How do OGLE-III and GAIA measure the mass of free microlensing black holes? What is the "hypothesized lower mass gap" between 2.5 and 5 solar masses? eventually links to Constraining the masses of microlensing black holes and the mass gap with Gaia DR2.
The angle of deflection of light passing a massive object is given by:
$$\theta = \frac{4GM}{r c^2}$$
where $r$ is the minimum distance from the mass that the light passes.
If two black holes pass by a line-of-sight to a distant object and their speeds and distances of closest approach $r$ both scale linearly with their mass, they produce the identical deflection magnitude and time dependence.
Question: How then can such an observation of a free-floating black hole be used to determine its mass? What additional information is necessary? I see from the article that parallax is involved, but without knowing the distance to the black hole I don't see how this is enough to work out a mass.

The introduction of the Wyrzykowski & Mandel paper gives the following information about estimating the lens mass.

In order to obtain the mass of the lens (Gould 2000a), it is
necessary to measure both the angular Einstein radius of the lens
($\theta_\mathrm{E}$) and the microlensing parallax ($\pi_\mathrm{E}$)
$$M = \frac{\theta_\mathrm{E}}{\kappa \pi_\mathrm{E}}$$
where $\kappa = 4G / (c^2\ \mathrm{AU}) = 8.144\ \mathrm{mas/M_\odot}$; and $\pi_\mathrm{E}$ is the length
of the parallax vector $\mathbf{\pi_\mathrm{E}}$, defined as $\pi_\mathrm{rel}/\theta_\mathrm{E}$, where $\pi_\mathrm{rel}$
is relative parallax of the lens and the source. The microlensing parallax vector $\mathbf{\pi_\mathrm{E}}$ is measurable from the non-linear motion
of the observer along the Earth’s orbital plane around the Sun.
The effect of microlensing parallax often causes subtle deviations and asymmetries relative to the standard Paczynski light
curve in microlensing events lasting a few months or more, so
that the Earth’s orbital motion cannot be neglected. The parameter $\mathbf{\pi_\mathrm{E}}$ can also be obtained from simultaneous observations of
the event from the ground and from a space observatory located
∼1 AU away (e.g., Spitzer or Kepler, e.g., Udalski et al. 2015b,
Calchi Novati et al. 2015, Zhu et al. 2017).

In particular, the Gould 2000a paper gives a good summary of the various relationships between the quantities. The Udalski et al. 2015b notes that the distance between the Earth and Spitzer (which would also apply to Gaia) means that Spitzer would see differences in the light curve, allowing the parallax to be determined.
Note that things get more complicated if the source is a binary, in which case a "reverse parallax" effect from the source's orbital motion, usually called "xallarap" needs to be taken into account — but that's a matter for another question...
The other relevant quantity is the angular Einstein radius of the lens. In their discussion of measuring $\theta_\mathrm{E}$, Wyrzykowski & Mandel reference Rybicki et al. 2018. That paper notes that precision astrometry can help measure $\theta_\mathrm{E}$ because microlensing also changes the apparent position of the source:

The positional change of the centroid depends on the
$\theta_\mathrm{E}$ and separation $u$. Contrary to the photometric case, the
maximum shift occurs at $u_0 = \sqrt{2}$ and reads (Dominik & Sahu 2000)
$$\delta_\mathrm{max} = \frac{\sqrt{2}}{4} \theta_\mathrm{E} \approx 0.354 \theta_\mathrm{E}$$
Thus, for the relatively nearby lens at $D_l = 4\ \mathrm{kpc}$, source
in the bulge $D_s = 8\ \mathrm{kpc}$ and lensing by a stellar BH with
the mass $M = 4M_\odot$, the astrometric shift due to microlensing will be about 0.7 milliarcsecond.

The bulk of the paper goes on to determine that these shifts should be observable by Gaia.
Another way to measure the size of the lens is to measure the lens-source proper motion by searching for the lens several years after the event, this has been done for a couple of exoplanet-hosting lenses but would not be possible for a dark lens like a black hole.


How did the orbits of Phobos and Deimos circularize? The satellites of Mars might be captured asteroids. But Phobos has an eccentricity of $0.015$, and Deimos of $3\times10^{-4}$. If they were captured, they likely were captured with rather eccentric orbits. How did their orbits circularize?

Astronomers are not still completely sure that these satellites are captured asteroids (see for example this paper). However, in case they are, it's true that their eccentricity should be much higher. There are several theories that provide scenarios for the circularization of the orbits (the most recent I found is this one, which invokes impacts on the moons causing them to start tumbling and exciting the spin), but more data is needed to test these hypothesis (see this last paper for a review of this).


Accretion discs emitting jets When there are accretion discs jets of particles or radiations will be emitted in the direction parallel to the rotational axis.

My beginner level astronomy textbook just says that it is due to magnetic field but does not explain why.

So can anyone explain the mechanism in layman's terms? Thanks in advance!

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

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

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

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


Why is the interstellar medium so hot? On this link, it states the following: "large assemblies of galaxies that are permeated by even larger amounts of diffuse gas. With temperatures of 10 million degrees or more".

How are these diffused (ionized) gas able to become so hot when they are in large distances from one another and have very little density?

What that ESA (European Space Agency) page titled Hot gas sloshing in a galactic cauldron that you link to describes are called WHIM (Warm–Hot Intergalactic Medium). They are not interstellar medium, but intergalactic medium gas. The difference in density is huge, with interstellar medium density at an average of $\rho ∼ 1\ ppcm$ (one proton per cubic centimeter), but the density of these WHIM being even a few orders of magnitude lower at $\rho ∼ 10^{−6}−10^{−5}\ ppcm$, or roughly 1 to 10 protons per cubic meter (NASA's Chandra X-ray Observatory quotes average density of 6 protons per cubic meter).

What is interesting about WHIM is that they are absolutely huge. We're talking of distances that extend across clusters of galaxies (so stretching multiple millions of light years), which means that even as tenuous as they are, account for a large portion of the baryonic matter of the Universe:

  Such matter is predicted to account for a sizable fraction ($∼ 50\%$)
  of all the baryons in the local        ($z < 1$ [redshift in the infrared
  spectrum]) Universe, and it is thus considered the best candidate to
  host the baryons seen at high redshift and missing from the low
  redshift census.

So now about their heat emissions, and why are they detected in the X-ray range in the first place (the ESA's article mentions the photograph featured there was taken by the ESA’s XMM-Newton X-ray observatory):

  Electrons and baryons in the WHIM are shock-heated during their infall
  in the dark matter LSS [Large–Scale Structures] potential well, and settle in
  filamentary/sheet-like structures surrounding LSSs.

I've added a few clarifications in quotes encapsulated in square brackets, but what this means is that parts of these WHIM interact with AGN (Active Galactic Nucleus) as the galaxies pass by, and the X-ray emissions of AGN excite Baryonic matter to a temperature $T ∼ 10^5−10^7 K$.

Quote sources:

The Cosmic Baryon Budget, Fukugita et al. 1998
The Cosmic Energy Inventory, Fukugita & Peebles 2004

Additional reading:

X-ray Discovery Points to Location of Missing Matter, NASA
Last “Missing” Normal Matter Is Found, Sky & Telescope
Studying the WHIM Content of the Galaxy Large–Scale Structures along the Line of Sight to H 2356-309 (PDF), L. Zappacosta et al., Harvard-Smithsonian Center for Astrophysics


Is this a coincidence? The smallest distance probed in any experiment so far conducted is $10^{-18}$m and the largest distance we can have is the radius of the universe, nearly $10^{26}$m. The ratio is $10^{44}$.

In another case, if we think about time, the shortest time studied is about $10^{-26}$ sec and the longest time is "the lifetime of the universe", which is estimated to be $13.7$ billion years, or nearly $10^{18}$ sec. The ratio is again $10^{44}$!

Is this(the ratio belongs is of same order) a coincidence or there is some deeper explanation(probably space and time are related etc.)? I am an undergrad student, don't know much about astronomy. I wondering about this question while studying quantum mechanics from the book The Quantum World, which claims that it is not a coincidence.

The book explains it in the following way -

"This is not a coincidence. The outermost regions of the universe are moving at a speed near the speed of light, and the particles flying about in the subatomic world are also moving at such speed. On both subatomic and cosmological frontiers, the speed of light is the natural link between distance and time measurements."

I haven't posted exactly what the book is saying because I wanted to hear what other people are saying about this and also because I haven't understood what the book is saying.

It's indeed a coincidence. The human limits on observables cannot have some deeper explanation.
In fact, we can claim many things like this, without having some sort of deep meaning.
The other thing is they are not "exactly" equal. Their ratios just happens to be in the same order. There are also uncertainties in all these measurements and the ratios are never equal to each other. If they were exactly equal in some sense then we may thought there is some deeper meaning.


Does anyone have spectra on exoplanet HD 219134b? How could one know either way? I have foraged through the internet to no avail.
Is there a way to systematically check if spectra from a given exoplanet has been reported, like a catalog or list of references, or is a literature search using likely keywords the only way to look into this? Thanks!

For an authoritative source of data on exoplanets, your best bet is the NASA Exoplanet Archive.  Their page for HD 219134b is here.
For planet spectra, the archive has a page for planets with transit spectroscopy, but searching that page for HD 219134 yields no results.  Note that transit spectroscopy is challenging, so there aren't very many planets to which this technique has been applied.
In general, if you want to search for literature data on a particular astronomical object (whether it's an exoplanet or not), two good bets are NASA ADS or Simbad.  For ADS, you would type object:"HD 219134b" in the search box.  For Simbad, you would do an object search for "HD 219134b", and then scroll down to the "References" section and click "Display".  The advantage of both of these types of searches over a general web search, or even a general database search with that string, is that different names for the same object are cross-indexed; for example if you searched one of those for "alpha Ori", you would also find papers that only used the name "Betelgeuse", and vice versa.


