Python controllable amateur astrophotography cameras? I would like to read further about any successes that amateur astrophotographers (or professionals) have had with USB CCD cameras that are controlled by Python scripts. I'd like to script the setting of exposure time, triggering of exposures, downloading and processing of images.

Ideally this would be computer platform-agnostic.

I've recently asked Python interface for DMK 23U274 USB camera for astrophotography but the more I read about that the more I think that it's going to be tough running that particular camera from a Python script.

Where can I read further about this being done, or at least what solutions are known to exist?

I use AutoHotKey scripts to control my observatory. As for the individual devices, cameras focusers etc I use the manufacturers software and control that software with Macro scripts.

I do some minor RS-232 control with AHK.

Below is a video of the scripts starting up my remote controlled observatory.

The video shows the two screens of the computer in the house as the scripts power up the observatory and start the four computers out there. Then load, position and make all the necessary software connection between the computers and equipment.

Example. The color bars of the cameras. The color bar shows the camera video and control connections are working and the script has turned on the color bars.

I use AHK to automate TheSky, MaximDL and a number of other apps. I have a script that will use Maxim to plate solve when finding objects.

I have no idea if or how Python fits in AHK but the question is here.

https://www.youtube.com/watch?v=UN4VoOKOcXo&feature=youtu.be

As I said you are watching the action from inside the house as the four computers in the observatory are powered up all the software is loaded.

You will probably need to watch it a few times before it makes much sense. What you see mostly are the windows of individual apps loading and being reposition. Individual setting in the windows are being changed but for the most part the screen is just too small to show everything.


How "fast" do astronomical events occur? I've seen quite a few news about astronomy but I'm not deep into knowing how it works or how people observe it. It got me thinking: if I want to watch a supernova for example, how long will I have to look for it to watch the whole "event"? Or events like NASA "seeing something coming out of a black hole", were they "looking" at the blackhole when something came out really quickly or did they observe it for months? Or maybe the birth of a blackhole?

Now, I do know black holes that quite a few trillion of years to vanish, I'm just curious how much time does it take for astronomers to observe such events.

A supernova (and many other interesting events) starts abruptly -- a flash of neutrinos lasting just a fraction of a second, but then different aspects of the event play out over minutes, days, hours or years.  The astronomy community has for some years been developing increasingly sophisticates ways to deal with such events early on. Instruments with a very wide field of view (gravitational observatories, neutrino observatories, gamma ray detectors on satellites) detect the events initially and bring first networks of robotic telescopes and then humans and still larger telescopes into the loop to try and get as much information as possible about the violent and fast developing early stages. See for example http://growth.caltech.edu/

Other things are slower. The "something coming out of a black hole" (not really that, just something coming from very close to a black hole) event takes a long time, this was just the first occasion when they'd observed that black hole with the right instrument.


Does the loss of mass create an observable change in a comet's orbit? Comets lose their mass through water evaporation due to close encounter with sun, so my question is does the loss of mass create an observable change in a comet's orbit? What kind of change that may be shrinking orbit or expanding orbit?

This problem was studied in Yu & Zheng (1995), who evaluated the effects of the change in the Sun's mass over time and the change in a comet's mass over time, for the case of Shoemaker-Levy 9, which had recently crashed into Jupiter. Given their mass model for the comet (Equation 6), they found that the Sun's mass loss created an increase in semi-major axis of about 8.5 centimeters per year, whereas the comet's mass loss created an increase in semi-major axis of about 10,000 kilometers per year.

Several things to note:

The comet orbited Jupiter prior to breaking up, so it did not come as close to the Sun as most short-period comets.
Mass loss rates can change over time, depending on the distance from the Sun.
Shoemaker-Levy 9 should not be considered a normal comet, in any sense, given its orbit and eventual destruction.

10,000 kilometers a year, though, is nothing to sniff at. Over the course of an orbit, that can be quite a lot - although keep in mind that longer orbits involve much larger semi-major axes - and I'd argue that it should be observable, given the correct calculations of how the orbit should evolve over time.


Is the age of the solar system equal to the age of meteorites? I've read articles that state the age of the solar system has been determined by measuring the amount of decay of isotopes in some meteorite, which has demonstrated that  the lead isotopes in the meteorite have cooled and locked into place some number of years ago.

However, if a yellow sun can not create heavier elements, wouldn't it be correct to say the age of an meteorite dated this way is how long it's been since the asteroid was ejected from a blue or heavier star? The sun may have started burning many years after that.

The radiometric dating of meteorites tells you when certain radioactive species were incorporated into that body. It is extremely difficult to see how that material could have got there at a later time, so the age you get is the age of formation of that body.

Meteorites do not form in stars. They require considerably cooler conditions, but they also require a certain density to stick together. Those conditions are found in a protoplanetary disk. So you might well ask, well maybe the meteorites were formed around another star and then somehow they have ended up in orbit around our Sun.

This is quite unlikely. Very similar ages have been determined for some tens of meteorites and that age happens to be just a touch older than the oldest rocks that can be found on the Earth and the Moon. There is no big age spread among meteorites. That age also coincides with the age for the Sun deduced from physical models (though this age is less accurate because of some remaining uncertainties in the physics, but it is not hundreds of millions of years out).

The obvious inference is that these meteorites were among the first solid bodies formed in the protoplanetary disk that ultimately formed the Earth and Moon and hence represents a lower limit to the age of the solar system. It depends where you wish to start the clock; the formation of the first solid bodies seems as good a place as any and models for disk formation and evolution suggest that these solid bodies do form within about a million years of the molecular cloud that formed the Sun beginning to collapse.


If the Mars polar caps were melted, how deep would the oceans be? If the Martian polar ice caps melted, given a surface temperature and pressure that allow for liquid water, how large would the oceans be? Would they match their ancient glory, or would they be smaller?

The oceans would be smaller because Mars has lost most of its water. From Wikipedia's article on the Martian polar ice caps:

  Evidence that Mars once had enough water to create a global ocean at
  least 137 m deep has been obtained from measurement of the $HDO$ to $H_2O$
  ratio over the north polar cap. In March 2015, a team of scientists
  published results showing that the polar cap ice is about eight times
  as enriched with deuterium, heavy hydrogen, as water in Earth's
  oceans. This means that Mars has lost a volume of water 6.5 times as
  large as that stored in today's polar caps.

  The water for a time may have formed an ocean in the
  low-lying Vastitas Borealis and adjacent lowlands
  (Acidalia, Arcadia and Utopiaplanitiae). Had the water ever all been
  liquid and on the surface, it would have covered 20% of the planet and
  in places would have been almost a mile deep.

So if all the water melted (and the atmospheric pressure allowed liquid water) the ocean would cover roughly 6% of the planet, and be roughly half as deep as the original ocean. This crude estimate totally ignores the topography of the Martian surface. In the comments, David Hammen writes:

  Given the marked unevenness of Mars's surface, if both of its ice cap melted completely the resulting water would cover a fraction of its northern hemisphere, plus a few depressions such as the Hellas Planitia (8 kilometers deep!), but to a depth much greater than 137 meters.

Heavy water is slightly heavier than normal water, so it evaporates slower and has a higher boiling point. So in a mixture of normal and heavy water the normal water evaporates at a slightly faster rate, which increases the relative concentration in the unevaporated water.

That same article mentions that the total volume of both Martian polar caps is around 3.2 million cubic kilometres. Most of that ice is water ice, but some of it is frozen carbon dioxide (aka "Dry ice"). The article doesn't state the proportions, but it says

  During each year on Mars as much as a third of Mars' thin carbon
  dioxide ($CO_2$) atmosphere "freezes out" during the winter in the
  northern and southern hemispheres.
  [...]
  Analysis of data showed that if these deposits were all changed
  into gas, the atmospheric pressure on Mars would double.

Note that not all of Mars's water is in its polar caps, some may be frozen underground, especially near the poles, and some underground water may even be in a liquid state, due to a high salt concentration.

For further details, please see the linked article.


Q: star scaling calculation If in our neighborhood of the Milky Way, the typical spacing between stars is 1 parsec, and the relative velocities between neighboring stars are of order 20 km/sec, what would the spacing between stars be in a scaled-down model in which stars are replaced by grains of sand (1 mm in diameter)? How fast would these sand grain sized stars be moving? [This is a quiz question from The Teaching Company's Introduction to Astrophysics by Joshua Winn.] The given answers are 44 km and 0.9 m/s.

I used the size of the Sun as a typical star (perhaps not accurate) and calculated the ratio of a sand grain's diameter to the Sun's diameter. I then multiplied this by 1 parsec (= 3.09 x 10^16 m) to get 21 km. This is about half the given answer so I thought maybe our Sun is not average sized.

I used the same calculated ratio and multiplied it by the give star velocity and got 1.43 x 10^-5 m/s. This is way off the given answer. What am I doing wrong? Is my calculation for the first part also wrong?

The average str is about half the size of the Sun. So I get 44 km.

20 km/s is equal to $6.47 \times 10^{-13}$ pc/s, but we've just established that a pc is about 44 km in your sand grain model. So the speed is $2.85\times 10^{-10}$ m/s.

Just a double check - to travel 1 pc at 20 km/s would take $1.5\times 10^{10}$s.

To travel 44 km at $2.85\times 10^{-10}$ m/s takes $1.5 \times 10^{10}$ s.

Check.

I have no explanation for you as to why it doesn't agree wth your answer or with Joshua Winn's. But I suggest you read his problem carefully, because I rather fancy he might have done something like say imagine 1 year is equal to a second or something... perhaps.


