gravitational energy and internal energy

Can the release of gravitational energy during planet formation in principle (is there enough) explain this much heat

Planets in general can have internal heat sources (like radioactive breakdown or a complex hot core).  During planetary formation there are thought to be multiple collisions with large proto-planets before things settle down to a more-or-less stable set of orbits and planets.  These collisions add energy to the planets as they develop and increase temperature (a lot), as does ongoing bombardment from boring-old asteroids.

Now all of this energy does basically come from gravitational sources, but as you can see it's a complex (messy) business and there are many significant events that can alter temperature.

It might be instructive for you to read about existing theories on the formation of Solar system as well as material related to the structure of the Earth.


Photographing Venus with a phone through a telescope

Phone cameras just aren't designed to take good photographs of astronomical objects. The sensor is too small and the configuration options (like exposure time) are minimal.

For astronomical photography, a better option is to use (the body of) a DSLR (or the analog version); the lens is provided by your telescope. Another option is a specialized CCD camera.


Think of dropping a pebble into a pond, and the circular pattern the resulting wave makes. Now take a photo every second, and stack these one on top of another, and you have a "cone" of sounds waves. The oldest image of the wave is also the biggest, and furthest back in time.


If two binary brown dwarfs began to exchange gas with each other, would they both ignite into stars?

It is possible for two brown dwarfs to form a contact binary. This would not, of itself, cause core temperatures in either to rise.

If the two brown dwarfs were to merge, the mass of the resulting body could be enough to generate sufficient heat and pressure in the core for nuclear reactions to start. If both brown dwarfs had a mass of about 60 Jupiters, the combined size of 120 Jupiters would certainly be big enough to form a red dwarf.


How do comets and other solar system bodies gain energy to exit the solar system?

The mechanism by which the path, speed and energy of a celestial object or an artificial spacecraft is altered by massive planets and bodies is Swing-by or gravity assist. Any encounter between a comet and a gas-giant like Jupiter is almost an elastic mechanical collision (though no physical contact occurs) and in any mechanical collision, energy exchange between the bodies undergoing collision takes place. In this case, when the comet C/1980 E1 passed closed to Jupiter, Jupiter passed on some of its mechanical energy to the comet, rendering its total mechanical energy positive and hence setting the comet free and no longer bounded to the solar system. It is the total mechanical energy of the Jupiter+C/1980 E1 that was conserved, not just that of the comet. But the small decrease in the mechanical energy of Jupiter would not have a significant impact on the massive planet.


How do a planet's moons and a planet's rings interact?

First, there are different types of rings. Using saturn as an example: there are icy particles, dust bands and more. These interact differently if they were to hit a moon, for example enceladus.

Every little particle in a ring is a „mini-moon“ and obeys the same laws (see kepler) as the big ones.
They technically all have their own orbit.

The main rings of saturn have a high enough particle density to exert gravitational influence on each other.

There are hypothesis that suggest that rings are made of shattered moons (like through collision) or another one that says that the rings are made of particles that „failed“ to make a moon in the first place.

So now, what happens when rings collide with moons is dependent on the type of ring.
Enceladus is a small icy moon in the so called E-Ring of saturn. This ring is small and made of water/ice, just like the moon. Because the ring is so very small and such a water/icy „cloud“ would  not really stay together, it is believed that it is „refilled“ by enceladus constantly.
The surface of enceladus was observed to be smooth, not really having craters one might expect, but this could also be due to geological activity. Enceladus also „geysirs“ out water all the time, so this might be how the ring is kept.

Edit: I recommend checking out this link: https://courses.lumenlearning.com/suny-astronomy/chapter/planetary-rings/


It's dust!

The black band is not absence of stars, but rather clouds of gas and dust — a significant component of almost all spiral galaxies$^\dagger$ — which block the light of background stars and luminous gas.

In the image, you see both individual stars, scattered all over, and the distinctive bright band of the Milky Way, most of the stars of which are so far away that they blend together. Most of the dust is in the plane of the Milky Way. The individual stars are closer to us than the bright band and the dust clouds, so what you see is the billions of stars in the Milky Way band, some of the light of which is blocked by dust, and then on top of this you see a few 1000s individual stars that are closer.

The components of galaxies

Galaxies consist of roughly 85% dark matter and 15% normal ("baryonic") matter. By far, most of the normal matter is hydrogen and helium, some of which is locked up in stars, and some of it in huge gas clouds, sometimes glowing (the pink clouds you see in the Milky Way image are probably hydrogen clouds being excited by the hard UV radiation from hot and massive stars, subsequently emitting H$\alpha$ light). A small fraction (1–2%) of the normal matter are heavier elements, by astronomers lazily referred to as "metals". Roughly 2/3 of the metals are in the gas phase, but the remaining 1/3 has depleted into dust grains, e.g. silicates and soot. This dust is mixed with the gas clouds and often become dense enough that they block light of stars.

A strong wavelength dependence of the dust's extinction properties

However, dust scatters and absorbs light with short wavelengths (such as blue and UV) much more efficiently than long wavelengths (such as red and infrared). So while it may effectively block visual and UV light, infrared light passes more or less unhindered. For instance, if you look at the molecular cloud Barnard 68 in red and near-infrared, it looks black (left in the image below), but if you look farther out in the infrared, you can actually see the background stars (right below):

Image taken from ESO.

Dark matter

Dark matter cannot be seen. It's… well, dark. It interacts with normal matter and light only through gravity. That means that if you could place a lump of dark matter in front of a star (you can't, really), you wouldn't block its light. It would pass right through. If the lump were big enough, it might gravitationally deflect the light, so the background star would look distorted to you, as would a lump of normal matter (e.g. a black hole).

$^\dagger$In contrast, the interstellar medium of elliptical galaxies tend to be much more gas- and dust-depleted.


How to compute for Distance Modulus' uncertainty/ies?

You can just use these upper and lower bounds to create an upper/lower bound for the distance modulus. The lower bound is $168-14.9 = 153.1$, and the upper bound is $168+27.5 = 195.5.$ You can calculate the distance moduli for these values to get upper and lower bounds:
$$5\log(153.1) - 5 = 5.924876 \\ 5\log(195.5) - 5 = 6.455734$$
Then we calculate the distance modulus of the base value: $$5\log(168) - 5 = 6.126546$$
To get the deviations, we just subtract the distance moduli from the original. Therefore, your distance modulus is equal to $6.126546_{- 0.20167}^{+0.329188} \text{ pc.}$ I hope this helps. If there are any issues with my answer, please notify me.


Given that a color index ($B-V$) of star A is lower than that of Star B, What can we say about the stars?

Color is a difference, not an absolute value
Being "blue" doesn't (necessarily) mean that a light source has a large flux in the $B$ band. A color is not an absolute value; it is the ratio between two fluxes or, equivalently, the difference between two magnitudes.
Being blue means "More flux in some short wavelength band (e.g. $B$) than in a longer-wavelength band (e.g. $V$)". Because of the stupid magnitude system, this then means a lower value in $B$ than in $V$, i.e. $B-V<0$.
Hence, the smaller $B-V$ value a star has, the bluer it is.
Intuitive example
So when you write

Star A is bluer than Star B means $M_B^A<M_B^B$.

it's actually not true. This mathematical expression only means that Star A is brighter (in the $B$ band) than star B; it says nothing about the color.
For instance, if you compare the flux of the Sun with the flux of a blue laser pointer, the Sun most definitely emits more blue light than the laser pointer, i.e. $B_\odot < B_\mathrm{laser}$. But that doesn't mean that the Sun is blue, because it also emits much more red light.
Astronomers are astronoming
In the above, I've followed the sloppy astronomer notation of using $B$ and $V$ both for the filters/bands/bandpasses, and for the absolute magnitudes. More clearly, we can also write the magnitudes as $M_B$ and $M_V$.
Not also that, in astronomy, the term "bluer" is often used interchangeably with "shorter wavelength", irregardless of the actual wavelength. Similarly, the term "redder" is used for longer wavelengths.
For instance, we could say that a red photon is bluer than an infrared photons. Similarly, a UV photon can be redshifted as it travels through the expanding Universe. If it doesn't travel too long, it might be observed as "blue", in the sense that its wavelength is around 4400 Å, but although it's not red, we still says that it has been redshifted.
Relation of color to temperature
For blackbodies, hotter objects are bluer and more luminous. This is a corollary of Planck's law, which relates the brightness $B$ per wavelength ("spectral density") of a blackbody to its temperature $T$.
If you integrate Planck's law you can show that the total power $P$ increases tremendously with temperature ($P\propto T^4$; Stefan-Boltzmann's law), while if you differentiate it you can show that the peak wavelength $\lambda_\mathrm{peak}$ is inversely proportional to the temperature ($\lambda_\mathrm{peak}\propto1/T$; Wien's displacement law).
In other words, 1) hotter objects emit more flux (i.e. are brighter), and 2) hotter objects emit a larger fraction of their flux at short wavelengths (i.e. are bluer).
The figure below (an annotated version of this) shows an example of the colors of a $5000\,\mathrm{K}$ (blue solid line) and a $4000\,\mathrm{K}$ (dark green solid line), respectively. The $B$ and $V$ fluxes are measured at $4400\,\mathrm{Å}$ (purple dashed line) and $5500\,\mathrm{Å}$ (light green dashed line), respectively.
The flux ratio between the $B$ and $V$ wavelengths are larger for the $5000\,\mathrm{K}$ object, so its magnitude difference is smaller.
In other words, it's bluer.
(I've omitted a factor 2.5)

Reddening
Stars are rather close to being blackbodies. However, a hot star could be reddened, e.g. by dust which preferentially absorbs blue wavelengths. Less hot stars may be  blanketed by metals absorbing light and re-emitting at lower wavelengths, effectively also reddening the star. Or an object could simply not be a blackbody, e.g. a blue laser pointer.
Comparison to velocities
If, for some reason, you don't know the value of $B-V$, but you know that "object A has a smaller $B-V$ value than object B", then you can't really say anything about their "true" color; you can just say that "A is bluer than B", although neither of these objects need be blue.
You can compare this to velocities. All velocities are relative, so "$v = 10\,\mathrm{km/h}$" doesn't mean anything, unless you specify a reference frame. But even given a reference frame, the statement $v_\mathrm{A} > v_\mathrm{B}$ doesn't tell you how fast A and B are; it just tells you that A is faster than B.


Absolutely possible. There's nothing magical about a black hole. The gravitational pull of a black hole reaches as far as gravity would for another object of the same mass.

If you replace the Sun with a black hole of the same mass, everything would continue to orbit it just as it currently does.

Anything with mass has a gravitational force itself, and a black hole will attract anything with mass. Again, it's the same as our star having an effect on the Earth, and the Earth having an effect on the Moon.