What is a parsec and how is it measured? Astronomical units of measurement are mostly pretty straight-forward:

Astronomical Units refer to the mean Earth-Sun distance (~150 million km or 93 million miles)
Light years are the distance light travels in a year (~9.46 × 10^12 km)

Another astronomical unit of measurement is the parsec.  What is a parsec and how is it measured?

A parsec is approximately 3.26 times a light year or 206,265 Astronomical Units (AU).

According to "Cosmic Reference Guide" (Caltech), the word 'parsec' stands for "parallax of one arc second", and is (according to the website):

  the length of the long leg of a right triangle, whose short leg is one astronomical unit when the angle between the Sun and the Earth, as seen from an object in space (a star for example), is one arcsecond

Illustrated (from the link above):


Less stars in the night, compared to 15-20 years back I am not sure if my question is true or not. But this is what I observe regularly. When I was young and when I look at the sky, there were too many many stars were there. But now, from few months I'm observing the sky (by eyes, not any telescope), I see very few stars. Perhaps, countable.

Is it because:

Stars are dead? (Not all) As I read somewhere, by the time you see the star, it would have been dead. Because, the light would've traveled so many years to reach earth. By the time, star would have completed 'X' light years and it is dead by now?
Current position of the earth in galaxy: As I read, the earth will not always traverse in the same path around the sun. So the current view from the earth to the sky, may not have any star? (This is my guess/imagination, I'm not sure)
Is it because of the city light/ pollution?

PS: Please be easy. May be the question is broad or simple, just want to know the brief answer and if its complete explanatory, I'm happy for that.

In my own case, if I see fewer stars than I did when I was younger, it's partly because my eyes have deteriorated with age. I don't know whether that applies to you.

The most likely explanation is increased light pollution. If you can manage to get to an area far enough away from city lights, on a cloudless and moonless night, you should see just as many stars as you did when you were younger.

  Stars are dead?

No.  All the stars you can see in the night sky are within a few hundred light-years of Earth. That means that you're seeing them as they were no more than a few hundred years ago. Most stars live for billions of years; the Sun, for example, is about 5 billion years old and is expected to live for another 5 billion or so years in essentially its current form. Some of the brighter stars have shorter lifespans (because they're larger and "burn" their nuclear fuel more quickly), but even they have lifespans of at least millions of years.

No naked-eye visible stars have "died" in the last several centuries. We do sometimes see stars explode (as novas or, more rarely, as supernovas), but I don't believe any of the stars that have done this were naked-eye visible before they exploded.

  Current position of the earth in galaxy

No.  The stars do move relative to each other, but not quickly enough for the motion to be visible over a human lifetime. Barnard's Star has the fastest proper motion of any star in the sky, but it only moves at about 10 arcseconds per year; it would take it nearly 2 centuries to move the width of a full Moon. And Barnard's star isn't even naked-eye visible. The stars you see in the night sky are in very nearly the same apparent positions as they were thousands of years ago.


Why is the moon that special? Is it really a well built coincidence? I find it funny that the moon is the only object in the solar system to almost perfectly fit the sun, resulting in the spectacular solar eclipses. No other object except maybe a few irregular moons like Pandora from Saturn would do that, and for them, they aren't round, and thus can't fully block out all light. The moon also has a perfect rotation that is almost exactly the same as Earth's rotational period, resulting in a tidal lock, which causes tides.

Is this a coincidence? Any speculation on why this happens? Is this the only observed place where such criteria play into such perfect coincidence?

If the "perfect rotation" you describe is the fact that same face of the moon always faces the earth, this is explained by the 'Giant Impact Hypothesis' which also explains the formation of the moon. The molten earth was impacted by a large mass causing a part of the liquid earth to be flung out into space creating the moon. For a time, the two celestial bodies to be (the earth and moon) were connected to one another by some bridge of mass until eventually the centripetal force of their rotation finally pulled them apart. In fact, the two bodies are still moving apart today.

https://en.wikipedia.org/wiki/Giant-impact_hypothesis


Is there a database for stars? I'm interested, is there any database of stars which I could browse on the internet?
It would be good, because I could read their parameters, features and maybe find a picture about it.
Thanks!

HD? Perhaps start with: VizeR. You can search for any star you like, regardless of catalogue.


Distance from the sun can equal a hospitable planet? It is said that planets are continuously moving away from the sun. At one time Mars was just the right distance to support life at least temperature wise?
Could Venus be far enough from the sun eventually to cool down?

Related:
Are planets moving away from the sun?

Is Earth losing orbit?

It's actually the opposite.  Venus, Mars and I'll include the Earth as well, are likely getting more solar energy per square meter, not less, over time, even as they slowly move away from the sun.

1.5 cm per year, from the Rob Jeffries' answer to the 2nd question you linked is too slow to make much difference.  That's 1 km every 66,000 years and about 70,000 km over the age of the solar-system at that rate.   Now, planets may have moved away at different rates, especially when the solar-system was young and orbits were more chaotic and when Jupiter was thought to be doing it's dance in and back out, so it's not precisely known how close Mars was to the sun, say, 3.7 to 4.1 billion years ago, but we do know that the Sun was a little smaller and not as hot at that time, perhaps about 25-30% less luminous overall, though it was also more active with solar storms.   But Mars probably got less solar energy 4.1 billion years ago than it gets now, and it will get more solar energy a billion or two billion years in the future, even though it will probably be a little further from the sun.

The primary factor is how much of an atmosphere it had.   Planets can't have liquid water oceans without a reasonably thick atmosphere and Mars is believed to have had an atmosphere long ago.   How close it was to the sun matters too, but the thickness and composition of the atmosphere matters more.  Heat from the planets interior also matters and all the planets had more internal heat after formation, in the 3.7 to 4.1 billion years ago range.

Its also not 100% certain that Mars ever had relatively permanent liquid oceans and what we'd consider moderate temperatures.   Mars' geology suggests it had floods, but flooding could have been caused by local or temporary heating from volcanic activity or large meteor impacts.   It might also have been driven by a a more extreme axis leading to greater seasonal variation with glacial melt every summer (687 days in a year so summer lasted longer), and perhaps re-freezing every winter.

The Mars Ocean Hypothesis is a good hypothesis, don't get me wrong, but it's worth pointing out that its not a certainty.  It's possible that the oceans were ice covered on top but liquid underneath from heat coming from inside the planet.   Nobody knows for certain how temperate Mars was 3.7 to 4.1 billion years ago.


Thought experiment about our location in the Universe [I am sorry if this isn't the best place for this - it was either here or maybe on the Worldbuilding SE.]

I recently re-read one of my favourite childhood sci-fi books (Chocky by John Wyndham), and in it a young boy is asked by an alien being, where in the Universe our planet is.

It occurred to me that this is almost impossible to describe.

Nowadays, with Wikipedia and the Internet, we can get some of the way by stating that we're in a galaxy that is approximately 10 billion years old, light takes approximately 100000 years to cross it, and which is part of a local group, itself part of a cluster, and that part of a super-cluster.

Unfortunately the alien might then ask a follow-up clarification question such as what is a year?

So we go to basics - in base 60 (as opposed to binary or base 10), our basic unit of time is called a second, and there are 60 seconds in a minute, 60 minutes in an hour and 24 hours in a day - which is the time our planet takes to rotate once completely. Our planet orbits a local star (that we call the sun) and each complete orbit takes approximately one year, and light takes around 8 minutes to get from our local star to our planet.

Unfortunately our alien might then ask, but how long is a second? - so we then get into atomic clocks, caesium 133 and phases of radiation, followed by the next inevitable question: what is caesium. So we have to get down to the Periodic Table of the Elements and decribe an atom, atomic weight, and how to define caesium in a way that a far off advance civilization could relate to - not to mention perhaps a tangent about Prime Numbers.

So if our alien was lucky enough to be in contact with someone like Professor Brian Cox, or a physyicist with astronomical leanings, how would such a description and conversation go?

It all depends on shared knowledge. What you are getting at is known in philosophy as indeterminacy of translation. But note that the alien is already talking to the human: there has to be a language present, and languages do contain a lot of assumptions about the world (tenses imply that there is time, and the question itself implies space, for example).

Now, consider a sequence of larger and larger scale maps of the neighbourhood, running from one showing the solar system to one showing the cosmic web. That seems to settle the "where" question using pattern matching fairly well...

...except that there is the obvious question of what the maps show. In this particular example it is stellar baryonic matter, about 0.002 of the total mass/energy budget. If the alien is made of dark matter this will not make much sense except that it would on the larger scales recognise the topology of the web of halo structures it was familiar with. If it knew about baryonic matter it would infer that this was a map of baryonic stuff. Similarly a gaseous alien might think we have left out the 90% of baryonic matter that is intergalactic plasma. But the point is, the right kind of map or explanation would latch onto something in the knowledge of the alien and allow them to start inferring more meaning.

Since this is not just sending a single message but a conversation there is a lot of room for testing and exploration:

"So you are telling me where you are using a map showing 'pulsars'. Are they [untranslatable1] or [untranslatable2]?"

"What is the difference between [untranslatable1] and [untranslatable2]?"

"[untranslatable1] is spatially localised but extends across time,
 [untranslatable2] is temporally localised but spread out in space."

"Then [untranslatable1] sounds more reasonable. We know them as compact stars that rotate fast and send radio waves."

"Are 'stars' also spatially localised and extend across time?"

...

It won't be easy, and it is quite possible that our map doesn't correspond to a known region (if the aliens are communicating in some weird non-local way; had they communicated in a local way they would have known the answer to the questions already). But pattern matching and interactive bootstrapping from the shared language and knowledge-base would be a way of getting around the indeterminacy of translation.


