Is Earth's 1g solid surface gravity unusually high for exoplanets? The surface gravities of the Sun’s planets are all close to 1g, 0.38-2.53 (about a factor of e, by chance). The cloud-top gravity of the gas giants is not too interesting here, but I suppose that the gravity on their solid surfaces (if any) does not tend to be higher. So called super-Earth exoplanets seem likely to generally accumulate light gas atmospheres, which decreases their expected solid surface gravity relative to their size as measured during for example transits (I suppose so anyway).

Are there good reasons to expect that in the zoo of planets in the universe, our Earth scores high in the surface gravity top-list? That it is unusual with solid real estate to stand on which has substantially (as in a factor of a few tenths) greater gravity than we have here?

Surface gravities from Wikipedia:

Mercury  0.38
Venus    0.90
Earth    1.00
Mars     0.38
Jupiter  2.53
Saturn   1.07
Uranus   0.89
Neptune  1.14

According to this paper, fig. 4, solid planets of arbitrary mass (up to 3,000 times Earth's mass) don't grow much about three- or four-times the Earth in diameter. That's because interior parts of the planet become compressed by the high pressure. Planets heavier than 3,000 times Earth's mass are in the transition zone to stars. In the case of rocky planets they would become substellar objects, similar in their physical properties (electron degeneracy in the core) to white or black dwarfs, not capable to start nuclear fusion due to lack of light elements in their core.

Surface gravity is proportioal to the mean density $\rho$, and to the radius $r$ of the planet: $g=\frac{4\pi}{3}G\rho r$, with $G$ the gravitational constant.

A planet with roughly the composition of Earth, and 3,000 times the mass of Earth would have about 3-times the Earth's radius, hence the 27-fold volume, and the 3000/27 = 111-fold density. Its surface gravity would hence be 333 g. That's close to the upper bound of surface gravity of what would be called a planet. With a pure iron planet we could go a little beyond. Any surface gravity below this upper bound is possible for a solid planet, at least in theory.

According to page 1284 of the paper:

  Massive solid exoplanets of
  hundreds to thousands of Earth masses may be able to form around
  massive stars (B and O stars; 5-120 solar masses) where the protoplanetary
  disk would contain enough heavy elements.

B and O type stars are rare (0.13% of main sequence stars) and short-lived (less than 100 million years; more accurately less than $10^{10}\cdot 18^{-2.5}$ years, for mass > 18 solar masses). Hence large solid planets of the described type will also be rare, although a priori not impossible.

The planet with the highest estimated surface gravity discovered thus far (March 8, 2014) is
CoRoT-Exo-3b (add "surface gravity" column to the exoplanets table, and sort by that column):

  CoRoT-Exo-3b has a radius of 1.01 ± 0.07 R_Jup and transits around its F3-type primary every 4.26 days in a synchronous orbit. Its mass of 21.66 ± 1.0 M_Jup, density of 26.4 ± 5.6 g cm-3, and surface gravity of logg = 4.72 clearly distinguish it from the regular close-in planet population, making it the most intriguing transiting substellar object discovered so far.

A surface gravity of logg = 4.72 means surface gravity equals $10^{4.72}\mbox{ cm}/\mbox{s}^2=52,480 \mbox{ cm}/\mbox{s}^2 = 53.5~g.$
This research has made use of the Exoplanet Orbit Database
and the Exoplanet Data Explorer at exoplanets.org.

CoRoT-3b on Wikipedia.


What percentage of Helium-3 is primordial vs produced in stars I tried to research but what I found is pretty limited.    A very tiny but non zero percentage of primordial matter was Helium-3 or 3He.

Stars produce 3He as part of the proton-proton chain but they also consume 3He.   It has about a 400 year half life in our sun.   From Wikipedia.