The spectral reflectance curve depends on the coatings applied.  For 'scopes used by people there is no point in maintaining reflectivity below 450 or above 700 nm.  You idea of "visible" range is, I fear somewhat optimistic.  You may be better off with an uncoated Al mirror (spectral curves easily found on the net) and giving up some signal in the photopic range.  -- or with a nice Ag overcoat, perhaps.
edit
Per request, some data on uncoated Al.
From   laserbeamproducts,

REFLECTIVITY OF PURE UNCOATED ALUMINIUM AT NORMAL INCIDENCE

Wavelength            % Reflectivity
248nm                      92.6
400nm                      92.0
532nm                      91.6
633nm                      90.7
800nm                      86.8
900nm                      89.0
1um                          94.0
3um                          98.0
10.6um                     98.7
20um                        99.0
100um                      99.4

An image published (by the author) at Researchgate


Why planet's orbit is not perpendicular or random ?

Short answer: conservation of angular momentum.

Long answer:

The origin of almost any planetary system is a sparse cloud. That cloud starts to contract due (typically) to a pressure wave crossing it.

The cloud fragments as it contracts, and each fragment is what we know as a pre-star cloud.

Since almost always there is some movement in the matter in each cloud, the cloud as a whole starts to rotate, very slowly. Contraction helps because, due to angular momentum conservation law, when the cloud contracts, the rotation accelerates.

Soon we get a protostar with the most contracted matter, surrounded by a protoplanetary disk composed with the less contracted matter. The rotation of the whole system is in the same plane, due to conservation of angular momentum.

The protostar becomes a star, and the protoplanetary disk becomes a bunch of planets. Each planet, in turn, orbits the star and rotates on itself, all in the same direction, based on which point of the protoplanetary disk started accreting mass.

Later on, interaction among massive bodies disturbs this shiny process by changing some orbits and by having some collisions, but that's minor.


Was the Universe expanding before the beginning of inflation?

This is likely unanswerable.  Inflation was in part a resolution to a fine tuning problem: without it, it seemed we needed the early universe in a very specific and precisely balanced state to get to where it is today, and there was no solid scientific way to explain why things were so perfectly arranged (other than to simply assert that they were).

With inflation, the state of the universe before the inflationary epoch is fairly irrelevant.  Mostly all it needs is that the region that inflates into what we will know as the observable universe had enough time to achieve thermal equilibrium (which we need to explain why the universe looks pretty much the same in all directions, exactly as if things well outside of light speed communication had nonetheless achieved thermal equilibrium at some point).  Curvature and such get "smeared out" by the inflation to give us the mostly flat and homogeneous universe we see today.

On a pedantic note, some researchers consider inflation and the big bang to be the same thing. At least in the sense that talking about "before inflation" is scientifically meaningless, so if the Big Bang is the (scientific) beginning then we might as well take it to be inflation.


The escape velocity of Mars is about $5.03 \mbox{ km}/\mbox{s}$ , that of Earth about $11.2 \mbox{ km}/\mbox{s}$ , that of Moon about $2.38 \mbox{ km}/\mbox{s}.$

The escape velocity is an important ingredient for the stability of a planetary atmosphere.
From this it's not surprising, that the density of the Martian atmosphere is between that of Moon and that of Earth.

The loss of most of the magnetic field of Mars makes its atmosphere less resistant to erosive solar wind.

(Actually things get more complicated, since sequestration and outgassing, meaning an exchange of atmosphere with rock, as well as sublimation, and resublimation of water and carbon dioxide take place. Water vapor is decomposed into hydrogen and oxygen by photolysis, hydrogen escapes to space due to its low molecular weight, and oxygen tends to be bound chemically to iron-bearing rocks. On the other hand solar wind provides a steady supply of hydrogen, which is bound by rocks in the top few micrometers; "ChemCam saw peaks of hydrogen and magnesium during the first shots that we didn’t see in subsequent firings.")

The history of the Marsian atmosphere isn't yet known in full detail. NASA's MAVEN mission is designed to find out more.


Why is the solar system often shown as a 2D plane?

This is not a coincidence at all, but a direct consequence of the way the solar system was formed.

The generally accepted model is that solar systems (including our own) form out of a Protoplanetary disc.  Gravitation causes mass to collapse around a protostar, which always has some angular momentum (as does everything).  Wikipedia explains it better than I can:

  Protostars typically form from molecular clouds consisting primarily of molecular hydrogen. When a portion of a molecular cloud reaches a critical size, mass, or density, it begins to collapse under its own gravity. As this collapsing cloud, called a solar nebula, becomes denser, random gas motions originally present in the cloud average out in favor of the direction of the nebula's net angular momentum. Conservation of angular momentum causes the rotation to increase as the nebula radius decreases. This rotation causes the cloud to flatten out—much like forming a flat pizza out of dough—and take the form of a disk.

And then, from this protoplanetary disk, planets form.  Consequently, they're all in the same plane.

As such, the inclinations of each planet's orbit are pretty close to that of Earth's:

Planet  Orbital Inclination
Mercury 7°
Venus   3.39°
Earth   0°
Mars    1.85°
Jupiter 1.3°
Saturn  2.49°
Uranus  0.77°
Neptune 1.77°


Since the Earth is approximately 109 times smaller than the Sun (regarding radius), and the distance between them is about 150 million km, the eclipse will be visible up to about 1.39 million km from the Earth. This distance will vary (1.36 - 1.41 mln km) because the distance Earth-Sun varies during the year. This is about 3.6 times more than the Moon-Earth distance.

At this position, the apparent size of the eclipsed Sun will be only 1.08% smaller than the one on the Earth. The Moon has the radius about 1/4 of Earth's, and it will be visible as such on average. However
when in front of the Earth, it will be apparently almost 2 times bigger (regarding radius) than when behind.

We could orbit Earth at such distance, because Lagrangian points L1 and L2 are a bit farther (1.5 mln km). This means also that at L2 we could not see full Solar eclipse, but only annular with magnitude 0.93.


The manual says the original eyepieces had 1¼ inch barrels and focal lengths of 25 mm, 20 mm, 12.5 mm, 8 mm (all reasonable) and 4 mm (not so much).
If I had to pick two replacements for a beginner, I'd get a 25 mm and a 10 mm.
You can probably get an entry-level Plössl by Meade, Celestron, or Orion for US \$20-25 from the Cloudy Nights classifieds or \$30-40 new.


How to manually locate a certain declination in the sky for a given location? (no sextant available)

Jacob's staff is probably all you need.  It is a pole with a cross arm. By positioning the arm at a certain point on the pole and sighting down the pole one can get a position on the sky.  A skilled user can achieve sub-degree accuracy with such a tool.

The position of the cross beam is easy enough to calculate by trigonometry, or by careful construction of triangles.
$\tan(\theta/2)= c/2l$
Where $\theta$ is the angle above the horizon, $l$ is the position of the crossbar from the eye and $c$ is the length of the crossbar.


What's the difference between a Newtonian telescope and a regular telescope

First off, I would not recommend any of these telescopes.  Of the ones you listed, the 114EQ is the least bad choice.  But, due to the poor quality of its mount, it's the better of the bunch.

Ok, before I get into the nuts and bolts, let me first offer a couple pieces of advice:

1.) Join an astronomy club or society.  In the US, membership usually runs $50 or less per year.  Most clubs offer guidance and advice to new members.  They also often conduct frequent star parties.  These are events where those who have them bring out telescopes and everyone spends the evening observing together.  At such events, you can see a lot of different types of telescopes of all different designs, prices, manufacturers, etc... and talk to owners about them.  Clubs also have other programs that can help those new to the hobby learn and better enjoy their observing time.

2.) The best telescopes I can recommend in the under $200 price range (which is where these seem to be) are the AWB OneSky 130 (also known as the SkyWatcher Heritage 130P outside of the US) or the Meade Lightbridge Mini 130.  Both are 130 mm Newtonian telescopes with fairly good optics for their price range and stable mounts.  Meade makes a 114 mm version, and so does Orion.  I would not recommend any telescopes smaller than that, as there's little they can show you that a set of binoculars in the price range wouldn't do just as well or better.

Now, your initial question was about the differences between them.  Please bear with me, I tend to be verbose and try to be thorough (in other words, grab a snack and settle in for the read).

First of all, let me clear up a common misconception many people who are new to amateur astronomy tend to have.  The primary purpose of an astronomical telescope is not to magnify, but to gather light.  The problem with astronomical objects is less one of apparent size (this is a secondary concern, and not unimportant, but not as critical) and more one of brightness.  For example, the Andromeda Galaxy, also known as object 31 in the Messier catalog (M31), has an an angular size of about 3° by 1.5°.  The full moon is about 1/2° wide - so M31 is a lot bigger in apparent size.  But it's very faint in comparison.  If you know where to look and have fairly dark skies, you can actually see it with your naked eye at times, but not very clearly.

All the light you see comes into your eye through your pupil.  The pupil of an average adult eye, when fully dark adapted, has a diameter of about 6 or so mm. That's about 28 or so square mm of light gathering area.   A modest 80mm telescope (80mm being the aperture) has an area of over 5,000 square mm - over 150 times as much light gathering area.  All that light is then funneled down and concentrated on your eye.  This makes dim objects much brighter.  Magnification helps, but it doesn't matter how big an object appears if you can't see it at all.

The light gathering ability of the telescope is dictated by its aperture - the size of the main optical element collecting light (either an objective lens or primary mirror).   As such, it's not really adjustable (well, you can adjust it downward by blocking the aperture, but you can't increase it).   Magnification, however, IS adjustable.  The magnification power of a telescope is calculated by dividing the focal length of the telescope by that of the eyepiece.  For example, if you have a telescope with a focal length of 1,000 mm and an eyepiece of 25 mm, you get 40x magnification. If you swap out the eyepiece for a 10 mm eyepiece, you get 100x.  Theoretically, you can magnify as much as you like, though you can only magnify so much before the image is a blurry mess (this is also a matter of aperture - larger apertures allow for higher magnification before blurriness sets in).

Thus, the most important consideration when choosing an astronomical telescope is usually the aperture of the telescope.

When most people who are inexperienced with telescopes hear the word telescope, they think of a refractor.  A refractor, or refracting telescope, uses a lens to collect light.  This lens is known as the objective lens (though sometimes referred to, erroneously, as the primary lens).  When light is passed through a substance, such as glass, it is slowed slightly by the substance.  An objective lens is shaped with a convex (outward) curvature.  When light passes through the thicker, center part of the lens it is slowed more than that in the thinner, outer portion of the lens,  This causes the light to refract, or bend. The symmetry of the lens causes the light to all bend toward a point along an axis through the center of the lens.  The point where the light rays meet is known as the focal point.  All the light it collects is concentrated here.  The eyepiece sits here to gather the light and reconstruct the image for viewing.

Refractors were the first telescopes made, invented in the early 1600's in the Netherlands.  People often mistakenly consider Galileo the inventor of the telescope.  In fact, the man most frequently credited with the invention is Hans Lippershey (though two other men sometimes are given credit).  Galileo was only the first to use it to examine the night sky.

Even in his day, however, there was a problem with the design.  When you refract light, you cause the colors to separate.  This is exactly how a prism works.  In fact, in a real sense, a lens IS a prism.  This is because different wavelengths of light, what we percieve as colors, are slowed at different rates.  Shorter wavelengths, which are bluer light, are slowed more than longer, redder, wavelengths.  What this means in a telescope is that the actual focal points of different colors of light are separated with bluer light coming to a focus a bit closer in than redder light.

