How to calculate the moon phase angle? I need to calculate the exact moon phase angle for each our of a day. I have my position in lat and long and a unix timestamp. At the end i would like to be able to caluculate the moon brightness with this formula: https://patents.google.com/patent/DE10113295A1

const moonPhaseAngle = Math.cos(( 2 * ( ( phase * 100 ) / 100 ) -1));

const moonBrightness = 10 * Math.sin( altitude ) * ( .5 + ( .5 * Math.cos( moonPhaseAngle ) ) );

This is my actual progress through web search.
Thank you for helping me!

I refer to the book : Astronomical Algorithms (Second Edition) by Jean Meeus.

To calculate the values of position angle of the bright limb and position angle you will need to calculate  the apparent RA and DEC of the Sun and the Moon.

Code the equations for the Sun RA and DEC using Chapter 25. Please note that L0 (equ. 28.2)  is in error To calculate T (Julian century you will need to compute Julian day (Chapter 7).

Code equations for the Moon RA and Dec from the equations given in Chapter 47.

For the calculation of Apparent RA and Dec you will need to compute Nutation and Obliquity - Equations in Chapter 22.

BTW. For all my calculations I have "standardized" on using the equations on page 144 for the variables : D, M, Mdash, F, and Omega.
Fyi compare the equations for M given in this book ie. equations 25.2 and 47.3

Chapter 48 provides a formula (48.1) for the illuminated fraction of the moons disc.

5.1 I used 48.2 and 48.3
Addition:
5.2 To use equation 48.5 you will need to obtain or write the atan code yourself to resolve the angle X in the correct quadrant - I wrote a routine: ATAN2(sin_tl,cos_bl).
I have coded all the above to run on my HP Prime graphing calculator and they have proved to be very, very accurate.
For the purpose of accurate determination of the moon phase angle you will need to calculate UTC(date and time) using your zone longitude or better still your local longitude.


Lunar Eclipse - Total darkness? Can a lunar eclipse completely 'turn off' the mooon, i.e. like a New Moon? Or is it always just a 'shading' of the moon (sometimes red in color)?

No.

Lunar eclipses are caused when the Moon is in opposition to the Sun. Normally this produces a full moon, but if the Moon is in exact opposition (considering incline of the Moon's orbital plane), all direct sunlight will be blocked from the Moon:

So if all the sunlight is blocked, how can we see the Moon? Well, the main reason is that sunlight is often dispersed in the atmosphere, and so it can reach the Moon. In addition, there's  airglow, which is when sunlight hits Earth's upper atmosphere and causes multiple chemical reactions, scattering light throughout the night sky.

Thus, all this light from Earth, called earthshine, reflects off the Moon and illuminates it. In addition, Earth removes and blocks parts of the sunlight's spectrum, leaving only the longer wavelengths. This causes the Moon to appear red. Lastly, because the Earth blocks off all the direct sunlight from the Sun (only diffracted sunlight and airglow reach the Moon), we can actually see Earth's shadow on the Moon.


Difference between first (focal length (fl.) 500 mm, ocular (o.) 10 mm) and second telescope (fl. 1000 mm, o. 20 mm)? (both 50 x) Magnification is defined as $f_{objectiv}/f_{ocular}$.
If I had the first telescope with the focal length of 500 mm, ocular 10 mm and second telescope with the focal length of 1000 mm, ocular 20 mm, is there a difference between them? Both of them has 50x magnification.

I want to know if there is any difference in the quality of image, etc.

Thank you for all!

The larger the entrance aperture, the greater the light-collecting capability.  Now, you didn't specify that, but that does matter in your final choice.

Next,  as the comments suggest, the choice of optics depends on the cost vs. quality trades.  You want not only a given magnification but also as little coma, distortion, spherical aberration, and so on.   At a very simple (and probably unrealistic) level, the longer lens pair (1000,20) might be easier to fabricate with low aberrations.  In real life, we quickly descend into multi-element systems, or cemented doublet designs, and so on.

I'm guessing you are talking about a Keplerian design (two positive lenses), in which case consider how much longer the physical telescope is in the 1000-20 case.


Why is the SETI survey limited to only Red Dwarf Stars? I was going through the official SETI website, when I came across this article on the same website which said :

Today, the SETI Institute uses a specially designed instrument for its SETI efforts – the Allen Telescope Array (ATA) located in the Cascade Mountains of California. The ATA is embarking upon a two-year survey of tens of thousands of red dwarf stars, which have many characteristics that make them prime locales in the search for intelligent life.

The part that I was not able to understand was :  The ATA is embarking upon a two-year survey of tens of thousands of red dwarf stars, which have many characteristics that make them prime locales in the search for intelligent life.
So my question is : Why is the SETI survey limited to only Red Dwarf Stars? Is there something special about them that I seem to be missing?
Also, is this survey still active?

Sun-like yellow dwarf stars are generally thought to be the best places for intelligent life to develop, if only because 100% of known life is in orbit about a yellow dwarf.  Larger, brighter stars don't last long enough for intelligence to evolve and red dwarf stars were thought to be too prone to solar flares.
However yellow dwarf stars are relatively rare, and the nearby ones have already been investigated.  Red dwarfs are far more common and so there are far more red dwarf stars in our neighbourhood, and perhaps solar flares are not as much of a problem as we previously thought.  So current investigations are focussing on those red-dwarfs that had previously been excluded.


Can we know whether any distant galaxy is moving away in same direction or in opposite direction w.r.t. our galaxy? The Hubble law states that the recessional velocity of a galaxy is proportional to its distance from us.
It takes redshift and angular sizes (for distance) of the galaxy to determine velocity, and thus rate of expantion.
But I guess the factor of direction in which the galaxy is moving may substantially affect the 'expantion rate' calculations. i.e. If the galaxy A is moving in same direction as ours, will have less relative speed and the galaxy B moving in opposite direction will have more relative speed. The actual local speed may be the same.
I wonder whether this fact can be considered when we determine the expansion rate?

Yes, it can. For instance the Andromeda Galaxy is moving towards us, due to local gravitational attraction but that does not affect the validity of the theory of the expansion of spacetime because one result based on local factors is perfectly compatible with a cosmological theory which, by definition, includes every galaxy.


How many exoplanets have been found around binary stars? This XKCD whatif talks about circumbinary and [the other kind of] exoplanets. How many such exoplanets have actually been discovered?

I tried to look it up at exoplanets.org, but didn't find a column mentioning if the star is a binary.

Go to the exoplanets.org database.

Top left hand dropdown menu - select "Stars"

You get a table listing all the known stars with exoplanets.

Click on the big plus sign in the top right hand of the table and select "Binary Flag"

This will now show a boolean flag that indicates if the star is known to be part of a binary (or multiple) system.

In the FILTER window at the top of the page, just type "BINARY" and the table will just show binary stars.

It is possible to export the table as a csv file using the "export" button at the top right. Counting the number of lines with the BINARY flag indicates that as of December 2019, there are 511 known exoplanets around binary stars.


Is there any way to find out what sky surveys are *currently active?* I've asked a question about active sky surveys that might intercept a 2% of 4π patch of sky in the next six months and potentially record this dim (+22 to +26 magnitude) object.

I've added a bounty with about one more day on the grace period, the only thing helpful was a comment that suggested I ask for a "database of surveys".

Are there any pages that list at least major active surveys? That would be a start at least.

If you really just want a list, Wikipedia has a pretty good page on astronomical surveys.

For some comprehensive information, including some tabulated values (such as sky coverage for each survey), try this paper from S. G. Djorgovski et al.


Has great eyesight been necessary for astronomers? In a different (but somewhat related) field, some baseball stars have been known to have "baseball eyes." That is, an exceptional ability to visually follow the trajectory of a 90+ mph baseball to a degree enjoyed by 0.1% (one in a thousand) of the human race, as discussed in "Moneyball" by Michael Lewis.

I would imagine that would also be an advantage for an astromer, all other things being equal. But has "history"shown this to be necessary? That is, have there been a large or at least disproportionate number of famous and successful astronomers that been identified with eyesight in the top 1% (or higher) of the human population? Conversely, have there been any noted astronomers with notably bad eyesight who have made contributions because their theoretical, intuitive, or other abilities were enough to compensate?

Johannes Kepler
Wikipedia:

"However, childhood smallpox left him with weak vision and crippled
hands, limiting his ability in the observational aspects of
astronomy."

He made great use of Tycho Brahes great systematic observations in his theoretical work. He did not need exceptional eyesight for his developments in optics and telescopes.


What Are Some Good Southern Hemisphere 7x50 Binocular Targets? I've been interested in astronomy for a while but only recently purchased my first set of binoculars (7x50). I've been learning various constellations as well as spotting some other objects with the help of my planisphere. I live in the suburbs of a big Australian city, but far enough out that the light pollution isn't too bad.

I've already been looking at objects like the Orion Nebula and Pleiades, but hoped some experts might be able to give me a list of objects that would be worth hunting down with my binoculars. Ideally some that would be visible now (it's Summer here), but some for other times of the year would also be fine.

For anyone else who might be asking a similiar question, I found a great resource for this. The PDFs found at the URL below contain a good list of binocular, naked-eye and telescope targets for each month of the year in both hemispheres.

http://www.skymaps.com/downloads.html


Does the moon have days? As like Earth, does the moon have a day (24 hours). If it has, how many hours are in a moon day?.

The answer depends on your definition of a day.

If you define a day as we usually define it in the Earth (time between the Sun is at noon or average time between sunrises, a 24 hours day in Earth), the length of a day in the Moon is the synodic period of the Moon and it takes 29.530589 days (29 d 12 h 44 min 2.9 s).

Anyway, if you define a day as rotation on own axis, that is, time between a far away star being at noon, you have what in Earth is called sidereal day (23 hours, 56 minutes in Earth). In the moon the day defined this way would be 27.321661 days (27 d 7 h 43.19 min 11.5 s).

Please notice that since the Moon is tidally locked to the Earth, rotation periods equal orbital periods.