  In the Sun, each helium-3 nucleus produced in these reactions exists
  for only about 400 years before it is converted into helium-4.[6] Once
  the helium-3 has been produced, there are four possible paths to
  generate 4He

My question is 2-fold.   Is the primordial amounts of 3He significant or insignificant compared to what stars produce and eject by coronal mass ejections or blowing up into nebulas, and are there specific stars, due to the internal heat and rate of reaction that product and eject more 3He into their mass ejections.

For example, would airless, rocky, magnetic field free worlds around red dwarfs be more saturated with 3He or would you find more around larger, hotter stars that undergo fusion faster.    If I wanted to go He3 mining, would I set my ship to a red dwarf system or a blue star system or a nebula?

It's not for a book or anything, I'm just personally curious, as 3He is potentially very useful stuff.

I'm aware that primordial isn't particularly easy to collect, as anything gaseous and primordial would spread out and only collect in large gravity wells, gas giants or larger.   But primordial 3He could, for example, exist in Jupiter or Saturn from formation, though the formation of their magnetic fields probably prevents them from absorbing any ejected from stars.  Hence the primordial vs star ejected ratio and the type of star that produced the most question are somewhat related, so I thought one question rather than two, but I can split into two questions if desired.

My question is 2-fold:

    1. Is the primordial amounts of $^3$He significant or insignificant compared to what stars produce and eject by coronal mass ejections or blowing up into nebulas, and ...

The $^3$He composition of CMEs can vary significantly, see: "Unusual composition of the solar wind in the 2-3 May 1998 CME observed with SWICS on ACE" (Jan 1999), by G. Gloeckler, L. A. Fisk, S. Hefti, N. A. Schwadron, T. H. Zurbuchen, F. M. Ipavich, J. Geiss, P. Bochsler, and R. F. Wimmer-Schweingruber, DOI: 10.1029/1998GL900166

  "Early work [Bame, et al., 1979; Schwenn et al., 1980; and Zwickl et al., 1982] showed that He and heavier elements are overabundant in CMEs and there is enhanced He$^{+}$. Observations from the SWICS instrument on Ulysses revealed some compositional differences in CMEs, such as a high O$^{7+}/\,$O$^{6+}$ ratio, indicating significant heating in the corona [Galvin,
  1997].

  ...

  SWICS is especially well suited to measure solar wind $^4$He$^+$
  and the isotopic helium ratio, $^3$He$^{++}/\,^4$He$^{++}$, as described by Gloeckler' and Geiss [1998a].".

    2. are there specific stars, due to the internal heat and rate of reaction that product and eject more $^3$He into their mass ejections.

Most of $^3$He was originally produced by stellar processes, but see also my answer above. Each star produces varying amounts at various times, see: "The Origin of Helium and the Other Light Elements" (Nov 4 1998), by G. Burbidge and F. Hoyle:

  4. D and $^3$He
  The light isotope $^3$He is produced in large quantities in dwarf stars where the masses are not large enough for it to be destroyed by $^3$He ($^3$He, 2$p$) $^4$He. It is also the case that there is a class of stars in which it has been shown from measurements of the isotope shift that most of the helium in their atmospheres is $^3$He. These stars include 21 Aquilae, three Centaurus A, and several others (Burbidge & Burbidge 1956; Sargent & Jugaku 1961; Hartoog & Cowley 1979; Stateva, Ryabchikov, & Iliev 1998). The stars are peculiar A, F, and B stars having He/H abundances that are $\sim \frac{1}{10}$ of the normal helium abundance. The $^3$He$/\,^4$He ratio can range from 2.7 to 0.5. These stars occupy narrow strip in the (log $g$, T$_{eff}$)-plane between the B stars with strong helium lines and those with weak helium lines that show no evidence for the presence of $^3$He. However, the detection of $^3$He from the isotope shift will fail if the $^3$He$/\,^4$He ratio is $\le$ 0.1. Thus, many of the weak helium-line stars may well have $^3$He$/\,^4$He abundance ratios far higher than the abundance ratio that is normally assumed to be present, namely, $^3$He$/\,^4$He $\approx$ 2 x 10$^{-4}$.

