"Table of Astronomy's" depiction of the solar system models

Ptolemaic system

Planets

In Ptolemy's system, the two missing symbols are Venus (♀) and Mars (♂), respectively. The arc in the bottom of the Venus symbol actually belongs to Mercury below (☿).

Stars

The two following spheres contain the stars, which were believed to be eternally fixed. The two next are, I believe, also containing stars, namely the zodiac constellations. The two last ones are the final boundary before the outside, which is the realm of God. The spheres are more discernible is this version (Petrus Apianus 1539):

Tychonic system

Planets and stars

In the Tychonic system, the two outermost rings again contain the fixed stars. The planets do not move between the spheres, but are fixed on the spheres, so the next sphere you ask about ("the third, not dotted, orbit") belongs to Mars (♂). The innermost one is Mercury, although its symbol (☿) isn't drawn in this version.

Earth's and Sun's orbits

The dotted orbit isn't Earth. In the Tychonic system, Earth is in the center, and the Sun and the Moon orbit Earth, but other planets orbit the Sun. This orbit isn't drawn in many other versions; I think it's the orbit that Earth takes around the Sun, if you considered the Sun in the center (which Tycho didn't). Although it's not exactly the same size as the Sun's sphere.

The Moon

The two stars underneath the Moon don't have anything to do with the Moon, I think, but belong to the spheres of Saturn and Jupiter. The reason I think this is that you see the same in some other versions, like this one from Tycho's De mvndi ætherei recentioribvs phænomenis (here taken from Graney (2010):

Copernican system

Saturn

Saturn's moons had not been discovered when Copernicus published his model in 1543. The drawing in your source is from 1728, at which time five moons had been discovered (Titan in 1655 by Christiaan Huygens; Tethys, Dione, Rhea and Iapetus between 1671 and 1684 by Giovanni Domenico Cassini). So since there are four rings, my guess is that they don't represent moons, but rather the rings, with the two midmost circles representing the Cassini Division, discovered in 1675.

Jupiter

Your listing of the moons of Jupiter is probably correct, since that's how they're ordered according to their distance from Jupiter.

Letters

I don't know about the  letters A, B, C, and D, but my guess is that they mark the direction in which the planets revolve around the Sun, which is correct if viewed from above the ecliptic plane, i.e. the plane with respect to which Earth has an axial tilt of some 23º.


Concerning the "Lithium test" for Brown Dwarfs

The "lithium test" for a brown dwarf involves measuring two things - the lithium content and the spectral type (a proxy for surface temperature) or luminosity. But you may also need to know (or assume) something about the age of the object. The method is useful because the tracks of low-mass stars and brown dwarfs run together very closely in the usual luminosity versus Teff plane of the Hertzsprung-Russell diagram. In addition to the considerable uncertainty in the placement of these tracks, one really needs to know the age of an object in order for its luminosity or Teff to lead to even a crude mass estimate and hence the Li-test is pursued.

Li is completely and rapidly burned in a fully convective, low-mass object, once the core temperature reaches about 3 million degrees. The time it takes for a contracting pre main sequence star (or some brown dwarfs) to reach this temperature is mass-dependent. It takes about 20 million years at 0.2 solar masses, about 120 million years at 0.075 solar masses (roughly the boundary between stars, and brown dwarfs that will never become hot enough to fuse hydrogen), and if the object is below 0.06 solar masses, it never gets rid of its Li. The physics of this process is well understood and hence it has become a very useful and trusted tool for mass and age estimation.

At the same time as this happens, the surface temperature and luminosity are also diagnostic of mass, but in an age dependent way. An M7 spectral type is on the brown dwarf limit at about 120 Myr, but a similar spectral type in an older and more luminous object would mean it was more massive.

Historically then, the Li test was applied to low-mass objects in the Pleiades cluster (see Rebolo et al. 1996), which has an age of about 120 Myr. Objects with spectral types cooler than M7 at that age should be brown dwarfs. This was confirmed by finding Li in their atmospheres.

The Li test is not a completely unambiguous indicator of whether an object of unknown age is a star or a brown dwarf. Both stars and brown dwarfs can have spectral types M6.5-L2, but a star would be older at the same spectral type. If you have measured the luminosity (or spectral type) of an object and know whether it has lithium or not, then this provides a constraint on both mass and age. The diagram below, from Basri et al. (1998), is I think as clear as it can be made.

The more-or-less diagonal curves are the tracks along which low-mass stars and brown dwarfs of the labelled mass evolve (getting cooler with age). The diagonal dashed region shows the area of the temperature versus age plot where Li is expected to be depleted. I have labelled the equivalent spectral types. The stippled (dotted) area marks where the presence of Li guarantees that you have a brown dwarf with mass $<$ 0.075 solar masses.

Thus an M6.5 object with Li could either be a brown dwarf or a star younger than about 140 Myr, but an M6.5 object without Li is a star. An M8 dwarf with Li is a brown dwarf with an age less than 200 Myr, but an M8 dwarf without Li could either be a more massive brown dwarf with age just over 200 Myr or an older low-mass star. At spectral type L0-L2 then an object with Li will be a brown dwarf if its age is less than 500-1000 Myr, but could still be a brown dwarf with mass above 0.06 solar masses with an age just a little older than this, or an even older low-mass star. Objects cooler than L2 are always brown dwarfs and always have Li - but this test may fail at much cooler temperatures (cooler than abut L8) because the Li atoms form LiCl and LiOH molecules, the atomic Li I line at 6708A disappears from the spectrum, and the molecular features are too weak/confused to be identified.

A similar diagram can be made with luminosity on the y-axis and actually, that is more secure, since Li depletion as a function of luminosity is much easier to
predict than the behaviour of effective temperature, which depends a lot on the details of the atmosphere. Of course to measure a luminosity you need the distance to the object though.

Nowadays, the presence of Li in a low mass objects atmosphere is more likely to be used as an age, rather than mass indicator. For example, if one sees Li in an M7 object, then it is likely to be younger than 120 Myr (and vice versa). The prediction of Li depletion, particularly as a function of luminosity or spectral type is thought to be more secure than predicting the position of such an object in the HR diagram. This forms the basis of the
lithium depletion boundary age determination method for clusters.

To sum up.

An object with a spectral type of M8 or cooler with Li is a substellar brown dwarf.
An object with Li but a warmer spectral type may be a brown dwarf if it is old enough.
An object younger than $\sim 150$ Myr without Li is a star.
An object cooler than about spectral type L2 is a brown dwarf.


The usual explanation of spiral arms involves density waves.
Nonuniform motion leads to matter alternately bunching up (boosting star formation) and spreading out.
Like cars on a congested road, stars move through regions of greater or lesser density over time.
The rotational smearing you anticipate is known as the "winding problem."
This Scientific American article
offers three explanations by professional astronomers.


Strange speck of light in the picture sequence of the approach to 2014 Mu69 (Ultima Thule)

It's probably a cosmic ray strike, a common artifact on New Horizons images.  The image, however, is of very poor quality which makes it difficult to analyse the exact nature of the spot.  I'll go into this more  at the end of the answer.
Possibilities:

Cosmic ray hit
A common occurance (see below); the most likely cause without evidence to the contrary.

Background star (e.g. a nova)
Novæ (and variable stars) brighten and dim over several days.  Were this a nova it would have shown in images from previous and later days.

Kuiper belt object
To not show in images from adjacent days it would need large relative motion and thus a sizable trail would be expected (none is evident).

Aliens
Possible?  Yes.  Likely?  Far, far, from it.  Not worth serious consideration.

Here's the New Horizons team's standard disclaimer:

This image contains one or more objects whose brightness exceeds the detector's saturation level. This sometimes produces a "tail" of bright and/or dark pixels to the right of the object. You may also notice a faint vertical white stripe passing through the saturated object; this is an artifact called "frame transfer smear" and is associated with the incomplete removal of signal produced when the image is transferred from the optically active region of the detector to the storage region of the detector. If the target is badly saturated, you may also notice a faint, X-shaped feature nearly centered on the object; these are optical diffraction spikes.
This image contains one or more streaks associated with cosmic rays passing through the detector. Nearly every LORRI image has at least one cosmic ray strike, but most are "single pixel" events (i.e., they only appear to be in single pixel and can easily be mistaken for stars). But sometimes a cosmic ray is energetic enough that it leaves a "trail" as it passes through the LORRI detector.

Example of cosmic ray hit and saturation trails:

Source: here (via "Learn more..." link), NASA/JHUAPL/SwRI
Note that despite being a PNG, JPEG compression artifacts can be seen in this image
The frame in question:

Source: Frame K-29 of this image taken from here, NASA/JHUAPL/SwRI/Henry Throop
Enlargement below is from frame K-29 of this image
The image in question is almost certainly a stacked composite of ~10 source images and appears to have been 2×2 binned (compare here).  To make faint objects more visible, the image was also likely contrast-enhanced.  The image also shows clear signs of lossy compression and quantization (from 12 bits per pixel to ~5), see enlargement below.  While these are (mostly) acceptable choices for the image's intended use it tends to obscure detail from the original images.
Crop showing artifacts:

Looking at the spot in question (enlargement below) a couple of things are noticable:
It is compact and relatively dim.  Its overall size in pixels is about half that of the stars and of MU69.  In addition, the drop-off in brightness moving away from its center is much more rapid than for the other objects.  This is consistent with a bright but compact object appearing in only one image in the K-29 stack.
This object also exhibits a less-blurry structure with two distinctly brighter spots or lines.  This is consistent with spray from a cosmic ray hit.
Crop showing mystery spot:

All in all, I think a cosmic ray hit is the most likely explanation.  Given the very poor quality of the image in question though it's difficult to give a definitive answer.  To make a proper determination one would need access to the yet-to-be-released raw imagery.

I've seen a paper with a more thorough discussion of New Horizons image artifiacts but don't seem to have saved a copy.  I'll add references from it when I track it down.


Which of the planets would be detected if they were exoplanets?

The distance moduli at 5, 500, 50,000 and 5,000,000 are -4.1, 5.9, 15.9 and 25.9 respectively. We add that to the absolute visual magnitude of the Sun, 4.8, to get apparent magnitudes of 0.7, 10.7, 20.7 and 30.7 respectively.

