Can the Keck Observatory be improved with a third telescope? Can the Keck Observatory resolution be improved by adding a third telescope? I imagine that that would be cheaper than building a completely new telescope?

Original Overview
The Keck telescopes mainly work independently of one another. Therefore, adding a third telescope wouldn't really have much of a point (aside from the benefits of having a telescope on top of a mountain with low light pollution and excellent atmospheric conditions). In the past, this might have been a slightly more appealing idea, because the Keck Interferometer (KI) ran from 2003 to 2012, and was a combination of the two telescopes (an effective increase from two 10-meter telescopes to one 85-meter telescope).
KI had some problems, though, that prevented it from living up to its full potential. For instance, four secondary "outrigger" telescopes to improve the interferometer were never built, even though they were an important part of early plans. It was eventually "mothballed" after several of the main scientific objectives were completed. Obviously, the interferometer could be reactivated, but you'd have to make a strong case for doing so. If you could get that to happen, and get permission to build another telescopes, many might choose to use the funds for these outriggers, rather than a whole new instrument.
Finally, building telescopes on Mauna Kea is controversial because of possible environmental impacts, as well as arguments that it endangers Hawaiian cultural and religious sites. There are objections enough over the planned Thirty Meter Telescope, never mind another Keck (which would, admittedly, probably be cheaper, although the two would use different wavelengths).
In short, a third telescope would help only if the interferometer was brought back on line - and even then, you'd have to make a case for both building a new telescope and for choosing it over the original outriggers.
Resolution
The angular resolution for a single telescope, to an order of magnitude, is
$$\theta\sim\frac{\lambda}{D}\tag{1}$$
where $\lambda$ is the wavelength of observations and $D$ is the diameter of the dish (adaptive optics, by the way, can make things a bit better). Each Keck telescope has a diameter of $10\text{ m}$ and observes at the optical and near-infrared ranges - so about $4\times10^{-7}$ to $4\times10^{-6}\text{ m}$. This approximation yields angular resolutions of about $0.0083$ to $0.083$ arcseconds. This approximation is accurate to within an order of magnitude.
Likewise, for an interferometer with two telescopes separated by a baseline of distance $b$, the angular resolution is approximately
$$\theta\sim\frac{\lambda}{b}$$
The Keck interferometer had a baseline of $85\text{ m}$ and observed at wavelengths of $2.2\times10^{-6}$ to $10^{-5}\text{ m}$. The approximation gives us resolutions of $0.0053$ to $0.024$ arcseconds - pretty close to the actual values.
What if we add a third telescope, identical to the other two and placed in a line next to them, $85\text{ m}$ from the closer one and $170\text{ m}$ from the farther one? The longest baseline is now $2\times85=170\text{ m}$, twice the baseline of the two-dish setup. We've now made the angular resolution twice as small as before. That's excellent. At $2.2\times10^{-6}\text{ m}$, the resolution is $1.3\times10^{-8}$ radians.
Let's think about observations at $2.2\times10^{-6}\text{ m}$, at the shortest wavelengths the interferometer used. Assume we want to build another telescope with the same angular resolution. Rearranging $(1)$ gives us
$$D\sim\frac{\lambda}{\theta}$$
Plugging in our numbers ($\lambda=2.2\times10^{-6}\text{ m}$ and $\theta=1.3\times10^{-8}$ radians), we find that $D$ should be about $170\text{ m}$ - which is of course the longest baseline of the three-dish interferometer. This is simply implausible at these sort of wavelengths. The largest single-dish optical telescope - observing in shorter wavelengths than this by a factor of about three - is the Gran Telescopio Canarias, at $10.4\text{ m}$ (although the Large Binocular Telescope is slightly better - $11.8\text{ m}$).
If you want to use the interferometer at optical wavelengths, you've still looking at a $\sim55\text{ m}$ dish, and that's currently maybe feasible. There are several extremely large telescopes being planned, notably, the Thirty Meter telescope I mentioned earlier and the European Extremely Large Telescope (E-ELT), coming in at around $40\text{ m}$. The E-ELT should be online as soon as 2-24 (!), which would be fantastic, and would have a comparable resolution to this three-dish interferometer.
Costs
Let's look at how much all of this might cost. The E-ELT is expected to cost a little over one billion euros (1.083 billion, to be precise), which comes out to around \$1-\$1.25 billion, in US dollars (that sounds like a lot, but space-based telescopes can be up to ten times that - though often with higher-quality images). Let's assume that our putative $\sim55\text{ m}$ optical telescope costs \$1.5-\$2 billion to build.
The cost of the instruments for the Keck Observatory ran about \$80 million. This includes the telescopes proper, as well as the cameras and other equipment needed to actually capture the images. Making improvements and other developments has run the total project to about \$200 million (not counting observing costs) We can be generous and say that building a third telescope - identical to the other two - and getting the interferometer system up and running might cost \$50-\$60 million. Maybe that's off, but it certainly can't be wrong by over an order of magnitude. At that point, the entirety of this new Keck Observatory has cost only about one tenth of what a single dish would cost.
Yes, it's likely that I'm wrong somewhere above - that I've made an error in the size of the single-dish telescope, or in estimating the total cost. But the discrepancy is still extraordinary.
Here's the problem, though: Nightly observing costs are high. One night using on Keck telescope costs \$54,000. The interferometer uses both telescopes, and, as an article I cited earlier notes,

it's complicated and expensive to link the giant eyes through a system of optical pathways for just a few dozen nights each year.

Interferometry isn't as simply as looking at an object with two different telescopes. It's complicated and costly. If we assume that operating the interferometer would cost around \$200,000 per night (is that gratuitous?), we find yearly operating costs of $\sim$\$73 million - assuming that it's used every night, which isn't likely. But even accounting for all the nights when the interferometer wouldn't be in use, that's probably still more than the estimated yearly costs for the E-ELT (50 million euros, or \$54.33 million). The single-dish telescope - again, operating a shorter wavelengths than Keck - would probably cost less per night than the new Keck Interferometer.
Perhaps the relatively low construction costs of the third telescope would offset all of this in the short-term (it likely would).
Here's the really obvious thing, however, that I've been ignoring: More dish means more light. An interferometer with an $170\text{ m}$ baseline is not the same as a single dish telescope of the same diameter. The three dishes are much, much smaller than the single one. While the resolutions are the same, the size of the dishes really does matter.
A dish with a diameter of $10\text{ m}$ has an area of $25\pi\text{ m}^2$. A dish with a diameter of $170\text{ m}$ has an area of $7225\pi\text{ m}^2$ - a difference of a factor of 289! If you want to make true comparisons between the new interferometer and a large telescope, you'll need more than three dishes. And that will make costs skyrocket.
Just for fun, imagine that we end up having 50 total dishes, with a baseline. If each costs \$30 to \$40 million to build (and here I'm being generously low), we quickly find that building 48 more dishes runs us to \$1.9 billion, or at least the cost of building our $55\text{ m}$ optical telescope.
Okay, so let's again assume that I made an error somewhere, and this figure is too high. Also, if you ended up building this interferometer (Where would you put it, by the way? Mauna Kea's crowded enough!), you could probably make each dish simpler than Keck I or Keck II. You still have to deal with high nightly operating costs - really high operating costs.
Science!
So, why isn't the interferometer up and running? According to NASA, it's because the interferometer's main target, observing certain circumstellar disks, is done:

NASA’s primary goal in developing and operating the two 10-m telescope KI was the characterization of faint dust disks around nearby main sequence stars.

For an overview, see Millan-Gabet (2011). As the authors note in section 3, the interferometer's primary observing wavelengths (about $10^{-6}$ to $10^{-7}\text{ m}$, the micrometer range) encompass the peak emission wavelengths of circumstellar dust. Additionally, the resolution at these wavelengths can resolve features of circumstellar disks.
As we determined before, it's hard to build a telescope of comparable resolution operating in the near-infrared. Remember, our single-dish telescope operates largely in the optical range, although it would of course have near-infrared capabilities, to an extent. Therefore, an interferometer like the Keck Interferometer is our best choice - and perhaps our only choice, in the foreseeable future. However, it's certainly possible to build extremely large telescopes operating at slightly shorter wavelengths, with roughly the same angular resolution (to an order of magnitude, really). Plus, they'll gather more light, and - as I found shortly before posting this, in an excellent answer by Rob Jeffries - limitations of interferometer instruments and atmospheric issues makes them less versatile than large, single-dish telescopes.
TL;DR
There are a few takeaways here:

Adding a third telescope would give you a resolution unmatched at those wavelengths by any other land-based telescopes.
This would likely be a lot cheaper.
However, you'd gather only a small fraction of the light that a much larger telescope would.
To gather more light, costs would skyrocket - as would technical problems.

So, economically and technically, it's really not feasible to add a third disk to Keck - at least, compared to building a single extremely large telescope.

This is a pretty long answer, and I'm guessing I made an error somewhere. I'll go back and do one more check in the next day or so. If anyone finds one, do let me know!


Planets just look like dots! Beginner telescope, hoping for some advice I'm very new to astronomy and got myself a cheap, used  700mm 70mm diameter refractor scope and am able to see (what I think are) good images of the moon. However when I try to view planets I see nothing but a very small dot! Pretty sure I was looking at Mars this eve (slight red colour) but with a 10mm lens its still just a dot. I can make out no detail. Same with Jupiter. Should I be able to see detail with the specs I have? Thanks in advance.

Most likely you will not see much detail except that they are a really tiny disc and don't wobble like stars.
The moon has an apparent diameter of 30' (half a degree) in the sky. Mars' apparent diameter varies between 3.5" and 25" (so about between 1/1000° and 1/120°). Thus at closest approach Mars is still about 60x smaller than the moon, currently at ~<10", it's about 1/180 compared to the Moon). How much detail do you see at such tiny fraction of the Moon in your telescope?
You can expect to see on Mars maybe some very slight colour variation between equator and poles. On Jupiter with an apparent diameter of ~50" (thus twice Mars' apparent size) you might see some slight colour changes with the cloud bands in its atmosphere and on Saturn you might see that it has rings.
All the images with details on the surface are both, taken with a camera and selected for (missing) atmospheric distortion and post-processed to enhance the features, and taken with a larger telescope where you can use a larger magnification.
But even with a larger telescope (longer focal length AND larger apperture), you WILL need to have good seeing; there are nights where even with a clear sky, also a larger telescope will not resolve features due to the seeing, thus the influence of Earth's atmospheric disturbances.