Why are only the poles of Mars frozen, despite too cold temperatures on lower latitudes? Actually the entire red planet is very cold, cold enough for ice and frost to form not only on the poles. Temperature rises above 0 deg Celsius in summer only. Why then are just the poles frozen, and not lower latitudes during most of the Martian year? Is it because most of these areas perhaps aren't within the triple point of water, unlike the poles?
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Just like a cold winters night on Earth does not guarantee snow, so does a cold Mars not guarantee ice everywhere.
In order for ice or snow to form, the moisture that is later solidifying has to be available at the cold location. Moisture in this case is $CO_2$ with traces of water, leading to an ice cap made of $CO_2$-ice, with a ring of $H_2O$-ice 'trailing' around its outer radius.
The atmosphere is in fact only cold and dense enough for $CO_2$ to solidify at the poles, so no mystery there. Otherwise the entire atmosphere would condense out into a thin layer of $CO_2$-ice. To understand the formation of the $H_2O$ ice, we have to take into account how the water is transported inside the martian atmosphere.
Mars possesses only one global circulation cell (see early modelling on this by Haberle et al., (1993), and a recent review on the martian climate for the interested reader by Read et al., (2015)).

((Figure from Haberle et al.,(2015))
The reason for the single cell is a combination of Mars' small size and rotation period, as outlined in this answer. The effect of the single cell is that the condensable moisture is only transported essentially pole-to-pole directly, with only small variations of ~10-15 degrees latitude during the martian year.
Hence, no other location on Mars than the poles can accumulate a significant enough vapour pressure in order to form ice.


Why is NEOWISE used for galactic observation? I've been studying images from the WISE and NEOWISE missions. WISE used infrared cameras to survey the entire sky. When its primary mission ended in 2010, it was repurposed to look for Near-Earth Objects (described here). What changes to the spacecraft and software were needed in order to change its mission profile? Does it use the same cameras and instruments now as it did when performing a whole-sky survey? My thinking is that a whole-sky survey requires different equipment than a survey to look for small, fast-moving (compared to the stars) asteroids.

After WISE ran out of cryogens to cool its detectors it was re-purposed to look for asteroids.  This was possible because the detectors for 3.4 microns and  4.6 microns could work at higher temperatures than the other detectors and passive cooling gets the spacecraft to 74 K.  It had already finished an all sky survey so there were no new deep space targets for it to study and, with only two detectors, the science would not be as significant, but it has been great as a detector of moving objects like asteroids since, for this, just one band-pass is enough.


trouble installing drizzlepac I'm trying to install drizzlepac to follow this tutorial on aligning hst images: https://spacetelescope.github.io/notebooks/notebooks/DrizzlePac/align_to_catalogs/align_to_catalogs.html
However, I get the following errors:
   ERROR: Command errored out with exit status 1:
   command: 'e:\ProgramData\Anaconda3\python.exe' 'e:\ProgramData\Anaconda3\lib\site-packages\pip\_vendor\pep517\_in_process.py' get_requires_for_build_wheel 'C:\Users\FAILTZ~1.DES\AppData\Local\Temp\tmpl00kr2fm'
       cwd: C:\Users\FAILTZ~1.DES\AppData\Local\Temp\pip-install-1l1f4o5x\drizzlepac
  Complete output (26 lines):
  Traceback (most recent call last):
    File "e:\ProgramData\Anaconda3\lib\site-packages\pip\_vendor\pep517\_in_process.py", line 280, in <module>
      main()
    File "e:\ProgramData\Anaconda3\lib\site-packages\pip\_vendor\pep517\_in_process.py", line 263, in main
      json_out['return_val'] = hook(**hook_input['kwargs'])
    File "e:\ProgramData\Anaconda3\lib\site-packages\pip\_vendor\pep517\_in_process.py", line 114, in get_requires_for_build_wheel
      return hook(config_settings)
    File "C:\Users\FAILTZ~1.DES\AppData\Local\Temp\pip-build-env-ey64g5zm\overlay\Lib\site-packages\setuptools\build_meta.py", line 147, in get_requires_for_build_wheel
      config_settings, requirements=['wheel'])
    File "C:\Users\FAILTZ~1.DES\AppData\Local\Temp\pip-build-env-ey64g5zm\overlay\Lib\site-packages\setuptools\build_meta.py", line 127, in _get_build_requires
      self.run_setup()
    File "C:\Users\FAILTZ~1.DES\AppData\Local\Temp\pip-build-env-ey64g5zm\overlay\Lib\site-packages\setuptools\build_meta.py", line 249, in run_setup
      self).run_setup(setup_script=setup_script)
    File "C:\Users\FAILTZ~1.DES\AppData\Local\Temp\pip-build-env-ey64g5zm\overlay\Lib\site-packages\setuptools\build_meta.py", line 142, in run_setup
      exec(compile(code, __file__, 'exec'), locals())
    File "setup.py", line 23, in <module>
      check_call(['git', 'clone', 'https://github.com/spacetelescope/relic.git'])
    File "e:\ProgramData\Anaconda3\lib\subprocess.py", line 358, in check_call
      retcode = call(*popenargs, **kwargs)
    File "e:\ProgramData\Anaconda3\lib\subprocess.py", line 339, in call
      with Popen(*popenargs, **kwargs) as p:
    File "e:\ProgramData\Anaconda3\lib\subprocess.py", line 800, in __init__
      restore_signals, start_new_session)
    File "e:\ProgramData\Anaconda3\lib\subprocess.py", line 1207, in _execute_child
      startupinfo)
  FileNotFoundError: [WinError 2] The system cannot find the file specified
  ----------------------------------------
ERROR: Command errored out with exit status 1: 'e:\ProgramData\Anaconda3\python.exe' 'e:\ProgramData\Anaconda3\lib\site-packages\pip\_vendor\pep517\_in_process.py' get_requires_for_build_wheel 'C:\Users\FAILTZ~1.DES\AppData\Local\Temp\tmpl00kr2fm' Check the logs for full command output.

This was using pip within an anaconda prompt. If I try to use conda, it can't find the package.
I'm extremely new to python, so I'm sure I'm doing something relatively basic wrong! But I've been bashing my head against this for a while.

As I understand it, you should not be using pip with Anaconda or you will get into a mess. Unfortunately, that's not the main issue here.
The workbook you linked has a banner at the top that mentions an initialization workbook, which is on this page. On that page it states that "Users unfamiliar with Anaconda/Astroconda should see the documentation."
I am unfamiliar with Astroconda, so I went to the documentation page which lists the following requirements:

System Requirements

64-bit Intel/AMD processor (x86_64)
64-bit Linux (glibc ≥ 2.12) or Mac OS X (≥ 10.7)
BASH or ZSH as your default shell environment (T/CSH is NOT supported)

I see from your command line that you are using this on Windows. That is not listed as a supported operating system: the software is Linux/Mac only.


If Alpha Centauri A's solar system exactly mirrored our own, what would we be able to detect? Suppose there was an exact replica of our solar system 4.4 ly away (people included). What would we be able to detect and with what telescope(s)? Which planets? Could we detect radio transmissions and/or any atmospheres?

I assume detection would be optimal if we were co-planar with the other star's ecliptic, so what would we see in the best- and worst-case (90°-view?) scenarios?

Post Script: a few months later I asked something like this during a von Karman lecture by Neil Turner.

This is a broad question and too broad for me to answer comprehensively. It should be broken down into doppler methods, transits and direct imaging; and that's before we get to questions of detecting Kuiper belts, radio emission etc.
I'll stick for the moment with what I know about detection of planets using the doppler wobble technique.
Doppler Technique
The reflex radial velocity semi-amplitude of a star for the case of a planet of mass $m_2$ orbiting a star of mass $m_1$, in an elliptical orbit with eccentricity $e$, and orbital period $P$ and with an orbital axis inclined at $i$ to the line of sight from Earth is:
$$  \left( \frac{2\pi G}{P}\right)^{1/3}\frac{m_2 \sin i}{m_1^{2/3}} (1-e^2)^{-1/2}.      $$
A (very) detailed derivation is given by Clubb (2008).
So I built myself a little spreadsheet and assumed that all the planets were seen optimally at $i=90^{\circ}$ (they could not all be seen optimally, but the smallest inclination would be about $i=83^{\circ}$ for Mercury, so it doesn't make too much difference) I'll also assume the mass of Alpha Cen A is about $M \simeq 1.1M_{\odot}$.
The results are

Planet
RV semi-amplitude (m/s)

Mercury
$8.3\times 10^{-3}$

Venus
$8.1\times 10^{-2}$

Earth
$8.4\times 10^{-2}$

Mars
$7.5\times 10^{-3}$

Jupiter
$11.7$

Saturn
$2.6$

Uranus
$0.28$

Neptune
$0.26$

The limits of what are possible are well illustrated by a planet around Alpha Cen B, claimed to be in a 3 day orbit and with a mass similar to the Earth (Dumusque et al. 2012, and see exoplanets.org). The radial velocity semi-amplitude detected here was $0.51\pm 0.04$ m/s, and some spectrographs, notably the HARPS instruments, are routinely delivering sub 1 m/s precision. Thus Jupiter and Saturn would be detectable, Uranus and Neptune are right on the edge of detectability (remember you can average over many RV observations), but the terrestrial planets would not be found (Earth detections would require precisions below 10 cm/s. Remember also that the weaker signals would have to be dug out from the larger signals due to the Jupiter- and Saturn-like planets.
However, there is a second limitation: to find a planet using the doppler method you need to observe for at least a significant fraction of the orbital period. Given that current m/s precisions have been available for only $\sim 5$ years, it is unlikely that Saturn would yet have been detected.
A picture that illustrates the situation can be obtained from the exoplanets.org website, to which I have added lines that approximate where RV semi-amplitudes would be for 10 m/s and 1 m/s precision (assuming the Alpha Cen A mass and circular orbits). I've marked on the Earth, Jupiter and Saturn. Note that few objects have been discovered below the 1 m/s line. Also note the lack of planets between the 1 and 10m/s lines with periods longer than a couple of years - the recent increase in sensitivity has yet to feed through to lower mass, longer period exoplanet discoveries.

