 or less per year).  Most clubs conduct freqeunt star parties, which would give you a chance to see a wide variety of telescopes in action and talk to their owners to find out more.

As Alphecca replied, the light gathering power of a telescope is proportional to the square of the radius of the aperture.  For example, your NexStar 130SLT has an aperture of 130 mm (just over 5").  If you use the formula for area (pi * r^2), this gives you an aperture area of about 13,273 mm^2 (square millimeters).  If you were to upgrade to an 8" telescope (203 mm aperture), you'd get an aperture area of about 32,365 mm^2, which is about 240% the light gathering area of your 5" scope.  If you go to a 10" (254 mm), you end up with about 50,671 mm^2, or about 380% as much as the 5" or 157% as much as the 8"

What does this actually mean?  That's a little more complex.

First, there is more light gathering.  Contrary to the common opinion of those not experienced in using telescopes, magnification is not the key concern with a telescope, light gathering is.  But there's more to it than just that, and I'll circle back around shortly.   In the case of light gathering, larger apertures are better.  The telescope, in a sense, acts as a funnel for light, collecting from a larger area and concentrating on a smaller area.  In this sense, a larger telescope can, quite simply, gather more light.

But there's a catch: it's gathering ALL of the light that hits it, not just the light from the target object.  It will gather the light of your target object and any stray light, such as light pollution.  To deal with this, we need to increase contrast to make the dim deep sky object stand out from the sky.  This is heavily influenced, then, by things such as focal ratio and your eyepiece selection.   I spend a fair amount of time on Reddit r/telescopes, and we've had some discussions about this. One of the other regulars there wrote a pretty good article on medium.com about this.

So, to put it mildly, there's more invovled than just getting a bigger aperture.  Though, in fairness, having the biggest aperture you can afford and handle as far as weight and size goes is always the best way to go - you will never have too LITTLE aperture.

The real questions then become what you want to do with it.  If you want to do astrophotography, you're going in entirely the wrong direction.  For visual observation, however, an 8 or 10 inch telescope is a good option.  I personally recommend an 8 inch for any beginner that can afford one.

But, based on what you said about being disappointed with the Meade 8", I think there is a question of expectation.  If you expect to see what you see in photographs, then you're just plain out of luck. Even a very large scope (say a 24" Dob) won't do that for you. Our eyes are simply not capable of long exposure imaging as are cameras.  But a good observer who has learned to train their eye properly can still see some pretty amazing sights.  And this is another good reason to join a club: to spend time observing with other, more experienced observers and learning how to truly see through the eyepiece.  Visual observation is not just a matter of pointing the telescope in the right spot and taking a look.  Observing requires spending time letting your eye and brain work together to really see what's out there.

I jsut said our eyes aren't capable of long exposures.  This is true, but it's also true that extended observing does do something that's not entirely unlike long exposure imaging.  As you spend time at the eyepiece, your eyes collect more and more light and your brain does start to synthesize a more complete image.  It starts to fill in gaps and show you more and more.  This is especially the case when observing from darker skies where there is less light pollution to muddle the view.  A great example in my experience is M51.  From my club's dark site, when it's up in the sky, I can almost alawys make out the two galactic cores, but frequently the spiral structure is hard to see.  But if I have a comfortable seated viewing position and spend a few minutes, my eyes pick up hints of the spiral arms, which my brain starts to fill in.  A good technique for developing this is sketching.  If you sketch the object, your brain process it differently and you start to catch hints and glimpses of more detail which end up in your sketch.

Your current telescope is a GoTo, as is the Meade you mentioned, and you talked about an 10" motorized Dob.  While GoTo technology and tracking can be very helpful, they are also often more of a hindrance than expected.  Most particuarly, the cost of the GoTo mounting inflates the overall cost of the instrument.  An 8" standard Dob can be bought or under \$500 (occasionally under \$400).  An 8" GoTo telescope is likely to run over \$1,000.  The GoTo equipment typically will at least double the price.  For the price of an 8" GoTo dob you could get a 12" manual Trus dob with almost 550% the light gathering area of your 5".  (Not to mention a larger aperture offers greater detail resolution capability for planetary observation).

In the end, my recommendation is to start by joining a club and spending some time learning more.  Then you can make a more informed decision on a purchase (additionally, many clubs have a thriving buy/sell/trade culture with members constantly selling off something to fund their next purchse - I've bought most of my equipment at very good prices this way).

Good luck and clear skies!


Is there a formula to calculate the angular speed at which the Sun apparently moves around the Earth?

If you don't have to have that precise value, then the angular speed of the sun is simply this:
$$\frac{360°}{24\text{ h}}$$
We can simplify this to SI units (we won't be using radians here, for simplicity):
$$\frac{360°}{24\text{ h}}=\frac{360°}{24\text{ h}}\times \frac{1\text{ h}}{3600\text{ s}}=\frac{1}{240}\frac{°}{s}$$

Let's suppose, that the angular speed of the sun is not constant on Earth. What would happen? The time between two same noons would be different. That would mean, that Earth would rotate with different angular rate on different points. We know that this isn't true, so the supposition is false. (But the Sun does rotate on different latitudes with different angular rates, so the angular speed of some star, looking from the Sun, depends on your location. Of course, this is theoretical and nobody has tried that yet.)
The angular speed doesn't depend on the latitude, but it depends on the current distance between the Sun and the Earth. For that you need true anomaly $\nu$, which is given by Kepler equation, which is transcedent (it can't be solved with finite steps = you have to use computer). Why does it depend on the current distance? If the Earth would not circle around the Sun and it would be stationary, the day would last for around 23 h 56 min. But it does circle around the Sun, thus the some point on the Earth has to wait a little more, to reach the culmination of the Sun, so it lasts for 24 h. The angular speed around the Sun is through the year changing, so the true day deviates from the mean day by a really tiny amount (i think that around 30 s).


Interaction of stellar winds in binary star systems

Systems like this are known as colliding-wind binaries (CWBs), and they do produce some interesting effects. When winds collide, they create shocks, which in turn heats up gas. The most notable result is x-ray emission, which tends to happen whenever you have non-negligible shock heating in stellar winds; there is certainly additional thermal and non-thermal (e.g. synchrotron) radiation at other parts of the spectrum. I don't know if this would lead to significant structure formation - bear in mind that hot stars with strong winds may also generate more prominent outflows and nebulae through unrelated mechanisms; Eta Carinae is a great example.
This actually poses a problem for people trying to study stellar winds in x-rays. We can derive a lot of information from spectroscopic observations of winds - composition, temperature distribution, mass-loss rates - for simple cases when x-ray emission arises from the line deshadowing instability (Owocki, Castor & Rybecki 1988), leading to embedded wind shocks (EWS).
Magnetic fields complicate matters by forming shocks as plasma is forced to travel along field lines, and colliding-wind binaries make it very difficult to study the individual winds because much of the emission is now coming from the CWB shock front and not purely the EWS model. I don't really know how folks studying CWBs disentangle the effects - when I was working on stellar winds, we mostly just ignored all magnetic and O+O binary stars in the sample!


