How can I hear (or at least detect) a pulsar at home? Scott Manley's video Using Relativistic Raytracing &X-Rays To See Detail on Surface Of Neutron Star talks about X-ray measurements using the NICER X-ray telescope attached to the International Space Station (Wikipedia, in SE: 2 & 3), results which can also be read about in arXiv and these recent papers. Since I'm on Earth I'll need to use radio waves instead.

I understand the signal will be weak and below the noise for a small setup so I'll have to collect a lot of data and do some type of folding analysis. But before I can even make an estimate of signal and noise levels I'll need to choose a few "strong" pulsars and go look up their radio spectral distribution and compare it to thermal and and sky noise spectra.

Also based on frequency I can then know better what kind of antenna can be used; if I can get away with an array of wires like the early days or I'd need a dish.

Is there something like a "Top 10 pulsars of all time" site which has a characterization of some "popular" objects that may have strong signals that would also list their period?

Are there other issues I might not have thought about yet?

Interesting questions - I believe I have answers to them.

You mentioned folding analysis - yes, you will need to do folding analysis to find pulsars - the majority of pulsars are found with what are called prepfold plots - here is some software that is very popular, but you can do this yourself. The general principle is that if you separate a signal at a time interval equal to the period of the pulsar and stack the resulting sections of the signal, you're effectively doing the equivalent of "image stacking" as you would do with images you take in a camera, which will boost the signal to noise ratio of the pulsar you're trying to look at.

The best bet for a small setup would be the Vela pulsar - here is a list of significant pulsars. The Vela is the brightest pulsar in the radio part of the EM spectrum (as per the link, which is from Wikipedia, so take its reliability with a grain of salt), so I would assume that's your best bet in trying to detect a pulsar.

As for antennas or dishes, I would think a dish would work best. You're going to want some decently sensitive equipment to detect a pulsar, so I would recommend a proper dish made for collecting radio waves.

And lastly, you would want to consider which frequencies the target pulsar emits at and check which ones your receiver can collect.

I think you've pretty much covered all the other issues; now you just need supplies and money :).


Converting Jy/beam to Jy? Maybe its a dumb question, but to convert Jy/beam to Jy, I just have to multiply it by the beam size in sr right?

Being $\Omega$ the beam size:
$\Omega = \frac{\pi \theta_{maj} ~~ \theta_{min}}{4 \ln{2}}$

Jy/beam $\cdot \ \Omega$ = Jy   ?

Actually to convert from Jy/beam to Jy/pixel you need to divide by the beam size.

Let's say you have a quantity of 1 Jy/beam, then

$\frac{Jy}{beam} \frac{beam}{\Omega}$, then to go from Jy/beam to Jy/pixel you would need to divide by $\Omega$.

The values of the beam major and minor axis must be in pixels.

Source: NRAO


How many rovers now on Moon? The Apollo missions took moving rovers to the Moon. How many moving rovers are presently on the Moon?

Three lunar rovers and the Chinese Yutu rover. None of which are operational. https://en.m.wikipedia.org/wiki/List_of_artificial_objects_on_the_Moon


What does G/IG signify in the galaxy names ESO 012- G 014 and ESO 293-IG 034? The NASA/IPAC EXTRAGALACTIC DATABASE (NED) lists galaxies from the European Southern Observatory (ESO) catalog as  G  (ESO 012- G 014) or  IG  (ESO 293-IG 034).

What does G/IG signify?

These identifiers correspond to the ESO/Uppsala Survey of the ESO(B) Atlas (1982). The identifier column (ESO) is associated with the following note:

Note (1)  : ESO designation is written FFF-oooNNN, where
FFF = Field (plate) number, see file "plates.dat"
ooo = Object class with
* = star ** = double or multiple star
SC = star cluster N = nebula
A = asteroid C = comet
G = galaxy IG = interacting galaxy
EN = emission nebula PN = planetary nebula
RN = reflection nebula SNR = supernova remnant
? = questionable object
NNN = number within field

So the "G" means "galaxy" and "IG" means "interacting galaxy".


When is the best time to observe M81 from the Northern Hemisphere? I have read that the best time to view M81 is during the spring. Considering its RA is ~10h and Dec. is +69°, wouldn't February 21st be the best time to view it since this is about when it transits my meridian at midnight or are there other considerations?

I'm observing from Cleveland OH, USA area; 41.5° N, 81.7° W.

You are correct that M81 transits the meridian at upper culmination just past mid-February in the eastern US. As long as the Moon is well below the horizon, as it will be in Feb. 2020, this will be the best time to observe M81, and its neighbor, M82. Since these galaxies are relatively bright, good views are available from late December through early May when they are 40+ degrees above the horizon, as long as, there is no interference from the Moon. I have taken some of my best photos of M81 & M82 in March and April, mainly because they are a bit farther to the W, avoiding some local light pollution here towards the NE.

It's likely that many think early Spring is the best time to observe M81, since they don't have to stay up so late; in the eastern US, M81 transits the meridian about 10 pm in mid-March and 9 pm in early April.

Edit:  Added answer to question from the comments.

The simplest and most accurate way to determine when an object transits the meridian at your location is to  use astronomy software such as, SkySafari, Stellarium, Cartes du Ceil, or any of the many other planetarium type programs. However, I started amateur astronomy many years before smartphones, tablets, gps, and even personal computers. So I learned to estimate when objects in the sky would be visible.

At my location, around 10 degrees of longitude to the east of yours, I remember that the meridian on January 15th corresponds to about 8 hours of right ascension. Since the sidereal day is about 4 minutes shorter than the solar day, RA at the same location and time increases about one hour in 15 days, or about 2 hours per month. Precession changes this RA slowly over time, however, it is a useful approximation that is roughly valid over decades.

Thus, M81 at 9h 56m RA, will transit my meridian near midnight around Feb 15th, and at your location a few days later. A month later M81 will be two hours farther to the west, so will cross the meridian earlier, around 10 pm.


Why is the Warm-Hot Intergalactic Medium (WHIM) so hot, and what is "collisionless shock heating"? The Phys.org article Researchers find last of universe's missing ordinary matter says:

Ordinary matter, or "baryons," make up all physical objects in existence, from stars to the cores of black holes. But until now, astrophysicists had only been able to locate about two-thirds of the matter that theorists predict was created by the Big Bang.
In the new research, an international team pinned down the missing third, finding it in the space between galaxies. That lost matter exists as filaments of oxygen gas at temperatures of around 1 million degrees Celsius, said CU Boulder's Michael Shull, a co-author of the study.

and refers to F. Nicastro et al, Observations of the missing baryons in the warm–hot intergalactic medium, Nature (2018). DOI: 10.1038/s41586-018-0204-1
The Wikipedia page Warm–hot intergalactic medium says:

Part of the gravitational energy supplied by these effects is converted into thermal emissions of the matter by collisionless shock heating

but the link to "collisionless" is ambiguous and I haven't a clue how a shock wave can get hot without collisions.
Is there a way to understand why is the Warm-Hot Intergalactic Medium (WHIM) is so hot, and what the "collisionless shock heating" process is that appears to be heating it?
Are the temperatures associated with the motion of the atoms and with the excited states of the atoms similar? With only of order 1 atom per cubic meter, collisions must be so infrequent that it seems to me that it might not be so hard for these two temperatures to diverge.

It should probably be added that the article includes a glaring error of the type you often see when the science writer apparently did not take an elementary astronomy class (this is why we have such classes!).  When the article states that the "lost matter exists as filaments of oxygen gas", you can be sure that Michael Shull never said any such thing, because that language clearly asserts that the "lost" matter is oxygen.  Of course what is meant is that the previously undetected baryons exist primarily as free protons (and electrons), which is merely evidenced by the presence of detectable oxygen that is simply assumed to be in the normal ratio we find between oxygen and hydrogen.  We could also quibble that calling it "lost" is hyperbole, a bit like saying the brain matter of the writer was heretofore "lost" because it couldn't be seen by looking into his/her ears, and was only recently "found" in some recent MRI.   Astronomers have long expected the WHIM to be there, and knew it would be hard to see, so it was hardly "lost."  But it is certainly nice to have greater certainty that it is there.

And not only is there a crucial assumption of the oxygen-to-hydrogen ratio, the OP is astute to ask how well we can even know the oxygen content itself.  All we see is the light from the oxygen, so we have to model how emissive it is to know how much oxygen is there.  That's not always easy at low densities where the emissivity can differ from what you'd get in local thermodynamic equilibrium, as was pointed out in the question and the first answer also.  So not all the uncertainties have been resolved, but it's certainly a nice step forward.

One final point regards why the gas is so hot.  This is related to the fact that low density gas cools itself very inefficiently, because cooling generally requires collisions between the particles (which generates the light).  Heating, on the other hand, can occur in various ways that might involve interactions with something other than the particles, say with cosmic rays or with magnetic fields.  Those latter elements can be present even if the density is very low, leading to high temperatures.  There is also a radiative cooling instability above about 10,000 K, where gas that is heated above that starts to cool less efficiently (as hydrogen gets all ionized), and responds by being compressed by the ambient pressure, which drives the temperature up even more.


Binary star not exceeding Roche Lobe? Is it possible that a binary star doesn't exceed its Roche lobe and so when the second star becomes a red giant, instead of accreting some of its plasma onto the white dwarf that it becomes a white dwarf itself which causes there to be no supernova?

Of course this is possible. Whether it will fill its Roche lobe depends on the masses and radii of the stars and the distance between the stars. If the distance is big enough then the second (less massive) star will not fill its Roche lobe and continue evolving as a single star would. The (large) majority of binaries will fall in this category, only close binaries will evolve into cataclysmic variables.

For instance the Alpha Centauri/Proxima Centauri system will not evolve into a system where mass transfer takes place via the Roche lobes as the distance between Alpha Cen A/B and Proxima Centauri is much too big.

Even when the stars are very close, a supernova might not occur if the masses of the stars are low. The supernova occurs when the mass of the existing white dwarf increases (because of the accretion) to a mass surpassing the Chandrasekhar limit (1.4 solar masses). If the transferred mass is too low then no supernova will happen.