Most planets are detected by the Doppler wobble in the parent star or if the planet transits in front of the parent star. Some are detected by direct imaging, but these are planets much larger than Jupiter, orbiting at much greater distances than Jupiter (although see below).

Both the main detection techniques require relatively bright stars. The Doppler wobble technique relies on measuring small velocity shifts. These would be of the order of 10 m/s caused by Jupiter, but only 7 cm/s caused by the Earth. To claim a detection requires observing the star over two orbits (at least). Current technology would easily detect Jupiter, probably Saturn, but possibly not the Earth, if you observed for long enough. But to collect enough photons requires a star brighter than about 13-15 mag. So only your closest distances are possible.

The transiting technique has found (large) planets around stars a little fainter than this, but these tend to be in short period orbits. No Jupiter-like (in terms of size and orbit) have been founds by transits and no Earth-like ones either. You could just detect a Venus-like planet for your closest two distances. There is also the factor of orientation to consider. The probability of alignment such that a transit occurs is something like stellar radius/planet orbit radius. For Venus this about $6\times 10^{-3}$, and even more unlikely for the more distant planets.

So my conclusion is yes, you would detect planets at the first two distances using the Doppler wobble technique (providing you observed for 20 years or so). But you wouldn't see anything by looking for transits, unless you were very lucky and spotted Venus, again at your closest two distances.

More information and detail
.
If Alpha Centauri A's solar system exactly mirrored our own, what would we be able to detect?

Note added.

There are currently instruments probing Alpha Centauri for planets, by direct imaging. This corresponds to your first distance. I need to check their sensitivities...

There is also a possibility in the near future (with SKA) of detecting radio emissions from an Earth-like planet/civilisation at the first of your distances. See https://astronomy.stackexchange.com/a/10638/2531


Surface conditions of lonely (rogue) planets

Rogue planets have two formation mechanisms: Independent formation and ejection.

An independently formed rogue planet would have condensed out of the nebular material by itself and not formed from a young star's protoplanetary disk. We understand how individual stellar-mass objects condense and which have been tested by observation, but I'm not aware of any tested studies which predict rate at which Earth-massed objects independently form.  (We certainly have not observed a large population of them, and the gravitational micro-lensing studies which have been done would have detected them if they existed.)  But since Brown Dwarfs are fairly common, it seems safe to assume that there is a population of super-Jupiters that formed by direct condensation.

The planets formed by ejection should range in size from meteors right up to super-Jupiters and should be broadly similar to the size distribution of bound planets. (If small planets tend to be closer to their star, there may be some bias against them being ejected.)

Planets are mostly ejected early in a system's life, but it can happen at any point in a star's lifetime -- planetary system dynamics never becomes perfectly stable.  And if there's a near passage of another star or a large rogue planet, ejections can occur even after billions of years of stability.  (See Fritz Leiber's "A Pail of Air"!)  But the population of ejected planets is probably very heavily weighted towards planets ejected soon after formation.

The distinction between independently formed and ejected planets is important because when a planet forms it is very hot from the gravitational energy released in formation, and most likely initially inimical to life forming.  But it cools (the surface much faster than the interior) and eventually life can form.

If the planet remains in orbit around a star, the surface temperature drops asymptotically towards an equilibrium temperature where the sum of the star's radiational input and the heat leakage from the still-hot interior balances the IR radiation leaked by the planet into space. Frequently, that will be in a temperature range where life can form on the surface.

Once a planet is ejected, the equilibrium temperature will be much lower. For example, for Earth at 4.5 GY old, the Sun's radiation dumps 3000 times as much energy onto the Earth's surface than does leakage from the interior, (see https://en.wikipedia.org/wiki/Earth%27s_internal_heat_budget for details.)  The time it takes for an Earth-sized rogue planet to cool to Earth's current temperature is on the order of 10 MY and after that it would just keep on cooling down to a surface temperature on the order of 30K.

So the question of life forming on an Earth-like planet is one of whether or not there is time before the surface layers freeze solid, since that would appear to halt the formation of life.

All bets are off for larger planets which (a) cool more slowly and (b) probably have enough more water to have subsurface liquid oceans even after the surface freezes solid.

For planetary bodies where life does have time to form, the only source of energy would be the heat leaking from the interior either directly due to the heat gradient (a very diffuse source), or indirectly from the equivalent of Earth's deep sea vents.  The latter seems more likely since the steeper heat and chemical potential gradients are a lot easier to exploit.

Hal Clement wrote two scientifically excellent stories set on such planets.  One, Star Light, has high-gravity aliens working with humans to explore the surface of Dhrawn, a Brown Dwarf.  The other was short fiction and had humans meeting intelligent life living on an closer to Earth-sized rogue planet. (I think it was "Sortie" and sequels, but I'm not sure.)

In any event, it seems likely that there would be a great variety of different kinds of rogue planet that might potentially support life.


Is a three body gravitating system doomed to collapse?

Since you are talking about real bodies with some extent, yes. The case for point-particles is somewhat different.
Any dissipative system will (by definition) gradually lose energy, and that means that the positions and velocities of the constituent bodies will be confined to (hyper)surfaces in phase space with less and less energy, $E=V(\mathbf{x})+K(\mathbf{v})$. Potential energy $V(\mathbf{x})$ decreases with distance between the bodies and there is a minimum possible separation giving a minimum potential energy, and kinetic energy is always positive $K(\mathbf{v})>0$. So as $E$ declines the distances and velocities have to decline accordingly. Typically things getting close move faster, with kinetic energy increasing as potential energy decreases, but since there is a limit to how much potential energy can decrease before things collide there will eventually be a collision (or, in systems allowing it - not the case here - losing all kinetic energy and ending up in perpetual rest).
(Point particles can cheat in Newtonian gravity by performing weird and physically unrealistic collision or oscillating orbits since they can reduce their potential energy arbitrarily far, see the Painleve conjecture)
In practice dissipation effects are tiny in planetary systems, and for most planets their orbits will instead be disrupted by rare stellar encounters, chaotic resonances with other planets, or mass-loss from the star when it leaves the main sequence. Planets near stars may be dragged in because of tidal effects, but this matters most if one is a few stellar radii away. In the truly long run gravitational radiation will force any planet to spiral in, but the timescales are staggeringly long.
The number of bodies does not change things, but for many-body systems ($N\geq 3$) some bodies can gain enough energy to permanently escape while the rest merge.


How do we know that Singularity is not an object with reasonable radius?

We don't know, but general relativity implies that no force can stop the object from shrinking to infinitesimally small.  Quantum Mechanics is more uncertain.

I just want to point out that 10 solar masses would have a radius 29.53 km at the event horizon, not 1 KM.   A Neutron Star is pretty close to its event horizon already.

The problem with "some distinct size inside the event horizon" is that things change when the acceleration is greater than the speed of light.   If the acceleration is strong enough that the gluons are drawn towards the center faster than they an move (speed of light), then it's entirely reasonable that the structure wouldn't hold.   Another way of saying this is that the gravity becomes stronger than the Neutron Degeneracy Pressure and there's no known force that would prevent continued shrinkage once the Neutron Degeneracy Pressure is overcome.

The Pauli exclusion principal says that two neutrons can't occupy the same space, but this can be overcome somewhat by the weight and pressure approaching infinity.   The neutrons can get closer and closer as their momentum approaches infinity and that happens at the singularity.

So, both relativity and QM (the two best theories for dealing with this question) say that a ridiculously small object may exist at the singuarity, not meters, but maybe the size of an atomic nucleus or size of a quark or even, possibly zero size, but without a quantum field theory of gravity it's impossible to say with much confidence.

Other possibilities exist like space might get stretched into some very strange properties, where virtual particles become real but fall back towards the singularity, similar to hawking radiation but without the escape, or, the angular momentum could form into a ring or band, not a point but all that is just guesses.    The mathematics of the two main theories both say that a very tiny point is entirely reasonable and the argument for what you suggest, an object with a distinct radius of say 1 meter, is far less supported by current theories.

That's my layman's answer anyway.   I can delete this if I got too much of the physics wrong, but I thought I'd give you a quick answer.


Filters for solar eclipse and solar feature viewing

You mustn't use an eyepiece filter. It is very very dangerous. In my opinion the best is buying an Astrosolar paper, that isn't expensive and you can see very well an eclipse. If you want to see solar features you must spend more and you need an H-alpha filter, but the view with that is amazing


Are there any good DSO API's for python out there?

Given you haven't specified which kind of DSOs you want or fully given the list of properties, it's difficult to recommend a specific source.

I would suggest searching for catalogues on VizieR, which allows you to search with keywords and/or various predefined categories. Bear in mind that you may need to use several different catalogues for the various types of DSOs (galaxies, globular clusters, supernova remnants, etc.) and not all of them will have all of the data you need. In particular, distances are not present in many catalogues. In the case of galaxies, you may need to convert from redshift or recession velocity depending on the source you're using. The VizieR results indicate sky coverage of the catalogue in the oval-shaped icons, which will give you some idea of which are all-sky catalogues and which are surveys of a small region of the sky.

Once you've found a catalogue on VizieR that contains the information, you can use astroquery to retrieve the data (details of the VizieR query interface), which imports the data in a form suitable for processing using astropy.


How to calculate moon overhead and underfoot time?

At the accuracy you need, the calculation can be quite simple:  It takes 12 hours and 26 minutes for the moon to go from "overhead" to "underfoot", and 12 hours 26 minutes to go from "underfoot to overhead".

So if you notice that the moon is overhead at 20:00 one evening, it will be overhead at 20:52 the next evening,  then 21:44, 22:36 and so on.  (Or, even easier,  approximate to "50 min" and use "add an hour, minus 10 min"

The actual motion of the moon is rather complicated, as the moon's orbit is elliptical and perturbed by the sun. If you need more accurate values you need to use a complex method of calculation, which is what the websites are using.

If you only have rise and setting time, the culmination will be roughly in the middle

The Solunar calculator page contains some obvious errors (such as on 5th May)


What is it that I have been seeing in the sky?