Why doesn't the fusion process of the sun speed up?

Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less?

Yes, at least over human timescales. You could reasonably expect the fusion rate within the sun to be the same today as a few thousand years ago or into the future, give or take some small fraction.

  Why does this not speed up, i.e. one fusion event creates energy for two fusion events, etc.?

The energy released by fusion is quickly distributed as thermal energy in the centre of the sun, and the temperature difference between surface (around 6000K) and centre (estimated 15 million K) drives an energy flow from hot to cold.

  Does every collision of atom cause a fusion event, or is the probability small for a fusion event to happen thus it's not a runaway reaction?

Fusion in the sun is not a runaway nuclear reaction (like a critical mass of uranium in a fission reaction).

It is possible in theory to have runaway fusion events, but the pressure and temperature for these to happen are not approached in the core of the sun. For stable stars like the sun, the forces and energy flows are in equilibrium - if the core grew slightly hotter, then the pressure would increase and the star expand slightly against the force of gravity to compensate. Interesting things happen when stars fall out of equilibrium and runaway fusion ignition can happen in some scenarios.

In addition, this equilibrium point moves during the lifetime of a star as its mix of elements changes due to fusion. This is predictable for many stars and forms the basis of the main sequence stars in the Hertzsprung-Russell diagram

  I have heard that the probability for fusion event to happen is only 1 in 10^12 for every collision

I don't know the accuracy of that, but it seems reasonable. The definition of "collision" becomes somewhat arbitrary in such a hot dense environment. If you only include approaches close enough to make the strong nuclear force dominate the interaction, the ratio could be higher.

Another fact that I found interesting in the same area is that the power density from fusion - i.e. the Watts per cubic metre of substance - in the sun is roughly the same as that found in a typical compost heap. It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.


Why is Enceladus the only geologically active moon among its neighbours?

This puzzle is something called the Mimas paradox. The two bodies are nearly twins, but Mimas is substantially less active than Enceladus. There are currently several proposed solutions:

Mimas cooled quickly, before it entered its resonance with Tethys (which contributes to tidal heating, just as Dione does for Enceladus). A "hot start" from rapid radioactive decay (of aluminum-26 and iron-60) shortly after Enceladus' formation could have allowed its subsurface ocean to stay liquid.
Enceladus is rockier and therefore has a higher density. This would have led to quicker cooling, and would also mean that tidal dissipation could be higher for Enceladus than for Mimas. It does seem a bit paradoxical - one would imagine that more rock would mean less water ice and therefore a smaller subsurface ocean - but this isn't a problem.

It's worth noting that Tethys is thought to be made almost entirely of ice. If the rocky-Enceladus hypothesis is true, then that could be a major factor in the lack of geological activity on Tethys. However, Tethys likely once had a subsurface ocean. Dione currently does, and experiences tidal heating; however, its semi-major axis is large enough that the effects are not as strong as on Enceladus.


What is the difference between these various Angular Resolution formulas?

$2.06 \times 10^5$ is the radian arcsecond conversion, equal to $180 \times 60 \times 60 / \pi$ .
The $1.22$ is a geometrical factor applicable to circular apertures, as outlined in Wikipedia.


All star names are unofficial. A few stars have ancient names (such as Sirius) all other stars are referred to by their position in a catalogue or star atlas.

As well as I can tell, Al Sadira means "the ostrich", or perhaps in context "The (riverbank) ostrich" indicating a type of tree that grows by rivers (baby name sites offer "lotus tree").

There may be a list of Arabic star names that lists this star as Al Sadira(h) but the name is not in common use and won't be recognised by most people.

The IAU name, Ran, from that of a Norse river goddess, is a recent innovation, and doesn't seem to be much used. The Bayer Designation "Epsilon Eridani" is by far the most well known.


How can Mercury's sodium tail be imaged?

Yes, it is possible; see https://twitter.com/sevospace/status/1328082050889490433
However, M. Voltmer used a “custom-made sodium filter,” without giving any details…


Why does this nucleogenesis table not label s and r-processes for all elements (not-labmade) past Fe?

If you go back to the sourced article in your graphics by Jennifer Johnson it has a more verbose colour legend. For the grey part it explicitly states

very radioactive elements. Nothing left from stars.

and her figure caption more verbosely

Each element in this periodic table is color-coded by the relative
contribution of nucleosynthesis sources, scaled to the time of Solar
System formation. Only elements that occur naturally in the Solar
System are shown; artificially made elements and elements produced
only through radioactive decay of long-lived nuclei are shown in gray.
The data plotted in this figure are available in table S1.

And that pretty much sums it up:
The colour gives the source of the elements as found on Earth.
Thus the colour does NOT indicate how the elements can be formed in general, but the origin of the isotops as found on Earth (or present on formation of Earth, thus still long after nucleosynthesis in their progenitor stars). The lifetime of the longest isotope of technetium is 4.2 million years, and of promethium of only 17 years. Thus basically all of those grey-coloured elements are decayed long ago and none of the naturally occuring specimen on Earth are anymore of primordeal origin but result of fission of higher-order elements.
EDIT to add: As to the 2nd question in your posting which is quite distinct from the title

"why the neutron capture processes begin at A = 38. I thought anything with an atomic number greater than 26 has to be formed by neutron capture?"

Those elements can actually be formed by fusion of the lighter elements in the core of the supernova due to the high density and temperature (basically similar to how we can create heavier elements in the lab, colliding heavy ions) - but that fusion will consume energy instead of releasing energy.


They can, and some do. These stars are called variable stars, because their luminosities as observed from Earth vary over time, often (though not always) in a regular period. Here are some broad categories:

Pulsating stars, where fluctuations lead to increases and decreases in size or temperature, which in turn produce changes in the star's brightness. Example: The oft-cited Cepheid variables.
Eclipsing binary stars involve two stars orbiting each other. When one passes between the other and Earth, the combined luminosity of the system appears to decrease, even though the stars' intrinsic luminosities are probably constant. Example: Algol.
So-called eruptive variables (often lumped together with cataclysmic variables, which are different) may have irregular outbursts caused by flares or other phenomena. Example: Luminous blue variables (LBVs).

The change in brightness and the length of the variations depend on the type of variable star in question. The luminosity can vary by anywhere from a fraction of a magnitude to many magnitudes. Likewise, the period (if there is one) can range from hours to years.


What decides the direction in which the accretion disk spins?

Stellar systems are born from clouds of turbulent gas. Although "turbulence" means that different parcels of gas move in different directions, the cloud have some overall, net angular momentum. Usually a cloud gives birth to multiple stellar systems, but even the subregion forming a given system has a net, and non-vanishing (i.e. $\ne0$), angular momentum.