How do we know that light is redshifted/blueshifted and not the original light of a star/galaxy? If a star/galaxy is moving towards us, then its light is blueshifted, and if it is moving away from us, then its light is redshifted. How do we exactly know that the color we are seeing is not part of the fuel burnt by the star/galaxy, and is instead caused by the Doppler effect?

If you had a simple slit spectroscope, and looked at an incandescent light, you'd see a smear of light with red on one end and blue on the other. This is because the filament is producing light by glowing from being heated.If you looked at one of those orange colored sodium vapor street lamps, instead of a smear of color, you'd see a group of lines. This light is produced by ionizing the gas.The lines represent specific frequencies of light coming from the lamp. You could add a horizontal scale and find that the lines represent a specific frequency in the light spectrum. If the street lamp was coming at you at a significantly high speed, the individual lines would be shifted in frequency towards blue, but would still have the same pattern. Conversely, if the light was moving away from you, you'd again see the same pattern of light lines, but their frequencies would be shifted toward red.This is what's being measured when the spectra of stars and galaxies are measured: not just what the color looks like, but whether the spectra of things like hydrogen, helium and iron are shifted in frequency towards red or blue.So, it's the specific line patterns produced by these elements when they are ionized in stars that helps us identify them. Comparing the light frequencies of locally ionized elements to the frequencies coming from distant stars that tell us if the stars themselves are approaching or receding.


Why is LUVOIR actively heated? From the Wikipedia page about LUVOIR design:

  To enable the extreme wavefront stability needed for coronagraphic observations of Earth-like exoplanets, the LUVOIR design incorporates three principles. [...] . Third, the telescope is actively heated to a precise 270 K (26 °F) to control thermal disturbances.

I always thought the chiller a telescope mirror, the better. Why is LUVOIR heated then?

The main reason appears to be to prevent any icing up and contamination of the mirrors and other optical coponents with water and other "non-volatile residues", which would be extremely detrimental to the UV performance.

Emissivity at 270K is not too much of an issue since the instrument only works out to 2 microns.

The issue is briefly discussed in section 8.1.10 of the LUVOIR final report.


Is any consensus forming on the solution to the "Lithium Problem"? The "Lithium Problem" relates to the fact very-low-metallicity stars appear to have a Li/H ratio approximately one third of what would be expected. The ratio should be the same as the prediction from Big Bang nucleosynthesis theory. A possibly related problem is that the ⁶Li/⁷Li ratio is several orders of magnitude too high.

No. There is no consensus.

The discrepancy between the predicted big bang nucleosynthetic abundance of Lithium 7 and the measured value can be summarised as follows.

If we take what we know about the the baryonic mass density of the universe and the Hubble constant, we get a self-consistent picture between the cosmic microwave background, observations of galaxy recession etc. and the estimated primordial abundances of Helium and Deuterium.

The problem arises because these same cosmological parameters predict a primordial lithium abundance of $3\times10^{-10}$, when expressed as a ratio to the hydrogen abundance.

On the other hand, measurements of the Li abundance present in the photospheres of the oldest stars (a.k.a. "halo stars") in our Galaxy suggest that the primordial abundance was about $1.2\times10^{-10}$.

The factor of 2-3 difference between these numbers is about 4-5 times the measurement precision. This is the so-called "Lithium problem".

The potential solutions are reviewed by Fields (2012). They fall into the following categories.

Astrophysical solutions - that we don't understand our measurements of the Li abundances because of an imperfect understanding of the atmospheres of low metallicity stars; or that we don't understand interior mixing mechanisms that mean at the photosphere, we see material that has been mixed upwards from the interior where the Li has been depleted in nuclear reactions.
Nuclear physics - maybe the details of the reaction rates and cross-sections in the big bang model are awry? There are still some sizeable uncertainties here which remain to be nailed down, but are seen as rather unlikely solutions.
Additions to the standard big bang model. This includes things like inhomogeneous nucleosynthesis in the early universe - i.e. that it was clumpy even at this early stage. Other possibilities include the possibility that the equilibrium reactions in big bang nucleosynthesis were upset by the decay of massive dark matter particles.

Thus there are lots of ideas to solve this problem, and other ideas which suggest it is not so much a problem, but that we can't do the measurements properly.

EDIT: Update 18/11/19 from the "Lithium in the Universe" conference in Frascati, Rome.

Having sat through some review talks on the problem, I can summarise progress as:

Explanation 2 (Nuclear Physics) is dead. All avenues appar to be explored and the likely remaining uncertainties are at the level of 10%.

However, the following possibilities were also discussed.

Perhaps the fundamental constants have changed with time resulting in somewhat different binding energy differences between different nuclei? Scherrer & Scherrer (2017) for example discuss a scenario where the mass difference between 2 He nuclei and a 8Be nucleus changes with time in such a way that lots of 8Be is produced during big bang nucleosynthesis, but decays back into He later on. This does not alter the primordial He abundance inferred from present-day observations, but the removal of He at early epochs results in a lower production of 7Li.
Inhomogeneous magnetic fields, contributing significant energy density in the early universe may have led to temperature fluctuations at the epoch of nucleosynthesis. Such fluctuations could reduce the predicted Li abundances a bit, but they also change the predicted deuterium abundance and this wouldn't fit with the now quite precise measurements of the primordial D abundance and the Planck value of $\Omega_b h^2$.
It has been hypothesised that magnetic fields can play a role in filtering how much primordial Li makes it into the structures that form galaxies. i.e. There is a sort of chemical differentiation caused when there is a gradient of the magnetic field strength at right angles to the magnetic field. This accelerates ions down the gradient and possibly away from forming structures. Since Li is easily ionised in the early universe (compared with H and He), this might lead to the gas that formed the first stars being Li-poor.

At the end of the meeting, participants were asked (for fun) to vote on what they thought to most likely solution was. #1 Was first, with #3 a distant second.


Eclipse Changes Directions? How could the lunar eclipse of January 20, 2019 begin from bottom to top and then finish from left to right?  Here's the video of it's progression's directions.
https://www.youtube.com/watch?v=A2sTkuZht4s

Because we are standing on a rotating globe.  We are changing direction as the Earth spins. This appears exaggerated in the video as the horizon is not visible, so there is no way of seeing how the camera was rotated to view the moon as the eclipse progressed.

Relative to the stars, the moon appears to enter the Earth's shadow from one side and leave it from the other,


Is the rotation of the Sun and the rotation/orbit of the Moon around the Earth a coincidence? While looking at sunspot information in connection with Amateur Radio, I found that the Sun rotates on a period from 27 days to 31 days.  Its rotation is differential, at the equator it spins at a rate of about 27 days, while at the poles it's closer to 31 days.

Earth's moon also rotates at a rate of 27.3 days.

I suspect this is nothing more than coincidence, but was wondering if there is more than a coincidental connection between these two rates of rotation, similar to how tidal locking forces the moon's rotation and orbit around earth to be the same rate.

This is purely coincidental.

Why? Because the moon used to rotate faster and is continually slowing down. Long story short, you just happen to be at a point in history where they two numbers roughly coincide. A few million years ago it wasn't true, and a few million years from now it wont be either.

And just to cover all my bases, there's nothing to suggest that the slowing orbit of the moon can have any reciprocal effect on the sun's rotational period.


Are there good animations of the movement of proximate stars over long time periods? After reading about Barnard's Star and how it's approaching us 140 km/sec, I'm curious if there are any visualizations of how the nearest stars to Sol will be moving over the next 100k years or so.

I'm amazed that Barnard's Star, currently 6 light years away, will be only 3.8 light years away in about 9,800 years. It's really moving!

This image should give you a good idea of the distances of stars relative to the Sun over the next 80,000 years (and the past 20,000 years).

Image courtesy of Wikipedia user Lambiam under the Creative Commons Attribution-Share Alike 3.0 Unported license.

I've created a modified version below. The nearest star's track in time is in red, and the four blue boxes indicate the five transitions (the second box contains two):

Proxima Centauri remains the nearest star to us for about 25 thousand years, at which point Alpha Centauri passes it. ~7.5 thousand years after that, Ross 248 swings in closer. It then moves away after about 10 thousand years, when Alpha Centauri moves just ahead of it for a short period of time. Gliese 445 becomes the nearest star for about another ~7.5 thousand years, until it wings away and Alpha Centauri again becomes the closest star. 80 thousand years in our future, Ross 128 nudges ahead.

A few interesting things to note are that for most of this time, the nearest star is closer to the Sun than Proxima Centauri is now - indeed, there are times when quite a few stars are closer simultaneously - and that this lasts for some time as a brief spike.

However, this is only a short-term pattern. In the past, many stars have come much closer - even inside the Oort Cloud.

Also, to address the mention of Barnard's Star: It will swing close to us within ~ 10,000 years, but will just as suddenly swing away.

As an aside, a cool animation shows the movement of stars that are near the Sun throughout their past trip around the galaxy.

Animation courtesy of Wikipedia user Henrykus under the Creative Commons Attribution-Share Alike 3.0 Unported license. Data and original information from the European Southern Observatory.

The above animation was the result of 1,000+ nights of observation over 15 years by astronomers working at the ESO. They observed or calculated the ages, velocities, and compositions of 14,000+ F and G stars. They then used the information to create orbital trajectories for these stars.


How to calculate right ascension of Greenwich? If I know specific time, t0, how can I calculate RA of Greenwich?

I think you mean to ask, "What is the local sidereal time of Greenwhich?".

To calculate the GMST (Greenwich Mean Sidereal Time):

  To an accuracy within 0.1 second per century, Greenwich (Mean) Sidereal Time (in hours and decimal parts of an hour) can be calculated as GMST = 18.697 374 558 + 24.065 709 824 419 08 * D , where D is the interval, in UT1 days including any fraction of a day, since 2000 January 1, at 12h UT (interval counted positive if forwards to a later time than the 2000 reference instant), and the result is freed from any integer multiples of 24 hours to reduce it to a value in the range 0–24.