The International Space Station can appear as bright as Venus, typically crosses the sky in about 6 minutes, and passes directly over the UK several evenings this week.
Heavens Above provides a table of ISS passes visible from a given location.
Their chart of a July 25 pass is consistent with your revised description.


What is the dielectric constant of a star?

Disclaimer
This not (yet) a full answer, since I only  focus on approaching part 1 of the question, namely the calculation of the dielectric constant of the corona.
Dielectric constant of the corona
The Sun's corona

... extends millions of kilometres into outer space and is most easily seen during a total solar eclipse [...]  measurements indicate strong ionization in the corona and a plasma temperature in excess of $1\,000\,000 {\rm K}$ much hotter than the surface of the Sun.

So let's dig a bit into plasma physics.
According to I. M. Podgornyi's article Determination of the Dielectric Constant of a Plasma, in the absence of a magnetic field,  the dielectric constant $\varepsilon$ of a plasma can be written
$$ \varepsilon(\omega) = 1 - {{4\pi {e^2}n} \over {{m_e}{\omega ^2}}} \cdot {1 \over {1 - i{{{v_{col}}} \over \omega }}}$$
where $\omega$ is the angular frequency of the electromagnetic wave, $e =  1.60217662 \cdot 10^{-19} {\rm C}$ is the charge of an electron, $m_e = 511 {\rm keV} = 9.10938356 \cdot 10^{-31} {\rm kg}$ is the mass of an electron, $i$ the imaginary unit.
The frequency of collisions $ν_{col}$ with charged particles is given by
$${v_{col}} = {{{{2.10}^{ - 5}}{n_i}Z} \over {T_e^{3/2}\left( {\rm eV} \right)}}.$$
Here, we have $T_e \approx 80\ldots90 {\rm eV} = 10^6 {\rm K}$ the temperature of a plasma given in electron volt, and (average?) $Z$ the charge number.
The next step towards an answer would be to figure out all parameters in this equation, in particular the average composition of the plasma (in terms $n$ and $Z$).
I did not yet manage that (yet), see also my question Physical properties of the stellar corona?
References
The following publications might be helpful for determining $n_i$ and $Z$ for some or all consitutents of the coronal plasma:

Mercier, Chambe: Electron density and temperature in the solar corona from multifrequency radio imaging could help us with $n_e$.

Bo Thidé: Radio Studies of Solar-Terrestrial Relationships shows a graph with the radially dependent plasma frequency of the solar corona at a given temperature.

V. G. Ledenev, V. V. Tirsky, V. M. Tomozov: High-frequency waves in the solar coronal plasma looks like saying something about $\omega$.

Related

What is the temperature of the solar atmosphere (the corona) and how is it measured?


Maximum Amplitude of a Lissajous Orbiting Object in a L4 or L5 Position

Is there a ceiling for stable L4 or L5 masses?

Your link says orbits about Lagrangian points L4 and L5 are dynamically stable  so long as the ratio of the masses of the two main objects is greater than about 25.

Earth < 4% Sun = Stable L4,L5 points

http://en.wikipedia.org/wiki/Giant_impact_hypothesis says that the Lagrangian object would destabilize once above 10% of the Earth's mass.

L4 or L5 object < 10% Earth = Stable object


What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?

From this page, we can see that it is 3.38°.

  Additionally, what is the orientation of Mercury's orbit (long axis) w.r.t. the Milky Way?

This is a bit harder to estimate precisely, but we can refer to this view from the IRAS satellite to get a rough figure of 63° between the ecliptic and the plane of the Milky Way (for more information on how to calculate that, check out Aitoff projection). Now just add/subtract Mercury's 7.01° of inclination to ecliptic and you have your result.

EDIT: Because the angles of 63 and 7 degrees may not be along the same axis, there are ways to be a bit more precise: you can take Mercury's coordinates in the ecliptic coordinates system, and transform them into galactic coordinates. If you do that with only 3 different points on Mercury's orbit, you can find the angle made by the plane defined by those 3 points and the horizontal galactic plane.

You can use this website to get your 3 sets of ecliptic coordinates for Mercury, and that website to transform them into galactic coordinates. You can transform those into regular Cartesian coordinates since the website that gave you the ecliptic coordinates also gives you the distance from Mercury to the Sun! Once you get there, it's easy to get the angle made by that plane and the horizontal: good luck!


Is it safe to watch solar eclipse's reflection in water?

If it's not safe to look at the sun's reflection in the water normally then it's no safer during an eclipse because even if 1% of the sun is showing this is more than enough light to seriously damage your eyes. The best way to view the sun, if you haven't ordered your eclipse glasses yet, is to poke a hole in a piece of paper and project an image of the sun onto another, like so:

There are instructions on how to do this here on the JPL website.


"Who saw" the binary neutron star merger first? What was the sequence of events? (GRB/GW170817)

The initial Fermi trigger can be found here, and the following sequence of alerts that were sent out by the LIGO Scientific Collaboration/Virgo Collaboration (LVC) and various electromagnetic observers following-up the event can be found in the GCN circular archive here. This doesn't quite give the whole story of the time line of events, but is a good start and the linked "Multimessenger" paper is fairly comprehensive in giving the complete story.

As shown in the figure, the signal arrived first in the gravitational wave detectors, with gamma-rays arriving at Fermi and INTEGRAL approximately 1.7 seconds after the observed merger time. However, the online analysis software on-board Fermi was quickest off the mark in detecting the burst and generating an automated trigger just 14 seconds after the signal arrived (I don't know any more of the internal details of Fermi triggering, how the alerts get sent out to people, or what subsequent manual interventions are required). Within a little over 7 minutes the online automated software searching for gravitational wave signals using compact binary coalescence templates (see, this paper and this paper) had produced a candidate using data from the LIGO Hanford detector alone (LIGO Livingston data had been automatically vetoed by the software due to the presence of a glitch [Figure 2 of this paper], and Virgo data hadn't yet propagated to the location where the analysis was run) - this automatically notified (via email or text) a number people in the LVC that something interesting had happened. Less than 10 seconds after the gravitational wave candidate had been recorded, an automated code called RAVEN (see, e.g., Section 4.1 of this paper) noted the temporal coincidence between the Fermi trigger and the gravitational wave candidate. After being notified of the gravitational wave trigger, various people within the LVC initiated a conference call and started manually looking at the data and saw the obvious chirp-like signal in time-frequency representations of the data. Roughly 33 minutes after the gravitational wave trigger, and 40 minutes after signal arrival, it was decided (people were involved) to issue an announcement (the first entry in the GCN list here) that there was a joint Fermi gamma-ray burst trigger and gravitational wave trigger.

Regarding signal arrival times at the various gravitational wave detectors: it arrived first at Virgo, followed by the LIGO Livingston detector, and finally at the LIGO Hanford detector.


How did Kepler determine the orbital period of Mars?

He knew the orbital period of Earth. You can do this by observing the sun's motion in the sky over a year. (And then you know that the orbital period of the earth is a year.)

Now from earth, you can observe Mars from one Opposition (https://en.wikipedia.org/wiki/Opposition_(planets)) to another. You can find out the time ($T_\text{relative}$ it takes for Mars to do this, using which, you find the relative angular velocity between the orbital motions of earth and Mars ($\omega_\text{relative}=2\pi/T_\text{relative}$). This angular velocity is the difference between the angular velocities of earth and mars. Using this and knowing the orbital period of earth, you can find the orbital period of mars.

$2\pi/T_\text{relative}=2\pi/T_\text{earth}-2\pi/T_\text{Mars}$

Hope this is clear.


What are the significant assumption of the Inflation Theory?

The most significant assumption is that very shortly after the Big Bang, the universe expanded very rapidly for a short period of time. Much faster than normal expansion due to Big Bang. It was an exponential expansion and occurred around ${10}^{-32}$ seconds after Big Bang and lasted for tiny fraction of second. The universe continues to expand but not at that exponential rate. Yes it is still a theory and is accepted by many scientists.


Why haven't more captured small moons been found?

Encounters between small planetesimals and the larger core of planets (especially Jupiter) transfer momentum between the two. But momentum is conserved. Momentum is:

$p = m v$, where $p$ is momentum

So, smaller planetesimals gain more velocity than a larger planet core, for the same momentum exchange. Thus, smaller planetesimals are preferentially boosted to escape the planet, either to be scattered outward to the outer system, or inward to the sun. Some will have the right velocity and aggregate with to core.

Since smaller planetesimals get a larger boost than larger ones, say 1000x rather than 20x (just an example, not real numbers), then they are much more sensitive to initial velocities. Thus a much smaller range of initial velocities will have that exactly right velocity to be captured by the planet, as either a satellite or aggregating with the planet. In the case I just made up, 1000/20 i.e. 50 times narrower. Thus, for an equivalent distribution of velocities, the smaller planetesimals in this example will be 50 times less likely to be captured.

As for dust, beyond it will eventually clump together into larger rocks due to electrostatic attraction, a lot will be sucked into the growing core, or scattered outwards. But it will exert a net drag on the core (see the Grand Tack hypothesis) and gain momentum, and through velocity, energy (a.k.a. heat). And this energy will be averaged out through elastic collisions. Eventually, the dust will either aggregate into larger bodies, sucked into the planet, or scattered into the sun or outwards from the solar system.


In general, don't think of black holes as material objects like stars or planets. We can ascribe properties like mass, charge and angular momentum to them, but if you're near a black hole and reach out to touch its event horizon, you won't feel anything (assuming you're still alive). Your hand will simply pass through empty space, and then into the black hole. Think of black holes instead as regions of spacetime, with the curious property that if you enter one, you can never exit.

I'm guessing your question is motivated by the fact that neutrinos interact extremely weakly with normal matter. They have very little mass and no electric charge, and mainly influence other particles through the weak nuclear force - which is, of course, weak, although not as weak as gravity. The problem with taking this assumption further and concluding that a neutrino might pass through a black hole unharmed is that black holes aren't "normal matter". They're sections of spacetime, and while they may contain matter, that matter is not at all important when interacting with other particles. The only thing here that is important is that they are sections of space which have only one-way openings. Once inside, nothing can leave.