South-eastern sky above the tree line just now, NE Cape Cod, I saw a very bright, twinkling "star" I live in New England, Cape Cod to be exact.

I saw a very bright, twinkling object. I watched it for several minutes and it never moved. It was in the east just above the tree line, what might it be?

I just did the following:

opened in-the-sky.org,
manually chose the location New Bedford, MA (close to Cape Cod) because I'm somewhere else (you wouldn't have to do that)
selected planetarium

and looked in the southeast and found that the brightest star in the sky Sirius is currently only 20 degrees above the horizon.  At that low altitude, and being a star (not a planet) it would definitely be bright and twinkle, and not appear to move like a plane.

It is rising, and moving diagonally to the right and up. Check back in an hour or so to confirm that if you are interested, or check in the next several days to confirm it's in the almost the same place each night.


Why do the gas giants in the Solar System have comparatively large orbits compared to the inner planets? Ever since I observed the depictions of the Solar System, I was obsessed with the question of why the gas giants (outer planets) have very large orbits, compared to the planets that are closer to the Sun. Is it because of their mass or something I have ignored?

The current explanation for this is something called the frost line (which changes over time). At greater distances from the Sun, a body will receive less and less radiation, and so it will be colder than if it were closer to the Sun. Eventually, conditions become cold enough for volatiles to condense into grains. These volatiles make it possible for large-scale accretion to happen, and thus giant planets are formed. They may become gas giants (like Jupiter and Saturn) or ice giants (like Uranus and Neptune), depending on the composition of the protoplanetary disk at a given distance.

A graphic (from here) gives a better summary, although it's copyrighted.


How many things are wrong in this "artist view" of the TRAPPIST-1 system? There is a this poster on the NASA site

(

that irks me. Of course its an artist's fantasy, but since it is on a NASA site, I see people considering it as scientifically accurate. So, beyond the two obvious "In which direction(s) is the sun", and "OMG these two planets are crashing into each other! Before we crash in the debris!", what else is wrong? Tides? Roche's limit? Atmospheres ripped out?

Also, since 1e is an intermediate planet, it is possible to have the other six planets all conveniently placed in the same area of the sky?

The 2 big planets are probably f & g and they don't look up to scale to me. While f is the size of Earth, or almost 4 times bigger than the Moon, but it's also further away, ~ 1.3 million kilometers, as opposed to 400,000km for the Moon. So I would say f will be somewhat larger than the moon but not anywhere near what the poster shows. G will appear even smaller than f because it's further away.
I haven't looked at the orbits in details but I think that sort of event should be possible. Trappist-1e being intermediate planet makes no difference.
As for "where is the sun" - it appears to be somewhere to the right, probably below the horizon. That makes sense because any base would probably be along the terminator.


What are the objects that are visible from a city? I live in a light polluted city of the northern hemisphere and I would like to know what I can see with my Skywatcher BK 707AZ2. I have already seen Jupiter (and its moons), Saturn, Mars, M45, the Double cluster, M42 and, obviously, the Moon.

M31, M13 and maybe M33 are also probably visible, but what else?

We can't know this without knowing what the sky is like for your city. However, you can determine what the limiting magnitude is for your location, then you can set planetarium software to only display catalogued stars that are brighter than this limit. You can do this for naked eye objects, but you could also do this using your telescope to find the limiting magnitude for your equipment also.


number density of Dark Matter Halos Versus number density of Galaxies Can we count only one dark matter halo for each Galaxy ?

Indeed, I am trying to estimate the total density of galaxies in given volume and I have often seen on physics exchange and other forums the estimation of dark matter halos distribution (as a function of the mass of halo or as a function of Luminosity).

If I could count one galaxy per halo, this would make the things simplier but I think that things are more complicated, especially since a halo and the galaxy associated to this halo have not the same mass, so distribution mass is different (mass of halo is not linear towards the mass of galaxy, isn't it ?).

If someone could help me on this issue, regards

No. You need to specify structure formation and galaxy evolution for details, but in general, you will expect many galaxies in a dark matter halo. If you need a relation, the halo occupation distribution might be a start. The Wikipedia page references three good texts on the matter.


What direction does 0° in the J2000 epoch point towards on January 1, 2000? Is 0 degrees in the J2000 Epoch 'right ascension' is a line from Earth to the sun on January 1, 2000? Or is 0 degrees the direction towards the ascending node 'equinox'?  Which is NOT towards the Sun on January first...

Or is 0 degrees the direction towards the ascending node 'equinox'? Which is NOT towards the Sun on January first...

The answer is yes, more or less. From JPL's HORIZONS, the location of the Sun at Noon Terrestrial Time on 1 Jan 2000 is, ignoring atmospheric effects, a right ascension of 18 hours, 45 minutes, and 9.36 seconds. This is nowhere close to a right ascension of zero hours (or 24 hours, same thing).
The right ascension of the Sun with respect to the center of the Earth is zero at the March equinox.


Photometric redshift - empirical methods (machine learning) I am currently learning about estimation of photometric redshifts with machine learning methods (or empirical methods in general).
These methods use the knowledge about the photometry and the spectroscopic redshift of many galaxies in order to infer a mapping between the photometry and the redshift. Then, based on this mapping, redshifts can be estimated for the photometry of other galaxies.

I've read that for empirical methods it is crucial that the training data (i.e. the data from which the mapping of the photometry to the redshift is inferred) represents the galaxies for which estimated redshifts are desired in the future. I understand that this is crucial, but in what sense does the training data represent a certain distribution and in what sense are other galaxies represented or not represented by this training data? How do I know if a galaxy is well represented by the training data so that I can estimate a redshift for the galaxy?

Would the galaxy have to be from the same region? Does it have to have the same mass? What are the factors to look at, if I want to know whether a galaxy is from the same distribution/is well represented by the training data?

This was too long for a comment, but is not a real answer since I'm not completely sure, but:

My guess is that "representing the galaxies" refers to the "type" of galaxy that you intent to observe, where by "type" I mean e.g.
Lyman-break galaxies (Steidel et al. 1996),
Lyman $\alpha$ emitters (Partridge & Peebles 1967),
sub-millimeter galaxies (Blain et al. 2002),
(U)LIRGS, etc.

These terms all refer to selection methods (i.e. observational techniques), and hence also to physical differences. The closer the training data are to the observed sample, the better your algorithm will be at linking their redshifts to their photometric properties.

Mass is just one property, there's also e.g. dustiness, star-burstiness, stellar population, age, and others. "Region" is probably less important, but since the clustering of galaxies also affect their properties (e.g. their morphology, Dressler 1980), it could potentially influence the result.


In 31.5kyr, Epsilon Eridani and Luyten 726-8 will be < 1ly close; but how distant from the Solar system? Greetings! Based on Wikipedia and more precisely this paper, it is said that, in about 31,500 years, the stars Epsilon Eridani and Luyten 726-8AB will "meet" at a very close distance (less than 1 light year); I found that anecdote very interesting in the context of some SF writings I'm trying to do (not in English, quite obviously!) to bring some "subtle consistency" in it, but I was unable to either answer nor calculate with my very little knowledge of astronomy this simple question: to what distance will those stars be from the Solar System in 31,500 years?

Thank you very much for your help!

P.S.: oh, a friend of mine told me that my question could lack of useful information, such as the current distance of each of those stars from the Solar System, so here there are:

Current distance from Epsilon Eridani to Solar System: 10.5ly
Current distance from Luyten 726-8 to Solar System: 8.7ly
Other info about Epsilon Eridani: declination 9.46° south of the celestial equator (no idea if this can help) | Right Ascension 03h 32m 55.84496s | Radial velocity +15.5 ± 0.9 km/s
Other info about Luyten 726-8: declination –17° 57′ 01.8″ | Right Ascension      01h 39m 01.54s | Radial velocity +29.0 km/s

[EDIT]
As suggested in the comments, I added the Right Ascension and Radial Velocity for those two stars and I'm looking for the "Velocity vectors" (even if I have no idea what this means)

[EDIT 2]
Added links to the Extended Hipparcos Compilation (XHIP) Epsilon Eridani and Luyten 726-8AB pages with all sort of info I don't understand, but there are things such as "Heliocentric Velocity" on 3 axis, maybe it can help. (Got the XHIP thing thanks to the following thread: Position and velocity vectors of nearby stars?)

These stars are currently moving away from us, in the case of Epsilon Eridani, at 15.5 km/s. In 31500 years they will be further away than they are now, the distance can be calculated by applying Pythagoras' theorem to the distances and velocities you mention.

It turns out that Eps Eri will be about 12.2 light years from Earth, and Lutyen 726-8 will be 12.6 light years.

They won't be particularly close together in the sky, I estimate about 3 degrees, and they will be even dimmer than they are now, though Eps Eri will still be a naked eye object.


Strange Sighting, Help Needed! I live in Hyderabad. Around a week ago, I, with my mother went to observe the planets from my roof I had observed Jupiter and Saturn till they went down below the horizon and then I was ready to move my telescope on Mars which was about just above my head. When I looked at it I saw a really strange 5 dotted flying thing fly a little towards the left of Mars. I asked my mother, and as she had sight and was not wearing spectacles, She didn't see it as 5 dots, but a thin straight line.
I know that it cannot be a rocket or jet plane as it left no trail, and it can't be  a plane or helicopter as it made no sound. I read a similar post about 2-3 weeks ago about a U.F.O collecting stars, but this was not at all like it. Could anyone tell me if it was anything known or if that is really an Undefined Flying Object?

