How does an Echelle spectrograph put such a long, high resolution spectrum into a square-ish format?

Echelle spectrometers operate in the higher spectral orders.  (CRIES+ uses orders 10-60 for IR light, I think.  Not sure about ESPRESSO, but you can count them...)

These high orders are highly overlapping, so, for example, if the input light is dispersed by the Echelle grating horizontally into a spectrum, then the light at a given point along the spectrum is has components contributed from several different orders.  But, since each contribution is a different wavelength from each order, the orders can be spread out vertically using a second grating, the "cross disperser", conveniently stacking them up for the camera.  You can see these components in this diagram of the ESPRESSO spectrograph.

TL;DR:  Each horizontal line you see is a portion of a single spectral order, and the orders are spread vertically by a cross disperser grating, oriented roughly perpendicularly to the first (Echelle) grating.

I found this presentation with more details & the equations.


How to calculate phase angle of a satellite?

Draw the triangle Sun-Earth-Satellite, we will first find the angle Sun-Earth-Sat.

The angle between the Sun, the observer, and the satellite will be the angular separation between the Sun and the satellite on the observer's sphere. Here is how you could calculate that:

You have horizontal coordinates for both bodies so the easiest way to go from there would be to look at the spherical triangle Zenith-Sun-Sat, the angle at zenith will be the difference between azimuths, and the Zenith-Sun and Zenith-Sat lengths will be $90^{\circ}-h_{Sun}$ and $90^{\circ}-h_{Sat}$, respectively.
Now using the cosine formula for spherical triangles, one may obtain the following formula:

$$\cos^{−1}(\sin(h_{Sun})\sin(h_{Sat})+\cos(h_{Sun})\cos(h_{Sat})\cos(A_{Sun}−A_{Sat})).$$

Now use the sine theorem to find the angle E-Sun-Sat (the sine of this angle divided by the sine of the one we calculated will be equal to the ratio of distances from the Earth to the satellite, and from the satellite to the Sun, respectively), and to find the third angle, subtract the two from $180^{\circ}$.

Note: if you do not know the distance from the Sun to the satellite, I am certain you may use the distance from the Earth to the Sun as the error is probably negligible.


What is wrong with my calculations of Venus' orbital period?

Kepler's laws state that a planet sweeps equal areas in equal times as it moves in its elliptical orbit. It doesn't state that different planets will sweep the same area.

The "equal areas" law can be derived from the "conservation of angular momentum". In fact dA/dt = L/(2m)  (where A is the area, L is the angular momentum and m is the (reduced) mass).

Different planets will sweep out different areas. To calculate the period you used Kepler's third Law: $T^2 = k a^3$ (T= orbital period,a = semi-major axis). If, for convenience you take a in AU and T in Earth Years, then the constant $k=1$.

For Venus, a = 0.72. so $T=\sqrt{0.72^3}=0.61$ or about 223 days.

Hyperphysics has a section on Kepler's Laws


A high end photographic tripod would probably work but it will be difficult to make small tracking adjustments. So a proper astronomical mount would be much better.

An EQ2 mount would just about manage a scope of that size however an EQ3-2 would be a lot more stable and have the advantage that you can either get a goto version initially or upgrade at a later date.

If you are on a tighter budget and/or don't want an equatorial mount then something like a Celestron Heavy Duty Alt/Az Tripod or Skywatcher AZ3 Alt Az Mount would be suitable.

All of the mounts mentioned are available from numerous online retailers but shop around as prices can vary.


What is the origin of the term "plane of the spirit level"?

A spirit level is an instrument used to find a horizontal or vertical plane. In surveying, the spirit level is used as a reference. Surveying instruments like the theodolite have a spirit level built in. This is used to set up the instrument, after which you can measure angles relative to that reference plane.
This makes sure you have a repeatable reference, and not one that depends on the incline of the local surface.
As to the origin of the name:
it's called a level because it is used to find a level surface (i.e. one that's horizontal). It's called a spirit level because the liquid inside is usually a spirit (i.e. a mix of alcohol and water).

Early tubular spirit levels had very slightly curved glass vials with constant inner diameter at each viewing point. These vials are incompletely filled with a liquid, usually a colored spirit or alcohol, leaving a bubble in the tube. They have a slight upward curve, so that the bubble naturally rests in the center, the highest point. At slight inclinations the bubble travels away from the marked center position. Where a spirit level must also be usable upside-down or on its side, the curved constant-diameter tube is replaced by an uncurved barrel-shaped tube with a slightly larger diameter in its middle.
Alcohols such as ethanol are often used rather than water. Alcohols have low viscosity and surface tension, which allows the bubble to travel the tube quickly and settle accurately with minimal interference from the glass surface. Alcohols also have a much wider liquid temperature range, and won't break the vial as water could due to ice expansion. A colorant such as fluorescein, typically yellow or green, may be added to increase the visibility of the bubble.

why "spirit":

The term "spirit" in reference to alcohol stems from Middle Eastern alchemy. These alchemists were more concerned with medical elixirs than with transmuting lead into gold. The vapor given off and collected during an alchemical process (as with distillation of alcohol) was called a spirit of the original material


Peculiar Amplitude of Earth's Orbit in Z Axis

I can reproduce your JPL Z coordinate plot if I use Earth's heliocentric position relative to the J2000 ecliptic.
Naturally the amplitude has a minimum around the year 2000:

Relative to the ecliptic of date, which accounts for precession, the heliocentric Z coordinate of the Earth-Moon barycenter is much smaller:

The IAU 2006 precession model has two components, detailed in Capitaine et al. 2003:

precession of the ecliptic, a 47"/century change in the plane of Earth's orbit
precession of the equator, a 1.4°/century change in the plane and axis of Earth's rotation

Fifty years of ecliptic precession amount to 23.5" or 0.000114 radian, and the first plot is consistent with that.
The classical precession of the equinoxes (the intersection of the two planes) is a combination of both components.

Williams 1994 examines Earth's equatorial precession (dominated by the Sun and Moon) and finds that Venus's effect is larger than Jupiter's.
When looking at fluctuations in Earth's year length, I noticed a cycle similar to Venus's synodic period.
I expect that adding Venus to your system would reproduce some ecliptic precession.


Summary: Small telescopes will not show any visual detail in a galaxy. The website What Can You See With Different Telescopes is a good reference. (Thanks to JamesK for referencing this website in a comment to Is the Celestron astromaster 130 EQ Model able to see nebulas and galaxies?) Keep in mind that observing an object visually is much different than photos you may have seen. A camera can take a long exposure to "build up" an image of a faint galaxy. The eye does not behave this way.

Most galaxies are faint when viewed visually. Therefore, details can be difficult to see. (In some cases, it is more a matter of "barely detecting" some detail compared to being able to easily "see" some detail.) For some amateur astronomers, being able to locate an object and see it is a victory. Being able to see some detail in a galaxy is a dream (for a larger telescope).

Here are some of the factors that will affect how much detail you can see in a galaxy:

How large is the telescope? The main purpose of a telescope is to collect light. The more light that is collected, the brighter the object will appear. A 60 mm telescope (2.4 inch, the size of the objective) is small and will not gather much light. Therefore, only the brightest portion of a galaxy will be visible. Fainter details will not be visible. (Perhaps the dust lane in M64 or M104 will be visible if the sky is really dark.)
How dark is the sky? Since galaxies are faint, seeing them well requires a dark sky. Can you see clouds at night because they reflect light? If the answer is no, you can only "see" the clouds because you notice that "stars are missing", then you have a dark sky. If the outline of the clouds are easily visible due to nearby light pollution, then seeing details in "faint fuzzies" will be impossible. What is the magnitude (the brightness) of the faintest star that can be seen? Both of those questions will help to judge how dark or bright the night sky is at your location.
Experience. Experienced observers can see more detail than inexperienced observers. Part is the difference is knowing to take time to study an object instead of viewing it for 5 seconds and saying "oh well".
Does the galaxy have any detail to be seen? (This factor should be obvious but is a good reminder.) Some galaxies are featureless blobs, such as elliptical galaxies. Other galaxies that theoretically have detail may be seen from an angle that makes the detail hard to see, such as a spiral galaxy seen from edge on. Spiral arms can be faint and may be indistinguishable even though the galaxy can be seen due to the combined light of the nucleus and arms.


Simulation and its necessity in astronomical research

I agree with @uhoh that you don't have to an expert, but above-average knowledge of coding is definitely useful, bordering on "a must". Not for writing huge programs with 1000s of lines, but for writing smaller pieces of code that help you in everyday tasks.

As uhoh says, you can very well find your place in a group where other people are in charge of numerical modeling, and focus instead on, for instance,
interpretation of observational results,
physical modeling,
or even the technical aspects of telescope operation.

Software

You ask about what software is used: Some of the more popular if you're working with observational data are IRAF for reducing and analyzing the data and SAOImage/ds9 for visualizing. A long list of astronomical software can be found here.

On the other hand, if you're modeling galaxies or the interstellar medium you may want to familiarize yourself with stellar population synthesis codes such as STARBURST99, semi-analytical galaxy formation models such as GALFORM, or radiative transfer codes such as Cloudy.

Coding

For coding, if you are constructing large codes that require many hours of computation — possibly on a supercomputer — you will probably want your code to be fast. Then languages like FORTRAN and C are often used. For programs that don't necessarily need to be fast, comprehensive languages such as Python are very popular. For smaller tasks, shell-scripting can be very useful.

Even if you don't run simulations, you will most likely eventually need to automatize your work. For instance, rather than reducing 100 images one by one, you create a pipeline that does all the biasing, flat-fielding, cosmic-ray-removing, etc. in one go. And rather than going through a catalogue of a million galaxies one by one to look for those that match your preferred criteria, you create filtering software that finds them for you.

Supercomputing

Not all simulations are run a supercomputer. Huge cosmological simulations, galaxy formation simulations, etc. usually are, because you can usually start a simulation and then let it run for three months without interfering. But if your simulations are smaller, and possibly require that you check the process all the time, it's often sufficient/easier to run it on your local computer.