In conclusion: only Jupiter would have been so far found by the doppler technique.
Transit techniques
I'll also add a few comments about the transit technique. Transit detection will only work if the exoplanets orbit such that they cross in front of the star. So high inclinations are mandatory. Someone who is better at spherical trigonometry should use the published data for the solar system to work out how many (and which) planets transit in some highly optimal orientation. Given that the planets have orbital inclinations with a scatter of a few degrees, then some straightforward trigonometry and a comparison with the solar radius, tells you that these orbits will generally not all transit for any particular viewing angle. Indeed a number of the Kepler-discovered multiple transit systems are much "flatter" than the solar system.
The Kepler satellite is/was capable of detecting very small transiting planets thanks to its very high photometric precision (the dip in flux is proportional to the square root of the exoplanet radius). The picture below, presented by the NASA Kepler team (slightly out of date now), shows that planetary candidates have been discovered that are down to the size of Mars. However these tend to be in short period orbits because a transit signal needs to be seen a number of times, and Kepler studies this patch of sky for about 2.5 years (when this plot was produced).
So from this point of view, possibly Venus would have been seen, but none of the other planets could be confirmed.
However, there is a wrinkle. Alpha Cen A is way too bright for these kinds of studies and way brighter than the Kepler stars. You would have to build a special instrument or telescope to look for transits around very bright stars. Some of this work has been done by ground-based surveys (mainly finding hot Jupiters). A new satellite called TESS (Transiting Exoplanet Survey Satellite, launched in April 2018) is a two year mission, focused on finding small planets (Earth-sized and bigger) around bright stars. However, most of its targets (including Alpha Cen) are only observed for a 1-2 months, so only the inner parts of their planetary systems will be probed.


How do astronomers at SETI know if they received intelligent radio signals from other planets? Do they expect to see wave modulations like the ones we use in our technology? what do they expect the waves to be like exactly?

I'm not sure what you mean by "wave modulations".

Loosely speaking, there's 2 things to look for regarding alien intelligence/technology.   One is a signal of some kind and the other is a very large or very energetic structure.

The most obvious place to look for a signal is in the water-hole.  Good article on that here.  Because of the relatively low interference in that bandwidth, it's a good region to send a signal from one solar-system and hope it gets picked up by another.   Stellar distances are so large that this still requires a very large output and very sensitive receiving devices.   People have been listening for water-hole signals for decades, with no luck, outside of the extremely inconclusive "Wow" signal.

The other thing they can look for is mega-structures but that's much harder.   We have a hard enough time seeing entire planets in other solar-systems and the idea of an alien technology building something as large as a planet, but distinguishable . . . a bit of a stretch.

Planets are mostly those are observed by seeing wobbles or changes in the star's orbit or brightness as the planet passes in-front of them.   To see an alien structure and recognize it as something manufactured, not natural, it would need to be very large.  See silly article.  Outside of a structure as large as a dyson-sphere, we don't have the means of seeing an alien structure.

Finally, and to your question.  There is no technological "wavelength" to look for, light is light.  We transmit our cell-phone signals at about 2 gigahertz but an alien species might use 5 or 8.  There are ranges that are best for traveling between atmosphere and satellite and regions of comparatively low interference that are best for communication across a solar-system, but at the distance of several light-year's it's difficult to overcome the background radiation.   We could recognize an intelligence signal if we saw one.  Any information or language carrying signal would have a different pattern than natural signals, but the hard part is finding the signal in the first place, over such vast distances.

Basically, Seti is limited to looking for the extremely large and so far, (at least as far as I know), nothing even remotely close to definitive has ever been observed.   They don't have the means to observe most technology, assuming there is technology to be seen.  Only the most enormous mega-structure could maybe be observed, and only under the right conditions.

As our telescopes get better they might find something.  The atmospheric signature of a technological alien society might become visible to the James Webb telescope, but only for planets that transit their stars, not all planets and that might only offer possibilities, nothing definitive.

Signals like our television, cell phone and satellite broadcasts, those would be extremely hard to pick up from light-years away.  The signals are too weak and over light-years of distance, the background radiation of space would grow too loud by comparison.   Likewise the visible light from the night-side of Earth, while we can see that from the space station orbiting the Earth, observing that kind of light from light years away would require a huge telescope, many times larger than anything we have.   I don't have the knowledge to give you all the specifics on how hard to see or how big a telescope but much bigger than anything we have.   An Alien race with a telescope a few miles across might be able to observe the light from our cities on the night side of Earth.   We don't have the ability to observe anything like that ourselves.


How does energy (from fusion reactions still inside the sun) still have gravitational attraction? In this answer they say:

  While the conversion of mass matter† to energy in the Sun's core now represents a loss of mass proper matter, it turns out that that energy (trapped in the Sun and slowly diffusing towards the surface) will have the same gravitational attraction as the matter it came from until it actually escapes the Sun!

Now I expect that when mass is converted to energy, it has become an electromagnetic wave that can propagate through matter or empty space.

My question is: How does energy (from fusion reactions still inside the sun) still have gravitational attraction?

I think it is best not to think of energy having a gravitational attraction, rather that energy, like mass, (since they are equivalent keeping in mind c) will curve spacetime.  Electric charges, magnets, and electromagnetic fields will curve spacetime, Scientific American. So whether it is mass or energy, they both will affect the curvature of spacetime, and the sun's total impact on spacetime will not change.


Why 2014-2015-2016 were the years when the number of discovered exoplanets increased the most? So, if we look into the cumulative number of exoplanets discovered by year, you can appreciate a huge rise in the number of discovered exoplanets in the years 2014 and 2016:

Credit to NASA Exoplanet Archive
Also, I don't know if it's by coincidence or not, but the number of publications with the word "exomoon" (a satellite of an exoplanet) seems to increase dramatically by 2015:

Credit to app.dimensions.ai. I might be wrong, but I thought that we would see this rise in the years following 2010, like 2011 or 2012, when the Kepler telescope was beginning this hunt, and in the case of the publications, Kipping (2009a, 2009b) were very important around that time too, so is there any reason of why 2014-2015-2016 seem to be the most important years?

The big increase is due to the Kepler spacecraft. Kepler is responsible for the majority of the discovered exoplanets (https://exoplanets.nasa.gov/keplerscience/), so its data releases will have a big effect on the planet’s discovered by year. It’s easy to see in the plot attached (From: https://link.springer.com/article/10.1007/s12145-020-00464-7 ) how the years you mentioned match the announcement of discovered exoplanets using Kepler’s and K2 (same spacecraft).
On why it is ~2014 and not 2010/2011/2012, from my understanding that's when the campaigns' data started being released (https://keplerscience.arc.nasa.gov/k2-data-release-notes.html#k2-campaign-1).


Can the theory of multiverse explain dark matter? If there are many universes, they must have a gravity effect on each other. Since dark matter is measured through gravity, but we don't know the mass behind it, isn't it possible that it comes from another universe?

This is a bit of a speculative question, but I can answer it.

Dark matter has been observed in galaxies, and the distributions of dark matter in galaxies have also been measured. It seems clear that the matter is firmly in our universe - we just can't detect it with electromagnetic radiation.
*However, there are ideas (extremely speculative) that dark matter is from a different spatial dimension. It sort of explains why gravity is so weak (it "leaks off" through the other dimension(s)). Note, though, that this highly, highly speculative.

There are ideas like this in string theory, too. Some say that we are trapped on a brane, a section of space-time with three spatial dimensions. However, there are one or more additional dimensions. All of these dimensions are known as the bulk. Gravity can "leak off" through these extra dimensions. Some theories take this to the extreme, and say that there exist many universes (of a certain number of dimensions each, but with less dimensions that those of the bulk) that are branes, just like ours. These have been considered the source of phenomena such as dark energy - in which case your scenario is within the realm of plausibility.

All of these ideas, however, are speculative and have no evidence in their favor.


How tightly clustered are the CHIME-detected FRB repeats? In the past few days, journalists have been playing the usual "X could be aliens" game. This time, X is repeating FRBs from a galaxy 1.5 billion light years away. (Or is it 1.5 billion years ago? The proper/conformal time distinction isn't clear in pop sci.) I couldn't find the galaxy's name anywhere, but then I suppose a galaxy that distant might not have been named, even if it's now of some interest. Still, the claim of "repeated" bursts (regardless of their cause) made me wonder: what is our tightest empirical upper bound on how far apart the detected bursts' sources are?

I'm prepared to believe we can see 1.5 billion light years away, clearly enough to pin two events to the same 100,000-light-year wide galaxy, with no others within 2 million light years of it (these are plausible example numbers based on comparing the Milky way to Andromeda). They're small fractions of 1.5 billion light years, but not so small we couldn't resolve an image that clearly. On the other hand, I sincerely doubt we can prove any two bursts that distant are from the same star. I presume what's notable about repeat FRBs is they're so close together in time we expect the source to be common, but I'm curious as to how tightly observations constrain that.

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


What is the coolest white dwarf known? Okay, so in this ArXiv report from 2014, scientists discovered the coolest white dwarf, with a temperature of below 3000 K. However, as this report was from over 6 years ago, I think this information may be outdated. So, as of right now, what is the coolest white dwarf, and where can I find more information about this?