[source]


What type of camera instruments are used to record optical images in space telescopes like HUBBLE & JWST.? I am just curious about photos taken by space telescopes

What kind of type of camera systems are used in space telescopes like HUBBLE and JWST?

Are those cameras systems specially built for that purpose?
Is there a significant difference between cameras we use and of those?

What type of lenses are used?

The HST and JWST are both mirror telescopes. There are many instruments on each, but the best known is the wide field and planetary camera (wfpc1), and its successors (wfpc2 and wfc3). These are all CCD cameras, and as such are based on the same basic technology that is found in many digital cameras, but there the similarity ends.

The camera is placed at the primary focus of the telescope. The telescope, therefore, acts as the main "lens". There are secondary optics, particularly in wfpc2 and 3 there are optics to correct the miscurvature of the main mirror. These optics are relay mirrors, not lenses.

Each camera is a one-off. Designed and built in-house by NASA. Compared to modern cameras, the resolution of the CCD is not so high. The original WFPC had an 800x800 pixel CCD, the WFPC2 had 4 800x800 pixel CCD sensors (at different pixel densities) If these seem low, remember that Hubble was designed and built in the '80s.

The cameras have to deal with a lot of things that the cameras we use don't. There are high-energy cosmic rays that can damage CCDs, and servicing required a shuttle launch. The CCDs are sensitive in a wide range of frequencies, from infrared to ultraviolet.

The latest camera, installed in 2009, has two light paths as shown, separating the visible from the IR imaging. As noted by Sean Lake in a comment, CCDs are read out by moving charge across the detector, infrared arrays read out individual pixels. IR arrays have more problems with stuck pixels or variable sensitivity across the array.

Schematic from the space telescope science institute.

So the cameras on Hubble, and the cameras that will be placed on JWST are distant relations of commercial digital cameras. They have no lenses, can see in a wider range of frequencies. The camera and its housing are also about 2m wide. However, the basic technology of converting an image into a digital file is shared between commercial digital cameras and the HST


How much competition there is for jobs in astronomy compared to other fields of science? I will soon finish a master's degree in computer science. I have been interested in astronomy for some time, which has included playing around with a Newtonian reflector telescope and reading about CCD-cameras.

I checked some local universities, and was amazed that they have only 10 or so graduates each year (1-2 get a PhD and the rest a master's degree). How many jobs are there for these graduates? Compared to other fields, how tough is the competition in astronomy? Is it different for some relatively new areas, e.g. space neuroscience or astrobiology?

I am considering to apply for an astronomy PhD, so any input would be helpful.

The competition for permanent positions in astronomy is very tough.  The field as a whole produces roughly ~200 Ph.Ds per year, but there are usually only a handful (say ~10) tenure-track positions that open up every year.  So perhaps ~5% of Ph.Ds end up in tenure-track positions in astronomy.  There are more permanent positions in astronomy that aren't tenure track positions, but not too many more, probably enough to support about ~20% of Ph.Ds.

As for non-permanent positions in astronomy, there are plenty of those.  The funding situation for postdoctoral positions is such that, while there is still a great deal of competition for open positions (stemming partially from the fact that almost everyone applies to almost every open job --- I exaggerate, but only slightly), there are enough open positions, that almost all astronomy Ph.Ds who want a postdoc will be able to get one.  But those positions only last for an average of three years.

However the job prospects for astronomy Ph.Ds who decide to leave the field of astronomy (like myself) are quite good.  Those holding an astronomy degree have one of the lowest unemployment rates of anyone (around 0.3%).

Here are a few links to some papers on the state of the job market in astronomy:

http://arxiv.org/abs/0712.2820
http://arxiv.org/abs/0805.2624
http://arxiv.org/abs/1512.02223


How can we tell if the asteroid 2013 TV135 will hit Earth in 2032? Some astronomers in Ukraine have estimated that there is a small likelihood (1 in 63,000) that the asteroid 2013 TV135 will strike Earth in 2032. (source)

Don Yeoman of NASA JPL says:

  with more observations, I fully expect we will be able to significantly reduce, or rule out entirely, any impact probability for the foreseeable future.

How can we improve our estimation of the potential threat posed by 2013 TV135?

That is, what process and techniques are used to calculate and increase the accuracy of our collision threat estimation?

The answer to the first question is: more observations.

  That is, what process and techniques are used to calculate and increase the accuracy of our collision threat estimation?

You just have to know the orbit.

But the orbit of asteroids is very complicated to predict, because they are influenced, sometimes dramatically, by the planets and by smaller bodies, like other asteroids, that we might not be able to see.

On top of this we cannot measure directly the orbital speed of an asteroid. We can only observe changes in the angular position. If we know the distance (which might not be easily measurable) we can at least put a lower limit to the speed.

The basic steps to study the orbit of an asteroids are:

find it and try to get a first guess about the position and velocity (this requires multiple observations far enough in time to be able to measure an angular movement)
plug the position and velocity info (and likely their errors) into a computer simulation that simulate the solar system dynamics. Fast forward the simulation to predict the future orbit.
observe again the asteroid and compare its position with the simulation. Re-estimate position and velocity. Go to point 2.

As the orbit can change significantly a continuous interplay between 2. and 3. is necessary.

Once you have an orbit you can estimate the probability for the asteroid to pass near enough to the Moon or Earth to be deflected and hit us.

But thankfully the solar system is huge, we are tiny and moving between the much much larger Sun and Jupiter :D


Mars vs Venus: the retention of atmospheres in relationship to Earth Correct me if I'm wrong, but the factors than enable a planet to retain or lose an atmosphere seem to be: 1. Magnetic Field, 2. Solar Winds, 3.Weight of Gases, 4. Planetary Temperatures. 5. Distance from the Sun In regards to solar winds and planetary temperatures.

A strong magnetic field usually helps to retain all the gases, even the lightest as on Earth.  The magnetic field prevents solar winds and radiation from reaching the atmosphere and exciting the gases enough to escape the gravitation planet.
The Strength of the Solar Winds have the ability to excite gas molecules enough to escape the gravity of the planet. Their strength is determined by the planets proximity to the Sun. Their effects are mitigated by the strength of the magnetic fields and molecular weight of the different gases.
Molecular Weight of the Gases and the gravity of the planet. While Venus has a much stronger solar wind than Mars and almost no magnetic fields, it's not a question of maintaining an atmosphere, as much as maintaining what kind of atmosphere.  The lighter gases like Hydrogen, Oxygen and Nitrogen are not present on Venus, but CO2 and heavier gases are.  The heavier gases are much less easily excited due to their heavier molecular weight and thus remain, unable to escape the gravity of the planet.  That is not to say that the heavier gases aren't aren't blown away to one degree or another, but aren't blown away in more significant amounts.  Thus Venus has an atmosphere of these gases.
Planetary Temperature as well as the lack of a molecular field has a great deal to do with the lack of an atmosphere on Mars.  The lighter gases still escape but the heavier ones, more specifically CO2, remain, but mostly in frozen form.  The lower temperatures are also thought to be responsible for some remaining water, stored in the form of ice below the surface.
Distance from the sun determines the strength of the solar winds and planetary temperature.

From what I see, these factors, interacting, determine the ability of the planets to retain an atmosphere, and of course, what kind of atmosphere.  While my writing may have some issues, I'm not so much concerned with that this point(I can work on it), as I am to know if I have an accurate and comprehensive understanding of all the known factors that determine a planets atmosphere.  Thank you.

One thing we have to keep in mind:
All atmospheres escape, always. Only the degree to which this happens is different on different planets.

I do appreciate your question a lot, as somehow people always forget about the existence of Venus when asking "what keeps an atmosphere in place?". As we will see, the classic story of "a magnetic field protects, that's all" is more of an urban myth.

Also all factors you have mentioned play a certain role. But let's pick them apart one-by-one, starting from the back in order to tell the story of escaping particles.

1. Distance
As you've correctly noted, distance plays a role indirectly by determining solar wind and irradiation at a given distance, so this factor is out as a direct cause.

After all, when modeling escape from planetary atmospheres one would rather take solar wind strength and UV-irradiation as free parameters rather than distance. This would make certain that one can compare planets around different stellar types, or stellar ages with each other.  Thus, distance is usually not considered a free parameter in the literature.

2. Planetary Temperature
Temperature is not equal to temperature. The surface of a planet has a variation in temperature from equator to poles $\Delta T_{\rm EP}$.
Then again, the temperature varies with altitude above the planet. Similar to the sun, the temperature difference from surface to high up in the atmosphere/corona is $\Delta T_{SC}$ and one finds $\Delta T_{SC} \gg \Delta T_{\rm EP}$.
On Earth temperature variations go (very roughly) from $+40^{\circ}C$ to $-60^{\circ}C$, so $\Delta T_{\rm EP} \approx 100 K$, while the temperatures in the exosphere at 500km height go up to $T_{exo} = 1000K$.

Modern measurments and calculations for Mars and Venus give low exospheric temperatures of around $300K$, due to efficient cooling to space of $\rm CO_2$ (see this presentation by Coates et al. for Venus and a recent review by Lammer et al. (2008))

For the escape of particles into space, it is the lower boundary of the exopshere, the exobase, that determines escape rates. Particularly, it is the temperature and the density at the exobase that determines escape rates, as from here the fast tail of the Maxwell-Boltzmann distribution can directly head off into space. This is a particular escape process called Jeans-escape. Below the exobase particles that travel upwards at escape speeds will encounter on average more than one particle and thus be scattered without escape. This is btw. also the definition of where to find the exobase.

This is to illustrate that neither the surface temperature of planets, nor the variations thereof play any role for atmospheric escape. Very strictly this is not true, as the atmosphere of a star and possible young rocky planet can get hot enough to launch a hydrodynamic wind into space, greatly increasing escape rates. For the sun we call this phenomenon the Parker solar wind.

Now let's take a look how particles can escape from the exobase.

3. Particle mass
High up in the atmosphere, at around 100km height, there lies another atmospheric boundary, the homopause. Here mixing of particles due to large-scale motions of the air becomes so inefficient, that particles start forming layers according to their masses.

