Usage of $\sim$, $\approx$, $\simeq$, and $\cong$ in observational astronomy? My understanding is $\sim$ generally means "on the order of magnitude of" e.g. $T \sim 10^5$ K

$\approx$ is obviously "approximately equal to" so for example one might write $d \approx 400$ pc rather than $d=4 \times 10^2$ pc

$\simeq$ and $\cong$, are where I am more confused as their usage is rarer and less consistent.

The below IAU resolution, for example, seems to use $\simeq$ for truncation $m_{\mbox{bol}} = -26.83199\ldots \simeq -26.832$, but I have also seen $\simeq$ used as "approximately equal to" in other works.

http://arxiv.org/pdf/1510.06262

How should one be using $\sim$, $\approx$, $\simeq$, and $\cong$?

$\simeq$ and $\approx$ both mean "approximately equal to". I don't think $\cong$ is used so often, but if I read it, I would interpret is as the same as the two others.

$\sim$ in principle means "of the order of", i.e. correct to within an order of magnitude. However, astronomers tend to be rather slobby sometimes (we call all elements except H and He for "metals", we call it "gas" even when it's ionized and should be called "plasma", we omit factors of order unity if we can get away with it, we mix up terms like "flux", "flux density", and "intensity", and so on), and thus $\sim$ is very often used to mean "approximately equal to". It is even sometimes used as "proportional to", but (usually) only when there's no risk of misunderstanding (e.g. $L_\mathrm{UV}\!~\sim\!~\mathrm{SFR}$, meaning that the UV luminosity is proportional to the star formation rate, omitting a factor of $10^{27}$).

Journals usually have rather strict standards for style and nomenclature, so they might not accept a $\sim$ for a $\propto$, or even for a $\simeq$. But if you read a conference proceeding or lecture notes or some other non-refereed text (even a paper on arXiv waiting to be refereed), you best be prepared for anything.


Could a cryo-volcano be the reason behind this colour difference in Iapetus's hemispheres? Iapetus's hemisphere facing Saturn is dark, whereas the opposite one is bright.

Could a cryo-volcano be the reason behind this colour difference in Iapetus's hemispheres?

Reposting my own answer from Space Exploration on nearly identical question:

The color dichotomy of Iapetus is due to the darker half, the Cassini Regio, being a result of the moon's accumulation of the dust in the Saturn's largest, yet extremely tenuous, diffuse dust ring called the Phoebe ring depositing onto the moon's leading hemisphere, while the trailing hemisphere's surface remains uninfluenced by this source of dust.

Phoebe ring was only observed first in late 2009 by the Spitzer telescope in the infrared range, and is believed to have been caused by bombardment of micrometeorites of Phoebe, another Saturn's moon that the prevailing theories suggest is a captured Kuiper belt originating planetesimal orbiting Saturn in a retrograde, 173° to the ecliptic inclined orbit. It is the largest discovered ring of the Saturn, stretching from calculated 59 to 300 Saturn's radii (observed between 128 and 207 Saturn's radii by the Spitzer telescope), is tilted 27° to Saturn's inner rings and with an orbital inclination of 175°, suggesting its origins and again retrograde orbit:

    The Phoebe ring was discovered by NASA's Spitzer Space Telescope, detecting it in the infrared spectrum. (Source: Wikimedia)

Iapetus, being in a prograde orbit with a 17.28° inclination to the ecliptic, is directly in the path of this retrograde dust ring at an average distance from Saturn of 215 Saturn radii. Since it's tidally locked to Saturn, it would always face same side towards any debris and dust in its path. Joseph A. Burns, Cornell's Irving Porter Church Professor of Engineering and professor of astronomy said in a press release for Cornell Chronicle:

  "The ring of collisional debris that has come off Phoebe and its
  companion moons is out there, and now we understand the process
  whereby the stuff is coming in. When you see the coating pattern on
  Iapetus, you know you've got the right mechanism for producing it."

So this is the theory. Iapetus is in a prograde orbit intersecting at inclination Phoebe dust ring in a retrograde orbit around Saturn, and since Iapetus is tidally locked and always hitting this dusty path head on, only one hemisphere (the darker one, called Cassini Regio) is covered in Phoebe's dust, while the other hemisphere (the lighter one, called Roncevaux Terra) remains free from it, unveiling this Saturn moon's true surface.

    Artist's conception showing nearly invisible and largest Saturn's Phoebe ring (Source: Wikimedia)

Same mentioned Cornell Chronicle press release explains why the transition from the dark hemisphere to the whiter parts of the moon isn't seamless, but rather a mottled, patchy array of bright and dark spots:

  Small, white craters that dot Iapetus' darker half indicate a veneer
  of dark dust, only meters deep, covering a white, icy surface that
  matches the rest of the satellite. The imaging data also revealed that
  all the materials on the leading side are much redder than the
  shielded and brighter trailing side -- another indication that the
  leading side's dust came from elsewhere.

  The pattern, the scientists say, supports a theory described in a
  companion paper in Science that the darker parts of the moon tend to
  heat up when struck by sunlight, encouraging the ice to evaporate
  underneath.

  This causes any dark spots to get even darker, creating the mottled
  look.

           Mosaic of Iapetus taken by Cassini probe during NASA's Cassini Solstice mission, showing the bright trailing Roncevaux           Terra hemisphere with part of the dark area Cassini Regio appearing on the right. (Source: Wikimedia)


Do telescopes exist that reflect the incoming light more than three times along their length? Refractors only use the length of the telescope once, reflectors twice, catadioptric telescopes like those of the Schmidt-Cassegrain design three times. Have telescopes been built that reflect the incoming light at least once more from one end of the telescope to another?

Say, by adding a tertiary mirror which reflects the light forward again, towards a small quarternary mirror in front of the secondary, which ultimately directs the light though a small hole in the tertiary (and primary) mirror towards the eyepiece?
Note: The above description makes this question different from the similar-sounding Do triple or more times reflecting telescopes exist?

If your willing to accept more than 2 discontinuous mirrors, the Three Mirror Anastigmat has 4 passes along some/most of the overall tube length. An early working prototype example (which I've actually seen in person many years ago) was built at the University of Cambridge by Dr. Roderick Willstrop. The Institute of Astronomy has a page on the Three Mirror Telescope which includes the following optical diagram:

The 0.5 meter 3 Mirror Telescope (3MT) has a focal length of 0.8 meters, in an overall tube length of 1.2 meters producing a very compact telescope with a large field of view (5 degrees in diameter) with good image quality (<0.33" across the whole field of view). The quality of the site at Cambridge was not ideal for building on the progress of the prototype and a proposal for building a larger version at a better site was declined by the UK science funding agency of the time (SERC, which became PPARC and then STFC) in favor of buying into the Gemini collaboration of two 8 meter telescopes.
Some constructed or in construction examples of three mirror anastigmat include:

James Webb Space Telescope (see Figure 2 of this page on the JWST Telescope)
The Simonyi Survey Telescope of the Vera C. Rubin Observatory which will carry out the Legacy Survey of Space and Time (LSST) has M1 and M3 ground into the same piece of glass (see Figure 7 of the LSST overview paper) with a separate secondary M2 mirror and prime focus camera
The planned ESO Extremely Large Telescope will use 5 mirrors in a folded three mirror anastigmat design along with 2 flat mirrors which will have 5 bounces along all or part of the overall "tube" structure before exiting to a Nasmyth focus; see the not very clear diagram below:    from the ESO ELT optical diagram page, somewhat better diagram in  Figure 4 of Hippler 2018.


Why did the Chelyabinsk meteor explode? In February, a meteor entered the Earth's atmosphere over Chelyabinsk, Russia and exploded, producing a shockwave blast that damaged hundreds of buildings around the explosion site in Russia.

But why did it explode? Why didn't it simply burn instead?

Same reason as the explosion at the Tunguska event years ago: Pressure, due to boiling water.

Meteors usually cme into our atmosphere at incredible speeds. At these speeds, the meteor is highly affected by the viscosity of air and is heated by compression (and a little bit of friction).

This starts boiling the ice present in the meteor, leading to a pressure buildup. If the conditions are right, it can just explode.

See also: http://www.nature.com/nature/journal/v361/n6407/abs/361040a0.html , https://physics.stackexchange.com/q/76045/7433


What good evidence exists for the 9th planet as spoken of by Caltech? I'm sure Caltech has answered this in some way, but I think it's a good question have on this site.
What good evidence exists for the 9th planet as spoken of by Caltech?
As I see it, Caltech has apparently looked at the orbital parameters of many Trans-Neptunian-Objects (TNOs) and noticed that the lines between their perihilion and aphelion (I don't know what the technical term might be) all run more or less in the same direction. This is evidence that some large body has gravitationally "synced" these planets over millions of years.
Except that alone isn't good evidence. Here are two stats first:

Known TNOs: about 1,750 according to the Minor Planet Center's list.

Estimated existing Kuiper-Belt Objects: 120,000+ according to Britannica. (That number is only estimating objects of 100-km diameter or more, but that's fine for our purposes.) Note: 120,000 is only for KBOs. There are other TNO sets such as the oort cloud (and the scattered disc if you define it separately), so I expect the number to be much much bigger than 120,000.

Given that we've only measured about 1% or less of TNO orbits, how can we be sure we have a good random sample to say that a massive body must have aligned these orbits? In other words, all the other orbits might be randomly distributed, which would be compelling evidence that some giant body is not "syncing" orbits or else way more than 1% of them would be "synced".
If I recall my statistics class correctly, 1% is not good enough for a random sample. The border-line good enough area starts at 3%, but there's still a catch: It has to be a random sample. There are still observation-bias questions here because obviously we've only mostly discovered the "nearby" TNOs, because they are easier to discover when they're "nearby". That certainly would not be a random sample even if we had 3% of them.
TLDR: There are at least 100,000 distant objects, of which less than 1% of them have some kind of aligned orbit as far as we know today. Is this actually good evidence that something has aligned them?
EDIT: Another analogy that might help. You have a bag of 100 marbles. You reach in and pull out 5 from the top of the bag (notice I didn't say any random 5!). They turn out to be 4 red ones and 1 black one. Is this actually good evidence that about 80% of the whole bag is red?