How can the Sun block an X-ray telescope from observing a QPO galaxy "for several years"? Phys.org's Black hole's heart still beating  says:

  X-ray satellite observations spotted the repeated beat after its signal had been blocked by our Sun for a number of years.

and links to the Open Access PNAS paper (Jin, Done and Ward 2020) Reobserving the NLS1 galaxy RE J1034+396 – I. The long-term, recurrent X-ray QPO with a high significance. (also arXiv)

I can't imagine any kind of satellite orbit (Lagrange-point associated, Earth orbit, Heliocentric) for which the direction of the Sun would prevent a suitable orientation "for a number of years" assuming that number is greater than 0.5.

Why was this so?

related in arXiv: Re-observing the NLS1 Galaxy RE J1034+396. I. the Long-term, Recurrent X-ray QPO with a High Significance

It is not that it is blocked by the Sun, but that the duration of time for which it can be continuously observed was too small to be useful for the investigations performed in the paper.

The object was observed by XMM-Newton, which has a highly eccentric ($e \sim 0.8$), 48-hour orbit arond the Earth. This is designed so that the satellite can stare for relatively long periods at the same object, whilst being beyond a lot of the geocoronal background that interferes with its detectors. Whilst doing so it has to avoid looking (i) close to the Sun; (ii) close to the Earth; (iii) at the Moon.

The "visibility" of a target is defined as the longest continuous time it can be observed before the satellite has to break-off and point somewhere else. This can be as long as 140,000 s (there's a deadspot where they switch the detectors off when near the Earth), but can often be much shorter (or zero) at any paticular time. For efficiency reasons, the satellite operations team don't want to schedule too many short-visibility observations, since this wastes valuable observing time and spacecraft resources, and scientifically you often don't want to use short-visibility windows because you want to observe some sort of variability over a certain duration (e.g. observing over a full eclipsing binary orbit, or looking for variablity on a variety of timescales and requiring a long continuous time-series to get enough signal-to-noise in the Fourier spectrum).

Since the position of the Earth, Moon and Sun change with respect to each other, the visibility of an object will change from orbit-to-orbit of the satellite. Furthermore, because the orbit of XMM-Newton changes with time in quite a complex way (it precesses and the eccentricity has changed a little bit over the years), then the exact visibility of an object will not repeat from year-to-year and some objects will "come into season" during different years. In some cases sources can go through epochs where they are not continuously visible for long periods of time. For this particular position in the sky there appears to have been an unfortunate gap of about 7 years (that is actually the gap between observations, but I haven't checked whether that was completely determined by lack of "visibility" or what the maximum "visibility" was during those 7 years) when the object was not continuously observable for the $\sim 70,000$ seconds required by the science.

I managed to find this old plot, produced not long after XMM-Newton was launched. It shows the maximum visibility versus position on the sky for orbits 3-432 (about the first 2.5 years of operation). This shows that there is a fairly substantial patch of sky which was not continuously observable for more than about 50,000 s for the whole of that period.

There is a more modern interactive map here. This was for "AO-19" (May 2020 - May 2021), and is used by astronomers proposing observations to check how long an object is continuously visible for (as a fraction of the maximum possible time of around 140,000 s; actually there is a detailed web calculator to do it for any position). You can see that in this 12 month period there are big areas that cannot be observed for very long.

NB: There's nothing special about the object coordinates. Ecliptic coords 144, +28, though I would think that high ecliptic latitudes are less likely to be afflicted by this problem, since you would always be pointing at a large angle from the Sun.


Eyepiece diameter for small focal length planetary viewing on a large dobsonian? I have a 12" dobsonian with a 1500mm focal length and want to do some higher magnification planetary viewing. Currently my highest magnification is ~80x via my 19mm eyepiece (calculated here), but I would like to do some viewing at higher magnification.

My telescope is fitted for 2" diameter eyepieces, but I have an adapter for  1¼" eyepieces as well. I'm seeing a plethora of affordable 1¼-inch 9mm eyepieces, but very few 2" diameter options and very pricey at that.

At this point, I see the following options, by increasing price:

1¼" diameter 9mm eyepiece
3x Barlow lens
2" diameter 9mm eyepiece

I've read that the 2" eyepieces yield greater field of view advantages. Is that still true when the goal is a narrow focal length on solar system objects?

Would using a Barlow lens on my existing 2" eyepieces retain this advantage in a meaningful way?

In summary, is there any significant value in keeping to the 2" diameter eyepiece capability of my telescope when doing planetary viewing with low focal length given the magnitude of the cost difference?

I've referenced a "9mm" eyepiece here merely as an example, I'm sure most anything in the 5mm-12mm range would be fine.

I would recommend a quality 1¼" 9 mm or 12 mm eyepiece as the optimal solution.

You already have a 19 mm so something close to that such as 15 mm would bring small difference, so take a step or two down to a 9 mm or 12 mm Plössl lens.

While a 2" eyepiece usually offers better eye relief (from bigger lens) the cost is also substantially more, and if you compromise on quality by buying a cheaper eyepiece the results will be less than satisfactory - lack of clarity, distortions from imperfections in the glass, etc.

Extra magnification can also be obtained by using a Barlow, but you are also adding extra lenses to the optical path which will slightly reduce the amount of light you are getting at the eyepiece, but also any imperfections will be magnified.

Also, with Dobsonians (usually) not having any tracking mechanisms you will be manually tracking your object of interest, and extra magnification means that it will be moving more quickly across your field of view as well as making it a little more difficult to keep it smoothly in the eyepiece. These are also arguments in favour of a 2" eyepiece over a 1¼" eyepiece, but either way, good eyepieces can be kept and re-used when/if you upgrade your telescope.


Looking for star database I am looking for a particular set of information to essentially make a star map for a game.

What I need:
All stars within 11,462 light years of Earth.
Their stellar classification.
If they have any planets around them (hypothetical or otherwise)
Their position using an x, y, z coordinate relative to earth or galaxy center (I don't really get declination, etc)

I need the extrapolated position/distance data, not the catalogs themselves.

The best one I can find is HYG, HabCat, and HYGHag is limited to to like 120,000 of 1,2 billion estimated stars... Are there any better?

Best chance would be the Hipparcos catalogue. The first set of Gaia data will be released Mid-September 2016, but I don't know if it will be more accurate than Hipparcos already.

  All stars within 11,462 light years of Earth.

That won't be easy. These catalogues have a magnitude (brightness) cutoff. Brighter, more massive stars can be seen further away than smaller, darker ones.

  Their stellar classification.

Spectral type should be in there.

  If they have any planets around them (hypothetical or otherwise)

Searching for Exoplanets is an ongoing project, and there is no complete catalogue in any sense of the word yet for any radius. Most Exoplanets were found with the Kepler mission, but that one looked in a fairly small part of the sky. What exactly do you mean by hypothetical planets?

  Their position using an x, y, z coordinate relative to earth or galaxy center (I don't really get declination, etc)

You'll have to convert that yourself. If you know how to program, there are libraries for coordinate conversion. I would argue, though, that an Earth-centred spherical coordinate system is preferential for most applications.

Distances are hard to measure. Hipparcos and Gaia use the parallax method. This gets less accurate with increasing distance. Also, distances are naturally measured in parsec, but the conversion to light years is trivial.

The Gaia catalogue will have $10^9$ stars with positions and velocities. It will also be more accurate than Hipparcos. But the measurements are ongoing, and the final data release is planned for 2022.


If I could see from a Gravitational-Time dilated place to somewhere far which is not dilated, would I see fast moving people? Gravitational time dilation like the one shown in the movie Interstellar, causes time to move slowly for someone within the high gravity reference frame. Now let's say that from such a place, I was able to view through a telescope somewhere far away which is not gravitationally dilated. And assume that I am able to view the near live video stream of it. Would those people look fast moving?

a Gravitational-Time dilated place to somewhere far which is not dilated

"Dilated" or "not dilated" are not absolutes. It's always a relative thing. "Dilated" relative to me, but perhaps "not dilated" relative to you. There is no dilation that happens "in general"; it's always relative to another place/observer/reference.

  Would those people look fast moving?

Yes.

Now, everything coming out of that place would seem like it's "speeding up time" relative to you. If you're looking at them through a telescope, all light would be blue-shifted. If you're receiving a TV transmission, the frequency of the carrier would also be higher.

Keep in mind, this time dilation is not an illusion. It's not just "how you see things". Their flow of time, as measured by you, is indeed faster. This affects everything in that place, as measured by you.

But it's also relative. Some other observer may be in a place where time flows differently from yours - they would look through the telescope and would see a different rate of change. And it's not an illusion for them either.

Time dilation is real. What's not real is a unique time, ticking the same for everyone - that's the actual illusion.

Time is a local thing. (more precisely, the rate of ticking for a clock is a local thing)


Dark enough to see stars during total solar eclipse? I was fortunate enough to see totality during the eclipse of 2017 August 21.
It did leave me wondering about the 1919 Eddington experiment, though.  During totality, it only got dark enough to see a few of the brightest objects like Venus and Jupiter.  It was more like twilight than night that way.

So how were they able to see and photograph stars in the 1919 eclipse?  Do skies get darker during some total eclipses than others?  Was there more light scattering in my case, from being in a populated area?  Or do the right optics let you photograph stars during twilight-like conditions?

One thing to double check is if you can see stars easily at night from the same location. If there are city lights nearby, combined with humidity or dust you can have a bright sky due to terrestrial effects.
Another thing to consider is the time needed for dark adaptation. If you walk from a brightly lit room out under a dark sky, you can't see many stars right away. There needs to be time for your eyes to adapt to the dark. If you were looking at the screen of a camera or phone or another light source, that would further slow down the adaptation process.
The astrometric data for the 1919 eclipse test of General Relativity was done with a telescope, not the naked eye. Stars are unresolved, which means that no matter what magnification, they are still small points of light who's size (visual or on film) is determined by other effects, like diffraction, aberrations and focus as well as the resolution of the retina or the film. In short, they stay small. However, sky brightness is reduced by magnification squared. If you expand a patch of sky by 10X then each point is 100 times dimmer.
Both of these are offset by the increase in light from the larger aperture of the telescope, but its mostly the sky brightness that decreases with magnification.
The interesting stars were the ones closest to the Sun, because they would have the largest deflection, which was really small, only of the order of one second of arc. There the problem was the brightness of the solar corona, more than the sky. If you look closely at the photographic negative below, you can see pairs of faint horizontal lines. In the middle of each pair is the position of a star to be observed. Even with the telescope and extended photographic exposure, the images are faint.

above and below: 1919 eclipse negative, from here. Original from: F. W. Dyson, A. S. Eddington, and C. Davidson, "A Determination of the Deflection of Light by the Sun's Gravitational Field, from Observations Made at the Total Eclipse of May 29, 1919" Philosophical Transactions of the Royal Society of London. Series A, Containing Papers of a Mathematical or Physical Character (1920): 291-333, on 332.