Parcels moving in opposite directions will collide, and friction will cause the gas to lose energy, such that the cloud contracts. Eventually subclouds moving in one direction will "win over" subclouds moving in other directions such that everything moves in the same direction, keeping the original angular momentum (minus what is ejected e.g. through jets).

This means that the central star will rotate in the same direction as the circumstellar disk and that, in general, the planets that form subsequently, will also not only orbit the star in the same direction, but also spin in the same direction around their own axes. This is called prograde rotation. Sometimes, however, collisions between bodies may cause a planet or asteroid to spin in the opposite direction. This is called retrograde rotation and is the case for Venus and Uranus.


What makes a really good observatory site, besides altitude?

• Not near an active or dormant volcano (but Mauna Kea seems to disprove this?). This is kinda a bummer because a lot of tall mountains seem to be volcanic.

There's nothing intrinsically wrong with dormant volcanos, as Mauna Kea and the observatories in the Canary Islands demonstrate. Not sure where you would have gotten that idea.

  • Somewhere with clear and/or dry weather for as much of the year as possible

Clear weather is essential -- you can't observe when it's cloudy!

Dry weather is very good, especially for infrared observations (water vapor blocks a lot of infrared light).

  • Not near major light pollution like cities

Yes.

  • Cold weather is better than hot weather? Not sure I understand that; if it's uniformly hot or cold then I dont see how it makes a difference.

Cold climates tend to have drier air. (Antarctica is an extreme case.) Of course, being on top of a mountain means colder air, which is why Mauna Kea is good despite it's being in the tropics.

  Am I missing any? Or are these wrong in some way?

You also want stable air with good "seeing", which rules out places with turbulent air and lots of wind (and is another reason mountaintops are good: they're usually above the most turbulent layers of the atmosphere).
Quoting from this page (which has a good discussion of the general topic):

  Seeing ... requires at minimum a lack of extra turbulence at all atmospheric levels, and seems to be best satisfied in the convergence zones just outside the tropics, at latitudes about ± 30º. Also, minimal local turbulence is often associated with mountain peaks that reach into the otherwise undisturbed oceanic airflow, as on islands or coastal ranges (and given the direction of the planet's rotation, this generally favors western coast ranges).


Why are only some of the geologic features in Mars' Noctis Labyrinthus region named?

Nomenclature on planetary bodies is meant to ease and standardize communication.  If an object is referred to often, or if it is important for someone's research, then the scientist(s) involved can submit a name request to the International Astronomical Union (IAU), or through some other body that submits names to the IAU (such as the United States Geologic Survey's Astrogeology branch in Flagstaff, AZ).  After checking for various criteria*, the name may or may not be approved.  If it is approved, then the feature is named.  If it is not approved, then the scientist(s) can resubmit that name or a different name.

I am not sure if it is a full list, but the USGS's Gazetteer of Planetary Nomenclature lists a lot of the themes on different bodies.  Based on that, cavi and tholi would be named based on nearby albedo (brightness) features or craters.

*Criteria for naming are extensive.  Almost every type of feature (such as an impact crater) on many bodies has a theme, and that theme must be followed.  If the name does not meet that theme, then it will be rejected.  Other criteria are that the name cannot be offensive or induce a strong negative response (such as naming something "Satan" could induce a negative response in many people); the name cannot be used elsewhere in the solar system already (though some historic exceptions exist); the feature must be well defined (a vague region on a surface would not, therefore, qualify); the feature cannot be named after someone who is alive or has died less than three years ago; and there cannot be a preponderance of names that are biased towards or against gender, nationality, region of the world, etc.


How far have individual stars been seen by radio telescopes?

Some types of red giant stars are susceptible to intense water and silicon oxide MASER activity in their outer envelopes. This results in intense radio emission. Examples of such maser activity have been observed in the Small Magellanic Cloud,at a distance of $\sim 60$ kpc  (e.g. van Loon 2013).

 Click for full size.

IRAS 05298−6957 (van Loon 2013): Note the logarithmic wavelength axis; the water maser peak lies at about 18 cm / 1670 MHz.


Hypothetical planet jump: Will it perturb star systems?

There is essentially no room in our habitable zone for another Earth-sized planet.  The exact timescale for instability would depend on the details of where you put it, but the system couldn't survive long term.
Most investigations of the long-standing question of solar system stability (without any additional planets) find that the system is stable in the long term with about 99% probability, e.g. this page by Scott Tremaine and this discussion on Wikipedia, but that there is some chance of instability even with the current set of planets, and no room for another.
It appears that many extrasolar planetary systems are "full" as well, although selection effects make it hard to know whether there are planets we aren't seeing in systems that aren't yet known to have multiple planets.
The "fullness" of these solar systems suggests that it may be that even more planets form initially, but then some are ejected until the system reaches a stable configuration.
Interestingly, in our own inner solar system, the place where there is the most dynamical space for another planet is inside the orbit of Mercury, say with a semimajor axis of 0.1 AU or so.  That might seem counterintuitive, since the spacings between Earth (1 AU), Venus (0.7 AU), Mercury (0.4 AU), and a planet at 0.1 AU would all be about 0.3 AU, the same in absolute terms.  But what matters most for stability is the ratio of the semimajor axes, so Mercury to an inner planet would be a factor of 4, whereas Earth/Venus is only 1/0.7 = 1.4.
Such a planet isn't in the habitable zone, and it likely wouldn't be stable on very long timescales with our current configuration of other planets, but we do know that many (perhaps even most) extrasolar planetary systems have planets at those very small orbital radii - see the image below, made from data from the NASA Exoplanet Archive.  There are strong selection effects there (close-in planets are the easiest to detect with both the transit and radial velocity methods), but the high frequency does tell us that such planets are relatively common and can have stable orbits.


How can I find the radius of a exoplanet using doppler spectroscopy

In general, you can't. The spectroscopy tells you about the properties of the star and the properties of the orbit and gives some information about the mass of the planet ($M \sin i$, where $i$ is the orbital inclination).
If the spectroscopy were super-precise and with extremely high signal-to-noise and high time resolution, then you might be able to map the Rossiter-McLaughlin effect with such precision that you can estimate the planet radius. However, this would only work if the planet is transiting in front of the parent star - in which case you might as well have used transit photometry in the first place.
Another possibility is to search in the spectra for the Doppler-shifted scattered starlight from the planetary atmosphere (e.g. Martins et al. 2015). However the amplitude of this is tiny depends on both the radius of the planet and its albedo/atmospheric composition.


What are the white reflections on Ceres?

According to the Wikipedia page linked by @BowlOfRed, they're ice and/or salt.

  a material with a high level of reflection, and [the scientists] suggested ice and salt as possibilities.

  ...