How fast would an observer have to travel in order to be able to study an entire star's life in one human lifetime? Humans live for about 80 years. A star, however, lives from 1 million to trillions of years, depending on its mass. If humans want to have a detailed study of the life of a $1 M_\odot$ star, how fast would they have to travel, and would it be viable?

We want it so that $1 \text{ year at the speed of x% of c} = \dfrac{10^{10}}{80} \text{ stationary years}$. The Lorentz factor (in terms of c) is $\gamma=\dfrac{1}{\sqrt{1-v^2}}$. Solving for $v$ with $\gamma =  \dfrac{10^{10}}{80},$ we get  $v=0.999999999999999968c = 299792457.999... \text{m}\cdot\text{s}^{-1}$. This is definitely impossible, as the energy requirements are astronomically large: $$5572282108577709241 \text{J}\cdot c^{-2} \cdot 125000000 =696535263572213655125000000 \text{J} = 7.75 \cdot 10^9 \text{ kg of mass converted to energy}$$
This is definitely unrealistic, and is a probably impossible feat to complete. It would be better to study more and more stars instead of exploiting time dilation to view a single star for billions of years.


How to show that the Jeans Criterion for Mass, Radius and Density are equivalent? The gravitational collapse of a gas cloud can be described by the Jeans Criterion for mass, radius and density of the gas cloud, which is (c stands for cloud):

$$M_J = (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{1/2} $$

$$R_J = (\frac{15kT}{4 \pi G \mu m_H \rho_c})^{1/2} $$

$$\rho_J = (\frac{3}{4 \pi M_c^2}) (\frac{5kT}{G \mu m_H}) ^{3}$$

I wanted to show that they are equivalent and simply used

$$\rho = \frac{m}{V} $$

on the criteria above an rearranged the equation.

However this doesn't work. What am I doing wrong?

Thank you for your help!

What you want to do is use the relation between Mass, Radius, and Density. The proper expression should be

$$\rho_c = \frac{M_c}{(4/3) \pi R_c^3}$$

Your three Jeans conditions are $M_c > M_J$, $R_c > R_J$, and $\rho_c > \rho_j$. By using the relation above, you can transform any of the three conditions into one of the others.

For example, going from $M_c > M_J \rightarrow R_c > R_J$ is done as follows.

$$M_c = \frac{4}{3}\pi R_c^3 \rho_c > (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{1/2}$$

Now, solving for $R_c$ gives

$$R_c^3 > (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{3/2}$$

$$R_c > (\frac{15kT}{4 \pi G \mu m_H \rho_c})^{1/2} \equiv R_J$$

The other transformations follow in a similar manner.


What is the distance that the Moon travels during one orbit around the Earth? Also, does it always take the same amount of time, or does it fractionally differ on each revolution?

The Moon has an orbital eccentricity of 0.0549, so its path around the Earth is not perfectly circular and the distance between the Earth and the Moon will vary from the Earth's frame of reference (Perigee at 363,295 km and apogee at 405,503 km), see for example second animation explaining Lunar librations in this answer.

But its orbit can be said, in an oversimplified manner, to be periodic, with no significant apsidal precession (not really true, but somewhat irrelevant for my following musings here to be still close enough), so we can calculate its orbital length based on its quoted average orbital speed of 1.022 km/s and orbital period of 27.321582 days.

So, plugging our numbers in a calculator, $l = v * t$, we get the Moon's orbital length of 2,412,517.5 km (or 1,499,070 miles). Should be close enough. Source of all orbital elements of the Moon is Wikipedia on Moon.


Astronomy online degrees I have a masters degree in computer science and statistics but would like to advance my knowledge in astronomy, ideally in getting some kind of online degree.

Does anybody have experience, suggestions or recommendation where I can do what without having to be physically present at a university?

Also, I would like to get view on what specific topics in astronomy it would make sense to focus if my skills are mainly in computer science and data analysis?

Try https://www.coursera.org/courses?query=astronomy

Online courses from several universities. Some are on a schedule with online lectures, etc. Others you can just do yourself at any time.

Also there are extensive libraries on YouTube, for instance from this guy Michel van Biezen.
http://www.youtube.com/playlist?list=PLX2gX-ftPVXWddG6sGq6H-3cN0IQzfQmG

That link goes to the first of his playlists about astronomy. His astronomy section has about 20 parts!


How can Earth-Sun Lagrange points L1 & L2 even be semi stable considering the moon? I know that the Earth-Sun Lagrange L1, L2, and L3 points are not considered stable over longer periods, especially when compared to L4 and L5... But, with the moon orbiting the Earth in the general path of the ecliptic, it would seem that the first time the moon got in between the Earth and the object at the L1 or L2 points, it would be enough to perturb the object a good bit, and the next lunar month would be even worse, and so on.

It seems that if the moon was taken out of the equation, the L1 and L2 points would be much more stable. And, as a result of this line of thinking, I would think that the L3 point would be much more stable than L1 and L2, since the moon would be so much farther away and unimportant, so its effect could be ignored.

Am I wrong in thinking this? Is the moon's influence not large enough to be a major factor? And, just to clarify, I am talking specifically about L1 and L2 with respect to a large 3rd body, like the moon, instead of some idealized 2 body only system.

The L1, L2, L3 points are not stable. (full stop) Small deviations grow exponentially, even in the perfectly circular restricted three-body problem. In reality, we have a non-circular many-body problem, when (1) these points are not strictly well-defined and (2) deviations from the simple case have to be accounted for to obtain the exact trajectories (and to keep any artificial satellites near these points).

The L1 & L2 points are attractive for space missions exactly because they are unstable, since this prevents the accumulation of debris (natural and artificial in origin) and hence reduces the probability of collisions.


Did uncertainty principle triggered big bang? So i want to know why world changed from the empty state ( no materials, energy and ...) from to the it's staring state in the big bang shape?

based of this answer's to this question:
big-bang-thoery-what-triggered-it-and-why
they said:

  You may think of the pre-Planck and the Planck epoch as a state of the
  universe, which is dominated by quantum fluctuations. They don't need
  a reason or a cause to occur. They are a consequence of Heisenberg's
  uncertainty principle.

So,Did any law like uncertainty principle caused this to be happened or any somethings?

Or

This is one meaningless question?!!!

Thanks a lot.

I think there is a real question in there, just not the one that you think.

Gerald isn't saying that the uncertainty principle played a role in the big bang.   He's saying it played a role in and shortly after the Planck era which came after the big bang.

On the big bang - nobody knows.   Really, that's the answer.  "Was the big bang this" answer "We don't know".   Period.   End of discussion.

We don't know what came before the big bang, We don't know if space was empty or not empty.  We don't know what physical laws may have applied.  Nobody knows.   The initial, basic theory was that the big bang happened at a point or a singularity, which if you think about it, doesn't make much sense cause gravity wins at a point that dense, but the big bang at a point idea does resemble, kinda/sorta a really crazy particle that, by something kinda like quantum tunneling, could become a universe - er, maybe.   String Theory brings other dimensions into this particle becomes a universe - idea, and again, maybe, but nobody knows.

A more modern model, but still, we don't know, is that the universe was infinite at the time of the big bang expansion, and it grew, not from a point but it expanded from infinite into infinite, which works mathematically and it makes (I think) a bit more sense than the super-particle model.  Still the answer is "We don't know".   The answer is always "We don't know".   There's also 4 dimensional models or two time dimension models (those are my favorite when you have x-axis and y-axis time), which are creative ways to avoid the singularity but . . . I'm repeating myself.  Nobody knows.  The answer to what happened at the big bang is always nobody knows.  We don't know if it was empty or not, we don't know if there was some 4th dimensional crossing of membranes which caused the 3d universe we live in.

Now, on modeling back to the big bang, the experts can run the model backwards and apply known laws and have a theory that works to a time shortly after the big bang, which still isn't certainty, but it's a reasonable hypothesis using the laws of physics as they work today.  But it doesn't work all the way back to the big bang.

So, the paragraph that you quote, I think there's a solid question in there, but the quote doesn't mean what you think.

  You may think of the pre-Planck and the Planck epoch as a state of the universe, which is dominated by quantum fluctuations. They don't need a reason or a cause to occur. They are a consequence of Heisenberg's uncertainty principle.

  So,Did any law like uncertainty principle caused this to be happened or any somethings?

Gerald is talking about a period AFTER the big bang and what the models tell us about that period when the universe was, (theoretically), very dense and very hot and governed by quantum mechanics.

Perhaps a better question, which I'm not qualified to answer but might be worth asking, is, "how did the uncertainty principle apply during and shortly after the Planck Era" - at least, that's the question I take away from the paragraph you quoted.   Gerald didn't say the uncertainty principal applied to the big bang, only to the Planck Era.


What is a pupil slicer, and how does it work with anamorphic optics in VLT's ESPRESSO Echelle spectrograph? This excellent answer points to ESPRESSO, - Echelle SPectrograph for Rocky Exoplanets and Stable Spectroscopic Observations. From there I looked at the Instrument Description and Performance page.

The page describes three parts or th esystem:

Coudé Trains (transporting light from each of the VLT's four telescopes)
Front Ends (corrections, stabilizations, fiber pick-up)
Spectrograph (fiber output, pupil slicer, Echelle grating, collimator, dichroic splitter, Red and Blue cameras)

I'm wondering what the pupil slicer is, and how it works exactly.

  At the entrance of the spectrograph, an anamorphic pupil slicing unit (APSU) shapes the beam in order to compress the beam in cross-dispersion direction but not in main-dispersion direction, where high resolving power needs to be achieved. In the latter direction, however, the pupil is sliced and superimposed on the echelle grating to minimize its size. The rectangular white-pupil is then re-imaged and compressed by the anamorphic VPH grism.

I'm familliar with the basics of how an Echelle spectrograph works, and as far as anamorphic optics, I understand this to mean that not all of the elements are not cylindrically symmetric so that the size and divergence changes in one plane is different than in the perpendicular plane. (Classic example is Anamorphic format).

Question: What is a pupil slicer, and how does it work with anamorphic optics in VLT's ESPRESSO Echelle spectrograph?

below: The Spectrograph itself. Light enters from the right from an optical fiber, and the anamorphic pupil slicer (whatever that is) forms a diverging beam to illuminate the entire Echelle grating (beam shown in gray) via the collimator. Everything else in this drawing is post-grating.