Negative of the 1919 solar eclipse taken from the report of Sir Arthur Eddington on the expedition to verify Einstein's prediction of the bending of light around the sun. For positive version, see Image:1919 eclipse positive.jpg.
Original caption:
In Plate 1 is given a half-tone reproduction of one of the negatives taken with the 4-inch lens at Sobral. This shows the position of the stars, and, as far as possible in a reproduction of this kind, the character of the images, as there has been no retouching. A number of photographic prints have been made and applications for these from astronomers, who wish to assure themselves of the quality of the photographs, will be considered as as far as possible acceded to.

below: Screen shot from the March 2009 Physics Today article Testing relativity from the 1919 eclipse—a question of bias. The telescope on the right, in the square tube is the 4-inch instrument that provided the photographs used in the analysis. Heliostat mirrors driven by clocks were used to reflect light from the region near the Sun into the fixed telescope.


Why do the solar system planets go rock-gas-ice instead of rock-ice-gas when moving away from the sun? The sun and the solar wind seem to do a good job of fractionating lighter materials to the outer solar system and leaving heavier materials in the inner solar system. So we end up with rocky/metallic planets in the inner solar system and the gas giants and ice giants in the outer solar system.
But why are the gas giants (Jupiter and Saturn composed mainly of Hydrogen and Helium - the lightest materials) closer to the sun than the ice giants (Uranus and Neptune composed of water, ammonia, methane, etc.)?

Prelude
It is now generally accepted in the planet formation community that planets form as a side-product of the star formation process in so-called protoplanetary discs.
Protoplanetary discs have initial masses of few to tens of percent of their stellar host masses, are relatively cold (T<150K in about 95% or more of their mass, which is outside the water iceline for a standard MMSN model) and are hence mostly detected in the infrared. The radiating infrared component is the 'dusty' component (first published detection and confirmation via the IRAS satellite in 1984-1985) making up about 1% of the mass, the other 99% being H/He gas.
Those discs are accretion discs, i.e. they loose angular momentum through various processes, which leads to mass infall into their host star. The dust settles into the midplane. For the case of turbulent accretion, dust and gas will be well-mixed and accrete relatively uniformly into the star, while in the case of disc-wind-driven accretion, H/He in the upper layers of the disc flows over the midplane and provides the accretion rate. Disc accretion rates can be too much for what the star can actually accrete and the excess mass is ejected vertically in jets that can exist throughout the disc lifetime, their mass decretion rates correspond typically to 1-50% of the disc accretion rate.
I mention the water iceline solely as point of reference, as its exact effect on the physics of planet formation is heavily debated, it currently cannot be observed, and icelines in several other molecules such as $\rm CO, CO_2, N_2,...$ could be playing roles too.
Planet formation
Our solar system originated very probably in one of those protoplanetary discs. We cannot follow the formation process over the disc lifetime, as this takes between 1-20 Myears (median value 3-5 Myrs, depending on the survey) and hence as often in astrophysics, we rely on snapshots and exoplanet statistics to try and puzzle together the physics.
50% of all exoplanetary systems harbour several rocky super-earths at radii interior to the water iceline. 6-10% of all stars possesses cold gas giant planets (giant planets at semi-major axes >0.5 AU) and 0.5-1% possesses hot gas giants (giant planets at semi-major axes <0.1 AU).
While from this our solar system seems to have an unusually low mass in the terrestrial planet zone, nonetheless physics seems to prefer to build rocky planets interior to the water-iceline. Those processes must happen in the protoplanetary disc phase and possibly shortly after gas removal (<100Myrs, it is poorly constrained which fraction of its final mass Earth possessed at disc dispersal).
Rocky planets are also thought to form beyond the water iceline. However in those regions of the protoplanetary disc, the mass reservoir is enormous and rocky planets can achieve runaway gas-accretion before their parent disc disperses.
Reaching runaway gas-accretion consists of two steps: First, after the rocky, multi-earth-mass planet is formed, it aquires an atmosphere that is hydrostatically connected to the disc via its own gravitation. This atmosphere cools slowly via Kelvin-Helmholtz cooling. The contraction allows for more mass to flow into the planetary domain, forming a massive atmosphere. Should this atmosphere reach a mass important enough for self-gravity to help the contraction further, the planet accretes more the more it cools and it cools more the more it accretes, hence runaway accretion is reached.
The architecture of the solar system
With all this, we can formulate the standard explanation for the architecture of the solar system:
Jupiter and Saturn are standard cold gas-giants that underwent a phase of rapid core-assembly and subsequent runaway gas-accretion. Uranus and Neptune grew far out in regions of low disc gas density (or small dust populations, increasing the core-assembly and cooling time) and hence were stuck in the hydrostatic gas accretion phase until the disc dispersed. The "ice" in ice giants hence refers to the solid component making up 60-80% of their mass, and not that they missed runaway accretion, which would make for a clearer name.
Now the other question is, why planets at small radii seemed to have evaded runaway gas accretion, in our solar system and at least 50% of exoplanetary systems. A candidate mechanism is "gas recycling", i.e. replenishment of entropy to protoplanetary atmospheres which prevents their contraction. This is possible close to the star because the gas is very dense, replacing cooling with advection as dominant entropy transport mechanism.
Summarizing
The broad strokes of the solar system architecture can be understood in terms of physical mechanisms which have been shown to work in simulations. However, when applying those same mechanisms in order to form synthetic populations of planets, those synthetic populations are usually inconsistent with the observed ones.
This is work in progress and needs future missions to the ice giants in order to measure their detailed heavy element abundances and use the latter to distinguish between competing formation scenarios, of which I have presented merely one.
The physics presented here is therefore very different from a simple 'heavy elements sink in solar wind' picture, which to the extent of my knowledge, was never considered as candidate for a planet formation model. Merely Laplace in the 18th century considered a similar sounding model of yours, of an extended solar atmosphere that centrifugally breaks into rings in order to form the planets. With my prelude as above, this model is now known to be incorrect however.


How long should it take for us to observationally determine if Caltech's Planet Nine is really there? Caltech just released a report that says there is possibly a large (10 earth mass) planet in a remote orbit (10 - 20 thousand year) that explains a lot of observations of Kuiper Belt objects. How long could it take for this to be confirmed or refuted observationally?

See this video

Those are model calculations, which hint to the existence of a possible body of about 10-times the mass of Earth. Calling this a discovery would clearly be premature. The confidence level is just a little above the "evidence" level of 3 sigma, under the assumption, that the discoveries of the KBO objects leading to the inference aren't observationally biased.
It's a long way to direct observation, since precise orbital data haven't been inferred.
Another option is, that there has been a planet. But it might have left our Solar system. Hence predicting a date for direct observation doesn't appear reasonable at the moment.


What is the probability of an astronomical body hitting you? It appears that there is already a calculated ballpark for meteor fatalities, but what about other factors such as space trash, comets, and other miscellaneous items in space?

Is it possible to obtain a ballpark estimate?

Comets are part of the meteor calculation.

Space junk is a minor concern. Most satellite and rocket bodies are fairly flimsy and are destroyed by the atmosphere. Sometimes chunks do come down. But by the time they reach Earth they are not falling fast enough to form a significant crater.

However being directly hit could be fatal.

  According to Mark Matney, a scientist in the Orbital Debris Program Office at NASA's Johnson Space Center in Houston, the odds that any of the 7 billion people on Earth will be struck by a piece of the soon-to-fall satellite is 1 in 3,200. "The odds that you will be hit are 1 in several trillion

In '97 a piece of debris from a Delta rocket hit Lottie Williams in Tulsa. It caused no significant injury. http://www.livescience.com/33511-falling-nasa-satellite-uars-risk.html


What is the reason for magnetic braking? What is the reason of Magnetic Braking in contact binary stars?
Just dark spots?
I mean hot spots can have this effect on binaries.

The magnetic braking in contact binary stars AFAIK is assumed to be similar to the magnetic braking mechanism that spins down the rotation of single stars.

A stellar wind couples to the large scale magnetic field and maintains this coupling out to the Alfven radius (where the thermal pressure and magnetic pressure are roughly equal). At this point the wind decouples, stops co-rotating with the star and hence removes a certain amount of specific angular momentum.

The situation will be more complicated in a binary system (entangled, asymmetric fields etc.) but the basic principle will be the same. Many contact binaries have fast rotation and strong, dynamo-generated magnetic fields.


What happens if one day aliens want to talk to us. How can we talk to them? Always I wonder about that if some day alien wants to talk to us, then how can we talk to them? They didn't know any of the language on the Earth.

What will be the basic building blocks for it?

Some basics:

Whether we listen for or send signals into space, something to keep in mind is that space is full of EM radiation of all wavelengths, from very long wavelength background radiation, to a range of microwave, infrared, visible and UV light from stars to gamma-rays from accretion disks and collapsing stars.   Any signals we send and perhaps any signals we could receive would be like raindrops in an ocean.   There's an obvious range to listen called the hydrogen line, which is the emptiest region of EM radiation in space.