We call this phenomenon chromatic aberration, or CA, and it can be seen in telescopes as ablueish blur or fringe around objects you're viewing.  It degrades the quality of the image.

There are two main methods for reducing its effects.  The first is to increase the focal length of the telescope. A lens with a longer focal length has less pronounced curve: the difference between the thinnest and thickest parts of the lens are less dramatic than in a lens with a shorter focal length. This means there's less dramatic bending, which leads to the focal points of the colors being closer together.  If you look at early telescopes, like those used by Galileo, they tend to be very long indeed with high focal ratios.  The focal ratio is the ratio of the focal length to the aperture.  My 8" (203 mm) Schmidt-Cassegrain Telescope (SCT) has a focal length of 2,030 mm, making its focal ratio f/10.  My 8" imaging Newtonian has a focal length of only 800 mm, making its focal ratio f/3.9.  I believe some of Galileo's early scopes had ratios like f/20 - the tube was 20 times longer than it was wide.  Mostly this was to reduce chromatic aberration (though it was also to increase magnification)

The other method of reducing CA is by using multiple lenses with different shapes and made of different glass materials.  The most common option is to use two complementary lenses that work together to shift the focal points of bluer and redder light to bring them closer together.  This is called a doublet or achromat.  Using three lenses makes a telescope a triplet and and is often called an apochromatic refractor.  Sometimes even more lenses are used.  But this comes at a price. Each lens modifies the light, and while correcting, can also introduce distortion.  It also requires 2 surfaces (front and back) per lens to be precisely configured with a specific curvature.  This is costly and time consuming.  Lastly, lenses are heavy, and with a refractor, they are mounted at the front, which puts the weight at the front, and makes larger and longer such scopes much more difficult to properly maneuver. The largest refractor ever made was only about 49 inches, and the largest used for research was only 40 inches.

Galileo and his contemporaries understood the problem and that it could be solved by using a mirror instead of a lens.  Unfortunately, in those days, mirrors were much more difficult to make.  They were made from solid metal, which is much more difficult to precisely form and polish.  Most often,they were made of a metal known as speculum metal, which tarnishes very easy.  It wasn't until Sir Isaac Newton came along about 60 years later that a reflecting telescope was first created.

Newton used a mirror with a parabolic curvature to gather light.  It sat at the end of an open-ended tube. Light gathered by this mirror was bent and directed toward a secondary mirror, which, mounted at a 45 degree angle, would send the light out the side of the tube to the waiting eyepiece.

Because a mirror of this kind is all metal, there's no glass to pass through, so the glass does not separate the colors of light.  This eliminated chromatic aberration (well, there was still some in the eyepiece, but not much).  But, again, mirror technology was a problem.  Still, a few people adopted the design, and even built very large reflecting telescopes.  William Herschel's telescope was about 48 inches in diameter.  It was much easier to maneuver such a telescope, as the weight was all at the rear of the instrument (not to say it was easy, but putting a 4 foot glass lens at the end of the 40 foot tube would have been exceedingly difficult to work with - very, very top-heavy).  Herschel's mirrors, however, tarnished easily.  He actually had multiple mirrors, so one was being polished while another was in use, that he could just swap out as needed.

In the mid 19th century, the process of depositing a thin layer of metal, originally silver, but now usually aluminum, onto glass was developed.  Glass, while heavy, is much easier to configure into a precise shape.  Once done, the reflective coating could be deposited onto the glass mirror blank.  Because the coating is on the front-side of the mirror, light doesn't pass through the glass, just bounces off the coating.  This made Newtonian reflecting telescopes much easier to produce.  Only the one surface, the primary mirror, has to be precisely configured, and this made them more cost-effective.  And the weight and balance issues made making larger and larger telescopes easier.

Newton's design wasn't the only.  Laurent Cassegrain and James Gregory both designed, but probably never constructed, reflecting telescopes relying on mirrors with a hole in the center.  The light is collected by the primary, sent to the secondary, and sent back toward the primary to pass through the hole.  In Cassegrain's design, the secondary is flat.  In Gregory's, it was curved.  Both allowed for a shortening of the overall physical length of the instrument.

In more recent times, the use of mirrors and lenses together lead to the development of the Catadioptric telescope.  The most common such designs are the Schmidt-Cassegrain Telescope (SCT) and Maksutov-Cassegrain Telescope (Mak).

Just as refractors suffer from Chromatic Aberration, Newtonian telescopes suffer from an aberration called Coma.  Coma appears as distortions of the stars around the edge of the field of view.  They appear to lengthen and have comet-like tails (hense the name, coma).  As with Refractors, the effect is reduced by increasing focal length.  In modern telescopes, corrective lenses are often used.

Catadioptric telescopes resolve this partly by using a different kind of mirror shape, a spherical mirror.  The curvature of the parabolic mirror used in a Newtonian design varies along its profile.  The curvature of a spherical mirror is a constant curve.  This makes them easier to manufacture and reduces or eliminates coma, but it introduces another problem: spherical aberration.  Catadioptrics overcome this by using the combination of lenses and mirrors to correct the distortions.  However, the numerous surfaces then require precise configuring, which increases the cost to manufacture.

Each of these designs has its pros and cons. There isn't one "all-purpose" option that's best for everything.

As far as amateur telescopes go, however, the Newtonian design tends to be the most popular.  This is largely due to the average cost per inch of aperture.  For example, an 8" Newtonian can be found in the neighborhood fo $300 USD for the telescope tube alone.  An 8" SCT tube alone is likely to cost about three or four times that.  Very few 8" refractors are available, and those that are would likely cost several thousand dollars.

Now, as to the telescope you chose.  The Celestron PowerSeeker 127EQ is actually not a true Newtonian.  It is what we call a Bird-Jones (or Jones-Bird, depending on who you talk to) Newtonian.  A true Newtonian has a parabolic primary mirror and a flat secondary mirror.  The only other optical surfaces are in any eyepieces used.  The Bird-Jones Newtonian uses a spherical primary mirror.  To counter the spherical aberration inherent in the design, a small correcting lens, which is not much more than a Barlow lens, is included in the optical path (usually mounted inside the focuser mechanism).  This allows for a more compact design (the PowerSeeker 127 is only 508 mm (20 inches) long, but has a focal length of 1,000 mm (almost 40 inches).  But it also decreases the quality of the view.  The optics on this telescope are pretty poor.  Worse still, it is extremely difficult to collimate.

All telescopes require the optical elements to be properly aligned to produce the best image.  This is collmation.  In the case of refractors, it rarely needs adjustment.  Most SCT's require it only occasionally, and Maksutovs rarely.  But Newtonians require it frequently.  Newtonians will have a set of adjustment screws or knobs on the back, behind the primary mirror, and on the front as part of the support that holds the secondary mirror (usually referred to as the spider).  Most telescopes use a 3-screw system for adjusting the tilt of the mirrors.  It can be a chore at first, but with a little practice, it becomes fast and easy to do.  It also requires a tool to insert into the focuser.  Traditionally, this is a collimation cap or Cheshire, which work like an eyepiece, but do not have lenses.  They help you visualize the alignment of the optics so that you can adjust it properly.  You can also get a laser collimator which uses a low-power laser to guide you.  But none of these work with a Bird-Jones scope.  The correcting lens makes them very, very difficult to collimate.

Furthermore, the mount on the PowerSeeker line, particularly the 127EQ, is not very stable.  It wobbles and shakes and makes for difficult observing.

Sadly, the reviews on sites like Amazon make it out to be a great instrument.  But these reviews themselves are flawed.  Amazon's servers use review rankings to determine the display order of search results.  If an item has higher ratings in its reviews, it appears toward the top of the list of results when you search.  Most buyers don't look past the first page, and rarely much further than 2 pages.  So if a seller wants to sell his or her product, it needs to be on the first page, preferably right at the top.  This leads a lot of companies and individual sellers to push customers to add reviews.  Some companies also falsify reviews (though it would be difficult to prove it) in order to game the system.  As such, the reviews on Amazon (and many other sites) are nearly useless to the consumer.  And the PowerSeeker telescopes are a prime example.

On multiple other sites I spend time on, this telescope is outright hated.  It is sold by Celestron (a chinese-owned company) to unsuspecting buyers.  In the long run, I think it causes more harm to amateur astronomy than good - because users often end up discoraged by its poor peformance and cheap construction.

If you've already purchased it, I'd strongly recommend returning it and getting something different (e.g. the two scopes I listed at the beginning).  If you haven't yet, then please do not.  The two telescopes I listed are significantly better in nearly every way for a similar price.  Even the slightly smaller 114 mm versions are better.  But the PowerSeeker line is nearly entirely composed of low-quality equipment that is not worth purchasing.

But even before purchasing, joining a club is the best thing you can do.  Some clubs even have loaner telescopes that members can borrow to learn before they buy.

I wish you luck with whatever you end up purchasing.  Clear skies!


Red shift is used for measuring the distance to very distant stars (galaxies mostly, in fact). The secret is to use spectral lines. Specific elements when very hot emit light at very specific colours and you can spot the pattern of those colours when the whole pattern has been shifted towards the red and see how far it has shifted.

For closer stars, there are other techniques, using types of star whose brightness can be worked out in some way, or, for the closest, by observing how the position of the star compared to more distant ones changes as the Earth moves from one side of the Sun to the other.


plotting stars in cartessian coordinates using the stars ra and dec produces the wrong location of the stars

A photograph of stars is essentially a
gnomonic projection
of part of the
celestial sphere.
If $\vec P$ is computed for the photograph center (RA, Dec) in the same way as $\vec S$,
and the north celestial pole is
$\hat{\mathrm k} = (0, 0, 1)^{\mathrm T}$,
then the photograph has basis vectors perpendicular to $\vec P$ and each other,
$$\hat{\mathrm u} = \frac{\vec P \times \hat{\mathrm k}}{\| \vec P \times \hat{\mathrm k} \|},
\hat{\mathrm v} = \frac{\hat{\mathrm u} \times \vec P}{\| \hat{\mathrm u} \times \vec P \|}$$
and you can project $\vec S$ onto the photograph as
$$\left( \frac{\vec S \cdot \hat{\mathrm u}}{\vec S \cdot \vec P},
\frac{\vec S \cdot \hat{\mathrm v}}{\vec S \cdot \vec P} \right)$$
scaling as needed.


How often do stars pass close (~1ly) to the Sun?

Here is the very study you are looking for by Bailer-Jones (2014).

Using the re-reduction of the Hipparcos astrometry, he has integrated orbits for 50,000 stars to look for objects that might come or might have come close to the Sun.

The K-dwarf Hip 85605 is the winner on that timescale, with a "90% probability of coming between 0.04 and 0.20pc between 240,000 and 470,000 years from now".

The next best is GL710 a K-dwarf that will come within about 0.1-0.44pc in 1.3 million years.

On a statistical basis, some work has been done by Garcia Sanchez et al. (2001). They estimate, using the Hipparcos data, that encounters within 1pc occur every 2.3 million years. However, the Hipparcos data isn't complete for low-mass, faint objects. Making a correction for this, the authors estimate a $<1pc$ encounter every 100,000 years. The probability for closer approaches scales as the inverse of the square of the separation. i.e. the timescales for approaches closer than 0.1pc is 100 times longer.