Fraction of circumpolar stars to all stars visible in a location

The answer depends on how detailed you want to be. For example,

Do you want (a) an approximate answer by assuming the stars are uniformly distributed, or (b) to use the real distribution of stars?
Do you want to know the answer based on (a) the number of stars visible throughout the entire year, or (b) just a given night, or (c) just an instant of time?
The magnitude limit of your sky affects how many stars are visible. More stars are visible from a dark site in the country than in a city.
The other factor that has a small effect on the answer is whether you want to include atmospheric extinction or not. Stars near the horizon appear to be fainter than their published magnitude due to the atmosphere. Some stars close to the southern horizon (in your example) are not visible because they do not rise high enough.

Combination 1(a) and 2(c) [uniform distribution, an instant of time] can be calculated based on the area of sky that is circumpolar versus the area for half of the sky. (As @zephyr indicated, you need to use steradians to calculate the area of the sky.)

Combination 1(a) and 2(a) [uniform distribution, entire year] is also easy to calculate from the area of the sky that is circumpolar, and the area of the entire sky minus the area of the sky that never rises (declination less than -55 for your example).

Any other combination requires using a catalog of star positions to "count" how many stars are visible at a given instant of time. In other words, based on the day and time (or the local sidereal time),

calculate the altitude of each star.
if above the horizon, calculate if it is bright enough to be visible given the sky conditions and (optionally) atmospheric extinction
if visible, tally the star as circumpolar or not
after counting all of the stars, perform the final calculation of the ratio.

Such a calculation would be easy to perform in a spreadsheet using a catalog such as the Yale Bright Star Catalog (which may not be complete down to the faintest stars visible). I did a similar calculation many, many years ago. See the May 1994 Sky & Telescope magazine, pages 86-88.


Where does the radiation in space come from and can we observe it?

Cosmic rays consist of both electromagnetic radiation (i.e. photons) of different frequencies (radio waves, IR, light, UV light, x-rays, gamma rays), as well as charged particles (protons, electrons, maybe even ions of light elements), and other stuff like neutrinos.

The vast majority of the radiation we encounter around earth will be from the sun, because it is so very close and basically a large radiating blob. Usually with isotropic (equally in all directions) radiating sources, the radiation intensity falls of with the square of the distance. That means radiation diminishes very, very fast. Go twice as far from the sun, and you only get a fourth of the radiation.

The EM radiation from UV and up (X-rays and gamma rays) is probably the most harmful. The earth's magnetic field shields us from these rays, but interplanetary travel will not have this benefit. X-rays and gamma rays may also come from supernovae and other stellar objects, which are far away, but will probably be much too faint to have an effect on astronauts. However, it can be picked up by sensitive specialized telescopes and satellites.

The charged particles may be a problem to spacecrafts and electronics onboard them, but can probably be dampened by shielding in the spacecraft, as to protect the astronauts.

Neutrinos are I think of no concern, since they hardly interact with other matter.

As an amateur, you will have problems detecting UV and above. Mainly because we are mostly shielded from this kind of radiation by the magnetosphere and the atmosphere.

You could detect particle radiation, by taking photos of northern lights, though... :)


Can Two Stars of One Same Type Orbit Each Other?

Certainly, the figure only uses different masses because that's the more general possibility (after all, the two masses won't be exactly the same).  In fact, if you look at the distribution over mass ratio q in binary systems in Figure 1b in https://arxiv.org/pdf/1304.3123.pdf, it seems that the cumulative number of binaries with mass ratio less than q is more or less equal to q (where q is smaller mass over larger mass).  That means the probability of mass ratio between any q and q+dq is about equal to dq and nearly independent of q.  So the chance of the mass ratio being between 0.9 and 1 is about the same as being between 0 and 0.1, for example.  In some sense, that's like saying all q are equally likely.  And of course if q is near 1, then the stars will have the same age and be in the same evolutionary state, so will be the same stellar type as well.


The axis type ‘FELO’ is regularly gridded in frequency but expressed in velocity units in the optical convention. The unit here is   $m/s$.

This means that the wavelength/frequency has already been expressed as velocity corresponding to a Doppler shift around a reference wavelength. This velocity is given by
$$
v = c \ \frac{\lambda - \lambda_0}{\lambda_0  } ,
$$
where $\lambda$ is the measured wavelength and $\lambda_0$ the reference wavelength in a rest frame.

The HEL term on the axis means that velocities have been transferred to the solar systems  barycenter (HELiocentric). This means, the velocity of earth around the sun and its rotation has already been removed.

Your distribution, therefore, has a width of about 200 km/s.
And the projected rotational velocity of the galaxy $v_\mathrm{rot} \sin(i) \approx 100\,$km/s, where $i$ is the galaxies inclination.

The offset of 550 km/s probably comes from the relative speed between our sun and the galaxy, but that is hard to tell without knowing which galaxy it is.

Some more information about frequency and velocity formats can be found in this paper


Total rotational angular momentum estimates for Jupiter?

A better estimate might be to use the moment of inertia of a $n=3/2$ (fully convective) polytrope, which will be a good approximation in an object like Jupiter, even when it approaches electron degeneracy in its interior.
You can look this up and it is given by $I = kMR^2$, with $k=0.205$ (almost exactly a factor of two smaller than a uniform sphere because the mass is concentrated towards the centre).

If we use a mass $M$ of $1.898 \times 10^{27}$ kg, the equatorial radius of Jupiter ($R=71500$ km) and the rotation rate determined from its magnetosphere ($P= 9.93$ hours), then the angular momentum ($2\pi I/P$) is $3.496 \times 10^{38}$ kg m$^2$ s$^{-1}$.

If instead, we use the volumetric average radius of $R=69900$ km, this reduces to $3.34\times 10^{38}$ kg m$^2$ s$^{-1}$.

This simple approach neglects the more complicated equation of state of Jupiter, it's non-sphericity and the probable presence of a solid core. Consideration of these, together with constraints provided by the measured even harmonics of the gravitational field (referred to as $J_2$ and $J_4$, led Helled et al. (2011) to suggest that $k=0.264$, with an uncertainty below 1%. Combined with the average radius, this gives $4.30\times 10^{38}$ kg m$^2$ s$^{-1}$.

More recent work by Ni (2018) has used refinements to the gravitational harmonics from Juno measurements, along with a more sophisticated interior model to suggest $k=0.274$ with 0.5% uncertainty.


Would bacteria on incoming meteors burn before impacting?

Any bacteria present in a meteor would not necessarily burn.  According to the Bad Astronomer's "Bad Addendum" to another unrelated point, "A Meteoric Rise", meteors sure do get extremely hot, but that outside part usually ablates away, leaving the inner part that was extremely cold for who knows how long, which may not have warmed up much during the fall.

Bad Addendum: many people think that a meteorite, after it hits the ground, is very hot and glows red. Actually, meteorites found shortly after impact tend to be warm, but not hot at all! It turns out that it certainly is hot enough to glow while it is in the part of the atmosphere that decelerates it the strongest, but any part that actually melts will be blown off ("ablated") by the wind of its passage. That leaves only the warm part. Even more, the meteor is slowed down so strongly as it moves through the atmosphere that the impact speed is typically only a few hundred kilometers an hour at most. Only the very large (and we're talking meters across) meteors are still moving at thousands of kilometers an hour or more when they impact. Small ones aren't moving that fast at all. Not to say you'd want to be under one: a car in New York was struck by a small meteorite and had a hole punched through it, and the whole back end crushed in. Ouch!

This American Meteor Society resource seems also to indicate that meteorites can fall to the ground without being extremely hot.

Are meteorites “glowing” hot when they reach the ground?

Probably not. The ablation process, which occurs over the majority of the meteorite’s path, is a very efficient heat removal method, and was effectively copied for use during the early manned space flights for re-entry into the atmosphere. During the final free-fall portion of their flight, meteorites undergo very little frictional heating, and probably reach the ground at only slightly above ambient temperature.
For the obvious reason, however, exact data on meteorite impact temperatures is rather scarce and prone to hearsay. Therefore, we are only able to give you an educated guess based upon our current knowledge of these events.

This study, "Bacterial Spores Survive Simulated Meteorite Impact", seems to indicate that it's possible, though unlikely, for a bacterium to survive the pressure shock from such an impact.
So it's plausible that an extremophile bacterium could somehow survive this impact by being on the inside of the meteoroid, surviving the ablation process as a meteor, and not even get warmed up to the surrounding atmospheric temperature when it crashes as a meteorite.
There is nothing to indicate that this has actually happened.  But what a discovery it would be if something like this were to happen and be confirmed.


If stars are uniformly distributed with density $\rho$ per cubic parsec, out to a distance $d$ parsecs, then that volume contains $4\pi \rho d^3/3$ stars spread over the whole sky.
The angular size of the whole sky is 41,252.96125 square degrees, so you would divide by that number to estimate the stellar density in stars per square degree.
This assumes that you can see all the stars. In practice you cannot, because of dust extinction and because stars become fainter the further away they are. You can account for these in any calculation but the details depend on the luminosity function of the stars, the distribution of dust and the instrumental properties of the telescope/detector used to take the picture.


Half of the Milky Way's baryonic mass is from other galaxies - when did this happen?

I think what you want to concentrate on is the bottom panel in your first figure, which shows the amount of accretion from different sources as a function of redshift (and thus time; see below). This shows that the contribution from "intergalactic transfer" (green cuve = material in winds outflowing from one galaxy that ends up in another) was almost zero until around $z \sim 3$ (age of universe $\sim 2$ Gyr), peaked at around $z \sim 2$ (age of universe $\sim 3.5$ Gyr) and then declined slowly from there, so that it's now about 10% of the peak rate.

"$z$" is the redshift, which is being used as a proxy for time: $z = 0$ corresponds to the present day (age of universe = 13.7 Gyr), $z = 1$ corresponds to an age of about 6 Gyr, $z = 2$ is about 3.3 Gyr, and so forth. You can use Ned Wright's Cosmology Calculator to convert individual redshifts to ages of universe (click on the "Flat" button to use a standard cosmology).

("Gyr" = gigayear = 1 billion = $10^{9}$ years, in case that's not clear.)


What can we expect from the first accurate image of a black hole?

Since the astronomers are using radio telescopes and not optical telescopes, I'd like to point out why they are doing so - The centre of the Milky Way is a very dusty place. Wavelengths from the millimeter to optical get easily absorbed by all this dust, so it's very difficult to see the centre of the galaxy in the optical spectrum. But radio waves do not get absorbed, and the centre of our galaxy is a very strong source of radio waves. Therefore looking in radio will give us the clearest picture of what's happening over there.