Even so, across light years, signals would still be faint, so the first signal that's sent needs to be an easily recognizable signal and math is very good for that.   Pi in binary or a series of repeating prime numbers wouldn't happen naturally, so signals like that are a great way of saying "We are here".

Once the first signals of prime numbers or other math are exchanged, and we see what direction the signal comes from and we know where to listen and where to send, then the communication can begin, presumably using more traditional language.


What do the words "p-type" and "s-type" mean? As explained in various sources (such as that answer), a planet in an s-type orbit orbits one of the stars of a binary star system, whereas a planet in a p-type orbit orbits both stars. A synonym for a p-type orbit (at least in binary star systems, not sure about N-ary systems in general) is "circumbinary", which is a self-explanatory term.
However, what do "s-type" and "p-type" mean, respectively?
Are "S" and "P" acronyms of anything?

"Satellite type" and "Planet type".
The terms seem to have been coined by Rudolf Dvorak in 1982 paper "Planetenbahnen in Doppelsternsystemen"

Due to the fact that quasiperiodic orbits exist around stable orbits 3 different types of possible planetary orbits are found: S-types (satellite type orbit around one primary), P-types (orbits around both primaries), and L-types (librating orbits in the vicinity of L4 or L5).

It's not immediatly clear from this whether "P-type" means "planetary type" or "primary type" (and the full paper would be in German and doesn't seem have been digitised).  However in an English paper from 1986 Critical orbits in the elliptic restricted three-body problem.  Dvorak is explict:

There are three types of stable motion for a third massless body that are interesting as regards this problem: the P-type (planet type) surrounding both primary bodies, the S type (satellite type) orbiting one of the primaries and L-type (Librator type) librating around the Langrangian equilibrium points ...


Could diatom blooms affect albedo of an exoplanet? And perhaps display a seasonal variation if the exoplanet has an axial tilt, providing proof of an Earth-like world supporting at least oceans with microbial life?

First off, I am rather certain that by diatom bloom you refer to algal bloom which is a seasonal change of sea color, at least here on Earth, the only system we have observed yet. Your question is indeed rather hypothetical, but from a astronomy/ physics points indeed an interesting one.
Next, let me summarize the assumptions I am reading out of your question:

You assume an exoplanet that is mainly covered by water. Such a waterworld is not only existing in movies, but also a common toy system in climate science (of Earth).

The exoplanet is not only habitable, but also has some kind of algae living in its ocean. A strong assumption indeed, but for our calculation, it gives us numbers for how the color and thus the emissivity of the planet changes.

Indirectly, you also assume our hypothetical waterworld has an atmosphere. Why? An waterworld exoplanet would otherwise have already quickly evaoprated evaporated into space. In terms of greenhouse calculation later on, we will assume that the atmosphere consists only of water vapor. This means that our hypothetical exo-algae on waterworld would not necessary be aerobic. After all, we are after the physics here.

Could algae bloom affect the albedo of an exoplanet? Definitivley yes, but the question is, how much. And if we could observe that from Earth would be a whole different question.
What do the assumptions mean in numbers?
The albedo of open ocean water (on Earth) is $\alpha_O = 0.06$. Concerning the algal albedo $\alpha_A$, I found e.g.

a poster "Algal blooms: how are they harming models used for climate management?" by Ian Jenkinson
an article John Raven: The possible roles of algae in restricting the increase in atmospheric CO2 and global temperature
another article by H Gordon, M Jacobs: Albedo of the ocean-atmosphere system: influence of sea foam.

In short: Algae do not only change the color of the ocean (which color depends on the involved species), but also may produce foam (with $\alpha_A \approx 1$). The latter seems to have the largger effect since $\alpha_A \gg \alpha_O$. One of the above mentioned papers suggests a 10% coverage of the ocean with foam, an assumption which sounds appealing to me.
Some calculations
For a planet with radius $r$ in thermal equilibrium, the incident radiation equals the black body radiation:
$$ \pi r^2 E (1- \alpha) = 4 \pi r^2 \sigma T_{\rm eqi}^4 \Longleftrightarrow
T_{\rm eqi} =\sqrt[4]{\frac{E (1-\alpha)}{4 \sigma}}$$
In this formula, $E$ is the solar constant ($E_{\rm Earth} = 1368 {\rm W}/{\rm m^2}$), $\alpha$ is the (overall averaged) albedo ($\alpha_{\rm Earth} = 0.3$) and $\sigma = 5.67 \cdot 10^{-8} {\rm W\cdot m^{-2}K^{-4}}$ the Stefan-Boltzmann constant. For Earth this leads to $T_{\rm Earth,eqi} = -18^\circ {\rm C}$ which shows that we have to include the Greenhouse effect, i.e. that the atmosphere absorbs part of the outgoing radiation. We do that by introducing a  factor $\beta$:
$$T =\sqrt[4]{\frac{E (1-\alpha)}{4 \sigma \beta}}$$
On Earth, all Greenhouse gases (${\rm H_2O, CO_2, CH_4}$) contribute to an average $\beta \approx 0.62$ which leads to a more realistic global average temperature of $T = 15^\circ {\rm C}$ for Earth.
For a water world (without algae) we may use $\alpha = 0.1$ which leads to $T_{\rm WW} = 32^\circ {\rm C}$ which compares to a snowball planet with $\alpha = 0.8$ and $T_{\rm SB} = -63^\circ {\rm C}$.


Could dark matter form coherent bodies the size of stars or planets? Is it possible that dark matter could collapse gravitationally, forming massive bodies similar to stars or planets?
When I say "similar", I don't mean to imply that they would undergo nuclear fusion or have solid surfaces, merely that gravitational collapse could create a much higher density of dark matter within some spherical region of a size similar to a star or planet.
My thought is that if there were, say, some billions of roughly solar-mass collapsed dark matter bodies in this galaxy, and they weren't associated with normal stars, we likely wouldn't be able to detect them. Gravitationally, they'd be part of the background noise; we can't, as far as I know, detect the gravitational influence of individual stars on each other over distances of light-years.
Is this plausible?
(What isn't plausible is that I'm the first person to think of this, so if there's existing research I'd be interested in seeing it.)
I understand that we don't know what dark matter is, and therefore this question might be unanswerable at this time.

No, it's not really possible, because if it could it wouldn't be dark.
To see why this is, think about some some collection of gravitating particles which is going to to collapse to a smaller, denser collection of particles.  To do this they need to lose a lot of energy: you can see why this is by thinking about where the average particle lives in the gravitational potential of all the particles.
Ordinary matter – clouds of gas and dust say – do this by getting extremely hot as they collapse, and then radiating energy electromagnetically.  Ultimately collapsing clouds of gas can get hot and dense enough that nuclear fusion can start and we get stars.
Dark matter, by hypothesis, doesn't interact electromagnetically, at all: if it did it would not be dark!  This means that the only way it has of losing energy is to put it into the average kinetic energy of the particles, and that means that collapse doesn't happen, because if you have a lot of particles with lots of kinetic energy they don't stay collapsed.  The virial theorem is important to understanding this.
So dark matter simply lacks a mechanism to lose energy since it can't radiate it away electromagnetically, and so does not collapse.
(In fact I think there are some mechanisms: dark matter could lose energy by gravitational radiation for instance, but this is an absurdly slow process except in really extreme cases.)
To answer the second part: it certainly was (perhaps is) a possibility that there are very large numbers of compact, massive dark objects which make up dark matter.  These things are known as massive compact halo objects: MACHOs (the alternative theory is weakly interacting massive particles which are, of course, WIMPs, and there are I believe also RAMBOs although I'm not sure what they are.  If there are not ARNIEs there should be (perhaps one kind of these will be CONANs?)).  These might either be failed stars, isolated planets or (I like this idea) small black holes: there's a wide range of possible BH masses below stellar mass BHs and the point at which they start becoming bright due to Hawking radiation.  In the case of small BHs these would probably need to be primordial (made in the big bang) as there's no plausible mechanism for them being the end-point of stellar evolution.
But it turns out that such objects are observable.  Because they are massive they will deflect light, and they can therefore be detected by gravitational lensing as they pass in front of more distant luminous objects (stars, really).  Well, you can do the statistics of how many of these things there would need to be (which is really a lot), and how often you would expect to observe lensing events from them as a result.  And then you can do a survey and look for lensing events.  And the results of such surveys (some of these are pointed to by the above Wikipedia article) are that there are not nearly enough MACHOs to account for dark matter, although I think there is still a little wiggle-room, depending on who you ask.


How did Meeus calculate equinox and solstice dates? In Astronomical Algorithms (2nd ed, ch. 27, 2009 corrected printing) Jean Meeus gives expressions to calculate the date and time (dynamical time, equivalent to Terrestrial Time) of equinoxes and solstices from the year -1000 to the year +3000. The expressions are accurate to 51 seconds or better for the years 1951-2050. First what Meeus calls the "instant of the 'mean" equinox or solstice" is calculated using a fourth degree polynomial; there are 8 expressions. There are different expressions for each solstice or equinox, and different expressions for the year ranges -1000 to 1000 vs. 1000 to 3000. Then two corrections are applied; the corrections are calculated the same way no mater which time period or equinox or solstice is being corrected. The first step is to calculate:

$$T = \frac{(\text{mean JD of event} - 2451545.0)}{36525}$$

$$W = 35999.373°T - 2.47°$$

$$\Delta \lambda = 1 + 0.334 \cos W + 0.007 \cos 2 W$$

Next, an additional correction is computed involving 24 periodic terms with various periods.

Can anyone explain, in general terms, how Meeus derived these expressions? I'm especially interested in understanding what the "mean" value represents?

The mean instant of an equinox JDE_{0} (or solstice) is the statistical mean calculated over a sample of equinox's instants (since they do not happen exactly in the same instant/day every year).

The first eight equations are derived by interpolation and the corrections (the 24 terms) are derived as a truncated iteration, I suppose from the JDE{N}-JDE{0,N}.