In order to reach the stability required for planet detection, the majority of high resolution echelle spectrographs are mounted off the telescope in a stable environment. This involves pressure and temperature stabilization, often by housing the spectrograph in a vacuum vessel.

In order to feed the spectrograph with light from the telescopes, and to reduce the effect of image motion causing variations in the slit illumination and corresponding variations in the spectral line shape which would mask the tiny radial velocity changes     we are trying to measure, optical fibers are used to feed the light. However there are two disadvantages to this: 1) the fiber is now round (although octagonal fibers are the new hotness for their better image scrambling properties) and not a long slit, and 2) due to focal ratio degradation, the output of the fiber is wider (lower f/ratio) than the fiber input which decreases the spectrograph resolution.

In order to combat these effects and increase the resolution, which is needed to measure the tiny 1/1000th pixel shifts caused by low mass planets, image slicers are typically used to reformat the output fiber into a new more slit-like shape. The classic example is the Bowen-Walraven image slicer which uses a glass plate and prisms to slice the circular fiber image into 2 semi-circles and stack them on top of each other. Higher efficiency versions using mirrors or glass slabs have been made more recently; this ESO SPIE publication has some examples of alternative constructions and the resulting image shapes. There is also this page which has more details, including construction of a prototype.

I am not an expert on optical design or anamorphic optics but I suspect this is done because the beam is not the same size and shape in the spatial/cross-dispersion direction compared to the dispersion/wavelength direction and so equal power in the optical elements in each direction is not needed. Given the size of the input beam from the 8.4m VLT that will feed ESPRESSO, and the need to get the whole instrument into an affordable vacuum vessel that will keep vacuum for a long time, I would imagine the designers employed as many tricks as they could in order to keep the size of the instrument down; the Optical Design section of the ESPRESSO Overview paper talks a little more about this.


How do stars affect the orbits of moons? I asked a question previous similar to this, but I'm wondering, can a star make a moon move closer to its planet or further away? How?

Short answer : yes.

Long answer : it's complicated.

This is a classic three body problem, and a real example would be the Sun-Earth-Moon system.

Unfortunately there is no general solution of the 3-body Newtonian gravity problem, so we have to rely on special methods for special cases and even these turn out to be quite complex.

The best answer I could find to your specific question applied to the Sun-Earth-Moon is this one on Yahoo.  As you can see the resulting effect is quite a complex set of periodic effects.  It's small but not negligible.

You can intuitively visualize the effect in such a system by noting that typically to a good approximation the planet-moon system will orbit the star at a considerably larger distance than the planet-moon distance.  The planet and moon themselves orbit around their center of mass.  The relative difference in gravitational forces between the position where the moon is furthest away and closest will give you a rough guide to the order of magnitude of the effect.  A back of the envelope calculation for Earth makes it about a 1% order of magnitude for the Sun-Earth-Moon.

Other systems will be different, but there will certainly be an effect.


What are other names for planetoids that aren't orbiting a solar system, but hurtling through space? I'm trying to think of a good word for an asteroid/planetoid that has no stable orbit but has been ejected from a system and is passing close to a sun. Any help?

See nnnnnn's comment below.

The NASA website, describing 'oumuamuam, uses the term "interstellar object." Extrasolar asteroid would seem to be another option.  I haven't seen a unique word different from asteroid that originated outside of the solar system.

It seems a simple prefix is sufficient to describe objects beyond the solar system, i.e. exoplanet.

Scienfitic American has an article which uses the term "interstellar object."

Also, see answers to:

Is intrastellar commonly used by astronomers to refer to objects within our solar system?
Adjective for things outside our solar system


why do we look at dark energy as an energy? We call dark matter dark matter, because of the way it acts on galaxies, but why do we look at dark energy as an energy? With $E = m c^2$, can't dark energy be mass as well? Why dont we look at dark energy as a mass, in the same way as we look at dark matter as being a missing mass (as halo around galaxies)?

Since relativistic physics almost always takes units of $c = 1$, there is no difference between units of energy density and mass density. Conceptually, however, the gravitational charge is energy, not mass.

In general relativity, a perfect fluid has no viscuous shear stresses and no heat flux, and can be described by the stress-energy tensor $T^{\mu\nu} = (\rho + p)u^\mu u^\nu + pg^{\mu\nu}$, where $\rho$ is density, $p$ is pressure, and $u^\mu$ is the fluid four-velocity. In the case of vanishing pressure, the result is called a dust, and is appropriate for modeling cold dark matter, galaxies on the cosmological scale, etc. On the other hand, a radiation fluid has $\rho = 3p$, and is appropriate for modeling incoherent electromagnetic radiation, or a fluid made of strongly relativistic particles (e.g., hot dark matter), and so forth.

Thus, in the specific context of cosmology, the matter-dominated regime has $p\sim 0$ and radiation-dominated regime has $p\sim\rho/3$. Note that in the latter case, the fluid particles have zero (or at least negligible) mass. In-between, we can talk about a fluid of particles that are not strongly relativistic, and hence have some mass.

On the other hand, the standard model of dark energy is a cosmological constant, which is equivalent to a perfect fluid $p = -\rho$. It does not fit into the above scale; it is not matter in the above sense, nor is radiation, nor anything in-between. Furthermore, it has the peculiar property of being exactly the same in every local inertial frame. This frame-independence enables us to think of the cosmological constant as an intrinsic energy density of spacetime, but it also makes it inappropriate to think of it as a collection of fluid particles that have some rest frame.

For your purposes, this important because $E = mc^2$ is only valid in a rest frame.


Does naked eye miss a lot between local stars and distant milkiness? When I look at the night sky I seem to resolve points which are local stars and milkiness at further regions of the galaxy. But what is between? I think there must be a lot of stars which are too faint to see individually, and yet not perceivable as part of a dense region of stars. Is this at all accurate? Does anybody know how much we fail to see when we think we are looking?

Even the Hubble telescope cannot "see" many of the stars and galaxies in the dark parts of space at first glance.  If they fix the Hubble on a dark patch for several days, the photons from extremely distant stars and galaxies will fill the image.

http://en.wikipedia.org/wiki/Hubble_Deep_Field

Also, the Earth's atmosphere will obscure and distort many stars that are visible from low Earth orbit and beyond.  Astronauts can see many more stars than we can.

Regarding your question, "how many are we missing", that's a tough one to answer. It depends on what level of detail you are talking about.


Can we simulate Earth's gravity in space? Earth gravity is 9.8 $ms^{-2}$

Gravity on ISS(international space station) is 8.2 $ms^{-2}$ approx.

As per this https://physics.stackexchange.com/questions/29929/gravity-on-the-international-space-station ISS never hit the ground because of it's vertical velocity.(free fall)

My question:

Can we create artificial gravity on ISS by using centrifugal force?
[Have anyone watched "The martian"? they have used this technique]

Simulating gravity in space basically means simulating weight, which requires acceleration. So basically, the question is, how do we create acceleration in space.

The easiest method for simulating gravity in space is by spinning the space station. In this case, the reaction force to centripetal force substitutes the force of gravity, which pushes the inner contents to be pushed against the outer edge, giving a sensation of weight. This method is theoretically sound and can be used to simulate gravity.

The problem with doing this in ISS boils down to two things- size and shape.

The best shape for a space station having a simulated gravity is something like a doughnut spinning about its axis, so that the occupants are pushed against the outer wall. The shape of ISS is nowhere near this.

If one excludes the solar panels and the truss structure, the ISS is around 45m along the axis of the habitable modules, if we spin along the axis of the truss structure, this gives a radial distance of ~22.5m.

For simulating earth's gravity in ISS, we will require,

$$\omega^{2} r = g$$

$$\omega = \sqrt{\frac{g}{r}}$$

For g = 9.8 $ms^{-2}$,

$$\omega = \sqrt{\frac{9.8}{22.5}} = 0.66 rad/sec = 6.3 rpm$$

The problem is that at this rotational speed, the head and foot of the astronomer will have different linear velocities. The head of a 1.5m tall astronaut will be moving at 13.86 $ms^{-1}$, while the foot will be moving at 14.85 $ms^{-1}$, never mind they are standing at the bottom of a barrel. This will lead to Coriolis effect, which will be uncomfortable.

So, while earth's gravity cannot be simulated in ISS due to various constraints, we can settle for reduced gravity (with reduced, ~2 rpm) in an ISS sized spacecraft.


How do we know about the origin of ʻOumuamua? How do we know that ʻOumuamua is interstellar and not otherwise? I fail to understand the logic that the extraordinary speed of the object proves it to be of interstellar origin, and that our sun's gravity failed to capture it.

I'll start with the second question first:

Also, do we exactly know about each and every object inside our solar system?

Each and every object? Of course not. That however is irrelevant to the question at hand. The vast majority of objects in our solar system are very, very small. Quite literally as small as dust. But because they're so small, they have essentially zero impact on the orbits of objects that are not as small as dust.

How do we know that ʻOumuamua is interstellar and not otherwise?

Because it's moving so quickly. ʻOumuamua's huge 26 kilometers per second excess means that is not possible that ʻOumuamua originated within the solar system.
I'll start with the two body problem, e.g., the Sun and some other object interacting gravitationally. The Newtonian gravitational interaction between two objects is a conic section: Either a circle, an ellipse, a parabola, or a hyperbola. The total mechanical energy (kinetic energy plus gravitational energy) for objects following a circular or elliptical trajectory is negative. Such objects are gravitationally bound to one another. Hyperbolic trajectories on the other hand have a positive mechanical energy. Parabolic trajectories form the boundary cases, where the total mechanical energy is exactly zero. Objects on parabolic or hyperbolic trajectories are not bound. They visit once and then they're gone. The excess velocity of an object on a parabolic trajectory (zero excess velocity) or a hyperbolic trajectory (positive excess velocity) is a constant of motion in the two body problem.
The N-body problem makes things a bit dicier. Long-period comets are believed to have originated from within the solar system. These comets appear because something perturbed their very long-period orbits. The perturbation make those long-period comets dive well inside Neptune's orbit. A few of those long-period comets appear to be following an unbound trajectory (non-negative excess velocity). A chance close encounter with a planet can make that happen.
This raises the question: Could ʻOumuamua be yet another long-period comet whose orbit has been perturbed to an extent that puts it on a hyperbolic trajectory? This would not be the first case of an object with an over-unity eccentricity. The answer to this question is no. A close encounter with any of the four known giant planets, or even with the conjectured Planet IX, or even multiple encounters, could not have given ʻOumuamua the excess velocity that has been observed for it.