  The high abundance of $^3$He in these stars has been attributed by G. Michaud and his colleagues to diffusion (Michaud et al. 1979 and earlier references). Whether or not this is the correct explanation, what these results do tell us is that stellar winds from such stars will enrich the interstellar gas with He in large amounts. This $^3$He is in addition to the $^3$He that will be injected from dwarf stars. The final abundance required is $^3$He$/\,$H $\; \approx \;$ 2 x 10$^{-5}$. It has been argued by those who believe that $^3$He is a product of big bang nucleosynthesis that there has not been time to build up the required abundance by astrophysical processes.

  However, not only do we not know what the rate of injection from stars is, but in the QSSC, the timescale for all of this stellar processing is $\sim 10^{11}$ rather than H$_0^{-1}$ $\approx$ 10$^{-10}$ yr. Thus, we believe that He may very well have been produced by stellar processes.

Another Wikipedia link not mentioned in your question is: "Helium-3 - Natural abundance - Solar nebula (primordial) abundance":

  "Solar nebula (primordial) abundance

  One early estimate of the primordial ratio of $^3$He to $^4$He in the solar nebula has been the measurement of their ratio in the atmosphere of Jupiter, measured by the mass spectrometer of the Galileo atmospheric entry probe. This ratio is about 1:10,000,$^{[43]}$ or 100 parts of $^3$He per million parts of $^4$He. This is roughly the same ratio of the isotopes as in lunar regolith, which contains 28 ppm helium-4 and 2.8 ppb helium-3 (which is at the lower end of actual sample measurements, which vary from about 1.4 to 15 ppb). However, terrestrial ratios of the isotopes are lower by a factor of 100, mainly due to enrichment of helium-4 stocks in the mantle by billions of years of alpha decay from uranium and thorium.

  Terrestrial abundance
  Main article: Isotope geochemistry

  $^3$He is a primordial substance in the Earth's mantle, considered to have become entrapped within the Earth during planetary formation. The ratio of $^3$He to $^4$He within the Earth's crust and mantle is less than that for assumptions of solar disk composition as obtained from meteorite and lunar samples, with terrestrial materials generally containing lower $^3$He$/\,^4$He ratios due to ingrowth of $^4$He from radioactive decay.

  $^3$He has a cosmological ratio of 300 atoms per million atoms of $^4$He (at. ppm),$^{[44]}$ leading to the assumption that the original ratio of these primordial gases in the mantle was around 200-300 ppm when Earth was formed. A lot of $^4$He was generated by alpha-particle decay of uranium and thorium, and now the mantle has only around 7% primordial helium,$^{[44]}$ lowering the total 3He/4He ratio to around 20 ppm. Ratios of $^3$He$/\,^4$He in excess of atmospheric are indicative of a contribution of $^3$He from the mantle. ...".

  [43] "The Galileo Probe Mass Spectrometer: Composition of Jupiter's Atmosphere" (Science 10 May 1996: Vol. 272, Issue 5263, pp. 846-849) by Hasso B. Niemann, Sushil K. Atreya, George R. Carignan, Thomas M. Donahue, John A. Haberman, Dan N. Harpold, Richard E. Hartle, Donald M. Hunten, Wayne T. Kasprzak, Paul R. Mahaffy, Tobias C. Owen, Nelson W. Spencer, and Stanley H. Way, DOI: 10.1126/science.272.5263.846

  [44] "Non-Lunar $\underline{^3}$He Resources" (Presented at the Second Wisconsin Symposium on Helium-3 and Fusion Power, 19–21 July 1993, Madison WI), by LJ Wittenberg - fti.neep.wisc.edu


Sum of the Masses of All Moons in the Solar System I'm looking for the total mass of the moons in the solar system in Earth Mass
this would be include the 67's moons of jupiter, the 62's of saturn, the 27's Uranus and the 14's for Neptune.

it's a crazy question I know, if you have some list of the object with mass in same range i can make the addition ^^

Here you can find a list of all the natural satellites in our Solar System. You can check one by one (good luck!) OR you can check this webpage, and just add the terms.