  On 9 December 2015, scientists reported that the bright spots on Ceres may be related to a type of salt, particularly a form of brine containing hydrated magnesium sulfate (MgSO4·6H2O); the spots were also found to be associated with ammonia-rich clays. Arizona State University scientists have proposed that the bright spots are best explained as resulting from briny water erupted from Ceres's interior that subsequently sublimated, leaving behind only the salt deposits. Near-infrared spectra of these bright areas are consistent with a large amount of sodium carbonate, (Na2CO3) and smaller amounts of ammonium chloride (NH4Cl) or ammonium bicarbonate (NH4HCO3). These materials have been suggested to originate from the recent crystallization of brines that reached the surface from below.


Is there any way a planet could form independent of a star?

Well you need to see the related question brown dwarfs and planets , because the answer to your question depends on how you define a planet.

If you demand that a "planet" has a rocky core then it seems very unlikely that a planet could form in isolation away from a parent star. The parent star is needed in order to differentiate the rocky material from the gas and allow it to condense.

On the other hand, if you wish to define a planet as simply an object below a certain mass (say the deuterium burning threshold at 13 Jupiter masses) then it seems very likely that such an object could form in isolation. They would be entirely gaseous, but there would be little to distinguish them from brown dwarfs at only slightly higher masses.

At present there are plenty of candidate "free-floating planetary mass" objects.
For example see Joergens et al. (2014); Liu et al. (2013); Zapatero-Osorio et al. (2000). Unless we have our understanding of the physics completely wrong, then it is likely that at least some of these are lower than 13 Jupiter masses. However, their origin remains unclear. It is possible they could all have formed around stars and then subsequently been ejected, but the presence of significant numbers of these objects in young star forming regions and the lack of $\sim$10 Jupiter-mass objects orbiting stars, suggests that there is an alternative formation scenario that can produce such objects in isolation.

What could these formation scenarios be? These low-mass objects could just be an extension to lower masses of the fragementation process that forms stars; they could be ejected embryos that started their lives in multiple systems; they could be "failed" stellar cores that could not accrete more gas because of photoevaporation by nearby massive stars; or they could form by gravitational instability around stars with unusually massive disks and be ejected by a close encounter with another star. These possibilities are reviewed by Whitworth et al. (2006) and Chabrier et al. (2014), and are all still thought plausible to some extent.


Impact on the rest of the solar system if Mercury or Venus were disassembled?

Not really. The mass of Mercury and Venus together are just 0.0027 of Jupiter's mass. So the effect is really small.
Now, in principle, there are resonances where the regular tug from planets affect the long-term evolution of the orbits of other planets. The exact details get very messy, but in some cases the presence of a planet stabilizes certain orbits - or destabilises them (this is very visible in the asteroid belt). But the planetary orbits look largely non-resonant. So if two planets went missing there would not be much of an effect.
It is likely that this is true for Jupiter too. What would happen is that the Trojan and Greek asteroids would drift out of their 1:1 resonance and end up in a loose belt, and the asteroid belt would also reorganise.
Overall, you don't need much mass for a Dyson sphere if you want to just collect energy. My paper on using one to settle the universe disassembled Mercury mostly as a conservative estimate to get more than enough material - I currently think you need a few largish asteroid masses, maybe Ceres (assuming it has the right elements, an important if). Conversely, Dyson element orbits will be subject to gravitational interactions from remaining planets that will cause precession and other possibly unwanted shifts one need to compensate for. But a Dyson sphere is mass-wise a microscopic thing compared to planets.


Are there many faint meteors that are too faint to see with the naked eye?

The answer is yes. The mass (and hence brightness) of meteorites follows some sort of power-law relationship such that there are many more small particles than big lumps of rock. This means that there are indeed many more faint meteorites than bright ones, though I am struggling to find any detailed study that goes below the magnitude range you can see with the naked eye. This paper by Kresakova (1966) seems thorough and representative. It suggests that the number of meteorites goes up by a factor of 3 for each unit increase in astronomical magnitude (i.e. decreasing brightness).

The paper referred to by Barry Carter in his comments is Cook et al. (1980), which finds that log of the cumulative number of meteors is proportional to about half the astronomical magnitude. i.e.
$$\frac{d \log \phi}{dm} \simeq 0.5 $$
This means that the number of meteorites increases by a factor of $10^{0.5}$ for each unit increase in magnitude - i.e. also about 3.

It is a separate issue as to whether you see more with night vision goggles on. I guess you would because these would have light amplification properties that would enable you to see fainter objects - that's the whole point of them.


In a sense, that is what a circumstellar disk is.

Source: European Southern Observatory (ESO)

These are usually most noticeable around young stars as protoplanetary disks, disks that form planets. In the picture above, the gaps in the "rings" likely represent forming planets, similar to how the gaps in Saturn's rings indicate the presence of moons.

Our own star has a few "rings", though likely not as visually impressive: the asteroid belt, the Kupier belt, the scattered disk, and the Hills cloud.

That said, there are differences between these structures and the ring systems we are familiar with from planets. For example, rings are made of solid particles whereas protoplanetary disks contain a lot of gases. Rings may exist around "mature" planets, whereas protoplanetary disks represent an immature phase of planetary systems. Protoplanetary disks are not visible to the naked eye, whereas ring systems may be. In technical terms, a circumstellar disk is not a ring system.


The Sun is not within the gravitational sphere of influence of any other star. The centre of mass of the solar system (which is very close to the Sun) instead orbits in the general Galactic gravitational potential. Because this has a roughly cylindrical symmetry (the Galaxy is basically a disk with a bulge in the middle), this means that it executes a (roughly) circular orbit around the centre of the Galaxy, taking about 230 million years to do so. At the same time it is wobbling in the vertical plane of the Galactic disc, up and down with a cycle of about 70 million years (see How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it? )

There is really nothing intermediate because the stars in the milky way form a "collisionless system" they don't really interact gravitationally on an individual basis. Stars can be influenced by perturbations in the smooth Galactic potential caused by massive star clusters, giant molecular clouds and spiral arms. This is thought to be why the velocity dispersions of stars around regular circular orbits increases with age.


Can I calculate the apparent position of comet NEOWISE C/2020 F3 from a topographic position with Skyfield?

The Skyfield docs don't show a way to import orbital elements directly, but it can read any BSP format ephemeris you have.
If you request a HORIZONS-generated binary SPK file for the comet, that might be enough to do the job.


Eccentric Earth Orbit and Day/Night Cycles

Your thinking is incorrect.    High gravity, or, more specifically, strong tidal forces lead to tidal locking not faster spinning.   Moons are a good example of this.   The Earth's gravity keeps the same side of the Mon facing the Earth at all times.   The Moon still rotates but it's rotation is in sync with it's orbit around the Earth, 1:1, so the same size of the Moon (the heavier/denser side) always points towards Earth.   Many moons are tidally locked to their planets as a result of the strong tidal force.  Pluto and Charon are tidally locked to each other.