On some research, most likely what I saw was StarLink train 12. Maybe due to some clouds on that part of the sky, I was unable to see the entire train. Thanks.
Also, I do not know if I am right, this is what I thought, when I read the launches page on this website.


Dark matter virtual objects hypothesis I've hypothesized for many years in my head that dark matter are the gravitational after effects of objects that have exploded. My understanding of physics is that when an object explodes, its center of mass remains in the same location. If that's the case, it seems that the gravity of that center of mass could remain. What observations or theories are there that counteract this hypothesis?

There are (at least) a couple of problems with your hypothesis.

The first problem is that dark matter is just that - dark, in other words it emits no light that we can detect. The exploded remnants of stars are not dark at all. Indeed the winds and superwinds associated with exploding stars are often extremely hot and luminous in the UV and X-ray region. Even when these winds have slowed and cooled, they will then form warm dust (because they are enriched in heavy elements) that is then detectable in the infrared part of the spectrum.

The second problem is that we know that the majority of dark matter is not just non-luminous "normal" matter. Estimates of the cosmic abundances of helium, deuterium and lithium produced in the big bang are sensitive to the overall density of this "normal" matter. These estimates tell us that only about 20% of the gravitating matter that appears to be required to explain the dynamics of the universe, the dynamics of galaxy clusters and the dynamics of galaxies, can be in the form of "normal" matter. But the atoms and particles released from an exploding star are just part of this "normal" matter budget - they have already been accounted for and so contribute nothing to solving this second dark matter problem.


What causes Earth's gravity? Why Earth is like a magnet? how can earth attract any object irrespective of what it is made of? What is the reason behind Celestial Body's gravity? i.e Sun,Jupiter,Moon etc.

Any object's mass warps the space around it towards the center of the object which causes other objects to 'fall' in towards that object or towards each other. Bigger objects have more mass therefore they warp space to higher degree causing more gravity. It doesn't matter what a substance is made of as long as it has mass. If it has mass it causes gravity.


Photometry extinction correction magnitude I am learning photometry and I am wondering is the uncorrected extinction instrumental magnitude always less than the corrected extinction instrumental magnitude when applying transformation coefficients?
For example suppose the uncorrected extinction instrumental magnitude is $m^A_V = 10.00,$ will the corrected extinction instrumental magnitude always be greater than $10$?

If an object has a measured brightness, then it's extinction-corrected brightness will always be higher. Hence the extinction-corrected magnitude will always be smaller than the uncorrected (observed) magnitude.


An image for the unified model of AGN In a paper I'm writing, I've referenced an image I got from the NASA site at https://fermi.gsfc.nasa.gov/science/eteu/agn/ regarding the Unified Model of AGN. I captioned it

The AGN unified model, showing the BLR and NLR, as well as different types
based on the observational angle. Credit: Fermi and NASA:
https://fermi.gsfc.nasa.gov/science/eteu/agn/

but my supervisor said that I should use proper referencing (I see that the image seems to be derived from this paper). The image from the NASA site is much clearer and easier to understand, and I'd much rather use it for clarity's sake.

What's the proper way to reference an image like this, which is derived from a paper?

Maybe I should just use the original image?

Or does anyone know of a better quality one from another paper, similar to the one from the NASA site?

EDIT: I should mention that this is a follow-up question to one I posted here.

I agree with your supervisor: If your paper is going to a peer-reviewed journal, you should cite peer-reviewed material. I also agree with you that the NASA version is clearer than the original figure, but if you're only going to refer to the image — i.e. not show a reproduction — I'm sure the reader can figure it out.

Alternatively, you could create your own (color) version with the annotations that are relevant for your work, and then write something like "…based on the schematic diagram of Urry & Padovani (1995)".


How was the mass of Venus determined? The mass of Venus seems rather complicated to determine to me:

Venus doesn't have any satellites, so you can't just apply Kepler's third law (like you would with Jupiter or Saturn for instance) to determine its mass.
The gravitational tug of Venus on the Sun is very small compared to that of other Jupiter or Saturn, so it seems like it would be difficult to extract what part of the Sun's proper motion is caused by Venus.
There are few asteroids with orbits close to that of Venus, so not many objects who might have their trajectory modified by Venus.

Knowing Venus' radius, and assuming it has the same density as the Earth, you can get a pretty close estimate of its mass (85% of Earth's mass with this assumption, when the actual value is 82%). But that's a pretty strong assumption (the density of Earth and Venus only happen to be close by chance) and a rather unsatisfactory "guesstimate".
Nowdays, there are a few probes that have flown by Venus, so by looking at their trajectory, you can infer what Venus' gravitational field looks like. But those fly-bys are pretty recent. Did we know about Venus' mass before those fly-bys?
How was the mass of Venus measured for the first time?

How was the mass of Venus measured for the first time?

In the mid 19th century, Urbain Le Verrier's predicted of the existence of a then unknown planet beyond the orbit of Uranus. He even predicted this planet's orbit. The discovery of Neptune based on his predictions was perhaps his greatest accomplishment.
Le Verrier then went on to investigate Mercury. He used observations of Mercury, Venus, the Sun (as a stand-in for the Earth) and Mars and calculated that Mercury should precess by 532 arc seconds per century based on Newtonian mechanics. Along the way, he had to (and did) estimate the mass of Venus. There was a problem here; the observed precession of Mercury's orbit is 575 arc seconds per century, 43 arc seconds per century greater than his calculated value. This led Le Verrier to conjecture that there was a planet even closer to the Sun than Mercury.
Despite the failure to discover the non-existent planet tentatively named Vulcan, Le Verrier's estimate for the mass of Venus was fairly close to the correct figure, within a couple of percent. Once the cause of this 43 arc second per century discrepancy was discovered by Einstein, the mass of Venus was determined with even greater accuracy. Of course, once probes were sent into orbit about Venus, its mass was determined with greater accuracy yet.
Reference:
Leverrier, M. "On the masses of the planets, and the parallax of the Sun." Monthly Notices of the Royal Astronomical Society 32 (1872): 322.


Why does in-the-sky.org say that Mercury is not visible from Taipei these days? I know that viewing Mercury is a challenge, but I'm fairly close to the equator and Mercury was at greatest elongation just a week ago.
In-the-sky.org says:

All around the world, Mercury is visible as an evening object, having recently passed greatest elongation east. From Taipei, however, it is not observable – it will reach its highest point in the sky during daytime and is no higher than 10° above the horizon at dusk.
12 Jul 2018, 12:00 CST   –  Mercury at greatest elongation east
27 Aug 2018, 05:48 CST   –  Mercury at greatest elongation west
06 Nov 2018, 22:59 CST   –  Mercury at greatest elongation east
15 Dec 2018, 23:17 CST   –  Mercury at greatest elongation west

Of course, since Mercury stays near the Sun it will always reach its highest point in the sky during daytime, probably around lunchtime!
But I don't understand why it would be viewable all around the world (presumably just after sunset since it is currently near maximum elongation East), but not from Taipei.
Question: Why does in-the-sky.org say that Mercury is not visible from Taipei these days?

The logic that In-The-Sky.org uses to decide whether events are observable or not doesn't currently work very well for Mercury and Venus. I have some plans to fix this in the long term, but always have far too many projects on the go.

As you correctly say, it's blazingly obvious that Mercury is highest in the sky at lunchtime, and the code I wrote to generate these descriptions should realise that (but doesn't, currently).

The statement that Mercury is going to be very challenging to observe from Taipei is basically correct. At this time of year, Mercury doesn't venture very far from the Sun, even at greatest elongation. The best time to see it in the evening sky is around late September. This is true regardless of where you live in the world, so the statement "all around the world..." in the web page is basically wrong.

As I say, I have plans to improve all of this, but the website is a one-man effort, done entirely in my spare time, so things don't always get fixed quickly :-)


Can Two Stars of One Same Type Orbit Each Other? This is a simple, textbook definition of a binary star orbit.  One low-mass star orbits a higher-mass star in a narrow but definitive center.  For example, the high-mass star would be our own sun--a G-type main sequence star--being orbited by a smaller dwarf, like a red dwarf or a brown dwarf.  But what about binary orbits of two identical stars--like our sun orbiting another G-type main-sequence star?  Is this sort of orbit possible?

Certainly, the figure only uses different masses because that's the more general possibility (after all, the two masses won't be exactly the same).  In fact, if you look at the distribution over mass ratio q in binary systems in Figure 1b in https://arxiv.org/pdf/1304.3123.pdf, it seems that the cumulative number of binaries with mass ratio less than q is more or less equal to q (where q is smaller mass over larger mass).  That means the probability of mass ratio between any q and q+dq is about equal to dq and nearly independent of q.  So the chance of the mass ratio being between 0.9 and 1 is about the same as being between 0 and 0.1, for example.  In some sense, that's like saying all q are equally likely.  And of course if q is near 1, then the stars will have the same age and be in the same evolutionary state, so will be the same stellar type as well.


If Mars were tidally locked to the Sun, what would the temperatures on its surface be like? Since there's very little atmosphere on Mars, it doesn't matter much whether the Sun is in the zenith or hanging just above the horizon: without the atmospheric "filter," the sunlight would have roughly the same effect on all sunlight areas.

What would the temperatures be like on the sunlit portion of the planet?

Tidal lock would mean only one side of Mars would be exposed to solar irradiation, so its temperature would be mostly defined by the equilibrium between the energy received from the Sun and the black-body radiation.

Because of the orbit eccentricity, solar irradiation reaching Martian surface varies between 493 and 717 W/m² (source). According to Stefan-Boltzmann law, the temperature required for the black-body radiation to match these levels will vary between 305 and 335 K (32°C an 62°C), respectively.

Such temperatures will be reached in the area directly facing the Sun (so the Sun is seen in zenith), and gradually decline towards the background radiation temperature of space (about 3K) when approaching the dark side (where the Sun is just barely above the horizon). Such temperature distribution is due to the angle of incidence, not to the atmospheric "filter" or lack thereof.


Galactic Coordinate System and distance If the Galactic Coordinate System is made up of two values, angles of longitude and latitude, how is distance measured? If you have those two values, you get a line from the sun in a certain direction. There might be multiple objects on that line, how can you specify one of them in the coordinate?