They mention in the article that they use a technique called VLBI (Very Long Baseline Interferometry) to image the blackhole.

The images you get from VLBI aren't images in the traditional sense, like you would get from an optical telescope. VLBI measures the phase difference of waves arriving at two different antennas (on two different continents, perhaps) and use this phase difference to infer the size of the source in the sky. Contradictorily, even though they can resolve sources of less than an arcsecond in size, it's very difficult to localize the absolute position in the sky to a very high degree.

So hopefully we will be able to resolve the event horizon of the black hole, but we will (almost definitely) not be able to see a picture like the one you posted.


Distance between stars Epsilon Eridani and Tau Ceti?

As answered by Wolframalpha, (as linked by chirlu) The distance between Tau Ceti and Epsilon Eridani is 5.45 light years.


Are there standard algorithms and procedures for creating unique sky maps based on latitude/longitude/date/time?

If you want to use the user's co-ordinates (longitude and latitude) to calculate the topocentric co-ordinates of an object to a not very high accuracy then you should be able to find a few software libraries (e.g. libnova) that do just that. Many use the algorithms from Jean Meeus' book Astronomical Algorithms. You can use catalogues such as the Bright Star Catalogue to find the equatorial co-ordinates of a star and then use the algorithms in this book to convert to topocentric co-ordinates (also called 'apparent place of a star').
Of course if you only want to calculate the 3D positions of stars, then it would be useless to use topocentric co-ordinates as the distances to the stars are so great that the error introduced by using geocentric (centre of the earth) instead of topocentric (position on earth) is very very small compared to the distance to the stars.

If you want to calculate the position of solar system object (planets, asteroids and comets) than you would need to be able to calculate the position of these objects to a very high precision. Meeus' book and software libraries derived from his book contain these algorithms. For planets you could use VSOP, but many software libraries include these. Meeus' book also includes algorithms to calculate the positions of the Moon and Pluto and the moons of the major planets. For asteroids and comets you can get the data needed for the algorithms from the IAU Minor Planet Center.

For high accuracy the algorithms provided in these libraries will not suffice and you may need to use professional software libraries such as NOVAS or SOFA. But then you are in milliarcsecond accuracy.


Why not both? (if you can afford them)

Both will do the job, but each has its strengths.

Cheshire eyepiece (or, more properly, Cheshire/sight tube combo, which is the popular combination nowadays):

It's easier to use when the scope is grossly out of alignment. Like, when you put it together for the first time, or after it dropped on the ground and rolled downhill for a minute. But can you do that with a laser? Yes.

It's easier to align the secondary mirror, and it tends to produce dead-on results for this purpose. But can you do that with a laser? Yes.

It is not immune from precision issues. Most parts of the Cheshire are usually well aligned (because they are machined on a lathe), but the crosshairs on the sight tube may or may not be perfectly centered. But can lasers also suffer from alignment problems? Yes.

Recommended Cheshire / sight tubes: the ones made by Catseye.

http://www.catseyecollimation.com/

They make much more than just Cheshire eyepieces, and all their tools are very high quality.

Laser collimator:

It's faster and arguably easier to use for normal daily collimation before you begin your normal observing sessions. But could you use a Cheshire instead? Yes.

A laser is only as good as its own alignment precision. A misaligned laser will not collimate your scope properly. But do Cheshire sight tube combos suffer from precision issues with their crosshairs? Yes.

Recommended laser collimators: the laser collimator and the tuBlug combo made by Howie Glatter. Unfortunately Howie has passed away recently, so either wait until someone picks up where he left off, or find a used laser/tuBlug combo somewhere on eBay or classified ads.

http://www.collimator.com/

Howie's lasers are essentially guaranteed to have zero collimation errors from factory. From anecdotes told by users, these devices tend to keep their perfect collimation even after being dropped to the ground repeatedly.

The tuBlug is not mandatory, but it makes primary mirror alignment much easier.

Keep in mind that you also need a precision center mark on the primary, for these methods to work well. Hopefully it's been placed there precisely from factory; otherwise there are ways to re-center it.

How to verify if the laser is centered:

Place it in the focuser. Do not tighten the screws too much, allow the laser to rotate in the focuser a little bit.

Turn the laser on. Watch where the spot hits the primary. Now rotate the laser in the focuser. If the spot describes a circle as you rotate the laser, then the laser is misaligned.

Different method: Make a V-shaped trough from 2 pieces of wood or aluminum. Placed the through horizontally on a sturdy table. Put the laser on the trough. Turn it on, projecting the spot on a distant flat object, like a wall or fence across the street.

Now rotate the laser in the trough and watch the spot. Does it stay put? Then the laser is fine. Does it move in a circle? Then the laser is misaligned - find the adjustments screws (if it has them) and adjust it until it's perfectly square.


Yes and no.

Yes, it's true that the apparent size of the Moon is 30 arcmin. It's true that the visual acuity of most people is 1 arcmin. So it's true that if you take the angular size of the smallest detail you can see on the Moon, and you put a bunch of those lined up straight in a row, you could span a Moon diameter with only a few dozen of them. In that sense, you are correct.

However, when you try to reproduce the situation on a computer screen, the comparison breaks down. First off, the eye doesn't see in "pixels". Like most optical systems, there's a point-spread function, that takes very tiny details and smears them up to a larger spot. The resolution of the eye is not the pixel size, but the size of the bell curve coming out of the point-spread function, and that has soft edges and is round, and it's everywhere and it's not fixed.

You assimilate the size of that larger spot with the size of a pixel on a digital screen, in your comparison. But that's not the same. The pixel grid in those thumbnails is fixed, so whatever falls between pixels is lost forever. Aliasing intervenes and creates artifacts that are not there in the original image. The dynamic range of the monitor is not the same like the dynamic range of the eye (the eye is much better). Color and brightness levels on the monitor are discrete, whereas the eye sees them as a continuum. Finally, the visual center in your brain is like a powerful computer that applies intelligent correction algorithms to the live image.

The list goes on and on. The bottom line is - all these effects combine and allow you to perceive a live image that is slightly more rich than those dead, frozen thumbnails that you posted. Not a whole lot better, but a little bit better. It's not like the eye can "work around" limitations, but it's more like you lose too much when you shrink a large image into a tiny fixed pixel grid on a computer screen.

It's very hard to reproduce reality on a computer screen. A much better way would be to take a 2000px by 2000px image of the Moon, put it on a big super HD monitor, and move it back to the point where the apparent size of that image is 30 arcmin. I know that doesn't sound satisfactory in the context of your original query, but it's a much better simulation.

Similar problems appear whenever you try to map the resolution of any continuous optical system (like a telescope) to a fixed digital grid (like a camera).

Let's say you're using a sensor with a pixel size of 4 microns. Let's say your telescope has a linear resolution in prime focus equal to 4 microns. You might be tempted to say - great, the sensor matches the telescope, right?

Well, not really. When that happens, you actually lose a bit of resolution. The image is good, but it's a little bit softer than it really should. See below an image of the Moon I took a while ago, with a system having exactly the parameters indicated above.

You can tell it's a bit soft, it's not really down to the pixel. Turbulence also plays a role, but part of the problem is that linear resolution is equal to pixel size.

Click the image below and open in new tab; if your browser shrinks it again to fit the window, left-click the big image to expand to full size - you must do this to see the full resolution image and notice the effects I'm talking about. The fuzziness is not visible on this small version here:

One way around that phenomenon, as an example, is to blow up the image in the telescope with a barlow until the linear resolution in prime focus is much greater than the camera pixel size, maybe 4x bigger. You do all your processing, and then you shrink it back, if you like, and you'll get a sharper image. Combine it with stacking multiple frames, and the overall quality can get pretty close to 100% the theoretical performance of the telescope.

TLDR: Continuous optical systems, and discrete grids of pixels, are very different things and cannot be easily compared.


Peebles and Fukugita's excellent "The cosmic energy inventory" estimates that about $10^{-4.3}$ of the total mass-energy is in the form of cosmic background radiation, and $10^{-5.7}$ in the form of radiation from stars.

The estimate that the perturbation due to plasma and other interactions of the background radiation is around $10^{-8.5}$, so clearly most photons have never been captured by anything and are still in flight.


First Telescope SkyWatcher Heritage 5" Tabletop Dobsonian Telescope

No telescope that fits your budget and observing habits is overkill.
As you may have read, the bigger the aperture, the more light it collects, and the finer detail it can resolve in steady air.

A 5-inch f/5 reflector (650mm focal length) is good for wide-field views of deep-sky objects and easy to transport to a dark site.
It should show the rings of Saturn clearly with a 10mm eyepiece (65x magnification); a shorter eyepiece could double that without losing clarity.
However, with such a short optical tube assembly, you might either operate mostly on your knees or put the mount on something less sturdy than the ground.
Also if the secondary mirror is exposed as shown here, it's vulnerable to dew, and stray light can enter the eyepiece and reduce contrast.

If you're primarily interested in planets, a 6-inch f/8 reflector (1200mm focal length) could be an even better choice.
A 10mm eyepiece gives 120x magnification, the primary mirror is less sensitive to collimation error, and the secondary mirror can be smaller for slightly better contrast.
On the other hand, you might want an additional eyepiece longer than 25mm for better views of certain faint objects.


Are there any planets or moons denser than Earth?

I feel it's a cheap answer but heavy Jupiters can get much denser than Earth because planets with Jupiter's mass stop adding size as they add more mass.   A planet with Jupiter's size and 10-12 times Jupiter's mass would be over twice Earth's density.

As far as Earth-like planets, there's a cool property of terrestrial planets, more mass means more tightly packed in their cores.  Basically a similar effect to the heavy Jupiters but not as pronounced.    If you double the mass and keep the element ratio the same, the density should increase.    For example, a planet like Mercury with a very high iron content, but much greater mass should easily surpass Earth's density.

Kepler 10b, appears to be a super-earth and its estimated density is greater than Earth's at 5.8 ± 0.8 g/cm3.  It has a mass of about 3.7 Earths.   The ± 0.8 g/cm3 offers some room for uncertainty, but some of the more massive terrestrial planets should be more dense than Earth.

I'd take this list with a grain of salt, but you can sort by density.