That said, tidal locking takes a long time.  Just passing close to a massive object wouldn't stop a planet from rotating.   It takes many orbits to create the tidal locking, usually millions or billions.  Because planets tend to be pretty uniform in density, gravity doesn't affect their rotation very much.

A highly eccentric orbit, the sun isn't close to the center, either.   A low eccentricity orbit, the sun is close to the center.

A highly eccentric orbit would affect seasons.  With no Axial tilt, the seasons would be entirely governed by the distance to the sun, not the position of the sun in the sky which governs seasons on Earth as it is.   To clarify, it's the Axial tilt that causes the seasons, not the distance to the sun.  But as you specified with zero axial tilt, then the distance would govern the seasons, but at Earth's current eccentricity, that effect would be small.   The greater the eccentricity the greater the seasonal variation in this hypothetical and the seasons would be planet wide, not alternating between the two hemispheres.

Summer would also be shorter than winter due to the Earth moving faster around the sun at the closest point,

The days themselves wouldn't be affected much.   The length of a typical day (Synodic) would vary by a minute or two based on the Earth's orbital speed.  It wouldn't be very noticeable but with a perfect 0 degree axial tilt, the 12 hours of sunshine would vary slightly based on orbital speed and position in the eccentric orbit. . . and the size of the sun in the sky and atmospheric refraction. . . . but only a couple minutes here or there with the Earth's current rotation speed.


Can any stars ever form supermassive black holes?

As you say, making black holes quickly in the early Universe is a major unsolved problem in astrophysics. There are various hypotheses, of which two roughly correspond to supermassive stars.  All basically involve trying to give the black hole a headstart in mass.  There isn't really enough time to grow a $100\,M_\odot$ black hole to $10^9\,M_\odot$, so the idea is to rather get something more massive than a few $\times1000\,M_\odot$. The most recent review I know of offhand is probably Volonteri (2010), but I'm not up to date on the literature.

Basically, imagine a few hundred thousand solar masses of gas collapsing into a primordial galaxy.  As the gas collapses, it potentially fragments, depending on whether or not it can cool efficiently.  If so, the fragments can presumably form stars, but in a very dense cluster.  Ultimately, either (a) the massive stars in the cluster collapse into black holes that then merge into a larger black hole, which can subsequently accrete its way to supermassiveness (supermass?); or (b) the individual stars first merge in the centre, creating a star of perhaps some thousands of solar masses, which would collapse into a massive black hole that could grow.

If the primordial galactic cloud doesn't fragment, we expect a sort of monolithic collapse.  Somewhere in the middle, gas will start to reach hydrostatic equilibrium: a protostar forms.  But it's a protostar that's potentially accreting several solar masses of material per year (or even faster).  So what happens next depends on whether that rapidly infall of material has to time to reach local thermal equilibrium with the protostar, or if it just piles up on the outside.

If the former, then the protostar can become very large: thousands or tens of thousands of solar masses, which, like the supermerger product above, presumably leaves a massive black hole in the end. If the small protostar evolves independently of the infalling gas, it's probably also big enough to leave a black hole, just a much smaller one: tens, maybe hundreds of solar masses.  But it's a black hole embedded in this enormous cloud of infalling gas, which potentially settles into an envelope around the black hole. This structure has been dubbed a "quasi-star", and the black hole inside can grow very rapidly in this cocoon. Eventually, the envelope will evaporate/disperse, leaving the now massive black hole to continue to accrete its way to supermassiveness.

Note that these formation mechanisms are expected to be particular to the early Universe.  Once you add even a small amount of metals to the gas, then star formation is expected to look much more like the "modern" Universe.  In fact, star formation in these scenarios is still far from settled.  The main reason is that you need to follow how the gas evolves from the scale of the protogalaxy all the way down to the protostars inside.  This is a range of scales something like $1\,\mathrm{AU}/10,000\,\mathrm{ly}\approx10^{-9}$, which is numerically very difficult.

And, finally, to answer the question directly, in neither of the supermassive star options do the stars really collapse in supermassive black holes.  SMBHs are so big that they can only have grown so large through accretion.  The supermassive stars would collapse into their progenitors (or "seeds"), which would subsequently grow to such large masses.


How much solar UV radiation would someone get who were on the surface of Titan?

You are correct that the incident radiation density is 100X less - it's a matter of computing the area of a spherical surface, which goes as the square of the radius. So even in the absence of atmosphere, and assuming you wear some kind of spacesuit that's transparent in the UV, you won't get much of a tan.


Calculating the black hole evaporation time

The derivation of life time formula for black holes:
It is not difficult to approximately estimate the life time of a black hole. Since Hawking radiation is indeed a black-body radiation, the rate at which the relativistic energy/mass of black hole ($M$) radiated away by its Hawking radiation can be computed by use of the Stefan-Boltzmann law as
$$- \frac{{dM}}{{dt}} = \frac{{dE}}{{dt}} = A\sigma {T^4},$$
where $\sigma  = 2.105 \times {10^{ - 33}}\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{{\rm{ - 3}}}}{{\rm{K}}^{{\rm{ - 4}}}}$ is the Stefan–Boltzmann constant, $A$ is the object's surface (here black hole's horizon area, $A=4 \pi r_s^2$, where $r_s=2GM$ is the horizon's radius), and $T$ is the (Hawking) temperature. Note that this is the rate at which any blackbody radiates energy, no matter what is that object. Also, in this formula, $- \frac{{dM}}{{dt}}$ means that when black hole radiates, its mass decreases. Having this together with the Hawking temperature of the (Schwarzschild) black hole, i.e.,
$${T_{Hawking}} = \frac{{\hbar}}{{8\pi G{M_ \odot }{k_B}}}\left( {\frac{{{M_ \odot }}}{M}} \right) = 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{{\rm{K}}^\circ },$$
we can evaluate the life time of the black hole ($M_\odot = 2×10^{30} \rm{kg}$ is the solar mass). For this purpose, we shall assume that the black hole initial mass ($M$) will eventually evaporate to nothing. So, integrating the Stefan-Boltzmann law, i.e. ${\tau _{{\rm{life}}}} =  - \int_M^0 {{{(A\sigma {T^4})}^{ - 1}}dM}$, yields
$${\tau _{{\rm{life}}}} = \frac{{256{\pi ^3}k_B^4}}{{3G\sigma {\hbar ^4}}}{(GM)^3} = (2.095 \times {10^{67}}\,{\rm{year}})\,{\left( {\frac{M}{{{M_ \odot }}}} \right)^3}.$$
This is much greater than the age of the universe!
In addition, I should emphasize that these computations have been performed in units in which the speed of light is set to unity ($c=1$). For example, in SI units, the Stefan-Boltzmann constant is given by $σ=5.67 \times 10^{-8}\,\mathrm {W\,m^{-2}\,K^{-4}}$), but in this answer we had $\sigma  = 2.105 \times {10^{ - 33}}\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{{\rm{ - 3}}}}{{\rm{K}}^{{\rm{ - 4}}}}$.
And your final question:

In which limiting case(s) does it hold?