The one from 2014 is still the record holder I believe - in the sense that it is reasonably convincing that the unseen companion of the pulsar PSR 2227-0137 is consistent with being a white dwarf with a surface temperature below 3000 K.
It is worth considering why such objects might be difficult to find.
(1) It is only the highest mass white dwarfs that have had time to cool below 3000 K in the lifetime of our Galaxy. The reason for this is that more massive white dwarfs are smaller and have much higher densities. This enables them to crystallise in their interiors at high temperatures and the heat capacity of a crystalline solid becomes much lower than a gas at the same temperature, meaning that they have less heat left to radiate away.  High mass white dwarfs are rare; they are the endpoints of the evolution of 6-8$M\odot$ stars, which are much less common than the 1-6$M_\odot$ stars that produce most of the lower mass white dwarfs in the galaxy.
(2) They are almost invisible. Temperatures below 3000 K are getting down into brown dwarf territory. Now we've found thousands of brown dwarfs, but these are (a) likely to be more common than massive white dwarfs (about 1 for every 5 stars) and (b) they will be about 20 times the radius of a massive white dwarf and therefore emit 400 times the flux at the same temperature. Thus it will be very difficult to find free-floating cold white dwarfs, especially if they are intrinsically rare.
(3) But can't we find them in binary systems? If the companion star is $>1 M_\odot$ then it will have lived and died in the time it took the massive white dwarf to cool and the system would have to be a double-degenerate. i.e. A neutron star + white dwarf or two white dwarfs. Many such systems are known, but I think we come back to the fact that massive white dwarfs are rare, and if they are more massive than say $1.15M_\odot$ it becomes difficult to distinguish them from neutron stars if all you have is the dynamics of the system. A more promising line of attack would be to look for them as binary companions to old K and M dwarfs. But this would imply that the original binary system would have had a very high mass ratio - a fair fraction of these sytems would have become unbound as the white dwarf progenitor lost its (and most of the system's) mass, and very high mass-ratio binary systems are not that common. My guess is that the Gaia astrometric survey may be a good source of candidate cold white dwarfs because it will identify the binary motions and secondary (really, the unseen primary) masses for many, many K and M stars. Any K/M star with an unseen companion that is more massive than $1M_\odot$ will be of interest.


Could the closest approach of stable orbits of two earth sized planets allow life to jump between? How close could two earth sized celestial bodies get in their nearest approach either as planets or moons around a larger planet?

Close enough for atmospheric interaction in a stable orbit at nearest approach, or for non-space-fairing species to somehow make the jump without cosmological assistance (aka asteroid impact)?

It seems just from using the roche limit of an earth to earth sized bodies any approach under ~20,000 km would be unstable in the long term, which is orders of magnitude beyond a normal atmosphere size.

From comments

no, you can't have orbits anything like that close. Nor can you have a large moon anything like as close as you are thinking. 2 earth sized planets near each other's Roche limits would cause catastrophic earthquakes. Independent orbits would need, probably a few million miles apart for stability. Orbiting each other you could work them to perhaps, earth-Moon distance or even closer assuming they were tidally locked to each other to avoid too much disruptive tidal forces.

BUT if this is for a story: don't forget the other direction. One could say it is perhaps generally agreed today that "panspermia" can happen: life - microbes - via asteroids/etc can travel large astronomical distances. If we learned today that Mars has life - and indeed it's the "same" as Earth life - we'd assume that it travelled via ejecta - the Mars-Earth distance is no problem for such a putative story

Comment converted to community wiki


Galactic Habitable Zone Do galaxies have habitable zones the same as stars do? Say in a galaxy with a very active nucleus producing a lot of heat and radiation, would there be a point at which no star's planets could harbor life do to the effects of the black hole? Also would there possibly be a habitable zone for galaxy clusters? If there were many galaxies with extremely active galactic nuclei condensed closely together would this possibly also hinder the evolution of life?

I'll be honest. I looked up Galactic Habitable Zone on Wikipedia. The idea was introduced in 1986 and expanded upon in the book "Rare Earth" by Brownlee and Ward. Essentially the idea is that in the bulge there is too much activity (supernovae, etc.) and in the halo of the galaxy there is not enough heavy elements (carbon, etc.) for life to exist. This was in reference to complex life "as we know it" (Check out another Peter Ward book entitled "Life As We Do Not Know It"). Stellar habitable zones are based on the presence of liquid water. This definition would greatly increase the theoretical "Galactic Habitable Zone". I think we need to look at a region and calculate all of the pluses and minuses of that specific region then keep our minds open.


How do I prevent or reduce shake when observing with binoculars? I have some 16 x 50 binoculars that my parents bought me years ago. Recently I have tried to do some basic lunar and planetary observations with them but I am seriously struggling with shaking. This was most noticeable tonight while attempting to observe the ISS.

While I accept that keeping the image of a moving target stable is going to be next to impossible, I would like to know how I could reduce my shaking while observing the moon and other objects.

Is anyone able to provide some simple techniques to reduce shaking?

Don't hold the binoculars in your hands. Humans are made of meat. wobbly wobbly meat. There's apparently devices that let you mount binoculars on tripods - (this google search would be a start). Those and a tripod would probably be helpful in decreasing shake.

I suppose it would affect mobility a little but that's a tradeoff.


Get orbital period of exoplanet from light curve using astropy.timeseries I am using astropy and I would like to calculate orbital period of an exoplanet by its star's light curve. I follow tutorial in astropy docs and I use data from Kepler in Nasa Exoplanet Archive.

There is KIC 10666592 b (expected period: 2.2 d) in tutorial, it works for me. But if I try another planet (e. g. KIC 10000941 b (expected period: 3.5047 d)), it gives bad results.

Load and plot light curve

from astropy.utils.data import get_pkg_data_filename, download_file
from astropy.timeseries import TimeSeries, BoxLeastSquares
from matplotlib import pyplot as plt
import numpy as np
from astropy import units as u

# First url is for KIC 10666592, the second one is for KIC 10000941 b.
URL = "http://exoplanetarchive.ipac.caltech.edu:80/data/ETSS//Kepler/005/755/19/kplr010666592-2009131110544_slc.fits"
#URL = "http://exoplanetarchive.ipac.caltech.edu:80/data/ETSS//Kepler/005/159/31/kplr010000941-2009166043257_llc.fits"

filename = get_pkg_data_filename(download_file(URL))
curve = TimeSeries.read(filename, format="kepler.fits")

plt.plot(curve.time.jd, curve["sap_flux"], "k.", markersize=1)

Apply periodogram

periodogram = BoxLeastSquares.from_timeseries(curve, "sap_flux")
results = periodogram.autopower(0.2 * u.day)

plt.plot(results.period, results.power)

Calculate orbital period

best = np.argmax(results.power)
period = results.period[best]
print(period)

$$\begin{array}{|c|c|c|} \hline
 & \text{KIC 10666592 b} & \text{KIC 10000941 b} \\ \hline
\text{Calculated period} & \text{2.2055172 d} & \text{13.595231 d} \\ \hline
\text{Expected period} & \text{2.2 d} & \text{3.5047 d} \\ \hline
\text{Result} & Correct & Incorrect \\ \hline
\end{array}$$

Why period for second planet is wrong? I'm doing something wrong? How can I get the correct result?

The second light curve you show has no obvious periodic behaviour and I cannot see any sign of a planetary transit. The period-finding algorithm appears to be working correctly.

The planet (if it exists) is supposed to be one of the smallest planetary candidates found by Kepler and will have a barely detectable transit (depth of order 0.004%). The small light curve variations you see are either from the star or remaining instrumental artefacts. If you wish to verify or otherwise the results published by Morton et al. (2016), who show no light curves, you will have to do considerably more work to "detrend" the other variability in the light curve.


Which binocular should I buy to be able to see Saturn and its rings? I am new to Astronomy and never had a binocular to look at the sky, I read somewhere that already with a 30x optical zoom should be possible to spot Saturn and see its rings, is that true?
I have a budget of up to 400 bucks, which binocular would be the best in this range to see Saturn and other objects in the night sky?

In your choice, you should also consider the binocular stand, your experience and how good your sight is. Also, with that money, you can also buy this telescope. With it, you can also see all the Messier objects. I have SkyWatcher 10 ", and I am quite satisfied with it. For cheap, but good telescopes, look at this site.
I haven't quite answered your question, but I think, that this will open more options.
Also, as @Mick has mentioned, this question is quite similar to yours, but you've said how to see Saturn and other objects in the night sky, so I searched for a link for these other objects. If I'd have 400 bucks, then I'll definitely go for a telescope.
Hope this helps!
P.S.
Try to buy as fast as you can because Saturn can be seen earlier and earlier in the evening. Now (middle of Juny 2020) it is observable at approximately 00:00—morning, but soon, it will be observable at 21:00!


Latitude, longitude and altitude of Ckoirama Observatory; where can things like this be looked-up? This answer to Is Starlink 1130 (Darksat) really dimmer? links to the March 16, 2020 paper in arXiv First observations and magnitude measurement of SpaceX's Darksat which finds only a modest decrease in brightness:

  Results. The calibration, image processing and analysis of the Darksat Sloan g’ image gives an estimated Sloan g’ magnitude of 7.57±0.04 at a range of 976.50 km. For STARLINK-1113 an estimated Sloan g’ magnitude of 6.69±0.05 at a range of 941.62 km was found. When scaled to a range of 550 km, a reduction of (55 % ± 4.8 %) is seen in the reflected solar flux between Darksat and STARLINK-1113.

The v1 preprint says

  ...using the longitude(24.1° west) and latitude (69.9°south) of the Ckoirama observatory...

which puts the Ckoirama observatory under about 4500 meters of freezing antarctic water (first image).

If I invert the latitude and longitude assignments it returns to the Atacama desert where it belongs. If I instead type "Ckoirama observatory" in maps.google.com I get a pin on a map near 24.0894 S, 69.9305 W but the satellite image only shows a hole in the ground. (second image)

click images for full size

The sidebar says:

Antofagasta, Chile W369+6Q Yungay, Antofagasta, Chile
astro.uantof.cl
+56 55 263 xxxx
Open now:  Open 24 hours

That link leads to http://www.astro.uantof.cl/research/observatorios/ckoirama-observatory/ which is very informative but has no latitude, longitude or altitude information. 5 Awesome Observatories Around The World mentions that Ckoirama is;

  the first observatory built by Chilean engineers

and links to Astrotourism in Chile but no specific information.

Question: What are the latitude, longitude and altitude of Ckoirama observatory? Is there a central location where coordinates like these can be looked up?