Is the local group bound to the Virgo cluster?

It would seem so. The local group is part of the Virgo cluster and as such is considered to be gravitationally bound. Although the Virgo cluster and the local group are currently moving apart, the mass if the Virgo cluster will likely slow and reverse the recession over time, with the local group ultimately merging with the cluster.

References:
http://astronomy.swin.edu.au/cosmos/V/Virgo+Cluster
http://heasarc.nasa.gov/docs/cosmic/local_supercluster_info.html


Most$^\dagger$ black holes are indeed expanding, but not because the Universe is expanding. Rather, their size (more precisely their Schwarzschild radius) increases proportionally to their mass, so they grow as they accrete more matter.

It is a common misconception that everything expands along with the expanding Universe. It doesn't. On small scales, i.e. black holes, planets, stars, galaxies, and even galaxy groups, gravity ensures that these things don't increase in size. Only on large scales, i.e. galaxy clusters and beyond, does the expanding Universe pull everything apart.

As for the theory about a universe inside, I… I don't even wanna go there. I'll just refer to HDE 226868's comment.

$^\dagger$Very small black holes will decrease in size due to Hawking evaporation.


I'm 8 hours from Greenwich, why is my sidereal time 10 minutes different from theirs?

I don't think the Google Search is passing any coordinates to the Heavens Above website to get the results the Search is displaying.
If you click through on the provided link in the search, Heavens Above gives the following default location:

Location:   Bristol
(51.4770°N, 2.5690°W)
Since Bristol is 2.57° West of the Greenwich Meridian, its Local Sidereal Time is about 10 minutes earlier than Greenwich.


What is the actual visual characteristics of a black hole?

This simulation (video link) has been made to show what a black hole might look like seen from different angles.
From APOD :

If the black hole was surrounded by a swirling disk of glowing and accreting gas, then the great gravity of the black hole would deflect light emitted by the disk to make it look very unusual. [...]  Surrounding the central black hole is a thin circular image of the orbiting disk that marks the position of the photon sphere -- inside of which lies the black hole's event horizon. Toward the left, parts of the large main image of the disk appear brighter as they move toward you.


Does all the exoplanets orbit the same direction?

All planets in a planetary system should orbit in the same direction. However, this direction varies among different planetary systems. The direction that planets will orbit, for any planetary system, depends on the molecular cloud that formed them. The reason for this is conservation of angular momentum.

Before a star and its planets exist, there’s just a cloud of disorganized gas and small molecules. The Solar System formed from such a cloud around 4.6 billion years ago.

On that scale, there is some small amount of rotation within the cloud. It could be caused by the gravity of nearby stellar objects, local differences in mass as the cloud churns, or even the impact of a distant supernova. The point is, all molecular clouds have at least a little rotation. The direction depends on a number of factors, so some clouds will turn clockwise, and others will turn counterclockwise.

In a large system like a molecular cloud, each particle has some angular momentum, and it all adds together across a very wide area. That’s a lot of momentum, and it is conserved as the cloud continues to collapse under its own gravity. That angular momentum also flattens the cloud, which is the reason why the Solar System is near-planar.

When the cloud finally collapses, it forms a star and shortly after planets. However, angular momentum is always conserved. That's why all the Solar System planets follow the same orbit, and why almost all of them rotate in the same direction. There's nothing to turn them the other direction, so they will continue spinning in the same direction as the original gas cloud.

There are a few exceptions, though. Whenever objects formed in such a way that sent them orbiting the opposite direction, they usually collided with objects going in the same direction as the original cloud. This destroyed any outlying objects or sent them in the same direction as the original cloud.

Still, two huge exceptions are planets Venus and Uranus. Uranus spins on an axis of almost 90-degrees (on its side). Venus meanwhile spins the opposite direction as Earth and the other planets.

In both cases there is strong evidence that these planets were struck by large objects at some point in the distant past. The impacts were large enough to overcome the angular momentum of the bodies, and give them a different spin. There are also a range of other theories; for example, some astronomers think that Venus may have been flipped upside-down. Point is, there were irregular events that happened to both of these planets.

Overall, what I described applies for all planetary systems. To answer your question, it depends on how their cloud rotated. The cloud that formed some planetary systems may have turned counterclockwise, just like for the Solar System (at least, it appears counterclockwise from above Earth's north pole). The cloud for others may have turned clockwise. That's why not all exoplanets systems will have the same orbital direction as Earth; however, typically all planets in a planetary system will share the same orbital direction.


How does a planet lose mass to its host star?

In the case of WASP-12b, at least, the close proximity to the star has actually deformed the planet so much that it is overflowing its Roche-lobe, the area around a planet or star that . We can show this mathematically by finding the approximate Roche-lobe of the planet using:

$\frac{r_1}{A}=.46224\sqrt[3]{\frac{M_1}{M_1+M_2}}$  for $\frac{M_1}{M_2}<.8$ (which it is)
where $A$ is the orbital separation and $r_1$ is the radius of the Roche-lobe around $M_1$.
Solving for $r_1$ we get $97788\pm 2318$ miles. The planet's radius is $77759 \pm 3909$ miles. Taking the best possible scenario, there would be $\approx$ 13,000 miles between the planet's surface and its Roche-lobe if the planet was a perfect sphere. It is unknown how much the star affects the planet, but it would not be a stretch to think that the major axis of the planet would be 13,000 miles longer than the minor axis, allowing for the planet's gas to be shoved out of the Roche-lobe and "sucked" into the star.


Estimating galactic dust extinction for medium band filters

The quantity you want is basically the extinction law, and is usually called $k(\lambda)$. An extinction law is a fit to several measurements of the extinction $A_\lambda$ in some direction (or an average of several directions).

Cardelli et al. (1989) provides different functional forms for the mean extinction law, parametrized in their Eq. 1 as
$$
\frac{A_\lambda}{A_V} = a(x) + \frac{b(x)}{R_V},
$$
where $x$ is the inverse wavelength in $\mu\mathrm{m}^{-1}$, and the coefficients are given separately for IR, optical, UV, and FUV in Eqs. 2, 3, 4, and 5, respectively. The total-to-selective extinction $R_V\equiv A_V/E(B-V)$ takes different values for different lines of sight, but usually lies in the range 2.5 to 6, with 3.1 being a typical value in the Milky Way.

To get the quantity you're interested in, simply convert your favorite wavelength to $x$, stick into Eq. 1, and multiply by $R_V$:
$$
k(\lambda) \equiv \frac{A_\lambda}{E(B-V)} = \frac{A_\lambda}{A_V} R_V.
$$


How can the observable universe shrink in a Big Rip?

I've found the answer to this is actually very interesting and (to me, at least) surprising — it is not just the light cone that shrinks, but the cosmic event horizon as well! To explain this, let us first look at a de Sitter universe. The scale factor $a$ of a de Sitter universe can be expressed as a function of time with:

$$a(t) = e^{kt}$$

where $k$ is some constant, depending on the specifics of our universe (in a de Sitter universe, $k = H$, the Hubble parameter). A universe dominated by phantom energy is actually pretty similar to a de Sitter universe, but the expansion rate gets exponentially faster. Thus, in a Big Rip universe $k$ is time dependent ($\dot{k} > 0$), so the scale factor can be written as:

$$a(t) = e^{k(t)t}$$

Why is this relevant? Well, the comoving distance to the event horizon is actually inversely proportional to $a$ and $k$:

$$d_{eh}(t)\propto\frac{1}{a(t)k(t)}$$

In a de Sitter universe (and our own universe, assuming $w = -1$), $k$ is equal to the Hubble constant, so the event horizon’s distance will remain constant (in a de Sitter universe, the $\dot{H} = 0$ because the deceleration parameter $q=-1$). In a Rip universe, $k$ is time dependent; as $k$ increases, the distance to the horizon decreases.

From this formula, we can see that the event horizon’s distance is inversely proportional to the scale factor; thus, if $a(t)$ reaches infinity, the event horizon’s distance will reach zero. This means that the Big Rip can be defined as either the scale factor reaching infinity, or the observable universe’s diameter reaching zero — these two events happen at the same time.


What should I expect to see with my telescope with what EP's?

A simple rule for the maximum magnification is: 2x the aperture in mm. In your case, that makes it 260x. But that's more like a guideline, since it's affected by so many factors. Speaking in general, trying to exceed it almost never provides any benefits, with some rare exceptions (high quality instrument, perfectly collimated, under a very stable atmosphere, observing close double stars - is one possible exception that comes to mind).

So eyepieces shorter than 2.5 mm (after dividing it by the barlow magnification, if any) would likely not be beneficial for you. Combined with your 2x barlow, that means eyepieces no shorter than 5 mm (+/- 1 mm, depending on some factors). In practice, such short eyepieces are pretty rare anyway.

There is also a longest usable focal length for eyepieces, which is the shortest usable focal length times 12, more or less, or about 30 mm in your case (plus or minus some change). Any longer than that and the exit pupil (the diameter of the ray of light exiting the scope) would be greater than the pupil in your eye, leading to loss of light and diminished performance. That's the theory. In practice, there's just not much benefit going much longer than that, as you will find out if you try.

So stick to the 30 ... 2.5 mm interval, give or take 10 ... 20%. These limits are not set in stone, but you will rediscover them on your own, given enough time.

A minimum of 3 eyepieces, and a maximum of 6 or so, should cover most use cases. But a lot of people just end up accumulating eyepieces over the years, or keep trading them.

Look into eyepieces with a wider apparent field of view. A typical Plossl gives you about 50 degrees AFoV. But the view is more interesting, and the scope is easier to use, in a wider AFoV.

As an example and a place to start, I've used the Explore Scientific 82 degree series, which provide 82 degrees of AFoV. Large objects like the Moon or some nebulae or clusters look spectacular in such a wide eyepiece.

It's not mandatory that you strive for a bigger AFoV, but it can be worth it. Also, lenses such as the ES82 series could be used even with high end telescopes, if you ever decide to upgrade.

130 mm of aperture is definitely enough for a lot of fun.