Kepler 131c which you mention appears to be a super Earth but its mass has a high margin for error.  I would add that a mass 8 times that of Earth and a radius smaller than Earth is probably impossible, so I'm highly skeptical of some of those numbers.


If a giant object would approach the Earth, would Earth start to turn faster or slower while the object starts to become nearer and nearer?

Could be, you'd be better to ask this on the Physics site, by the way.

No, the spinning of a perfectly round object is not affected at all, it just keeps spinning identically.

Let's say ..

say another Sun appeared, just like that.

Yes, the path that Earth is taking would go crazy.  We might fly off, head towards this new Sun, starting making a figure of 8, or some crazy path!

But the spinning as such of the Earth would be totally unchanged.

The "way we point" ("North up" so to speak) would be completely unchanged, and the speed we turn at (ie, 24 hours a turn) would be unchanged.

This might seem surprising but, well, you can learn all about it in Newtonian physics!

Once again, no change to the spin as such of the Earth.  the "direction we point" (ie the axis we spin around) and the time taken is totally unchanged if (for some bizarre reason!) we are "moved around in the heavens".

Note that there would be some absolutely tiny effects due to things like relativity (have a read!), the water on our planet sloshing around, and after all our Moon could be moved, bashed, etc.  These effects on the speed we are spinning would be extremely small over the short period of time you're considering, in the "sci fi" scenario you outline, as seen in the diagram.


How accurate should sun\moon ephemeris be for calculating eclipses?

I suspect this is fundamentally a question of error propagation.  My personal favorite reference is "Data Reduction and Error Analysis" by Philip Bevington.  Basically,  once you have the formulas in place,  such as

eclipse_time = f(sun_ephem, moon_ephem)

You take all the partial derivatives and "stuff" them into the error propagator formulas along with the known or estimated standard deviations (error/uncertainty) of your input variables.  The uncertainty in the output  just pops right out.


A Supermoon is when the point in the Moons orbit when it is at its closest to the Earth coincides with full or new Moon (see Wikipedia entry).

A blue Moon is (in this instance) the occurrence of a second full Moon in the same calendar month (see Wikipedia entry), and is unrelated to the coincident Supermoon.

Also the apparent size difference between a full Supermoon and a more usual full Moon is inlikely to be detectable by a casual observer. As can be seen in the side-by-side comparison graphic on the Wikipedia page reproduced below:

The apparent giant Moon you report is the well known Moon illusion where the Moon appears larger when observed near the horizon compared to when seen higher in the sky. The Moon illusion is startling if you are unfamiliar with it and very much more noticeable than a Supermoon to the casual observer.

While out in the morning of the 31st, walking the dog in the local recreation ground, between ~04:00 and 05:00 BST (03:00-04:00 UT) I observed the apparent size of the Moon and commented to the dog that we will probably see someone (in the media I thought) mistake the apparent size illusion I was seeing for the effect of the Supermoon. (I had also noted the same effect the day before at a similar time, but not on the 1st as we go out later at the weekend)


Are there currently observed systems of close (narrow) binary brown dwarfs?

You've got a lot of questions here, so I'll try to answer them one by one.

  Are there currently known double systems of gravitationally bound binary brown dwarfs?

Certainly there are. I don't study low-mass stars so I'm not up on the literature enough to know any good examples off the top of my head. Kozaky suggested the Luhman 16 system in his comment and a quick google search also brought up the triple star system Epsilon Indi with a central K type star being orbited (at ~1460 AU) by a binary brown dwarf system. Perhaps some other more knowledgeable person can suggest more systems of interest.

  Given their comparatively low gravitation, how long would they need to melt together to a larger object?

That depends on a variety of factors, including their respective masses and initial distance apart. As two objects orbit one another, they continually release gravitational waves due to their interacting gravitational fields. These waves carry away energy and to conserve energy, the stars must lose angular momentum, thus getting closer. The time necessary for two stars of mass $m_1$ and $m_2$, with an initial radius $r$ to fall into each other is given by

$$t = \frac{5}{256} \frac{c^5}{G^3} \frac{r^4}{(m_1m_2)(m_1+m_2)}$$

where $c$ is the speed of light and $G$ is the gravitational constant. We can apply some basic numbers and calculate an order of magnitude time. If I assume $r = 1\:AU=1.5\times10^{11}\:m$ and $m_1 = m_2 = 2\times10^{28}\:kg$ (10 times the mass of Jupiter) then I find the timescale for collision into one object is

$$t = 5\times10^{30}\:s \approx 100\:billion\:trillion\:years$$

This is obviously way longer than the age of the Universe and effectively means that two Brown Dwarfs will never collapse. This fact could have been seen from the equation. Notice that the mass of the objects is in the denominator. That means that with lower mass means longer timescales and the fact that the distance is to the fourth power means the distance is a much stronger controlling factor here. You need huge masses to offset that $r^4$ for you to get a collision on reasonable timescales. Even doing a best case scenario of $r = 0.1\:AU$ and $m_1=m_2=10^9\:kg$ still gives me $\sim100\:thousand\:trillion\:years$.

  What happens if binary brown dwarfs collide?

Not much? I don't know if anyone has ever really tried to work out the theoretical work for such a problem. Given the answer above, the only chance two brown dwarfs are going to collide is by a freak chance that they're on a collision course in space and not in a slowly degrading orbit, but the chances of that are pretty minuscule as well.

One situation that may be of related interest is to consider regions of star formation. While it is somewhat contested, some people believe stars can form "bottom up". That means smaller objects form first, then those coalesce into larger objects and on upwards until you end up with big, bright stars. In this scheme, you would first form many many brown dwarfs in a tight area and these would dynamically interact and collide with each other to form larger stars. This isn't the same scenario of two lone binary brown dwarfs colliding, but its the best you're going to get if you want to see colliding brown dwarfs.

In this scenario, I think that since brown dwarfs aren't fusing and their main energy output comes from core collapse (by some Kelvin-Helmhotz mechanism), it isn't going to be a hugely explosive result (comparatively) to smash two together. They will of course burst out with energy in the process, but it won't be as dramatic as two fully-fledged stars merging. And that's probably a good thing. If you had a star forming region with many brown dwarfs that were slowly coalescing into larger stars, you wouldn't want each merger to blow out huge energies as it would prevent further stars from coalescing.

  Is this anyway a matter of any interest?

From an academic standpoint, I'm sure it is. As I said, I don't study this topic, but understanding the affects of brown dwarf collisions would be very useful for fleshing out star forming theories. I'm sure one could think of numerous other academic benefits for studying such a topic.

From a practical standpoint, I can't say understanding brown dwarf collisions immediately benefits humanity. But as with any topic in science, you never know how knowledge gained today may help in the future.


How many stars are there in a Globular Cluster of 10^5 solar masses?

The mean mass of the initial mass function is ~0.3 solar masses (Maschberger 2013).  A GC with 10^5 solar masses will therefore have ~3 x 10^5 stars in it.  Since the most massive stars will already have died in a GC, the mean mass of the present-day mass distribution of the GC will actually be slightly lower, implying a slightly larger number of stars, but this will be a relatively small correction (certainly not more than a factor of 3).


Do moons experience periods of no sunlight when orbiting "behind" their planet?

This is basically a repeat of Aristarchus’ experiment of trying to figure out the distance between the Earth and the Moon from a lunar eclipse. See for example https://pwg.gsfc.nasa.gov/stargaze/Shipprc2.htm
So, we know that Iapetus orbits Saturn in 79.3215 days, at an average distance of 3,560,820 km. Saturn’s diameter is 116,464 km, and it orbits the Sun at an average distance of 1,433,530,000 km. The Sun’s diameter is 1,392,684 km.
Using this data, you can now use the formulas given on Wikipedia https://en.wikipedia.org/wiki/On_the_Sizes_and_Distances_(Aristarchus)#Lunar_eclipse to calculate the size of Saturn’s shadow at Iapetus’s distance. Knowing how long Iapetus spends in it is then only a matter of dividing the size of Saturn’s shadow by the circumference of its orbit, then multiplying by Iapetus’s orbital period.
I’ll let you do the math. We’re here to help find the answers, not give all the answers. ;-)


Could an asteroid impact cause radioactive fallout?

I'm not aware of any relevant uranium ore deposit, which is related to meteoritic material.

Dangerous fallout is by far the most caused by short-lived radioactive isotopes. Those isotopes are rare in meteorites as well as in rock on Earth.

Natural nuclear reactors, which would produce short-lived radioactive isotopes, don't occur any more on Earth, since the natural U-235/U-238 isotope ratio is too low, now. This is due to the half-life of U-235 of about 713 million years in contrast to 4,468 million years for U-238.

If an asteroid would impact into a uranium ore deposit, most of the molten ejected material would fall back to Earth as tektites.

Evaporated ejecta and aerosols (dust) would probably increase health risk for some time, until it's washed out from the atmosphere. But other, less-radioactive aerosols would be a health risk, too.


If I want to learn about astrophysics, should I study up on mathematics?

Yes, absolutely. At bare minimum, you'll need to study some geometry to understand how distances are calculated and how the shapes of orbits are related. Next, you'll want to study some algebra, so you can understand the likes of the inverse square law for how the intensity of light drops off as you move away from an object emitting light equally in all directions. Algebra also helps in understanding the magnitude system astronomers use to describe how bright objects are in the night sky.

After geometry and algebra comes trigonometry, especially spherical trigonometry, because of its use in dealing with how we describe the location of objects in the sky, and how to relate positions in one coordinate system to positions in another.

Finally, if you want the really deep understanding, you'll want to go for integral and differential calculus. As my high school physics teacher used to say, the difference between an ordinary formula and one that uses calculus is like the difference between prose and poetry - the calculus encodes deeper meaning and is applicable in more situations than any single formula that can be derived from it.


Why does it take so long to transmit an image from New Horizons to Earth?

New Horizons has just passed the Kuiper Belt Object (KBO) 2014 MU69 also known as Ultima Thule. KBOs form a belt of asteroids (the Kuiper Belt) from Neptune's orbit outwards and of which Pluto is the largest member of the Belt. During the encounter with Ultima Thule, all of the 7 instruments on New Horizons were gathering data (although not all at the same time) and the total data collected is expected to be about 50 gigabits of data (compared to 55 gigabits of data taken during the Pluto encounter in 2015).