What they have at present is best described as a theory based on data that is not precise enough to make a definitive statement.  They are telling people they think something is out there and roughly what they think it might be, but they really don't have accurate enough data to make any statements beyond that.

The fact that the possible range of distances they suggest is so large tells you all you need to know about the accuracy of the data being insufficient to draw reliable conclusions.

My biggest problem with this claim is that we have a historical lesson that suggests caution : Pluto.  Pluto was found when looking for an entirely different object ( coincidentally about the same mass as the new planet X, if memory serves me ).  It transpired that the orbital anomalies that led to predicting a Planet X 1.0 were later evaporated by more accurate measurements and theory.

Are we now seeing the same sort of error ?  Could be, might not be.  All we can hope is that the quest for an explanation generates either the result claimed or, if not that, better data, more data and perhaps more accurate simulations and theory.

I am reminded that, in the past, quite eminent and respectable scientists have claimed any number of things, which would later prove incorrect.  Cold fusion leaps to mind.  The problem with humans looking at data is that, sometimes, they see what they want.  I think this planet X 2.0 claim needs a colder and more objective, perhaps even cynical, examination, away from the media.


How easy would be to see the Moon from Mars? This question arises from "What planet is better than earth to infer solar system configuration?", where one of the answers pointed that it's easier to discover the Moon from Mars than discovering the Galilean satellites of Jupiter from Earth.

Then my question is:

How easy (or difficult) would be to see the Moon from Mars?
Could it be seen with naked eye?
Is actually easier to see the Moon from Mars than to see the Galilean moons of Jupiter from Earth?

I know there are favourable factors (Mars is closer to Earth than Earth to Jupiter, and Earth is smaller than Jupiter, so it's expected to obscure less its moon than Jupiter obscures its) and also unfavourable factors (the Moon is closer to the Earth than the Galilean moons to Jupiter, and Earth and Moon are backlit when seen from Mars). What is the overall result?

We have pictures of this, thanks to Curiousity!

(Source)

  “A human observer with normal vision, if standing on Mars, could easily see Earth and the moon as two distinct, bright “evening stars,” said NASA in a statement issued today.

So yes, you can see our moon from Mars.

The apparent brightness of our moon from Mars is +0.9. The apparent brightness of Jupiter's 4 biggest moons is between 4.6 and 5.6 under ideal conditions.  However, while our moon can be seen visually separate from Earth from Mars, Jupiter's moons can't be distinguished from Jupiter itself with an unaided eye.


Can Pluto and Neptune collide anytime in future? We know that, the orbit of pluto ambiguously cuts the orbit of Neptune. I want to know, whether the revolving speed of Neptune is equal to the revolving speed of pluto. If not, is there a possibility in future,  that the two planets, or rather two celestial bodies will collide in future?

It's hard to find a good 2 dimensional representation, but Pluto and Neptune's orbits don't actually cross.

In fact, the orbits (not the planets but the orbits) never get within 100 million miles (I remember reading that, but can't find a link right now).  This is because Pluto's orbit is on a different plane.   It's easy to represent in 3D, hard to draw in 2D, but when Pluto is effectively the same distance from the sun as Neptune it's "below" Neptune, looking at the solar system from a sideways point of view.   When it's on the same plane as Neptune's orbit, which it only crosses twice in one of it's years, it's further away from the sun.

Source_1 and Source_2 and Source_3

An astronomer diagrammed the 2 objects orbits for the next 4 billion years and they don't intersect during that time.  Source

But there is a quirk in the calculations.   Pluto's resonance with Neptune is probably long term unstable and because of this, it's orbit could be subject to variations that leave open a Jim Carrey's chance of hitting Neptune.   Source

It's very unlikely, because they're not on the same plane and they're not orbiting at the same period and they're currently in orbital resonance which they would need to fall out of to even have a chance, so it's a lottery ticket's chance of happening but it could theoretically happen, in a billion or couple billion years.

It's more likely that Pluto would fly past Neptune and have it's orbit changed, and even that's still a long shot.  The most likely scenario is that they never get close, at least over the lifespan of the sun, but if they do get close, Pluto could fly further out into the solar system or closer in due to gravity assist.   It might even hit Earth.   But, I repeat, this is crazy-long-shot territory, but I think it's worth mentioning that it's about as likely to hit one of the inner planets as it is to hit Neptune.  Jupiter being the largest, might have the best chance of being hit by Pluto.

The question is also answered here, but because everyone else was answering this question no, I wanted to point out that it's just barely possible but extremely unlikely.


Is the SETI's `needle-sharp at 982.002 megahertz` signal accessible? In April-May 2019, the Parkes radio telescope, in Australia, received a weird signal while surveying Proxima Centauri. Is this signal available, ideally in some digitalized form? Where could I download it?

If those data were available to the public, the Breakthrough Listen Open Data Archive would most likely have them.
Queries there currently return no 982 MHz data for any sources, and "PROXCEN" data only at 3009 and 3094 MHz from 2017-18.
This Scientific American article mentions work in progress by Sheikh et al.; they may have dibs on the 982 MHz data from 2019 until they publish their paper.


Is there going to be a really bad Solar Storm in 2014? Here are some Articles I've found.

Is it true? If it is, How big could it be?, what might be possible or will be possible?, what will happen?

Is it true?

Solar storms are like Earth storms: unpredictable until they are quite near.

  What will happen?

If there is a massive solar storm, we will experience power outages in long lines (e.g. U.S.A. mainland, Russia), communications disruption on satellite links (and again in long copper lines), and very nice auroras.


Does tidal heating imply orbit degradation? What compensate the energy loss in tidal heating? Is it orbital decay?

There is a wonderful post at Physics Stackexchange:

  Gravitational coupling between the Moon and the tidal bulge nearest the Moon acts as a torque on the Earth's rotation, draining angular momentum and rotational kinetic energy from the Earth's spin. In turn, angular momentum is added to the Moon's orbit, accelerating it, which lifts the Moon into a higher orbit with a longer period. As a result, the distance between the Earth and Moon is increasing, and the Earth's spin slowing down.

You can find more reading material and information in the original thread.


Why is the cloud around Eta Carinae shaped the way it is? According to (german) wikipedia, Eta Carinae has a cloud shaped like two cones with the points at the poles of the star as well as a disk around the equator. These matter clouds are mostly or only condensed stellar wind from Eta Carinae.

In my understanding, to create the dual-cone chape, the stellar wind provding the matter originates at the poles and this gives the dual cone shape (Ok?).

Why the poles? Stellar wind would, I guess, be mostly stellar matter so hot it reaches escape velocity, further pushed by the radiation. This should be isotropic. I can only guess that maybe Eta Carina has a magnetic field shaped similarly to earth with a north and south pole, nearer the equator the stellar wind is perpendicular to the field lines so the flightpaths of the charged particles are bent back inward. Near the poles the flightpaths and field lines are mostly paralell, so the magnetic field does not really act on the stellar wind. I included this guesswork on my part so you know where to start.

ETA: Turns out my guessing about stellar wind was wrong or mostly wrong, the cloud appears to be mostly shaped by eruptions in recentish history. But the question why they are shaped that way still stands.

It seems likely due to rotation.  The disklike "skirt" could be due to material that is orbiting the star that gets thrown out by the eruption, and the bipolar nature of the eruption can be because material is preferentially ejected from the poles of the star.  It may seem odd that the poles, where there is less angular momentum per gram, would be better at ejecting material, but this might be due to a greater role for radiation pressure in causing the eruption.  Radiation pressure produces forces that are proportional to the radiative flux, which can come out preferentially in the polar directions because this becomes the path of least resistance when rotation makes the shape of the star highly oblate.


Just after the Big Bang, was the space, in its entirety, filled with Hot gas? After the Big bang occurred, was the space filled with hot gas totally?

If there's any problem in my question please inform me. Thanks!

I'm not sure what you mean by "nebulae".  You could mean "matter in a gaseous or plasma state" or "cloud of low density gas and dust in space" (which is what astronomers mean by nebulae.)

The former case is believed to be correct, and it pretty well described in Wikipedia. I'll summarize. Immediately after the Big Bang (10-20 seconds or so) the universe was full of extremely hot and extremely dense matter -- hot enough that not only did atoms come apart, and nuclei come apart, but even protons and neutrons came apart into a quark-gluon plasma. This cooled into a stew of elementary particles, which cooled into mostly protons, neutrons and electrons, which cooled partly into nuclei. At this point we're maybe a minute into the expansion and the universe was full of a plasma that was still hotter than the center of a star. After around 300,000 years, the plasma cooled enough to recombined into atoms. Stars eventually started to form.

No astronomer would call any stage of this expansion a "nebula" -- they first formed later after the atomic gas left a milion years after the Big Bang started to condense into stars and patches of gas and patches close to vacuum.

Bottom line: Nebulae did not exist until millions of years after the Big Bang, but the universe was filled with hot gas much earlier in the Big Bang.


Does the ratio of daylight change from year to year? Over a year, the ratio of daylight changes, e.g. in Montreal there is more daylight in the summer than in the winter.

I want to know if, given a bad data sample, e.g. five years, one might notice a trend of increasing or decreasing daylight ratios.

Here's why I ask. I'm doing research on the Dead Sea scrolls, and some of the fragments contain ratios that a scholar has argued represent daylight. What I'd like to know is a trend that the community of Qumran might have noticed in their dataset. They had a theory that darkness was increasing. I doubt their data was exhaustive, but I wonder if around Jerusalem there might have been a period of decreasing daylight ratios over a period of a few years roughly around 100 B.C.E.

I don't think it's possible to have a change in the ratio of day to night when averaged on a yearly basis.  To do so would require either a nonlinear change in the Earth's rotation or a significant shift in the polar axis tilt within a year.
If the rotation were simply slowing down over time, the year-averaged ratio would be the same every year (only the number of days in a year would change).  Similarly, the only way a polar tilt could cause a change would be if the change in angle occurred at some (sinusoidal) rate not an integral ratio of the orbital period (year).
Either of those putative mechanisms can only change the per-year ratio by producing a day/night ratio in "summer" that is not the inverse of the "winter" ratio, and producing a different non-inverse the following year.  (I'm oversimplifying a bit, since the elliptic orbit produces a repeatable asymmetry in the Sun's zenith pattern).