Jupiter will show off its equatorial belts. On a good night you will begin seeing swirls in between the belts - I did that with 150 mm of aperture when seeing was good, and you're within a rounding error from it. You will also see the 4 big moons of Jupiter, but only as little dots. Use medium-high magnification, don't push it too high, because Jupiter is low contrast and it washes out easily at high mag - just try and see what works best.

Give it time and allow your eye to adapt. I believe your scope can track, so enable tracking, relax, breathe calmly, and just take the image in over time. In a couple minutes, your brain/eye system will begin to adapt, and you'll start to see more details. Allowing some time to pass is especially important with low contrast objects like Jupiter, where details are so close to the eye's performance limits.

Saturn will show you the Cassini division in its rings. You will also begin to see some equatorial belts on the planet, but just barely. High magnification is useful, but not necessarily the highest.

Mercury and Venus will show phases like the Moon, but no details beyond that.

Mars can be observed for a few weeks every 2 years, at oppositions; there's a big opposition this year (2018) in the summer. It requires the best seeing (the most stable atmosphere), excellent collimation, and the shortest eyepiece (highest magnification) your telescope will take. Whichever polar cap is pointing at you should be visible, as a tiny white dot sitting on the edge. I have seen the Hellas Basin as a large, bright, white spot when it was full of frost or fog (not sure which), but it melts away quickly. Syrtis Major is probably easiest to see, as a dark, almost black line.

http://www.nakedeyeplanets.com/mars-oppositions.htm

When seeing (air turbulence) is bad, Mars just doesn't work; it looks like an orange pill with no details. Either take a road trip to whichever part of the country has better seeing at that time, or use lots of patience.

Uranus and Nepture can definitely be seen in 130 mm. At high magnification, you will see them as tiny disks, clearly (but just barely) bigger than any star nearby. Uranus is a pretty blue-green shade, Neptune is a deep blue, almost indigo.

The Moon will provide endless entertainment, because there's so much to see. Zoom into a feature using high magnification, or zoom out and observe the whole thing at low magnification - it's fun either way.

For the planets and the Moon, what's important is the so-called seeing, which is the inverse of turbulence:

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

When seeing is good (turbulence is low) then you can push the magnification all the way up to the limit. When seeing is bad, you have to reduce magnification, or else the image is soft and foggy. Mars only works well with the best seeing; it is possible that seeing might be consistently bad for the whole duration of an opposition - or it could be consistently good. It's weather basically, so it does whatever it wants to do.

Use the Clear Sky Charts site to get a seeing forecast:

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

On that site, select a location closest to your home. Use the Seeing line in the chart for an estimate. Keep in mind this is weather forecast, so it's not always 100% accurate, but usually it works well enough.

When seeing is good, observe planets. When seeing is bad, observe DSOs (deep space objects, like galaxies, nebulae, star clusters).

Light pollution does not matter at all for the planets or for the Moon. You could observe these objects from the biggest city and they look about the same.

Another important thing for planets is collimation. Read your scope's manual and practice what they preach. A scope out of collimation may severely underperform. It could mean the difference between seeing and not seeing the Cassini division. See here for some examples:

http://www.astrophoto.fr/collim.html

There's a bewildering variety of tools and techniques for collimation. Start with the simple stuff (the manual) since it's better than nothing. Then keep reading. Here's a good page:

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

Or see here:

http://garyseronik.com/?s=collimation

A laser collimator should in theory be one of the easiest and most precise tools, but beware of lasers that are not centered. Plug the laser in the focuser, and turn it (spin it) in place. Watch the laser spot on the primary mirror. If the spot moves in a circle as you keep turning the laser, it's not centered. Some lasers have adjustment screws for centering, some don't. High quality lasers like the Howie Glatter series are very precisely centered from the workshop and don't require adjustments.

https://www.collimator.com/

https://www.catseyecollimation.com/

I do own more complex collimators like the CATSEYE system, but in 99% of cases I just use a Howie Glatter laser with a 2" TuBlug to do a quick check-up before each session. Any laser is fine as long as it's precision-centered.

Collimation is something you'll keep learning for a long time, so don't worry, do what you can at first, and slowly refine your technique over time. But keep your scope collimated. Once you're comfortable with the technique, it only takes 5 minutes.

Air currents in your instrument can also affect the planets and the Moon. These are due to your scope being at a different temperature from the air (because it was stored indoors). The effect is like observing under bad seeing conditions.

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

http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/

The cure is easy: take the scope outside 1 hour before you begin your session, and let it "breathe". A scope at thermal equilibrium with the air has no internal convection.

Your scope is probably not big enough to require a fan on the primary mirror, for forced convection, but you can definitely experiment with it if you like. At your aperture size this is low priority. Fix everything else first (collimation).

DSOs (deep space objects) should be visible in your aperture, especially given the dark sky you're enjoying. More aperture is better, but the level of light pollution is also important. You are very lucky to have such a great location for DSOs.

Pretty much the whole Messier catalog, over 100 objects, should look great in your scope under a dark sky. Among the easiest objects, my favorites are: M31 (the Andromeda galaxy), M13 (the Hercules Globular Cluster), M27 (the Dumbbell Nebula), M57 (the Ring Nebula), M42 (the Orion Nebula).

But since you live under an ideal dark sky, you could just plug a low magnification eyepiece in it and slowly browse the sky. When you see something interesting, stop and switch to somewhat higher magnification. Chances are you'll stumble upon many DSOs this way. I do this with binoculars (low magnification) when I go to Death Valley, and it's amazing what you can discover randomly this way.

Different DSOs require different magnifications, from low to high. You'll just have to experiment and see what works best in each case.

Seeing, collimation, thermal equilibrium don't mean a whole lot for DSOs, because they are fuzzy and low detail to begin with. But a very dark sky helps a lot.

You could also observe the Sun. Get a white light filter, the kind that fits on the top of your telescope, covering the whole aperture. Filters made from the Baader Solar Film are best, they are very good performers and tend to be cheap. Either buy the Baader foil and make the filter yourself, or purchase a complete filter.

https://www.astrozap.com/scripts/prodList.asp?idCategory=31

https://astrosolar.com/en/information/how-to/how-to-make-an-inexpensive-filter-cell/

Just make sure the size of the filter fits the diameter of the OTA (the main tube) of your telescope - NOT the aperture size, but the external diameter of the OTA.

This will allow you to see sunspots. You can also observe the transit of Mercury over the solar disk, which happens once every several years. I think the next one is in 2019. You can also observe solar eclipses this way.

If you need to use the finderscope when observing the Sun, then make a small filter for it, too. I made some cute little solar filters for my finderscopes from Baader film and white plastic tubes used for sprinkler systems. Cheap, but effective.

https://astrosolar.com/en/information/how-to/how-to-make-your-own-objective-solar-filter-for-your-camera-or-telescope/

Obviously, never point the scope at the Sun without the full aperture filter. And make sure the filter doesn't fall off. Or else all sorts of mayhem may ensue, up to and including someone losing an eye, or the house burns down, etc. Not kidding. Unfiltered, the solar energy coming out of the scope is like a plasma torch.

Buy a book called Turn Left At Orion. It will get you started with how to find and observe DSOs, complete with instructions on which time of the year is best for each object, how much magnification (this is a bit subjective), etc. It's a very good book, I highly recommend it.

https://www.amazon.com/Turn-Left-Orion-Hundreds-Telescope/dp/0521153972/?tag=amz-track-20

I would go with either a sturdy hardcover edition, or the spiral bound one - something that works well out in the field, not some snowflake paperback.

A planisphere is a good thing to have. You only use it 5 or 10 minutes each night, but it will quickly get you oriented in the sky when you do.

https://www.amazon.com/Guide-Stars-Ken-Graun/dp/1928771017/?tag=amz-track-20

When reading the book or the planisphere, use a red light at the minimum intensity that allows you to read. This will preserve your night vision. There are many red flashlights out there made for astronomers, that allow you to dial down the intensity. This is only needed for DSOs. Night vision does not matter for the planets or the Moon.

EDIT:

It's probably beneficial to get a small sky atlas too. Let's say you read in the news that a supernova has been discovered in a nearby galaxy, and it can be seen by amateurs. You could point the scope in its general direction, but you'll need a detailed local map of the region. The one point of light that is seen in the sky but is not seen on the map is the supernova. It also helps when observing other transients such as comets and asteroids (again, they stand out because they are not on the map). Finally, with some smaller DSOs it makes it easier to identify the object from the stellar background (and they should be marked on the map because they are permanent objects).

https://www.amazon.com/Sky-Telescopes-Pocket-Atlas/dp/1931559317/?tag=amz-track-20

You could also use various digital maps on a portable device, but I recommend that you master the paper maps first.

Once you own a handful of eyepieces, a barlow, etc, you should keep them together in a carrying box. Many astronomers use a pluck foam case, and they just shape the foam to hold each and every component. I've never used the dividers that come with the case, since optical components are small, lightweight, rather fragile, and should be surrounded only by foam.

https://www.harborfreight.com/18-in-x-6-in-x-13-in-black-aluminum-case-62271.html

Another item I've found useful is a tiny folding table. It doesn't have to be huge, just big enough to hold a book, the sky atlas, etc. It's better than making a pile of stuff around you on the ground.

A cheap, collapsible barstool or something like it would help with long observing sessions. Definitely beats hunching over the scope for hours. Just make sure it's the right height. Big dobsonian owners tend to make or buy so-called observing chairs, which are adjustable, but those are probably overkill for you since your eyepiece doesn't move around that much. Your scope has a geometry similar to my SCT, which doesn't really require a true observing chair, just anything to sit on that's the correct height. Now if you had a big dob then that would be different.

A pretty common improvisation is an empty box or frame that's 3 different sizes in all directions (length, width, height). That allows some height adjustment, and it's pretty cheap to procure or make. Between sessions you could use it to store the telescope or something.


The feature in the photograph ressembles a condensation trail.

Meteors usually move rather rapidly, mostly perish within a few seconds or less.
In rare cases they may leave long trails; the meteor would have been very bright (brighter than the moon) in those cases, moving much faster than an aeroplane.


Do we expect life to develop only in certain regions of the galaxy?