Since New Horizons is about another billion miles further out than Pluto was and 3 more years have elapsed, there is less power for the (tiny) transmitter and the signals are much weaker. The bit rate is about 1000 bits per second and so the 50 gigabits to transmit this will take 50e9 bits / 1000 bits per second = 50,000,000 seconds or about 579 days. Converting (roughly) to months by dividing by 365.25 and multiplying by 12 shows that it will indeed take about 19-20 months to transmit everything back. The first image at about 300 meters per pixel resolution and so about 100 pixels across the 30 km KBO, should be received on 2019 Jan 1. A second higher resolution image with about 300 pixels across the KBO is expected to be downloaded by 2019 Jan 2. There will be a press conference on 2019 Jan 2 when these images are due to be released and shown. (more details on what to expect when at Emily Lakdawalla's Planetary Society blog entry)

After the initial data download, they expect to perform some analysis to see which images have the best data with 2014 MU69 in the frame. Given the uncertainty in the position of 2014 MU69 and the high speed of the encounter, they had to shoot strips of images and not all will contain the target. These data will be prioritized in the downlink so they arrive on the ground first and can be analyzed first.

As mentioned by @luis-g there is also the Solar conjunction which will cause a 5 day period (according to PI Alan Stern in the 2019 Jan 3 press briefing) when reception of the data won't be possible. We would expect this to re-occur in January 2020 but these approx. 10 days don't make a large difference to the time taken which is dominated by the weakness of the received signal after the 15W transmission travels the ~4 billion miles and falls off due to the inverse square law, the corresponding low bitrate allowed by the need to have the transmitted data decodeable and the amount of data to transfer.


How tightly clustered are the CHIME-detected FRB repeats?

You'd be right to be a little suspicious if CHIME had just detected two events. Fortunately, that's not typically the detection criterion for a repeating FRB. In the case you're talking about, FRB 180814.J0422+73 (Amiri et al. 2019 for the CHIME/FRB Collaboration), six bursts were detected from one particular region, with only 18 total bursts detected in that pre-commissioning phase (so 13 sources total). That high a fraction of bursts coming from a very small part of the sky is enough to raise some eyebrows.
You're also right that CHIME's angular resolution is not enough to localize the bursts to within a galaxy-sized region at this distance ($\sim500$ Mpc), as CHIME folks will readily admit. To lend more credence to a repeating source, long-baseline interferometry is required, involving either follow-up observations or digging through archival datasets. The latter was the mode of choice for FRB 180814.J0422+73, as the group looked at old survey data from the Very Large Array (VLA) Sky Survey (VLASS) and the NRAO VLA Sky Survey
(NVSS). They found five sources in the region with fluxes close to what would be expected from this source. It's insufficient to say which one corresponds to the host galaxy, but it's likely that one of the five is responsible.
At the moment, the CHIME team is trying to improve its localization capabilities by adding "outrigger" telescopes in the United States and Canada, including one at the Green Bank Observatory. Ideally, this would reduce or eliminate the need for follow-up observations with other telescopes.


I'm not sure I see the point to this question, but if one of the orbiting objects is hollow and you allow for course correction, then yes, one object passing through the other is possible.

I don't see what force would keep the object with a hole in the middle properly positioned so it was in effect perpendicular to the movement of the other objects.   There's also problems with the orbits lining up just right rather than missing and giving each other a gravitational push, and with gravitational perturbations from the sun and the moon.   A system like that would be a lot easier and require less adjustment if the Sun and Moon weren't there.

You'd probably want both orbits to be over the equator to reduce fluctuations, but eliminating fluctuations would be impossible.  Every orbit would probably require some adjustment, and while it's theoretically possible, I'm not sure I see why anyone would want too - apart from perhaps it would look cool.


Why is there a black stripe in the Hubble images of Pluto?

That photograph is a composite of two images taken with different exposure times.
To be correct we'd have to say that the exposure of the two photographs is different, i.e. the outer photo was created by absorbing more light. In this case we can assume that the focal ratio (derived from Hubble's lens aperture) and the luminance of the scene (how much light is travelling in the lens' direction) are identical for both photographs, which leaves only the exposure time as a free variable when it comes to determining exposure.
This is necessary because we're photographing objects with very different brightness. For Pluto to show up a relatively short exposure time is required, but its moons reflect much less light and would need a longer exposure time to visible. As long as the sensor is exposed, Pluto would continue to increase in brightness to the point that it becomes washed out. Objects that are significantly brighter become over-exposed resulting in a loss of detail and fidelity, known as blown-out highlights in photography. In our case Pluto would turn into a solid white dot compared to the more detailed picture that is now possible. You can draw a parallel with false color images rendered from infrared: this composite is not what the human eye would see if it was capable of picking up this level of light and detail.
In another Hubble image, NASA explained the reason why composite imaging is used:

This is a composite image because a single exposure of the stellar background, comet Siding Spring, and Mars would be problematic. Mars actually is 10,000 times brighter than the comet, so it could not be properly exposed to show detail in the Red Planet. The comet and Mars also were moving with respect to each other and could not be imaged simultaneously in one exposure without one of the objects being motion blurred. Hubble had to be programmed to track on the comet and Mars separately in two different observations.
Source: Hubble Sees Comet Next to Mars

Very long exposure times are often necessary since relatively little light is coming our way from distant planets and stars. As the Hubble website explains for its Deep Fields images:

Hubble has made a series of very deep observations taken in very dark parts of the sky. Like using a long exposure on a digital camera, these long exposure shots (up to several weeks) reveal very faint details that are not normally visible in shorter exposures.
Source: "What are the Hubble Deep Fields?", Spacetelescope.org FAQ.

Wikipedia summarises a paper by Robert E. Williams and the HDF team, "The Hubble Deep Field: Observations, Data Reduction, and Galaxy Photometry" as follows:

Between December 18 and December 28, 1995—during which time Hubble orbited the Earth about 150 times—342 images of the target area in the chosen filters were taken. The total exposure times at each wavelength were 42.7 hours (300 nm), 33.5 hours (450 nm), 30.3 hours (606 nm) and 34.3 hours (814 nm), divided into 342 individual exposures to prevent significant damage to individual images by cosmic rays, which cause bright streaks to appear when they strike CCD detectors. A further 10 Hubble orbits were used to make short exposures of flanking fields to aid follow-up observations by other instruments.
Source: Hubble Deep Field, Wikipedia, retrieved 2014-12-09


Why is there no concavity in the orbit of the moon around the Sun?

Turner's expression for the radius of curvature ρ is correct, but in case (3), ρ = -0.988 a where the Moon is in the first or last quarter (θ = 7.5° or 22.5°).
Where the Moon is full (θ = 15°), ρ should be -0.749 a.

Case (1) models a spacecraft stationed at L1 (b = a / 100, ρ = -0.99 a) or L2 (b = -a / 100, ρ = -1.01 a), roughly 4 times the lunar distance from the Earth.

The case (2) example b value is too large for a stable orbit around the Earth.
Instead we can model a geosynchronous satellite with n = 366 and b = a / 3550.
Then ρ is +0.0218 a on the day side and -0.0314 a on the night side.
This trajectory is alternately concave toward and away from the Sun, resembling Turner's figure 1 but with shorter waves.

Between cases (2) and (3) we can find parameters such that, on the sunny side of the Earth, the ρ expression denominator approaches zero and the trajectory is momentarily straight.
This is nearly the case if n = 24 and b = a / 575, corresponding to a 15.2 day orbit at 0.677 lunar distance.

Any curvature in an object's trajectory is due to acceleration in the direction of the net force acting on it.
We know from Newton that
$$ F = \frac{G m_1 m_2}{r^2} $$

For the Moon, the Sun is ~390 times as far as the Earth but 330,000 times as massive, so the Sun pulls roughly twice as strongly: $F_\oplus$ ≈ 0.45 $F_\odot$.
At full moon, both Sun and Earth pull the Moon in the same direction, and the combined force is 1.45 $F_\odot$ toward the Sun.
At new moon they pull in opposite directions; the resultant force is only 0.55 $F_\odot$ but still toward the Sun.
While the radius of curvature fluctuates, the direction of curvature never reverses.

For a geosynchronous satellite at 0.11 lunar distance, $F_\oplus$ ≈ 38 $F_\odot$.
On the night side of the Earth, the resultant force is 39 $F_\odot$ toward the Sun.
On the day side, it is 37 $F_\odot$ away from the Sun.


I'll give a more grasp the concept answer if you don't mind.   Consider the discovery of the planet Neptune.
Neptune was discovered because it was observed that Uranus wasn't moving as fast as orbital mechanics suggested it should be moving.  By calculating how much Uranus had slowed down, an estimate was made for where an undiscovered planet should be and Neptune was identified very soon after.
Uranus is well inside Neptune's orbit, nearly twice as close to the sun and very far inside of Neptune's L1, but Neptune was still able affect Uranus' orbit enough to get noticed.  Props to Newtonian mechanics.  It works exceptionally well for a lot of measurements.
Loosely speaking, this type of orbital variation, L1 semi-stability or orbital perturbations like the slowing of Uranus' movement is a subset of the 3-body problem and that's actually more real.   Observed orbits don't precisely follow Newton's or Kepler's formulas because of other massive bodies and perturbations.  3-body problem math gets crazy complicated, but the specific example of L1 can be explained using basic arguments without much math.
Earth's L1 point is about 1% closer to the sun or about .99 Astronomical Units or "AU".   1% closer gives an orbital velocity of a little over 0.5% faster and an orbital period of a bit over 1.5% faster or an orbital period of about 359.8 days, about 5.4 days shorter.  That's what .99 AU circular Kepler orbit should do.
Consider an object at .99 AU, and a period of 359.8 days orbiting the Sun just inside the Earth's orbit - assuming both with circular orbits.  Now imagine that object passes the Earth.  It's relatively small so it doesn't change the Earth's orbit much, but passing that close to the Earth would give it a considerable gravitational tug, basically tossing it out of it's orbit.   .99 AU is too close for a 2nd Kepler orbit and as a result, it's quickly tossed into a new orbit.
Two bodies can't orbit a massive central body with just 1% variation in their distance unless both bodies are very low mass (for example, Janus and Epimetheus are a rare exception to this).
Earth doesn't have low mass, so an object with a Kepler orbit at .99 AU would get tossed about after just one pass.  In other words, if the object orbited as fast as your "gut" tells you it should, then it wouldn't be stable at all.
Now consider an object at .99 AU but orbiting the sun too slowly to maintain it's orbit.  Slower velocity means the object falls in towards the sun, speeding up as it falls, entering a new, shorter period and more elliptical orbit.  That's where the balance comes in.  The slower velocity means the object wants to fall towards the Sun but being between the Earth and the Sun, Earth's gravity pulls it towards the Earth.   The point where those two forces are equal is the L1 point, where it's no longer a Kepler orbit, but a 3 body orbit, slower than a Kepler orbit, but pulled in opposite cancelling ways by the Earth and the Sun.   The "too slow" orbit is still unstable, but it's considerably more stable than a Kepler orbit in the same location would be.
With Earth's perturbed elliptical orbit, L1 isn't really a point so much as a smallish region, but on a computer, if we give the Earth a perfect circular orbit around the Sun with no other planets, L1 becomes a point of perfect balance.  For actual orbits, L1 is always unstable.
Despite it's instability, Lagrange points are very useful low maintenance areas to station telescopes because much less energy is needed to keep the telescopes there.  This only works if the central body (the Sun) is at least 24.96 times the mass of the orbiting, less massive body (the Earth)


How can we calculate velocity of an object given the distance from an object in parsecs?