Equation for solar radiation at a given latitude on a given exoplanet? I'm trying to find equations that would help me determine the amount of solar radiation hitting a certain latitude on a certain planet given the following inputs:

the degrees of latitude of the location in question
this hemisphere's current season (winter or summer)
size of the planet
luminosity of the star(s) the planet orbits

Without taking into consideration wind, air pressure, or any atmospheric effects.

Ideally I would like to determine the average solar radiation of a given location in both the winter and summer.

My end goal is to determine the average surface temperature of a given latitude on a planet using the base solar radiation and the effects of wind, air pressure, and surface ocean currents.

I know it is possible to do this for the entire planet generally, but I would like some way of doing it for a particular latitude and season.

The rotational period ( or whether a body is tidal locked ) is a factor here.  A tidal locked body has a hot spot on one side and a cold spot on another.  Rotation may also affects average temperatures.

Season depends on orbital factors.  Simply saying it's one season or another does not tell you the distance you are from the star, which is the single most important factor ( combined with the power of the star ).

However in terms of making a very basic calculation you can use a black body radiation calculation and thermal equilibrium to get something.  You can see more on this at these links :

http://claesjohnson.blogspot.ie/2010/04/black-body-temperatures-of-planets.html

http://www.acs.org/content/acs/en/climatescience/energybalance/predictedplanetarytemperatures.html

It's a pretty good predictor considering it's simplicity.  Note the impact of atmosphere and ocean.  Note also that significant energy is being radiated by some bodies.

This slide show has several equations that may help you.  It also has an interesting model on the effect of an ice cap, which I'd not seen before.

http://www.math.umn.edu/~mcgehee/Seminars/ClimateChange/presentations/2011-2Fall/20111102Handouts.pdf


Place on Moon that always lit? Is there a place on Moon (sure it's near north or south pole) that is always lit by sun? It's to install solar panels if I will choose to make rancho there :)

What diameter of Moon Polar circle?
How high the hill or mast on the pole should be to be always lit?

I used some online calculators and found next:

  Axial tilt to ecliptic is 1.5424°, Lunar Radius 1738km

  Polar circle is 46.8km

  Spherical cap height 0.15km

Are my calculations correct? It shows that it should be near 150 meters high. Is any high enough hill exists on a poles?

P.S. Where I can check good 3d model of Moon with elevation (sure there is no sea level but ideal sphere)?

Thanks

This kind of location is called a peak of eternal light.

Apparently there is no certain identification of such a peak on the Moon, but there are definitely places that get more than 86% sunlight during the lunar year (ignoring eclipses). Kaguya did not find any permanent peaks, but there is apeak near Peary at the North Pole that gets 89%, and several places near the South Pole that have multi-month illumination.

Calculating the requirements for such a peak is complicated since the moon is definitely not spherical on the hundreds of meter scale; at the very least one needs an ellipsoid model. The south pole situation is complicated because there is a large basin around the actual pole.


A [very basic?] question about light scattering and colors Since the night two days ago, the moon here in Israel appears to be orange as it rises, and then gradually becomes white as it continues travelling across the sky.

This phenomenon comes in a great timing as I am starting to become interested in astronomy (I have no special knowledge in astronomy, just to make it clear).
So I googled up and came across this explanation.
I'll try to summarize: it says that it happens because of Earth's atmosphere, that tends to scatter short wave-lengths (such as blue and green) and let long wave-lengths (such as red and orange) pass through. When the moon rises, it's in the position where the light coming from it to our eyes goes through the thickest layer of atmosphere (as opposed to when it's high up in the night skies) and that gives the orange effect.
I totally get this explanation, but it raises a couple of questions for me, whose answers I couldn't find. Maybe you could help me out.

Why doesn't the orange moon phenomenon occur all the time?
It does not happen quite often. What's so special about the past three nights? I mean, the rising moon's light always has to go through a thick atmosphere layer, doesn't it?
Maybe that's not entirely astronomic, but, to complete the big picture - why does our atmosphere scatter short wave-lengths?
This explanation suggests that all light coming to us from the sun is (at least) slightly reddish. Is this measurable (if true)? Because to the naked eye, sunlight refrected through a prism shows no tendency towards red.

Thank you for your time reading the question, I hope you could rescue me from my ignorance...

There may be a particular effect. Smoke from wildfires or cities, fine dust in the atmosphere etc. can have the effect of making the moon appear redder. However it is most likely an example of observation bias.  You noted that the moon appeared orange, and on subsequent nights you were looking for this effect. In fact it happens each moon rise. You just didn't notice it before (or if you did, it didn't make enough of an impression on you.)
There are two forms of scattering that are applicable. Rayleigh scattering, and Mie scattering. There is also absorption of some colours. Rayleigh scattering occurs as the electromagnetic wave that is light causes the molecules of the air to act like little electic dipoles. The light then bounces of these electrically active molecules. Shorter wavelengths are affected more than longer ones.
The light emitted by the sun is very slightly yellow. It is slightly reddened by scattering through the atmosphere. However it still contains a mixture of all visible colours, and so a prism will still show a complete rainbow. The reddening is not apparent to the eye, but can be measured by instruments.


Real-time position of a distant celestial body How might it be possible to determine (via some specific online empemeris or via some well-understood algorithm) the observed position of some distant celestial body (e.g. Betelgeuse)?

To be concrete: suppose I wished to do what any number of mobile phone applications do, i.e. computationally model the observed position of stars and project them onto the screen using GPS coordinates and compass direction, so that an `overlay' of the night sky is produced. Then:

a) Where could I source the relevant data from?

b) What coordinate system would it be best to use?

The most robust, but generally not computationally feasible way would be a full N-body simulation of all the relevant interacting neighbours of the object in question.

On a more crude note, one could observe the eigen motion of the object (its velocity vector), its distance to us, and from that information (assuming the eigen motion is constant over the period between now and when the light was emitted from the object) one may infer the current position.

You can of course make a hybrid of the above methods: use the observed motion of the object, and use simulations to include the interaction with other objects.

Note that 'current position' is not unambiguously defined. Astronomers would often consider the observed position to be its current position, even though we are looking back in time. And, needless to say, on cosmological scales these concepts pose more problems.


Is it possible to use Hubble Telescope to observe Earth in the past? We can observe many astronomical objects and structures beyond our solar system,  however our observations are as the objects appeared in the past due to the speed of light. My question: "Is it possible to use Hubble Telescope to observe Earth as it was in the past" by turning Hubble's lens to a previous location of Earth in space?

The only way you can use Hubble to view the Earth "in the past" is to swivel it and point it at the Earth now. Assuming that this didn't break the telescope (it would) you would collect images of the Earth's surface that were approximately $500\times10^{3}/3\times10^{8} = 0.0017$ seconds old, since that is how long it takes light to travel from the Earth's surface directly below HST to its $\sim 500$ km orbital height above the Earth's surface.

The problem with what you suggest, is that although we might be able to pinpoint where the Earth was in the past with respect to its position now and the Galactic centre (actually that would be very hard in practice), the light that came from the Earth then has long since travelled beyond where the Earth is now.

For instance, let's say the Earth/Sun travel around the Galaxy at approximately 230 km every second and say we want to look at the spot where the Earth was 1000 years ago. Well, it turns out then that the Earth was at some location in space that is currently about $7\times 10^{12}$ km away (7 trillion). But the light that came from the Earth then, has already been travelling outwards at the speed of light for 1000 years, covering a distance of 1000 light-years, which is about $10^{16}$ km. So that light passed by our current location in space more than 999 years ago.


What does "O I" mean in the context of detected molecules? I've recently been analysing the Exoplanet Database, and have been looking at the 'molecules' field (molecules detected on the planet or in its atmosphere). They all make sense apart from "O I". It's not iodine oxide (which would be I2O), so I'm at a loss as to what it could be, especially since I'm very new to this field.

Any help is very much appreciated!

This is the standard astronomer notation for neutral oxygen. Where chemists would write
$\mathrm{O}$ (or $\mathrm{O}^0$),
$\mathrm{O}^{+}$,
$\mathrm{O}^{++}$, and
$\mathrm{O}^{+++}$ (or $\mathrm{O}^{3+}$),
for neutral, once-, twice-, and thrice-ionized oxygen, respectively, astronomers almost exclusively write
$\mathrm{O}\,\mathrm{I}$,
$\mathrm{O}\,\mathrm{II}$,
$\mathrm{O}\,\mathrm{III}$,
$\mathrm{O}\,\mathrm{IV}$, etc.

The same is of course the case for other elements. For instance, you may have an $\mathrm{H}\,\mathrm{II}$ region (i.e. ionized hydrogen) surrounding hot stars, in a larger $\mathrm{H}\,\mathrm{I}$ (i.e. neutral) envelope.

In other words, we start at 1, where chemists start at 0.


Can Hubble red shift be interpreted as time dilation? Can we interpret the de Sitter universe as a spherical cosmic horizon null surface of finite radius, centered at Earth, and containing the Hubble volume of space where time is dilated and radial dimensions contract closer to the edge in such a way that objects closer to the edge do not recognize that they are radially contracted?

Everything is attracted to the edge, but the total radius remains more or less constant and emits de Sitter radiation at finite temperature.

P.S. I know real universe is not de Sitter but it is usually asserted that it approaches de Sitter state asymptotically.