The truth is that Meeus does not explain anything at all (though he mentions interpolating the apparent longitude of the sun in 3 dates for higher accuracy). Without an explanation in the book, it is pretty difficult to guess how he did it.

You can find some more details about the iterative method here:

http://www.spacebanter.com/showthread.php?t=171014

I hope this helps :)


What is the long term fate of the gas giants? If I'm not mistaken, it is believed that the reason for such turbulent weather on the 4 outer gas giant planets is that the internal pressure is so great that it is generating heat, which is causing convection, which causes extreme weather.

Will these planets forever generate heat, or at some point will they freeze up?

What will Jupiter look like a trillion years from now?

The timescale on which Jupiter cools is reasonably well understood and predicted by the current generation of evolutionary models.

Jupiter's luminosity is provided mostly by gravitational contraction. For a planet that only contains gas governed by the perfect gas law, the appropriate timescale for this contraction (or indeed for the luminosity to fall significantly) is given by the Kelvin Helmholtz timescale.
$$ \tau = \eta \frac{GM^2}{RL},$$
where $M$ and $R$ are Jupiter's mass and radius and $L$ is its current power output (or luminosity), and the parameter $\eta \sim 1$. This timescale is a few $10^{11}$ years.

However, giant planets like Jupiter are not governed by perfect gas laws. The gas in the centre of Jupiter is dense enough that electrons become degenerate. Degenerate electrons fill the available energy levels up to the Fermi energy. Their consequent non-zero momenta of the electrons exerts a degeneracy pressure that is independent of temperature. As a result, the rate of contraction slows and the release of gravitational potential energy slows; the planet is able to cool and remain in hydrostatic equilibrium without the same degree of contraction.

One can express this change using the $\eta$ parameter. For Jupiter $\eta \simeq 0.03$ (Guillot & Gautier 2014) - i.e. the timescale for the luminosity to fade is 30 times quicker than the naive Kelvin-Helmholtz time and Jupiter's luminosity will scale as the reciprocal of its age and will fall by a factor of a few in $10^{10}$ years. In a trillion years, the luminosity of Jupiter will be lower than it is now by roughly a factor of 250.


Mass of the universe How do we calculate the mass of the universe taking into consideration dark matter?
Mass of visible matter can be computed by stellar method, but how do we calculate mass of dark matter which we don't see? Why do we assume that dark matter is more abundant than visible matter?

Check out the Wikipedia page on the missing baryon problem. It's how we compute the mass of ordinary matter. You can see there are methods of getting the mass of ordinary matter without adding them up. In other words, you can find ordinary matter even if they're invisible.

This is similar to how we find dark matter. Again, look at Wikipedia's article on dark matter, in particular the observational evidence section. Ignore the more technical methods if you're new to the field and just look at the first few:

With galaxy rotation curves, you know you need more matter to recover Kepler's laws. How much matter? You can tweak that number and therefore get an estimate for the dark matter needed. You can't prove that this dark matter isn't just invisible ordinary matter using this method, however.
The same goes for velocity dispersions.
Gravitational lensing can measure the total mass of an intervening cluster without knowing anything about the kinematics of the cluster.

If you're wondering how we know dark matter can't just be invisible ordinary matter, again look at the Wikipedia article, in particular this section.

Big bang nucleosynthesis constrains the amount of ordinary matter in the universe. If dark matter were just invisible ordinary matter, we also expect more helium, lithium, etc in the universe than we observe.
Gravitational microlensing constrains the amount of MACHOs (massive compact halo objects, such as black holes and invisible planets) we can have in a large mass range, covering most of the plausible candidates.
Data from the cosmic microwave background indicates a large fraction of matter does not react with photons.


Black holes and time relation Why Is it said that if someone is being sucked inside a black hole then he/she may see that time is running slowly?

General relativity can be used to discuss, theoretically, what happens to objects as they approach the event horizon of a black hole; it all has to do with your point of view. Outside viewers, those not affected by the gravity of the black hole, would view an object approach the event horizon slowly and pretty much never enter past the horizon. For the observers point of view time stops for the object. From the point of view of the object, its own time runs as normal, and it is the time outside of the event horizon that runs faster than normal. For more general information at a really easy to follow format you might check out What is Relativity by Jeffery Bennett.


Plant on large Asteroid? I'm just curious if it is possible for a plant to survive on a large asteroid. Some say no, some say it depends.

With the scenario of an asteroid at about 1 AU from the Sun, receiving a similar amount of sunlight to Earth, and with wet soil or something similar.

It might depend on what qualifies as a plant.   A thin film of rugged mold for example . . . just maybe.

The tricky part is that asteroids are too small to retain an atmosphere and you need an atmosphere to have liquid water which plants and all life (so far as we know) needs.   I don't want to dismiss the possibility that there's underground water and some kind of very sturdy life that can eek out a living on the airless surface of an asteroid, but I wouldn't bet money on it.  I think it's unlikely, maybe unlikely multiplied by about 5 orders of magnitude.    It's very close to impossible.

That said, Ceres is an asteroid and Ceres may have a liquid underwater ocean.   Life in an underwater ocean is thought to be possible.


Is the newly discovered exoplanet 'Teegarden b' habitable? A new exoplanet has been just discovered around the star Teegarden, at only 12 light years away.

TLDR version: at this point there are too many unknowns to determine whether this planet candidate is habitable.

The discovery paper for this system, Zechmeister et al. (2019) "The CARMENES search for exoplanets around M dwarfs. Two temperate Earth-mass planet candidates around Teegarden's Star" is available on the arXiv. Table 1 lists the properties of the star and table 2 lists the properties of the planets.

The planetary quantities we have reliable estimates for are the minimum masses, the orbital periods and the distances. The eccentricities are consistent with zero so I'm going to assume for the moment that the orbits are circular. The reason that minimum mass is relevant is that this is an example of a radial velocity detection, which results in a degeneracy between the inclination of the planet's orbit to the line-of-sight and the mass of the planet.

Habitable zone

According to figure 13 in the discovery paper, Teegarden b is too close to the star to be in the conservative estimate of the habitable zone, though it is in the optimistic habitable zone. The Earth Similarity Index (ESI) appears to be a fairly misleading metric (I am of the opinion that ESI is a bad metric that should be abandoned), from the discovery paper:

  In addition, planet b receives nearly the same stellar flux
  as the Earth and therefore has a nearly identical equilibrium temperature, but is outside the conservative HZ. As discussed by
  Kopparapu et al. (2014), this is caused by a runaway greenhouse
  effect that is due to water vapour starting at lower insolation for
  low-mass stars. This effect is neglected in the ESI calculation
  and leads to the curious fact that planet b has a higher ESI value
  than planets inside the conservative HZ.

The position of Teegarden b therefore suggests it may have undergone a runaway greenhouse effect, though this might be avoided if it can build up enough reflective cloud cover on the daylight hemisphere, or if it is a desert planet without enough water to destabilise the climate.

Even if a planet is located in the habitable zone, it is not necessarily habitable. An obvious counterexample would be the Moon.

Terrestrial or not?

The next question: are these planets terrestrial? Unfortunately we don't know: Teegarden b does not transit the star, and there has so far been insufficient coverage of Teegarden c to say whether it transits. As a result, we do not know the radii of the planets so we cannot tell whether or not the planets have densities compatible with being terrestrial. The minimum masses are low enough that these planets do likely fall below the transition region between rocky and non-rocky planets suggested by Kepler data, see for example Rogers (2015) "Most 1.6 Earth-Radius Planets are not Rocky" - using the mass-radius relationships of Zeng et al. (2016) "Mass-Radius Relation for Rocky Planets based on PREM" the transition is somewhere around 3–5 Earth masses.

Another point of comparison here is the other known system of low-mass planets around an ultracool dwarf star: TRAPPIST-1. In this case, despite the small size of the planets, the inferred masses indicate the presence of substantial amounts of volatiles, e.g. Dorn et al. (2018) "Interior characterization in multiplanetary systems: TRAPPIST-1" where the volatile contents of the planets are estimated to correspond to a few percent by mass, compared with 0.02% by mass for the Earth. A planet like this would be very different to the Earth, in an optimistic scenario it might result in an ocean planet. The habitable zone for ocean planets would be different to that for Earthlike planets, since the negative feedback of continental weathering would not occur.

Red dwarf stars

Red dwarf stars have a variety of problems when it comes to habitability. One issue is that they have an extended high-luminosity pre-main sequence phase which might result in runaway greenhouse effects and/or severe water-loss on close-in planets, see Ramirez & Kaltenegger (2014) "The Habitable Zones of Pre-Main-Sequence Stars".

An additional problem is that red dwarf stars have severe stellar activity that would erode an Earth-like atmosphere in a very short amount of time, as an example Garcia-Sage et al. (2017) "On the Magnetic Protection of the Atmosphere of Proxima Centauri b" estimate an Earthlike atmosphere at Proxima Centauri would be lost in less than 400 million years. Teegarden's Star is fairly quiet for its spectral type, but it is an old star and was likely far more active in its youth.

These issues are not necessarily fatal, outgassing from volcanism might be able to replenish the atmosphere to some extent after the stellar activity dies down, or if the planet was born with a large amount of volatiles it might be able to emerge as a habitable planet. Unfortunately we don't have sufficient data on habitable zone planets around red dwarf stars to know for sure.

A further issue is that the atmospheric chemistry on planets around stars like Teegarden's Star would likely result in substantial quantities of carbon monoxide accumulating in the atmosphere, see Schwieterman et al. (2019) "A Limited Habitable Zone for Complex Life". This would be lethally unpleasant for any human visitors (or other organisms that use haemoglobin as their oxygen carrier) caught outside without an air supply. Note that haemoglobin is not the only possible oxygen carrier: some groups of life on Earth use alternatives (haemerythrin, haemocyanin) that are less vulnerable to carbon monoxide poisoning, as noted by Howell (2019) "Comment on "A Limited Habitable Zone for Complex Life" (Schwartzman et al. 2019 ApJ 878, 19)":

  In short, there is a notable diversity of extant invertebrates on this planet that would be quite comfortable outside of the range of CO concentrations proposed by Schwieterman et al. to constrain the habitable zone. There is no a priori reason to rule out the possibility that these lineages could have given rise to intelligent life during the history of life on this planet. A significantly higher atmospheric CO concentration would have prevented the evolution of hemoglobin, but not necessarily prevented astronomically observable biosignatures from Earth.