You need three coordinates to represent a point in 3-dimensional space. Galactic coordinates are a 2-dimensional coordinate system on the surface of the abstract celestial sphere. Typically it is easier to measure position on the celestial sphere than it is to measure astronomical distances, so usually the systems are given in those terms. For 3D you need to supplement the Galactic coordinate system with distance as a separate coordinate, this is in fact the natural choice for a 3D spherical coordinate system.

Remember that the galactic coordinate system has its origin in the same place as equatorial coordinates (right ascension, declination), so the distance measurement is the same: you don't have to translate it relative to the Galactic Centre. To measure it, you use the usual techniques used to determine the cosmic distance ladder: parallaxes, standard candles (Cepheid variables, etc.), redshifts...


How can absolute value of azimuth exceed distance from pole? I'm sure I'm just doing something dumb, but:

Yildun has a declination of approximately 86.5. This puts it 3.5
degrees from the celestial north pole.
Per the snapshot below, Yildun's azimuth can be as high as 4.17
degrees.
Since the celestial north pole always has an azimuth of 0,
wouldn't this put Yildun 4.17 degrees from the pole, contradicting
the distance calculated from declination?

What am I missing?

EDIT: I thought I'd figured this out:

On Earth, one degree of latitude is not the same as one degree of longitude (except at the equator).
Analogously, in the sky, one declination degree is not the same as one right ascension degree.

I thought the situation for right ascension applied to azimuth, but it doesn't.

Both azimuth and declination are measured on "great circles" (declination is measured on a half circle, but same general idea), so degrees in azimuth should equal degrees in declination.

It turns out I was right when I thought I was wrong.

Consider the azimuth circle for 85 degrees elevation. It's very small. For example, 85 degrees elevation and 0 azimuth is only 10 degrees away from 85 degrees elevation and 180 azimuth.

Therefore, the azimuth circle for a given elevation is smaller than a great circle. It's only cosine(elevation) the size of a great circle.

At 35 degrees, the azimuth circle is cosine(35 degrees) = 0.819152 the size of a great circle, so 4.17 degrees in azimuth = 4.17 degrees * 0.819152 = right around 3.5 degrees on a great circle.


Can virtual particles be the source of dark matter and dark energy? I had this question for a long time. But otherwise the question basically sums the whole thing up.

The problem with associating virtual particles with either dark matter or dark energy is that, quite literally, virtual particles aren't real.  By that I mean, they are intended to change the effects of energy, like how the energy is transported and where it goes, but they shouldn't alter the amount of energy.  As such, virtual particles are just a kind of picture, a way to view the terms in a particular way of writing the equations.  Many physicists reject that picture, they think it gets taken too literally, while others embrace it wholeheartedly, but it shouldn't appear in the stress energy that is used in general relativity.

But of course, the problem is that general relativity may be wrong, especially since it does not seem very consistent with quantum mechanics.  So the idea is, maybe there is some way to correct general relativity where virtual particles do have a more direct role.  Then the people who tend to dismiss their relevance would have to reconsider.  But so far it's not at all clear that this will be the right way to fix GR, because you run into significant problems.

The problems you run into with dark matter being virtual particles is that most of the effects of dark matter are seen on completely Newtonian scales, like the gravity in a galaxy.  On that scale, one should not need GR, and no one is suggesting that we need to fix Newtonian gravity by including virtual particles, that would not be consistent with everything Newton did.  For example, we see no effects of dark matter within our own solar system, so why would we need to fix Newtonian gravity on the galactic scale but not the solar system scale?  We would need to do more than include virtual particles, we'd need a whole new kind of gravity (sometimes called "MOND", modifications to Newtonian dynamics").  If you are going to do MOND, it opens a full range of possibilities, there's really no reason to be constrained by the virtual particle concept.

The problems you run into if you try to associate virtual particles with dark energy is that the answer you get, if you proceed in the most straightforward way possible, is that the dark energy should either be zero (if you don't think virtual particles count in the energy of the vacuum), or something more than 100 orders of magnitude too large (if you think they do count).  So how do you take that virtual particle concept and whittle down its effects from something more than 100 magnitudes too large to something tiny but not zero?  It's not at all clear this is the best starting point!


What's the difference between a crater and a basin? When I was searching the name of the Moon's largest crater, many websites said that it was the South Pole-Aitken Basin. But in some definitions I've found, the word "crater" is used to describe smaller impacts on the Moon surface, and since the South Pole-Aitken Basin has a diameter of almost 2500 km, I'm not sure if you could say that the South Pole-Aitken Basin is the Moon's largest crater.

The French version of the Wikipedia article has a "nomenclature" section, absent from the English version. It says:

Lorsque l'énergie cinétique de l'impacteur est suffisante pour
atteindre le manteau à travers la croûte et provoquer des épanchements
magmatiques, on parle de bassin d'impact et non plus de cratère
d'impact.

Which roughly translates to:

When the kinetic energy of the impacting body is sufficient to reach
the mantle through the crust and trigger magmatic outpourings, the
term impact basin is used instead of impact crater.

However, there is no reference to back this claim, so I would take this with a pinch of salt...


Where can I download processed data of the Kepler telescope? I search for lightcurve Kepler data: time $t_i$, flux $F_i$ and measurement uncertainties $\sigma_i$ I need processed data (after removing trends, normalized, etc.).  Where can I find it?

What you are looking for is here. http://archive.stsci.edu/kepler/search_retrieve.html
This one is good for exoplanet light-curves https://exo.mast.stsci.edu/


How can neutron stars have gaseous atmospheres? Neutron stars can have small atmospheres. However, they also have extremely strong gravitationally pulls. Shouldn't the all the gas molecules be drawn to the star's surface, and become solids under the immense pressure?

Maybe I'm thinking about this the wrong way, but I don't see how it could be possible.

Gravity is only important insofar that it is capable of compressing the material to high densities. Whether that material is capable of solidifying depends on the competition between Coulombic potential energy and the thermal energy of the particles. The former increases with density, the latter increases with temperature. A dense plasma can still be a gas if it is hot enough.

A rough formula for the exponential scale height of the atmosphere is
$$ h = \frac{kT}{\mu m_u g},$$
where $T$ is the temperature of the gas, $m_u$ is an atomic mass unit, $\mu$ is the number of atomic mass units per particle and $g$ is the surface gravity, with $g = GM/R^2$.

For a typical neutron star with $R=10$ km, $M= 1.4M_{\odot}$, we have $g=1.86\times 10^{12}$ m/s$^2$. The atmosphere could be a mixture of ionised helium ($\mu=4/3$) or perhaps iron ($\mu = 56/27$), so let's say $\mu=2$ for simplicity. The temperature at the surface of the neutron star will change with time; typically for a young pulsar, the surface temperature might be $10^{6}$ K.

This gives $h = 2$ mm.

Why is this not a "solid"? Because the thermal energy of the particles is larger than the coulombic binding energy in any solid lattice that the ions could make. That is not the case in the solid surface below the atmosphere because the density grows very rapidly (from $10^{6}$ kg/m$^{3}$ to more than $10^{10}$ kg/m$^{3}$ (where solidification takes place) only a few cm in, because the scale height is so small. Of course the temperature increases too, but not by more than a factor of about 100. After that, the density is high enough for electron degeneracy, and the material becomes approximately isothermal and at a small depth the "freezing temperature" falls below the isothermal temperature.


Which data should I use to take the fast Fourier transform (FFT) to find time period for eclipsing binaries? I am trying to use Kepler Data for Eclipsing Binaries to estimate time period, and then other parameters such as mass, eccentricity, semi-major axis, distance, etc. of the system. I want to write code in MATLAB which will use FFT. The available data has the following columns:

bjd(date) | phase | raw_flux | raw_err | corr_flux | corr_err | dtr_flux | dtr_err

Does the data require any cleaning?
If I want to take the FFT of the data to find time period, which data should I use?

For this, you should create a lightcurve, a graph of brightness over time, to view the data. For Kepler data, the bjd(date) column is the time in BJD. The dtr_flux stands for detrended flux, meaning that it should contain "cleaned" data.

First, I would plot the dtr_flux column over bjd(time) to obtain a lightcurve. Depending on the type of eclipsing binary, you should see a minimum and maximum dip occurring periodically. The data also contains a "phase" column, meaning the data has been "folded" to allow you to plot each cycle of the eclipsing binary on top of each other. See http://www.southampton.ac.uk/~sdc1g08/BinningFolding.html

After that, you can take the FFT if you wish. Remember that since this is an eclipsing binary with both maximum and minumum brightness dips, the FFT will have a spike both at P and P/2.


Star revolving a star Are there any star system(planet system) where a star revolves around a star? Is it theoretically possible?

What you are asking is about the existence of binary star.

There are many binary stars in the universe. They are in fact more common  than single stars. Here are interesting explanations for the latest.


Is Venus visible throughout the night? I know that Venus is visible during sunset and sunrise, but is it also visible in deep dark night?

James K has provided a short answer "no". The longer answer is that because Mercury and Venus are in closer orbits to the Sun than Earth's, they will always be in roughly the same area of the sky as the Sun, whereas the other planets aren't so restricted. Since something is generally only visible if it's above the horizon, it follows that if the Sun is below the horizon, Mercury and Venus will either be close to or below the horizon.

The very furthest that Venus gets from the Sun (its "elongation") is 47.8°. If Venus is at its peak elongation around the time of the summer solstice, its angle above the horizon around sunrise (or sunset, depending on which side of the Sun Venus is on at the time) will be maximised. Add to this that since Venus is very bright at this elongation, it's actually visible even when the Sun is low above the horizon; and it remains visible for about three hours after sunset (or, if it's on the Sun's other side, it rises above the horizon about three hours before sunrise). In winter, on the other hand, the ecliptic is substantially lower and Venus's elevation at sunrise/sunset is therefore much less.