The de Sitter spacetime is maximally symmetric, and so having a full set of Killing vector fields, it must have a static form produced by a timelike Killing vector field.
One way to get without worrying about coordinate transformations is to take the $M\to 0$ limit of the general spherically symmetric lambdavacuum solution, which is the the Schwarzschild–de Sitter spacetime:
$$\mathrm{d}s^2 = -\left(1-\frac{1}{3}\Lambda r^2\right)\mathrm{d}t^2 + \left(1-\frac{1}{3}\Lambda r^2\right)^{-1}\mathrm{d}r^2 + r^2\,\mathrm{d}\Omega^2\text{,}$$
where $\mathrm{d}\Omega^2 = \mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2$ is the usual unit $2$-sphere. Except for minor notational differences, this is also identical to the static de Sitter slicing presented here in terms of different slices of a hyperboloid in Minwkoski $\mathrm{E}^{1,4}$, with $\alpha\equiv\sqrt{3/\Lambda}$.

In these coordinates, the timelike Killing field is simply $\partial_t$, which corresponds to a family of static observers with time dilation relative to the static observer at the origin given by
$$\dot{t}\equiv\frac{\mathrm{d}t}{\mathrm{d}\tau} = \frac{1}{1-r^2/\alpha^2}\text{,}$$
which diverges as $r\to\alpha$.

For geodesics, the Killing field also generates a conserved specific energy $\epsilon = \left(1-r^2/\alpha^2\right)\dot{t}$, which makes solving for radial geodesics not difficult. What's interesting about them is they take an infinite amount of coordinate time $t$ to reach $r=\alpha$, but only a finite amount of proper time $\tau$, and they have no trouble at all continuing beyond it. Additionally, time dilation for static observers becomes infinite there. This behavior is analogous to the event horizon of a black hole, except that the location of the de Sitter horizon is dependent on the choice of origin.

  Can Hubble red shift be interpreted as time dilation?

Yes. If we wish, we can think of the redshift due to a particle (e.g., galaxy) at some radial coordinate as a combination of two things: the special-relativistic time dilation of a galaxy with respect to a local static observer and the gravitational time dilation between static observers given above. In fact, the $\epsilon$ parameter above is exactly the (per-mass) energy of the particle as measured by a local static observer, i.e., the relative Lorentz gamma.


Can you create a black hole using kinetic energy? Imagine you have a particle accelerator that you can crank up to arbitrarily high energies.  Because of General Relativity, the particles get heavier and heavier as you dump more energy into them.  Will these particles ever become a black hole from this relativistic mass?  If so, what will an observer particle moving parallel to the black hole, just below that speed, see?  What I'm getting at is, kinetic energy is relative.  So therefore relativistic mass is as well.  So would these particles be black holes to some observers but not others?

You can't make a black hole simply by accelerating a particle, but it has been theorized that smashing some together with ridiculously high energies could produce micro black holes. This, as far as I know, has never been known to happen at particle accelerators, simply because they are not powerful enough.


Is Pluto still a dwarf planet? Recent news seem to suggest that astronomers are arguing about whether Pluto should become a planet again.

However, I cannot find an official source for this. Is this actually true?

Yes, Pluto is still a dwarf planet. According to the IAU website, it still fits the criteria for a dwarf planet, fails to meet the criteria for a planet, and still carries the "dwarf planet" label, whatever its future status may be. I'm sorry I can't provide a longer or more detailed answer, but this is really a yes-or-no question.


Total mass of light? I know that light is considered as energy, and as such has a mass.

So it is certainly taken into account in universes models and calculations.

So I'm wondering: How much of the total mass of the universe does light and electromagnetic waves accounts for? And how much does it contributes to its expansion?

Is it positive or negative? Since it is considered as mass only, it should decrease expansion somehow. Am I right about this?

First, some clarifications:

  I know that light is considered as energy

No, light is a group of photons and those have energy

  So it is certainly taken into account in universes models and calculations.

The energy of radiation is minimal, reaching about $0.01\%$ (yes, what you said is correct, kingW3) of the energy of the Universe, and that's because of two phenomena causing it to:

Decrease in absolute amounts: because of redshift.
Decrease in relative amounts: because of more dark energy, because it doesn't "dilute" with the expansion:

Think of it in this way: even if there is more space (because of expansion) the ratio between the dark energy and volume of the observable universe is constant (almost, this is more like a metaphor).
Another way to think about (and was one of the possibilities of what is dark energy) is to think of it as a property of space. So you would see that the ratio remains constant.

  Is it positive or negative?

If it were negative, we would be testing some kind of light propulsion engine or using it for building an Alcubierre warp drive. You don't see that happening. Also, light would be a tachyon (which are bad: if you ever see a theory produces a tachyon, throw the theory out of the window, as it's equivalent to saying it doesn't work)
If it were positive it would have mass, we wouldn't have discovered a lot of things, such as Special Relativity because of it traveling at subluminal speeds, and possibly it would hurt we could "feel it" (because of touch) it depending on the exact value of its mass.

  it should decrease expansion somehow

It does, but $0.01\%$ of the energy of the Universe clearly is negligible, and as I said, that number decreases over time.

The mass of light is equal to...?

Light is massless, meaning that its rest mass ($m_0$) is zero. But that doesn't mean it can't interact gravitationally: the reason objects have gravity is that they have mass (and momentum, if not you would have what I think is about $92\%$ less mass), and in this case, mass is equivalent to energy in the sense that it curves Spacetime similarly.

Other way to phrase that last sentence is as StephenG put it:

  [...] Energy (i.e. non-mass energy) can contribute to the distortion of space-time, but that's not the same as it having mass.

Summary of what I said, answering other questions:

How much of the total mass of the universe does light and electromagnetic waves accounts for?

About $0.01\%$.

And how much does it [energy of radiation] contributes to its [the Universe's] expansion?

Almost nothing.


When will the last total solar eclipse occur? Because the moon distance from earth grows, how long will it take, that it is no longer possible to see a total eclipse but only circular eclipses? And does it make a difference if you are on the mount everest or on sea level?

I will take a stab at it.  Assuming

the last eclipse happens when the moon is at perigee and the earth is at aphelion.
the moon's perigee is increasing by 4 cm per year,
the earth's aphelion does not change with time,
the sun and the moon are perfect spheres,
the radius of the sun and moon don't change, and
a total eclipse occurs when $r_{\mathrm{sun}}/d_{\mathrm{sun}} < r_{\mathrm{moon}}/d_{\mathrm{moon}}$ where $r$ is radius and $d$ is distance,

I get the following results:

The last total eclipse at sea level will occur in 721,587,917 years at sea
level  (obviously too many significant digits).
The last total eclipse at mount Everest will occur in 721,807,917 years.
The difference is 220,000 years = (height_of_Everest) * 25000 years/km.

Mathematica Source Code:

dSun = 152097000; rSun = 695508; dMoonNow = 357347; rMoon = 1737;
rEarth = 6371; hEverest = 8 + 8/10;
sol2 = Solve[ rMoon/(d + t 4/100/1000) == rSun/(dSun - rEarth), t][[1]];
Print[ t /. sol2 /. d -> dMoonNow - rEarth  // Round];
Print[ t /. sol2 /. d -> dMoonNow - rEarth - hEverest // Round];


How is mass curving spacetime? In the GR Einstein's interpretation was in terms of a deformation of the geometry of the space ending in a curvature of spacetime.

I've thought that mass could do this by interacting with other mass and causes gravitational waves which makes the space curved. But a prof told me that this isn't the case. For example the gravitational waves caused between the earth and the sun is just a very tiny effect which causes spiraling earth to the sun which is of little effect.

So mass is establishing the geometry of the space and the time. But the question remains how is mass doing that (curving spacetime)?

In general relativity, the curvature of spacetime is an interpretation of  the model for gravity: Starting with the assumptions of gravity being equivalent to acceleration, and the speed of light being constant, a mathematical framework that describes gravity is developed. The mathematical conclusion of this model is that the distances between points is dependent on the distribution of mass and energy. A simple way of describing this is to say that spacetime is curved.

In GR, the apparent force of gravity is explained by the curvature of spacetime. It is reasonable to ask if there is a more fundamental theory that can explain the curvature. String theory provides one possibility. In string theory it is possible to deduce the existence of a particle that would be a graviton, and the from that the field equations that describe gravity. In other words, string theory can predict exactly the same mathematical formulae that arise from curved spacetime in GR. However, on very small scales, string theory makes a series of corrections to the field theory, suggesting that the curvature of spacetime is actually an approximation, which is valid on the large scale, to a string theory of gravity.


How is the Lithium Depletion Boundary used to determine the age of a stellar cluster? According to my understanding of Soderblom et al. (2014), lithium ages of stars are determined as follows:

Determine lithium abundance from equivalent width measurement of Li$_{\mathrm{I}}$ transitions at 6708 angstroms for all stars in a coeval group.
Plot all stars in group on color-magnitude diagram, with each star labeled as Li-rich or Li-poor.
On plot, look for the boundary where the stellar population transitions from depleted to undepleted. That is the Lithium Depletion Boundary (LDB).
Use the luminosity of the LDB and the age-LDB luminosity relationship to determine the age of the stars in the group.

What is the age-LDB luminosity relationship, and what are its physical origins?

When low mass stars are very young, they are termed pre main sequence (PMS) stars. These PMS stars have larger radii than main sequence stars of the same mass, and energy transport in their interiors occurs primarily through convection. The convection ensures that the star is thoroughly mixed and chemically uniform.

As the PMS star radiates away its gravitational potential energy, it contracts. The virial theorem tells us that as it does so, its interior becomes hotter. Roughly speaking, the core temperature is proportional to $M/R$, where $M$ is the mass and $R$ the radius.

Nuclear fusion of hydrogen will not commence until the core temperature reaches more than 10 million K, however there are other fusion reactions that become possible at lower temperatures - namely deuterium burning at around 1 million K and then lithium burning at around 3 million K. The latter reaction is not energetically important in the star's life because there is not much lithium in the star to begin with (about 1 part in a billion), however this lithium can be observed in the photosphere of the PMS star (via the 670.8 nm Li I resonant absorption line) and the convective mixing means whatever we see at the photosphere also represents the abundance at the core.