There are plenty of big error bars in all this work - recall that an error of 1km/s in velocity leads to a 1pc error in position after 1 million years. All this should be nailed down really well by Gaia results in the next 2-3 years.


How is an exoplanet characterised as "Earth-Like"?

There is no unique definition of Earth-like and it depends on what characteristics are important in the context of discussion.

The most simple one is to just compare radii and masses. That is observationally sensible as radius is directly deducable from transits (assuming stellar size known) and mass is often an accessible properties for transiting planets when you have access to complementary radial-velocity data. It's also a good starting point if you want to add  further restrictions. The habitable planet catalogue uses this definition (considering only planets in the habitable zones around their stars, using a simple equilibrium temperature definition)

If your quest is actually for a 2nd Earth not only in surface gravity but other characteristics, one approach is outlined by the Earth similarity index (ESI). While it is not directly a measure for habitability either, it is a good proxy by comparing more essential parameters with Earth like mass, gravity, surface temperature. For a selection of parameters you define this ESI as

$$
\mathrm{ESI} = \prod_{i=1}^n \left(1 - \frac{x_i - x_{i,o}}{x_i + x_{i,o} }\right)^{w_i/n}
$$

where $ x_{i,o} $ is the Earth's value of a property and $n$ the total number of properties. This is basically a weighted geometric mean of the property differences. It is a good idea to consider 'internal values' and 'external' ones. The Planetary Habitability Lab uses surface temperature and escape velocity as external and radius and density as internal and use this to create a nice 2D scatter plot for known planets.


Measurement precision of celestial events

This doesn't answer your question, but has two screenshots, so that's something :)

If you run http://ssd.jpl.nasa.gov/?horizons with these settings:

part of the output looks like this:

This shows the Sun crosses the equinox point between the 8th and 9th second of 14:21 (which is 10:21 your local time). HORIZONS will even let you compute at half second intervals but not beyond that.

If you use the http://naif.jpl.nasa.gov/naif/tutorials.html library, you can compute the time down to the nanosecond or better, but this doesn't really help because we don't know the Sun's position, the Earth's position, or even the direction the North Pole is pointing to that level of precision.

So, not an answer, but some hopefully helpful comments.


Why isn't our local interstellar space colorised?

This is really more of a biology question, I think, so: The Mark-I Eyeball is far less sensitive to color than to intensity.  There are rods and cones in the retina. The neural signals from the rods are used to report intensity only. There are different spectral sensitivity curves for different cones (typically three types in humans), so the relative outputs from all cones are used to generate color information.  However, cones are less sensitive than rods, so for dim light such as that reaching us from nebulas, only the rods produce a neural output.


The motion of the moon, as measured by the Lunar Laser Ranging data is not as predicted by Newton. As the scientist being interviewed says, Newton's model was correct to the accuracy of the data available to him, and was good enough to land the Eagle on the moon in '69. However the Lunar Laser Ranging data can position the moon to an accuracy of 3cm.

Relativistic effects cause the moon to deviate from the orbit predicted by Newtonian mechanics by about 1m, mostly due to special relativistic effects such as Lorentz contraction. General Relativity accounts for a variation of 10cm. This is much less than the perturbation due to Jupiter (about 1km). However, general relativity does fully account for the motion of the moon, to the accuracy of the experiment.

  All tests are in agreement with Einstein's General Relativity which is used for the numerical integration of ephemerides. No variation of the gravitational constant is discernible. Source

Lunar ranging is ongoing at the Apache Point Observatory Lunar Laser-ranging Operation.


What if we are looking for the wrong signs of life on other planets?
Are there any fields of astronomy that look beyond the "near-Earth"
life comparisons and look for life in other (drastically different)
forms?

and

I'm always bothered by these reports because they are all predicated
on the premise that life on other planets will closely resemble our
own.

Short answer:
It doesn't matter what people think.   What matters is what we can observe.   Just because they will probably look for life on Earthlike planets first, doesn't mean they won't look elsewhere.  The answer will depend on what they find, not what anyone expects to find.
Long answer:
I think there are a lot of good comments below your question, but I want to try to address a few points.
On looking for life.
The scientists that look into space through our biggest telescopes aren't exactly "looking" for life on other planets because life is too hard to see.   Mostly, they look at stars.   Our best telescopes today can't see exoplanets at all.  Exoplanets are detected indirectly by variations in the stars they orbit.    If exoplanets are too far away most of the time to even be seen, looking for signs of life on them is obviously out of reach.   We're not actively looking for life because we can't.   We need much bigger telescopes for that.
So, basically "20 Earthlike planets discovered" is the upper limit of our technology right now.  That's today's cutting edge discovery.   I understand that it might be boring.  I know that they're just adding more planets to the directory, but that's the limit that our best equipment can deliver right now.
Almost nothing is known about all those planets, just the very basics, like how close they are to their star, and about how large they are and sometimes, about how dense they are giving us a limited idea of their composition and if they are hot, cold or perhaps in the Goldilocks zone.   If there's only 5 lines of data per exoplanet, the tag-line "Earth-like" often makes the best headline, not because NASA is focused on only Earthlike planets, but because there's not a lot of detail.
It's kind of like "I see a bird, but it's far away and I don't have my binoculars.  It's flying south", but I can't see what kind of bird, I can make out the size, to some extent, I can see that there's just one, but I can't make out the shape.   So, I write in my bird journal "One largeish bird flying south - too far away to see what kind".
That's basically where we stand with exoplanet observation right now.
This will change some when the James Webb Space Telescope is operating.   The JWST should be able to see star-light filtered through some exoplanet atmospheres, which will tell us a whole lot more than we currently know about some of those exoplanets.  Information such as how thick an atmosphere and what it's made of will tell us a lot, and give us some information on the climate of the planet.  JWST will be a HUGE (all caps for emphasis) step forward in the search for alien life, though it still won't observe life directly, but it will hopefully get a good read on some exoplanet's atmospheres, and from that, deductions can be made, such as abundant oxygen isn't expected without some form of photosynthesis.   Oxygen is too reactive to be likely to exist on it's own.
A combination of CH4 and O2 could be especially interesting and one of the first things to look for, but they also have to be careful.  There's always the chance of more than one explanation for a planet's atmosphere.  We shouldn't expect any kind of guarantee of discovery, but JWST should be an enormous step forward in what we can observe about exoplanets.
On reporting
Scientists in this field, their job is essentially to analyze and report the data.   Their job is not, for the most part, to speculate on life but to simply report what can be verified (and today, relatively little other than "I found a planet" can be verified).
Theoretical biology is great.  Don't get me wrong.  It's fun and it's real science, but the majority of telescope observation focuses on observe, analyze and report.  That might sound dry, but that's mostly how science works.  "20 Earthlike planets found" was impossible 10-15 years ago.  It really is cutting edge.
On Theoretical Alien Life:
Scientists, today (Neil deGrass Tyson among them) who would love to get a closer look at some of the Moons of Jupiter and Saturn, because they are thought to hold the best chance of having life beyond Earth in our solar-system.   The desire to find alien life, if it's out there, is one of the highest held goals of NASA and Scientists in this field.   Nobody's ignoring or dismissing anything.  They will likely point JWST towards Earthlike planets first, but that's only natural.
I'd like to also point a basic rule of astronomical observation.   "Weird" is actually good.   Weird leads to discoveries like dark matter and dark energy.  Like the Isaac Asimov "That's funny" quote, finding things they don't expect is often the best find.   Scientists love that.   If weird atmospheres are found by JWST, it won't matter if they are on Earthlike planets or not.  Anything weird or unexpected will be of great interest and closely looked at.
Like those theoretical Alien Mega-structure or Dyson sphere stars (which probably aren't Dyson spheres).  I'm highly skeptical that it's aliens, but I'm glad they're looking into it.  Anything unexpected is worth studying.
When JWST starts to get better data on atmospheres, then you'll see more interesting headlines than "Earthlike".  For now, and for the next year or so, "More Earthlike planets found", is as far as our technology can take us.
Finally, I want to point out that on theoretical biology, there are issues.  (The other answer covers this better than I did so I'll abbreviate), but there are reasons to suspect that life might need to be carbon and liquid water based, so planets thought to have liquid water should be the first places to look.
And ultimately, it doesn't matter what people think is out there.   What matters is what JWST (and telescopes that follow) will discover.   What we know about the Universe is based on what has been observed, not on the various prejudices and assumptions people made.   Sometimes what is observed goes against the scientific belief of the day, and that's what's great about science.  Evidence trumps belief.


Can the Sudarsky's gas giant classification be applied to ice giants?

That classification scheme originates from before we had much in the way of observational data on exoplanetary atmospheres. It's not clear that it's a particularly useful description of how hot Jupiters behave: recent models, e.g. Gao et al. (2020) seem to indicate that silicate clouds and hydrocarbon haze are the main factors at play on hot Jupiters, rather than silicates versus alkali metal absorption.
As regards is applicability to ice giants, these planets have substantially higher atmospheric metallicities than gas giants, so it's not clear that these models would apply. For a study of some warm Neptune atmospheres, there's Fortney et al. (2020), who note that the cooling history of the planet is important, and things like tidal heating at depth caused by an eccentric orbit can have significant consequences. Or for a more experimental approach, see He et al. (2020), who indicate that hazes can form in hydrogen-rich atmospheres even in the absence of methane in the initial gas mix. It's a lot easier to find studies on hot and warm planets as these are the ones that are accessible to observations, so they attract more attention: habitable zone terrestrials are an exception to this pattern.
In summary, it's complicated, and things have moved on quite substantially since the Sudarsky et al. papers.


How to listen to our world from 10,000 light years away?

As the above commentator says, our technology isn't sufficiently sensitive to pick up everyday radio broadcasts from so far away, and even if we could it wouldn't be worth sending them a message. The earliest we could hope to receive a reply would be in 20,000 years time, by which time there might be no humans left on Earth to receive it, as a result of nuclear war or some other catastrophe. If this extra-terrestrial civilisation were much, much nearer, tens of light years rathe than thousands, prospects would be very much better, but it is extremely unlikely that an advanced alien civilisation is as close as that.


Why isn't the moon's surface made of mini craters?

Nice question, unfortunately your analoguous experiment is not as analoguous as you might hope.
In your comparison image you are using both, not enough impactors, and a too tiny size distribution. Impactors on the Moon (or any other objects) follow a reasonably well-known size distribution where the smaller an impactor is, the more frequent collisions with those do occur.
The visible and measured / counted size distribution of craters on the Moon follows this probability law quite well (e.g. this paper (Cross, 1966), but many others get about identical results). The main difference is that in your comparison you miss all the very tiny impactors which smooth-out the surface again.
Further, the Moon is covered with loose regolith of about 45µm diameter. In the sunlight it is subject to charging and thus the Moon surface is covered by a tiny sheet of statically elevated particles. This leads to a re-distribution of the surface regolith particles, smoothing out the tiniest craters even more.
Even further, the landing sites of Apollo were carefully chosen to be the most level and crater-free places to ensure a safe landing. The amount of cratering in a particular place is also random, even when all the surface is covered by several layers of craters (this is a study for asteroids, but applies to other bodies, too).
And further more, your comparison experiment is conducted in air. That affects the formation of craters and in particular the distribution of ejecta due to friction with air, more so, the smaller the impactors get and the slower the ejecta are.