If the Sun is more than 47.8° below the horizon, then Venus must also be below the horizon. You can never see Mercury or Venus in "deep dark night" (e.g. around midnight), whereas Mars, Jupiter and Saturn are readily visible (and quite bright!) if they happen to be above the horizon at midnight. [NB I'm excluding the situation of summer at extreme latitudes, since that isn't "deep dark night"].

This diagram on Wikipedia provides a simple representation of elongation and angular separation, and shows the difference in how the "inner" and "outer" planets can appear in the sky.


How do astronomers know when the Sun will die? How it is possible for astronomers to find out when the end of the Sun's life will be?

Yes, astronomers could be wrong.  Part of being a scientist is always having to be ready to admit that you were wrong.

Astronomers have developed models of how the sun and other stars work. We understand them to be nuclear furnaces. These models may not be perfect, but they are well supported by the evidence and can predict much of what we observe from the sun and other stars.

Using these models we can expect the sun to exist for about 10 billion years, and it is currently about 4.7 billion years old. While the model could be wrong, it is probably not very wrong, and the sun will go on shining for a long time yet.


Why do some celestial bodies have atmospheres, and not others? Question is rather self-explanatory. An answer touching upon how/why atmospheres are formed in the first place would be ideal as well. An example of such astral bodies would be The Earth and it's moon. Earth has an atmosphere, while the moon does not. Why is that so?

The amount and kind of gases a body can trap depends on the object's surface temperature, and its density & radius (which refers to it's gravity).

An object with high gravity and low surface temperature will be able to hold more gases in it's atmosphere. In the case of the Moon, due to its low gravity it could barely trap an atmosphere of Xenon.

You can check this thanks to this amazing plot: http://astro.unl.edu/naap/atmosphere/animations/gasRetentionPlot.html

But this is not all you need to hold an stable atmosphere like the Earth does. The object will also need some protection against the solar wind. In the Earth we have a magnetic field that prevents the solar wind to reach our atmosphere, but in the case of Mars it has a weak and poor magnetic field to defend itself against the solar wind that rips it's atmosphere.


Is it possible to break apart a neutron star? I was inspired by this question on Physics, as well as this question right here on Astronomy. Neutron stars are tightly bound together as neutron degenerate matter. They're very massive and have a strong gravitational field. Is it possible to break one apart into sizable chunks? How would you do this?

The answers given are good and answer my question; I'll just clarify some one thing (based on the comments) for posterity.

I would define "broken" as when any significant amount of mass is removed from the neutron star, as in mass shedding, as Mitch Goshorn wrote. The resulting object, however, should contain a significant amount of neutron matter - that is, it should largely retain its prior composition.

It would appear theoretically possible (to some degree) through extreme applications of recycling to trigger mass shedding in pulsars.

Pulsars are rapidly spinning neutron stars, the fastest class of which are millisecond pulsars. The current belief is that they build up rotational speed through accretion, a process known as recycling. One study, Recycling Pulsars to Millisecond Periods in General Relativity (Cook, et al), explores the limitations of this process.

The following chart shows their results:

At the point where the dotted lines meet the two plots, you can see a reduction in mass at those energy levels. This is due to the angular velocity of the body creating instability which results in mass shedding - essentially mass at the equator of our neutron star being flung off the star due to the body's angular velocity.

Unfortunately, this is not exactly an easy process.

  The timescale to accrete the required rest mass, ~0.1 M☉, at the Eddington limit, ~10-8 M☉yr-1, is ~107 yr. This timescale is largely insensitive to the adopted nuclear equation of state. If other astrophysical considerations require a considerably shorter time scale, then the simple recycling scenario described here will have to be modified beyond the variations explored in this paper.

(Note however that the research here is actually attempting to avoid such instabilities, and they accomplish this by adding even more mass, such that the body can support even greater rotational velocity without encounter instability. Additionally, they're trying to create millisecond pulsars, but we don't need to do this as they exist naturally, so we could save ourselves a lot of time by (very carefully) approaching an existing millisecond pulsar)

I don't think this would exactly be breaking apart (despite Wikipedia's use of that exact verbiage to describe it), but it allows for the return of mass that was at one point in a neutron star.  Of course, chances are our theoretical neutron star miners are very likely to be the ones who put that mass on the neutron star to begin with. On the other hand, this (hopefully) accomplishes the task without reducing the object to a quark star or black hole.

  Cook, G. B.; Shapiro, S. L.; Teukolsky, S. A. (1994). "Recycling Pulsars to Millisecond Periods in General Relativity". Astrophysical Journal Letters 423: 117–120.


How big would the Earth look from the Moon? Imagine standing (somewhere) on the Moon's surface. How big would the Earth look from the Moon, compared to the image of the Moon we see from Earth? What would be the ratio between the diameters of each body? How would this ratio be calculated? Any help with answering such a calculation would be highly appreciated.

There are a couple questions here. The ones you have asked are:

What would be the ratio between the respective diameters?
How would this ratio be calculated?

But I think you meant to ask:

How much larger does the Earth appear when standing on the Moon than the Moon appears when standing on the Earth? (and the calculation)

For the answer to #1 (and #2), it is the ratio between the numbers (of the Earth's diameter of 12,742 km and the Moon's diameter of 3,474 km) which is $Earth:Moon = 3.668:1$ as @pela's comment mentions.

For the answer to #3, it is the same! The more interesting part is why. We know that the distance from the Earth to the Moon is the same as the Moon to the Earth, so it follows that when we are on one of them looking at the other we are gazing across the same distance. Next, we know that an object (approximately) appears larger in the same proportion as it is larger, so an object that is twice as wide or tall will appear twice as wide or tall. Since the Earth is 3.668 times as wide as the Moon, it will appear that much larger when viewed from the same distance, and we've already discussed why it is the same distance. So, that's it really. It is the ratio of the diameters.


At what depth below the Sun's surface does the density reach that of water? I'd like to write a short supplementary answer to A probe floating in Sun's surface but in order to do so I need to know at roughly what depth relative to the Sun's surface the mass density reaches 1 g/cm^3.

At the bottom of the photosphere the density seems to be either 1E-06 or 1E-09 g/cm^3 as discussed in the question What is the density profile within the Sun's photosphere? but how deep do you have to go below that to reach 1 g/cm^3?

The sun's density is 1 gm/cc at approximately 50% of the way down towards the core. if radius of Sun is R then at R/2 the density will be that of water.

@astrosnapper's comment links to this answer in Physics SE. A reverse image search of that unsourced graphic leads to

Thesis: Multi-spacecraft analysisof the solar coronal plasma
NASA: The Solar Interior
Whatever this is shown below
the link in that image http://www.sns.ias.edu/~jnb/SNdata/Export/BP2004/bp2004stdmodel.dat provides some raw data
the reference in that image is standard solar model BP2004 J. N. Bahcall and M. H. Pinsonneault, Phys. Rev. Lett. 92 (2004) 121301 What do we (not) know theoretically about solar neutrino fluxes? (also in ArXiv)


Why don't Australia/Russia have large optical telescopes? Australia and Russia have large land areas. However, I am only aware of one 3.9m telescope in Australia. As far as I know, there are lots of mountains in Australia and Russia.

So why are there not many large optical telescopes?

The reasons for each country may differ because of their latitudes.

Optical telescope sites ideally need to be somewhat remote (to avoid light pollution) yet still accessible for construction, engineering, and observing; high in altitude; dry; with exceptional "seeing" (stable atmospheric conditions); and clear weather. Neither Australia nor Russia really have any sites meeting all those criteria: their mountains tend to have some combination of too low altitude, poor seeing, and/or bad weather. (This is also why there aren't any really big, modern telescopes in Europe.)

This site has some discussion of the different criteria that go into choosing the best sites for new, large telescopes.


What are the reasons for referring to equinox J2000 or the equinox of the date? (Background: Trying to learn something about the mathematical and computational part of astronomy I have acquired a copy of Meeus' "Astronomical Algorithms" and implemented the material up to Chapter 33 in Java, with only one major sidestep: VSOP87, for which I obtained the full data set and "translated" the FORTRAN program.)

VSOP87 exhibits the reason for my question: you can compute rectangular/spherical heliocentric positions either for the reference frame J2000 or the one of the given date. Clearly, if I want to observe some planets tonight, I'll compute the data for today's equinox. On the other hand, there is the possibility of a transformation between equinoxes of different dates.

So why are these methods (here and for other methods, too) provided for both reference frames? Was it the (then) weaker computer performance?

Only a little later, iauPlan94 provides the result only for J2000 although Meeus in the chapter dedicated to the paper of J.L.Simon e.al. provides tables with data from their theory for J2000 and the equinox of the date.

What would be good reasons for today and the years to come to refer a computation to J2000?

Maybe you are interested in watching the occultation of a star by a solar system object. In which case you will need to compare their positions. Catalogues of stellar positions are most commonly given with J2000 positions.


What is the Habitable zone all about? I've a question on Habitable zone.

Does Habitable zone represent the region where we humans or the living beings of earth can survive?

Or is it a region where life can exist?

If it is second I want to know what are the parameters that are considered for existing of life.

Nothing that complex. Habitable zone merely defined a circumstellar volume where a planet (if any) can receive an amount of radiative energy to keep H2O in the liquid phase.

Yes, it has been based on the requirements for life from Earth, lacking any wider samples.

I encourage you to play around with interactive illustration of concept (Flash applet).


Is there a maximum temperature a planet can have? This may sound a little bit confusing, but I will try to explain. (I have done research on this topic already)
I have looked at the temperatures of planets and I was wondering, is there a point where a planet is at perigee and it's temperature cannot reach any higher? Let's just say the planet is tidally locked, and is currently at perigee. Would the planet be at a point where it's temperature can no longer be increased at the current distance from its star?

A planet in a stable orbit can only receive a finite maximum power from it's star and likewise can only generate a finite energy from internal heat (like our cooling core, which is a tiny power output compared to what we get from the Sun).  There's a finite limit to available power.

If that power is not radiated out (which is thermal radiation) the planet's temperate would rise without limit.  This never happens because hot things radiate heat (as EM radiation) to try and reach thermal equilibrium with their environment (which in the case of a planet is the very cold space around it).