Heavier atomic and molecular species will thus, once they reach the exosphere $\sim$500km be there in much smaller numbers than, for example Hydrogen. Thus, hydrogen escape rates will always be higher than other species.

This is the 'classical' picture, which cannot explain some mysteries involving the noble gases. Noble gases are chemically inert, meaning they don't form molecules and once in the atmosphere, they should just hang around there for billions of years. Except for those who escape. This is why noble gases are giving important datapoints when constructing theories of what happened to Earth's atmosphere.

An important problem is that Xenon is missing, relative to its lighter counterpart Krypton and the other noble gases. This shouldn't be if the classical picture is correct. I recommend reading the introduction in the article of Zahnle et al. (2018) for a very detailed picture of this "missing Xenon problem".

In order to address this problem and this article, let me introduce another particle parameter:

4. Ionization energies
So particle mass only doesn't explain escape rates. the article of Zahnle et al. (2018) (and references therein, this is not a new idea) however proposes that if escape rates are mostly given by ions, then a different picture emerges, that could reconcile escape via mass fractionation and missing Xenon.

The general idea is that because the ionization energies differ as we move through the periodic system of elements, it is much easier to ionize Krypton and Xenon. A mass of ionized hydrogen, or protons will be pre-existent in the high atmosphere. Ion-ion collisions are much more efficient in coupling species together than ion-neutrals.
So if we now assume that instead of the neutral hydrogen, the protons mostly escape, they'd have dragged Krypton and Xenon ions along.
Except that under Earth atmospheric conditions the Krypton ions quickly recombine to neutral Krypton, while Xenon doesn't. So the Zahnle article concludes that the missing Xenon is a signpost of past, dominant ion escape.

So if ions are this important for atmospheric escape, as opposed to the neutral species, we probably have to think more about the magnetic field lines that they follow. Finally we get to discuss the planetary magnetic field.

5. Magnetic field and solar wind

Some magnetic field lines connect to the field in interplanetary space. Ions travelling along those will be inevitabely lost to space and picked up by the solar wind. The strength of this effect is dominated by the geometry of field lines intersecting with the solar wind, and can lead to net protection or net erosion of the atmosphere.
This, among with other effects like pick-up and sputtering have been summarized in a comparative article on Earth, Mars and Venus in Gunell et al. (2018). Their key finding is

  While a planetary magnetic field protects the atmosphere
  from sputtering and ion pickup, it enables polar cap and cusp escape,
  which increases the escape rate. Furthermore, the induced
  magnetospheres of the unmagnetised planets also provide protection
  from sputtering and ion pickup in the same way as the
  magnetospheres of the magnetised planets. Therefore, contrary
  to what has been believed and reported in the press (Achenbach
  2017), the presence of a strong planetary magnetic field does not
  necessarily protect a planet from losing its atmosphere.

They find, that with all those complications of different escape processes and different ionized and neutral species, still for any single planet the ion escape rates can outcompete the neutral ones by a factor of $\sim 4$. Is this enough to explain the retention of the Venusian atmosphere and the escape of the Martian one? I think not. There is another factor that we've ignored so far.

6. Planetary mass

To close off this escape-story, I want to come back to the beginning of my answer, where I was happy that you've mentioned Venus. This is because Mars and Venus form a small set of a nearly identical case-study on atmospheric escape. Only two factors are different, while it is much harder to compare any other two planets in the solar system.

For the sake of comparing Mars and Venus to zeroth order, one could say that their atmospheres have identical composition, which is mostly $\rm CO_2$, as you have already stated. Both have no intrinsic magnetic field. Then, Venus has a hotter exosphere, stronger solar wind conditions, but still is somehow able to retain an atmosphere that has several thousand times more mass than the martian one.

If we now add to our perspective that a magnetic field doesn't play such a huge role for the retention of atmospheres, then the only remaining parameter that is different between Mars and Venus is the mass, which differs by a factor of about 10.
A factor of 10 is significant here, because if we go to planets now with 10 Earth-masses we already get into the regime of the Ice-giants Uranus and Neptune, which are able to hold on to their neutral hydrogen.
Understanding this is comparatively simple, as the totel escaping flux $\Phi_0$ is the integral of the Maxwell-Boltzmann-tail which goes very roughly as
$$ \Phi_0 \sim n(z_{\rm exo}) \cdot v_{\rm rms} \cdot \left( \frac{v_{\rm esc}^2}{v_{\rm rms}^2} + 1  \right) \cdot \exp(-\frac{v_{\rm esc}^2}{v_{\rm rms}^2}) $$
(source: Coates, or homework problems...)
where $n(z_{\rm exo})$ is the number density of a species at the height of the exosphere, $v_{\rm esc} = \sqrt{2GM/R}$ is the escape velocity of a planet with $M$ the mass and $R$ the radius and $v_{\rm rms} = \sqrt{2 kT_{\rm exo}/m}$ is the root-mean-square velocity of the MB-distribution at a given exospheric temperature $T_{\rm exo}$ for a given mean molecular mass $m$. So as there is an exponential factor involved, the Jeans-escape rate must go up quickly as planetary mass goes down.
So as ionic escape rates are a factor of a few times the neutrals, even if they dominate the scape, they're still bound to the exponential in the function.

More information can also be found in this, a bit older article by the same authors as the Zahnle paper, where they speculate also a little bit about escape from exoplanets, and the role of atmospheric chemistry for this effect.

Summary

The escape rate of ionic species can dominate over the escape rates of neutral species.
Ionic escape rates however don't react as strongly as one could think to the presence of a magnetic field. This leads to the interesting picture that the presence of a magnetic field is only a second order effect for determining escape rates.

The dominant parameter is then simply how deep the gravitational well of a planet is, that ions need to escape from. A escape rates go exponentially with planetary mass vs. molecular weight and available thermal energy at the exobase this is what determines escape rates most strongly.

Noble gases and their depletion can possibly resolve some mysteries of Earth's past, all the while telling us how much hydrogen has escaped from Earth.
As technology progresses, we might be able to address the same questions for Mars and Venus one day, but we're not there yet.

This was a long rant, please tell me if something is unclear.


What is the radiant intensity of the celestial sphere? Specifically, excluding the Sun and the Moon:
How much energy, as light and other frequencies reaches a point in space on the average, if you add up all the sources like distant stars, the Milky Way, distant galaxies, nebulae etc.? How much brighter is the space than "perfect blackness" on the average?

Say, we've developed a magical solar battery that has 100% efficiency, can absorb all electromagnetic frequencies, is spherical, fully omnidirectional and has $1 m^2$ of cross-section surface (56cm radius);  we sent it out halfway to Proxima Centauri and we measure how much power it produces - how many watts we'd get out of the battery utilizing all the starlight available?

Note: If "all frequencies" is not easily obtainable, the luminous counterpart to radiant values will be okay (only taking visible spectrum into account).

The electromagnetic energy density is dominated by the cosmic microwave background and the optical/IR background. This Physics SE answer, contains the plot below, showing the contribution of energy density at different frequencies. You can integrate under this "by eye" to see that the CMB contribution is the largest followed closely by an optical/IR bump.

A plot of $\nu I_\nu$ (proportional to energy density per logarithmic frequency interval), taken from Hill et al. (2018)

The CMB is approximately isotropic blackbody radiation with a temperature of 2.73 K and therefore a specific intensity of
$I_{\nu} = \frac{2h\nu^3}{c^2}\frac{1}{\exp[h \nu/k_B T] -1} $$

The integral of this over all frequencies is
$$ u = \frac{8\pi^5 (k_B T)^4}{15 (hc)^3} = 4.2 \times 10^{-14}\ \ {\rm J\ m}^{-3} $$

In terms of power, we just multiply by $c$ to get $1.3\times 10^{-5}$ Wm$^{-2}$.

Note that you can't harness this unless you have a receiver that is colder than the CMB.

You can do a rough calculation for the contribution of optical background light by noting that the dark sky at a good, dark astronomical site is around 22 mag per square arcsecond in the V band. Only about 0.1-0.2 mag of this could be attributed to light pollution at the best astronomical sites (see for example Benn & Ellison 2007). Noting that the zero point for the V-band magnitude scale is $3.6 \times 10^{-12}$ Wm$^{-2}$ per angstrom for $V=0$ and taking the optical band as 3000 angstroms, and noting there are $5.34\times 10^{11}$ square arcseconds in the sky, then a point in space will receive $10^{-5}$ Wm$^{-2}$ (or perhaps about half this, because a significant fraction of the dark sky brightness is caused by airglow).

This roughly agrees with what is shown in the plot. Contributions at mid-infrared and at shorter wavelengths than optical do not contribute appreciably.

It might be thought that discrete sources or the Milky Way might contribute more, but I think that isn't the case. The Milky Way surface brightness is only about a factor of three above the dark sky brightness, but of course occupies a tiny fraction of the total sky.

A way to see that discrete stars don't contribute much is to note that the optical flux from the entire dark sky is equivalent to about 1000 zeroth magnitude stars or $10^7$ stars at 10th magnitude. Both numbers are higher by orders of magnitude than the actual numbers of Galactic stars at these brightnesses.


Are there upper or lower mass limits for black holes? Natarajan & Treister (2008) describes a practical upper limit for black hole masses at $\sim 10^{10} M_\odot$. This is all due to the black hole's interactions with nearby matter.

However, is there a theoretical upper mass limit for black holes in general relativity? More specifically, do any solutions make note of this? Would this depend on whether the black hole described is eternal or time-variant, static or spinning, charged or uncharged, etc.?

Similarly, do any metrics make note of lower mass limits? Would it be possible for a black hole with the mass of an electron to exist (at any point in time, putting aside Hawking radiation)?

In classical General Relativity, black holes can exist at any size (mass) without any problem. The upper limit is given by the available mass of the universe and there is no theoretical lower limit.

As already noted in the question, quantum effects like Hawking radiation set up lower limits on stable black holes; the ones with too low mass will decay quickly into radiation.