Please, keep in mind that the latter website is kind of unknown, so double-check at least some of the masses, before to trust it.

Perhaps, you can cross check with this list as well, and see if other parameters (size, distance, etc.) agree.


Measuring parallax angle, same local time or the same time I was thinking about measure the distance to the moon using parallax angle. But I have a confusion. I have read couple of articles about that, but one thing is unclear. Should we take pictures of moon at the same (literally the same, for example, taking both pictures at UTC 16) moment or should we take pictures the same local time (for example, if the timezone difference is 2h then the time difference between two taken pictures should be 2h).
So which one is the correct method for it? I think it is the former one, but I cannot tell why it is not the latter.

The simplest way: take pictures at the same (ie 16UTC). The position of the moon in the sky will be very different for different observers, due to the cuvature of the Earth.  But this is not what you are measuring. You need to see the difference in position relative to the stars.
For more accuracy, instead of measuring the difference is position directly, you can time when the moon occludes a star. The absolute time when the moon will occlude a star will differ for different observers, and this can be converted into a difference of position of the moon, and parallax can be calculated from that.  It is easier to be accurate in timing than it is to measure position directly. With some clever maths you can just measure the length of the occulution rather than the absolute time, and this means you don't even need a accurate clock, you only need an accurate stopwatch (not so much of an issue now, but significant when these measurements were first being done)


Head on collision of two black holes Inspired by If two black hole event horizons overlap (touch) can they ever separate again?, I was wondering what would happen if two black holes collided into each other in a head-on collision.

In this model there's two black holes on a 3d Cartesian plane. Each black hole has a mass of 1 billion solar masses, and they're both going 0.9c. They're traveling on the z axis in opposite directions.

If both black holes are not spinning, one traveling positive z, the other in negative z, and collide at the origin in a perfect head-on collision what would be the result?

How long would a full merger take, and what would the resulting spin be? Would there be a large wave of gravitational energy, or some other emission?

I don't believe this to be a duplicate of Would LIGO Detect Head-On Collision? as suggested by @John Rennie or @uhoh beucase I'm more asking what's the effect on the BH themselves and what type of emissions are involved. Not necessarily ones detectable by LIGO.

Well the spin is easy. Your system has zero angular momentum, so the spin will be zero.

I think there will also be no gravitational radiation. for the slightly technical reason that gravitational waves only come from changes to the quadrupole moment of a system and in your system the quadrupole moment is always zero.

So everything ends up in the final black hole which will have mass about 4.6 billion solar masses, coming from the mass of the two colliding holes, and their kinetic energy.

I'm not sure exactly how long the merger would last, but it will be comparable to the Schwarzschild radii involved, divided by $c$, so around three hours (as seen by a distant observer).


Why are galactic nuclei yellow? So we've seen millions of galaxies, and all of them have a bright, yellow nucleus. What is the cause of the yellowness? Is it because of aging stars, redshift from the SMBH, or something else?

It's an observational fact that many disk galaxies have a color gradient with redder (or as you say 'yellow') stars in the central regions and bluer stars in the disk.
There are two things that can cause this central redness.

Age, older stars being redder than younger stars
Metallicity, with red stars being more metal rich than blue stars

Whether it's age or metallicity or both causing these gradients is topic of current research. It's a popular theory that galaxies grow inside-out, which causes a red center.


Can someone explain this diagram showing the spectral type distribution of bright stars Can someone explain this diagram?

The text is in Dutch, free translation: "You are given a graph: a histogram of the 10 000 most apparent bright stars. Explain 1, 2 and 3."

The fact that the number of counted stars of types O and M is small due to

O is bright but just very rare and,
M is common but quite dim.