So there must be a finite limit to the temperature any planet in a stable orbit can reach.

As @Cody indicated the actual point of maximum temperature may not be at periapsis.

Atmospheres can act in complex ways to raise the temperature of the surface significantly.  Venus is a prime example as this page demonstrates.  Without it's pressure cooker-like atmosphere we'd expect a temperature of around 240 K, whereas it's actually a rather staggering 740 K.  But all of that is solar powered and can't rise without limit.  But it illustrates the importance of atmospheric composition (and indeed the Greenhouse effect) on planets.

Update : I came across another page on the basic theory of planetary temperature while looking into something else. It may be of use so I've added it in this edit.


visualisation of galaxy density If I had a long stick that was millions of light years long, and with a width of about the Sun, and I then swiped it through the center of the milky way galaxy, what would the probability be of me hitting a star?

EDIT:

Ignore the speed of light. Assume it is possible to just swipe it through instantaneiously.
Assume that the stars in the galaxy are equally distributed. Obviously if you were to swipe right through the middle you would hit the galactic center.
The purpose of this thought experiment is just to help visualise the density vs size of the galaxy.

For your experiment, you can take it 2D, so let's ingnore the thickness of Milky Way.

So, we have 400 billions stars(the real number is between 100B-400B) in a circle with 100k ly(the real number is between 100k-160k).

The area of the circle is pi*50k*50k ~= 7800M sqly = 7.8 billions sqly.

We have then ~51 randomly stars per square ly.

But you do not sting the Milky Way with your stick, but cut through it.

If you swipe through the middle, you have 51*100k stars in a width of 1 ly.

So, you have a chance to hit a star per 1/5.100.000 ly which means 1.855.000km.

But the diameter of sun(of your stick) is 1.395.000km. Taking diameter of sun as medium diameter of a star(this is not true), you have a chance of 1.4/1.8 = 77% to hit a star. If you go with the "low density" numbers, 100B stars and 160k ly diameter, you'll get ~31%.

My conclusion is you often hit a star.


Why are telescopes positioned in Lagrange points? In this Wikipedia article about the list of space telescopes to be launched (which I assume is exhaustive), of the 11 telescopes yet to be launched, 6 will be positioned at the Sun-Earth L2 Lagrange point. Why is that so?

I read that it has something to do with lower fuel consumption, but I would like a more detailed answer.

tldr;

  L2 is a very stable thermal environment as well as good instantaneous sky visibility and high observing efficiency.

The main reason space telescopes are placed in an L2 orbit is because L2 is a stable thermal environment. Telescopes in Earth orbit can receive sunlight and earthlight in different directions, meaning that the telescope would have to shield in most directions if it wanted to keep itself cool. At L2, the Sunlight and Earthlight come from the same direction, meaning the telescope has to only shield itself from that direction. This is very important for infrared telescopes, as heat radiation acts as interference to measurements. Of the 6 space telescopes going to L2 listed, 3 are infrared telescopes.

For example JWST, one of the infrared telescopes, has to cools some of it's detectors to 7 kelvin to operate properly.

JWST L2 Halo Orbit

The other main benefit from being in L2 is that being so far away from Earth means that there is a much greater available field of view as the Earth blocks a very minor portion of the sky. This increases sky visibility and observation efficiency.

Other Benefits:
- Ease of communication
- Delta-V requirements low (2-4 m/s per year)


Half of the Milky Way's baryonic mass is from other galaxies - when did this happen? The recent Phys.org article Milky Way's origins are not what they seem explains the recent MNRAS article Anglés-Alcázar et al. 2017; The cosmic baryon cycle and galaxy mass assembly in the FIRE simulations DOI:(https://doi.org/10.1093/mnras/stx1517), also in ArXiv.

  Using supercomputer simulations, the research team found a major and unexpected new mode for how galaxies, including our own Milky Way, acquired their matter: intergalactic transfer. The simulations show that supernova explosions eject copious amounts of gas from galaxies, which causes atoms to be transported from one galaxy to another via powerful galactic winds. Intergalactic transfer is a newly identified phenomenon, which simulations indicate will be critical for understanding how galaxies evolve.

This is quite a sizable paper about quite a sizable simulation, and I'm having difficulty even getting started to read it.

Question I wanted to know, for a Milky Way-like galaxy, was most of the influx from other galaxies early on, or is this something that is predicted to happen somewhat steadily throughout the galaxy's lifetime?

Galaxy m12i in the simulation represents a Milky Way-mass galaxy, and so I show a screen shot of one of the figures for that galaxy below. Already I am stuck because I don't understand what $z$ represents. It seems to be related to time, rather than red-shift and $z=0$ has a special meaning, but I don't know if that's the present, or the beginning of the simulation.

below: Selected parts of Figure 3 from here. The selected column is data for galaxy m12i, which is representative of a Milky Way-mass galaxy in the simulation.

  Figure 3. Top panels: contribution of different processes to the evolution of galaxies m11 (left), m12i (middle), and m13 (right) from early times down to z = 0. Total stellar mass (top), ISM gas mass (middle), and gas accretion rate onto the galaxy (bottom) are indicated by the thick gray lines for each galaxy. Lines of different colors show the contributions from fresh accretion (purple), NEP wind recycling (blue), intergalactic transfer of gas (green), and galaxy mergers (ISM gas: orange; stars: red). All quantities represent average values over a timescale of ∼ 200 Myr. Bottom panels: fraction of stellar mass, ISM gas mass, and gas accretion rate contributed by each process as a function of redshift. Fractions are represented in terms of ranges, so that the contribution of each component at any redshift is given by the vertical extent of the corresponding color at that redshift.

I think what you want to concentrate on is the bottom panel in your first figure, which shows the amount of accretion from different sources as a function of redshift (and thus time; see below). This shows that the contribution from "intergalactic transfer" (green cuve = material in winds outflowing from one galaxy that ends up in another) was almost zero until around $z \sim 3$ (age of universe $\sim 2$ Gyr), peaked at around $z \sim 2$ (age of universe $\sim 3.5$ Gyr) and then declined slowly from there, so that it's now about 10% of the peak rate.

"$z$" is the redshift, which is being used as a proxy for time: $z = 0$ corresponds to the present day (age of universe = 13.7 Gyr), $z = 1$ corresponds to an age of about 6 Gyr, $z = 2$ is about 3.3 Gyr, and so forth. You can use Ned Wright's Cosmology Calculator to convert individual redshifts to ages of universe (click on the "Flat" button to use a standard cosmology).

("Gyr" = gigayear = 1 billion = $10^{9}$ years, in case that's not clear.)


why are galaxies different in color? whether its in movies,television,real-life or some planetarium different galaxies are shown in different color.

For ex:- Milky way is always seen in Blue where as Andromeda gives out a bit reddish color.

Is it because different galaxies contain different kinds of stars?
Does this effect the life the galaxy harbors...

^_^

See What do a galaxy's colors mean? Are they its true colors? This article should help answer your question. There's really nothing I can add to the article. It lists all the reasons I could think of.


Estimation of average rock and asteroid mass associated with different stars There is a lot of information on different star types and compositions.

Can we make estimates of the total mass of rock, ice, and non stellar matter that orbits stars based on star evolution models?

Take, for example, the sun - are there studies of the total mass of non-solar material associated with the sun and it's composition regardless of accretion. What about nearby stars?

Can we estimate how much non-incandescent material the galaxy contains aside from dark matter?

The composition of the gas from which stars and their planetary systems form is reasonably well known. About 1-2% of this gas is in the form of chemical elements heavier than Helium (the so-called metallicity of the gas).

A fraction of these "metals" - the iron, silicon, oxygen etc. is capable of forming dust and then accumulating to form "rocky" material.

So to answer your last question first, given the available reservoir it seems most unlikely that even 1% of the baryonic (normal) matter in the universe could be rocks, even if none of it were gathered into luminous stars and galaxies.

Forming stars are surrounded by circumstellar material from which planets and other rocks form. Observations of young stars suggest these discs can be as massive as 10% of the stellar mass, but more usually 1% or less. This means that as a fraction of the whole formed star system the rocks will be at most 1% of 10% (ie 0.1%) of the stellar mass. For a Sun-like star that means there's less than 330 Earth masses of rocky material around it.


I see a non flickering star. How is this possible? I looked outside my window one night, and the moon was shining brightly, and it wasn't a full moon. But thats not the point. Really close to it was something that looked liked a star. It was really bright, and it gave off a white light, but the more I looked at it, the more I realized that it wasn't flickering or twinkling the way stars do. A few minutes later, I looked at it again, and this time it was smaller, less bright, and farther away from the moon. I know its not a satalite because I've seen those before and they move faster, and they give off the same amount of light all the time. If anyone has any idea what this thing could be, please tell me, because I really want to know.

Stars are, to all intents, infinitely far away and hence present themselves as points - thus the small changes in the atmosphere's refractive index caused by different patches of density and temperature of air make them appear to twinkle.

Planets, on the other hand, are relatively close by and (with the exception of the Moon and the Sun itself) whilst we cannot resolve any solar system body into a disk with the naked eye, they are still larger than (most) variations in the atmosphere, hence they don't appear to twinkle or the twinkling is much reduced.

The evidence suggests you saw a planet, quite likely Jupiter or Saturn.


Was there ever a hypothesis about a planet called Biga? I just re-read a popular astronomy book that was published when I was a child, about 1973. It starts with an interview to Harold C. Urey where the interviewer asks about a hypothetical planet in our solar system called Biga.

Has a hypothetical planet ever been called "Biga" by its  proponents, in the same way there was a proposed planet "Vulcan", or is this "Biga" name a printing or translation error?

If it's a mistake and there never was a "Biga" proposal, which is the proposed planet with the most similar name?

This link contains such interview, in Spanish. I will translate and quote its most relevant text:

  Interviewer: At present, do we know all the planets of the Solar
  System, or is there the possibility of still unknown planets in our
  system?