If a white dwarf collides with a giant star, could it create a TZO? Thorne–Żytkow objects (TZOs) form from collisions between neutron stars and main sequence or giant stars. Ultimately, the neutron star becomes the "core" of the giant star.

However, could this also happen with white dwarfs, if the combined mass is below the TOV limit? White dwarfs are degenerate stars, just like neutron stars, so would they be able to replace the star core in the same way? Would anything happen to the white dwarf if the combined mass is beyond the Chandrasekhar limit?

Thorne and Żytkow's original paper on TŻOs actually opens with a comparison of TŻOs and the type of object you mention, with a  white dwarf degenerate core instead of a neutron star degenerate core. They note that the equilibrium states - essentially, stable configurations - of such combinations lie near the Hayashi track (actually acting a bit like AGB stars, in some cases), indicating high metallicity, as is the case with TŻOs.
These objects generate energy the same way TŻOs do: matter is accreted by the core, releasing gravitational potential energy, and the red giant envelope continues some fusion, although, of course, core fusion has been substantially disrupted by the arrival of the new degenerate core. The main difference in energy production are the ratios between nuclear contributions to luminosity and gravitational contributions to luminosity:
$$L_{\text{nuc}}/L\approx0.99,\quad L_{\text{grav}}/L\approx0.01\quad\text{for white dwarf core}$$
$$L_{\text{nuc}}/L\approx0.04,\quad L_{\text{grav}}/L\approx0.96\quad\text{for neutron core}$$
Why the difference? $L_{\text{grav}}$ is proportional to
$$\frac{GM_c}{R_cc^2}$$
where $_c$ refers to values for the core. The masses and radii of neutron stars differ drastically from those of white dwarfs. This becomes less important in the case of supergiant TŻOs (i.e. $M>10 M_{\odot}$), because convection cycles "burned" nuclear fuel back outwards into the envelope, and so energy ratios become more like those found in the case of a white dwarf core.
This difference in energy production ratios also means that the objects will remain in roughly stable states for different amounts of time; red giants with white dwarf cores can survive in equilibrium for at least an order of magnitude or more as long as TŻOs.
One interesting thing to note is that TŻOs and red giants with white dwarf cores may share some of the same problems when it comes to stability. The envelopes are expected to be composed similarly and act similarly, with the potential difference in nuclear fusion rates, and so the same dynamical instabilities are possible in both cases. However, Thorne and Żytkow state that they find this possibility unlikely.


What exactly is it that is being magnified 50 times in this gravitational lensing observation? In the Los Angeles Times news item Scientists get a rare view of a type Ia supernova magnified 50 times what exactly is magnified 50 times?

This supernova is really very far away. Is it somehow imaged by the gravitational lensing and being resolved — the actual supernova itself?

edit: I believe this is it, but it's behind a paywall (will confirm in due course): http://science.sciencemag.org/content/356/6335/291

above: From the Los Angeles Times. Photo credit: Joel Johansson

above: "This schematic image represents how light from a distant galaxy is distorted by the gravitational effects of a nearer foreground galaxy, which acts like a lens and makes the distant source appear distorted, but brighter, forming characteristic rings of light, known as Einstein rings. An analysis of the distortion has revealed that some of the distant star-forming galaxies are as bright as 40 trillion Suns, and have been magnified by the gravitational lens by up to 22 times. (Credit: ALMA ESO/NRAO/NAOJ, L. Calçada (ESO), Y. Hezaveh et al., edited and modified by Joel Johansson)" From the Los Angeles Times.

Reading through the full article, I see that the authors write

  we estimated the lensing amplification to be $\mu\sim52$.

This refers to the magnification factor. Essentially, it describes the solid angle of the image, as related to the solid angle of the source1:
$$\mu\equiv\frac{\theta}{\beta}\frac{d\theta}{d\beta}$$
Here's a diagram of the magnitudes of the angles:

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


Can we manipulate cosmic rays? Is it possible to send data using cosmic rays?

Like sending high and low energy cosmic rays, since they can travel farther. And this data can be converted to binary form that can be processed by computers.

I'm not an astronomer or studied any thing related to astronomy, I was just imagining if this is possible.

Most of the communication on Earth and to-fro Space takes place over low frequency visible, infrared, micro and radio wave spectrum of the electromagnetic spectrum.

The immediate benefits of using low frequency EM spectrum which I think of are as following:

They are not blocked by Earth's atmosphere and magnetic field. On the other hand, cosmic rays which consists mostly of high energy electrons, protons and ß particles carries electric charge and magnetic field. Hence, they will be hugely affected by Earth's magnetic field. Maybe setting up receivers and transmitters at poles helps a little.

2.Speed:  EM waves travel at light speed, cosmic rays always travel at lesser speed which brings me to the next point.

Energy needs: The energy required to generate radio waves can be provided by low power solar panels. However, to generate cosmic rays one would need a particle accelerator. Something like a portable LHC, which I don't think is feasible. Moreover, if you want to generate faster cosmic rays then you have to put in more energy.

I can only think of these constraints at this moment. I don't know if sending information via cosmic rays carries any advantage.


Milky Way as seen by the human eye: where exactly do those photons come from? I'm asking here about the diffuse glow seen by the human eye, when looking at the Milky Way.

Where do those visible photons come from exactly?

the surface of a star?
starlight bouncing off something (dust or a molecule)?
emitted by gas in the interstellar medium?
all of the above?

I believe that the answer is mainly the first (directly from stars), although there will certainly be some of each of the others. The diffuseness comes from the fact that there are many more relatively dim stars than rods in the retina of the human eye.  My argument for this is that magnified photographs of most of the Milky Way at least, break down into individiual stars, as seen in this image from astropixels.com


Day,Night, and Year in the Ptolemaic Model As far as I understand, the Ptolemaic model explains day and night by postulating that the whole celestial system revolves around the Earth once every day. Since we observe the sun to move from East to West over a day, the whole system would have to move in an East to West direction once a day. But over the year, the planets move (usually) towards the East and the constellations towards the West. So in this model, the planets on their different move opposite to the daily motion of the whole system? Whereas the sphere of stars moves in the same direction as the daily motion of the whole system? I ask since I just found this odd for some reason.

As I remember, the Ptolemaic model involved a series of concentric transparent spherical shells around the Earth.
The stars were attached to the outermost shell or sphere which revolved about once a day, but not exactly once a day since there was also a yearly cycle to which stars were visible at night.
The innermost sphere was the sphere of the Moon, which sometimes passed in front of the Sun and thus was closer, and the next innermost sphere was the sphere of the Sun.
The Sphere of hte Moon revolved about once a day, but not exactly, since there was a difference which added up to a full circle every 28.5 days.  So, if i remember correctly, the sphere of the Moon would revolve about 360 degrees minus 12.63 degrees every day, or about 347.37 degrees per day.
The sphere of the Sun made a full and complete revolution of 360 degrees in exactly one day.
And the spheres of Mercury, Venus, Mars, Jupiter, and Saturn also revolved about 360 degrees per day, but not exactly, the differences accounting for their various synodic periods.
Since the geocentric model was totally different from the actual heliocentric model, and the giant transparent spherical shells didn't really exist, there were a lot of problems fitting the directions to various planets at various times predicted by the system with the actual observations of the directions to those planets at those times.  Even though the ancient Greeks had to rely on naked eye observations, they were able to notice deviations of the planets from their predicted paths across the background of the stars.  Which meant that the theories had to be improved to make the predictions better.
So Ptolemy had to do a lot of thinking and explaining to account for those deviations and to make correct future predictions of the movements.
So Ptolemy took the basic geocentric model I described above and elaborated it with various modifications which are called eccentrics, epicycles, deferents, and equants.  And if that sounds complicated, that is correct.
There is a famous story that King Alfonso X The Wise (1221-1284), King of Castile, etc., and rival King of the Romans, once said of the Ptolemaic model of the universe that if the creator had asked Alfonso's advice during the creation the universe would be simpler.
And the Ptolemaic model was able to predict the apparent motions and apparent positions of the planets well enough to satisfy almost all European and Muslim astronomers for over a thousand years.


What is considered "contact" for contact binary white dwarfs, and roughly how long can they stay that way before merging? The question Why do they think that WDJ0551+4135 is “snowman shaped” i.e. a contact binary? has also raised the question of what a contact binary is or isn't, and how hard it might be to see one if it is, so let's explore this separately and then hopefully establish some usage guidance for the contact-binaries tag.

Questions:

What's considered "contact" in contact binary white dwarfs? Photosphere contact perhaps?
roughly speaking, how long can pairs like this stay in contact before merging? Millenia? Minutes? If the time is very short that means they are much harder to observe in a state of contact.

"bonus points:" Are there any contact binaries we can see by eye or with binoculars or a small telescope? (wouldn't have to be white dwarfs) Could we see one vary in brightness and/or color as the pair rotates?

From the BBC article cited in the linked question:

  Artwork: The object may be two so-called 'white dwarves' merging into a smaller denser star

The issue I see here is that as a white dwarf gets more massive, it gets smaller. So in white dwarf–white dwarf mass transfer, the accretor will shrink, i.e. it evolves away from contact. Meanwhile the donor will tend to expand and lose mass faster through the L1 point.

So it might not even be possible for white dwarfs to form a stable contact binary. Proving a negative is somewhat tricky though.

There has been research into simulating white dwarf mergers. Dan et al. (2011) "How the merger of two white dwarfs depends on their mass ratio: orbital stability and detonations at contact" show several scenarios in their figure 7. The 0.6+0.55 M⊙ collision looks a bit like the Mark Garlick artwork from the press release, but it's on the line where disruption occurs on a timescale of ~20 orbits so it wouldn't look like that for very long until it merges and (judging by figure 10) undergoes contact-induced helium detonation.

From what I can tell, the usual use of "contact" in regards to white dwarf binaries refers to when matter from the secondary impacts (contacts) the primary, either by direct collision or mass transfer.

In terms of visible contact binaries, the brightest of the W UMa variables is Epsilon Coronae Australis which varies between magnitudes 4.74 and 5 over the course of an orbit. The next one is the secondary of 44 (i) Boötis, which is somewhere close to magnitude 6 (the magnitudes listed for its entry in the General Catalogue of Variable Stars include the contribution from the primary star).


Brown dwarfs and planets As far as I know, a brown dwarf is a 'star' whose core never underwent a fusion reaction, so it never became a star.