How are the ESPRESSO Echelle spectrograph's calibration "lines" produced? The CBC News Technology and Science Q&A This ESPRESSO machine doesn't make coffee but scans the skies for habitable planets includes a nice description of ESPRESSO (Echelle SPectrograph for Rocky Exoplanet and Stable Spectroscopic Observations) and links to the YouTube video Echelle SPectrograph for Rocky Exoplanet and Stable Spectroscopic Observations.

In the spectrogram shown at about 00:55 in the video I've noticed that each band is accompanied by a finely space array of points. I am assuming that they are some kind of calibration, but not sure how they are produced.

Since the orders are changing, I can't really figure out if these are equally spaced in wavelength or in frequency.

How are these densely space calibration "lines" produced?

You can see the full spectrogram in this question.

above: Cropped subsections from a HD Frame grab from the video at about 00:55. Recorded with Python using this script, then extracting frames using this script in Blender.

You're right to think that this is a calibration technique. In use here are laser combs that use self-interference of short pulses that are reflected multiple times inside their own cavity to generate the spacing that you see.

The spacing then is mainly limited by the inverse of the pulse length, and thus by current laser technology. Sharpness of individual lines is usually increased by putting a Fabry-Pérot interferometer in sequence after the laser. For additional precision and estimation of laser stability, one uses often some secondary lamp with theoretically or laboratorially well-known lines (Th-Ar for example).

The technology demonstration for ESPRESSO was done with its precursor, HARPS.
Also A site with more details, the lowest poster looks pretty good in explaining the basics, I find.


Does the dimming of Betelgeuse present any observational opportunities? Recent photometry suggests that Betelgeuse has been dimming over the past three or so months, reaching the faintest V-band magnitude seen in modern observations. The changes are apparently even visible to the naked eye, and appear to correspond to a temperature drop of $\sim$150 K relative to maximum brightness. The second Astronomer's Telegram on the phenomenon ends with

  This continues to be an opportune time to carry out complementary measures of Betelgeuse while it is in its current low state and is unusually cool and faint.

This got me thinking. Obviously, continuing photometry will tell us more about the dimming itself. However, regardless of the mechanism behind the phenomenon, is there anything we can glean from the dimming about the star itself - e.g. composition, rotation, limb darkening, etc. - and its circumstellar environment (e.g. a binary companion, which I assume is unlikely, or circumstellar dust)?

Essentially, are there any observations of Betelgeuse that would be made easier by the recent changes? Or am I being overly naïve, and they're really too minor to be of any use?

In terms of "easier" - not really. There isn't anything made particularly easier or more optimal to measure. Theoretically, as you mentioned, "space stuff" like tiny planetoids or dust that are very close to the star itself might become more visible, but that's about it.

However, the reason we would want to get measurements now is to compare them to data collected when the star was at normal brightness and see differences, perhaps in spectroscopic lines, for instance. As well as that, if the star also returns to normal brightness, then we can observe changes that occurred side-by-side with the dimming or maybe even as a result of it.

Side note: Regarding the composition of the star, which you provided as an example - the reason we are unsure about when Betelgeuse will explode is because we can't actually know its true composition, since we can't see directly inside the star. If we could, we would be able to predict when it would finish fusing remaining heavier elements and then extrapolate when it might go supernova. All we can really see is what is coming off of the surface of the star, infer its age from other data, and then make an educated guess about when it will explode.


How does Io's atmosphere behave locally near volcanic plumes? Io's surface air pressure is about $0.3 {\rm mPa}$ but Io's atmosphere is strongly variable, depending on whether it's on the near side or far side of Io (relative to Jupiter) and it collapses at night. When it's heated up, it rises again, and so it also gets thicker in Io's volcanic plumes.
I'd like to know just what does happen with the atmosphere above a volcanic crater during an eruption? How much thicker does it get?
The eruption temperatures reach ${\rm 1\,300^\circ C} \, (2 \,400 {\rm F} )$.

My impression is that for answering your question, one would actually have to run simulations, ideally so-called global circulation models (GCM). If this is for a research project, the MIT GCM would be a good candidate to set up for the atmosphere of Io.
Of course, there are research groups working on that issue, e.g. the team of Prof. Goldstein at the University of Texas in Austin, and probably some more.
If you think this is like using a sledgehammer to crack a nut, you could also think about local plume modeling, I found a paper entitled A new strategy for the estimation of plume height from clast dispersal in various atmospheric and eruptive conditions which seems to come along with Matlab code:

Highlights [of the quoted paper]

A new revised numerical strategy for the determination of plume height is presented.
Accounting for wind, particle shape, gravitational spreading and climate zone.
Wind significantly affects the clast support envelope of weak and transitional plumes.
The new method is available through a Matlab package and a series of nomograms.

Edit/ Warning. Jean Marie Prival says

Being a bit familiar with the Rossi et al. (2019) plume model, I don't think it can apply here: it needs isopleth data as input, i.e., it needs maximum grain-size data from multiple locations in the field... (It is generally used to infer maximum plume height of past eruptions for which we don't have direct plume observations/measurements, but only tephra deposits.) Also, the model is parametrized for the Earth: things like particle density, tropopause height, etc., which are all different on Io

Last but not least, I would aim for some back-of-the-envelope estimates. For such a Fermi estimate we have to make some assumptions about the plume dynamics, and we also need some numbers. For that purpose, I found an Wired article by Erik Klemetti: Just How Hot is Io's Volcanic Surface? which might help us to estimate the state of the atmosphere, given by $T$, $p$, and humidity on Earth and probably ${\rm SO_2}$-content on Io.
Up to now I sketched 3 approaches to solve the problem in prinicple, without giving an answer. That's why I want to close my post by quoting the above mentioned Computational Fluid Physics Laboratory from U Texas which shows the result of their simulations.

For me it looks like the plume material shoots out of the atmosphere (and probably carying along parts of the atmosphere with it), before slowly setteling down on Io's surface. The plume height in the above picture is about ${145 \rm km}$, other websites report plumes of ${300 \rm km}$ height. It is a global phenomena which influences the overall atmosphere, probably causing wave-pattern all over the moon's atmosphere.
Related

I asked Characteristics of Venus' jet streams? which about the atmosphere on another object, but touches the same issue of running GCMs for other celestrial objects with atmosphere.
I also found The collapse of Io's primary atmosphere in Jupiter eclipse pretty interesting. The following paragraph supports my hypothesis that the volcanic plumes shoot out into the exosphere, as well as your statement, Ioannes, that the atmosphere is highly variable.

If the predominately ${\rm SO_2}$ atmosphere is in vapor pressure equilibrium with the large reservoir ${\rm SO_2}$ ice that covers much of Io's surface, as Io enters eclipse and is no longer heated by the Sun, the decreasing surface temperatures will cause the primary atmosphere to collapse out onto the surface [Saur and Strobel, 2004]. The atmosphere, with a scale height of 10 km, would transition from a collisionally thick atmosphere, with a mean free path (l) = 1 m, to an exospheric atmosphere (l ~100 km). Conversely, if the atmosphere remains constant during eclipse, it would point to a volcanically supported atmosphere.


How do radio telescopes gather information about visible light? Is the first image of a black hole that was captured by the Event Horizon Telescope what would be seen with a large enough optical telescope (or with the naked eye from a proper distance of the black hole)? If so, how do radio telescopes gather information about visible light?

How do radio telescopes gather information about visible light?

They don't; the picture you saw all over the news was a false color image, where frequencies are shifted from the radio spectrum to the visible spectrum (blue meaning shorter radio wavelengths and red meaning longer radio wavelengths) and/or the intensities mapped to different colors (instead of just affecting the brightness as with true color images).

Or, to quote an old StarChild question:

  The term "false color" is used to describe what astronomers (and others) often do to images to make them more comprehensible. Long ago, when radio astronomers first started generating images of sources, they wound up with essentially images that were just shades of gray - ranging from pure black to pure white. Each shade represented the intensity of the radio emission from a particular part of the object. Radio astronomers took their shades-of-gray images and converted them to color ones by assigning red to the most intense radio emission and blue to the least intense emission recorded on the image. Intermediate colors (orange, yellow, and green) were assigned to the intermediate levels of radio intensity. Black was assigned to places in the image in which there appeared to be no radio emission.


Is the Sun in our solar system moving or stationary? When I was small, I read that Sun is fixed at the center of the solar system and that all the other planets rotate around it.
But later I heard that even the Sun is not fixed; it moves. Is this true?
Why had people previously thought that the Sun is fixed?
Is this correct planets in helix path or spring like path?

The Sun moves, even in the context of the solar system. Gravity of the planets (mostly Jupiter) pulls the Sun out of position with respect to the centre of gravity of the solar system. This wikipedia entry explains it in a lot more detail, and also explains that their common centre of gravity lies outside of the sun.

This wobbling of a star due to planets orbiting them is also one method of how we detect planets around other stars.

Assuming that the Sun is fixed within the context of the solar system is a fairly good approximation. You need accurate and long term observations to detect the wobble.


Possible to make a cheap reflective objective lens with a mirror? Sounds kinda crazy and cheap but has anyone ever tried to take a flexible (plastic) mirror and attach a string to the back side then pull the string so it makes a concave surface? I realize it wouldn't be the most accurate device in the world, but if I wanted to build a telescope with a child I think it would be a cheap and fun way to make it. Thoughts?

That won't work because the shape will be too distorted to be useful. For astronomical purposes, a good quality mirror is needed since the objects you look at are dim and small. If you want a usable astronomical telescope then the optics need to be high quality

If you want to illustrate the properties of telescopes, a simple refracting telescope can be built using the lens from (low power) reading glasses as the objective and a (high power) hand lens as the eyepiece. I've also tried with a shaving mirror, but found the distortion to be too great to form a useful image.


Three-torus model of the universe I want to ask a question about the Three-torus model of the universe, as described in wikipedia:

https://en.wikipedia.org/wiki/Three-torus_model_of_the_universe

"is a proposed model describing the shape of the universe as a three-dimensional torus."