Near the pinned location, at 24.0893°S 69.9306°W,
Bing Maps shows a building like the one in the University of Antofagasta photos,  with signs of recent construction.

It's probably a matter of time until Google updates their satellite imagery in that region.

The Minor Planet Center maintains a list of observatory codes with positional information.
A related page there says:

  The KML file of observatory-code locations has been removed, following a request from some Spanish observers concerned about security at remote, unattended sites.


Do the "schwarzschild metric" or the "isotropic metric" fit the Event Horizon telescope data on Messier 87 better? The "Schwarzschild solution" in "Schwarzschild metric" looks like:

$$\text -c^2d\tau^2=-\left(1-\frac{2GM}{rc^2}\right)c^2dt^2+\left(1-\frac{2GM}{rc^2}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$

Using this metric you get certain values for the radius of the "Schwarzschild radius", the "photon sphere" and the "least stable circular orbit", among other things. When the Jet Propulsion Laboratory or other institutions are dealing with orbital dynamics in the solar system, they instead use the "isotropic metric" or at least a first order expansion of the isotropic metric to calculate relativistic effects on the orbits. Now the Schwarzschild solution in isotropic coordinates looks like:

$$ -c^2d\tau^2 = -\left(\frac{1-GM/2rc^2}{1+GM/2rc^2}\right)^2c^2dt^2 +(1+GM/2rc^2)^4[dr^2 +r^2( d\theta^2 +\sin^2\theta d\phi^2)]$$

Now in isotropic coordinates the "Schwarzschild radius", the "photon sphere", the "least stable circular orbit" among other things take other values.

Question 1: Now that we have the measurements from the "Event Horizon Telescope" on the black hole, can something be said about what coordinates that fits the data better?

Question 2: It might be that both metrics fit the data equally well if we change the mass of the central black hole. Do we get different values on the mass of the central black hole if we fit the data using the isotropic coordinates instead of the Schwarzschild coordinates?

According to the Schwarzschild solution in Schwarzschild coordinates the photon sphere is located at an "r-parameter" of $3GM/c^2$. According to this post when using the Schwarzschild solution in Schwarzschild coordinates the apparent radius ($r_{obs}$) of something residing in a spherically symmetric gravitational potential when viewed from infinity is: $$r_{obs}=r(1-\frac{2GM}{rc^2})^{-0.5} $$
According to this formula we will observe the photon sphere at a radial distance of $r_{obs}=3\sqrt{3}GM/c^2$. Now I assume astronomers estimate the  "observed radial distance", "$r_{obs}$" from the center of the M87 black hole to the photon sphere around the M87 black hole by measuring the angular size of the observed ring together with some estimate of the distance from earth to M87. If you use the formula for the photon sphere radius in Schwarzschild coordinates together with the formula for gravitational lensing enlargement you end up with a formula for the central mass of M87 as: $M=\frac{r_{obs}c^2}{{G3\sqrt{3}}}$.

Now if isotropic coordinates are used instead you will get another expression for the "r-parameter" of the photon sphere (I do not know what that expression looks like) and another expression for the "gravitational lensing enlargement effect" that makes objects in a spherically symmetric gravitational field look larger when observed from infinity.

What i specifically would like to know in the second question above is how much the mass of the central body will "change" if you keep $r_{obs}$ constant but you use isotropic coordinates instead of Schwarzschild coordinates and in both cases equate the "r-coordinate" with radial distance.

I am a bit confused about the concept that you can do various coordinate transformations of the "r-parameter" in the Schwarzschild solution and still get a solution that fits reality.

One of the things that can be confusing for beginners to GR is that there is no physical significance to the choice of coordinates we make. For example when studying static black holes we can use Schwarzschild coordinates, isotropic coordinates, Gullstrand-Painlevé coordinates, Eddington-Finkelstein coordinates, Kruskal-Szekeres coordinates and lots of others that I can't think of offhand.

But regardless of what coordinates we choose it's the same metric, just written in different ways. You ask about the Schwarzschild metric and the isotropic metric as if they were different things, but it's the same metric just written in different coordinates. It would be better to call them the static black hole metric written in Schwarzschild coordinates and the static black hole metric written in isotropic coordinates but, well, life is too short and even experienced physicists will just say Schwarzschild metric and isotropic metric to save time.

The point is that the coordinates are not directly observable. We choose some coordinate system and then do a calculation to compute something that is observable e.g. the trajectories of photons. And the results will come out the same. After all the photons don't care what coordinate system a physicist chooses to compute their paths.

The reason we have all these different coordinates is that it's often the case that some calculations are easier in some coordinates than others. For example the Schwarzschild coordinates are the closest to our intuitive grasp of the geometry around the black hole. If we were considering simple questions like the time dilation for an observer hovering near the black hole I'd use Schwarzschild coordinates to do this because they make the calculation easy. But Schwarzschild coordinates don't work well at the event horizon. If I was explaining why light can't escape from a black hole I'd use Gullstrand-Painlevé coordinates.

To return to your question, I'm not familiar with the method the JPL team use for computing trajectories in the Solar System so I can't comment on why the calculations would be easier in the isotropic coordinates than in the Schwarzschild coordinates. Presumably the computational method they use is easier in isotropic coordinates, but the observables that they calculate will come out the same regardless of which coordinates are used.

In any case neither coordinates are useful for the EHT calculations because Messier 87* is rotating not static and the spacetime geometry around it is described by the Kerr metric not the Schwarzschild metric. The Kerr metric can also be written in lots of different coordinate systems - I don't know which coordinate system the EHT team used.


Reason for a correlation between Hot Jupiters and higher metallicity in Kepler data This question is from an amateur data analysis I did in May 2015 of data from Kepler.nasa.gov/Mission/discoveries. There were 80 Jupiter size planets in the Kepler "confirmed" planet table on this website (through 4/1/15) that had host stars with metallicity data. Of these, 24 stars have [Fe/H}>0.2 and 40 have -0.1<[Fe/H]<+0.2. This is a much higher percentage than the sample as a whole. Would this be strong evidence that high metallicity measurements are due to accretion? Or, do the metallicity measurements given reflect abundance deeper into the star than accretion can count for (ie. higher metallicity stars are more likely to have Jupiters in general)?

This is a well known, well researched phenomena. Yes, there certainly is a correlation between metallicity and the likelihood of observing a hot Jupiter.

There are two classes of explanation.

(1) The correlation is real and due to the fact that it is easier to form planetary embryos from metal-rich material in the core-accretion model of giant planet formation.

(2) The observed exoplanets are those that weren't swallowed. The hosts are metal rich because they did swallow others.

Finally, it was thought possible that there could be an observational bias. Planets are easier to find and measure around high metallicity stars. This applied mainly to the Doppler technique, because high metallicity stars have stronger spectral lines. Transit detection should be much less affected.

Most think that some form of (1) is going on. (2) is argued against because the correlation still appears in stars with a wide variety of convection zone depths into which planetary material would have mixed. I'm not sure it is dead though, there have been a number of claims for abundance enhancements that look like the product of planet engulfment.

I'll add some references later. Nice work though to discover the correlation if you knew nothing of it previously.


Missing piece in my telescope! I purchased a National Geographic 76/700 EQ telescope and there was a piece missing. I'm wondering if it's needed or if I should return it, if you could also tell me what it does that'd be great.

Here's an image of what is missing :

It looks likely to be one of the bolts for adjusting the polar alignment of the scope. There are usually two opposed hand operated bolts for fine adjustment of the left/right pointing of the mount on the tripod (so it points north/south - you're expected to position the tripod so its roughly correct to start with), and another two opposed bolts (might only be one in some designs) for tilting the top part of the mount  up/down so that the rotation axis in the top part (parallel to the telescope tube in the photo) (which pivots around the silver bolt left and level with the top of the red circle) points at the celestial pole (how much tilt this needs varies with your latitude).

If that bit is missing - (it looks most likely to be the one you screw in to tilt the top part of the mount upwards more) - you won't be able to set the scope up properly.

If you check the instructions on how to set the mount up, you should be able to tell which bit it is.

(The reason why it's important, is that if you set the mount up properly, then when something starts to drift out of view due to the earth's rotation, then you can keep it in view by only adjusting one of the slow motion controls).


How would the combined masses of the Centaurs or the scattered disk compare to the combined mass of the asteroid belt? I'm interested in finding the combined mass of the Centaurs and the scattered disk. I looked around but I wasn't able to find any estimates. This seems to suggest that either the combined masses of the Centaurs or the scattered disk are negligible in comparison to that of the asteroid belt, or nobody ever thought of making an estimate. All I could find was that the asteroid belt is approximately 0.04 moon masses and that the Jupiter Trojans are approximately 1/5 of the mass of the asteroid belt. My feeling is that these combined masses should be way more than that of the Jupiter Trojans.

Thank you in advance, and yes, I tried Wiki.

According to Sheppard et al. (2000) "A Wide-Field CCD Survey for Centaurs and Kuiper Belt Objects", the total mass of the centaurs is likely around 10-4 Earth masses.

Gomes et al. (2008) "The Scattered Disk: Origins, Dynamics, and End States" estimate the mass of the scattered disc as between 0.01 and 0.1 Earth masses. (Going by the abstract, I don't have a copy of the book to check the details.)

As you noted in your question, the asteroid belt is around 4% of the mass of the Moon, or about 0.0005 Earth masses. So the centaurs have a smaller or similar mass to the asteroid belt (bearing in mind the uncertainties on the values), while the scattered disc is considerably more massive.


Are there any good images of Sirius B at apastron in the Sirius system? Sirius B has a distance to Sirius A of 11 arc seconds at apastron; are there any good images of it at that point in its orbit?

HST took pictures close to periastron in 1997, when the separation of the stars was about 3.7 arcsecs. See Bond et al. (2017). There are no pictures exactly at periastron.

HST took pictures close to apastron in August 2016, when the separation was 10.7 arcsecs (see the same reference). There are no pictures from HST closer to apastron but some may be taken over the next year or so.