Conundrum involving distance to object, universal expansion, age of universe, etc

The 13 billion lightyears distance of the quasar mean 13 billion lightyears light-travel distance. In other words, the light took 13 billion years to reach us, independent of the distance the quasar is now away from us.

The current proper distance (a chain of rulers would measure) is called comoving distance; it is much larger than the light-travel distance for large redshifts.

The quasar is moving away from us in an accelerated fashion, and is now moving away faster than light. This means, light which is emitted now (in the sense of cosmic time) by the galaxy, the quasar may have evolved into in the meanwhile, will never reach us.
The light, the quasar emitted 13 billion years ago, left the quasar just in time to leave the regions of the universe which later have been receding faster than light from us, to eventually reach us.

As a thought experiment, imagine walking with 5 km/h (our playground speed of light) on a rubber band which is expanding with a constant rate of 1 km/h per meter actual length of the band. (This leads to an exponential acceleration of the distance between the two marks on the band). If you start closer than 5 meters away from your goal, e.g. from a start mark (our quasar) 4.50 m away, you'll finally reach the goal mark. The start mark will soon be further away from the goal mark than 5 meters, therefore receding with more than 5 km/h from the goal shortly after you left the start mark.
At the moment you arrive at the goal mark, the start mark will be much further away (comoving distance) than the distance you needed to walk (light travel distance). And you've been walking a longer distance than the (proper) distance between the marks was at the time you started walking.

Btw.: Acceleration is only felt as a force, when the velocity to the local rubber band (mataphoric space-time) is changed.

Example calculations with a protogalaxy of redshift $z=11.9$:
Based on the Cosmology Calculator on this website, the cosmological parameters $H_0 = 67.11$ km/s/Mpc,
$\Omega_{\Lambda} = 0.6825$ provided by the Planck project, and the scale factor $d(t) = d_0 / (1+z)$, setting $\Omega_M = 1- \Omega_{\Lambda} = 0.3175$, the
age of the universe is $13.820$ Gyr, and the comoving distance of the protogalaxy is $d_0 = 32.644$ Gly.

The age of the universe, we see the protogalaxy (at redshift 11.9), was 0.370 Gyr, light-travel distance has been 13.450 Gly, proper distance was 2.531 Gly.

After the protogalaxy has been emitting light 0.370 Gyr after the big bang, the light travelled towards us through space of redshift beginning with 11.9 shrinking to 0; the light arrived at us 13.820 Gyr after the big bang. The comoving distance (to us) of the space traversed by the light started with 32.644 Gly shrinking to 0. The remaining distance, the light needed to travel, started with 13.450 Gly shrinking to 0. The proper distance between the protogalaxy and us started with 2.531 Gly increasing to 32.644 Gly due to the expansion of spacetime.

Here some intermediate states described by
a couple of tuples, consisting of

redshift $z$,
according age $t$ of the universe (Gyr),
comoving radial distance (at age $t$) of the emitted light, we can now detect from the protogalaxy (Gly),
remaining light travel distance of that emitted light (Gly),
proper distance of the protogalaxy at age $t$, according to $d(t) = d_0 / (1+z)$:

$$(11.9,  0.370, 32.644, 13.450,  2.531),$$
$$(11.0,  0.413, 32.115, 13.407,  2.720),$$
$$(10.0,  0.470, 31.453, 13.349,  2.968),$$
$$( 9.0,  0.543, 30.693, 13.277,  3.264),$$
$$( 8.0,  0.636, 29.811, 13.184,  3.627),$$
$$( 7.0,  0.759, 28.769, 13.061,  4.081),$$
$$( 6.0,  0.927, 27.511, 12.892,  4.663),$$
$$( 5.0,  1.168, 25.952, 12.651,  5.441),$$
$$( 4.0,  1.534, 23.952, 12.285,  6.529),$$
$$( 3.0,  2.139, 21.257, 11.680,  8.161),$$
$$( 2.0,  3.271, 17.362, 10.549, 10.881),$$
$$( 1.0,  5.845, 11.124,  7.974, 16.322),$$
$$( 0.0, 13.820,  0.0  ,  0.0  , 32.644).$$

The Hubble parameter, meaning the expansion rate of space per fixed proper distance, is decreasing with time. This allowed the protogalaxy to recede almost with the speed of light, although it was just about 2.5 Gly away from us (proper distance) in the time, when it emitted the light we detect now.
Nevertheless distant objects in this space accelerate away from us, since their increasing distance is multiplied with the expansion rate of space.


Why do we define the Milky Way's rotation in a left-hand coordinate system?

The problem, as far as I can tell, is that the North and South Galactic Poles were defined well before we had any idea of which way the Galaxy rotated. It turned out that the Galaxy's angular momentum vector (in the standard right-hand sense) actually points in the same direction as the previously defined South Galactic Pole. Since there's a separate convention for having "north" point "up", you end up with the backwards rotation definition.

William Herschel seems to have started this back in the late 18th Century; I assume his idea was that whichever Galactic pole was in the North Celestial Hemisphere should be called the "northern" pole. (No one back then had any clear sense that the Galaxy necessarily rotated at all.) By the time the actual direction of Galactic rotation was figured out in the 20th Century, the traditional orientation was too well entrenched to be worth upending.


Does rotation affect gravitational lines of force

An spherical electric charge has the same electric field lines whether spinning or not. The difference between those two cases is entirely in the magnetic field. Thus, one should expect as similar thing to happen for gravity.

The parametrized post-Newtonian formalism, weak-field GTR has the metric
$$\mathrm{d}s^2 = -(1+2\Phi)\,\mathrm{d}t^2 + 2\mathcal{A}_j\,\mathrm{d}t\,\mathrm{d}x^j + (1-2\Phi)\delta_{ij}\,\mathrm{d}x^i\,\mathrm{d}x^j$$
where $\Phi\equiv -U$ is essentially the Newtonian gravitational potential, while $\mathcal{A}_j\equiv-\tfrac{7}{4}V_j - \tfrac{1}{4}W_j$ in terms of the other PPN potentials. For the four-velocity $U^\alpha\equiv{\mathrm{d}x^\alpha}/{\mathrm{d}\tau} = (U^0,\vec{U})$, the geodesic equation for time-independent $\Phi$ and $\mathcal{A}$ becomes, to linear order in $\Phi$,
$$\frac{\mathrm{d}\vec{U}}{\mathrm{d}\tau} = U^0(\vec{G} + \vec{U}\times\vec{H})\text{,}$$
where the gravitoelectric field is $\vec{G} = -\nabla\Phi$ and the gravitomagnetic field is $\vec{H} = \nabla\times\vec{\mathcal{A}}$, here $\nabla$ being used in the ordinary sense of $\nabla_i = \partial_i$, i.e. with respect to the Euclidean metric $\delta_{ij}$. If $\Phi$ and $\mathcal{A}$ are not time-independent, then $\vec{G} = -\nabla\Phi - \partial_t\vec{\mathcal{A}}$, paralleling electromagnetism, but there will be an extra term in the analogue of Lorentz force that has no electromagnetic counterpart.

Much the same thing can be done in any stationary spacetime, including the rotating Kerr black hole. See also Costa and Natário (2014) for a much more general treatment of several gravito-electromagnetic analogies.

References:

Costa, L. F. O, Natário, J. "Gravito-electromagnetic analogies". Gen. Rel. Grav. 46, 1792 (2014)[arXiv:1207.0465]


How to compute the sun's (upper) transit

Here is the page code for my BASIC-to-php implementation of the algorithms to calculate the  (geocentric) RA & Dec of the Sun for a given date from Jean Meeus' book Astronomical Algorithms as well as calculating the angular distance between two celestial objects.  I can't remember the accuracy of the calculations and I didn't make the corrections for observer lat & long.

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=ISO-8859-1" />
 <base href="sionnagh.com" />
<title>Mick Todd Distance Calculator</title>

<?php include("mtheader.php"); ?>

Mick Todd Distance Calculator

<?php
$earthincl=23.44; //Earth's axial tilt
$sollat=0;

$fdate = $_POST["fdate"];
if ($fdate=="") {
 $fdate=getdate(date("U")-date("Z"));
 $fday=$fdate[mday];
 if (strlen($fday)==1) {$fday="0".$fday;}
 $fmonth=$fdate[mon];
 if (strlen($fmonth)==1) {$fmonth="0".$fmonth;}
 $fyear=$fdate[year];
 //$fdate=$fdate[mday]."/".$fdate[mon]."/".$fdate[year];
 $fdate=$fday."/".$fmonth."/".$fyear;
}
else {
 $fday=substr($fdate,0,2);
 $fmonth=substr($fdate,3,2);
 $fyear=substr($fdate,6,4);
}
$fra1 = $_POST["fra1"];
$fra1 = strtolower($fra1);
if ($fra1<>"") {
  if (strpos($fra1,"h")==0) {
  $frad1=$fra1;
  $fra1=converttohms($fra1);
  }
  else {
    $frad1=converthtod($fra1);
  }
}
$fdec1 = $_POST["fdec1"];
$fdec1 = strtolower($fdec1);
if ($fdec1<>"") {
  if (strpos($fdec1,"d")==0) {
    $fdecd1=$fdec1;
    $fdec1=converttodms($fdec1);
  }
  else {
    $fdecd1=convertdtod($fdec1);
  }
}
$fra2 = $_POST["fra2"];
$fra2 = strtolower($fra2);
if ($fra2<>"") {
  if (strpos($fra2,"h")==0) {
    $frad2=$fra2;
    $fra2=converttohms($fra2);
  }
  else {
    $frad2=converthtod($fra2);
  }
}
$fdec2 = $_POST["fdec2"];
$fdec2 = strtolower($fdec2);
if ($fdec2<>"") {
  if (strpos($fdec2,"d")==0) {
    $fdecd2=$fdec2;
    $fdec2=converttodms($fdec2);
  }
  else {
    $fdecd2=convertdtod($fdec2);
  }
}

function JulianDay ($fday, $fmonth, $fyear, $fUT) {
    $JDpartsign=(100*$fyear+$fmonth-190002.5)/abs(100*$fyear+$fmonth-190002.5);
 $JD = 367*$fyear - floor((7*($fyear+floor(($fmonth+9)/12)))/4) + floor((275*$fmonth)/9) + $fday + 1721013.5 + $fUT/24 - 0.5*$JDpartsign + 0.5;
 return $JD;
}
function converttodms($inval) {
 if ($inval==0) {
   $fdsign=1;
 }
 else {
   $fdsign=$inval/abs($inval);
 }
 $inval=abs($inval);
 $fddeg=floor($inval);
 $remain=($inval-$fddeg)*60;
 $fdmin=floor($remain);
 if (strlen($fdmin)==1) { $fdmin="0".$fdmin; }
 $remain=($remain-$fdmin)*60;
 $fdsec=round($remain*100)/100;
 if (strpos($fdsec,".")==1) { $fdsec="0".$fdsec; }
 $fddeg=$fdsign*$fddeg;
 $fdms=$fddeg."d".$fdmin."m".$fdsec."s";

    return $fdms;
}
function converttohms($inval) {
    $inval=$inval/15;
    $fddeg=floor($inval);
    $remain=($inval-$fddeg)*60;
 $fdmin=floor($remain);
 if (strlen($fdmin)==1) { $fdmin="0".$fdmin; }
    $remain=($remain-$fdmin)*60;
 $fdsec=round($remain*100)/100;
 if (strpos($fdsec,".")==1) { $fdsec="0".$fdsec; }
    $fdms=$fddeg."h".$fdmin."m".$fdsec."s";

    return $fdms;
}
function converthtod($inval) {
  $posh=strpos($inval,"h");
  $posm=strpos($inval,"m");
  $poss=strpos($inval,"s");
  $temph=substr($inval,0,$posh);
  if ($posm) {
    $tempm=substr($inval,$posh+1,$posm-($posh+1));
  }
  if ($poss) {
    $temps=substr($inval,$posm+1,$poss-($posm+1));
  }