This estimation has been derived for nonrotating black holes. For rotating (Kerr) black holes, it is expected the order of magnitude of this estimation is approximately valid, at least in the case of slowly rotating black holes. I guess, the computation in such cases would be more difficult.
Edit: In comments, it has been discussed that there are other approximations. Of course, for example, the Hawking temperature for astrophysical black holes would be extremely small, even much smaller that the CMB radiation (see more details in my answer here, please), so black hole can absorb more matter and then they grow more and more! In order to derive that formula, you should omit this challenging part. Another approximation comes from the number of particles (scalars, photons, vectors, fermions). However, the Hawking temperature (e.g., according to surface gravity definition) is same for every type of particle, and the number of species only adds a numerical factor to the Stefan-Boltzmann law, but still the order of magnitude of the final answer is approximately valid. In addition, the particles scaping from the black hole need to pass trough the potential and this change the resulting spectrum by a greybody factor ($\Gamma(\Omega)<1$). In this answer, which is very common in GR books, I utilized the Hawking radiation of Schwarzschild black hole with $\Gamma(\Omega)=1$. For more (another) details, also see the following related SE links/papers/books:

Is Hawking radiation detectable?

https://www.amazon.com/Introduction-Quantum-Effects-Viatcheslav-Mukhanov/dp/0521868343

https://arxiv.org/abs/2011.03486

https://arxiv.org/abs/1711.01865


How close would the orbit of Mars have to be to that of the Earth to induce a collision?

It's more likely that perturbations by another planet might induce a collision.

In other systems with bodies in almost the same orbit, what happens is like a game of orbital catch-up: one overtakes the other, then they swap places, so in your (rather vague) scenario, that might well be the outcome instead of an outright collision.

Some of the Saturn ring moonlets interchange positions, for example.

The solar system is stable over a timeline of millions of years (as far as we know), but there have been instabilities in the past, and due to "butterfly effects" instabilities could occur in the future.
(See also https://plus.maths.org/content/rocks-suggest-theres-chaos-solar-system)

Have a play with a simulator such as orbitsimulator.com !

http://orbitsimulator.com/gravitySimulatorCloud/simulations/1433364162018_Solar%20System.html


Does a set moon sky equal a new moon sky?

After the moon has set, the sky is dark. There is a little "lunar twilight", but this is pretty insignificant. Once the moon has set you don't need to worry about the moon. The sky will be dark at midnight.


Is the Sun in our solar system moving or stationary?

The Sun moves, even in the context of the solar system. Gravity of the planets (mostly Jupiter) pulls the Sun out of position with respect to the centre of gravity of the solar system. This wikipedia entry explains it in a lot more detail, and also explains that their common centre of gravity lies outside of the sun.

This wobbling of a star due to planets orbiting them is also one method of how we detect planets around other stars.

Assuming that the Sun is fixed within the context of the solar system is a fairly good approximation. You need accurate and long term observations to detect the wobble.


Is it true dark matter was discovered whilst looking at the rotational speed of galaxies?

What you're possibly missing here, judging from your questions, is the link between the mass distribution of a galaxy and the rotation velocities.

The simplest way to obtain a prediction for a tangential velocity component is to look at the Kepler problem, i.e. a central mass, dominating another smaller mass, orbiting it.
Newton's force-law $\vec F = m \dot{\vec{v}}$ relates acceleration and acting force, which is given by gravity.
All acting forces then sum up to change the velocity of a test-particle $\dot{\vec{v}} = \sum_i \frac{\vec F_i}{m}$.

Then we get a number of cases, depending on how many summands there are, and how they are distributed spatially:

Kepler problem: Only one central $F_i$ exists, which dominates the system's behaviour. For a circular orbit one gets the well-known tangential velocity $v_t = \sqrt{\frac{GM}{r}}$, with $M$ being the mass of the central object and $r$ the distance from it.
A Galaxy: In a galaxy the visible mass distribution is very different from a stellar system. It is spaced out more equidistantly and thus would give rise to a peak in $v_t$ and then a decrease again, after one gets far away from most of the mass. This is however not observed, and calculating the measured velocities of visible mass backward to the true mass distribution leads to the postulation of dark matter.

A related problem is the virialisation of galaxy clusters. When a bunch of mass particles interact with each other gravitationally for long enough, in this case galaxies, their averaged kinetic energy $<T> = <\frac{1}{2}mv^2>$ becomes half the available gravitational potential energy $<V>$. Thus, under the assumption of virial equilibrium one can measure the total mass in a system by measuring the velocity dispersion. Because galaxy clusters are very old objects, it is usually assumed they display this equilibrium. Departures from this are still ongoing research and for reasons I don't remember the milky way is usually not assumed to be virialised.
Zwicky 1933 realised that the Coma cluster then needs to contain as much as 400 times the observed mass, to explain it's observed velocity dispersion.


Looking back in time by looking further away

Before the advent of telescopes, we could only look back in time from a few years (for nearby stars) to a few thousand years (for the most distant stars visible to the unaided eye). In addition to this, a handful of galaxies are visible without binoculars, letting us look back a few million years.

The first telescopes allowed us to see much farther, like hundreds of megaparsec, so we could look back hundreds of millions of years. As technology improves and our telescopes become better, we are able to observe more and more distant objects, and thus look further and further back in time.

The farther away an object is, the less light we receive, and hence the bigger telescope and better detector we need in order to see it. So, being able to look further back in time than previously is not about the light from more distant objects eventually reaching us, but simply a matter of developing the tools necessary to see it.

However, even the most distant galaxy we will ever see (which are at earliest a few hundred million years after the Big Bang, i.e. roughly 13.5 billion years ("Gyr") ago), are not farther away than the cosmic microwave background, which was emitted 380,000 years after Big Bang, i.e. ~13.8 Gyr ago, and which we have been able to detect since the mid-1950's. If we ever manage to build an efficient neutrino detector, we may be able to look back to seconds after the Big Bang, when these particles decoupled from matter and began traveling freely through space.


What colours can rocky planets (or moons) have

I'll begin by saying that for exoplanets, the answer to your question is an active area of research, and currently there are no observational constraints on what rocky exoplanets would look like to the human eye.

Also, this may be a lot more complicated an answer than what you want for a computer game.

If you want to hunt around for this in the astronomical or planetary science literature, what you're trying to do is determine the reflectance, or albedo spectrum, of the planet. That's what tells you how much of the incoming light is bounced back off the surface for you to see as function of wavelength. Note that this assumes the planet is not hot enough to thermally emit in the optical, which will only start to matter if it's >3000K or so. This albedo spectrum is determined by the surface properties of the planet: what it's made out of, how shiny it is, the scattering properties, etc.