There's a short answer here and a long answer.

Short Answer: It isn't special.

Long Answer: Yes and No.

The Local Bubble really isn't so special with regards to where it's placed in the galaxy. Location-wise, it's perched on the inner edge of the Orion Arm, one of the Milky Way's spiral arms. The Orion Arm, too, isn't incredibly special. It's relatively small - it's often referred to as the "Orion Spur" - and near 1/4 of the way out from the galactic center. Not too far away, but not too near. The donut-shaped area 4,000-10,000 parsecs out from the galactic center is, however, sometimes called the Galactic Habitable Zone. In theory. life as we know it could exist quite nicely any where in this area (I should note that the scientific community is divided on whether the Zone is important, with some saying that it is arbitrary).

But let's dive in deeper. Actually, we're in a decent spot in the "Arm." We're on the edge of it - not too near large groups of stars that could effect our system via gravity, or near any supernova candidates. And the "Arm" itself isn't really a spiral arm. It is, as I said earlier, more of a spur, or bulge. There aren't as many stars to watch out for. Also, we're near the corotation circle - i.e., we orbit the galaxy at such a distance that we move at a similar speed to the arms, reducing further the likelihood of crossing them. Is this necessary for life? Nope. But does it help? Oh, yes.

There are, of course, other factors that make this area a nice spot: No black holes nearby, gamma-ray bursts, etc. But could life survive elsewhere? Yep. These same characteristics can be found in many other places in the galaxy - and life can probably survive in even more dangerous spots.

By the way, I'd avoid using the term "nebula". Technically, we're not in one, or anywhere near one. The term can refer to a multitude of objects - planet-forming areas, "clouds" surrounding white dwarfs, and other rather photogenic places - none of which we inhabit.

I hope this helps.


What is the exact meaning of vicinity of Black hole or a disk? the same as inside?

From the Oxford English Dictionary:

  "vicinity" just means "the area near to or surrounding a place"

so "in the vicinity of a black hole" just means "near to a black hole."

Other dictionaries say "immediately next to" so in response to your comment about vagueness, treat vicinity as the immediate neighbourhood of the black hole. You will not find an exact meaning - it is used as it is in English. There is no specific distance measurement. You don't have a distance from radius which would stop being in the vicinity.


Can you test solar eclipse glasses with a remote control?

I was considering a similar line of thinking.  However, there were a few hangups which I believe prevent yee-average-human from testing the glasses.

Many sites suggest a transmittance of 0.0032 (welding shade 12) to be the minimum to be safe (and typically recommend 13 or 14, which has even lower transmittance).  That means you're only seeing 0.3% of the light get through the goggles.  Your test would need to have enough sensitivity to pick up that tiny amount of IR light, and to not only receive a signal but be able to meaningfully measure its amplitude. You would also need to ensure that UV light is blocked as well... and all the other IR bands (I am not confident as to how wide-band protection one actually needs).

Even then, you'd have to deal with the issue of having built your own contraption to test a device which must protect your eyes from a permanent damage that you literally cannot see.  I'd want to be quite the confident electrical engineer and systems test expert to make sure that the tests I do actually test what needs to be tested.

Note: I have a related question asking how one can track down the ISO certifications and see if your particular vendor has legitimate glasses or not.  I run into the same issue: how can you trust that this device performs as indicated, because your eyesight depends on it.


How do solar winds affect the atmospheric composition and density of planets?

Do these particles contribute to the planets' atmospheres? Or do they
  do more harm than good? (define: good = contribute).

That depends on the size (mass/escape velocity) of the planet.   See atmospheric escape and pretty picture

For the 4 inner planets, the solar wind strips gas from the planets.  This process if fairly straight forward.   It has more to do with heat than traditional "blowing", and when the when the outer atmosphere gas molecules reach a certain temperature they can achieve escape velocity.   Earth mostly just loses hydrogen and helium and very little of it's heavier gases.   Venus over time has lost most of it's Methane and Water Vapor, the key difference between Earth and Venus is that Earth has a magnetic field.

Mars has lost most of it's atmosphere and what atmosphere Mars retains may be related to the thawing and re-freezing of it's icy poles, and Mercury's atmosphere is so thin that it's atmosphere actually is little more than a loosely retained collection of solar wind particles.

The 4 outer planets don't lose much to the solar wind and the Solar wind (I would think) adds to rather than takes away from their total mass, but this addition is very small compared to the mass of a planet.

  Do solar winds contain any larger atoms than helium. For instance,
  Oxygen?

Yes, but only trace amounts.  The Solar wind is 92% hydrogen and 8% helium.  Source.  Very close to the universal average ratio.   Solar particles are so hot that when they're ejected from the sun the electrons and atomic nuclei are largely separated so you get a stream of charged particles, electrons and protons being the most common and Alpha Particles (Helium Nuclei) 3rd.  This means that any planets with a magnetic field, those particles get caught in planet's magnetic field.   That's why a magnetic field is considered very important in preserving a planet's atmosphere.

Some of the captured solar wind particles would probably get deposited into the Planet's atmosphere over time but I'm not clear on that process and I don't know the percentages.

Getting back to the Sun, it has about 2% "heavy" elements in it, see article, and by "heavy" I mean, Oxygen, Carbon, Neon, others, but gravity draws the heavier elements towards the center of the sun.   Also, the helium the sun creates is largely formed near the center of the sun, so the outer parts where the solar wind gets ejected has a more standard/universal balance of about 92% hydrogen by element.

  My intuition on this issue is mixed since we have Mars with a very
  thin atmosphere, but Venus with a very thick atmosphere, neither
  having a very strong magnetosphere to protect them from solar winds

Venus mass is sufficient to make a difference.  Atmospheric escape has to do with escape velocity from the planets and Venus' escape velocity is over twice Mars'.   Source.  Venus' atmosphere is also sufficiently thick that it generates it's own magnetic protection of sorts, (I've read that, didn't see a good article to reference just now).

Jupiter has an enormous magnetic field, which casts a pretty big net, much larger than the planet itself and that captures solar wind particles, but compared to the mass of Jupiter, the captured solar particles, even with the large magnetic net, is teeny-tiny and makes no real difference in the mass of the planet.

Interestingly, when the sun approaches it's red giant stage, it's ejections will significantly increase, so it's possible that, 5 billion or so years from now the sun will feed Jupiter a stream of particles in a measurable way and Jupiter will grow larger and perhaps, visibly glow due to the particles it receives.  Jupiter might be a fun planet to watch when our sun goes red giant, perhaps from the safe distance of a colony on Saturn's moon titan.   But that's just speculation.   :-)

It will never turn into a star, it's far far far too light for that, but Jupiter could still be fun to watch (from a safe distance) 5 billion years from now.   (My answers always tend to be too long). . .

Corrections  welcome.


Einstein's general theory of relativity explains time dilation. Wikipedia provides a concise summary:

  time dilation is a difference in the elapsed time measured by two
  observers, either due to a velocity difference relative to each other,
  or by being differently situated relative to a gravitational field. As
  a result of the nature of spacetime, a clock that is moving relative
  to an observer will be measured to tick slower than a clock that is at
  rest in the observer's own frame of reference. A clock that is under
  the influence of a stronger gravitational field than an observer's
  will also be measured to tick slower than the observer's own clock.

As one experiences more gravity, time seems to flow slower compared to (say) Earth. That means that "standing" on Jupiter, which isn't possible due to the gaseous surface, will cause you to move through time faster; but you do not age slower. Suppose your lifespan is 80 years. On Jupiter you still would live 80 years, but the time that has passed on Earth during your 80 years would be slightly more than that.

The gravitational time dilation on the Moon is the opposite: the weaker gravity means that an Earth clock ticks very slightly slower than a Moon clock.


Is there a limit to how hot a star can be?

Yes, there is a limit. If the radiation pressure gradient exceeds the local density multiplied by the local gravity, then no equilibrium is possible.

Radiation pressure depends on the fourth power of temperature. Radiation pressure gradient therefore depends on the third power of temperature multiplied by the temperature gradient.

Hence for stability
$$ T^3 \frac{dT}{dr} \leq \alpha \rho g,$$
where $\rho$ is the density, $g$ is the local gravity and $\alpha$ is some collection of physical constants, including how opaque the material is to radiation. Because there must be a temperature gradient in stars (they are hotter on the inside than the outside) this effectively puts an upper limit to the temperature. It is this that sets an upper limit of around 60,000-70,000 K to the surface temperature of the most massive stars, which are dominated by radiation pressure.

In regions of higher density or higher gravity, radiation pressure is not such an issue and temperatures can be much higher. The surface temperatures of White dwarf stars (high density and gravity) can be 100,000 K, the surfaces of neutron stars may exceed a million K.

Of course stellar interiors are much denser and consequently can be much hotter. The maximum temperatures there are controlled by how quickly heat can be moved outwards by radiation or convection. The very highest temperatures of $\sim 10^{11}$ K are reached at the centres of core-collapse supernovae. Ordinarily, these temperatures are unattainable in a star because cooling by neutrinos can carry energy away highly effectively. In the final seconds of a CCSn, the density gets high enough that neutrinos become trapped and so the gravitational potential energy released by the collapse cannot escape freely - hence the high temperatures.

As to the last part of your question, yes there are astrophysical masers found in the envelopes of some evolved stars. The pumping mechanism is still debated. The brightness temperatures of such masers can be much higher than anything discussed above.


What is the overlap between the Gaia and the Hipparcos catalog?

As part of the Gaia DR2, cross-matches to several catalogues including HIP2 were included. The main reference for this is Marrese et al. (2018) "Gaia Data Release 2. Cross-match with external catalogues: algorithms and results". The resultant matches are included in the tables gaiadr2.hipparcos2_best_neighbour and gaiadr2.hipparcos2_neighbourhood available via the query facilities in the Gaia Archive.

Unfortunately the match is not particularly complete and it is not entirely clear why. From the paper:

  While we expect to find Gaia counterparts for most of the Hipparcos2 sources, with the exception of the brightest ones, the
  cross-match results include only about two-thirds of them. This
  means that according to the adopted cross-match algorithm, only
  about two-thirds of the Hipparcos2 objects have a Gaia counterpart that is compatible within the position errors (i.e. have at least
  one good neighbour). Hence the Hipparcos2 cross-match results
  clearly show an issue that needs to be investigated.