Here is a raw HST image from August 2016, taken from the HST archive  (freely available and easy to find).

There are many photographs of the binary taken at around 11 arcseconds separation. Table~3 of the same reference above lists a stack of observations taken between 1970 and 1975 when the separation was around 11 arcseconds. One of these images is shown below; Sirius B is the faint object below and to the right of Sirius A. The other symmetrically spaced blobs are image artefacts.


Metal distribution in our solar system The sun dominates in our solar system.
I wonder whether in every aspect the sun plays the most important role in our system.

For example, all kinds of metals are mainly located in the sun instead of the rest combined together in our system?

Is there any complementary relation between the sun and its children (all the planets, asteroids etc..)?
I mean: for one kind of metal, if the sun has more, her children should have less.

According to Abundance in the Sun of the elements, at least Li and Tb abundance of the sun are very low ($\approx1e-8$). Furthermore, these abundances are from the solar atmosphere which is near the surface of the sun.

The metal content of the Solar system is completely dominated by the Sun. The Sun contains $\sim1\%$ of 'metals' (in astronomical language anything but hydrogen and helium is a 'metal'), but all the other bodies of the Solar system combined have less mass than that. So even if they were only made of metals (but the outer planets are mostly made of H and He) the Sun would still dominate the metal budget.
The Sun does not dominate the angular momentum of the Solar system, which is dominated by the orbital angular momentum of Jupiter. The angular momentum due to the spin of the Sun is rather modest. This is easy to estimate: the orbital angular momentum of a planet is $\sim m\sqrt{GM_\odot a}$ which increases with semi-major axis $a$, while the spin of the Sun contributes $\beta \omega M_\odot R_\odot^2$ with $\omega$ the Solar spin frequency, $R_\odot$ the Solar radius, and $\beta\sim0.1$ a factor depending on the inner structure of the Sun. The ratio of the latter to the former is
$$
\beta \frac{M_\odot}{m} \frac{R_\odot^2}{a^2} \frac{\omega}{\Omega}
\sim0.01
$$
For Jupiter, the first factor is $\sim10^3$, the second $\sim10^{-6}$, and the third $\sim180$ is the ratio of the Solar spin frequency (once in 25 days) to the orbital frequency $\Omega=\sqrt{GM_\odot/a^3}$ of Jupiter (once in 12 years).


Why would a tidally-locked rocky planet have a first-order spherical harmonic surface temperature distribution? The new Letter to Nature Absence of a thick atmosphere on the terrestrial exoplanet LHS 3844b (also ArXiv) analyzes the thermal infrared light curve from the system (about 4.5 to 5.5 um). The planet is assumed to be tidally locked, so lack of asymmetry in the curve is cited as evidence that there is not thermal inertia due to a thick atmosphere, which is what one would expect for this planet.

In the beginning of the paper the authors say:

  We fit the extracted light curve with a simultaneous model of the astrophysical signal and the instrument behavior. The astrophysical signal consisted of a transit model and a first-degree spherical harmonics temperature map to represent the planet’s thermal phase variation.

and later:

  In addition to the spherical harmonics model, we also tested a sinusoid model, which has been commonly used to fit other phase curve data.

I am thinking that the incident flux at a given point on the tidally-locked planet would be

$$I \sim \max (0, \cos(\theta))$$

where $\theta$ is the static zenith angle at a given point, and so the temperature would be something like

$$T \sim I^{1/4} \sim \max (0, \cos(\theta))^{1/4}.$$

Why do they use a first-order spherical harmonic model instead? Is it related to the thermal conductivity of rock?

It probably wouldn't. But when studying these things, you don't want to go in assuming you know more than you do or you might bias the analysis.

I am thinking that the incident flux at a given point on the tidally-locked planet would be
$$I∼\max(0,\cos(\theta))$$
where θ is the static zenith angle at a given point, and so the temperature would be something like
$$T∼I^{1/4}∼\max(0,\cos(\theta))^{1/4}.$$

Unfortunately this assumes that the planet is a blackbody with zero heat redistribution over the surface, e.g. by winds or currents in a magma ocean, and no sources of heat on the nightside such as tidally-driven hyperactive volcanism which is something you certainly do not know starting out.
Spherical harmonics are a generic set of basis functions over a sphere, so it does make sense as a fit which does not assume any physical processes operating or their relative importance. In fact, this is noted in Louden & Kreidberg (2018) "SPIDERMAN: an open-source code to model phase curves and secondary eclipses" (reference 15), which is referenced when they talk about using a spherical harmonic model. Conveniently, the paper shares the lead author with the LHS 3844 b paper, so presumably this does reflect some of the thought that went into the LHS 3844 analysis. A relevant quote from that paper:

A useful and physics-independent model is a sum of spherical harmonics. This method was used for the case of the phase curve of HD 189733b by Majeau et al. (2012). An example map generated by SPIDERMAN is displayed in Figure 4. The main observational features of a phase curve, including the offset hotspot, can typically be recovered with a only the first spherical harmonic, with the centre offset from the substellar point. (Cowan et al. 2017) explore the effects of odd harmonics in phase curve data, and find that these can correspond to weather features in the planet atmosphere.

(emphasis mine)
Kreidberg et al. (2019) do note that a simple sinusoidal fit produces unphysical negative temperatures on the nightside which can be corrected using odd harmonics, hence their choice of the first-order model.
Physics-based models probably would start to introduce a whole bunch of poorly-constrained parameters, probably overkill for the first analysis: this is mapping the terrain, figuring out how the terrain ended up like that is another stage entirely. This is a Nature paper, so brevity is par for the course.


Does this amateur observation hold a record even including robotic searches for supernovae? The new paper in Nature A surge of light at the birth of a supernova (doi:10.1038/nature25151) describes the fortuitous capture of a supernova "Shock Breakout" - the earliest rise in brightness of a supernova only about 4 hours old. An open access epdf is linked in the Washington Post article if you click from inside it: A self-taught astronomer spotted something no scientist had ever seen which is a good read itself.

  His discovery, reported Wednesday in the journal Nature, is a landmark for astronomy. Buso’s images are the first to capture the brief “shock breakout” phase of a supernova, when a wave of energy rolls from a star’s core to its exterior just before the star explodes. Computer models had suggested the existence of this phase, but no one had witnessed it.

Is this the earliest that such a phenomenon has been "seen" by a person, or does this observation also hold the record after including all of the automated searches for supernovae by robotic telescopes as well?

below: "Astronomer Victor Buso poses in front of the telescope with which he witnessed the birth of the supernova 2016gkg. (Victor Buso)" From here. Click/open for full size.

No, supernovae have been observed earlier and with better cadence than this one.

The Kepler Satellite has been observing galaxies at a 30 minute cadence from months before the supernova explosion to months afterwards.  Most of these have been Type Ia which do not have detectable shock breakouts, but there was one SN with a breakout published in the Astrophysical Journal before this one: "Shock Breakout and Early Light Curves of a Type II-P Supernovae Observed with Kepler".

The Nature paper linked in the question provides 4 other SN that were also detected about as early.  I think the new thing here is that it was discovered so early by an amateur astronomer.


What is this "Table of astronomy" about? I found this picture while looking for some tattoo ideas and would like to know what these sketches are supposed to mean.

Im especially interested in figures 1, 21, 53 and 56.

It appears to be a scan of page 164 of the Cyclopædia (an Universal Dictionary of Arts and Sciences), from 1728. Higher Resolution

Fig. 1 and many others on that page seems to be diagrams of various trigonometric relations, used e.g. to convert between coordinates of celestial objects. But I'm not sure about this particular one.

Fig. 21 is a armillary sphere, i.e. a physical model showing the position of objects in the sky. Because this particular model has Earth in its center, it's also called a Ptolemaic sphere. Had it been centered on the Sun, it would have been a Copernican sphere, seen in Fig. 22.

Fig. 53 is a quadrant, which is  an instrument used to measure angles, e.g. between stars, or longitude and latitude.

Fig. 56 is a horary quadrant, which is used for finding the time of the day, using the Sun. This particular version was made of wood, paper, and brass by Henry Sutton around 1658.

Additionally,

Fig. 2 shows a diagram of the diurnal motion, i.e. the apparent motion of celestial objects around Earth.

Fig. 3 is titled "Aspect Trine Tetragon". Aspect is the astrological term for the angle between planets. This diagram shows a triangle ("Trine"), a square ("Tetragon"), and a pentagon.

Figs. 11 and 12 are micrometers (not to be confused with the length unit $\mu\mathrm{m}$), which is a device used to measure lengths very accurately. It was invented by the British astronomer William Gascoigne.

Figs. 13–20 are diagrams of the phases of the Moon and drawings of its surface. It seems the person who drew these diagrams accidentally swapped "1" and "8" in Fig. 18, since the figures run like 13, 14, 15, 16, 17, 81, 19, 20.

Fig. 34–38 illustrate how Lunar eclipses occur.

Figs. 43, 44, and 45 show the positions and orbits of the planets, Earth, and the Sun as depicted in the Ptolemaic (geocentric), Copernican (heliocentric), and Tychonic system, respectively. The latter used the mathematics of the Copernican system, but held Earth in its center since that's what Tycho Brahe believed.

Fig. 57 shows a gnomon, which is the part of a sundial casting the shadow.

Fig. 58 shows a globe. Fig. 59 seems to show how the spherical sky can be mapped onto a flat map by dividing it into 12 elongated stripes, each going from the south pole to the north pole, with a central width of 2 hours, or 30º. One of the stripes (around R.A. ~ 15 h, it seems) is shown in Fig. 60 with the constellations in that region.