You can't, not very well anyway. That's simply because the velocity of an object isn't calculable from the Hubble constant until the so-called Hubble flow starts to dominate (which in turn only happens at cosmological distances, i.e. very far away).

That said, if you are in the regime where Hubble flow dominates, you can do this easily. The Hubble constant is approximately $70 km/s/Mpc$. Just convert your distance into megaparsecs, and multiply. You end up with a velocity, which is the one you're looking for (this neglects peculiar velocity, which will affect everything in the sky).


If you use the equation of the Schwarzschild metric.
$$ds^{2}=-(1-\frac{2M}{r})dt^{2}+(1-\frac{2M}{r})^{-1}dr^{2}+r^{2}d\Omega^{2}$$
(equation 11.1 from Schutz's book A first course in GR).

If we think the neutron stay at rest in $r=r_{S}=2M$ then between the born and dead of the neutron $dr=0$ and $d\Omega=0$, and $ds$ will be the proper time, which I will call $\tau$ (I'm folowing the convention $c=1=G$).

Putting this on the metric we have

$$  d\tau^{2}=-(1-\frac{2M}{r})dt_{obs}^{2}$$

$$  \frac{d\tau^{2}}{dt_{obs}^{2}}=-(1-\frac{2M}{r})$$

and if we put $r=r_{s}$ (where the neutros was), we have

$$  \frac{d\tau^{2}}{dt_{obs}^{2}}=0$$

Then $\Delta t_{obs}=\infty$ .

I couldn't write a comment that's  why I answered, but I think you had some mistake using the red shift equation to get the wavelength difference.

Can you show your calculus? May be I can help.


Has lunar opposition surge ever been observed from Earth? From Earth orbit?

While I couldn't find any quantitative information about views from Earth or orbit down to zero phase angle (especially since views direct from Earth are limited in phase angle due to Eclipse), this website gives an unreferenced data point suggesting Apollo astronauts observed that a zero-phase full moon is approximately 30% (0.2 magnitudes) brighter than we can ever observe from Earth.
This site cites a value stating that the moon's brightness at zero phase angle may be as much as 14x brighter than an ideal diffuse reflecting disk, with 40% of the increase within the last 5˚ of phase angle (which cannot be observed from Earth).
Finally, there is one data set I have found which seems to show the opposition surge effect, but as noted, it is not what I would consider complete, and only has a few data points outside of Earth's viewing capability. That said, the source does not provide a value for maximum reflectance based on their model.

Robotic Lunar Observatory data, Reflectance vs Phase Angle


Is there any difference between parallel universes and multiverse?

I'd like to see the notion "parallel universe" restricted to universes similar to each other in some sense. This applies best to the many-worlds interpretation of quantum theory. Two universes in this sense can be almost identical, just different in a couple of quantum states.

Two such parallel universes could be interpreted (in the simplest case) as two elements of a Hilbert space. A Hilbert space defines a metric as a measure of distance between two elements (here applied to universes).
This may be seen in some analogy to points (universes as analog) in an Euclidean plane (many-worlds multiverse as analog). Here the notion "parallel" can make some sense. At least, it's possible to compare the items in some way (by the distance between two points/universes).

Talking of universes as parallel in the very general case of items in a multiverse, when there is no similarity at all between these items, would be a bit too lax in my eyes.


When did it take place, and what was the cause of the first blackhole in the universe?

We don't know for sure.

It is possible that random fluctuations in the density of matter of the early universe created black holes that could be much smaller than regular black holes formed by stellar collapse. These would have formed in the moments after the big bang.

Otherwise the first stars formed about 400 million years after the big bang, and some of these could form black holes after their (short) lives. Stars over 250 solar masses (which could form in the early universe) would produce large black holes when they collapsed.


Why does Astronomy still use the Anno Domini system for Time Synchronization

What makes you think that it is "obviously not 2013 on Earth"?

In actual calculations, astronomers use the Julian day, which is a decimal representation of time. A Julian year is exactly 365.25 days of 86,400 seconds each. Astronomical coordinates are usually written in the J2000 epoch, which allows us to compensate for Earth's axial precession.

Our ability to measure and record time is limited only by the accuracy of our most precise clocks. There's nothing fundamentally wrong or imprecise about it.


What happens when an Ultra Massive Black Hole cannibalize another?

From this article:

  “Bigger black hole masses are in principle possible – for example, a
  hole near the maximum mass could merge with another black hole, and
  the result would be bigger still. But no light would be produced in
  this merger, and the bigger merged black hole could not have a disc of
  gas that would make light.”

The theoretical size limit has to do with the ability to form an accretion disk, not the hole itself.   While it's probably an extremely rare occurrence due to space expansion, if a 50 billion solar mass black hole was to merge with another 50 billion solar mass black hole, you'd have a 100 billion solar mass black hole.    There might be some interesting spiraling in towards each other and some gravitational waves made in the process.   It would be cool to study, but in such a scenario, there's no force that would prevent the two black holes from merging into a bigger black hole.


Will the nebula of Betelgeuse be visible to the naked eye? How bright, how large, how soon, for how long?

There is an article here that describes the visible effects quite well. In essence, within a week or so, it would be comparable in brightness to the moon and therefore visible during the day.

  Betelgeuse would then start a phase of final, rapid dimming and again reach its current brightness level after possibly three years. After six years, it would be too faint to see with the naked eye. This would forever alter the visual appearance of Orion and we might need to think of another object the remaining constellation might represent.

This means that after 3 years, Orion would go back to "normal", and after three more years, we wouldn't see it at all with the naked eye - so no, not centuries or decades.

As for the pulsar question, I highly doubt that the jets would be visible... pulsars are very, very dim optically. Their pulses are best observed in the radio spectrum, so even if the emissions did hit gas clouds, the resulting "light" would likely be in the radio spectrum. It would also likely be spinning incredibly fast, since it would be young, meaning that even if there was some optical light from the jets hitting the gas, it would just appear constant to us given just how fast the pulses are - at least with the naked eye, as you asked.

Now, with other equipment than our eyes, we could likely get some sort of light curve or time vs. intensity profile which would make that pulsation apparent. This would also best be done in radio waves, which are the best bet when dealing with pulsars. (There are a few exceptions, as you mentioned with the Crab pulsar being visible in some other frequencies - see here for more details.)


Do low-mass red giants last longer than high-mass red giants?

Higher mass stars will have shorter lives.  Even though they have more fuel for nucleosynthesis, they burn this fuel much quicker than lower mass stars.
Generally, you should think of "red-giant" as an evolutionary phase and not a particular type of star.
Looking at the Scheller et al. (Aston. Astrophys. Suppl., 96, 269, 1992) data given in one of the comments, you want to compare the times denoted by points 5 and 6: ascent of the (first) red giant branch.  For a one solar mass star, the elapsed time ascending the giant branch is 634 million years.  By comparison, the time for the same phase for a five solar mass star is 0.289 million years.
When looking at the Hertzsprung-Russell chart (surface luminosity vs effective temperature), the RGB tracks at lower masses are all bunched up.  Therefore, it is very difficult to discern stellar mass (and age) from just luminosity and temperature.  By looking at chemical composition through spectroscopy, it is possible to infer mass by the different types of nucleosynthesis.


What is the definition of the "solar mixing ratio"?

An answer which explains it possibly, but remains speculation as I don't find the term explained:

Taking the pristine abundance of atoms, from which the sun formed (thus solar mixing) gives you a ratio of H and He atoms of 92% hydrogen to 8% helium. Converting that to the place and phase where we have molecular hydrogen (H2), gives a molecular mixing ratio of 83% hydrogen (H2) and 13% helium (He) - which is pretty close to the values given there. They probably use another source with slightly different numbers for the pristine elemental abundances than this wikipedia article.

Thus I consider it likely that they compare molecular abundances. They compare the abundances or ratios you would see, if you took the pre-solar nebular without any chemical segregation and formed the giant planets with the one measured by their spectrometers when actually looking at the atmospheres of the planets.


Are there more stars than all the words ever spoken by humans?

Number of stars in the observable universe
There are about 100 to 200 billion galaxies in the observable universe. Galaxies range in size from a few billion to hundreds of trillions stars. Using 100 billion galaxies and 1 trillion stars in a galaxy yields 1023 stars in the observable universe as a rough order of magnitude estimate. (After getting this result, I found multiple sources that give answers ranging from 1021 to 1024 stars in the universe.)
Number of words ever spoken
We're talkative, but we also sleep and eat and listen. Assuming we talk for the equivalent of 3 hours per day and we speak at about 150 words per minute, or about 10 million words per year. I recall 100 billion as being a rough estimate of the number of modern humans who have ever lived. Many of those people died during or shortly after  childbirth. Assuming an average life expectancy at birth of 25 years (generous), that means humans have spoken 2.5×1019 words since we first gained the gift of gab.
That might be off by a factor of ten, but it's not off by a factor of ten thousand. There are a lot more stars in the observable universe than the number of words ever spoken by humans.


The key finding why we think Earth's water came from Asteroids (big rocks) and not Comets (small rocks) is the Deuterium/Hydrogen ratio that we can measure in several sources.
When a star forms, it has an initial value of D/H that came from the nucleosynthesis in its progenitor nebular / star.