So I was wondering if, apart from orbiting a star, is there any difference between a massive planet and a brown dwarf?

I heard that they found an exoplanet 5 times the mass of Jupiter, and say a brown dwarf was in a binary pair where the other is a star: What stops it from being classified as a planet?

I guess what I am really asking for is a clearer definition of a brown dwarf.

Stanley, there really isn't a very clear definition and this is still a keenly argued point.

Definitions include:

Browns dwarfs burn deuterium. In models this happens if they are more massive than about 13 times Jupiter. The weakness of this that we think isolated brown dwarfs could condense from a gas cloud that are less massive than this; and young brown dwarfs won't have got around to fusing deuterium.

Planets must form from the disk around a star. This is ok, but: brown dwarfs may also form from the disk and it is also possible for planets to be tidally stripped from their stars and be found alone in space.

Planets must have a rocky core. This used to be thought definite, but now we think maybe sometimes planets can collapse from a gas instability in the disk in some circumstances, without the need for a rocky/icy core. It is true that brown dwarfs should not have a rocky core. However as an observational definition this is fairly hopeless since we can't even tell yet if Jupiter has a rocky core.

A flavour of the controversy can be gleaned from reading between the lines of the IAU statement on the definition of planets vs brown dwarfs.


According to my calculations Jupiter's moons do not follow Kepler's 3rd law - Why is that? I'm asked to collect data on the orbital properties of the four Galilean satellites of Jupiter and show that they obey the same scaling as in Kepler's 3rd law.

My approach for moon Io:

Online, I found that the distance from Io to Jupiter is 422,000km ~= 0.00282089577 AU
which is

$2.82\times10^{-3} AU$

Io's orbit around Jupiter is 1.77 Earth days; 1.77/365 ~= $ 4.85\times10^{-3}\, \mathrm{Earth\ years}$

By Kepler's 3rd law, $P^2(\mathrm{Earth\ years}) = a(AU)^3$

Thus

$(4.85\times10^{-3})^2 = (2.82\times10^{-3})^3$ which is obviously not true even just by looking at the powers of 10 after distributing the exponent on each side.

Kepler's third law is that $R^3/P^2$ is a constant. However it is not a universal constant; it depends on the mass of the body that is being orbited.
$$\frac{R^3}{P^2} \simeq \frac{GM}{4\pi^2},$$
where $M$ is the mass of the orbiting body (assuming that $M\gg$ the mass of the moons).

Depending on your level of sophistication, you could try plotting $R^3$ against $P^2$, or better, $\log R$ against $\log P$, to demonstrate consistency with Kepler's third law.


Earth Versus Catastrophic Meteor How evasive is the Earth to Catastrophic Meteors?

Google says the Earth is approximately 92.96 million miles from the Sun. It also says the suns radius is 432,474 miles.

Therefore, it is 93,392,474 miles from the Earth to the center of the Sun:

92,960,000 + 432,474 = 93,392,474 miles

I've heard the Earth's orbit is elliptical, but to keep things simple, let's just say the orbit is a perfect circle. Therefore the radius of this circle is 93,392,474 miles. This would mean the diameter of our orbit would be 186,784,948 miles:

93,392,474 x 2 = 186,784,948 miles

The circumference of a circle is π x the diameter of a circle. Therefore, the number of miles in Earth's orbit would be about 586,504,736.72 miles:

3.14 x 186,784,948 = 586,504,736.72 miles

So, the Earth travels about 586,504,736.72 miles in one year. Since there are 365 days in year, that would mean that Earth travels about 1,606,862 miles in one day.

586,504,736.72 ÷ 365 = 1,606,862.292383562

That would mean the Earth travels about 66,953 miles per hour:

1,606,862.292383562 ÷ 24 = 66952.595515982

And this would mean the Earth travels about 1,116 miles per minute:

66952.595515982 ÷ 60 = 1115.876591933

Lastly, this would mean the Earth travel about 18.6 miles per second:

1115.876591933 ÷ 60 = 18.597943199

Google says the diameter of the Earth is 7,917.5 miles. So, as fast as the Earth is traveling, a catastrophic meteor has about a 7 minute window to hit target Earth.

7917.5 ÷ 18.597943199 = 425.719119329 seconds
425.719119329 ÷ 60 = 7.095318655 minutes

What are the odds of a catastrophic meteor hitting the Earth? Or, maybe a better question: What are the odds of the Catastrophic Earth hitting a poor meteor!?

There is nothing wrong with your maths, but it isn't relevant to the question of the frequency of impact.

Now you ask about comets, but do you also want to include asteroids, since these are rather more common.

You also ask about catastrophic, and this is a poorly defined word. A Tunguska scale event is a local disaster, but has no significant global effects.

The Torino scale classifies objects according to their threat. A "10" on the Torino scale signifies a object carrying an energy of more than 10000 Megatons of TNT and capable of causing a global catastrophe. Such events occur, on average of less than once in 100,000 years.

Within the next 100 or so years the probability is considerably lower than $10^{-5}$ in each year, as many of the potentially hazardous objects have been observed, and their orbits shown not to intersect with Earth. So while a naive estimation would suggest that the chance of an impact in the next 100 years is about 1 in 1000, the actual probability is rather less than that.

If you restrict to only comets, the probability is reduced further.


Can stars be observed from space by x-rays, near infrared and radio wavelengths? I have a doubt.

Can stars be observed from space by x-rays, near infrared and radio wavelengths?

Yes, and not only from space but from the Earth surface too. Stars emit in almost all wavelengths depending on their surface temperatures. The hotter the star is the shorter (higher energy) wavelengths it'll emit.

You can try this simulator to check this:

http://astro.unl.edu/naap/blackbody/animations/blackbody.html


Exoplanet naming conventions I'm looking for any exoplanet nomenclature, official, semi-official, proposed, or just a good idea, beyond what can already be found on Wikipedia at

http://en.wikipedia.org/wiki/Exoplanet#Other_naming_systems

Or via the obvious Google searches.

This is a vaguely sci-fi related request.  Lowercase Latin alphabet isn't going to do it once we get out there, and I like the proposal for Greece-roman mythology but there are simply too many planets now, and will be more over time.

As a side issue I'm already familiar with Ian Banks Culture addressing nomenclature, so I accept that humanity being what it is, we can expect about a hundred or so "Vulcan" orbiting various different stars, much as there are 37 "Greenville"s in the USA, currently.

A pointer to a really great fictional idea or naming scheme aside from the schemes mentioned about might be useful.

Interesting technical (non literary) naming schemes beyond the most obvious and trivial ordinal systems could be interesting.

I realise that the following is not a great answer to the question, but was a bit too much to fit into a comment.

The official word on exoplanet naming is

http://www.iau.org/public/themes/naming_exoplanets/

The rules for the scientific designation of an exoplanet are clear and I think will remain so in the published literature for the forseeable future.

However, here is an earnest contribution to the literature which sees the need to change things, especially to cope with planets in multiple star systems.

http://adsabs.harvard.edu/abs/2010arXiv1012.0707H

But perhaps the big news is that the IAU is setting up to allow "popular" names to be given to exoplanets and are running a competition to name a batch. These names will not replace the scientific designations, but will be recognised by the IAU.

http://www.iau.org/news/pressreleases/detail/iau1404/


Far side of the moon question Do all / most / any other moons orbiting a planet synchronize their rotation so as to become tidally locked with it, or is the earth a special case?

Moons are not always tidally locked.

They are formed with a certain spin angular momentum that will be dragged by asymmetries on the hosting planetary body. Those asymmetries then backreact onto the moon and drag it's spin to become synchronized with the orbit angular momentum.
How rapidly this process happens is roughly a function of the tidal forces
$$F_{Tid} = G \frac{m_{Moon} \cdot m_{Planet}}{r^3} d$$
exerted by those asymmetries. Here $d$ is the moon size, $r$ the distance between planet and moon and the other quantities are masses and the gravitational constant.

The time $t_{lock}$ this locking takes can be estimated by comparing the excess angular momentum $L_{ex} = r^2m_{Moon}(\omega_{actual} - \omega_{orbit})$ with those acting forces:
$$t_{lock} \approx L_{ex} / \left(r F_{Tid} \right) \approx \frac{r^5}{ d \cdot m_{Planet}}$$

The latter equation makes it fairly obvious that this is not a process that happens at a certain magical orbit or at a certain time. It is an asymptotic process, and the only coincidal thing is that we observe any given moon at a time where it may not have lost enough spin angular momentum yet.
This should hopefully also clarify some confusion from the comments:

Yes, the tidal locking time is also dependent on the mass of the host-planet, but much more sensitive to the planet-moon-distance. Note that this derivation is independent of the moon's mass, and only dependent on it's size. The literature knows much more sophisticated variants of this, that also incorporate a (realistic) weak dependency on the moon's mass.
For Hot Jupiter exoplanets it is thought they'd be in tidal lock when the asteroseismologic age of the system is bigger than the derived tidal lock time.

Concluding
1.) Any moon would be tidally locked if we could wait long enough.
2.) There are much more effects / torques to consider for a detailed understanding than I did here.


Is the time lapse considered when estimating the age of the universe? We say that our universe is 13.7 billion years ago. During the big bang, it doubled at least 90 times in trillionth of a second (as given here), and other topological statements.

The question is which time are we talking about. Is the time-lapse considered? If yes, how?

As relativistic effects will cause clocks to run slower, a frame of reference must be chosen when considering the time of the "big bang". There is a natural and convenient choice of reference frame, based on the cosmic microwave background. The cosmic background appears to be extremely red-shifted light, indicating it is receding from us very fast, due to the expansion of the universe. If we choose a frame in which the CMB is receding equally fast in all directions, we have a convenient frame of reference. It is called the Comoving frame.

Now that we have a frame of reference, we can talk about time and distance in a way that all observers that share this frame can agree. In the comoving frame, the "big bang" occurred about 13.8 billion years ago.

To directly answer the question: The time measured is the time in the comoving frame and relativistic time dialations (time-lapse) are considered.


Is a black hole heavier than the star from which it was created? I did some googling around and couldn't get answer to a simple question of black hole mass compared to the star from which it was created. It is well known that giant stars burn very bright and relatively very quickly. They end up their lives as red giants and eventually collapse into a black hole.