The puzzling bit in this diagram is the sudden dip from a local maximum F (why is there a maximum) to G (why is G a local minimum). The drop from K to M is explained above - that being that M class stars are just very dim (is this all there is to say here?).

It seems that if there wasn't a dip at 3, and the graph would just continue, that the local max at K can be explained as being there just because there is a dip at G. If there wasn't a dip it wouldn't be a maximum and it'd just look like a bell-curve.

So

Why is there a dip at G?
Why does it rise so smooth from O to F? Is it just because B - F are more common?
Why is there a maximum at F?

It's all to do with the relationships between mass, spectral-type and luminosity and the initial mass function of stars.

I think your explanation of points 1 and 2 are completely correct. O and B stars are rarely born and short-lived; so even though they have enormous luminosities relatively few make it into a list of stars ordered by apparent brightness. A further important reason is that although they are immensely luminous, the more distant examples that might have made it into the list, cannot make it into the list because they are in the Galactic plane (they are young and still close to where they were born) and thus severely affected by extinction.

M-dwarfs are indeed the most common type of star, but not common enough to overcome the fact that their luminosities are much, much lower than hotter stars.

The dip in G stars is actually a peak in the K-stars caused by giants. Most low(ish) mass stars will evolve towards more or less the same point in the HR diagram forming the red clump. These stars are luminous, older and not concentrated in the galactic plane, can therefore be seen to great distances and produce a peak in the number counts.

So you can see this using actual data rather than a cartoon, I producedthe HR diagram for the brightest 10,000 stars in the revised Hipparcos catalogue (about V<6.7). You can clearly see the red clump of giants. Now collapse this distribution onto the B-V axis and you can see there is a local minimum at $0.6<B-V<0.8$ corresponding to G-stars.

Absolute colour-magnitude diagram for the brightest 10,000 stars

The corresponding B-V distribution for these stars. G-stars have $0.6<B-V<0.8$


Does the Milky Way's movement through space have spacetime implications? The way I usually see spacetime explained, it says that if you just sit still, you are moving at 0kph and so you are moving at maximum speed through time (ie. time for you is flowing at maximum speed).  However, sitting still we're moving at the speed of our galaxy through space, which is about 600km/s.  The speed of light is about 300,000km/s.  So for simplicity's sake, let's say the galaxy is actually moving through space at 1,000km/s.  Does this in fact mean that if I sit still on Earth, time for me is actually flowing 1/300th slower than if I were sitting still with respect to the universe?  And does it also mean that if the Milky Way were travelling at 99% the speed of light, time for me would be flowing at 1% the rate of the flow of time for someone that was sitting still with respect to the universe?

Your question has a misunderstanding at its root:  You say, "So for simplicity's sake, let's say the galaxy is actually moving through space at 1,000km/s". The problem here is that the whole point of Relativity is that the idea of movement with respect to space is meaningless -- the only thing that has meaning is movement with respect to other bodies.

Time dilation is also relative. If you're moving at 1000kps relative to Fred, Fred sees your clocks as running slower, and in his "reference frame" they are.  But you, in your reference fame, see Fred's clocks as running slower! (This is the essence of the famous Twin Paradox.)

You're being led astray by your intuitions about space and time. (And don't be ashamed about that. Newton, who was probably the greatest physicist ever, even greater than Einstein, and Kant (one of the great philosophers) and many, many others had the same intuitions. Yet experiment has taught us that they were wrong.)  Neither space nor time is absolute, and an understanding of Special Relativity (which is much easier to understand than it was to discover in the first place!) is required to ask as well as to answer your question.


What happens to the energy from a GRB? A gamma ray burst produces somewhere around 1044 J of energy, according to Wikipedia. Only a small portion of that energy will be absorbed by other objects.

The rest of that energy will go on to travel the universe indefinitely I presume?

How do we know how much of the universes total mass-energy is in "in transit" in the form of gamma rays and other forms of radiation?

No energy is in transit as GRBs, just like Gamma rays that started in a GRB.