  $fdms=($temph+$tempm/60+$temps/3600)*15;

  return $fdms;
}
function convertdtod($inval) {
  $posd=strpos($inval,"d");
  $posm=strpos($inval,"m");
  $poss=strpos($inval,"s");
  $tempd=substr($inval,0,$posd);
  if ($posm) {
    $tempm=substr($inval,$posd+1,$posm-($posd+1));
  }
  if ($poss) {
    $temps=substr($inval,$posm+1,$poss-($posm+1));
  }
  $fdms=$tempd+$tempm/60+$temps/3600;

  return $fdms;
}

function SolarCoords($fday, $fmonth, $fyear, $fUT) {
//Julian Day of 1991/ 5/19 at 13 UT         JD = 2448396.04167
//Julian day of 2000/01/01 at 12 UT         JD = 2451545.0
//number of Julian days since 2000/01/01 at 12 UT               -3148.95833
//number of Julian centuries since 2000/01/01 at 12 UT  T = - 3148.95833/36525
//used by the algorithm for L                                                           = 0.086213780

    $k = 2*pi()/360;
  $T = (JulianDay($fday, $fmonth, $fyear, $fUT)-2451545.0)/36525;
    //echo "\$T: ".$T."
";
 //mean anomaly, degree
 $M = 357.52910 + 35999.05030*$T - 0.0001559*$T*$T - 0.00000048*$T*$T*$T;
    //echo "\$M: ".$M."
";

    // mean longitude, degree
    $L0 = 280.46645 + 36000.76983*$T + 0.0003032*$T*$T;
    //echo "\$L0: ".$L0."
";

    $DL = (1.914600 - 0.004817*$T - 0.000014*$T*$T)*sin($k*$M) + (0.019993 - 0.000101*$T)*sin($k*2*$M) + 0.000290*sin($k*3*$M);
 //echo "\$DL: ".$DL."
";

    // true longitude, degree
    $L = $L0 + $DL;
 if (abs($L)>360) {
        $divsign=($L/360)/abs($L/360);
  $div=floor(abs($L/360));
  $L=$L-$divsign*$div*360;
  if ($L<0) { $L=$L+360; }
    }
    $L=round($L*10000)/10000;

    return $L;

}

?>

<form METHOD="POST" ACTION="mtdistcalc.php" name="mtdistcalc">

Date <input type="text" name="fdate" id="fdate" size=10 value="<?php echo $fdate; ?>"> <input type="button" value="Now" onclick="fdate.value=settoday();">

Object coordinates (#1) RA <input type="text" name="fra1" id="fra1" size=15 value="<?php echo $fra1; ?>">  Decl <input type="text" name="fdec1" id="fdec1" size=15 value="<?php echo $fdec1; ?>">

Object coordinates (#2) RA <input type="text" name="fra2" id="fra2" size=15 value="<?php echo $fra2; ?>">  Decl <input type="text" name="fdec2" id="fdec2" size=15 value="<?php echo $fdec2; ?>">

<input TYPE=SUBMIT VALUE="Calculate distances">

</form>
<?php
echo "

Date:	".$fday."/".$fmonth."/".$fyear." "."0000UT
";
echo "
Julian Day: 	".JulianDay($fday,$fmonth,$fyear,"0")."
";
echo "
Ecliptic latitude of the Sun: 	".$sollat."&deg; (assumed)
";
$sollong = SolarCoords($fday,$fmonth,$fyear,"0");
echo "
Ecliptic longitude of the Sun: 	".$sollong."&deg; (".converttodms($sollong).")
";
echo "
Earth's axial inclination: 	".$earthincl."&deg;

";

//echo "\$feclong ".$feclong."

";

$deltaSol = round(asin(sin($sollat*M_PI/180)*cos($earthincl*M_PI/180) + cos($sollat*M_PI/180)*sin($earthincl*M_PI/180)*sin($sollong*M_PI/180))*180/M_PI*10000)/10000;
$alphaSol = round(acos(cos($sollong*M_PI/180)*cos($sollat*M_PI/180) / cos($deltaSol*M_PI/180))*180/M_PI*10000)/10000;
if ($sollong>180) {
 $alphaSol = 360-$alphaSol;
}
echo "

Sol

RA: 	".$alphaSol."&deg; (".converttohms($alphaSol).")

Dec:	".$deltaSol."&deg; (".converttodms($deltaSol).")

";

//spherical angular distance
//cos(theta) = sin(delta_a) sin(delta_b) + cos(delta_a) cos(delta_b) cos(alpha_a-alpha_b)
if ($fra1<>"" AND $fdec1<>"") {
  $theta1 = round(acos(sin($deltaSol*M_PI/180)*sin($fdecd1*M_PI/180) + cos($deltaSol*M_PI/180)*cos($fdecd1*M_PI/180)*cos(($alphaSol-$frad1)*M_PI/180))*180/M_PI*1000)/1000;
}
if ($fra2<>"" AND $fdec2<>"") {
  $theta2 = round(acos(sin($deltaSol*M_PI/180)*sin($fdecd2*M_PI/180) + cos($deltaSol*M_PI/180)*cos($fdecd2*M_PI/180)*cos(($alphaSol-$frad2)*M_PI/180))*180/M_PI*1000)/1000;
}

if ($fra1<>"" AND $fdec1<>"" AND $fra2<>"" AND $fdec2<>"") {
  $theta3 = round(acos(sin($fdecd1*M_PI/180)*sin($fdecd2*M_PI/180) + cos($fdecd1*M_PI/180)*cos($fdecd2*M_PI/180)*cos(($frad1-$frad2)*M_PI/180))*180/M_PI*1000)/1000;
  echo "
Angular distance between Object #1 and Object #2 is: ".$theta3."&deg;
";
}

if (($fra1<>"" and $fdec1<>"") or ($fra2<>"" and $fdec2<>"")) {
  echo "

&nbsp;	RA	Dec	Solar elong.
";
}
if ($fra1<>"" AND $fdec1<>"") {
  echo "
Object #1:	".$fra1."	".$fdec1."	&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;".$theta1."&deg;
";
}
if ($fra2<>"" AND $fdec2<>"") {
  echo "
Object #2:	".$fra2."	".$fdec2."	&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;".$theta2."&deg;
";
}

?>


After a bit of searching in Aladin. I found that the other star is V* V2170 Ori.
By comparing their respective Simbad pages one can see the following:

LL Ori is actually slightly further away.
V2170 Ori has a GAIA parallax of 2.84 while LL Ori has one of 2.55. Simply using $D=1/p$, this translates to a distance of 0.35 kpc and 0.39 kpc
LL Ori and V2170 Ori have about the same proper velocity of 1.65 and 1.69 mas/yr. So, that should affect the shock size.
LL Ori is a lot brighter.
It has an apparent V-Band magnitude of 11.5, while  V2170 only has 18.7.
LL Ori is hotter.
Its spectral typ is K0-K3e, while V2170 is of spectral type M4.7e.

It is therefore not unexpected that LL Ori has a significantly stronger stellar wind and thus a larger bow shock.

Since the distance is basically the same, you can just have a look at this image with scale to estimate the size of the other bow shock.


What are these three black cap-like pieces?

I am currently using a Celestron Astromaster 130 EQ and I also have a similar type of caps.
Starting from left to right, the first two are caps used to cover eyepiece holder in case the lens is not connected with it, and the last one is a cap for the lens. So I think they might have given you one extra cap as a lens holder.


How big should be a piece of neutron star to reach Earth's mass?

The question has no answer. The minimum mass for stable neutron matter is about 0.1 solar masses (see my answer here) and so is much more massive than the Earth.

As neutron stars become less massive they also become less dense, so your question is unanswerable at that level too, since we cannot put a figure on the density of something that cannot even exist in principle.

If your question is merely what radius is a spherical object that has a similar  density to a typical neutron star, can you not do
$$\frac{4\pi r^3}{3} \rho = M_{\rm Earth}$$
to find $r$?


They certainly can. A ring is often formed around a celestial body when its gravity rips apart another smaller celestial body. The Sun is really massive, so it could destroy any object that is not dense enough. Just Google about the Roche Limit for more informations (and better explanations).

Now, take a look at our solar system : You have the asteroid belt between Mars and Jupiter, and you also have the Kuiper belt beyond Neptune. Would you consider these being rings ? They sure are not as smooth as Saturn's, but to my mind, they still are rings.


Local Standard Time Meridian - LSTM (Units)

In a time, the word "standard" normally means the civil time, recognized by the government in the place of interest. I'm not sure about other countries, but in general conversation in the USA, "standard" also means that daylight saving time is not in effect. "Daylight" time (as in "Eastern Daylight Time") means the civil time, recognized by the government, and daylight saving time is in effect.

These can be expressed in the ISO 8601 format; the standard gives the example "15:27:46+01:00" meaning the time zone is one hour ahead of (east of) Greenwich.

In astronomy, time can also be the local time of the observer, ignoring the existence of time zones. This is local mean time, and is obtained from Universal Time by adjusting for the longitude of the observer. Usually the exact form of Universal Time that would be used would be UT1.

Local apparent solar time is based on the actual position of the sun; it is the time given by a sundial. It can be about 15 minutes ahead of or behind local mean time.


Absurdly cheap and (potentially) very fun!
The absolute cheapest telescope to make would be extremely hard to make, you'd be doing something for the first time, but it would at least be fun!
Make your objective lens out of ice, and find a way to polish it smooth. Don't make it as big as these though. You can then use an eyepiece out of a pair of old binoculars or a toy telescope or use a dime-store jeweler's loupe for an eyepiece.
From Has anyone ever tried to make a simple telescope using ice?
These are solar collector projects, a telescope objective would have to be polished to a much better shape.

Here is a functioning camera lens made from ice for example, from this answer there:

click images for larger
Source:  Mathieu Stern
Chromatic abberation is a problem, so use a filter that removes blue and green (I guess you'd call it a red filter) and look at things that are mostly hydrogen.
From Alex Kozik's Orion Nebula spectral composition we can infer that the Orion Nebula would still have the shape of the Orion Nebula in the presence of chromatic aberration if got rid of green and blue light.