Suppose that the star is mid-life and has some mass M. Over the time it burns its fuel, turns into a red giant and after supernova, turns into a black hole. This black hole has some mass B. Why B >> M?

According to some simple physics laws, mass could not be created nor destroyed.

The mass of a black hole is always much less than the star from which it formed.

A very large star such as Eta Carinae (which is expected to form a black-hole some day) is about 100 times that of the sun, Having already lost as much as 50 solar masses, blown off the star by powerful outbursts in the past.

The future of the star is not fully understood. It is possible that it will collapse in a supernova, in which most of the rest of the star will be blasted into space, forming a supernova remnant, and the core of the star will collapse. The black hole formed may have a mass of more than 3 times the sun, but much less than the mass of the original star.

So while the black hole is not heavier, it is much denser. A 3 solar mass black hole would have a diameter of about 20km.


Is it possible to calculate distance between two objects? For example: I want to know the distance between Alpha Orionis (Betelgeuse) and Beta Orionis (Rigel). Is this possible? You only know the distance from Earth. May be this is a math problem and not an Astronomy question.

If you know the distance of each star from earth, and the angle between the stars, then you can us the cosine rule to find their distance from each other.  See this for detail
https://en.wikipedia.org/wiki/Law_of_cosines


How large was Mercury before it shrunk? There's a theory that the reason Mercury has such an enormous iron core is that it was once a much larger planet before it got impacted by an object, resulting in most of its mass getting blasted away. If this theory is true, how large was Mercury before the impact? Could it have been the size of Earth or Mars, or maybe even a super-earth?

If we look at the planet's cores and I'm going to ignore liquid vs solid and focus on size overall.

Mars:  Core estimated 1,794 +/- 65 km radius.  The planet is 3,390 km radius.   About 53% of the planet's radius is its core.   Mars also has more sulfur in it's core and more Iron in it's mantle than Earth, suggesting that it probably didn't mix as well as Earth did, but I'm not sure that would significantly effect the size of it's core.

Venus:  Core estimates have some uncertainty, but by this article, it's core is thought to be about 3,000 km, about 49.5% of it's 6,052 km radius.

Earth's core is about 3,400 km and it's radius 6,371 km, about 53.4% of it's total radius.    If we use the 49.5%-53.4% as a guideline, Mercury's 2,440 km radius (85% core, so it's core is about 2,074 KM), so a rough estimate using the other 3 planets as a guideline, 3,880 - 4,190 km radius.    That puts it 500-800 km larger than Mars in radius, roughly 2,000 km smaller than Venus in radius.

That's obviously just a rough estimate, but assuming it was similar to the other 3 inner planets, it's probably in that range.   I should probably also account for compression.  The more massive the planet the greater the compression of it's core, but that would probably only vary a couple percentage points off the estimate and it wouldn't make a big difference.


How dark/bright are polar nights? During the polar night directly at the pole the Sun sets once and—after a period of twilight—stays down for months at a time.
The only sources of non-human-made light are the stars, the auroras, and the sunlight reflected by the moon.
Yet when I see various photos of, for example, Amundsen-Scott base during this period, the sky (and the ground illuminated by it) is often quite bright. Here is a timelapse of August, which should be way outside even the civil twilight for this period: https://youtu.be/cf14SL7m8Ys
Is there an issue with photo-sensitivity of the cameras?
Are these sources of illumination actually bright enough to turn the middle of the night into something more twilight-like (if not day-like) when perceived by a human eye? If so, how much of the polar night is actually dark, percentage-wise?

That's a timelaps video, most likely created from a sequence of photos. It's easy to choose sensitivity and ISO settings on normal cameras such that the night sky at the darkest night is over-exposed. The moon light looks like sunlight - as it simply is reflected sunlight.
E.g. see https://cloud.planetmaker.de/index.php/s/yJ7AXWG4dEC6csW - it's a an exposure of several seconds taken several hours after sunset in November in Kiruna, Sweden... the over-exposed part in the center are headlights of a car, the rest is illuminated by the nearly-full moon at that time.
As children of modern civilisation living in urban areas, we are simply not accustomed anymore to how bright even nights can be when you had sufficient time to adopt - and cameras can do that instantly. In arctic (or generally snowy) regions, the snow adds a lot to the overall brightness as it reflects the little available light at night extremely well.


Dynamic Method-Please explain How exactly, with steps and math shown, do you derive the masses of asteroids via perturbations? please, no links, because I have searched and searched for years and can't find anything.

I mean like asteroids- how are their masses determined exactly? What equations are used?

The mass of Vesta was determined by observing its effect on 197 Arete. The model used is that Vesta will cause the eccentricity and semi-major axis to change. By observing the asteroid before and after their encounter, and determining the asteroid's orbital elements before and after, one can find the angle of deflection. Model the motion as a hyperbolic encounter and one can find the combined mass. Since 197 Arete is much smaller than Vesta, this gives a good estimate of the mass of Vesta.

The mass of asteroids can also be found by their perturbations of Mars, providing there are sufficiently accurate observations. Viking radar data (with an accuracy of 7m) allowed for good estimates of the mass of the major asteroids. The Dawn mission to Vesta and Ceres made very high accuracy measurements of the gravity field around them and so their mass.

The math involved is essentially just Newtonian mechanics. You know the positions and velocities before the encounter. There is an energy conserving encounter with a force of $$F = \frac{m_1 m_2 G}{r^2}$$ Finding the angle of perturbation allows you to find the reduced mass $$\frac{m_1 m_2}{m_1+m_2}$$


Io: Why so much sulfur? Io is the only body in the Solar System whose crust is dominated by sulfur and its compounds. Carbon and nitrogen are also very common in the universe but we don't see them or their compounds in Io's crust. Why not?

Io is not dominated by sulfur. Io is mostly silicate rock and iron. Sulfur is a thin coating on the surface and sulfur dioxide makes up most of the atmosphere, but that's because sulfur dioxide is a volatile. In other words, it has a much lower melting and boiling point than silicate rock. So while the rock remains solid, the sulfur dioxide is gas.

Carbon is not common on Io's crust for the same reason that it is not common on Earth's crust. Carbon may be common in the universe, but it binds to oxygen to make a gas. When the solar system was forming most of the carbon was in the form of CO or CO2 and would not easily have coalesced into a solid body. Silicon is much less common than carbon, but Earth and Io are made of silicon because silicon makes solid stuff at the relevant temperatures.


How certain are we about the universe's flatness? The universe is thought to be flat: $\Omega = 1 \pm 1\%$. As I understand it we can determine this by measuring triangles against the CMB.

Yet during inflation dark energy made the universe grew much larger than the radiation universe, of which the CMB is the limit. Isn't it possible then that at the "smaller" scale of our visible universe it appears flat, but at the universe's full size it's curved? A bit like a square of 1 meter sides on the surface of the earth also looks flat, while at larger scale the earth is curved.

What are the chances that the 1% uncertainty could lead us to a universe with $\Omega \ne 1$?

It's entirely possible that our seemingly flat observable universe sits inside a much larger and very curved universe. Since there is, by definition, no way to ever measure anything beyond the observable universe, such possibilities are essentially non-scientific musings.


Is it accurate to compare comets to clouds and rain? I'm trying to avoid an opinion-based question, so before I outline the comparison I'm proposing, I will qualify the specific facts that yield this comparison. By focusing answers on the relative accuracy of those facts, hopefully we can avoid primarily opinion-based answers.

Facts

Over the course of millions of years, planetary bodies, especially those closer to the sun, tend to lose their water and atmosphere as it is blown away by solar wind, and radiated or boiled off by other forces, unless they have a strong magnetosphere to repel the bulk of that solar wind, or enough gravity to hang on to their matter. On a much faster time scale, this also happens to comets as they approach the sun; thus, the tail.
Over the course of millions of years, bombardment of planets by comets can increase the water and gas on a planet, or at least counter-balance the gradual loss of the same (depending on other factors like proximity to the sun, size, and magnetic protection).
The Oort cloud is composed mostly of icy bodies (including solids that would be a gas at Earth temperatures). And it is from this region that comets "fall" toward the inner solar system.

Also Fact?
So I started wondering what ultimately happens to the water and gas that gets blown off of planetary bodies. I would guess that given the outward pressure of solar wind and solar radiation, the direction of such material is generally outward, away from the sun. Well, where to? Media coverage of the departure of Voyager 1 from our solar system into the realm of interstellar space alerted me to the idea that there is an outer edge to solar influence, and I would guess that this is where the lighter molecules, like water and gases would tend to land. And indeed there is a collection of similar material out there at that solar radius.

Given these facts, is it accurate to view the Oort cloud as behaving similarly to a terrestrial cloudscape? Our clouds form at the altitude where water tends to deposit after it evaporates from the heated surface of the Earth. It gathers there and coalesces to a form which is then overcome by the opposing force of gravity. Then it returns from whence it came. Similarly, I imagine water and gases ultimately being driven from the heat of the sun to an “altitude above its surface”, where ultimately, over a much longer time scale, that material coalesces until other forces eventually drive them back toward the hot sun. Is this an accurate understanding? Is this comparison justified?

Dating cometary material is tricky, just because you have to go to space to get it. But the dates from the "stardust" mission show that comets are very very old. As old as the oldest asteroids, and date back to the origin of the solar system.

Now while there is evidence that material that formed in the inner solar system when the sun was young, and the planets still forming, was carried to the outer solar system. There is no suggestion that this process is ongoing. There is no astrononomical water cycle.

Although the planets might lose a little water to solar wind. It is very very little. Earth is protected by a magnetic field. Mars has lost all its  water already, as has Venus, and Mercury probably never had much.


How far from Betelgeuse is its habitable zone? Betelgeuse is a red supergiant or hypergiant in the constellation Orion. It is one of the biggest and brightest stars seen from the Earth, and is expected to soon go supernova. The habitable zone, or Goldilocks zone, is the distance from the parent star where the temperature allows for liquid water to exist on a planet's surface (in case the planet has an atmosphere of at least ~0.0061 atm (triple point of water)). How far is the habitable zone of Betelgeuse, in astronomical units? For comparison, the Sun's habitable zone is at about 0.7-1.7 astronomical units (other than Earth, Venus and Mars are on its rims).