I'm not sure i understand the maths of three-dimensional tori, but I want to ask:

Does this model posit that the shape of the universe is that of a ring doughnut? If so, is the universe contained inside the ring doughnut, or is it supposed to exist only on the surface?

Or does the model say that the universe has the shape of a higher dimensional version of a ring doughnut? If so, does it exist inside the higher dimensional ring doughnut, or on its surface?

A three-torus is the boundary of the solid three torus, just like the two-torus is the surface of a solid donut. You can imagine it as a cubical room where each wall/ceiling/floor is a portal to the opposite-facing wall (i.e. the wall to your right is a portal that sends you to your left), but preserves orientation (when you walk out of the portal, your heart is still on your left).

You can also think of it as a world where your position is described by three coordinates (like x, y, and z in Euclidean space), but each coordinate corresponds to an angle on the unit circle. If you go far enough in one coordinate, you loop back to where you started (360 degrees is the same as 0 on a circle).


What happened to the 2017 proposal on redefining planethood? Is this information available? In 2017, Alan Stern et al. submitted a geophysical planet definition to the IAU for review which states
“A planet is a sub-stellar mass body that has never undergone nuclear fusion and that has sufficient self-gravitation to assume a spheroidal shape adequately described by a triaxial ellipsoid regardless of its orbital parameters.”
Obviously, the IAU didn't accept it (as yet) but does anyone know what exactly the IAU stated about the draft and whether it is still being debated or entirely discarded etc.?
I'm just asking if it is possible to know the IAU's reaction. While users may have opinions on how reasonable the proposal is or isn't, please don't post opinions as answers.

In the linked article by A. Bouchard and another by J. Daley,
the words "in the journal Lunar and Planetary Science" link not to a journal article
but to a poster in a K-12 education session
at the 2017 Lunar and Planetary Science Conference.
First author Kirby Runyon told Universe Today in 2017 that he would not submit this geophysical definition to the IAU process:

We in the planetary science field don’t need the IAU definition...
If [the geophysical definition] is the definition that people use and what teachers teach, it will become the de facto definition, regardless of how the IAU votes in Prague.

It was not presented at the 2018 IAU General Assembly in Vienna.
I don't know where Bouchard got "has been submitted to the IAU for review."


The specifics of two Earth-like worlds orbiting each other? I'm currently looking into writing a story or two set on twin worlds, which I came to find out in my research is apparently not a new idea (Double planets?
 Rocheworlds?  Am I close?).  But I want it to be as scientifically plausible as possible, so I'm here fact-checking.  I guess my main question going forward is: is the description of my worlds, as they are, scientifically plausible, and if not, what should I change?

Specifically, these two worlds are very similar to our own Earth.  To simplify things, let's say they have the exact same or relatively similar atmosphere and living conditions, so that humans and Earth-based life can live on it without much or noticeable acclimation.  If you were to hop a ship from Earth and landed on one or the other, getting off the ship you would notice little to no difference than you would on Earth.  Similarly, the short trip between each world shows little to no difference, although one has slightly more gravity than the other.

The only real difference between the two worlds themselves is that one is slightly smaller than the other, but by an almost negligible amount (a few thousand miles); the difference isn't as dramatic as, say, Earth and Mars orbiting each other, maybe Earth and Venus being the better comparison.
Let's also suppose that they are also far enough away from one another that they aren't dramatically effecting one another's gravity or tides, etc., but are still visible in one another's skies, prominent enough to be important to the culture of the other.  I initially wanted them to be very prominent in one another's skies, but I suppose being as close as the moon is to our world would be too close: I'm curious as to how far away one would have to be and still be caught in the other's gravity (I suppose the smaller of the two would have to be the one caught in the bigger one's gravity?  I plead ignorance on that one).

I'm in love with the idea that they broke off from one another in the same way the Earth and the moon supposedly did billions of years ago, in an impact event, and they slowly came to orbit one another (although this idea has recently been challenged, is my understanding).  From what I've read, though, it might be more plausible that one formed naturally in the disk formation of that planet's solar system, and a rogue planet got caught in its gravitational pull.  I like that idea less, but I'm not so proud I can't just get over it.

In my head, they have their own day and night cycles, just like Earth.  They may not have identical hours in a day or days in a year, etc., but its similar.  They same face of the planet is never quite facing the face of the other: it's not like the moon, which is locked to one side always facing us and the other always away.  From what I've researched, though, it's probably more likely that the planets would be facing one another on one side constantly, and their day-night systems would be greatly affected by that.  Again, not in love with that idea, but if it has to go, it has to go.

That's all I have for now.  I would try to explain how they rotate one another better, but I can't: I'm terrible in that area, and I've forgotten a lot of the terms you use in describing this.  Please let me know what your thoughts are.

A couple thoughts:
Tidal lock depends on the age of the system, so you can have it or not as you wish, depending on how long they came to orbit each other -- which is to say, of course, orbiting a common epicenter very nearly halfway between the pair.
Depending on your whim, you might want the orbit plane to be in or 90 degrees out of the ecliptic, the latter leading to interesting effects on tumbling.  I'd recommend staying in the ecliptic, which is more stable and allows each of the planets to rotate about an axis which allows night/day similar to Earth.  Notice that there will be a dramatic bunch of total solar eclipses of decent length, on both planets in turn.

As to their spatial separation, take a look at Newton's Laws.  Nearly any separation is allowed, though if they were very close they'd have suffered severe gravitational distortions (of shape) , messed up each others' atmospheres, and almost certainly have crashed together with horrific results.


How many space telescopes are currently active? Hubble, Chandra, XMM-Newton, Kepler,... I can name a few off of the top of my head, but how many space telescopes are there in all?
Related

What are the next planned space telescopes?

More than 20 if the Wikipedia's List of Space Telescopes is accurate. I extracted the active ones, and removed duplicates (to the best of my knowledge):

Swift Gamma Ray Burst Explorer
AGILE
FGST
IKAROS
NuSTAR
Astrosat
Insight (Chinese: 慧眼)
Спектр-РГ (Spektr-RG)
The famous Hubble Space Telescope, HST, see hst
STSat-1
IRIS
Hisaki
Lunar-based ultraviolet telescope
BRITE
NEOSSat
TESS, see tess
CHEOPS, see cheops
Odin
IBEX
AMS-02
DMPE
Chandra, see chandra
XMM Newton, see xmm-newton
NICER (Mounted on the ISS).

Interestingly, the majority is satellite based, but there is e.g. also a moon-based one.
In addition to that, there are space crafts whose Instruments have been temporarily or permanently been repurposed to astronomy. For instance, you could look at the Moon with METEOSAT.


Is the Astronomy community still concerned about the lumpyness of matter distribution in the universe? A decade or so ago, when I was still a science undergrad, one of the open questions in astrophysics was to explain the uneven distribution of galaxies in the observable universe.  That is, why did the universe look lumpy instead of evenly distributed.

Is this still an open question, and if so who is working on it?

The "lumpy" distribution of galaxies in the universe is now well understood.  In the current cosmological paradigm, the very early Universe was seeded with fluctuations in the distribution of matter and energy.  Some regions of the universe happened to be slightly more dense than average, and some happened to be slightly under-dense.

Over time, these perturbations evolve.  Gravity, for instance, will cause the initially over-dense regions to become more over-dense.  When galaxy formation occurs, it usually happens in the dense regions where there is more material with which to form galaxies.  Therefore, galaxies will tend to trace (to some degree) the underly fluctuations in the matter density.

In short, the lumpy distribution of galaxies results from (a) the initial density fluctuations set down in the early universe, and (b) the evolution of these fluctuations according to well known laws of physics.  Cosmologists frequently model the lumpy distribution of galaxies using, for instance, numerical simulations or theoretical models; we find that these models agree extremely well with the observed distribution of galaxies in our Universe.  In fact, measurement of the galaxy distribution is used as a powerful tool to test and constrain our cosmological models.

Of course, the above answer doesn't explain the process responsible for laying down the initial density perturbations.  One theory that attempts to address this question is called Inflation.


Is there a MATLAB equivalent of PyEphem? Many people on the Astronomy stack exchange use the PyEphem  package for the Python  programmimg language to perform astronomical computations.  Are the underlying numerical routines (from XEphem) also available for MATLAB?

Disclaimer: I've never used this toolbox and so cannot speak to the comparison of this toolbox with PyEphem.

There are some ephemeral and the like functions within Matlab, for example planetEphemeris, earthNutation, or moonLibration.

Unfortunately these functions are only accessible within the Aerospace Toolbox and do not come with a standard Matlab license. You'll need an entirely separate license for the Aerospace Toolbox to use these functions in your Matlab code.

If it is easy for you to obtain this license (be it from some university or organization), then by all means go for it. Otherwise, your best bet may be free packages like PyEphem.

I would also like to echo the comment made by David in that you should definitely learn Python. It is an amazingly simple, yet robust language that can do just about anything. And its free!

If you're really dead-set on using Matlab, but the PyEphem package is the only thing that does what you need, you can always consider porting between the two languages. Matlab has several tools and capabilities to interface with other languages. You can read more about using python packages natively in Matlab right here. I will say, this would probably be much harder than just going with Python outright though.


What are the chances of a star colliding with another during a galactic collision? My specific example for the question is the future collision of the galaxies Milky Way (our own galaxy) and Andromeda in a couple billion years. The star in question is obviously the sun in this case. I want to know the chances of a collision with another star and if it's significant or not.

Let's see what we get from some back-of-the-envelope estimates.

Imagine throwing one star (e.g., the Sun) at the other galaxy. How likely is it we'll hit a star in the other galaxy? Well, it's basically proportional to how big a target each star in the other galaxy is (its cross-sectional area) compared to the size of the whole galaxy, multiplied by the total number of stars in the target galaxy.

Let's assume it's the Milky Way-Andromeda scenario, so each galaxy has about 100 billion stars, and each star is roughly the same size as the Sun (some are much larger, most are smaller). The actual target area for an individual star is a circle with twice the star's radius (we're counting one star just grazing the other as a collision). Let's also assume the stars are more or less evenly distributed in a circular disk. Since "100,000 light years" is a common (and not completely crazy) estimate of the Milky Way's size, that's a circle of radius = 50,000 light years (about $10^{16}$ meters).