In a protoplanetary disc, as dust grows to rocks grow to planets your nebular gas will be trapped with initial D/H ratios in gas giant atmospheres. But the way to get Water onto Asteroids and Comets (those don't have substantial mass to retain atmospheres themselves!) is sublimation and maybe adsorption.
The latter two processes are strongly sensitive to the gas-mass and therefore different D/H ratios from the protosolar ones are expected.
And in fact we find them to be different:

This was in the news lately, as ESA managed to touch down and measure 67P's D/H ratio which gave another hint on the asteroidic origin of water on earth.

This finding, however does not resolve the question:

There maybe other isotopical tracers like D/H that give a hint on the history of water.
There could have been a decrease in D/H and subsequential increase again, or the other way round.
We know that the protosolar nebula must have had massive amounts of water (Oxygen is the third-most abundand element in normal stellar fusion!), so again the question is, why did Earth retain so much water, while the lighter planets Venus and Mars retained the heavier element $CO_2$
...

I could continue this for quite some time, but the bottom line is: We only have hints, not definitive answers.

To refer to the rest of your question: A hot plume in a colder atmosphere will not necesarily leave the whole gravitational potential well.


Where does energy at the beginning of a star's lifecycle (before any nuclear reactions) come from?

This is basic thermodynamics.

When you compress a gas, you inject energy into it. Think of the pump you use to inflate the tires on your bike. It takes some force to move the piston, right? That effort is not wasted, but goes directly into the air in the pump. Now the air has more energy.

But what happens to a gas when you put energy into it? It's molecules jiggle around faster. Well, faster jiggling is basically the definition of higher temperature. By putting more energy into the gas, you raise its temperature.

You can actually tell that the bike pump is getting warmer if you pump quickly and forcefully - this is something you can experience yourself.

Same with stars - the whole star is the "bike pump", and gravity is the one who's pushing the piston. Due to compression (shrinking) under gravity, the gas gets hotter and hotter. It turns out a star has A LOT of gravitational energy, so the gas can get VERY hot.

In your terms, yes, it's the acceleration that molecules experience falling into the gravity well of the star that makes them move faster. Faster moving molecules = higher temperature. Pretty straightforward phenomenon, really.

Historically, gravitational compression was thought to be the main source of energy for the stars, before the discovery of nuclear physics. Helmholtz and Lord Kelvin proposed this hypothesis in the 1800s.

The pressure-temperature relation of any gas was originally known as the Gay-Lussac law. Now we know it's just a particular case of more general phenomena (ideal gas law) tying together pressure, temperature, volume, and various kinds of energy.

A spectacular application of the p-T relation is the so-called "fire piston" or "fire syringe", which can ignite small pieces of cotton or paper by just hitting a piston really hard (extremely strong compression = big temperature rise). Search Youtube for some videos like this one:

https://www.youtube.com/watch?v=4qe1Ueifekg


Is the water underneath Europa's ice cap potable?

According to this 2007 paper, the current research as of the time of their own research had a huge range in possible concentrations of $\text{MgSO}_4$, magnesium-sulfate, with over four orders magnitude (approximately times $30,\!000$) differences between the extreme ends of the predictions.  It conducts its own analyses and near the end of the paper makes some analysis on habitability.  They say (with slight formatting modifications for units by me):

If the ice and liquid water layers on Europa fall within
the limits of Fig. 2 (A = 0.7) then, by standard definitions
of “freshwater” environments on Earth [broadly meaning
$<3$ g salt per kg
H${}_2$O (Barlow, 2003)], Europa’s ocean would be a
freshwater ocean, though admittedly more salty than most terrestrial lakes. Indeed, in this case, the putative global ocean of
Europa could be more like the mildly saline environment of
Pyramid Lake, Nevada than like the Earth’s ocean. While the
drinking water regulations of the U.S. Environmental Protection Agency recommend no more than 0.25 g of sulfate per
kilogram of water, adult humans can acclimatize to drinking
water with nearly 2g MgSO${}_4$ per kg
H2O without much discomfort
(EPA, 2004; CDC-EPA, 1999). Animal toxicity (the lethal dose
for 50% of the population) is in the range of 6 g MgSO4 per kg
H2O
(CDC-EPA, 1999), but most livestock are satisfied provided the
total salt concentration is less than 5 grams per kilogram of water (ESB-NAS, 1972). If we assume the low amplitude regime
for our solution (A < 0.8) then it is possible that human or beast
could drink the water of Europa.

However the best estimates for that parameter $A$ they had were from magnetic field observations, which put $A=.97\pm .02$.  In this case the article concludes that the subsurface ocean would then be very salty.  The most salt tolerant organisms we know of could potentially survive in the environment.  However, such organisms evolved into such salty niches from less salty ones, rather than having evolved directly within them.  Current evidence suggests that life as we know it is unlikely to be able to arise in such a salty medium.


Of course you would need to specify who the person is - an Olympic athlete? Let us assume so and then you can scale downwards accordingly.

So an Olympic high jumper can jump hard enough to raise their centre of gravity about 2m off the ground.

Let us assume this is a ballistic problem. The athlete actually gives themselves sufficient upward speed to get their centre of gravity from about 1m to 2m in the Earth's gravitational field. Using the usual equations for uniform acceleration, the required initial velocity is $v \simeq \sqrt{2gh} = 4.4$ m/s.

Now let us assume that the athlete could deliver something similar on an asteroid. This is doubtful, because getting a good run up, while wearing a spacesuit, is probably not going to happen. But if it were possible, then we just equate the jump velocity to the escape speed
$$ v = \sqrt{\frac{2GM}{R}},$$
where $M$ and $R$ are the mass and the radius of the (assumed spherical) asteroid.

Thus we do not have separate constraints on the mass and radius of the asteroid, only on their ratio. If
$$ \frac{M}{R} > \frac{v^2}{2G},$$
then even an Olympic athlete couldn't jump into space.

To get something more definite we would need to assume a density, $\rho$, for the asteroid. This depends on what type of asteroid you are talking about, but could be between 1500 and 5000 kg/m$^3$ (Carry 2012).

If we assume (spherical asteroid again) that $M = 4\pi R^3 \rho/3$ and substituting this for the mass, we get a constraint on the asteroid radius such that someone is trapped if:
$$ R > 4.2 \left(\frac{v}{4.4\ {\rm m/s}}\right) \left(\frac{\rho}{2000\ {\rm kg/m}^3}\right)^{-1/2}\ {\rm km}$$

You can mess about with this and assume a different take-off speed (you might want to argue, as MarkP does, that you could work up a larger horizontal speed by just running fast - but I doubt that in a "milligravity environment" - the surface gravity is of order 0.002 m/s$^2$) or different asteroid densities to modify the answer (trans-Neptunian objects or comets have densities lower than 1000 kg/m$^3$). Or you could come up with an equivalent constraint on the mass by substituting for $R$ in terms of the mass and density instead.


Why do planets move in an elliptical orbit?

Not sure if you're looking for a more mathematical answer or just the "why", but to answer the why, I'll start with some history on this.

Everyone who worked out a model for the Solar System, from Aristotle to Copernicus, liked circles.  Even though Copernicus correctly reasoned that the Earth moved around the Sun and not the Sun around the Earth, he continued to use circles in his models of the motion of the planets.

After Copernicus, Tycho Brahe, funded by the King of Denmark, had the best equipment at the time for observing the motion of the stars and planets and he was able to make star charts that were ten times as accurate as anyone before him. Brahe used equipment like this mural quadrant, and a large private observatory to take extremely accurate records.

Kepler, who was a better mathematician than Brahe, desperately wanted to get his hands on Brahe's star charts and the use of his observatory and equipment (so much so that when Brahe died, there were rumors that Kepler had poisoned him, though that probably didn't happen). When Kepler finally had everything at his disposal, he was able to work things out and study the Solar System more accurately. However, he still didn't know why the planets moved in ellipses; he'd only worked out that the ellipses fit the movement so well that it almost certainly had to be true, but he had no idea why.

Kepler, in fact, didn't care for ellipses.  He liked circles better, but he couldn't deny that ellipses worked.  Source.

Nobody knew why planets moved in ellipses until Isaac Newton was asked that question and had to invent calculus to answer it.  Calculus explains why planets orbit in ellipses, and that's the real answer.

If "calculus" isn't a satisfying answer, a way to kind of explain it would be to throw a penny out of a space shuttle (which isn't a good idea, but let's say you do). As the penny falls towards the Earth, it falls faster and faster (if we ignore air resistance) until it hits the ground.

Now, if you fling the penny from the space shuttle at a much faster speed and at a different angle, so that it only gets near Earth and misses the planet, it would actually start to orbit Earth. It would fall faster and faster until it passes the Earth, and then, like shooting a bullet into the air, the penny will slow down as it flies past the Earth.

According to Kepler's second law, the penny's fastest speed is at the point closest to the Earth (the perigee). That's essentially how objects in orbits work: as they move closer to the body they orbit, they accelerate faster and faster. Our penny will get so fast that, once it comes around the planet, it will be flung very far away, which will then slow it down. This is what creates an elliptical orbit.

Its motion is like a spring, falling toward the planet then flying away, but at the same time, orbiting in a circular motion with the spring motion, with 1 period per orbit. That motion of moving closer and then further in each orbit forms an ellipse.

It makes the most sense if you think of the velocity being greatest at the closest point and lowest at the furthest point. The low velocity moves it closer while the high velocity moves it back out further.  The total energy of the object in orbit (kinetic energy plus potential energy) remains constant.


Where might a semi proficient amateur analyst participate in meaningful astronomical efforts

Those sites where one views a light transit curve or tries to classify a galaxy type are not very interesting to me. I feel I could investigate things more in depth than that.

There is a lot more to many citizen science projects than that! I'd recommend having a look through all the projects available as there may be one in an area you are more interested in : Wikipedia: list of citizen science projects

Zooniverse is a great collection of Astronomy related projects. They are not just toys or made for school children, they are real science projects where you will be contributing to important and current research. Members who show they have advanced skills often get given more credit in the results weighting algorithms, or upgraded to more difficult tasks. People have even spotted things professional scientists had missed, and been able to bring new things to the attention of the scientific community and be credited on published papers.