The Li burning reaction is extremely temperature dependent (like $T^{20}$ or thereabouts), so it turns on like a switch once the core reaches the appropriate temperature (e.g. see Bildsten et al. 1997). The time it takes for a PMS star to reach this core temperature basically depends on its mass. More massive, and hence more luminous stars contract faster and reach the Li burning temperature quicker. Once they do so then the Li in the star is rapidly and thoroughly consumed by fusion. The relationship between the age at Li destruction and the luminosity of the PMS star at that time is the age-LDB luminosity relationship that you refer to.

The result is that if you look at a bunch of stars in a cluster (assuming they all have the same age), then the more massive, more luminous PMS stars will have destroyed their Li, whilst the lower mass, lower luminosity stars will still contain their original Li content. The luminosity at the reasonably sharp transition between these two regimes is known as the LDB.

LDB ages are arguably the most accurate way to find the ages of stars in clusters. All age determinations are to some extent dependent on what physical ingredients are in stellar evolution models, but the sensitivity of the age-LDB luminosity relationship to various uncertainties is quite weak (e.g. Burke et al. 2004) - we basically understand the physics of a contracting, fully convective ball of gas quite well. The LDB ages can also be precise, because the very sharp turn-on of Li burning and its rapidity should lead to a sharp transition between stars with low luminosities that have Li and stars with only slightly higher luminosities that don't.

Interestingly, in the last couple of years it has become apparent to us (and others) that there are some model ingredients that are not completely understood, namely the effects of dynamo generated magnetic fields and dark starspots. Both of these may lead to the suppression of heat transport, either throughout the star or just at the surface, slowing the PMS contraction so that at a given age, the PMS star has a cooler core temperature. This might delay the onset of Li burning and mean that the currently determined LDB ages are underestimates by 10-20% or so (e.g. Jackson & Jeffries 2014; Somers & Pinsonneault 2015 ).


Warm jupiter vs hot jupiter, fluffyness If Jupiter was orbiting at 1AU, replacing Earth, but everything else in the solar system remained as it is currently, how much would the increased heat from the sun increase Jupiter's radius?

In other words, how much would Jupiter fluff up if it was closer to the sun?

Would that be close enough to classify it as a warm jupiter, or hot jupiter?  or would it still just be a standard jovian planet?

Radius inflation seems to mainly occur in planets with irradiation in the range 105–106 W/m2, see Sestovic et al. (2018). In contrast, at 1 AU from the Sun, the irradiation is 1361 W/m2, which is well below the threshold at which the inflated gas giant radii are observed. It seems unlikely to me that there would be a significant increase in the radius of Jupiter if it were moved to a 1 AU orbit.

As regards the hot/warm Jupiter classification, the term "hot Jupiter" seems to mainly be used to describe the population of giant planets with orbital periods below about 10 days or so. Your Jupiter at 1 AU would not fall into this category.


How large could a second moon of Earth be? Question in the title. Let's say that I have a lot of garbage- mined-out asteroids, spent nuclear fuel, satellite debris, and just general trash. Also safe ways to get these into Earth orbit. Instead of sending this into the Sun, Jupiter, or out of the Solar System's plane, what if I took some random quasi-satellite, put it a bit behind the Moon on its orbital path, and started piling garbage on it?

How large could it get before bad things start happen to Earth? For example, tides drowning the coasts. I'm not going to consider collisions, because of (insert megacorp here) Moon Tethers™.

Your question doesn't work because it's based on a false premise - that you can have a stable 2nd moon in orbit. You can't. If you accept the fact that the 2nd moon won't be stable, the size it becomes dangerous is the size that you wouldn't want to fall to Earth - if it's garbage and loosely held together, 100-200 feet in diameter would probably be safe enough and make a big flash and a lot of noise if it hits the Earth, but not cause too much damage.

Space garbage disposal is pretty easy - because there's so much empty space. In densely packed regions like low Earth orbit with lots of satellites, you'd want to avoid dumping garbage, but in the vastness of the Earth's Hill sphere, letting garbage just float in empty space wouldn't be all that dangerous, unless there was a lot of it. If there's enough manufacturing that you can't let garbage float, or if there's toxic minerals like lead or plutonium that you don't want to let fall to Earth, then creating a garbage dump in space makes sense, but it'd be way smarter to just use the Moon as a garbage dump than to build a 2nd one.

If the Earth-Moon system could hold an additional moon in a stable orbit, then your question on how massive could be answered, but there's no practical reason and no benefit towards creating a garbage moon, it would have an unstable orbit and require periodic re-adjusting or it would be dangerous. Just make a garbage dump on the moon.  It's way smarter.

Would the L4 and L5 Lagrange points be unstable?

Doing a bit more research, it seems that Earth-Moon L4 and L5 are stable. NASA did some research into it. Perhaps that makes my answer not correct to this question.  An L5 garbage moon would be possible. - See @Andy256 answer here.

As noted in the comments, a moon-sized object is a lot of material. The Moon is 20 times the mass of the asteroid belt. The number of comets and asteroids that would need to be captured, mined and discarded into the L5 orbit to create a 2nd moon is huge. We're not likely to create anything but a tiny moon out of captured objects because the Moon is comparatively enormous. I rather like the idea of using L5 as a place to capture, place and mine asteroids.

There's still a size limit though. If it grows to a few times bigger than the Moon, it's no longer the Moon's trojan, the Moon becomes its trojan and there's a mass limit for a stable trojan orbit, perhaps about 2 lunar masses, after which, the Moon's orbit would become unstable - and could be a world-ending scenario.

(old update below)

That's a tough call. The Earth-Moon system is far from ideal for stable L4/L5 Lagrange points because the Moon's orbit wobbles due to solar tugging. The mass ratio of 81 to 1 is large enough, with about 26 to 1 being the minimum required for stable L4/L5 saddle points, but the eccentricity and lunar apsidal precession could be problematic for long term stability.
I'm not smart enough to say for certain whether they would be stable or not.  Perhaps another question, or it also fits here if someone else wants to answer.


Can the moon turn red even without an eclipse? My friend posted a picture of the moon as if there was an eclipse because it was so red but there wasn't a scheduled eclipse at the time (it was last night, from the Philippines).

I'm a little concerned because how else is that possible? It was in Manila so I guess there was a lot of smoke in the air but I don't know if that's the reason behind the moon's redness.

Yes, pollution from (for example) wildfires can cause the moon to appear red. It is normally not noticed unless the moon is low in the sky.

Earthsky magazine is reporting that red sunsets and red moons are being reported across the USA. caused by wildfires. Even when the moon is quite high in the sky. I believe that fires from logging in the jungles of Borneo and the Philippines can have the same effect.

This has also been answered on https://earthscience.stackexchange.com/questions/12223/why-were-both-the-sun-and-the-moon-red-today


What are the differences between a Black Hole and a Supermassive Black Hole From what I understand, the mass of a black hole should be nearly infinite, how much more massive can something get?

Is the name to be literally interpreted such that a Supermassive Black Hole just has more mass?
Or rather, is a Supermassive Black Hole just a regular Black Hole with nearly infinite mass that is larger in diameter?
If the difference is in fact a change in diameter, how are the changes in size with the retention of the immense mass reflected in the Supermassive Black Hole's gravitational field?

Stellar mass black holes form from the collapse of massive stars at the end of their lives. You can then find them scattered throughout galaxies, just like you find massive stars. They typically have a mass a few times the mass of the sun.

Supermassive black holes are found at the centers of galaxies. They typically have mass of millions of Suns.

Recently they have started to discover Intermediate Mass Black Holes which blur the lines between a stellar black hole and supermassive black hole. The typically have a mass in the range 100 to one million solar masses.


Why there are less massive stars than low mass stars? What factor determines the selection of star mass?

And how we know that the number of massive stars is less than the number of low mass stars?

One popular argument for the Initial Mass Function is that the mass spectrum -- i.e., the fact that there are more low-mass than high-mass stars -- is inherited from the density spectrum of molecular clouds. In other words, stars form from the collapsing cores of molecular clouds, and the distribution of cores is already biased towards there being more low-mass cores than high-mass cores (this is in fact observed in molecular clouds). So you naturally end up with more low-mass stars than high-mass stars.

Of course, the natural next question is: why is the distribution of clouds cores biased towards low-mass cores? The usual suggestion is that it is the result of supersonic turbulence, and simulations of supersonic turbulence for conditions similar to those of a molecular cloud tend to produce density distributions similar the observed distributions.


Publish data in the virtual observatory I study Computer Science and I need to understand perfectly how is the process of publishing data to the virtual observatory, since I will start a project that seeks to develop another platform to enter data into the virtual observatory (in a more intuitive way if it's possible)

If I understand correctly I believe that this is currently done by means of http://cdsweb.u-strasbg.fr/publication-support#yourDataInVizieR and trough other web pages like http://www.cadc-ccda.hia-iha.nrc-cnrc.gc.ca/en/ or programs like http://aladin.u-strasbg.fr/

Does anyone understand the topic or where can I find documentation?

I feel a bit overwhelmed because in these web pages I see there are many concepts that I do not understand well catalogs, VOTable, I do not even understand very well what a virtual observatory, I do not want to ask my "thesis" adviser every 5 minutes about basic concepts.

Any help will be well received.

"Virtual Observatory" is a catchall phrase basically for a place any scientist can publish their data (ie. tables in their papers) or for hosting databases (such as the latest releases of major survey measurements) which comply with TAP protocols and so can be easily communicated with using a variety of astronomical data analysis programs.