How is the polar angle of a pulsar beam determined? I just read the question "What is a typical polar angle of a pulsar beam?" and I was surprised that it had an answer. I thought there was little real information besides the timing of the pulses, and it seems like any given timing could be consistent with a beam with any cone angle.
As I think about it, if you know the angular width of the beam as it leaves the pulsar, then you could infer something from the relative durations of the pulse and the time between pulses. Is that it? Or is it something else?
Note: I misunderstood the first question to be asking about $\alpha$ in the notation of the accepted answer, but they were really asking about $\rho$, the angular radius of the "lighthouse beam". It is easier for me to understand how that could be estimated.

The simple answer is that yes, we can determine the pulse width if we know the angular radius of the emission cone and a couple other geometric quantities about the pulsar and its orientation relative to us.
Pulse profiles are complicated - with some exceptions, they're more than just quick spikes or smooth Gaussian curves. Most involve multiple components added together into complicated features; the beams have structure, and that structure leads to a fingerprint of sorts. Some switch between different emission modes over various cycles. There are plenty of fascinating nuances to study.
Nonetheless, even with these complicating factors, we can indeed define a pulse width. It depends on three things: the angular radius of the emission cone, $\rho$, the angle between the magnetic and rotational axes, $\alpha$, and the impact parameter, $\beta$, which defines the angle between the magnetic axis and the line of sight of the observer.
After a mess of spherical trigonometry, we can arrive at an expression for the pulse width, $W$ (see Gil & Han 1981, or any other work on emission geometry):
$$\sin^2\left(\frac{W}{2}\right)=\frac{\sin^2(\rho/2)-\sin^2(\beta/2)}{\sin\alpha\sin(\alpha+\beta)}$$
Therefore, the beam size, the location of the magnetic axis, and the orientation of the observer do uniquely determine the pulse width, as you guessed.
Obviously, it's unlikely that an observer can know these additional parameters entirely a priori, but that doesn't mean we don't know anything about them. You might expect that $\alpha$ would be randomly distributed, and it's possible that pulsars are born as such, but it is expected that $\alpha$ decreases as the pulsar ages due to the same torques that increase the spin period (Young et al. 2010). The two axes align on timescales of $\sim10^{6\mathrm{-}7}$ years. If you could determine the age of the pulsar, and you assumed that it was born with $\alpha\approx90^{\circ}$, you could make some estimates about the likely distributed of $\alpha$ in the present day.


How would Earth's climate differ if it's axis were tilted around 90 degrees like Uranus? As the title states, what would the effect on Earth's climate be if it's axis were tilted approximately 90 degrees like Uranus?

I'm specifically wondering about the effects on seasons, temperature, and life on Earth.  However, I'm happy to hear about other effects as well.

This is a good question, and has been somewhat a topic of study by a few.  So below is some consequences of if Earth had a tilt like that of Uranus (for this explanation, I am disregarding the wobbles in the Earth's current axis, just focussing on the 90 degree axis):

According to the article Not All Habitable Zones Are Created Equal (Moomaw, 1999), each day and night would be 6 months each (including an epically long dusk and dawn), with daytime temperatures reaching up to 80C, but the night time side may not reach freezing - owing to the time to take to cool down from the 6 month day.  Parts of the equatorial regions would be, however, permanently encased in ice.

Pretty much like this diagram of Uranus' rotation (from Source: University of Hawaii:

A major consequence of this according to the author is:

  In such an environment, life could almost certainly still appear, but it would have much more difficulty evolving into forms that could survive such grotesque temperature extremes -- which would greatly slow down its evolution into more complex forms, maybe by billions of years.

A great consequence of this, particularly the 6 month night is that photosynthesis would have been stilted, if it could have started at all.  This has a great consequence on oxygen levels of the atmosphere.

Interestingly, the authors claim that if the Earth had an axial tilt of 90 degrees, but at 210 million km from the sun, then:

  its climate would be positively balmy -- the equator would be 11 deg C (52 deg F), and the poles would never rise above 46 deg C (115 deg F) or fall below 3 deg C (37 deg F). Earth would have no ice anywhere on its surface, except on some of its highest mountains.

According to the article High Planetary Tilt Lowers Odds for Life? (Hadhazy, 2012), has a great way of putting it:

  "Your northern pole will be boiled during part of the year while the equator gets little sunlight," said Heller. Meanwhile, "the southern pole freezes in total darkness." Essentially, the conventional notion of a scorching hell dominates one side of the planet, while an ultra-cold hell like that of Dante's Ninth Circle prevails on the other.

Then, to make matters worse, the hells reverse half a year later. "The hemispheres are cyclically sterilized, either by too strong irradiation or by freezing," Heller said.

They also describe that if life were to evolve, extremophiles (specifically, thermophiles) would be dominant - seasonally.


How often does the supermassive black hole in the Milky Way become a quasar? My current interest is black holes and quasars. I was wondering how often they erupt--specifically ours? Would it be blindingly bright on Earth? Finally, would there be any effects that would alter our life on Earth that we know of?

To answer your question, the black hole in the milky way was never a quasar, and as it currently stands will never become one.

Info about how earth and life on earth could be affected if this ever happened, can be found here.


How was the acceleration calculated for this CME (Coronal Mass Ejection)? I am looking at data used in a project to get the velocity and acceleration of a CME (Coronal Mass Ejection). I understand how the velocity was calculated, and for the acceleration, I get the v2-v1 part of the formula for acceleration that was used (ex. 306.5-331.2), but what time intervals (t2-t1) were used to calculate the accelerations?

The average velocity in the interval 8:05 to 8:36 is 331.2. So, use $v_1=$ 331.2 at $t_1=$ 08:20:30 under the assumption that acceleration is constant.

The average velocity between 8:36 and 9:27 is $v_2=$ 306.5, so use $t_2=$ 9:01:30

Acceleration is $(306.5-331.3)/(41\times60)=-0.01\ $km/s$^2$.

The same calculation gives me $0.029\ $km/s$^2$ and $-0.038\  $km/s$^2$.

Is this just arithmetic error on the first value (using 51m instead of 41m), a small rounding issue on the second and the third I agree with.

Other than that, then some unstated alternative interpolation scheme has been used.


The most feasible hypothesis about Theia The formation of the Moon had many hypothesis over the years, most of them involving a Mars-sized body called Theia, which presumably collided with the Earth. But even within the Theia hypothesis, is not very clear the conditions of that collision. So, what is the most plausible hypothesis about Theia today?

Theia fully disintegrates and the remnants debris form the Moon in a head-on collision(the Big Impact hypothesis)
Theia-Earth suffer a glazing collision and Theia manages to escape to space, but some of the remnants of the collision form the Moon (Reufer, 2012)
Other hypothesis, such a double Theia system

All the hypothesis have problems, but to me the most believable is 2, because it solves the majority of the problems of the other ones, although there's not much consensus on that one, the only who defends it is Reufer.

For example, some research shows that only 2% of Theia actually contributed to the formation of the Moon (I don't remember the reference), so 1 doesn't seem too much feasible because according to that one Theia fully disintegrates.

According to new simulations, a better model seems to be one where the Earth had a magma ocean at the time of impact, see here:

We perform density-independent smoothed particle hydrodynamic collision simulations with an equation of state appropriate for molten silicates. These calculations demonstrate that, because of the large difference in shock heating between silicate melts and solids (rocks), a substantial fraction of the ejected, Moon-forming material is derived from the magma ocean, even in a highly oblique collision. We show that this model reconciles the compositional similarities and differences between the Moon and Earth while satisfying the angular momentum constraint.


Does the Sun have any atmosphere? Does the Sun have any atmosphere of its own? Is it just like the atmospheres that the planets have.

Yes the Sun has an atmosphere.

Disclaimer: I'm not sure if you meant this, but your question implies the Sun is a planet. It, of course, is a star and not a planet. Just wanted to make that clear.

What is an atmosphere?

When you ask if the Sun has an atmosphere, you're actually asking a tricky question. What do you mean by atmosphere? How do you define the boundary of the Sun, above which anything is considered an atmosphere? This is pretty easy for planets like Earth since they have a nice solid surface. But the Sun is a giant ball of plasma heated to thousands of degrees. There's no easy or clear division between the "surface" and "atmosphere". Any discussion of the Sun's atmosphere involves defining what we mean by the surface of the Sun.

Optical Depth

That being said, astronomers have come up with (arbitrary) ways to define the surface of the Sun. One common metric is to use optical depth. Optical depth is a unitless number that defines one's ability to "see" through a gas (or plasma). An optical depth of 1 or higher means the gas is opaque and cannot be seen through. An optical depth less than one means the gas is transparent and can be seen through.

However, when you have something the Sun or even fog, the optical depth varies with the distance you're looking into that object. I'll talk about fog since it's familiar, but the same idea applies to the Sun's atmosphere. Say you're standing in a forrest and its very foggy out. There's a tree 1 meter away from you that you can see. You could measure your optical depth, $\tau$, of the fog between you and tree and might find that $\tau = 0.15$. Since $\tau$ is less than one, that implies you can see the tree, but the value of $\tau$ also implies how well you can see it. If $\tau = 0$, there's nothing between you and the tree to impede your ability to see it. Let's say there's another tree that's 5 meters away. Now there's more fog between you and the tree and while you can still see it, it is harder to see it. The optical depth of the fog between you and the tree 5 meters away might be $\tau = 0.75$. It's still less than one, implying the tree is visible, but because there's more fog between you and the tree, the optical depth is higher. Finally, there may be a tree 10 meters away with so much fog between you and the tree that the optical depth is $\tau = 1.5$. You can't see this tree because there's too much fog in the way. Hopefully you now realize that anything which is at a distance where $\tau > 1$ is not visible to you. That effectively defines a "surface" around you precisely when $\tau = 1$. Anything beyond that point is not visible and anything closer is visible.

If you're talking about the Sun, you can look at the Sun, but you'll only see light which originates from a point where $\tau < 1$. There are countless photons bouncing around inside the Sun, but you can't see them because they're in an opaque part of the Sun. Astronomers use the optical depth as a metric for defining the "surface" of the Sun.