Once you have an albedo spectrum, you then can multiply it be the incoming radiation from whatever primary star you've chosen. That's what will make a planet seem slightly more orange around an M-dwarf, and more "normal" around a Sun-like star.

To get to how a person would see it, you'd then need to then multiply that resulting reflected-light spectrum by the spectral sensitivity of the human eye, and then convert that final observed spectrum into a color.

For the planetary albedo spectrum, there aren't a lot of sources in the astronomical literature. The best I can find is this paper. You'll probably have more luck looking through Solar System planetary stuff.

For the stellar spectra, you can try here, but that presumes you're comfortable with astronomical FITS files. For stars hotter than >4500K (the Sun is 5800K), you can probably just assume that the incoming radiation into the planet's surface is blackbody of whatever temperature you want.

Unfortunately, I am not familiar with the spectral sensitivity of the human eye, or how to go from an input observed spectrum to a single color value.


For your experiment, you can take it 2D, so let's ingnore the thickness of Milky Way.

So, we have 400 billions stars(the real number is between 100B-400B) in a circle with 100k ly(the real number is between 100k-160k).

The area of the circle is pi*50k*50k ~= 7800M sqly = 7.8 billions sqly.

We have then ~51 randomly stars per square ly.

But you do not sting the Milky Way with your stick, but cut through it.

If you swipe through the middle, you have 51*100k stars in a width of 1 ly.

So, you have a chance to hit a star per 1/5.100.000 ly which means 1.855.000km.

But the diameter of sun(of your stick) is 1.395.000km. Taking diameter of sun as medium diameter of a star(this is not true), you have a chance of 1.4/1.8 = 77% to hit a star. If you go with the "low density" numbers, 100B stars and 160k ly diameter, you'll get ~31%.

My conclusion is you often hit a star.


Just after the Big Bang, was the space, in its entirety, filled with Hot gas?

I'm not sure what you mean by "nebulae".  You could mean "matter in a gaseous or plasma state" or "cloud of low density gas and dust in space" (which is what astronomers mean by nebulae.)

The former case is believed to be correct, and it pretty well described in Wikipedia. I'll summarize. Immediately after the Big Bang (10-20 seconds or so) the universe was full of extremely hot and extremely dense matter -- hot enough that not only did atoms come apart, and nuclei come apart, but even protons and neutrons came apart into a quark-gluon plasma. This cooled into a stew of elementary particles, which cooled into mostly protons, neutrons and electrons, which cooled partly into nuclei. At this point we're maybe a minute into the expansion and the universe was full of a plasma that was still hotter than the center of a star. After around 300,000 years, the plasma cooled enough to recombined into atoms. Stars eventually started to form.

No astronomer would call any stage of this expansion a "nebula" -- they first formed later after the atomic gas left a milion years after the Big Bang started to condense into stars and patches of gas and patches close to vacuum.

Bottom line: Nebulae did not exist until millions of years after the Big Bang, but the universe was filled with hot gas much earlier in the Big Bang.


Whatever difficulty you're having, the surest way to get past it is probably an in-person tutorial with an experienced telescope user.
Find a local astronomy club via
the Astronomical League,
the NASA Night Sky Network,
Sky and Telescope,
or Go Astronomy.
Then bring your scope to a public observing session, and most likely someone there will be willing and able to help.


Does nucleosynthesis happen near a black hole or during its formation?

After a google search the answer is yes, in theory. There is nucleosynthesis associated with the Black Hole accretion disks, see here, here and here. These sources look at the nucleosynthesis in the hotter parts of the accretion disks surrounding black holes, and how this is effected by the neutrinos produced.

It is also proposed that its possible for very small mass (10e-9 Mass of sun) black holes to be devoured by a neutron star (see here) and produce many of the heavier elements. This method starts with primordial black holes which can be of the required mass compacted into a region the size of a single atom. This tiny black hole could crash into a neutron star and initiate the r-process for nucleosynthesis.


"Maybe I should sum up their luminosity and then, using Stefan Boltzmann law, calculate the temperature? But does the law work for red giants? And can I sum the luminosity?"

This answer is not about placing the object on HR diagram, but is about the characteristic temperature as a blackbody.

It seems like you have a point source which you know that it is a combination of two stars: one blue, and the other redder. You can apply two-component blackbody: $F_{\lambda,total} = C_1 \times F_{\lambda,1}(T_1) + C_2 \times F_{\lambda,2}(T_2) ; T_1 \ne T_2$ where $F_{\lambda,total}$ is the observed specific fluxes from the point source, $F_{\lambda,1}, F_{\lambda,2}$ are blackbodies with temperature $T_1, T_2$ respectively, $C$s are normalization constants.

Blackbody approximately (to the first order) works with stars including red giants as long as you include a portion of the SED where the line absorption/emission is negligible. For example, some stars have strong absorption at less than 4000 A due to metals, therefore that portion deviates significantly from the blackbody.

You can sum fluxes, since fluxes are normalized for the luminosity distance. Since luminosity $L$ is $L = F \times 4 \pi D_L^2$ where $D_L$ is the luminosity distance, the total luminosity is $L_1 + L_2 = 4 \pi D_{L,1}^2 \times F_1 + 4 \pi D_{L,2}^2 \times F_2 = C_1 \times F_1 + C_2 \times F_2$. Therefore, you can sum the luminosity when $D_{L,1} \approx D_{L,2}$.


Would it be possible to detect a magnetic field around an exoplanet?

You usually look for electronic synchrotron emission in the radio, which corresponds to the magnetic field strength expected for giant planets. There has been a single detection claimed with LOFAR by Turner et al., (2020) around $\tau$ Bootes, but it's very noisy and hence disputed.
Other effects like Zeeman-splitting have been considered for detection, but require clean, high-resolution signals from planets. So far this is too difficult to perform due to stellar contamination.
Magnetic field searches around brown dwarves have been more successful, but those are not planets, of course.


How do satellites impede current telescopes?

Satellites, even in geostationary orbits, move with respect to the background stars and make "trails" on telescopic images that are tracked at the sidereal rate.
Removal of these can be as straightforward as taking the median of a set of exposures.
What I mean by this, is that in a sequence of exposures, the satellite trail will pollute different pixels in each exposure. Therefore if you take a sequence of exposures, tracked to keep stellar positions identical, then take the pixel-by-pixel median of the sequence of frames, the presence of the satellite will be removed. i.e. Each x,y pixel in your stacked image is formed by taking the median of the values of that pixel (scaled by the exposure time if necessary) in each image. It is essentially the same procedure by which cosmic ray strikes are removed.


How would Alpha Centauri A appear from the surface of Alpha Centauri Bb?

I'm trying to imagine what a hypothetical observer on the surface of
  Alpha Centauri Bb would see in the daytime sky. I know it would vary
  depending on the orbit of b around B, as well as the position of B and
  A in relation to each other.