Nobody knows the limits of the Sun's habitable zone, or how broad or narrow it is.
Here is a link to a list of various estimates of the inner, or outer, or both, edges of the circumstellar habitable zone of the Sun:
https://en.wikipedia.org/wiki/Circumstellar_habitable_zone#Solar_System_estimates
Note that one well known estimate, that of Hart et al. in 1979, makes the habitable zone very narrow, while another well known estimate, that of Kasting et al in 1993, produces a conservative habitable zone that is several times as wide as Hart's and an optimistic habitable zone that is wider still.
And there are other estimates with the inner edge of the habitable zone closer to or farther from the Sun than Kasting's and with the outer edge of the habitable zone closer to or farther from the Sun than Kasting's.
So if you study all of the original papers where those limits of the habitable zone were proposed, you can decide which ones are most convincing to you, and then use the habitable zone for the Sun to compare with Betelgeuse to determine Betelgeuse's habitable zone.
Taking the safe route, I will assume that a planet would have to receive exactly as much radiation from Betelgeuse as Earth receives from the Sun in order to be habitable.
According to Wikipedia, Betelgeuse has a luminosity about 126,000 times that of the Sun.
https://en.wikipedia.org/wiki/Betelgeuse
Since the square root of 126,000 is 354.96, a planet orbiting about 354.96 AU from Betelgeuse should receive the same amount of radiation from Betelgeuse as Earth receives from the Sun and thus should be within the circumstellar habitable zone of Betelgeuse, no matter how broad or narrow that habitable zone is.
But Betelgeuse is a variable star. Its luminosity varies between about 90,000 to 150,000 times the luminosity of the Sun, so the distance from Betelgeuse where a planet would receive exactly as much radiation from Betelgeuse was Earth receives from the Sun would vary between 300 AU and 387.298 AU.
Possibly a planet could orbit Betelgeuse with a somewhat elliptical orbit so it was closest to Betelgeuse when Betelgeuse was least luminous and farthest from Betelgeuse when Betelgeuse was most luminous and so constantly receive the same amount of radiation as Earth receives from the Sun.

Betelgeuse is classified as a semiregular variable star, indicating that some periodicity is noticeable in the brightness changes, but amplitudes may vary, cycles may have different lengths, and there may be standstills or periods of irregularity. It is placed in subgroup SRc; these are pulsating red supergiants with amplitudes around one magnitude and periods from tens to hundreds of days.[8]

https://en.wikipedia.org/wiki/Betelgeuse#Variability
The semiregular nature of Betelgeuse's variability means that nobody could even begin to design an orbit around Betelgeuse that would enable a planet to constantly receive the same amount of radiation as Earth receives from the Sun.
So planets in the habitable zone of Betelgeuse would receive significantly varying amounts of radiation from Betelgeuse as the star varied.  Whether that would make life impossible on those hypothetical planets is unknown.
I may add that there are still more problems with having a planet with conditions suitable for life orbiting Betelgeuse; the answer by antispinwards mentions some of them.
So the probability that there will be lifeforms on any planets orbiting Betelgeuse when Betelgeuse becomes a supernova and destroys all its planets seems to be very, very, very low.
I note that humans can not survive in the majority of the biosphere of Earth.  Humans can't survive in most places where other Earth lifeforms can survive.
A discussion of the conditions necessary for humans or similar beings to survive can be found in Habitable Planets for Man Stephen H. Dole, 1964, 2007.
https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf
Here is a link to a list of the nearest stars and brown dwarfs to the Sun, stars within a distances of 5 parsecs or 16.3 light years:
https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs
There could be worlds with some kind of life in some of those star systems.
And what would happens to those hypothetical worlds with life orbiting stars within 5 parsecs or 16.3 light years of the Sun if the Sun became a supernova?  The Sun will never become a supernova, but if it did the planets of those nearby stars would probable receive so much radiation during the supernova event that  their oceans and atmospheres would boil away and their surfaces turn to red hot lava, and all life on them would die.
And a supernova might be deadly to life on worlds at a much greater distance.  I am not very familiar with the distances at which a supernova would wipe out all life on a planet.
Betelgeuse is expected to have a supernova explosion in less than 100,000 years.  During that time many stars will get closer and closer to Betelgeuse and then start to get farther and farther from Betelgeuse, as those stars and Betelgeuse orbit around the center of the galaxy.
Here is a link to a list of stars which, according to the calculations of astronomers, have passed within 5 light years of Earth in the past three million years or will pass within 5 light years of Earth in the next three million years.
https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs
And it seems to me that any alien astronomers on a planet orbiting a star near Betelgeuse would be very displeased to learn that their star is getting closer and closer to Betelgeuse and is likely to be only a few light years from Betelgeuse when Betelgeuse supernovas.


Does anybody recognize this kind of prism? Supplemental to this answer to Did Edmund Scientific 4¼ inch Newtonians have 90° prisms as secondary mirrors in the 1970's? I was poking around a bit further and ran across this "1960's Edmund Scientific 'Space Conqueror', 3" reflector" with a prism diagonal at http://www.retrotechnology.com/glass/edmund.html

Question: Does anybody recognize what kind of prism this is? It doesn't look like a simple right-angle prism, but possibly something that has additional reflections and does some additional flipping or rotation.

Is there any chance that it isn't positioned correctly, it doesn't look like it is centered along the eyepiece holder's axis, but the image has a strong perspective component and we can't see the business end of the prism.

My guess is that it's not a prism (or at least doesn't function as one optically).  It looks like it is a right triangle in one cross section.  The diagonal face would be silvered to act as a star diagonal. If aligned correctly, the black plate it's glued to that attaches to the support arm would be perpendicular to the tube axis.

It looks like it's rough beveled on other sides, maybe to get an ellipsoidal cross section?


Are black holes spherical during merger? I've been thinking about black holes, specifically during the final moments before two merge.  I'm wondering if black holes, or I guess more specifically their event horizons, are always spherical.  It seems to me that in the moments before two merge, their respective event horizons will be stretched, somewhat like how the Moon causes our ocean's tides. I have drawn a (poor) diagram of what I think they may look like.  Notice how the event horizons are closer to the singularity on the inner side, this is because the gravity from each black hole is in opposition.  The event horizons are further from the singularity on the outer side because the gravity from each black hole adds up.

No need to guess. There's solid research done in this field. Even Wikipedia has some info:

  As two black holes approach each other, a ‘duckbill’ shape protrudes
  from each of the two event horizons towards the other one. This
  protrusion extends longer and narrower until it meets the protrusion
  from the other black hole. At this point in time the event horizon has
  a very narrow X-shape at the meeting point. The protrusions are drawn
  out into a thin thread. The meeting point expands to a roughly
  cylindrical connection called a bridge.

https://en.wikipedia.org/wiki/Binary_black_hole#Shape

There are research papers with images showing the results of calculations of the shape of the event horizons during merger. Here's an example:

The image above is taken from this paper:

On Toroidal Horizons in Binary Black Hole Inspirals

  We examine the structure of the event horizon for numerical
  simulations of two black holes that begin in a quasicircular orbit,
  inspiral, and finally merge. We find that the spatial cross section of
  the merged event horizon has spherical topology (to the limit of our
  resolution), despite the expectation that generic binary black hole
  mergers in the absence of symmetries should result in an event horizon
  that briefly has a toroidal cross section.


Why is there a difference between the cosmic event horizon and the age of the universe? According to wiki the cosmic event horizon is 5 Gpc but the age of the universe is about 13,7 billion years. Usually light from far galaxys needed at most 13,7 billion years to reach the earth so the event horizon should be the same. But in wiki the event horizon is 5*3,2billionly=16 billion ly. So is this difference just a matter of rounding or old data or is there something else what causes the differences?

https://en.wikipedia.org/wiki/Cosmological_horizon#Event_horizon

From the link you provide:

  The particle horizon differs from the cosmic event horizon, in that the particle horizon represents the largest comoving distance from which light could have reached the observer by a specific time, while the event horizon is the largest comoving distance from which light emitted now can ever reach the observer in the future. The current distance to our cosmic event horizon is about 5 Gpc, well within our observable range given by the particle horizon.

Your confusion somes from the fact that you are mixing to different horizons:

The particle horizon is the sphere centered on us that has a radius equal to the distance that light can travel in 13.8 Gyr (the age of the Universe). That is, light that was emitted when the Universe was born at a point on that horizon, reaches us today. Note that, because the Universe is expanding, the distance is not 13.8 Glyr, or 4.2 Gpc, as one might naively expect, but in fact 46.3 Glyr.

The cosmic event horizon is also a sphere centered on us, which is the boundary inside which light, if it is emitted today, may still reach us sometime in the future. If it is emitted outside this horizon, the expansion of the Universe ensures that the light will never reach us.

As time goes, we will see light that came from farther and farther away, and thus the distance to the particle horizon is increasing. Meanwhile, the accelerated expansion of the Universe ensures that light emitted from distant galaxies in the future are able to reach us only if they are increasingly closer. Thus fewer and fewer galaxies will be inside the event horizon. But since space is at the same time expanding, the boundary will asymptotically reach a finite size (~17 Glyr).

You can see the particle and the event horizons as blue and red lines in the plot in @Pulsar's great answer here.

The 16.5 Glyr that the distance to the event horizon is today is sort of a coincidence. It has nothing to do with the age of the Universe. It only depends on the future expansion of the Universe, which in turn depends on the densities of the components of the Universe ($\Omega_\mathrm{b},\Omega_\mathrm{DM},\Omega_\Lambda$, etc.). If the Universe had been dominated by matter (or radiation), then there would be no event horizon: No galaxy, ever-so far away would not be visible to us, if we just had the patience to wait. A galaxy is 10,000 billion lightyears away? Just wait long enough (exactly how long depends on the actual density).

However, our Universe happens to be dominated by dark energy, which accelerates the expansion without boundaries. This unfortunately means that the light leaving today from a galaxy 17 Glyr away will be carried away by the expansion faster than it can travel toward us. In contrast, the light emitted today from a galaxy 15 Glyr away will travel in our direction, but will nonetheless initially move away from us due to the expansion. However, its journey toward us makes this expansion rate smaller and smaller (since the expansion rate increases with distance from us), and after a period of time it will have traveled so far that it has overcome expansion and starts decreasing its distance from us and eventually reach us after 100 Gyr or so.