Keep in mind the above description is highly simplified, almost to the point of being wrong. The optical depth is a useful metric for defining a surface, but it doesn't imply there's an exact radius for the surface or even that the surface is constant for every wavelength. There are a lot of other factors that make this much more complicated than I'm describing here. Hopefully you get the general idea though.

The Sun's atmosphere

For the Sun, the atmosphere would be anything above the surface. Nominally the surface is defined as the point where the $\tau = 2/3$ (despite what I said above, and for reasons I won't go into here). The atmosphere above this surface is complicated and difficult to study. The atmosphere, just above the surface, is violent, turbulent, filled with outbursts and magnetic fields, and extremely hot. Below are some pictures of this region of the atmosphere.

 Left: Image of the corona during a solar eclipse. Right: Image of the corona from SOHO. An occulting mask has been placed over the Sun.

The atmosphere of the Sun extends far beyond that though. In fact, Earth is currently moving through the Sun's atmosphere. It's very tenuous out near Earth, but still exists. The Sun's atmosphere hitting our planet is the reason for aurora. Beyond the lower portions, the atmosphere is generally referred to as the Solar Wind. This solar wind actually extends far out, beyond Pluto even. Exactly how far is difficult to define, but estimates place our Sun's atmosphere extending out to about $\sim230\:\mathrm{AU}$. At that point is the bow shock, where our Sun's atmosphere slams into the interstellar medium surrounding us.

Our own Sun's atmosphere is hard to study on a large scale since we're inside it, but we have been able to observe this bow shock around other stars, as shown below.

LL Orionis bow shock in Orion nebula. The star's atmosphere collides with the nebula flow. Hubble, 1995


Questions about Hubble's brand new view of an interstellar comet (video) This GIF is made (via giphy.com) from the new NASA Goddard video Hubble's New Image of Interstellar Object and if I understand correctly many (if not all) of these frames are from https://archive.stsci.edu/proposal_search.php?mission=hst&id=16009 (from this answer).

I have some questions about this set of images. I've plotted the start times below.

There are four groups that last about 45 minutes, each followed by a break of either 45 minutes or 90+45 minutes, is this related to the HST's orbit?
I notice a slow rotation of the star tracks over the duration of the roughly seven hours of data, it looks like roughly 20 degrees, (see GIF or the individual images) and I can't figure that out at all. Is it a rotation of the HST around its own axis?
Individual preview images (example) appear to be rhomboidal rather than square, why?

TL;DR version, Yes, yes, distortion. In more detail:

HST's orbit is approx. 90 minutes so this will cause the breaks. Some more information is given in the Orbital Constraints section of the Cycle 27 guide for proposers.
Since HST is having to move to track the moving comet to keep it in the (small) field of view of the instruments, it will likely need to roll the spacecraft. This is to keep suitable guide stars in the Fine Guidance Sensors, to keep the Sun on the solar panels for power and to keep the pointing constraints of not looking to close to the Sun, Moon or Earth satisfied. These are discussed in more detail in the Orientation and Roll constraints section of the primer/guide for proposers mentioned earlier.
This is due to the geometric distortion of the optics in UV/Visible (UVIS) channel of the WFC3 instrument as described in more detail in this section of the WFC3 Instrument Handbook,


What is the area of the Summer Triangle? I'm pretty sure spherical trigonometry comes into use here but I don't know how to calculate the area. I'm pretty sure Vega, Deneb, and Altair's declination and right ascension figures need to be converted to coordinates, but again, I have no idea how.
Is there a formula for a general case like this? Not necessarily just the Summer Triangle, but a formula for the area enclosed by any 3 (or more) stars on the celestial sphere?

The area of a triangle enclosed by 3 stars on the celestial sphere, in square degrees, is given by:
$$ A = \frac{180}{\pi}\times E$$
Where E is the spherical excess and is equal to the sum of all angles of the triangle minus 180°. Problem is, we don't know the angles of the Summer Triangle and instead we have to use an alternate equation for the spherical excess given by:
$$ E = 4\tan^{-1}\left({\sqrt{\tan\left(\frac{s}{2}\right)\times\tan\left(\frac{s-a}{2}\right)\times\tan\left(\frac{s-b}{2}\right)\times\tan\left(\frac{s-c}{2}\right)}} \right) $$
Where a, b, and c are the angular distances between 2 stars and thus the sides of the Summer Triangle in degrees, and $s = \frac{a+b+c}{2}$.
I highly recommend you just go to Wolfram Alpha and just type "Angular distance between Star 1 and Star 2" and get all values for a, b, and c and then you could calculate the spherical excess and subsequently the area.
But if you insist on giving yourself a headache and finding the values yourself then check the next section. Otherwise skip to the bottom.

The angular distance between 2 stars on the celestial sphere is given by:
$$\cos^{-1} \left ( \sin(\text{D} _{1})\sin(\text{D} _{2})+\cos(\text{D} _{1})\cos(\text{D} _{2}) \cos(\text{R} _{1}-\text{R} _{2})\right ) $$
Where:

$\text{D}_{1}$ and $\text{D}_{2}$ are the declination values of both stars in degrees.

$\text{R}_{1}$ and $\text{R}_{2}$ are the right ascension values for both stars, also in degrees.

To convert a star's declination from DMS to degrees, divide the arcminutes by 60 and arcseconds by 3600. For example Vega's declination is 38° 47' 1.28". Converted to degrees only, that becomes:
$$ 38 + \frac{47}{60} + \frac{1.28}{3600} = 38.7836° $$
The exact same formula is used to convert RA to degrees, only difference is you multiply it by 15 at the end. Vega's right ascension is 18h 36m 56.33s. Converted to degrees only, that becomes:
$$ \left(18 + \frac{36}{60} + \frac{56.33}{3600}\right) \times 15 = 279.2347° $$
You could just spare yourself all this mess by directly getting the angular distances from Wolfram Alpha or you could calculate them yourself. I recommend the former.

Once you plug in all the numbers, you get the spherical excess $E = 7.296508$ and the area of the Summer Triangle:
$$ \bbox[5px,border:2px solid #C0A000]{A = 418.059 \text{deg}^2} $$
It would cover 1.0134% of the celestial sphere and be the 40th largest constellation in the sky, bigger than Capricornus but smaller than Aries.


Yet another Question about the expansion of universe A galaxy 13 billion light-years distant is 94% of the way to the big bang. The universe was 800 million years old when the light was emitted. That galaxy today is traveling away from us at 94% of the speed of light roughly. So 13 billion years ago it was 750 million light years away from us. Assuming the expansion is NOT greater than light speed, that photon did NOT go backward. This is also evidenced by the fact that the photon arrived at earth. It seems to me the photon took 13 billion years to cross 750 million light years. What am I doing wrong?
   Question two. If this is right then please explain how we go from 750 (or 800) million light years to 46.5 billion light years.

Neither the amount of time it took for light to reach us emitted long ago from a distant galaxy nor the distance to that galaxy are directly observable. What is observable is the redshift, how much light has been shifted toward longer wavelengths.

The amount of time it took for that redshifted light to reach us can be calculated from the redshift $z$ by assuming various parameters regarding the universe (regular matter vs dark matter, the Hubble constant, whether the universe is open, flat, or closed, etc.), and by assuming that that galaxy was and our galaxy is more or less moving with the local Hubble flow. A distance value can be calculated simply by multiplying the calculated time by the speed of light.

Even though the second calculation is much simpler than the first, it is this second calculation that is rather misleading. There's an underlying assumption in this calculation that the universe is not expanding. The universe is expanding, so it's not quite valid to multiple very long times such as 13 billion years by the speed of light to get 13 billion light years. In fact, that distant galaxy is about 29 billion light years from us now and was only about 3.5 billion light years away when the light was emitted.


Tidally locked Venus, is it possible and consequences? Suppose, Venus becomes tidally locked. Will its dark side cool enough so that the CO2 from the atmosphere to precipitate in liquid form to make an ocean?
Also, I wonder, how close Venus is to becoming tidally locked? Does it accelerate or slow down its rotation?

It may not be possible for Venus to become tidal locked
I don't think we know if it's possible for Venus to become tidally locked. Correia et al. 2008

expect the equilibrium rotation to differ from the synchronous motion

for planets like Venus with thick atmospheres relatively close to the Sun.
This might be best illustrated with a graphic from Auclair-Desrotour et al. 2016:

The gravitational torque is driving towards a tidal lock, but the thermal atmospheric derived torque is in opposition.
Observations may show that Venus is slowing down, according to this esa article. However, according to the article,

the planet could have weather cycles stretching over decades, which
could lead to equally long-term changes in the rotation period

A tidal-locked Venus might not have big day/night temperature variation
This article: The Gale Winds of Venus Suggest How Locked Exoplanets Could Escape a Fate of Extreme Heat and Brutal Cold suggests that the super-fast winds of Venus could occur even if Venus was tidally locked.  These winds exceed 100 m/s, which, according to the article is 60 times the surface rotation speed.  Even though Venus has an incredibly slow rotation rate currently (~243 days) the difference between night and day temperatures is very small.  We wouldn't expect much more deviation if Venus became tidally locked, as the heat transport is primarily driven through various other transport mechanisms than the Sun.  A relatively small amount of the Sun's energy makes it to the surface of Venus.
Edit:
In the comments, uhoh asks about the super-fast winds of Venus:

Is this mechanism keeping Venus' back side warm discussed anywhere in
a peer-reviewed state[...]?

As a quote from Horinouchi et al. (the "SR" in the quote refers to Super-Rotation, which is the super-fast winds):

SR may also occur in tidally locked exoplanets—those that always face
the same way toward their host star and are therefore heated only on
one side. Zonal (east-west) flow around the rotation axis, including
SR, can transport heat from the dayside to the nightside of those
exoplanets

Showman and Polvani go into greater detail in their paper: Equatorial superrotation on tidally locked exoplanets.