My calculations are based on data from these wiki pages:

http://en.wikipedia.org/wiki/Alpha_Centauri

http://en.wikipedia.org/wiki/Alpha_Centauri_Bb

The planet Alpha Centauri Bb orbits the star Alpha Centauri B on an orbit with a semi-major axis of 0.04 AU. That's 25x closer to its star than Earth is to the Sun.

Alpha Centauri B has 86% the diameter of the Sun.

Therefore, ACB would appear in the sky of ACBb as a huge, extremely bright and hot disk, about 21x bigger than the Sun as seen from Earth (25 x 0.86). The heat would destroy and melt many materials.

ACB looks a bit more orange than the Sun, but the color difference is small.

ACA and ACB orbit each other, being separated by as little as 11 AU, or as much as 35 AU. ACA is about 22% bigger in diameter than the Sun.

Therefore, ACA as seen from ACBb would appear as a very small "sun" (about 1/10 the size of the Sun as seen from Earth) when at the closest approach, or basically an extremely bright dot, with no apparent size, so shiny it would hurt the eye, when at the maximal distance.

ACA would provide essentially no heat, and very little light, when compared to ACB (or even when compared to what we get from the Sun here on Earth). But it would be very similar in color to the Sun.

At the closest approach, ACA would look very similar to the Sun as seen from Saturn. At the maximal distance, it would look like the Sun seen from Neptune.

Summary:

ACB is a huge fireball of doom, 20x bigger than how we see the Sun, a bit of an orange hue, scorching hot, melting and igniting everything. ACA is a tiny disk, or even just a dot, yellow-white, very bright, providing essentially no heat and much less light than ACB.

ACA would grow bigger, then smaller again, following a cycle of 80 years.

ACA and ACB would get closer together in the sky, then drift further apart, on a cycle following ACBb's "year" - which is only 3.2 days in our terms. A very, very short "year".

It is not unlikely, although this is speculative, that ACBb is tidally locked to ACB, because they are so close together. If that's the case, ACB would only be visible from one side of ACBb, and would appear to remain "fixed" in the sky.

Here's an artist's impression. Keep in mind that ACB is a giant fireball, so this is more like a fisheye lens image (horizon to horizon in one snapshot). The size of ACA is overestimated. The Sun as seen from Earth would be the size of the letter B in that image.

  I know that at times B and A would overlap and appear as a single sun,
  but what about when they don't?

Well, do they? Do we know that for sure? For the overlap to happen, the orbit of Bb would have to be relatively coplanar to the mutual orbits of A and B. This seems likely, but I'm not sure if it has been established as a fact.


Is there evidence of super-heavy elements in the x-ray and gamma-ray spectrums of neutron stars?

The crusts of neutron stars will contain "super-heavy", neutron-rich nuclei. This is an inevitable consequence of the high density material, the accompanying degenerate electrons (that block $\beta$-decay) and what we know about nuclear physics.

However, the only things that contribute to a neutron star's observable spectrum are materials within a few cm of the neutron star surface. It is expected that this will be made of highly ionised "iron-peak" nuclei (Fe, Co, Mn etc.) and perhaps hydrogen and helium that has been accreted from the interstellar medium.

The "super-heavy elements" won't be present until the densities are much higher than in the surface layers. In a low-density material, these exotic nuclei will simply undergo a series of $\alpha$ and $\beta$ decays or even fission. In high density materials, the electron Fermi energy can exceed the maximum possible decay energy of any beta electrons and stymie the decay process.

e.g. For a surface density of order $10^{9}$ kg/m$^{3}$ (at which point nothing very exotic is created), the electron number density will be about $3\times 10^{35}$ m$^{-3}$.
Soft X-rays emitted at $10^{6}$K, will undergo Thomson scattering with a mean free path of $1/n_e\sigma \sim 50$ nm. No (detailed) information about the interior can emerge in such circumstances.


What is the maximum distance measurable with parallax?

Quick Google reveals a couple simple analyses.  For example,

  The Andromeda Galaxy, M31, is the nearest major galaxy to the Milky
  Way. The distance to M31 has been measured using other techniques to
  be
  2.5⋅10^6 light years , or 7.6⋅10^5 parsecs. Using the slightly modified parallax formula, we can find the necessary parallax angle to
  measure the distance to Andromeda. $ p = \frac{1}{d} =
> \frac{1}{7.6*10^5} parsec = 1.3 *10^{-6} arc-seconds $

  This is an incredibly small angle. For comparison, the resolution of
  the Hubble Space Telescope is 0.05  arc-seconds, so even Hubble would
  not be able to detect the necessary angular shift of the nearest
  galaxy to effectively use parallax as a measure of its distance.


The processes by which planets can be removed from their parent stars are discussed in some detail by Davies et al. (2014). These include direct ejection through interactions with other stars in dense birth environments; the ejection of planets due to planet-planet interactions, again usually taking place fairly early in a star's life; the later ejection of planets after an instability due to a passing star. There is also a possibility of exchange interactions, i.e. a planet will swap stars during an interaction. Again, this is most likely to happen in a dense birth environment rather than later in a star's life.

If the ejection of the planet happens while they are still in the dense environs of their birth, then recapture by another star is possible. A very recent paper by Wang et al. (2015) explores this very possibility using N-body simulations, but there are other papers to look at by Parker & Quanz (2012), or Perets & Kouwenhoven (2012).
These papers seem to agree that the process of recapture is not common (but not rare either) and will happen sooner rather than later, before the cluster disperses. The latter of these papers has some particularly eye-catching results: they find that if the number of free floating planets is comparable to the number of stars then about 5% of stars recapture a planet. The capture tends to put the planet in a wide, eccentric orbit (hundreds of au or more). Capture is more likely for massive stars, but they claim that even brown dwarf-planet or planet-planet pairs could be formed in small numbers.

Once the planet has left the cluster environment then, as is the case of star-star interactions in the Galactic field, the probability of encountering another star and being captured is very remote.


Is there a way to know the time according the moon appearance?

It is not easy.

Knowing the time means (roughly) knowing the position of the sun. If you are in the Northern Hemisphere, then the sun is (roughly) South at midday, East at 6am, West at 6pm and so on.

The moon orbits the Earth, so the angle between the sun and the moon is always changing. It may be possible so reason as "The moon is at first quarter, so the angle between the sun and moon is 90 degrees. I can use the position of the moon to determine the position of the sun (below the horizon) and so estimate the time."  There are problems with this, for example when the moon is nearly full, it will move through quite a large angle relative to the sun, with only a very small change in the amount of illumination.

With careful measurements and an almanack, it would be possible to find the local solar time. This wouldn't account for daylight saving time, or civil time that differs from solar time. There are much easier methods: get a sundial during the day and an hourglass for the nighttime. Or a watch.