So: 100 billion stars in the target galaxy, each with target radius $\sim 2 R_{\odot}$, gives us a total target area of $10^{11} \times \pi (2 R_{\odot})^{2} \approx 10^{30}$ m$^{2}$.

The area of the target galaxy is $\pi R_{gal}^{2} \approx 10^{42}$ m$^{2}$. So the chance of our Sun hitting a star in the other galaxy is $\approx 10^{30} / 10^{42} = 10^{-12}$ -- or about one in a trillion.

The odds of any star from our galaxy not hitting a star in the other galaxy would be $(1 - 10^{-12})^{10^{11}} \approx 0.90$.

So there's only about a 10% chance of one (or more) of the galaxy's 100 billion
stars hitting a star in the other galaxy. And the chances of any one particular star (like our Sun) hitting a star in the other galaxy is about one in a trillion.


What are the challenges for the building and data analysis of the CHIME telescope? The  Canadian Hydrogen Intensity Mapping Experiment (CHIME) has discovered the second repeated fast radio burst recently. However, its structure is simple apparently. I wonder why we did not build a telescope before. What is the challenge? I just know that its data volume is large.
It is difficult to update the telescope?
There are 4 cylinders and 1024 antennas. What determines the distribution of the 1024 antennas? I mean why is it not just one longer cylinder?

There are many radiotelescopes that consist of large arrays of antennas. We usually think of an array individual dishes and VLA and ALMA are two of the most commonly recognized.

But if you want to keep an eye on a huge patch of sky in order to try to catch a very rare event, you want a wide field of view more than ultrahigh spatial resolution. And if what you are looking for is super-hot science, you want to build your device quickly which means a modest budget.

So wide field, lower resolution means a smaller baseline, and CHIME is of order 100 meters on a side. but building 1000 small say 3 meter dishes or 1000's of log-periodic dipole arrays and trying to steer them all would be a mess.

So instead CHIME built four, simple, parabolic cylinders and told all those antennas to be nice to each other and share!

Each antenna picks up signals from a huge area of the cylinder. If you include all directions it's like 25x25 meter section. The antennas are spaced only about a wavelength or so apart, but that's plenty far to do interferometry with the signals in a computer.

The ArXiv preprint The CHIME Fast Radio Burst Project: System Overview goes into great detail about the design and optimization of the antenna, considering what it is supposed to do and the necessity to make it straightforward to build an have a reasonable budget.

  Figure 1. Photograph of the CHIME telescope on 15 September, 2016, looking North-West. The shipping containers housing the X-Engine and CHIME/FRB backend can be seen adjacent to the right-most cylinder. The receiver huts containing the F-Engine are beneath the reflectors and cannot be seen here. The DRAO Synthesis Telescope (Kothes et al. 2010) can be seen in the background. See Table 1 for detailed properties
  of CHIME.

The Field of View in the North-South direction (the axis of the parabolic cylinders) is about 110 degrees, but in the East-West direction it's of course very narrow, only 2.5 to 1.3 degrees.

I've borrowed the following text, images and links from my question How does Canada's radiotelescope CHIME manage polarization-dependent apodization of the fast f/0.25 optics?

Here's a picture of some of the antennas, looking down into the trough:

below x2: from The cloverleaf antenna: A compact wide-bandwidth dual-polarization feed for CHIME

Fig. 1. At left, the shape of each petal consists of two perpendicular straight
lines, two 45 degree circular arcs with radius R and one half an ellipse. W is
the major axis of the ellipse and L is the length from the intersection of the
straight sides to the outer edge of the ellipse. The shape is illustrated here for
the adopted values of gap, R, L, and W. Each of the four tabs shown at the
centre is connected to one side of a vertical microstrip transmission line and
in each case the full width of the adjacent petal is connected to the other lead.
At right, CST simulated currents for one linear polarization at 600MHz are
shown. Note the small asymmetry in the curent distribution near the centre
becasue of the tab geometry.

Fig. 3. A linear array of eight cloverleaf antennas installed at the focal line of
the CHIME Pathfinder at the Dominion Radio Astrophysical Observatory in Penticton, BC, Canada. The picture is taken through the wire mesh reflective surface (mesh spacing 19 mm) illustrating a photonsview of the antennas and ground plane. Notice that each feed has an image-feed in the ground plane, 1/2 λ away at the passband centre frequency. Notice also the four slots cut to remove dielectric material from the gaps between the petals.

below: From https://chime-experiment.ca/instrument

  CHIME consists of four adjacent 20m x 100m cylindrical reflectors oriented north-south. The focal axis of each cylinder is lined with 256 dual-polarization antennas, each of which receives radiation from a large swath of sky that nearly stretches from the northern horizon to the southern horizon. This gives CHIME its enormous field of view. The CHIME antennas are custom-designed to have good sensitivity from 400 to 800 MHz, in both linear polarizations. This gives CHIME its large frequency coverage. Signals from the antennas are amplified in two stages, using low-noise amplifiers developed by the cell-phone industry. This novel application of consumer technology makes CHIME affordable. The 2048 signals from these inputs (256 antennas x 2 polarizations x 4 cylinders) are fed to the F-Engine for the initial stage of digital processing.


How to calculate the time of Ekadashi days? Is there an official reference algorithm for calculating the Ekadashi days?

I found the textual description of Ekadashi calculation on wikipedia but can't really make sense of it in order to write the code.

I already have the (java) code for calculating the new/full moon phases, and originally assumed that Ekadashi would be as easy as appending 11 days to the time of new and full moon but apparently this approach is not providing correct results.

Update:

Based on James's answer the answer is 11 lunar days and 26 lunar days after new moon (not gregorian days).

The required conditional adjustments are described in chapter 1.4 of Vaisnava Events Calculation which don't match the ones described in the wikipedia article.

It is not clear that it can be calculated in a simple way

The length of the lunar month varies between 29.3 and 29.8 days, due to perturbations from the sun. Since ekadashi is based on the actual astronomical moon, you would need to know the length of the current synodic month in order to work out when ekadashi is.  You can predict the motion of the moon very accurately, but the calculation is beyond this answer.

However on average, each lunar month is 29.53 days, so ekadashi is the day which is 259.86 hours and 614.22 hours after each new moon. The calculation is
24 hours/day * 11/30 * 29.53 days

And New moon last occurred on 4th Feb at 21.03, you can add 29.53 days to find the next new moon etc.

The irregular motion of the moon is why the Christians choose to use an "average moon" and an "average sun" for their calculation of Easter.


Temperature gradient in stars It is a well known fact that in stars, there exists a temperature gradiënt. The
observational reason is because we perceived spectral lines in the otherwise continuous spectrum of a star. If this wasn't the case, and thus if the temperature were uniform, then the absorption and the emission would cancel each other out and we would observe a net effect that results in a continuous spectrum. Now my question is: what is the physical explanation for this temperature gradient?

You're talking about the temperature gradients in the atmosphere of a star, that's where the emission and absorption lines are originating. This is a much more complicated matter than the star itself.

The temperature of a star must increase from its surface downwards.
This is a simple result of the fact that the stellar structure is given by pressure gradients balancing the force of gravity. We can express this via the law of hydrostatic equilibrium
$$\frac{\partial P}{\partial r} = -g(r)\rho  $$
which is a pretty good approximation to reality in most cases.

The pressure gradient then translates into a temperature gradient, as the pressure originates in the thermal motion of particles, formalized as the law of ideal gases
$$ P = \frac{\rho k_B T}{\mu} $$.

So wherever there's non-zero gravity and no other force in a gaseous medium, there will be temperature gradients.

In an atmosphere, things depend on how efficient the gas is at cooling. In very thin media, like in Earths or any planets / stars exosphere cooling is inefficient enough to make temperature gradients vanish.


Confirm: We can only notice planets if their orbit passes in front of their star and our view If we can tell a planet exists by the dimming of the brightness of a star when it passes in front of our view of the star does that mean we cannot see a planet if it does not cross in front of our view?

If planet of a distance star goes around the star rather than in front of it than we cannot see that planet. Correct?

The answer by choeger provides a correct outlook, but I wanted to provide a bit more detail on some exoplanet detection methods that have actually been used to find planets we couldn't see. Now I interpret "see" in the manner posed by the question in that "see" references observing a transit of the star by the planet. The questioner is correct that we cannot "see" the planet in this manner if the alignment is not perfectly correct. So how else might we detect the planet?

Radial Velocity Method - We can try to detect its gravitational effects on the star through a method known as the Radial Velocity method. Both the star and planet(s) will be orbiting their common center of mass. That means if we watch the star, we can see it orbit that center of mass as well. If we see the star "wobbling", we can infer what mass must exist in that system to cause the wobble. I should point out that we are only able to see radial wobbling (that is, wobbling of the star towards and away from us) which means that while we can still detect planets which are not directly aligned with the star as they would be for the transit method, they are much harder to detect. I also want to point out that the answer by choeger mentioned tidal forces, but that is actually not something we can use to detect planets or measure their effects on their host stars.
Direct Observation - This is exactly what it sounds. With some clever techniques and the right conditions, we can take a picture of the planet itself. One pretty famous example was the direct imaging of Fomalhaut b. Shown below is an image over a few years as it travels through its orbit.

Microlensing - This uses the concept of gravitational mass acting as a lens to background, luminous objects by bending spacetime, and consequently, the path of light through space. If a planet passes in front of a background object (not its host star), it will cause a slight magnification of that background object. I should say that this requires a chance alignment and so it is not a repeatable measurement, meaning there is often little chance that we can actually confirm that we saw a planet. Below is a light curve of a background object and you can see it grow brighter as the main star passes in front of the background object. Pay attention to the little blip on the right which indicates the planet also passes in front of the background object.


Why does the moon not leave its orbit around the earth due to the gravitational force from the sun? See additional information below added after the answer to the question.

When the moon is between the earth and the sun, the gravitational force on the moon from the sun is greater than the gravitational force from the earth, hence the moon should continue its path towards the sun instead of staying in its orbit around the earth.