Summary

It is too early to be able to say whether the planets around Teegarden's Star are habitable. There are plenty of reasons to be sceptical but worlds like Teegarden b are still promising enough to be worth investigating in more detail. It is unfortunate that it is not a transiting planet, but similar systems like TRAPPIST-1 should be helpful in constraining what the possibilities are.


Why (the heck) is the basic angle of GAIA 106.5°? This answer to Why does the Gaia space telescope have two main mirrors says:

According to the GAIA FAQs which does an excellent job: http://www.cosmos.esa.int/web/gaia/faqs:

Why is there an angle of 106.5 degrees between Gaia's 2 telescopes?
The choice of the so-called basic angle of GAIA was a non-trivial one. On the one hand, it should be of order 90 degrees to allow simultaneous measurements of stars separated by large angles on the sky. On the other hand, it should not be a harmonic ratio of a 360-degree circle (e.g., 60 deg, 90 deg, or 120 deg). Taking these considerations into account, acceptable ranges for the basic angle are 96.8 +/- 0.1 deg, 99.4 +/- 0.1 deg, 100.5 +/- 0.1 deg, 105.3 +/- 0.1 deg, 106.5 +/- 0.1 deg, 109.3 +/- 0.1 deg, 109.9 +/- 0.1 deg, etc. Accommodation aspects identified during industrial studies subsequently favoured 106.5 deg as the value finally adopted for Gaia.

I don't understand any of that.
Questions:

"On the one hand, it should be of order 90 degrees to allow simultaneous measurements of stars separated by large angles on the sky." Well any large angle is a large angle, why should it be 90?
"should not be a harmonic ratio of a 360-degree circle" I can imagine for instrumental reasons, but why are "acceptable ranges for the basic angle are 96.8 +/- 0.1 deg, 99.4 +/- 0.1 deg, 100.5 +/- 0.1 deg, 105.3 +/- 0.1 deg, 106.5 +/- 0.1 deg, 109.3 +/- 0.1 deg, 109.9 +/- 0.1 deg, etc." and what does "etc." mean here? Is there a pattern I'm not seeing?

Any angle with an uncertainty of +/- 0.1 deg can be reached with some rational number with a denominator of 1800 or less, that's as far as I've gotten.

I did not find anything conclusive on the web, and I am yet waiting to receive feedback from ESA's contact page. In a German speaking IT forum there is something a bit less vague than the FAQ cited in the question:

The reason for the 106.5 degrees is not easy to see. It has to do with the fact that we are (also) doing an astrometric solution on a great circle and for this all angles 360x(n/m) [n, m = small integers] have to be avoided because otherwise one would have a congruence state that would hinder the solution. Values other than 106.5 degrees would also be possible. 90 degrees would be ideal for parallax determination, but unfortunately 90=360x(1/4).

This said, why not simply choosing 91° as $\gcd(360,91)=1$ (with $\gcd$ being the greatest common divisor)?
Maybe it is really only a numerology of the engineers, just like the Mars parachute Morse message. An argument for that would be that Gaia also has also exactly 106 CCD sensors. Or it is simply a logistical issue on how to mount things on the spacecraft. I am really curious what the official ESA statement will be.


What are Flamsteed numbers? Can you tell me about Flamsteed numbers? I just know that it is a specific number used for identification of stars. But what is the theory behind it? In what way is the number given to stars?

Flamsteed designations are assigned by John Flamsteed. He was an English Astronomer.

Flamsteed Designations (Flamsteed Numbers) are numbers assigned to the stars for the purpose of identification and cataloguing.
Flamsteed designation contained 2554 stars.

Flamsteed designations for stars are similar to Bayer designations, except that they use numbers instead of Greek letters.

[ Bayer designation is a stellar designation in which a specific star is identified by a Greek letter, followed by the genitive form of its parent constellation's Latin name. The original list contained 1,564 stars.
For the most part, Bayer assigned Greek and Latin letters to stars in rough order of apparent brightness, from brightest to dimmest, within a particular constellation. Since in a majority of constellations the brightest star is designated Alpha (α), followed by beta for the 2nd brightest and so on.
But this was not the rule always followed by Bayer.
Somtimes by position of a particular star in that constellation, sometimes the order of rising up in the sky (the star rises up early will get earlier number), sometimes he follows mythological stories or data or sometimes he even uses his own way of assigning the number.]

In Flamsteed designations, Each star is assigned a number and the Latin genitive of the constellation it lies in.
e.g. 52 Cancri is star in the Constellation Cancer and assigned a number 52 according to star's right ascension. i.e. there are 51 stars in the Cancer constellation to the West direction of 52 Cancri star or there are 51 stars that rise before/earlier of 52 Cancri star.

The numbers were originally assigned in order of increasing right ascension within each constellation.
[Right ascension is the angular distance measured eastward along the celestial equator from the vernal equinox (it is one of the point of intersection in the space of two imaginary circles: Celestial Equator and Earth's orbit around the Sun. Our Earth is at this point on 21/22 March each year while orbiting around sun.) to the the position of the star. It is measured in degrees  but for simplicity and ease of use, mentioned in the form of hours. That the 360 circle is divided into 24 hours (1 hour = 15 degrees). ]

But due to the effects of precession (Top like rotation of Earth's axis of rotatation in the space mataining a constant angle of 23.5 degrees with verticle) they are now slightly out of order in some places.

Flamsteed catalogue don't cover all the stars visible from Earth with naked eye. Flamsteed's catalogue covered only the stars visible from Great Britain, and therefore stars of the far southern constellations have no Flamsteed numbers.

Reference/Information Source/Quoted From:

1) List of constellations for a list of constellations and the genitive forms of their names: https://en.wikipedia.org/wiki/88_modern_constellations

2)Bayer Designation: https://en.wikipedia.org/wiki/Bayer_designation

3)Flamsteed Designation: https://en.wikipedia.org/wiki/Flamsteed_designation


Topography / elevation data of Mars For a design project I'm looking for elevation data for Mars' surface - the idea is to replicate a surface area of Mars as a physical model, using the elevation data to create a very rough 3d model of that particular area's topography.

Where can I find such data?

Try this link http://geo.pds.nasa.gov/missions/mgs/megdr.html
This should have the data you're looking for, though you may need to process it a bit to get it into a format hat will be usable.


How to visually tell the difference between a planetary nebula and a supernova remnant? If we see a nebula by looking through a powerful telescope, how can we tell whether we are looking at the remnant of a supernova or at a planetary nebula?
Thanks

A Supernova Remnant contains a Black Hole or a Neutron Star while a Planetary Nebula contains a White Dwarf.

Also, Supernovae Remnants are likely to have great velocities, so Doppler is another plausible choice. - Not through telescope though.


How can the temperature of stars can be determined by their color? How do you determine the temperature of a star by color?

I think you are talking about color index and temperature. So, I will briefly explain under that circumstance.

Color index, e.g., B-V is the difference in brightness (magnitude) of the light observed by filters B and V. (The following discussion can be generalized to other color indeces, e.g., U-B or g-i). Filters B and V have effective wavelengths 4450 (bluer) and V 5510 (redder) A (see Wikipedia), respectively, therefore B-V increases when the object becomes red. (Recall that magnitude is more negative for being brighter). The temperature which is derived from the color index is called "color temperature;" this is different to blackbody temperature, or kinetic temperature, or other types of inferred temperatures (see [this 2).

For the color temperature, this is very observational inference. To map from the color index to the color temperature, it is inferred from other well-known object. See Wikipedia for this.

Another way to work around is to transform the color index into the flux ratio, then applying the blackbody function as the inference. For example, from B-V >>> flux(4450A)/flux(5510A) >>> blackbody temperature corresponding to the flux ratio.


Does a week represent something in astronomy? I know that a day represent the time for the earth to make a full rotation around its own axis and a year is the time for the earth to make a full rotation around the sun.

But does a week represent something? And a month?

Thank you.

ps: Not sure what tags I need to use, feel free to add/remove them

The concept of a month derived from the amount of time it takes for the moon to cycle from new moon to new moon (which is roughly 29.5 days). The modern month has experienced changes from this original concept due to trying to fit a standard number of months within a solar year.

If you divide the lunar month into quarters, each quarter is approximately 7 days long. This is believed to be the origin of the seven-day week.


Is it true that we can estimate the mass of a star more accurately if it has a companion star? I remember I hear a theory that if a star has a companion star, we can know the mass of the star more easily, is that true? if it is true, while I understand that by measuring the period of the companion star, we can know the mass ratio of 2 stars,but how can the companion star also help to measure the mass of parent star?

In general, masses can only be precisely estimated for stars in binary systems (asteroseismology and some other indirect techniques can yield estimates too).

There are two broad types of binary where this is possible - astrometric binaries, where the motion of both stars can be seen on the sky and where the distance to the binary is known (e.g. Sirius AB); or eclipsing binaries, where the stars eclipse each other and their motions are inferred from their line of sight velocities, measured by spectroscopy (e.g. YY Gem).

You are correct, that the relative motions of the stars only yields their mass ratio. What is needed to get the individual masses is something that fixes the absolute sizes  of the orbits and the orbital period, which can then be used with Kepler's third law to give the total mass of both stars and hence combined with the mass ratio to give the masses of the individual components.

In astrometric binaries you can measure the angular size of the orbits and calculate an absolute size using the known distance. In close (unresolved) binaries, that size is inferred from the velocities and orbital periods, but also requires knowledge of the inclination of the orbital plane. This is known quite accurately (is close to 90 degrees) only for eclipsing binaries.