  H.C. Urey: We certainly know the nine usual planets. Recently there
  has been much talk of another object moving in space and supposed to
  be close to the Solar System, but I am not sure that it is a real
  hypothesis, and personally I think it is better to wait before
  deciding whether it is Not of a planet. I consider it sufficient to
  say that there are Mercury, Venus, Earth, Mars, Jupiter, Saturn,
  Uranus, Neptune, Pluto and the asteroidal belt.

  Interviewer: When talking about this object, are you referring to the
  so-called planet Biga?

  H.C. Urey: Yes, that's what I meant, but I'm not sure of its
  existence.

Please, remember:

1- This interview mas made about 45 yeras ago.

2- I'm not asking if this hypothesis is correct or even possible. Only want to know more about this hypothesis which was well known by H. C. Urey in the 70's.

3- I'm not asking if the name "Biga" is a correct proposal of a name for a planet in our Solar System accordind to current naming rules.

If another planet in our solar system was to be named, I think that Biga wouldn't be the first name to come up, all of the known planets in our solar system are named after Roman deities and Biga is not a deity but rather a cart on which mostly Luna is depicted. I have never heard of this proposal and if the 'rules' are used for naming, it wouldn't be the first candidate.


How can I id this celestial without data? Anecdotal. Also, I'm using the term 'celestial' loosely; to mean something shiny in the skies.
I'm based out of Nagpur in Central India. Sleeping outdoors on account of summer over the last few days I've made a weird observation. No matter what time it is, there is a fairly bright body visible in the NNW skies at an elevation of perhaps 20 degrees. From what little I gather a GSO object would be really dim, or even invisible. A LEO couldn't be in GSO unless it was in powered flight. I'm utterly flummoxed. The only reason I haven't seen it before is because earlier in life I would sleep outdoors on the balcony where this particular field of view is blocked by other structures around. Sleeping on the ground floor there is a field of view not available on the first floor. Adjacent buildings blocking the view on the first floor also act as a reference that the position of this object has not changed no matter what time of the night it is.
My location (from Google Maps) is 21.17103958702249, 79.08452532230748
The general direction where this object is visible is due 21.171644779275333, 79.08410335986589
How can I id this object?

My proposal in case of real time simulation is Stellarium software or an excellent android app like Sky safari.


How does gas in accretion disks of supermassive black holes create orbital torques on smaller black holes within the disks causing them to migrate? This answer to What enhances the capture and merge rates of pairs of small black holes orbiting around supermassive black holes? links to Migration Traps in Disks Around Supermassive Black Holes which includes the following sentence in the introduction:

Migration toward trapping orbits may be such a mechanism. Objects orbiting within differentially rotating disks exchange angular momentum with the gas around them as they orbit, which results in a torque, typically causing the objects to migrate.

Question: How exactly do density gradients in accretion disks of supermassive black holes create torques on smaller black holes orbiting within the disks? Is it through simple gravitational forces, GR effects (maybe some "spin-orbit coupling" thing?), aerodynamic drag, accretion by the small black hole in a velocity gradient, something else, or some combination of these?
One sentence in the linked answer is

These "migration traps" (Bellovary et al. 2016) are caused by torques exerted on the orbiting objects (in this case black holes) caused by gas accretion.

and I think it means that the orbital torque (torque on the small black hole about the SBH's center) comes from gas accretion by the small black hole. If this is the case, is it possible to elaborate a bit?

Not really my area, but this question is probably related to the planetary migration in circumstellar disks. In this case, the migration is caused by gravitational interactions between the planet and the gas in the disk. There are two explanations for this effect

Impulse approximation: consider a parcel of gas in the corrotating frame. If the gas is close enough to the orbit of the planet in an internal orbit, it will eventually reach the planet and its gravitational pull will move it to an external orbit. From the change of velocity in the gas, you can calculate the torque exerted to the planet. This approach is relatively simple in terms of math, but uses a lot of assumptions which are not true. However, it can be used to get an order of magnitude estimation.

Resonant torques: Decompose the torque exerted to the planet in the sum of individual torques at resonant locations in the disk. Much more complicated, but better results.

You can find the complete deduction in Astrophysics of Planet Formation by Armitage, or the compact version in this paper.
There are surely differences between planets and black holes (GR corrections, larger accretion rate, etc.), but I'd assume the physical principles are basically the same


Is there still a possibility that Makemake is not ellipsoidal but asteroid-shaped? In 2008, Makemake and Haumea have been named under the expectation that they would prove to be dwarf planets. Well, did they prove to be ones? Haumea obviously did since we know its shape which is a pretty weird ellipsoid. What about Makemake? Did we learn meanwhile whether Makemake is in hydrostatic equilibrium and/or whether it has an ellipsoidal shape? Or is there still the possibility that it's no dwarf planet according to the IAU's rules?

Intention to answer

"Is there still a possibility that Makemake is not ellipsoidal but asteroid-shaped?"

Probably not under reasonable assumptions.
From Ortiz et. al. (2012) "Albedo and atmospheric constraints of dwarf planet Makemake from a stellar occultation" https://www.nature.com/articles/nature11597

Our preferred solution that fits the occultation chords corresponds to a body with projected axes of 1,430 ± 9 km (1σ) and 1,502 ± 45 km, implying a V-band geometric albedo pV = 0.77 ± 0.03. This albedo is larger than that of Pluto, but smaller than that of Eris. The disappearances and reappearances of the star were abrupt, showing that Makemake has no global Pluto-like atmosphere at an upper limit of 4–12 nanobar (1σ) for the surface pressure, although a localized atmosphere is possible. A density of 1.7 ± 0.3 g cm−3 is inferred from the data.

Concerning the irregular shape of Haumea there are some uncertainties, it might not be a homogenous body but one with a ring and it has satellites.
From Ortiz etl a. (2017) "The size, shape, density and ring of the dwarf planet Haumea from a stellar occultation" https://www.nature.com/articles/nature24051

Secondary events observed around the main body of Haumea are consistent with the presence of a ring with an opacity of 0.5, width of 70 kilometres and radius of about 2,287 kilometres. The ring is coplanar with both Haumea’s equator and the orbit of its satellite Hi’iaka. [...] The occultation by the main body provides an instantaneous elliptical projected shape with axes of about 1,704 kilometres and 1,138 kilometres.

The paper's also on Arxiv it seems.


Basic celestial data complete beginner here, only got into star gazing a few days ago using my friend's telescope and I don't really understand at all what any of the measurements mean.

Anyway, talk about jumping in the deep end, but I'm poking around Nasa's Intergalactic Database, but sadly it's all greek to me: http://ned.ipac.caltech.edu/

Here's a galaxy: NGC 7590

Here's its basic data:

EquJ2000.0 RA

EquJ2000.0 DEC

Velocity km/s

Redshift / z

Magnitude / Qual Filter

Separ arc min

Now I'm afraid I don't really understand what any of these measurements and their values actually mean.

I remember from school what Red Shift is, its the wavelength increasing becoming
'stretched' making the light seem to appear more reddish the faster an object is accelerating away from us. But what is that z unit?

I'm afraid I really don't have a clue regarding anything else... spear arc min, magnitude / qual filter?

Is there a book for a brief understanding of these units? Not out to solve problems regarding celestial mechanics, I just want the NED to make a bit more sense to me so I can visualize these objects.

If you browse further down the page, there are details for each of the top table data.

So RA and Dec give you coordinates (Right Ascension and Declination, like longitude and latitude). There is a designation J2000.0 which tells you the zero point to which the coordinates relate (remember, the Earth wobbles, so any coordinate system tied to Earth changes over time, hence a zero point), in this case the year 2000. The Equ. in front of that means equatorial, which ties is directly to the earth axis (poles & equator; the latter explains the name).

Redshift (z is the symbol for redshift) for a distance estimate.

Qual probably stands for qualification, for example the type of galaxy. It's likely not related to the filter. The type of galaxy would be the classification, which is further down the page.

Magnitude is the brightness, and Filter the filter it is measured in (most astronomy is done in a filter, to obtain a brightness in a specific wavelength range. Eg, red, blue, green, uv, infrared, H-alpha, Oxygen filters).

Separ arc min: the separation (or more likely, the diameter) in minutes of arc (so in 1/60 of a degree). You can't really measure the actual size (unless you trust the distance estimate), but you can measure the size as a angle projected on the sky. That's what it says here (e.g., the sun and the full moon are about 0.5 degree, or 30 arcminutes, "wide").

As mentioned at the start, browse down to get more in depth information, which may shed a bit more light on what the top table abbreviations actually mean and indicate.


Which Planet/Star is it? I live in north India. About 24° latitude. I can see a very Bright star in sky. Its towards east and high up. Is it Venus , Jupiter, star or anything else?

Its not Sirius. I used Orion belt for that.

Date: 24 March , 2016

Time : 8 :00 p.m.

At 11: 00 pm.
Its directly upwards. Angle b/w moon and sun ~60°.

The site in-the-sky.org has a wide variety of functions and options. In Planetarium mode, I chose a random city at about 24N and in the middle, which helps to get the correct UTC + 05:30 India Time Zone, and then just put in the time and date and turned on alt/az grid.

So it is likely to have been Jupiter, as you suspect. Below are two screen shots - 20:00 and 23:00 local time, with Jupiter climbing from 41 to about 73 degrees above the horizon.


Is there anything like a supermassive white hole? How are they formed? And where are they present? The idea of a 'super massive white hole' is sometimes discussed regarding GR,SR and the destination of material 'lost' in a black hole. Is there any good theoretical or observational evidence for these SWH, or are they merely metaphysical speculations?

White holes are a theoretical construction of General Relativity.

Despite extensive searches, nothing has been found: not a supermassive white hole, not a stellar-sized white hole, nothing. Thus, the answer to your question is: on our current knowledge, they don't exist anywhere.

It is unclear how could they be formed. The geometry of the Kerr-solution (i.e. rotating black holes) suggests a white hole could exist on the "other side" of a black hole, but currently it is pure speculation.