Could the universe have evolved differently? Given the initial conditions at the Big Bang, could our universe have evolved any differently? In other words, did the initial conditions predetermine every stage in the evolution of the universe to this day, or - if we "roll the tape backward" and play it again - could the universe be entirely different on the macro scale (eg. no stars and galaxies)?

Given chaos theory, things could have been different. Given the laws of physics, there are limits to how much difference there could have been.

To put it bluntly, a subatomic particle could have zigged instead of zagging, causing a fusion event that led to the formation of a star not to take place in one spot but another.
But it would not prevent stars from forming.


Can exoplanets be found using neutrino detectors? I read that KamLAND can detect geoneutrinos produced by thorium and uranium decay in Earth's crust.
Could a larger detector detect neutrinos from other planets in the solar system or perhaps even exoplanets?

The answer is presumably no.
Let's assume that the nearest possible exoplanet is in the Proxima Centauri system, and it presents an electron antineutrino luminosity identical to that of Earth. The flux from the exoplanet at the KamLAND detector would then be weaker than the flux from Earth by a factor of roughly
$$\left(\frac{\text{Radius of Earth}}{4\;\text{light-years}}\right)^2\approx2\times10^{-20}$$
thanks to the inverse-square law. So the intrinsic signal strength is likely to be essentially zero.
More concretely: the KamLAND collaboration's 2013 measurement of the Earth's geoneutrino flux was approximately $\sim3.4\times10^6\;\text{cm}^{-2}\text{s}^{-1}$, which isn't a lot considering the low probability that a given neutrino will interact with the detector. Indeed, the group reported only $116^{+28}_{-27}$ events in their period of monitoring - and some of that was during a time where nearby nuclear reactors were off and therefore not producing neutrino noise!
Any signal from an exoplanet would be tiny and easily lost in the background of all of the other possible neutrino sources.


Question about extreme space distortion and creation of a new dimension In all the illustrations regarding space distortion, I find an extra dimension that depicts space distortion. Is there any mathematical equation that proves the creation of an extra dimension ?

Thanks.

Just because there's an extra dimension on a diagram doesn't mean that it's real. No such thing has been created. It's just an artifact of embedding a manifold with a non-Euclidean geometry into a Euclidean space.

One of the biggest stumbling blocks of intuition for people learning general relativity is that the physics only cares about the intrinsic geometry. For example, imagine an ordinary ball in three (Euclidean) dimensions, and take its surface. In jargon, the surface is the two-dimensional sphere $\mathrm{S}^2$, and it has geometric properties that can be described by reference to it alone, e.g., lengths of curves drawn on it, angles between intersecting curves, that if one starts at some point and goes in a single direction, eventually one would be back to the starting point, etc.

One should consider the two-sphere as a valid geometry by itself, and the fact that it can be pictured as the surface of a three-dimensional Euclidean object as purely incidental--it $\mathrm{S}^2$ can be embedded in the Euclidean space $\mathrm{E}^3$ in that simple manner, yes, but that's a fairly arbitrary choice. It can be embedded in other spaces as well, or not embedded in anything at all.

For Riemannian manifolds (the purely spatial geometries), there are some nice mathematical results regarding embedding into Euclidean spaces, but nevertheless, they're usually impractical and not relevant to general relativity. Even worse, general Lorentzian manifolds (spacetimes) have no comparably 'nice' results regarding embedding spacetime geometries into flat pseudo-Euclidean spaces $\mathrm{E}^{n,m}$, so it's doubly useless to worry about it, except perhaps in the cases where the spacetime is extraordinarily simple.

In the end, such extra dimension(s) are just artifacts of making a picture. They're not physically real, so they're not 'created' in any physical sense.


Catalog of Galaxies in the Visible Universe I have seen the Gaia project from here which is about mapping the stars in the Milky Way galaxy. But I'm wondering if there is any open data on the positions (or something like "positions" since it probably not calculable now), or "coordinates", of the galaxies in the visible universe, as opposed to the stars. I wonder how animations such as this (a time=1m30s point it shows the galaxies zooming out from the Milky Way) get their data on the positions of the galaxies.

There are lots of galaxy catalogs.

If you're only interested in nearby galaxies, there are things like the Updated Nearby Galaxy Catalog; downloadable versions can be found here.

A classic all-sky galaxy catalog is the Third Reference Catalogue of Bright Galaxies (RC3, published in the early 1990s), which you can query here; if you follow
the "FTP" links, you can find links to download the actual catalog.

(The HyperLEDA catalog that Dr. Chuck mentioned started as the RC3, but has various
bits of data and re-classifications added more or less continuously since.)

A more recent catalog, based on the near-infrared observations of the 2 Micron All-Sky
Survey (2MASS) is the 2MASS Galaxy Redshift Catalog, with about one million galaxies.

Even larger catalogs (though not covering the whole sky) can be found among the data
products of the Sloan Digital Sky Survey, e.g. here

Many more catalogs exist, though some do not have accurate redshift information (meaning their distance from us is unknown or uncertain), and many are only for parts of the sky, sometimes very small regions (e.g., catalogs from Hubble Space Telescope observations).

(As for that video, I suspect very little is based on real data once you get beyond the Andromeda Galaxy...)


Terrestrial Exoplanet Skies – I've Built a Visual Sky Chart. Is it Accurate? I'm an artist (and science enthusiast) and I've been trying to find a comprehensive resource that would help me clearly identify likely sky colors (as perceived by human vision) for exoplanets that have atmospheres with a similar chemical composition to Earth. It's trickier than I expected. I've cobbled together several resources, and built what I hope is a decently accurate chart of apparent sun color along with sky color.

UPDATE: I have revised this post and added a new, improved sky chart (below). Science references for sky chart: Ref A, Ref B

How accurate is my chart? Is this a fair representation of sky/sun on alien worlds with heavily Nitrogen/Oxygen atmospheres? In what ways could I improve it?

The chart is not meant to account for things like dust, the look of the sky at sunrise/sunset, or other atmospheric effects. This is meant to be a boilerplate for the baseline look of the sky during the day. However if you'd like to comment on how the sky may change during things like sunset, or with effects like volcanism, I'm game!

Also, I'm more interested in relative color, than getting absolute color 100% right. The image was built in a vector program in RGB mode, so assuming your screen is calibrated in the standard way, we are probably seeing almost the same thing.

Here are the key ideas I've gleamed from my reading that I'm using to build this. I consider most of them tentatively held, and am very open to input:

Earth-like atmospheres would tend to be light to dark blue at the zenith, due to the scattering of low-wavelength blue light. If the planet's sun was very hot, the sky would look a deeper blue, while cooler stars would give the sky a lighter blue to almost white look. When the sun gets to 3000k and below, the sky starts to take on an orange/brown tinge.
Like on earth, the horizon is the lightest in color and the zenith the deepest.
Denser atmospheres would appear brighter (more washed out) and the primary color in the spectrum more "pure" (I'm unsure what the term "pure" means exactly when it comes to optical perception...would it look whiter?). Likewise, thinner atmospheres would be less bright than earth's and the colors more "pure."
With increasing pressure the sky color at the zenith becomes increasingly yellow. In my image this means that an earth-like sky at 10x earth pressure would appear blueish/green near the zenith.
At lower temperatures I'm assuming the sun would appear tinged by the color listed under "star temperature." Otherwise you'll probably only see the star's color when it's near the horizon.
I'm assuming when you get down into K and M class suns, the surroundings on the planet would take on a progressively redder tinge due to the decreasing prevalence of blue wavelength light.
I suspect the gradient of color from horizon to zenith will be steeper/gentler in some atmospheres. I've guessed that the gradient would be more apparent on a high G world (on the right).

References:
Reference #1
Reference #2

Original Chart:

I think the star,its halo and cloud can't be darker than the sky color even early M type still look blindingly bright orange-red on the sky not dim orange-red.
The sky color, if heavier i think it will look more "desaturate"(brighter too but not bright to the point it's brighter than the star) and thecolor shift toward the red side(you are correct) that dense sky will be affected by the color of starlight very intensely to the point it's almost have the same color of the starlight
example: 10 bars sky on a planet around K5V sun would look light orange-ish grey also the cloud would look bright orange, same color for the star's halo.
example2: 10 bars sky on a planet around A5V sun would look bright pale teal/cyan with yellowish white cloud same color for the star's halo.
example3: 5 bars sky on a planet around A5V sun would look powder blue with pure white cloud, same color for the star and its halo.


Martian dust storms-what was the smallest in area, or shortest in duration storm detected? I understand localized Martian dust storms can develop into global wide events. All searches I do produce results that focus on local storms expanding.  Is there any information on what was the shortest duration or smallest area dust storm recorded?

SOLAR SYSTEM/SUN, ATMOSPHERES, EVOLUTION OF ATMOSPHERES | Planetary Atmospheres

Numerous dust storms occur each Martian year and are generally classified according to size. From the smallest to largest, they are dust devils (<10−1 km2), local storms (∼103 km2), regional storms (∼106 km2), and planet-encircling storms (>106 km2).
Local dust storms are also quite common. Based on Mars Global Surveyor (MGS) images, as many as 2000 local storms occur each Martian year. This gives a daily-averaged rate of 2–3 storms per Martian day. They have typical lifetimes of less than several days.
Regional storms have been observed at nearly all seasons, but are most frequent during southern spring and summer. Most regional storms develop within ±30° of latitude, although there is a distinct bias toward the southern hemisphere. Regional storms can last from days to weeks. These storms can drift a significant distance from their original location, and new satellite storms can develop that are quite remote from the original center.


How can I improve the Field of View of my home-made telescope? I made my telescope, with two convex lens of focal length 1000mm and eyepiece (plano-convex) with 25mm focal length. But I can see only a very small field of view, with inverted image. Do I need to increase the diameter lenses for more area capture, and can I use concave as eyepiece for upright image?
Or does some different solution exit?
I can read all information related to telescopes and binoculars, ie. how lenses work, which lenses combination are for which purposes (terrestrial and astronomical), and increase or decrease area and magnification.
Please note here I am talking about only refracting telescope, and binoculars.
note: This is a follow-up question based on comments below this answer to my previous question.