When the moon is at 90 degrees to the earth relative to the sun, the moon is moving almost directly towards the sun and then the gravitational force on the moon from the sun is roughly double the gravitational force from the earth, hence there is nothing to stop the moon's trajectory towards the sun and it should continue towards the sun away from the earth. What makes the moon "turn" back towards the earth and away from the sun at the point where it is right between the earth and the sun when at this point right when it turns, the gravitational force from the sun is roughly double the gravitational force from the earth?

I am aware of the movement of the sun and also of the fact that the moon is moving on a different plane to that of the earth and I don't think that can explain my question above.

Thank you for your time, I would like to understand this situation.

Additional information:

The answer does not address the question, let me explain. Although it is true that the moon orbits the sun as described in the answer, this does not explain why the sun does not pull the moon out of its orbit around the earth.

For an object to change direction a net force has to be applied to it according to Newtons laws. a=F/m applied to the moon from the point in its orbit where the moon gets closer to the sun than the earth is to the sun, shows a force vector on the moon towards the sun applying twice the force as another force vector on the moon towards the earth. This results in a combined net force vector on the moon pointing away from the earth and this net force vector will progressively point further away from the earth and more towards the sun as the moon moves closer to the sun and therefore should pull the moon out of its orbit around the earth and pull it towards the sun.

Newtons laws must be true on the three separate objects; the sun, the earth and the moon in the example above.

I have a Master of Science in Engineering and know my physics. I have some friends who are convinced they can prove that the earth is actually flat and that a search on "flat earth" will prove it, and they have asked me the question in this thread to prove to me that the heliocentric model is not true. I took up the challenge, however when looking at this simple example I have not been able to find an answer that is in accordance with normal physical laws.

I hope somebody can come up with an explanation that is in accordance with physical laws.

The speed of the Earth (and Moon) around the Sun is $\approx 29.8 \frac{km}{s}$.
The speed of the Moon around the Earth is $\approx 1.05 \frac{km}{s}$.

The result is that the Moon has a nearly perfect circular orbit around the Sun, the effect of the Earth is only a perturbation in it:

In fact, the "waves" are much smaller (400.000km Moon orbital radius to  150.000.000km Earth orbital radius)

This is mainly because $R_{Earth-Moon} \ll R_{Earth-Sun}$, and $M_{Moon} \ll M_{Earth} \ll M_{Sun}$. In general, 3-body gravitationally bound systems tend to be unstable, i.e. one of the components tends to fly away with time. But this system seems stable, because the ratio of the masses are big.

The same can't be said from the whole Solar System, (it is a notoriously unsolved problem) although also this seems stable in human time scales.


Are black holes considered part of baryonic content of universe? Are black holes (like those detected by LIGO, for instance) part of baryonic content of universe or part of dark matter? Why?

Black holes formed after the epoch of primordial nucleoynthesis would be part of the baryonic matter, since they would be mostly formed of baryons. It is possible or even probable that every black hole must contain some non-baryonic matter too, since there is nothing to prevent it being accreted, in small amounts, directly into black holes. However, the cross-section for that is very small because dark matter falling towards a black hole has no means to lose angular momentum. Unless there is almost a direct hit within the photon sphere then it won't be accreted. Normal matter can form an accretion disc and lose energy and angular momentum through viscosity, radiation and other processes.
Since the fraction of dark matter that is in black holes formed since the big bang is negligible, then the approximation that their mass consists only of baryons is hardly a big deal when it comes to doing the mass budget for the universe.
Primordial black holes on the other hand, that were already present at the epoch of cosmological nucleoynthesis, would not be counted as baryonic matter and it would be possible that they could contribute significantly to the required non-baryonic component, if they exist.
Could there be dark matter black holes? is essential reading.


Why won't the Milky Way colliding with Andromeda affect the Solar System? According to Wikipedia

  Such an event would have no adverse effect on the system and the chances of any sort of disturbance to the Sun or planets themselves may be remote.

I'm imagining that this is because the distance between stars is so big, it'd be highly unlikely for any star to actually collide.

But the Oort Cloud is 2 light years in diameter. Given that the average distance between stars in the Milky Way is about 5 light years, that's almost half way. If Alpha Centauri had an Oort Cloud of the same size, we'd already almost be overlapping each other.

Assuming that we're not the only star with an Oort Cloud, isn't there a high probability that the Sun will actually be inside the Oort Cloud of a passing star?

I would imagine this would because the Sun and our planet would be bombarded with lots of comets which have been thrown off their orbits by either the gravity of the Sun or the passing star.

Outside my field, but collecting information from the Wiki article, I can make an attempt nevertheless:

You're right that we will probably pass through Oort-like clouds of other stars. But the largest of the planetesimals of the Oort Cloud is measured in kilometres, whereas the distance between these objects is measured in AU, so the total cross section of object compared to that of the cloud is still miniscule. Smaller objects are larger in number, but still it's a very low probability.

Gravitationally, the clouds also won't affect us. The total mass of the Oort Clouds is probably of the order of 10 Earth masses, i.e. much smaller than the Sun.

The outer parts of the Oort Cloud is only loosely bound to the Sun. I suppose the best evidence that we won't have any major collisions is that then we would already have similar objects bombarding us from our neighboring stars' clouds.


can we use connect a long pipe with space station and use vacuum for fuel transfer? We all know that we need a lot of fuel to escape from earth. Even 90% fuel of an spaceship is wasted in escaping the earth which definitely affect our power to explore the universe. we have to do a lot of calculations and even a small failure can waste a lot of money. for ex- the Manglayan Indian mission to Mars due to under performance of the engine was almost lost. As it a cheapest mission to Mars. They have a very limited fuel. So why can't we connect our space station with a long pipe and Make our space ship refuel at that point. I know a lot of energy will be required to pump the fuel up. But I was thinking about a better option. why can't we use vacuum to transfer liquid. Like if we have a pipe and we suck one end and the water automatically travels. Is it possible or what are the technical difficulty here.

A pump can lift water about thirty feet. That's the height where the column of water in the pipe weighs the same as an equivalent column of air (and that's why deep wells use submersible pumps: they push the water up from below, instead of pulling from above and relying on atmospheric pressure to provide the lift). While fuel would be less dense than water, it's still much more dense than air, and the pressure of the atmosphere simply doesn't provide enough force to raise that column of fuel very far.


calculating rising time from culmination time I want to know how to calculate the rising time of a star, if
culmination and setting times are available with me?

Say Culmination time is 03:00 on 28-Nov-2018
and Setting time is 05:30 on 28-Nov-2018.

For a star, the time from rising to culmination is the same length as the time from culmination to setting. From knowing the time of culmination and setting, you can calculate that difference and subtract it from the time of culmination to obtain the time of rising.

This method is generally accurate for the planets and Sun, but it is not always accurate for the Moon. Culmination refers to when an object is highest in the sky. For a star, this is when the object is on the meridian. For the Moon, there can be a significant difference between the time the Moon crosses the meridian (transits) and when it culminates. See Transit and Culmination (although the images on that site were not working today).

P.S. I have never seen tables that give the time of culmination for the Sun and Moon. Times are generally given for transits (when the object is on the meridian).


Is there a clear-cut difference between rocky and gaseous planets? A gas giant planet is large enough that it retains a lot of hydrogen and helium. A rocky planet is one with a solid surface. That's the rule of thumb to distinguish between the rocky and gaseous planets.

But gaseous planets can have a solid core and rocky planets can have a very thick atmosphere. So is there a clear-cut difference between the two categories of planets (perhaps linked to different formation histories), or is there a continuum of planets that goes from more rocky to more gaseous? If it's a continuum, where is the limit set between the two types?

The currently-detected planets do not show a clear distinction between rocky and gaseous planets. While there seem to be somewhat two ensembles, the rocky planets of Earth-size and "super-earths" and the gaseous planets (Jupiter-size), there is a broad transition between them. The transition is roughly where we find Neptune and Uranus. Play around with plotting various parameters at exoplanet.eu

It's still subject to debate whether this apparent depletion in the 'ice giants' (cold, gaseous, but with heavier gases like Methane etc instead of Hydrogen and Helium as main constituents) is real or due to observational bias.

In brevity:

The argument for "is real" is that it is more likely that in the inner nebula, inside the snow zone you have enough material to form terrestrial planets, but being too hot and too few to grow to a size that a sizable atmosphere can be retained. Outside the snow zone you have plenty of more solids (all the ices), so that you form cores which become big enough to also accrete an atmosphere.

However, the counter-argument is that you do find planets of all sizes in all kind of stellar distances, you have various initial conditions with different densities of solids in protoplanetary discs, and thus any final result is possible with possibly various probability. The actual observations seem to support this.

Generally, in current literature and on conferences, planets are often distinguished further than just rocky and gaseous: you have terrestrial, super-earths, neptune-likes and the gas giants, and often then somewhat separated even the so-called hot jupiters and hot neptunes, the gas giants which migrated inwards after formation.

EDIT: one possible distinction which could be made between rocky and gaseous is looking at the atmosphere : if the atmosphere gets dense enough at the lower levels that the distinction between gas and liquid vanishes (thus the critical point reached) , it's a gas planet, otherwise a Rocky planet with atmosphere.


Is there a webpage that shows the night sky but can filter out dim stars? This webpage is very good for viewing the night sky and learning the names of the stars
Timeanddate - Astronomy
However, it shows too many stars. Because I live in the city, I can only see about 50 stars. If I could filter out all of the dim stars, that would help me focus on just the top 50 brightest stars. If I want to learn the names of the stars and how to identify them I have to start with the brightest. Yes, you can scroll over the star and it will tell you how bright it is, but that is very time-consuming.

I think your requirements can be met by Stellarium. It is a freely available open-source planetarium software available for PC, and can be used offline.
There is also a web version, which you can try out here.
You can filter stars out by pollution levels, as illustrated in the stellarium wiki.
Here is an Astronomy Stackexchange answer on matching the magnitude of stars to what is visible from specific cities.