What are the possible solutions to the Red Supergiant problem? I have recently come across this so called "Red Supergiant problem" in the literature, a phrase that was coined by Stephen Smartt in 2009 in reference to why red supergiants with masses ∼16-30M⊙ have not been identified as progenitors of Type IIP supernovae.

As I understand it astronomers have observed red supergiants with masses between 8-25M⊙ but only those below ∼16M⊙ have been proven to have undergone a Type IIP supernovae. With the minimum mass required to produce a Wolf-Rayet star at least 25-30M⊙ these red supergiants arent massive enough to be evolving into WR stars either.

There could be systematic errors with the models or calculations and it is well known that calculating the mass of a supernova progenitor post-eruption can be quite a difficult task, but assuming these calculations are correct it leaves a big hole in our understanding of stellar evolution. So my question is, what can explain this gap?

Solutions to the Red Supergiant Problem can be either observational or physical, and to date, both types have been proposed. Recent data and improved computer modeling have both helped make the later stages of red supergiant evolution clearer, leading to possible solutions to the problem.

Walmswell & Eldridge (2012) suggested an observational solution, namely, that circumstellar dust resulting from mass loss via the strong stellar winds of a red supergiant could lead to strong extinction in some cases, making it hard or impossible to detect the resulting supernovae. Based on data collected on 18 supernovae, they found an upper limit for the mass of a Type IIP  supernova progenitor (at 90% confidence) to be around $27 M_{\odot}$, which would explain the dearth of detections of progenitors in the $16$-$30 M_{\odot}$ range.

The observational difficulties are moot if there is a physical reason behind the problem, i.e. if there is some mechanism preventing these red supergiants from undergoing Type IIP supernovae. Yoon & Cantiello (2010) suggested that red supergiant stellar winds could be increased to "superwinds" by envelope pulsations. This would lead to periods of larger mass loss rates than expected, followed by periods of smaller mass loss rates. This would correspond to oscillating around states of instability due to the pulsations, and would be prominent only in red supergiants of masses greater than or equal to about $19M_{\odot}$. The result could be drastic, with stars losing well over half their masses. The star could then evolve toward a Type Ib or Type IIb supernova.

Smartt et al. did consider the possibility of envelope loss in the paper you cited, but dismissed it as being unfeasible for a star in the mass range. However, it appears that they did not consider the superwinds induced by pulsations, making this solution feasible once more. It's also worth mentioning that the authors suggested direct collapse to black holes with at the most a very faint supernova; this idea had been explored in the past.


Superscript-subscript uncertainty notation While reading publications in astrophysics I've encountered this kind of notation (fictitious example):
$$M = 6.1^{+0.6}_{-0.1} \times 10^{30} \textrm{ kg}.$$

Several examples are here, p. 2.

I assume that the superscript and subscript refer to the right and left uncertainty of the main magnitude. I have three questions:

Can someone explain to me what these three numbers (main magnitude, superscript, subscript) refer to, in terms of the probability distribution for the true value? For example, are they the mean and 10% and 90% quantiles? Or maybe the median and 25% and 75% quantiles? Or have I misunderstood them completely?
Can someone provide early (or the first) examples of this particular superscript-subscript notation? I only knew the "$x \pm \triangle x$" notation.
Can someone provide a written reference – textbook, manual, article, or similar – where this notation is explained?

I checked this question, but it didn't really help me. I also checked the Guide to the Expression of Uncertainty in Measurement, published by the Joint Committee for Guides in Metrology (JCGM), formed by various organizations (BIPM, IEC, IFCC, ILAC, ISO, IUPAC, IUPAP, OIML). It is a standard for nomenclature and practice in reporting uncertainty, and makes important distinctions, for example between "coverage intervals" and "confidence intervals" (see JCGM 101:2008 definitions 3.12–3.16). But I can't find anything about the superscript-subscript notation there.

Cheers!

When quoted like this they would usually be the most likely value (peak of the probability distribution) or the median value (where half the distribution is above or below), and then the superscripts and subscripts would be increments that would contain $\sim 34$% of the probability distribution. In the M87 papers, the numbers are the 50th percentile (median) and lower (upper) limits corresponding to the 16th (84th) percentile.

The total range corresponds to about a 68% confidence interval. The reason for this is just that it corresponds to $\pm$ one sigma in a normal (Gaussian) distribution.

It is good practice to say what your error bars mean, but in the absence of that I would assume the above.


Why do the planets orbit in the same direction? Theoretically, planets would have an approximately equal chance of going one way in their orbit or another but in reality, this is not the case (at least in our solar system). Why is this?

The same reason (almost) all of them rotate in the same direction: because of the conservation of angular momentum.

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

On that scale, there is some small amount of rotation within the cloud. It could be caused by the gravity of nearby stellar objects, local differences in mass as the cloud churns, or even the impact of a distant supernova. The point is, all molecular clouds have at least a little rotation.

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

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

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

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

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


Is the Milky Way Visible from Nebraska? Is the Milky Way Visible from Nebraska?  If so, where is the best place to view it, and also what would be the best time of night to see it?  I know this is probably a very novice question, just trying to get a better understanding of what I can see.  If it helps, I'm south of Omaha.

Well, because the axis of the rotation of the Earth is not the same as the axis of rotation of the disk of the Milky Way (and also because we're transforming a 2-dimensional spherical map into a 2-dimensional cartesian map), the path of the disk of the Milky Way galaxy looks something like this:

So, there is actually a wide range in declination that the Milky Way can be seen at. The range of declination you can see depends on your latitude (for a review of RA and declination, coordinates used in the celestial coordinate system, see this post). For example, here in Philadelphia (just about $+40^{\circ}$ declination), I'd be able to see from from $-50^{\circ}$ to $+90^{\circ}$ in declination. For Nebraska, find the latitude of your location. To calculate the lower limit, add your latitude (which will be positive since you're in the northern hemisphere) to $-90^{\circ}$ (mine was $+40^{\circ}$, so: $-90^{\circ} + 40^{\circ} = -50^{\circ}$). To find the upper limit in declination, it's even easier. Since you're in the northern hemisphere, you can actually see the north celestial pole. This means that the upper limit is simply the maximum it can possibly be, which is $+90^{\circ}$. The larger the latitude the more circumpolar your night sky gets.

The good news is that you should definitely be able to see it in Nebraska. The only question during what season will you see it at its best? Take a look at the constellations which contain parts of the Milky Way and find the season you can see it in - now you know where to look.

The last thing I want to add is that you need nice dark conditions. Local weather conditions or sources of light pollution may very easily hide the Milky Way. From Philadelphia, even on perfectly clear nights we have no chance of seeing it. If you've got a city on one of your horizons, try to plan around that - either go to a darker location or try looking opposite in the sky from any sources of light pollution.


Is the age of the Universe really 13.8 billion years? Ok, I know this has been asked by a lot of people, but my reason for asking this question is a bit different. Please read further.
I was watching a video by Fermilab (Start at 6:30, at 8:30 he mentions how we cannot ever see light emitted from objects > 14bn light years away). In it, the presenter Dr. Don Lincoln talks about the speed of light and how some things are faster than that. Expansion of the Universe can be faster than the speed of light, because galaxies 1 megaparsec away appear to be moving at the rate of 68km/s, galaxies 2 megaparsecs away at 136km/s, and so on. So the velocity of the galaxies doubles every megaparsec. If we do the math, the speed of the galaxies moving away surpasses the speed of light at 14 billion light years away.
14 billion is just 200 million give or take more than the current measurement of the age of the Universe.
My question is, at 14 bn light years away the galaxies are moving faster than the speed of light and we can never see anything beyond that ever. Does this mean that even after say 5 billion years (assuming we still exist), if somebody uses the same technique to measure the age of the Universe, they will end up with the same number that is lesser than 14 billion?

The age of the universe is not calculated based on the size of the visible universe. The age of the universe is being calculated based on the fact that the laws of nature have no direction. This means that you can use the laws of nature to predict future behavior, but also assume previous behavior. Based on calculating backwards with the laws of nature, for instance, general relativity, physicists can to a pretty certain degree calculate backwards how the universe started.
The fact that the universe is expanding and the size of the universe is what it is, has no relation to the age of the universe. Please, look at another video from Don Lincoln about the size of the universe and its age. Here it is explained that the universe already had an extra large expansion making it bigger than the age.
What will happen, though, is that the visible universe, which is smaller as the actual size of the universe, will be more empty over the course of time. In the 5 billion years you mentioned, there will be more galaxies outside the visible boundaries of the universe, because they are moving away from us faster than the speed of light. The expansion of the universe, hypothesized due to dark energy, makes the universe bigger, not older.
If the laws of nature don't drastically change, and we use these laws in 5 billion years to calculate the age of the universe, we should end up with the right number, 14 billion plus 5 billion years.


Maxwell stress contribution to $\nabla \cdot \mathbf{P}$ in the Navier-Stokes equation for fluid in stars I was reading through the following extract outlining how the Maxwell stress contributes to the $\nabla \cdot \mathbf{P}$ term of the Navier-Stokes equation for fluids in a star. Here $\mathbf{P}$ is a general symmetric stress tensor.

I am struggling to see how the estimate $\vert \nabla\cdot \mathbf{P}\vert/\rho \approx 3\times 10^{-3}$ ms$^{-2}$ was derived using the information given in the passage. Can someone please provide a quick derivation of this result?

$\nabla B^2 \sim B^2/l,$ where $l$ is a length scale on which $B^2$ varies.

In SI units, $B^2 \sim 4\times 10^{-8}$ T$^2$ and $\mu_0 = 4\pi \times 10^{-7}$, so an order of magnitude for $\nabla \cdot P_B$ is $4\times 10^{-8}/(8\pi \times 10^{-7}\times 10^{6}) \sim 10^{-8}$ Pa/m.

The density of the solar photosphere (at optical depth unity in the visible spectrum) is around $\rho \sim 10^{-4}$ kg/m$^3$.

Thus $\nabla \cdot P_B/\rho \sim 10^{-4}$ m/s$^2$.

So no, I can't quite reproduce your number.