Amateur observational astronomers are also useful members of the scientific community. Some beautiful and intriguing images of object have been taken by skilled amateurs, who don't have the pressures of short observing times.

Are you focused on research or on generally aiding science?

As someone with a lifetime of experience, you would be invaluable as a volunteer at outreach and engagement activities. You could ask your local University or even city council if they are running any science outreach projects you could collaborate with. Again, this does not just mean school children, although that can be a part of it. Lobbying for science literacy in politics, organising science fairs, giving talks to interested groups, etc.

I will try to think of more research related examples later today. It would be helpful to know what country you are based in, as my experience may be very different to yours.

Later edit: If you are interested in exoplanets but want more autonomy of research than the citizen science projects afford, you may be interested to know that NASAs Exoplanet Archive is a publicly available archive of data from exoplanet missions along with the tools to work with the data. Happy planet hunting!


Distance between objective lens and eyepiece

tl;dr: 1025 +1/-3 mm
If you were to focus at infinity, then the primary focus is 1000 mm from the primary lens and you would put the eyepiece lens one of its focal lengths away from that point to make a new parallel beam. That's 1025 mm. If your eyes can not focus at infinity without glasses, then wear them so that you can see the parallel beam and image it on your retina.
If you don't have your glasses or would like to try without, then your eye needs a diverging beam which means the eyepiece should be a little less than its 25 mm focal length from the primary focus, i.e. a tiny bit closer to the objective, perhaps 1 or 2 mm.
Now you mentioned that objects are over 1 kilometer away from your telescope. If they were exactly 1 km (1,000,000 mm) then the primary focus would be farther from the objective by a little bit.
The thin lens equation 1, 2 is
$$\frac{1}{d_{img}} = \frac{1}{f} - \frac{1}{f_{obj}}$$
so your image distance would then be 1001 mm, and so you'd have to move your eyepiece away from the objective by 1 mm to correct for that.


How easy would be to see the Moon from Mars?

We have pictures of this, thanks to Curiousity!

(Source)

  “A human observer with normal vision, if standing on Mars, could easily see Earth and the moon as two distinct, bright “evening stars,” said NASA in a statement issued today.

So yes, you can see our moon from Mars.

The apparent brightness of our moon from Mars is +0.9. The apparent brightness of Jupiter's 4 biggest moons is between 4.6 and 5.6 under ideal conditions.  However, while our moon can be seen visually separate from Earth from Mars, Jupiter's moons can't be distinguished from Jupiter itself with an unaided eye.


Any cheap (less than 100 euro) telescope for observing Jupiter?

Jupiter is pretty easy. A little refractor with a 50 mm aperture (diameter of the primary lens) will show you the disk of Jupiter and the biggest two equatorial belts, and the four galilean satellites. Or Saturn and its rings - that's doable too. The image will be tiny, but you'll see those features - assuming the instrument is not junk and it can take a decent magnification.

However, that's all that you'll see at that level of investment. No small equatorial belts on Jupiter, or inter-belt swirls. With the Great Red Spot shrinking and becoming washed out each year, you may not even see that one. Also, Saturn will be disk + rings, that's it; no Cassini division, no ring shadow, or anything else.

If that's okay, then keep reading.

Availability of instruments depends on where you live. I'm more familiar with the market in North America, but EU can't be too different.

Surprisingly, a very decent cheap telescope is the Galileoscope.

http://galileoscope.org/

Yes, it's a DIY kit that you assemble yourself. But the primary lens is a surprisingly good achromat doublet; the design makes sense for an instrument like this; it accepts standard 1.25" eyepieces used for bigger telescopes; and the price is within your limits.

As an amateur optician and telescope maker, I was pleasantly surprised by this instrument. It works well up to about 100x, which is the theoretical limit for its aperture.

What will be difficult with this scope - and with any cheap refractor - is the mount. For astronomy you need a rock-solid mount, which is hard to do. The Galileoscope can be plugged into a standard photographic tripod - maybe you can find a good one second-hand. The site also gives suggestions for DIY mounts. Try different things and see what works. I've seen many commercial scopes, more expensive than this, that were basically unusable due to really poorly thought out (or poorly executed) mounts. Beware.

If the DIY route is not appealing, try some ready-made scopes. Here are a couple suggestions of vendors (I'm not affiliated with anyone):

http://www.telescope.com/Telescopes/Best-Telescopes-for-Kids/pc/1/459.uts

http://www.optcorp.com/telescopes/all-telescopes.html

Sort by price and keep digging. The mount will be a weak point, always, in this price range. Those refractors with tall, thin tripods tend to be very wobbly. It's a true hell to push the scope a bit this way, only to have the tripod rebound the other way - very annoying.

You may have better success with one of those short, stubby 76 mm table-top reflectors. Tweak the mount until it moves relatively easy, but not too easy. Hold it with both hands and guide it to track the planet. It may actually work better than one of those tall refractors. OTOH, the focal ratio for this reflector is only f/3.9, which is awfully short (it's hard to control aberrations for such a short instrument, unless you invest in high-end optics).

If you do get a refractor, get one of those with short legs. The nice thing about these cheap refractors is that they have a long focal length (large focal ratio), which helps keep aberrations low.

Frankly, there will be some limitations in this price range no matter what. It would help if you could buy from a store with a good return policy - just in case things don't work well for you.

Finally, you say "the Moon is not challenging"? It can be very challenging, once you get into exploring the tiny craters and so on. Just a suggestion. ;)


How far can a planet with living beings get before we cannot see the living beings?

The simple answer to this would be anywhere that is physically unreachable. If there happened to be Earth-like life out there in our solar system, we would likely be able to detect it, then land a probe and confirm.

When it comes to planets outside our solar system, then it becomes difficult. If there was no trash orbiting (and thus no way to notice slight dips in light etc) then we would likely be limited to studying the atmosphere using the transit spectroscopy method (more on this HERE) where we can detect the gasses and chemical makeup of an exoplanets atmosphere. There may possibly be other things to look for in xray or infrared spectrums, such as localised hot-spots if there is intelligent life, although depending on distance and size of the planet, this could be difficult.

Overall, no matter what we detected, if it was to far away for us to physically be there, then we would only be theorising. We could say 'This planet is a great candidate for life' and things along those lines and present papers on findings. We would, however never be able to confirm it.


Would a comet break up during a close pass of the Earth

It wouldn't break up.

As my Granddad used to say "Close is only good in horseshoes". If the comet doesn't impact the Earth will be able to pass by unchanged. 200000 miles is a long way, about the distance of the moon from Earth. Asteroids have come this close before (eg 2011 XC2).

Comets can be broken up by close encounters with the sun, but the sun is rather hotter than the Earth [citation needed]. 200000 miles is also well outside the Roche limit for the Earth-comet system, and at the speed that comets travel relative to the Earth, tidal effects won't have enough time to act.

So the comet will pass the Earth quite happily. It is a rather small comet, and even the comet's coma won't interact with the Earth. After closest approach it is possible that the Earth could pass through the comet's tail, which would cause a meteor storm.

Even at 100000 miles, there is not much effect on satellites nearly all of which orbit at 22000 miles or less (22000 is geostationary orbit) It is plausible that they could be damaged by meteoroids.

The dust that comes of comets is all quite small. If you are looking for a "near miss that still causes damage" try this. The comet makes a close pass of Jupiter that causes it to split into two parts. The larger part misses Earth by 100000 miles but the smaller part (a 300m chunk that wasn't noticed until too late because story) hits Merchtem 10km outside Brussels


Are the radial spokes in Saturn's rings reliably visible via ground-based telescopes

Bryan (2007) gives a number of reasons why O'Meara's discovery was largely discounted:

As you stated, the behavior was entirely inconsistent with Keplerian predictions of motion.
While O'Meara was able to reproduce his findings, no other independent observers could.
The detections were done entirely visually, rather than with numerical measurements.
There had been previous observations (i.e. in the 19th century), but in different places in the rings (the A and C rings, not the B ring). These have never been confirmed, and it is still believed that these were actually illusions.

Reliable observations from ground-based telescopes, especially those of amateurs, have arisen within the past decade or so. Since 2007, 67 "candidate" observations have been made of the spokes, many during the 2009 Saturnian equinox (keep in mind that the spokes may be a seasonal phenomenon). One additional factor that made observations difficult was that the spokes seemed to have vanished, even from space probes, in the years after the Voyager observations, making it impossible for ground-based telescopes to see them.


More recent data and simulations of "Milkomeda", the collision of the Milky Way and Andromeda galaxies?

Since the second data release (DR2) of the European Space Agency's Gaia mission there has been a revolution in astrometry, including measuring the motion of the Andromeda Galaxy.

On February this year van der Marel et al (also in ArXiv) published interesting results on that matter by using Gaia's DR2 measurements. The results reveal that the collision is going to happen 600 million years later than the previous estimate (in 4.5 Gyr instead of 3.9 Gyr). Also Andromeda appears to have more tangential motion than previously thought and thus "the galaxy is likely to deliver more of a glancing blow to the Milky Way than a head-on collision".

Also interesting, M33 (Triangulum galaxy) is going to make its first infall towards Andromeda and might interact gravitationally quite a lot with it. The absence of stellar streams between Triangulum and Andromeda show that this is the first time they are going to meet each other.

Thanks to Gaia the measurements are now so precise that for the first time we are able to notice even the minute rotation of both Triangulum and Andromeda galaxies astrometrically (and not only by using doppler shifts).


What is the significance of the discovery of a pulsar flipping between radio and x-ray emissions?

The significance of this discovery, according to the European Space Agencies web article "Volatile pulsar reveals millisecond missing link" is the first time a pulsar in a crucial transitional phase between has been observed, and that this explains the origin of mysterious millisecond pulsars.

The significance is as stated by the ESA (bolding mine):

  This bouncing behaviour is caused by a rhythmical interplay between the pulsar's magnetic field and the pressure of accreted matter.

What this tells us about pulsar dynamics is

  When accretion is more intense, the high density of accreted matter inhibits the acceleration of particles that cause radio emission, so the pulsar is not visible in radio waves but only through the X-rays radiated by the accreted matter. When the accretion rate decreases, the magnetosphere expands and pushes matter away from the pulsar: as a consequence, the X-ray emission becomes weaker and weaker, while the radio emission intensifies.

The article also contains contact details of scientists involved