There are many many virtual observatories, here, for example, is the German Virtual Observatory, which is run by the excellent people in Heidelberg: http://www.g-vo.org/

They have a link describing the data publishing process here: http://www.g-vo.org/pmwiki/About/PublishYourData

Usually what this entails is putting your data, such as tables from your paper, into a universal format, like VOTable. This type of table is readily accessed by astronomy data analysis tools such as TopCat (or Aladin, like you mentioned): http://www.star.bris.ac.uk/~mbt/topcat/

Programs such as TopCat can also pull data from these virtual observatories using the TAP protocol which is described a bit here in reference to the GAVO: http://dc.zah.uni-heidelberg.de/system/tap/run/info (note: this link is displayed incorrectly) and a bit here: http://www.star.bris.ac.uk/~mbt/topcat/sun253/TapTableLoadDialog.html

The main purpose of the VO type databases is to allow people to query them while referencing data they have on their computer by allowing SQL (ADQL) queries (crossmatching, selection cuts, etc.) The TAP protocol also allows programs such as Aladin and TopCat to communitcate, so you can do a crossmatch of your data in TopCat with a catalog on the VO and the data will be instantly displayed visually in Aladin as well. Vizier, as far as I know, is just a repository of databases and does not allow this type of querying and interaction.


How can we know the nature of the companion in a neutron star binary? The companion of a neutron star could be a main sequence star, white dwarf, neutron star or black hole.

We detected a pulsar-pulsar binary several decades ago. But if one of the pulsar does not radiate towards us, we may be not able to tell it is a pulsar or a black hole. If the companion is not a main sequence star, how do we know the nature of the companion in a neutron star binary?

Of the binary systems with one single neutron star, is that possible there is another neutron star or a hidden black hole?

One way to figure out if one (or both) of the objects is a black hole, neutron star, white dwarf, or other compact object would be to try to measure its mass. For example, a neutron star and a white dwarf are both compact stellar remnants. However, there is one decisive factor that determines which type of stellar remnant a progenitor star will become: the remnant's mass.

The Chandrasekhar limit ($\sim1.39M_{\odot}$) is the maximum mass of a white dwarf; the Tolman-Oppenheimer-Volkoff limit ($\sim1.5-3.0M_{\odot}$) is the maximum mass of a neutron star. The large uncertainty in the latter means that it's not particularly useful for identifying compact objects that could be neutron stars or low-mass black holes.

So if you know the mass of one of the objects - and let's assume that it's well within the mass range of one of the classes of stellar remnants - you might be able to figure out which type it is. How would you measure the mass? Well, you could study the orbit of its companion star to try and determine the stellar remnant's effect on its orbit.

Another way would be to study gravitational waves emitted by the system. These waves can only be emitted under certain circumstances - for example, in a system of binary neutron stars - see, for example, the Hulse-Taylor binary system, also known as PSR B1913+16. The power radiated as these waves, as well as the orbital decay, depend on the masses of the objects. While detecting gravitational waves is an incredibly difficult task, there are several detectors planned or already in operation, such as eLISA and LIGO.

You could also attempt to look for other effects that characterize certain stellar remnants. For instance, a certain pattern of radio waves would characterize a pulsar (white dwarfs and black holes are not normally strong radio sources). However, if the objects are not luminous, this may not be effective.


What are Thorne-Żytkow objects? Last night I was randomly surfing astronomy pages and came across this.I searched up Wikipedia and other sites,but it doesn't exactly give me enough information about their characteristics.

I'll try to add a bit of context to the Wikipedia article, though the major references are all available there.  The article covers things like the potential formation mechanisms but I guess doesn't really put them into scientific perspective.

Consider, for a moment, a red giant of one solar mass, far up the red giant branch. It has a deep convective envelope, at the bottom of which there is a shell of hydrogen fusing into helium, all surrounding an inert, highly degenerate helium core.  The core at this point is much like an isolated white dwarf.  So, back in the late 1970's, Kip Thorne and Anna Żytkow tried to model (Thorne & Żytkow 1977) what would happen if you replaced the white dwarf with a neutron star.  Such hypothetical objects thus became known as Thorne-Żytkow objects (TZOs).

TZOs have mostly remained theoretical exotica, at least in part because it's very difficult to distinguish them from "ordinary" red supergiants. The main giveaway would be the presence of unusual elements in the atmosphere, produced by the much hotter nuclear reactions happening in the fusion shell around the neutron star core. There have been just a handful of papers published about them, almost all by theoretical groups, in the intervening 40 years.

The only candidate that I think anyone takes seriously is HV 2112. As far as I know, Levesque et al. (2014) found the object serendipitously as part of a survey of red supergiants in the SMC.

PS, following comment:
The consensus view is that a single star couldn't form a TZO because if the core collapses into a neutron star, models predict that the envelope will be ejected.  That is, single star's only evolve into "naked" neutron stars.  Under this assumption, the only way to form a TZO is to have a neutron star collide with or accrete enough material from another star. This could happen if two wandering stars collide. Alternatively, it could happen if one star in a binary collapses into a neutron star.  In particular, it's known that the collapse process is probably not symmetric, so the neutron star gets a "kick" in some direction. If the neutron star is kicked into the companion, it will presumably undergo some viscous drag and settle into the centre.


Gallery of 'actual images' from space? Where can a gallery of actual unaltered photographic images taken in (or of) space be found? Specifically ones that are untouched, not colorized (not necessarily black and white, but they usually are), and taken by natural light photography? Pictures and videos claiming "actual image" are few and far between.

E.g., this:

Not false color:

Both of these pictures are from NASA's (Voyager) Saturn Images gallery. Some of the other ones there are listed as false color, some aren't (but obviously they are). Or maybe not so obviously, hence the question: what does it really look like out there?

I've a pretty good idea of what Saturn looks like IRL, because I've seen it in a telescope (exactly like the first picture, except it's more colorful - absolutely nothing like the second). For most other celestial objects, I have no such baseline.

The title of the website I'm looking for would be along the lines of: View of our solar system through the eyes of a human. Decidedly, not containing any pictures from the HST, as all of them are photoshopped.

You can use one of the digital sky surveys. Examples include:

Sloan digital sky survey SDSS
ESO online digitized sky survey

Their images contain visible (red or blue) wavelengths as well as infrared. All images are monochromatic, as are almost all professional astronomy photographs. You could build your own composite color image from the different channels.

Here is an example image from the ESO DSS2 archive in red wavelength:


Why do planets and satellites in the Solar system look so wildly different if they came from more or less the same matter? First, the planets. We have Mercury, which is rocky, no atmosphere. But then we have Venus, which is completely different: thick atmosphere, very hot, geologically active. Then Earth - blue, full of water. Mars, the opposite: red like nothing else. Jupiter and Saturn are fairly similar. Then Uranus and Neptune, fairly similar but still differ in color between each other and also totally different in color than the two gas giants.

On the other hand: satellites. Let's analyze satellites of Jupiter and Saturn.

Ganymede and Callisto fairly similar, but then Europe, total opposite: completely icy. And then Io, again something completely different: strikingly yellow.

Saturn's moons: mostly rocky, but then, something completely different: Titan, with a thick atmosphere like no other satellite and oceans of liquid methane.

If during the formation of the Solar system there was a protoplanetary disk of matter, wouldn't it be pretty homogenous and therefore give rise to similarly looking planets? I understand that gas giants can't look the same as rocky planets, but why are there differences even between similarly sized rocky planets? Granted, there are wildly different temperatures throughout the Solar system, depending on the distance from the Sun, which probably explain some of the differences.

But then what I especially don't understand are the differences between the satellites. If say Jupiter had a disk of matter orbiting it, which eventually formed into satellites, wouldn't at least that "local" disk around a planet be fairly homogenous? But nevertheless it developed into wildly different satellites. For example, how did the "yellowy" thing get concentrated on Io, and not equally distributed on all the Jupiter's moons?

This questions can be split in two; for planets and satellites.
The diversity of planets reflects in part the diversity in terms of chemical composition of the protoplanetary disk. We know that UV radiation from the sun can dissociate complex molecules or even very simple ones; for example, when UV rays split water molecules the result is free hydrogen and oxygen atoms. Since hydrogen is extremely lightweight they can be transported easily on the flux of stellar winds. So water, to keep with that example, if close to the sun could end up been dissociated and depleted from the region of the disk, but above the so called "snow line" UV radiation from the Sun was so weak that this couldn't happen as often and thus water molecules (which are very heavy in comparison to single hydrogen atoms), stayed there. That only explains the dichotomy between interior and exterior planets in terms of water content, and even then, some processes (like the late heavy bombardment) might add some water to the interior (as it happened on Earth). But this reasoning is not only for water, carbon dioxide, ammonia, methane and hundredths of different molecules have their own "frost lines". Closer to the sun the carbon can't be methane is a volatile gas that gets quickly pushed outwards, but at some tenths of AUs methane can remain in stable conditions and can even condensate into liquid droplets.
All of this just to say that the protoplanetary disk was NOT homogenous in terms of chemical composition, and was not homogeneous in terms of density or pressure. The thermal and chemical gradient across the nebula ensures some diversity and complexity for the entire planetary system.
Here you have a beautiful diagram showing how different chemical compounds could condensate in different temperatures and pressures on the protoplanetary disk.