From James Kaler's1 The Spectrum of the Orion Nebula:

1Professor Emeritus of Astronomy, University of Illinois

But seriously folks
If we throw out any hope of an electronically driven and computer controlled equatorial mount and just want to look at things as they slowly drift then spend most of your budget on an f/8 spherical mirror, either a second hand one from a discarded telescope or one out of a catalog like an Edmund Scientific from the 1970's. Seriously I don't know if there are catalogs where you can order premade mirrors any more, but it was fun to dream!
You may have to invest in a telescope mirror making kit. There are an abundance of resources on how to make home-made mirrors, and as long as you are f/8 or longer you don't need to worry about figure correction (making it a parabola instead of the natural spherical shape it will want to have)
Next, put your mirror in a Dobsonian mount made from plywood or scrap metal, depending on your materials and particular skills and tastes.
You'll need a diagonal mirror or prism to send the light out the side and a stick or spider to hold it near the center of the aperture. If you are only going to use one eyepiece then you can find a simple way to hold it in place, but you've got to do some reading about how to align the whole thing.
That would be an excellent follow-up question here!
You can also take a cue from Sir William Herschel and instead of using a diagonal in the center of the aperture, just peer over one edge with the eyepiece and (since your telescope will be a lot smaller) use a diagonal mirror or prism attached to the edge. You'll have more distortion, but if you increase your f/ number (the one below was f/10) and stick with low magnification, you'll see some nebula, at least as opposed to not seeing them. :-)
I used a small front-surface mirror, some alligator clips, an eyepiece from an old pair of binoculars, a cardboard tube (a la Clyde Tombaugh) and a weak concave magnifying mirror for my first 10 cm Herschellian reflector telescope and had a truly thrilling time!
From here:

click for larger

left: Source, right: "Two Dobsonian telescopes on display at Stellafane in the early 1980s." Source


.mt files are the output files (or possibly intermediate ones as the final products in the ESO archive have .fits extensions) produced by the ESO-MIDAS data reduction system for the (in this case) FEROS echelle spectrograph. The individual echelle orders have been extracted from the original 2D CCD images, flattened, wavelength calibrated and joined together into a 1D spectrum.
The .mt files are FITS files however the fact that they are 1D can cause some issues for normal image viewers; the ESO archive has some help on how to setup various tools (including fv) for displaying 1D spectra. In this case, since the file is a standard FITS file, regular FITS tools e.g. AstroPy, IRAF, MIDAS etc will work to convert the file and the ESO archive link above has info on how to do this in Python, IDL and IRAF.
One relatively pain-free option could to use Splat-VO to do the plotting, investigation and conversion. (When loading the .mt spectrum you will need to either rename the extension to .fits or tell Splat-VO the file format is FITS as it won't be auto-recognized). Once it's loaded in, you can convert to a variey of formats, including ASCII/text.


A point of light moving between the stars

It's almost definitely not alien, if that's what you're asking.

A satellite in low Earth orbit (around 200 km altitude) would have needed to move about 60m in the time you noticed it for the movement to have been visible. So, given that satellites at this altitude generally travel at around 7km/s or 7000 metres per second, it is not inconceivable that it could have been a man-made satellite correcting its orbit.


Formulae for gravitatitional equilibrium

Mass = $\text{atomic mass} \times \text{number of atoms} = 4.4400513610370694 \times 10^{30} \mathrm{\ kg} \approx 2.3M_☉$

Herein lies the problem: your units and exponents are messed up. The mass of your iron isotope is $55.934936 \mathrm{\ u}$, and there are $7.938⋅10^{31}$ atoms. Do the math, and you'd get a total mass of $4.4400513610370694 \times 10^{33}\mathrm{\ u}$. That converts to approximately $7372878 \mathrm{\ kg}$. Nowhere near the mass of the Sun!


If you don't mind a layman's explanation, the first thing to understand is why the accretion disk is flat and that's a product of conservation of angular momentum, explained very well in this video.

If we begin with a flat, rapidly rotating (or orbiting) disk of very hot matter, too hot for electrons to stay bound to atoms, so it's both hot and electrically charged, you get gonzo magnetic fields, which (by a process I don't understand well) creates magnetic field files that extends in twists both above and below the disc, see pretty pictures:

So, as the matter from the disk tries to fall inside the black hole, if there's too much matter, it can't all squeeze inside at once because mostly, black holes are quite tiny and the pressure and magnetism shoots a lot of the matter back out, primarily through the middle of the magnetic field lines.

That's a kind of simple explanation anyway.  Correction is welcome.


Dealing with damp and dew whilst stargazing at night or in the early morning

Condensation will build up because the temperature of your optics has fallen below the dew point.

You simply have to keep the temperature your telescope above the dew point, which can be quite difficult on humid nights. Dew shields can help, as can dew heaters (which literally just gently heat your telescope and/or eyepiece).

  Hypothetical: "I'm looking at the graph of dew point vs air temperature on the link above. I don't understand how my optics can get colder than the air temperature!

At night, our source of heat (the Sun) has disappeared. The unlit side of the Earth (and everything on it) begins radiating its latent heat away into space (which has a temperature of ~3 Kelvin). Your telescope is trying to get into thermal equilibrium with outer space.


Air temperature range recommendation for observations in a summer night with an entry level telescope?

The air is only part of your problem here.  There are several factors involved.

it is absolutely true that the air can cause a loss of detail in the image you see.  But even if there's no atmosphere between you and the moon or anything else you're looking at, there is only so much detail you can see.  Technically speaking, you can magnify as much as you want ad infinitim.  But there is only so much USABLE magnification  you can get out of a telescope.  This is due to the diffraction of light.

When the light from the moon or any other object enters your telescope tube, it begins diffraction.  If you want to better understand the physics behind it, there's a fairly easy-to-undersatand Khan Academy video about it.   Essentially, what is happening is interference patterns form among the light waves and the result is a blurring of detail.  This blurring is less noticable at lower magnifications, and gets worse the more you magnify.  While the specifics are a little different from the result of blowing up a picture, the effect is similar.  I put together this example of the effect.  In my example, I started with a picture I took of the moon in May or June and each image in the series is doubled in size then cropped.  I go from a full image of a waxing gibbous moon to zoom in on the crater Herschel.  As the magnification increases, you start to see the image getting fuzzier and fuzzier.  In the case of a picture, this is because there is only so much data to work with, depending on the geometry of the image sensor in the camera. Technically, it is also the case with the light that there is a fininte amount of information/data to work with, but the mechanics are a bit different.  In the end, the result is the image blurs past a certain point and you can't get a clearer image out of it.

(Technically speaking, using speckle interferometry, a camera CAN increase detail when multiple images are captured and analyzed together, but there's still a finite limit, and this doesn't apply at all to visual observing.)

Your telescope has an aperture of 114mm, as you indicated.  I did a quick search for Seben 114mm telescopes and the only ones I found showed a focal length of 1,000mm.  As it turns out, it appears this is not a true Newtonian telescope, but a Bird-Jones Newtonian (or Jones-Bird, depending on who you ask), which is actually a type of catadioptric design and has some inherent issues, which I'll circle back around to.

They eyepieces I found listed that come with this telescope include an SR4mm and a H6mm. The SR4mm is a symmetrical Ramsden design eyepiece with a 4 mm focal length.  The H6mm is a Huygens design with a  6mm focal length.  The Huygens eyepiece design was invented in about 1662 by Christian Huygens, and was an improvement upon the original basic single-lens eyepiece design of the first Lippershey and Galilean telescopes, and Kepler's minor improvements in 1611.  Huygens added a second element and produced an eyepiece that provides better eye relief (the distance from the eyepiece to your eye, and less chromatic aberration (the separation of colors of light caused by refraction.  These eyepieces, however, have fairly narrow apparent fields of view and perform poorly in telescopes with focal ratios under f/10.  The next major upgrade in eyepiece design was Jesse Ramsden's invention in 1782.  Ramsden's eyepiece provided better apparent fields of view, but decreased eye relief and increased chromatic aberration.  When used in longer focal lengths, this is less of a problem,b ut still a problem.  The Ramsden eyepiece is also susceptible to "ghosting" which is caused by reflections within the eyepiece.  These eyepieces aer fairly inexpensive to manufacture, however, which makes them common in lower-cost telescopes such as yours.

laying aside, for the moment, the inherent issues with these eyepieces, the sizes you specified are 4 mm and 6 mm.  Based on the design types, I am going to assume an apparent field of view (AFOV) of about 40° for both.  WhenI do the calculations, I find that the 6 mm gives you a magnification of about 167X with a TFOV (true field of view) of about 14 arcminutes. The 4 mm gives you 250X magnification with a TFOV around 10 arcmin.

As I mentioned above, however, aperture limits usable magnification.  In particular, it limits the angular size of details you can resolve.  A 114 mm aperture can provide details no larger than 0.93 arcseconds in size.  At the distance to the moon (about 384,400 km) this means the smallest details you can see are about 1.73 km in size.  But this is the theoretical limitation.

In the mid 1800's, William R. Dawes produced a method of calculating how far apart two stars must be to be able to determine that they are, in fact, two separate stars. His formula, which provides a measure in arcseconds we call the Dawes' Limit, is one of the most commonly referenced ways to describe how small a detail you can see with a given telescope aperture.  For your telescope, that would be about 1.02 arcseconds.  By this measure, you could not see details any smaller than about 1.9 km in size on the moon.  A third measure, known as the Raleigh Criterion, which is also measured in arcseconds, suggests the smallest details visible in a 114 mm telescope would be about 1.21 arcseconds in size or 2.25 km at the distance of the moon.

Based on these formulae, then, if there was a particular crater on the moon that was about 2km in size, you could probably just make it out.  You could magnify this as much as you like, but you won't see more detail, just a blurrier and blurrier hint of a crater.  Eyepiece size/design won't change that, nor will your focus ability.

Depending on who you ask, for a given inch of aperture, you can only get about 25 to 60x of usable magnification - meaning that anything over that level is going to be so blurry as to not be worth viewing.  On a typical night, that would be toward the lower side, and on a night with very good air and sky conditions, toward the higher end.  Based on this, I'd say on a typical night you shouldn't expect more than about 135X magnification, and on a really nice night, maybe 225X.  Technically, the 4mm gets you much higher than that, and with the Barlow lens, double that (to around 500X) but there's absoloutely nothing you can do to improve the sharpness of the image - the limits of physics dictate that the telescope simply cannot do it.  Anything over 135X on an average night is likely to be fuzzy and get fuzzier still as you try to increase magnification.

But wait, there's more!