As a result, they also provide a table of cross-matches based on a 1-arcsecond cone search, available on the Gaia DR2 known issues page. This cross-match is a lot more complete, but there are still a few objects that are present in both catalogues that are missing. Several of these are nevertheless present in SIMBAD, e.g. 85 Pegasi = HIP 171 = Gaia DR2 2853900872207150720.


Your question is very unclear.  As long as the orbit is bound (i.e. the total energy is negative), then there is a unique closed orbit solution (see Bertrand's theorem).  There will be no 'transient' part of the solution.


A double star consists of two stars of equal brightness at equal distance from the Earth. What is the apparent magnitude of the combined object?

The magnitude formula is $$m = -2.5 \log_{10} \left(\frac {F}{F_{0}}\right)$$
Where $F_0$ is the flux of magnitude 0 star.
So if you have two stars of equal brightness, the flux is doubled, and you get
$$m = -2.5 \log_{10} \left(\frac {2F}{F_{0}}\right)\\
= -2.5\log_{10}(2) -2.5 \log_{10} \left(\frac {F}{F_{0}}\right)$$
The first value is about -0.75, so the double star is 0.75 magnitudes brighter.  If each object is 8.34 then the double star appears as a 7.59 object.


If we count Avogadro's Number of stars that are closest to Earth, how big that space would be?

It's going to be all (or perhaps nearly all) of the observable Universe. Roughly speaking, there are several hundred billion stars in the Milky Way. And extrapolating the number of galaxies in deep Hubble images suggests something like a hundred billion galaxies in the observable Universe.  Put these together and you get about $10^{22}$ to $10^{24}$ stars in the observable Universe.  Avogadro's number is $6.022\times 10^{23}$, which is in the upper end of that range.  So, given current estimates, you'd need at least most, if not all, of the observable Universe to get $N_A$ stars.


In layman's terms, why do we put the "time" in space-time?

To answer your point about why "spacetime" as opposed to space and then time - photons (and other massless particles) travel through spacetime on "null geodesics" - in other words they do not experience time at all. But they are not in some separate class as particles at approach ever closer to the speed of light travel on paths through spacetime that are ever closer to null geodesics.

In other words the four dimensional geometry is inextricably linked and "space" and "time" are a unified whole.

In theory humans moving also experience this effect but the speed at which we move is far too slow for this ever to manifest itself to our senses. But we do see its impact in high precision timing - such as with GPS satellites.


What would a black hole merger look like?

I'm pretty far from an expert, but there are a lot of moving parts to this question and it's not easily answered, partially because we don't have a good definition of precisely what a black hole is.   Special relativity tells us that it's a point singularity surrounded by an event horizon from which noting can escape.  Quantum physics tells us that we don't know what it is, but hawking radiation can escape the event horizon and information is conserved on the outside . . . er, I think.   A recent gravity wave discovery (see special relativity), indicates that they lose a pretty fair portion of mass when they merge.

  To ship 1 it would appear that ship 2 would be moving slower to the
  point it would seem like ship 2 stopped moving the closer to the black
  hole ship 2 gets.

This is the common explanation, but I'd like to add some detail to it.   This only works if the ship falls direct into the black hole with zero tangential velocity.

If we imagine a ship that's both reflecting visible light and radiating infrared light as it falls into a black hole, it speeds up very fast as it falls into the black hole, until the time dilation takes over and it appears to slow down, at which point it's very close to the black hole.  It's velocity is a significant fraction of the speed of light - somebody else can do the math if they like, but If memory from another question is correct, about 1/3rd the speed of light when the slowing down takes over - to the observer.  To the ship it continues to accelerate.

The ship continues to red-shift however, because the light that escapes the ship is red shifted both by the SHIP's increasing velocity combined with the gravitational red shift, so as a visual, you'd lose sight of the ship pretty quickly.

Something else happens because of a quirk with relativistic gravity.   An object reflects and emits light in all directions, but as it gets close enough to the black hole, the only light that can escape is almost entirely perpendicular.   Light can only escape the photonsphere of a black hole if it's close to perpendicular, and maybe my wording could use improvement there.  This question has a very nice answer on orbital velocity around a black hole.  Tangential light gets considerably bent or trapped entirely.  What this means is you pretty quickly will lose sight of anything that isn't directly perpendicular between the ship, the center of the black hole and where you are viewing from.   To actually continue to see the ship you'd need numerous cameras spread around the black hole and the view would need to be reconstructed.

And even if not for the photonsphere, the fact that parts of the ship would be closer to the event horizon than other parts, and that light would travel from the ship to your eyes at slightly different directions, you'd likely see some distortion, similar to the high velocity simulations you might have seen on relativity tv shows, or this website gives some examples, or when Randy Johnson throws a baseball.

And lets not forget, any object would likely break apart or spaghettify due to tidal forces, unless you had a really big black hole where spagetification wouldn't be an issue and the warping would be less but the photonsphere would still be a problem.

My point is, the story of the ship remaining visible is a nice simplification, in reality it's oversimplified.  It ignores tangential velocity and it's frequently not mentioned that it would both spread out and quickly become so red shifted that it couldn't be recognized.     The particle nature of light would suggest that in not too much time, the ship would be reduced to individual photons flickering and you'd need an enormous number of telescopes (assuming you could build a telescope to capture light that far red shifted), and an enormous amount of time to capture enough photons, to even track the ship at all.   In practical terms it disappears pretty quickly.   You might see it slow down and red shift and stretch for a while, but not very long.

And if the ship falls into the black hole with tangential velocity, well, that's even worse for keeping a visual track on it, and almost everything in space that falls into something else, has tangential velocity.  The idea that the ship appears to freeze kind of in place and fade is probably not accurate at all but if any experts here want to correct me on that, I invite correction.

I wanted to point out this complexity before giving a layman's answer to your question.

The best way I can personally imagine what it would look like when two black holes fall into each other is to think of the photons outside the event horizon and what happens to them.

This is probably a pretty good estimate of what it looks like from ship 2.    I think images like this are computer generated and not bad.  One thing I disagree with is that these images are always drawn with a kind of ring around the black hole.

The black circle that you see is actually the photonshpere, not the event horizon and it's not completely black because some light with a tangential velocity can escape the photonsphere, but it's probably pretty dark, unless something very energetic is going on inside.

Also, those black holes above are drawn with a kind of light-circle around them.  That's probably not accurate either.   The entire space is warped, so there's no light circle.   We imagine seeing a circle around an object that blocks a light because light can wrap around an object and we think of things as having edges, but that's not accurate.   There would be warping around the photonsphere, extending outwards, but not a circle of light.  I suspect there's a good chance you wouldn't see one black hole pass infront of the other, you'd just see it disappear, like black on black, then reappear as it warped space behind it when it was no longer infront of the other.

I don't object to the liberties taken to make videos like that more visible, but if you're asking what it would look like, I felt those corrections were worth pointing out.

Onto your questions:

  because of the time dilatation is this gif picture below a good
  rendering of how the merger would appear to ship 1?

(give posted above as well).

It's an OK rendering to give amateurs an idea of what it would look like.  Time dilation is largely irrelevant because the black holes would orbit each other very very fast, unless they were very large black holes.   That's another difficulty with your question.  The answer changes somewhat with 2 supermassive black holes, like is expected to happen when Andromeda and the Milky Way's black holes are expected to merge, and with stellar mass black holes.

Stellar mass black holes would orbit each other very quickly and the time dilation would slow it down a little bit, but not enough where it would appear slow.   The interesting stuff happens when they get very close.  Take the collision that was detected by Ligo, when two roughly 30 solar mass black holes merged.    A 30 solar mass black hole has a radius of about 90 km and if the two objects are orbiting each other close enough where you'd want to park a space ship and watch them collide, figure about 5 or 10 times that distance, say 900 km apart.   Objects travelling at relativistic speeds around a (3.14 x 2 x 900), 5,652 km track, at relativistic speeds, you're looking at several orbits per second, give or take, that would likely speed up as they got closer.   Time dilation doesn't catch up fast enough.   You might get some red shifted slowed down bits and pieces, but like the space ship, the idea of watching the black holes kind of freeze on each other is probably not what it would actually look like.

A few other things to keep in mind.   As the black holes spiral towards each other, they lose mass, so their event horizons actually shrink, which would be hard to observe until the very last moment where most of the gravitational wave energy is lost.

And, the black part is actually the photonsphere, not the event horizon.   It could look to the viewer that the black holes are touching when it's just their photonspheres shaking hands, so to speak.   If it's a small black hole and a bigger one, the small one could actually disappear into the larger one's photonsphere before the merger actually happens and the enormous pulse of gravitational wave energy is released.

So, the video probably isn't a very accurate representation of what it looks like, even allowing for the slow-down, but it's not a bad one either.   I don't think it's meant to be accurate as much as it's meant to try to show what happens. It's like when they give Jupiter all these colors that it doesn't really have.  It helps people like us "see" Jupiter, but an actual picture would be less colorful.   It's OK to sacrifice some accuracy in order to make something more visual.

  I imagine that 2 naked black hole(an infinity dense point with out a
  surrounding black hole) merging would look like a blur from the speed
  of them,

OK, here's where I'm in over my head, cause this is hard.   I imagine that you're doing a thought experiment, not actually proposing "naked black holes", that is, if you could see the two singularities spiral towards each other, what would it look like.

First thing is, you can't see a singularity.   Imagine you want to see one, so you sail your ship into a black hole.  Sure it's a suicide mission but you will be the first person to see a "naked singularity" - WRONG.   No light can travel from the theoretical point singularity outwards, so as you're inside the event horizon, not much changes.  You see stuff that fell in with you, with different variations of tangential velocity.  You see things stretched and warped by the curvature of space but you can't see the singularity because it doesn't send any photons in your direction.   You could maybe observe some hawking radiation but even that is iffy, because unlike the border of the event horizon where one particle can escape and one can leave, inside, both particles would be drawn to the singularity and probably annihilate each other like good virtual particles are supposed to do.   You'd see space, star light behind you, and stuff around you but you can never see the singularities (assuming that's what happens anyway, we don't have a good answer for what happens inside a black hole).