If you remember that all that energy is in the form of radiation, it obeys the same laws as ordinary light: the farther, the dimmer (inverse of square of the distance law).

So if you see stars dimmer as they are farther from you, same happens to the RGBs. Radiation from the stars expands infinitely (becoming dimmer and dimmer) until absorbed by clouds or just becoming dimmer than the background.

Same happens to gamma rays from GRBs. It is just that since rays are more energetic, they are less easily absorbed, and since GRBs themselves are more energetic, the becoming dimmer than background distance is bigger.


On the cosmological principle Just more of a conceptual question on the mutual inclusivity of the cosmological principle. That is to say, I was wondering if it were possible to have a Universe that were isotropic but NOT homogeneous OR a Universe that were homogeneous but NOT isotropic.

My spidey sense is telling me that it is possible that a Universe could be istropic but not homoegenous.

Surely, by default, if a Universe is homogenous then it follows that it has to be isotropic?

I was wondering if it were possible to have a Universe that were isotropic but NOT homogeneous OR a Universe that were homogeneous but NOT isotropic.

Most people will be happy with the UC Berkeley definition that says homogeneous means "looks the same at every location" and isotropic means "looks the same in every direction". And some will know that as per Ned Wright's article, these attributes aren't quite the same:

He says "the figure above shows a homogeneous but not isotropic pattern on the left and an isotropic but not homogeneous pattern on the right".  However as far as I know pictures like this just don't apply to our universe scattered with galaxies.

  My spidey sense is telling me that it is possible that a Universe could be isotropic but not homogeneous.

My spidey sense is telling me some guy 46 billion light years away might say the universe is neither isotropic nor homogeneous. Because when he looks up, half the night sky is black or something.

  Surely, by default, if a Universe is homogenous then it follows that it has to be isotropic?

I agree with the gist of that. IMHO if an observer sees a homogeneous universe, he sees an isotropic universe too. Yes, one can find hypothetical scenarios wherein the universe is homogeneous but not isotropic. But they're only hypothetical. And let's not forget that it's only an assumption. If you lived in a forest would you assume that the world was covered in trees? Which look the same at every location and in every direction? It isn't a particularly scientific  assumption. For all you know some guy lives near the edge of the forest. I think you're better off saying we just don't know.


How do I see the milky way? I've travelled to farms at night, and away from civilation. I don't really know alot about astronomy, but my goal is to see something like this:

Stars, lots of them, but that "thick purple cluster of gas right in the center". I want to see that.

I assume that on a very very dark day, we would be able to see this, but sometimes I only see stars, just lots and lots of stars, something like this:

I want to see this "thick cloud cluster, which I assume is the milky way". How do I see this? How on some days do I only see just stars and stars, when I really want to see this main beauty?

"thick purple cluster of gas right in the center". I want to see that.

First off, the human eye is a wonderful device, but it's no match for modern cameras. We can't see colors at such low levels of light. We're all colorblind in those conditions. Only cameras continue to see color when illumination is low.

As a result, when looking at objects such as nebulae or galaxies, including our own, all you see is a black-and-white image. Even with telescopes, the human eye still doesn't see colors in these objects.

There is exactly one single exception - the Orion nebula, which is by far the brightest of all. Some observers, after a bit of practice, can begin to distinguish a few shades of color there.

Planets, and some stars, are different. These objects are much more bright. The human eye has no trouble seeing color where color exists here. E.g. Jupiter is a slight shade of yellow, Betelgeuse is quite clearly red, etc.

  purple

I'm pretty sure the Milky Way is not purple. That's probably just a processing artifact. Go on Astrobin and search for "milky way" for a lot more examples - they tend to not look very purple usually.

https://www.astrobin.com/

  I want to see this "thick cloud cluster, which I assume is the milky way". How do I see this? How on some days do I only see just stars and stars, when I really want to see this main beauty?