Also accretion of planetesimals is more energetic closer to the Sun (meaning that breakups can occur more frequently and is difficult for a planet to grow large), while in the outer regions planets can increase in mass with regularity since collisions with other planetesimals are performed at lower relative speeds (because of how two similar orbits have a difference in periods that gets larger when you get closer to the Sun and thus larger relative velocities). This coupled with the gravitational interactions of the protoplanets and the early disk (see planetary migration and nice model etc...) allow for different accretion rates and the accretion of materials of different composition of what was found in the original place of formation of a particular planetesimal. This also helps to keep a wide variety on planetary masses.
A wide variety in planetary masses is the starting point for a larger variation as the planets evolve in time and diverge from their initial conditions. A rocky small planet (Mercury) might have less heat trapped inside than a larger one (Earth), due to the smaller energy released by smaller accretion rates. Thus it might get cold rapidly and a magnetosphere due to a melted interior can't happen. The absence of a magnetosphere allows for solar wind charged particles to erode your atmosphere by sputtering. Instead on a planet like Earth, the larger mass has led to a melted interior that in turn generated a magnetosphere that lasted for billions of years, in Mars it lasted some time but now is almost gone so the atmosphere has also been almost destroyed. On Earth the presence of an atmosphere leds to all sorts of chemical erosions, and phenomena. Also, it's melted interior coupled with the specifics of its chemical composition and thickness of the crust allow for a mechanism called plate tectonics. Tectonics can't happen on Venus because the crust is not as thick (due to different composition) and thus it doesn't break apart in plates but just deforms and folds in a complex behaviour that is unique to Venus.
Also collisions with planetesimals can alter the future evolution of similar planets. Venus was probably very similar to Earth (similar mass, very similar composition and not so dissimilar temperatures as one might think) but their paths diverged completely as tectonics on Earth recycled the lithosphere and on Venus the carbon dioxide got more trapped in a greenhouse effect, and because Earth had a collision with another planet that have us our Moon, which is a mechanical stabilizer while a random collision with Venus (with different impact parameters) led to an extremely slow rotation and long days (but no moons). Longer days mean different insulation, and that changes drastically the climate of a planet. On Mars days are similar to Earth's but since it is smaller and the atmosphere has gone many things are very dissimilar from Earth. Also, Mars has no tectonics (the crust is thicker than Earth's and moves like a monolithic object) so there are few volcanoes and they grow huge (while on earth a single hot-spot generated by a mantle plume drills several holes on the crust as the plate moves on what is called a volcano chain) and there are stress faults like Valles Marineris (unique in the Solar System) that on Earth would have been relaxed by tectonic movements.
To see how different the evolution of two planetary objects can be just by making them different mass take a look at our Moon. It has the same chemical composition (it is a chunk from Earth in fact), it is basically at the same distance to the Sun as Earth, it lives in the same interplanetary environment (same solar radiation, solar wind, impact rates etc...), and still it is completely different. This is all due to mass! The moon can't retain a large atmosphere as Earth because it has less gravitational pull. The same temperature for our atmosphere there means that the particles reach escape velocity easily and start to escape from the gravity well. With no atmosphere, not internal heat, the moon lacks almost any type of erosion in across billions of years of evolution. Processes of erosion on Earth have made the diversity of geological formations to explode in comparison to those found on the Moon. Even then the moon has its own peculiarities and dynamical features unique to it.
Now we are getting closer to the satellites question. They in fact should look almost the same, since they are formed from very very similar material in extremely similar conditions. And indeed we believe that moons originally were very similar (for example the 4 Galilean moons). But Io is to close to Jupiter and the other moons interact with it in such a way that the geological processes are completely different. Water and volatiles evaporated quickly as it heated up by tidal forces from Jupiter. These tidal forces were not as strong in Europa since it is farther away, thus it only melted part of the icy crust creating an ice analog of plate tectonics which generated a plethora of diverse formations. Satellites evolve. Enceladus shoots jets because of tidal interactions and orbital resonances with other moons. Some moons like Japeto have a dual colored surface because of the material sprayed by Enceladus landing on on one of its sides. Some moons like Triton have nothing to do with the other because they formed in another region of the solar System and later got trapped by the gravitational pull of a planet (Neptune in this case).
As I mentioned before. Atmospheres (density, composition and pressure) are largely dependent on the mass of the planet or moon. Look at this graph:

It shows the velocity of gas molecules in relation to temperature of the gas. For larger temperatures gas molecules move faster. In a planet with low mass, the escape velocity is lower than one with a larger mass. Thus a planet closer to the sun (at higher temperature) needs to be larger in size if it wants to preserve the same gas molecules in its atmosphere as a planet that is farther away (colder). You can see why Earth's atmosphere could trap and retain water, oxygen, carbon dioxide, ammonia, methane nitrogen and other gases while it is unable to trap hydrogen and helium (because they are lighter and thus for the same temperature they can move as fast as needed to escape Earth). Meanwhile, the Moon, which has the same heat coming from the Sun as Earth, since it is less massive it can't retain almost any gases (maybe a bit of Xenon). Titan, is a huge moon thus it can retain many gaseous molecules like Nitrogen and Oxygen (those in turn make the pressure high enough to retain also volatiles like methane in a liquid form on the surface). But why doesn't Ganymede has the same atmosphere as Titan if they are basically the same size? Because Ganymede is closer to the Sun, larger temperature means that the molecules move faster and thus they escape its attraction easily.
As you can see the complex processes of atmospheres of a moon or a planet change everything (erosion, recycling processes, chemical corrosion, etc...) and in turn that diversity in atmospheres comes from a diversity of masses and distances to the Sun.
I think the Solar System is a chaotic system, dynamically, geologically, chemically etc... Chaos means that for an small difference in initial conditions the system will evolve in exponentially diverging different states. Planets and moons might have started as similar objects but history and the chaotic dynamics of the system have evolved into completely different environments. Not only that, but the truth is that planets didn't started as equals but were very different from the beginning, so imagine how far away is Venus to become a Titan, or an Io to become an Earth.
Also there are processes and conditions that are specially well suited for divergence. For example: Earth is very dynamic while Mars, Venus, Mercury, the Moon and others are totally not. Why? because on Earth water can exist in 3 different states of matter. We can find liquid water, water vapor and ice in different regions and seasons. And that is because Earth is at an average temperature and it's atmosphere has just the right pressure to allow this. Earth's conditions are very close to the triple point of water (where all three states of matter coexist), that's why we have a water cycle on Earth, with rivers and glaciers eroding the landscape and clouds regulating the climate.

Mars, Venus, Mercury, all have temperatures and pressures were this can't happen, not only on water but on many compounds present there. You know where this can happen? On Pluto! This was very surprising, Pluto shows a variety of terrains and geological features that exceeds all expectations. Now we know this is because Pluto is extremely dynamic (as Earth) and a lot of erosion and geo-chemical processes can occur, but this is not because of water (since Pluto has low pressure and low temperatures), but because of Nitrogen and Neon! Both elements have their triple point inside to Pluto's range of conditions and thus Neon rivers, Nitrogen glaciers and hazes are expected on this dwarf planet.
It is indeed an interesting question. How incredible are the laws of nature that allow for extreme variety even between brothers. I wonder how a planet around any other star might be, our simplistic categories of hot Jupiters, mini-Neptunes, super-Terras, etc... are just so primitive and restrictive. What wonders await us in this complex and diverse cosmos is beyond our comprehension.


How does the Moon look like from different latitudes of the Earth? I have heard, that on the northern hemisphere, you can estimate a direction of south if you imagine a line going through the tips of the border between illuminated and shadowed part of the Moon, and where this line crosses the horizon, that is estimated direction of the south (and vice versa from the southern hemisphere).

This inspired me to think about how much information could one get simply by observing the Moon. I think you can estimate the latitude of your location. I live in the northern hemisphere and never have been anywhere else, so it's all just an imagination. That's why I need somebody to confirm or correct my conclusions.

I think the angle of the tilt of the Moon will be different as seen from different latitudes. Specificaly near the poles you would see the line between illuminated and shadowed part of the Moon very verticaly. At the equator horizontaly. And somewhere in the middle between the equator and the pole tilted under some angle.

You would also see the Moon much lower on the horizon when standing near the poles and going near the zenith when standing on equator.

I also think you can use observation of the phase changes througout a month to find out whether you are at the southern or the northern hemisphere.

So this is what I imagine the Moon looks like. I have searched the internet for relevant information, but nobody seems to discuss the whole thing in detail. There is only one picture somewhat depicting the tilt of the Moon as seen from 3 different latitudes of the Earth and that's it.

EDIT
A software for simulation of the skies (Stellarium) have been recommended to me by @barrycarter and after few hours playing with it, I can clonclude, that the graphics I provided above are accurate.

An interesting observation made while seeing the Moon during daytime. In the real life at the surface of the Earth this observation would not be possible, but the software allows it. Since the Sun is no longer positioned "under horizon", it illuminates the Moon from different angle, than depicted in the grapics.

I made few screeenshots from view from the France (northern hemisphere) of this situation to illustrate what I mean.

It is possible to see, that at nighttime, the moon looks exactly as in the graphics, but as soon as the Sun rises, the illumination direction changes.

Here are some answers to your various comments and questions:

  you can estimate a direction of south

True, but it would be a very rough approximation at best. In order for it to be accurate, the bright limb of the Moon would need to face either due east or west on the celestial sphere. (In astronomy terminology, the position angle of the bright limb would need to be 90 or 270 degrees.) In reality, the bright limb of the Moon faces the Sun, which can be north or south of due east/west relative to the Moon. Therefore, the cusps are not north/south, so the cusps can point to the "left or right" of due south on the horizon.

The second problem is using two points just 1/2 degree apart (the two cusps) and extending it to where it meets the horizon (along a great circle, too!). Depending on your location and the Moon's declination, that distance is most likely to be much larger than 1/2 degree. At 40 degrees N latitude, the moon can be approximately 27 to 73 degrees from the southern horizon. (These numbers are when the Moon is on the meridian.) The accuracy of extending such a short line over such a long distance of is going to be very low.

  I think you can estimate the latitude of your location

Technically, you can use any astronomical object to determine your latitude. Depending on how you do that, the Moon is probably the worse object to use. In order to estimate your latitude, you need to know the declination of the object (as well as other data or measurements). Because the Moon orbits the Earth, the declination changes "quickly". The Sun or stars are better because their declination moves more slowly or are fixed, respectively. Polaris is the easiest object to use if you are in the northern hemisphere.

  I also think you can use observation of the phase changes throughout a month to find out whether you are at the southern or the northern hemisphere.

I agree.

  An interesting observation made while seeing the Moon during daytime. In the real life at the surface of the Earth this observation would not be possible, but the software allows it. Since the Sun is no longer positioned "under horizon", it illuminates the Moon from different angle, than depicted in the graphics.

The Moon is visible during the day, so I do not understand the second sentence. For the rest of the comment, I suggest that you choose a day when the Moon is closer to the Sun and run the simulations. For example, choose a day when the Moon is within 3 or 4 days of New Moon, and watch the Sun and Moon as they cross the sky. The bright limb of the Moon always faces toward the Sun (along a great circle), so the angle of illumination on the Moon changes throughout the day -- just like Stellarium shows. (The Moon's image for Oct 2017 at 8 hours looks a little off to me, but it may just be the difficulty of estimating the great circle direction from the image. Displaying the lines of right ascension and declination may help.)