I could present the Maths if you really like, but these are quite standard results.


Why shouldn't you look at the Sun through a camera using eclipse glasses? One of the NASA approved eclipse glasses manufacturers has the following warning on one of their eclipse glasses:

  Do NOT look through a camera, telescope or binoculars with ANY eclipse shades on!

I can see how the enlarged image of the Sun through a telescope or binoculars might overwhelm the eclipse glasses' filters.

But how would looking at the Sun using a camera's viewfinder or LCD screen reduce the effectiveness of eclipse glasses (assuming you're not using any zoom function)?

Electronic viewfinders (such as the ones using LCD or OLED screens) are excluded from this discussion. There's no danger of eye injury there, except perhaps danger of damaging the camera's sensor if exposure is very long and the lens aperture is big.

It's the classic or optical viewfinders that are targeted by the warning.

  I can see how the enlarged image of the Sun through a telescope or binoculars might overwhelm the eclipse glasses' filters.

A few observations.

It's really the larger aperture of the instrument that is the problem here. The pupil of your eye is only a few mm in diameter. The objective lens or mirror of a telescope could be anywhere between dozens of mm to hundreds of mm (or thousands for very large instruments). The ratio of areas between telescope aperture and eye pupil is even greater - it's the square of the ratio of diameters.

Let's say your eye's pupil is 2 mm in diameter. Let's say you use a 50 mm aperture telescope (small refractor or binoculars). The diameter ratio is 25x. The area ratio is 625x.

All the light captured by the very large area of the instrument is funneled into your eye through your pupil. With the instrument, now you're getting 625x more energy from the Sun, compared to the naked eye view. It's already dangerous to look at the Sun with the naked eye - with the instrument it's 625x more dangerous. And this is with a very small refractor.

There is a class of "solar filters" that are made to be mounted on the instrument's eyepiece, or after the eyepiece. THESE ARE VERY DANGEROUS! All the increased, focused energy of the Sun is now absorbed by the filter, which can warp, melt, crack, or burst into flames. A filter failure at this point is likely to injure the user. White light filtering for solar observations must always occur ahead of the instrument, not after it.

As for why classic, optical viewfinders were included in that list: it was out of an abundance of caution. Most of those viewfinders do not capture more light than your eye does, but some do. It's better to be safe than sorry. You can't expect everyone to be able to tell whether their viewfinder is dangerous or not. So, in a warning addressed to the general population, just tell them to stay away from it.

Now, if you install a full-aperture solar filter (like the Baader solar film) ahead of the viewfinder, then it's safe to use - provided the filter is attached firmly to the camera and cannot be blown away by some random gust of wind.


Estimates for "undetectable" planets in extra-solar systems As far as I've been able to determine, it appears there are some significant  limitations on our ability to detect exo-planets that are below certain sizes or beyond certain distances from their stars.

It seems that Uranus and Neptune are about as small as we can detect unless the planet is relatively close to its star, and Earth is near the lower size limit we can detect, regardless of proximity to the star.

In other words, our bias toward detecting large planets, and planets close to their stars, means I have been able to find reasonably good data regarding how many of those kinds of planets are generally present in a system.  But since we can't seem to tell (by direct observation) how many small planets are generally present in a system, or how many planets are present that are just too far from their star for us to detect, I'm struggling to find data regarding how many plants of those types are common in planetary systems.

For example: our system has 8 planets, 4 rocky, 2 gas giants, 2 ice giants.  But chances are that if we were observing (with current technology) our own Solar system from some other solar system, we'd probably only see 2 to 6 of them.  We'd almost certainly see Jupiter and Saturn, and we'd almost certainly NOT see Mercury or Mars (they're just too small), Venus and Earth "might be" close enough to the sun for us to see them even though they are on the small end of what we can detect, and Uranus and Neptune also "might be" detectable, even though they are small-ish for their distance from the sun.

My specific question:  How many planets are in a 'normal' or 'average' planetary system, according to humanity's best scientific theories available so far?(beyond what we can simply detect.)  In other words, where can I find reasonable scientific research data sets, models, estimates, theories, evidence, etc., that describe the likelihood of "undetectable" (or very hard to detect) planets being present in a planetary system?

This is my first question on Astronomy stackexchange, so please be gentle, but don't hesitate to provide constructive criticism if I'm doing anything wrong.

This is a partial answer. It is partial because this is an active area of research, with this very topic as one of the prime questions, second I've only chased down the estimates from one method; microlensing, because this method is not biased toward large (in size and mass) planets in close orbits around small stars.

The microlensing method is biased toward more massive planets in Jupiter like orbits i.e. cold planets. It is biased toward distant planets about stars much closer toward the galactic center. This is important because stars there have lower concentrations of heavy elements need to make planets. It is equally sensitive to planets round all sorts of stars, and even free floating planets. It can detect down to Earth mass planets.

Here is a link to a technical summary of the method. The part you are most interested in is section 4.1.3 pages 23. Microlensing Searches for Exoplanets Yiannis Tsapras
This poster paper has a summary plot Fig 3 of planet occurrence rates from various methods Poster: Planet frequency from microlensing observations Arnaund Caasan
There is also a link to an archive of papers on the subject: NASA archive: Planet Occurrence Rate Papers

I do a quick summary of the main results: (microlensing perspective).
Estimating the biases, combined with actual detections it is possible to estimate (or put limits on) the numbers of planets and how theoccurence depends on mass, distance from star.
Snow line == distance from star where water remains frozen. In our solar system it is between Mars and Jupiter. Jupiter's moons have lots of ice. Dimmer stars this is closer in.
1 AU = Earth-Sun distance

Number of planets beyond snow line (Jupiter and beyond in our solar system) is 7 times greater than close in hot planets. In the distance range 0.5 to 10 Au of 20% of stars have Jupiters. 50% have Neptunes, 60% have super earths. There are many stars with more than one planet. So there is about one planet for every star in the galaxy. At least!. About 1/6th of stars with planets have a solar analog (massive planets like Jupiter and Saturn with room for small rocky planets close in).


Distance from W(West) to vernal equinox I am pretty new to astronomy and the project I am currently working on requires some knowledge of equatorial coordinate system.
As long as I have been researching this question, I understood that vernal equinox is the origin of this system. At this moment I am stuck with task of finding distance on celestial sphere between west(on horizon) and vernal equinox.
So, how do you count this distance?

The right ascension of the point on the horizon and due west is LST - 6 hours, where LST is the Local Sidereal Time. The LST can be calculated from knowing the date, time, and longitude. The declination of the west point is 0 degrees. So the angle between that right ascension and 0 hours right ascension (the vernal equinox) is just the difference between the two.


Does all the exoplanets orbit the same direction? In one stellar system like our milkyway, all the planets orbits in the same direction. But are exoplanets orbitting in the same direction as we (earth) do, or is there a higher percentage which orbit in one or the other way? If so what could be the reason for that?

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.


What are the correct symbols for vernal and autumnal equinox? What are the correct symbols for vernal and autumnal equinox?

I have seen people use both ♈ (\Aries) and $\gamma$ (\gamma) to denote vernal equinox. However, I have never seen the symbol for autumnal equinox. Should it be either ♎ (\Libra) or $\Omega$ (\Omega)?

Are ♈ and ♎ the conventional symbols to use?

Are $\gamma$ and $\Omega$ sometimes used just because they resemble the above symbols?

An alternative name for vernal equinox is First Point of Aries. Is there a similar name (First Point of Libra) for autumnal equinox?

The Wikipedia article on astronomical symbols says:

  The zodiac symbols are also sometimes used to represent points on the ecliptic, particularly the solstices and equinoxes.
  Each symbol is taken to represent the "first point" of each sign.
  Thus, ♈ the symbol for Aries, represents the March equinox; ♋, for Cancer, the June solstice; ♎, for Libra, the September equinox; and ♑, for Capricorn, the December solstice.

The article cites Roy and Clarke, Astronomy: Principles and Practice.
This question copies an example usage of ♈ and ♎ from that book.
Stellarium uses the same symbols if you set View: Markings to show the equinoxes and solstices.

$\gamma$ is a poor substitute but clear enough if ♈ is unavailable.
Since $\Omega$ stands for an orbit's longitude of ascending node (angle ♈-☉-☊), it would be a confusing substitute for ♎.

"First point of Libra" is archaic but correct.
These names and symbols represent tropical zodiac signs, not IAU constellations.
♎ is fixed at ecliptic longitude 180° regardless of constellation boundaries.
Its location in Virgo this century doesn't change its name or symbol.


What is "superficial gravity" I have seen the term superficial gravity used and it seems to be equivalent to surface gravity

seen, e.g., here http://arxiv.org/pdf/1701.02295

Is there any difference between superficial and surface gravity? When should one be using one or the other?

The term "superficial" means "on the surface," so "superficial gravity" is the same as surface gravity. As @pela pointed out in the comments, the author of the article is Portuguese and in that language the word for surface is 'superficie.'


Does mean solar time and sidereal time sometimes indicate the same time? I'm trying to wrap my head around the different definitions of time. Since mean solar time depends on the Sun, and sidereal time depends on the stars, and since the position of the Sun relative to the stars changes over the course of the year, does this mean that the difference between the two times increases and decreases over the course of the year?

Once a year, mean solar time and sidereal time will be the same. Is that right?

The 3m56s mean difference between solar and sidereal days is due to the Earth's orbital motion around the Sun.
This adds up to 1 day per year; for every 365¼ solar days there are 366¼ sidereal days.

Sidereal time equals the right ascension of what's on the celestial meridian at that time.
The Sun is on the meridian at 12:00 apparent solar time.
These align at the September equinox, when the Sun is at RA 12h.
Mean solar time is ~7.5 minutes behind apparent solar time on that date and aligns with sidereal time ~1.9 days earlier.