While you couldn't see it, you could model it.  Two theoretical point singularities would, in a decaying orbit, spiral into each other.   The gravitational waves released inside the black hole would be enormous, but, somehow, all that energy inside the black hole would likely remain trapped inside the black hole.   The gravitational wave energy that's lost probably all happens before the theoretical singularity crosses the other black hole's event horizon, or before the event horizons touch and merge, so to speak.

But if you imagine there are two point singularities in this scenario, I see no reason why they wouldn't just very quickly spiral into each other.  To get a better picture of this, we'd need an accepted model for quantum gravity, which we don't have.  There are also models of black holes where the matter accrues on the outer edge of the event horizon, where time stops and there is no central singularity.

So the real answer to your question is that nobody knows what it would happen, but if we imagine two point singularities when two black holes merge, the laws of special relativity suggest that they would fly into each other very very fast, perhaps a tiny fraction of a second.  Gravitational energy would be emitted as they spiral into each other, as defined by special relativity, but I don't imagine that energy would escape, it would remain inside the event horizon, but on that, I'm just speculating.   There's many reasons why this question probably can't be neatly answered.

There's also a kerr black hole possibility where the point singularity is more of a rotating hoop or ring in order to retain it's angular momentum, but I've always had a hard time seeing how that meshes with losing energy via gravitational waves . . . but I'm not going to try to explain that one, cause it's above my pay grade.

Apologies if I waxed poetic a little too much, but I think this question almost works as a thought experiment, though it's probably impossible to answer.


The Milky Way Galaxy is a large spiral galaxy with some characteristic features worth mentioning:

1) The bulge - This refers to the collection of tightly packed stars located in the central region of the galaxy.

2) The spiral arms or the disk - This region extends from the inner region of the galaxy (where it meets the bulge) to outskirts of the galaxy, and contains stars, dust, and lots of gas. It's also very thin. The reason why a disk forms is largely thought to be due to slight asymmetries in the accretion of material early on in the galaxy's history. Over time, and due to conservation of angular momentum, the material collapses down into a disk. Incidentally, this is where the sun lives.

That ribbon is the disk of the Milky Way. It looks cloudy because of the dust and gas which scatters light from the rest of the galaxy. The reason why gas accumulates here is because as things begin to collapse into a disk, the gas collides with itself, and sort of 'sticks' together. In other words, where a bunch of stars may very happily pass through a bunch of other stars without many of them colliding or being disturbed, gas has a much harder time doing this. If you want to learn more about what happens when galaxies and clusters collide, take a look at the Bullet Cluster.

The majority of the other stars you you see which are not part of that structure are much more local. The reason why the rest of the sky looks rather transparent in comparison is because you are looking out of the plane of the galaxy. There is simply much less stuff (namely gas and dust which obscures your sight in optical wavelengths) that you are looking through. Because of this, most optical telescopes look in these directions when studying the rest of the universe.

For more about the classification and structures of galaxies, see the Hubble sequence.


Have all Wolf-Rayet stars evolved off the main-sequence?

Wolf-Rayet stars have long been a subject of controversy in astronomy. Observationally, they are the class of stars that astronomers identify as being very luminous, having very little hydrogen, and are mostly composed of helium, carbon, nitrogen, and oxygen. This class of stars remained an enigma until the 1980s, when Wolf-Rayet stars started to be considered as the end result of high-mass stellar evolution (due to better modeling and observations, specifically improvements in spectroscopy and better telescopes). Nowadays, WR stars are suspected as the likely stellar progenitor to black holes and high mass neutron stars (depending on the mass and metallicity, i.e. see Heger et al. 2003).

Does this mean that these stars already have burned their hydrogen and have evolved off the main sequence, or are these stars still on the main sequence but are classified as WN5h as a result of its high surface temperature and luminosity?

Regardless of whether you interpret the Wolf-Rayet phenomena as the end result of high-mass stellar evolution, a Wolf-Rayet star is off the main sequence because it is a type of core-helium burning star: it exhausted the hydrogen in its core and thus is not on the main sequence.
Wind mass-loss of Wolf-Rayet stars is strong for their relatively small size. That is, the mass loss rate is actually comparable to that of main sequence O-type stars (the likely stellar progenitor of Wolf-Rayet stars), however the Wolf-Rayet star has smaller radius and smaller radius of gyration than main sequence stars so the comparison becomes complicated. Such mass-loss rates depend on the star's luminosity (and thus mass), as well as the metallicity. The star WR R136a is located in the Tarantula nebula, which is not a high-metallicity environment, meaning the mass-loss rate for this star is not particularly high, which is consistent with it still having such a large mass. Recent analysis of this nebula has actually revealed that the existence of such very high-mass stars might not be as rare a once thought, implying there may be many stars similar to R136a in the cosmos.


GW from merging of neutron stars and black holes

I don't think we have a clear answer yet.  LIGO hoped to detect black hole (BH) mergers and expected to detect neutron star (NS) mergers, yet the BH mergers are observed more frequently, partly because they are louder and thus LIGO can see them from further away and partly because the 20-40 solar mass BHs LIGO observed are more common than we expected.

BHs this size can be formed by the supernova implosion of a massive star, but BHs formed this way should be quite rare  They can also be formed by the merger of two smaller BHs -- which is exactly what LIGO was observing!  It seems likely that these larger (but still stellar mass) BHs were produced by previous mergers.

So in the end, the frequency of LIGO detections depends on two things: The real distribution of objects in binaries: NS-NS, NS-BH, and BH-BH and the loudness/brightness of the gravitational waves emitted by the merger.

In general, black holes form from more massive stars than neutron stars, and that means rarer stars. (The moreso since the more massive a star is, the quicker it sheds mass through stellar winds.)

So NS-NS systems come from lesser-massive stellar pairs.  NS-BH systems form the same way, but with one of the original stars being more massive than would lead to a NS.  On the other hand, the BH-BH systems we're detecting probably come from BHs that themselves were the result of mergers in very dense stellar environments.

Consequently, we wouldn't expect that NS-BH mergers would be especially correlated with BH-BH mergers, since they precursor systems are formed through different mechanisms.  But we would expect NS-BH mergers to be rarer than NS-NS mergers since the precursor systems are rarer.

There's a bit of handwaving in that!  I'll try to find some papers on the subject -- there must be a bunch.


Can a tidally locked planet have their own habitable zone?

As you suggest, it might be possible for a habitable corridor to exist along the stationary terminator. But there are ideas around for more than that. The planet might be exposed to tidal forces the volcanism of which warms the far side. It might have a thick stormy ocean or atmosphere which evens out the surface temperature (All but the smallest planets have atmospheres). It might have habitable moons which when tidally locked to the planet rotate regularly versus the star. And if it, like Mercury, has an eccentric enough orbit it might rotate relative to its star although it is tidally locked. Planetary diversity is huge.


How many times can iron be detected for the first time in an exoplanetary atmosphere?

Neither of these, interestingly enough, is the first time iron has been detected in an exoplanetary atmosphere. Other groups (Hoeijmakers et al. 2018, cited by both papers) have detected absorption lines of iron and other metals in the day-night transition zone of exoplanets, during transit. Ehrenreich et al. 2020, the new ESO paper, did something similar.

Now, Pino et al. 2020 are referred to in the spaceref.com article as having "directly demonstrated" the presence of iron in an exoplanet atmosphere for the first time. What I believe "directly" in this case refers to is the detection of iron emission lines, rather than iron absorption lines. This is what makes Pino et al.'s observations different: they observed KELT-9b shortly before it passed behind the star, allowing them to view the hot, day side of the planet and thus iron emission.

In other words, I think the spaceref.com article is playing fast and loose with its wording - Pino et al.'s results don't constitute the first time iron has been detected in an exoplanet atmosphere, just the first time iron emission lines have been detected. They got around that by characterizing emission lines as "direct"$^{\dagger}$, implying absorption lines are "indirect". Whether you agree with that is up to you, I think.

If you had to write an article labeling one study as the first detection of iron in an exoplanetary atmosphere, you could make a good argument in favor of Hoeijmakers et al., but I think the better option might be to simply not try to describe any of these in that manner. Some nuance here wouldn't be a bad thing.

$^{\dagger}$Hoeijmakers et al. refer to their observations as the first "direct detection" of iron, adding a layer of complexity and irony.


How to calculate geocentric conjunction time and moon altitude at a given time

I assume you have an algorithm to compute position of the Sun, since you can compute rise and set times.  You will need to compute the position of the Moon in a similar manner.  Then, convert both positions to geocentric ecliptical coordinates (ecliptic latitude and longitude) and use some numerical bisection method (root finder) to find where the difference of the longitudes of the two objects has a root (i.e., becomes zero).  The corresponding JD provides the moment of the geocentric conjunction.
Calculation of the altitude is a standard procedure - you'll need to convert your ecliptic coordinates to equatorial (right ascension and declination) and then to horizontal (azimuth and altitude).  Finally, correct for the horizontal parallax to convert your geocentric coordinates to topocentric coordinates for the location of the observer.


What constellation is closest to the Zenith with RA=12 and Dec =47.6?

With Stellarium, it appears that the point with coordinates RA=12, DEC=47.6 is in Ursa Major.

In the screenshot from Stellarium below, the constellation boundaries are in red. Ursa Major is at the bottom, Canis is on top. The blue grid is the equatorial coordinates.


what does it mean of different eyepieces?

The 4, 24 and 30mm eyepieces give different magnification (equal to the the focal length of the telescope in mm, divided by the number for the eyepiece).
You can use eyepieces from other brands, but make sure they have the same size barrel (the bit that slides into the telescope).  I don't know the Gskyer brand, but the eyepiece size will be either .965 inch or 1.25 inch.  Provided you get the same size, any other brand will work.  It's probably best to get familiar with what is provided by Gskyer before buying anything else.

You will likely find that using lower magnification at first gives the best outcome.