There are two main obstacles to doing astronomy at the bottom of an ocean of air: atmospheric turbulence (a.k.a. "seeing") and light pollution.

Seeing affects observing high resolution objects such as planets and double stars when seen in telescopes. It's caused by air eddies and vortices that muddy the image. It's of no interest here so we can ignore it for this discussion.

Light pollution affects observing faint objects such as galaxies (including our own), nebulae, and star clusters. It's caused by light from artificial sources (city lights, industrial lights, cars, etc) going up, being reflected on particles in the air, then going back down. It makes the sky seem to "glow". It drowns out those objects which are not very bright - and galaxies tend to not be bright.

There's only one way to avoid light pollution: go far away from human settlements and facilities. The desert and the open sea are not light polluted.

Google "light pollution map" for some websites that will help you choose a dark site. Here's one example out of many:

http://www.jshine.net/astronomy/dark_sky/

Recently I've visited Death Valley. The surrounding area has essentially zero light pollution. Even driving 30 minutes out of Ridgecrest you're already under a very dark sky. I took the opportunity to show my kids how a dark sky really looks like.

The Milky Way is very clearly visible. Wherever you look, there are stars upon stars. Take a pair of binoculars, point them at random places in the sky, and even more stars are visible. There are star clusters that you could chance upon just by sweeping the sky with the binoculars, and they look like someone threw a handful of diamonds on black velvet. Little shooting stars are visible often. It's extraordinary.

It should be noted that the Milky Way is not visible during the winter months (December – February). It's best to try to watch it in June – August. Even later in the year (September – November) it's still visible in the West, just before it sets, a short time after nightfall. If you're an early riser, you can see it before daybreak in the spring (March – May).

You need to wait until the sky goes completely dark after sunset. That means 1.5 hours after sunset, or more. To determine the precise moment of sunset, go on Google or Wolfram Alpha and search for "sunset today". Then add 90 minutes to the result.

You will also need to time your visit to a night when the Moon is not visible. The Moon must be below horizon, otherwise it would cause some "light pollution" of its own. A few days before and after the new moon are the best time for this.

You could go on Wolfram Alpha and search for "next new moon" or even "new moon in september" and it will give you the date. Several days plus or minus that date should be fine.

http://www.wolframalpha.com/

TLDR: A moonless night in the summer, in a desert or some other unpopulated area, 1.5 hours after sunset or later.


Simulate an orbit with orbital elements I want to simulate the orbits of the planets from our solar system. I want to use orbital elements to calculate the current position(xyz) at a time t. The simulation doesn't have to be too exact, but the initial position of the bodies should be somewhat realistic.

Doing some research on the calculation I stumbled upon the following formula:

R,X,Y,Z-Heliocentric Distances
TA - True Anomaly
N - Longitude of the Ascending Node
w - Argument of the Perihelion

R = a * (1 - e ^ 2) / (1 + e * Cos(TA))
X = R * (Cos(N) * Cos(TA + w) - Sin(N) * Sin(TA+w)*Cos(i)
Y = R * (Sin(N) * Cos(TA+w) + Cos(N) * Sin(TA+w)) * Cos(i))
Z = R * Sin(TA+w) * Sin(i)

Source: https://www.physicsforums.com/threads/calculating-elliptic-orbits-in-cartesian-coordinates.712979/

For testing purposes I calculated the true anomaly with the help of the following calculator: http://www.jgiesen.de/kepler/kepler.html
It needs the mean anomaly and eccentricity which I took from Wikipedia. Doing some more research on the mean anomaly, I now believe that the time at which I want the position will be somehow fed into the calculation of the mean anomaly along with the time of the initial position.

Can somebody clarify for me on how to correctly calculate the position at time t with the above or a different formula. With the above formula I also think I need the initial values at a specific time.

I also want to say that I am not an astronomer and I don't have a clue on how to handle the calculation or which formula to use. The result of the calculations should mimic our solar system as closely as possible.