As I mentioned, your telescope isn't a true Newtonian.  A true Newtonian uses a mirror with a curvature defined by the formula of a parabola.  These kinds of mirrors are a bit harder to manufacture, so many manufacturers of lower-priced telescopes cheat a little and use a mirror with a spherical curvature.  The spherical curvature is a constant-radius curve, and this makes it MUCH easier to manufacture.  However, it introduces a problem known as spherical aberration  Luckilly, spherical aberration and most other aberrations are decreased by increasing focal length.  The Bird-Jones design uses a shorter-focal length spherical mirror and adds a correcting lens (almost identical to a Barlow lens) in the path of light to correct the errors. If you look into your focuser without an eyepiece, you'll probably notice a lens between your eye and the secondary mirror - this is the Bird-Jones corrector.  It adds more glass to the optical path. The more glass, the more chance for distortion and further aberration.  Additionally, this is not likely to be a very high quality optical component, making it more likely to produce a poorer image.  But worse, it makes the process of collimation very difficult indeed.

I'm guessing, though I may be wrong, you have little or no understanding of collimation.  Simply put, collimation is the proper alignment of the optical components in the telescope.  If the mirrors and lenses are not properly aligned, it introduces further errors in the path of light, which manifest themselves in image aberration and distortion.  When collimation is off, the image quality degrades and sharpness is lost.  Unfortunately, this built-in lens makes collimation more difficult, which means it's likely that your collimation is off somewhat, which will FURTHER degrade your image quality at higher magnifications.

So, what can you do?

Honeslty, not much.  While warmer air will cause issues inside the tube of the telescope, this can be reduced through acclimatization - letting the telescope come to ambient temperature.  If you can collimate your telescpe accurately, that will improve your image.  And replacing your eyepieces with better designs (even lower-price Plossl eyepieces are significantly better) will help. Even then, you're not going to get sharp detail at higher magnification.  It is simply not possible. To get sharper detail, you need a larger aperture and better air conditions.

My personal recommendation for a beginner telescope is an 8" Dobsonian. Though somewhat more expensive, the 8" aperture provides a lot more detail.  And a Dobsonian with a true Newtonian OTA is far easier to collimate and get a clearer image out of.

For now, keep using your 114, but accept its limitations.  And start looking to your next scope.

I would also recommend finding a local astronomy club you can join and learn more about how scopes work, how to maintain them, and ways to better enjoy your observing sessions.

Good luck and clear skies!


Would a spacecraft just go "through" a gas giant?

In short, No.

Side detail:
Uranus and Neptune consist likely of 20% gas and 80% rock, coming from simple density considerations. They have large inner cores with masses around $\rm 12-14 \; m_{earth} $, and something like $\rm 2-3 \; m_{earth} $ of gas on top of them.
Jupiter and Saturn are true gas giants. They both have $\rm 5-20 \; m_{earth} $ of solids in them  and the rest of the $\rm 320 \; m_{earth} $ (=Jupiter) and $\rm 95 \; m_{earth} $ (=Saturn) are gas.

However even if Jupiter and Saturn were 100% gas, you couldn't fly through them.

The main reasons for this is

Enormous density, pressure and temperature. A gas mass of several tens of Earth masses will collapse under its own gravity to enormous central densities, temperatures and thus static pressures.
The densities that this gas will be compressed to, can have densities comparable and exceeding that of water, which is a good rule-of-thumb for the center of Jupiter (typical calculations show gas densities around $\rm 20 g/cm^3$, source).
The static pressure will crush your spacecraft, the high density will near instantly slow you down to zero due to strong friction, and the high temperatures will just melt or evaporate your spacecraft.

Just because something is gaseous, doesn't mean you can soar through it, as you would on a calm day with a glider on a light breeze.

There are other factors that complicate even getting there:

Re-entry heat. The cold upper atmospheres of the gas giants increase the mach numbers $v / c_{sound}$ at re-entry by up to a factor of two. As recently discussed on space exploration the re-entry heating that a spacecraft experiences scales as the 8th power of the re-entry speed or more. So a spacecraft would experience around $2^8 = 256$ times more heat load. The engineering challenge to survive this already would be considerable.
I bring this up, because as @uhoh pointed out, it is unclear whether you want to just slice the atmosphere or fly through the $r=0$ coordinate of the planet.
Core erosion. Funny enough, the solid core is probably not even there anymore. Recent work has shown, that under the enormous pressures given at the center of a gas giant a phenomenon called 'core erosion' will take place. In this, typical rock-composing elements prefer to exist in a gaseous phase rather than in a solid phase at those high pressure, thus the solid core will dissolve into the surrounding gas.
This will further increase friction when trying to fly through the gas.


Yes. One of the features that helps this scenario is the gravitational bending of light close to the neutron star, which allows for a larger fraction of the surface to be visible at a given time.

Many pulsars have double-peaked light curves, for example here's the light curve of the Crab Pulsar. In the pencil beam model these peaks correspond to the beams from the two poles.

According to Annala & Poutanen (2010) "Constraining compactness and magnetic field geometry of X-ray pulsars from the statistics of their pulse profiles", the ratio of single-peaked to double-peaked pulsar light curves puts a most likely value for the maximum inclination of the magnetic axis relative to the rotational axis of 40°±4°.


What do the lines in this Event Horizon Telescope and Global mm-VLBI Array infographic represent?

I cant see a definitive statement, but I can see some infiormation and make some reasonable suggestions:

The two colours represent the two different collaborations. Blue for the EHT, yellow for the GMVA. We can see this by following the links on the source page of the image. The GMVA page says:

Participating Stations:
in Europe: Effelsberg (100m (Ef)), Onsala (20m (On)), Pico Veleta (30m (Pv)), Metsahovi (14m (Mh)), and Yebes (40m (Ys))
in USA: VLBA (8x25m) - BR, NL, PT, LA, FD, KP, OV, MK (HN, SC are not equipped with 3mm receivers) and the GBT (100m)
ALMA: for VLBI experiments involving phased ALMA the sensitivity depends on the number of antennas which are phased. For N=50 antennas and a typical phasing efficiency of 0.90 a gain of 1.3 K/Jy is obtained, and the effective diameter is ~ 80 m.

This matches the yellow lines.
This page describes the EHT array and more or less matches the blue lines.
So now what are the lines? I'm pretty sure they are "baselines". That is they connect pairs of telescopes whose signals are directly combined in great detail to produce information about the  image. The data from the various baselines is then combined in a later step. This wikipedia article describes some of the physics. This paper refers to the individual baselines used for the EHT and their contribution to the image.


Calculate latitude and longitude based on date and sun

What you are looking for is the navigation method used by ships and aircraft before the advent of GPS.  It requires not only an instrument for measuring the angle between the sun or a star (such as Polaris) and the horizon, but also an accurate time measurement, and charts that can be used to interpret the numbers.  And of course an accurate chronometer -- you'll need to know the time at Greenwich Obervatory, in London.

The traditional instrument used for determining this angle is the sextant.  Here is an article describing the use of a sextant for deterining one's position:

http://www.ehow.com/how_7562747_use-nautical-sextant.html

I suppose that an algorithm for doing what you are asking is available where good sextants are sold - as to the charts, well, perhaps there are downloadable charts somewhere.

Also, here is a good article on determining latitude and longitude by the stars.  Latitude is what a sextant will tell you, longitude is determined differently.  For that you need two clocks: one set to GMT; one set to your local UTC time (NOT your timezone time).  The method is described here: How to Calculate Longitude.  Which brings us to the question of how to calculate your local solar time, which is described in this article How to Calculate Solar Time.  Clearly you can't use this for navigation - it's convenient for human economics, but not finding your location.

Just to make it clear why you can't use local civil time for determining longitude is that the solar time is nearly the same on both Hawaii and Kodiak islands, but Hawaii's civil time is one hour after Kodiak's.  And China stretches some 5,026 kilometers across the East Asian landmass, which is about 4 hours of solar time, but it all has the same civil time.

This question might better be asked in the Sailing SE.  Oh, wait, there isn't one, yet.


How to calculate how much of the Earth's (or another planet's) surface is visible?

The Earth's surface which is visible when you look at the planet from a certain distance is a spherical cap in terms of geometry. Here it is, in blue:

$A$ is the position of the observer, $H$ is the distance from the observer to the surface of the sphere, $O$ is the center of the sphere,$r$ is the radius of the sphere, $AB$ is the distance to the true horizon, $\angle ABO = 90°$,  $\angle \theta$ is the angle between the rays from the center of the sphere to the apex of the cap (the pole) and the edge of the disk forming the base of the cap.
The area of the cap $A_c$ can be found according to this formula:
$$A_c=2 \pi r^2(1-\cos \theta)$$
$\cos \theta$ in the right triangle $ABO$ is the ratio of the adjacent cathetus $OB = r$ to the hypotenuse $OA = r+H$, that is, $\cos \theta = \frac{r}{r+H}$, so
$$A_c=2 \pi r^2(1-\frac{r}{r+H})$$
The percentage ratio $R_\%$ of the visible area $A_c$ to the full area of the sphere $A_s$ is $$R_\%=\frac{A_c}{A_s}\times 100\%$$
Since the full area of the sphere is $A_s = 4\pi r^2$, we have:
$$R_\%=\frac{2 \pi r^2(1-\frac{r}{r+H})}{4\pi r^2}\times 100\%$$
$2 \pi r^2$ above and below the line are canceled, so the final formula is like this:
$$\bbox[7px,border:2px solid red]{R_\%=\frac{1}{2}  \times (1-\frac{r}{r+H}) \times 100\%}$$

Answering your question, from the altitude of 3,000 mi (4,800 km) one can see $21.4842\% \approx 21.5\%$ of the Earth's surface.
I have typed it in the Google calculator here, so you can use it, just substitute the Earth radius of 6371 km and the altitude 4800 km for the numbers you would like.
Also, here is a graph of how the visible area of the Earth ($r=6371 km$) changes with the altitude, scale and drag the graph with the mouse.
As for finding the altitude from which the given percentage of the Earth's surface is visible, here it is, it is just a transformation my final formula so that $H$ is on one side of the $=$ and all the rest is on the other, only $R$ here is just a ratio, no percent, so you will put 0.38, not 38% into the formula:
$$H=2r/(\frac{1}{R} - 2)$$
The calculator says that to see 38% of the Earth's surface you need to be 20,174.83 km (12,536 mi) above the surface.


Most of the stars in the Hipparcos catalogue do not have a common name.

In the main catalogue file you will also find the Henry Draper (HD) number of the star (if it has one) at columns 391-396. You can use this ID to find the name in the Bright Star Catalogue.

The Bright Star Catalogue contains all stars brighter than magnitude 6.5 (naked eye stars) together with the HD number (columns 26-31) and a Name (columns 5-14), which in fact is the Bayer (greek letter - constellation) or Flamsteed (number - constellation) name.

Using both catalogues you can get the Bayer or Flamsteed name from the HIP identifier. I do not know of any catalogue to get the common name.

The distance of the star can of course be found in the Hipparcos catalogue, which is the reason the Hipparcos catalogue exists. Columns 80-86 give the trigonometric parallax $\varpi$ in milliarcseconds. You can get the distance from: $$d=\frac{1}{\varpi}$$ where the distance $d$ is in parsec and the parallax $\varpi$ is in arc seconds. For the Hipparcos catalogue you should divide the parallax by 1000 as the parallax is given in milliarcseconds.

You can get the constellation for any position in the sky from the algorithm and data in http://cdsarc.u-strasbg.fr/viz-bin/Cat?VI/42