There basically are three main properties which influence what you see with your telescope:

The aperture defines how much light you get, thus how sensitive your setup will get. The bigger, the more light you collect, thus the fainter objects you will be able to detect in principle. There's the somewhat sense-free saying that you can compensate (clear) aperture only by clear aperture.

The focal length is the primary factor which drives your field-of-view. You can somewhat vary it by the choice of your eye piece. The magnification is the ratio of the two focal length, of your primary optics and the eye piece: $m = \frac{f_{tube}}{f_{eye}}$. Thus the bigger this magnification is, the more details you can see, the easier you resolve things which are close to eachother. However, phyics strike here and the size of the aperture limit how much magnification makes sense due to the diffraction ocuring at it. The so-called useful magnification (also see here, thus maximum magnification) is by rule-of-thumb about twice the aperture in mm. You can increase or decrease your focal length on an existing telescope with either a Barlow lens or focal reducer respectively. Both, of course, result in some optical degradation, as any optical system does.

The actual field-of-view is also determined by the eye piece properties and its opening angle. This can vary in a limited range only and is usually around 50°. See also this telescope calculator by Sky & Telescope.

Taken aperture, focal length of tube and focal length of eye piece, you derive one additional property which immediately affects what you see: the so-called opening ratio or f-ratio which is the ratio of focal length of your tube and the clear aperture. This determines how bright you will perceive a certain object as that defines how much light from an object is projected per unit area. Telescopes with identical f-ratio will show identical objects at the same brightness - independent of their actual field-of-view. Thus this number really defines how well you can see faint objects like nebula or faint / distant stars.
If I should put it in three sentences, this is it in short: your choice of focal length determines how big an area you see of the sky. The longer, the smaller your FOV. The ratio of aperture and focal length defines how faint objects you see.
As to the inverted image: astronomical telescopes usually have that, it keeps the system simpler. There are terrestrial telescope which employ an additional lens. Taking the words of uhoh's comment: The nature of real images is that they are upside-down and reversed left-right. It is possible to use a 1:1 relay lens to re-image your focal plane again but it will make your system more complicated to build and to hold everything in the right place. You can also use a prism to flip it in one direction, or two prisms to make an image inverter. If you don't want to ask a new question about using the relay lens, you can just read about it in wikipedia.


Is there any point on earth where the moon stays below the horizon for an extended period of time? When I was teaching a class why the earth has seasons today, I mentioned how the poles of the earth experience months of daylight and darkness. Then one of my students asked whether the moon disappears as well.

Using an orrery and a string, we looked at the orbit of the moon around the earth and tried to find a place where the moon was below the horizon for an entire orbit. In our attempts, we could find no such place, though some students speculated that if you were in the right place at the right time, the moon could make a full orbit while partially obscured by the horizon.

Is there a place on the earth where the moon cannot be seen on for long periods of time, like the sun in northern latitudes? Or does the fact that it orbits around the earth make this impossible?

Depends on the interpretation of your question...
The best places not to observe the moon are the north and south pole. On the north pole you will only be able to see objects above the celestial equator. As the moon orbits the Earth in one month its orbit is inclined from the celestial equator. This inclination is almost the same as the inclination of the ecliptic (path of the Sun) with the celestial equator. The ecliptic crosses the equator at two opposite points on the celestial sphere. This means that for about half its orbit, the Sun, and as the moon's orbit is near the ecliptic, also the Moon, will be above the ecliptic and therefore visible from the north pole.

That being said, the Moon does not follow the ecliptic precisely as the moon's orbit is inclined from the ecliptic by about 5°. The inclination of the ecliptic is 23°, so during very special circumstances the maximum altitude of the Moon above the horizon on the North pole will be 18° during one month. The duration that the Moon will be above the horizon on the north pole will be about 10 days (a guess) about 13.6 days (edit:see comments below).

So if we interpret your question as: Is there a place where the Moon will be below the horizon for a long period of time (> 1 month), then the answer is NO.

But if this happens near June, then the Sun will also be above the horizon at the north pole (for six months), and as the Moon will be close to the Sun (as it follows more or less the ecliptic), it will be very hard to see the Moon during that time. So if you specifically ask whether the Moon will not be visible for an extended period of time, then the answer is YES.

And there are of course also places with perpetual cloud cover ;-)


What are those long, smooth, longitudinally-opposing areas from pole to pole on Mercury? Look at this model and spin it - Mercury 3d model

Are those artifacts of stitching together images, or does Mercury really have these opposing, huge pole to pole lines of smoothness?

During the Messenger mission, there were (among many others) two global images of Mercury taken.
One is the so called "Low Incidence Angle Mosaic" which is created from combining images taken at local noon when the Sun is high in the sky and the amount of shadows is reduced. This image is useful to see different types of surfaces.
The other one is the "Morphology Mosaic", taken at a higher incidence angle at local morning / afternoon. This shows enhanced shadows and therefore provides a better view of craters, but there are some areas were not photographed at the right time.
The Mercury 3D model is filled with images of the Morphology Mosaic.
You can see the same smooth areas present in the mosaic if you view it in higher resolution at the link below.
https://messenger.jhuapl.edu/Explore/Images.html#global-mosaics
And, comparing the two mosaics, the smooth looking area is a copy of the low incidence angle image that has been used to replace missing images. So, the only difference between the areas is the incidence angle of the light, making many of the craters almost disappear. The resolution of the image of both areas is the same, but shadows are not.
This is also described in the image caption provided by NASA:

Any remaining gaps in the global mosaic were filled by using images
obtained in support of the high-incidence imaging campaign.

https://photojournal.jpl.nasa.gov/catalog/PIA17230


Would a black hole evaporate via Hawking radiation before you fall past the event horizon? It takes an infinite amount of time for something to fall past the event horizon of a black hole from the perspective of someone outside the event horizon. Black holes also evaporate after a finite amount of time from an outsider's perspective  due to Hawking radiation.

Does this mean that you would never actually reach the event horizon if you fell into a black hole because the black hole would evaporate?

I've asked this question to a couple of physicists a few days ago. Great minds think alike, huh?

First, bear in mind that Hawking radiation is only hypothetical. It is not theory. If we trust that hypothesis, this is what we can get.

In general relativity, black holes can be described through a number of approximations. For example, the Schwarzschild solution for a black hole describes it as an eternal object — not something that exists for some times, and doesn't exist for others. According to this solution, the event horizon must have always existed, and must remain existing eternally.

Schwarzschild black holes approximate black holes very accurately, but as you can tell, they fail to explain how a black hole can form, and (assuming Hawking radiation is real) they don't explain how one could eventually evaporate.

Of course, that solution won't help us. I've kept looking for one that accurately describes an evaporating, creatable black hole, but I've found nothing. The conclusion I've come to, along with those I've asked, is that our question has a major problem: Hawking radiation is explained via quantum field theory.

Thus, you can't simply use a GR solution for a black hole; you'd need some unholy mixture of quantum field theory and general relativity (keep in mind that both GR and QFT are incompatible in many situations).

In the end, it all comes down to how little we really know about black holes. It's not really possible to determine which solution is the best, and our inability to reconcile QFT with GR poses a big problem. The best answer I could give is "nobody really knows what would happen if you kept approaching a black hole".

We don't know if we would reach the event horizon, we don't know if the black hole will evaporate. We simply don't understand them well enough to know what solution would work, or how we would put QFT into it. If we somehow managed to find an approximation that properly combines GR and QFT, I assume (but don't quote me on this) the situation you described would be possible.

If it is possible, by the way, then we could confidently say that a black hole of any size could rip you apart through tidal forces. Tidal forces become weaker as the black hole's size increases, so one would assume a large enough black hole wouldn't rip you apart.

However, if we take Hawking radiation into account, and if your proposed scenario were indeed correct, the black hole would shrink as it evaporates. Since it would get smaller at a faster rate as we approach the event horizon, it would soon be small enough to rip us apart.


Are large body orbits self correcting? If a rogue planet disrupts Earth's orbit slightly to cause a long winter or summer will the Earth's orbit normalize back close its original orbit over time? Could a Planet X or some other elliptical orbit large body revolving around the sun effect Earth's orbit by being inside the inner solar system?

No. Basically, there is no such thing as a stable orbit in a realistic solar system.  All planetary orbits are unstable due to gravitational interactions between planets, but how long it takes the instability to make a big change in a planet's orbit varies a lot.  It can be a few of the planet's years, thousands, millions or billions. (Earth's orbit has been stable, so far, for 4+ billion years.)  But stability going forward can't be guaranteed beyond at most a hundred million or so years.

The main source of instability is resonances. A resonance is where two planets are in orbits where the ratio of their periods is a simple fraction like 1/2 or 2/3 or 5/3 or whatever.  Because of the simple relationship between periods, planets repeat the same configuration in space repeatedly, and gravitational interactions can build up. (Without the resonance, the interactions are just as strong, but in random directions and thus mostly cancel.)

Resonances can disrupt an orbit or can stabilize it. The Hilda group of asteroids are in a stable 3:2 resonance with Jupiter, for example, and Pluto is in a 2:3 resonance with Neptune.  Both of these resonances act to stabilize the orbits.  But the Kirkwood Gap in the asteroid belt is a 2:1 resonance with Jupiter which tends to keep the gap empty by disrupting the orbits of any asteroids that stray into the gap.

After the interaction, Earth will be in a new orbit and that orbit may also be stable over an extended period or it may be one where interactions with other planets quickly render the new orbit unstable.  There's no way to tell without detailed orbital information on the encounter.

In any event, there is no restoring force to push Earth back to its old orbit.


What do the months and hours mean in Tomas Filsinger's Map of The Universe I have had one of these in my room for decades:

I don't have the booklet that seems to usually come with it, can someone explain to me what the months (with days below it) and hours mean? I imagine it has to do with the position of the earth and therefore the part of the universe that is possibly observable, but I get confused when I try to get the months in the equation.

Thank you very much

The hours show the lines of right ascension, where right ascension is one of two coordinates used to specify a point on the sky. (The other coordinate is declination, measured in degrees from the celestial equator.) Right ascension and declination are similar to longitude and latitude on the Earth.

The month and day show the right ascension of the Sun on a given date. The dotted circle passing through the zodiacal constellations (Pisces, Aries, etc.) is the actual path of the Sun (the ecliptic). Therefore, the position of the Sun on any day can be determined from the chart.