Thanks,
Rene Hollander

As it happens, I was writing code for the exact same transformation last night using this memo as a reference. I haven't yet tested my implementation but the document is quite clear about what each step is doing, which pleased me greatly.

You'll have to understand the terminology to some degree; "epoch" means a reference time for which you have the orbital elements, and "considered epoch" means the time at which you're evaluating the position, for example. Other points requiring caution: the equations inconsistently drop subscripts; the use of the Newton-Raphson solver isn't particularly clear (i.e. how many times should it run)? Make sure you get the units right also.


tidaly locked celestial body in three planet system If a Planet have 2 Moons (so suppose system of 3 bodies) and we say "One Moon is tidally locked to Planet" how can we say that it is tidally locked when Baricenter of 3 bodies has its trajectory?

Tidal locking just means that one side of the Moon will face the planet all the time. The period of rotation around its own axis is the same as the period of rotation around the planet. It is not absolutely "locked" exactly in that configuration, but on average the two periods will be the same. Another moon or two could possibly also be in the same condition, and there wouldn't be strong interference between them. There could be some slow processes that could work against both remaining tidally locked, or some slow processes that kept them that way, but it's a complicated problem


Why does Titan's atmosphere not start to burn? If Titan has a methane atmosphere and seas of methane, then why doesn't some meteoroid / chemical reaction ignite and blow the whole thing?

Titan is one of Saturn's moons. Titan has a dense atmosphere, at about 1.5 bars. It also seems to have lakes of liquid methane.

For a conventional combustion, you would need a good Methane-Oxygen mix. Every combustion is basically an oxidization. Apart from very energetic events, such as SL9 hitting Jupiter, you will need quite a bit of oxygen to get a nice, explosive fireball in Titan's atmosphere.

According to Wikipedia, you need two Oxygen molecules for every Methane molecule you want to burn.


When is Earth closest to the Sun? When is Earth closest to the Sun?

Well, you say, perihelion of course.  But perihelion is when the Earth-Moon barycentre is closest to the Sun. Not Earth itself.

The barycentre is on average 4,671 km from the Earth, which should mean that even if the orbit of the Earth-Moon barycentre around the Sun were perfectly circular, the Earth-Moon orbit around the barycentre would cause an oscillation in the Earth-Sun distance by twice that amount on a monthly basis.  That should be enough to move Earth's closest approach to the Sun away from perihelion.

Considering this, how many days from perihelion may Earth's closest approach to the Sun be?

Question inspired by: What place on Earth is closest to the Sun?

Earth's perihelion timings and distances are based on the distance between the center of the Earth and the center of the Sun. They are effectively the times and distances of the closest approach between the two bodies, not the barycenters. As we can see on Fred Espenak's astropixel page about perihelion and aphelion, the time intervals between two successive perihelions tend to vary between 363 and 368 days.

The data on astropixels seems perfectly valid. With JPL Horizons, for example, the next perihelion is calculated to be on 2020 Jan 5 at 07:47:42, at a distance of 0.98324356482 AU (147,091,144 km) which matches astropixel's values.

I analyzed the perihelions of the 21st century to see how much the actual timing can differ from an idealized "average" perihelion moment (based on the anomalistic year length). Perihelions this century remain within 1.4 days of that average. The differences from the average are mainly due to the gravitational effect of the Moon.

Using all perihelion times in the 21st century, and the average perihelion interval (anomalistic year, 365.2596 days) I constructed the following graph showing these differences:


Why is 3GHz the best frequency at which to measure the sun's microwave output? I have read that 3GHz is the best frequency at which to measure the sun's microwave output, but that measurements are made at 2.8GHz for historical reasons, however I can find no explanation of why 3GHz should be the best frequency.
Could anyone tell me please?

From "The 10.7 cm solar radio flux ($F_{10.7}$)",

It has become clear that wavelengths in the region of 10 cm are best for monitoring the level of solar activity because solar emissions at these wavelengths are very sensitive to conditions in the upper chromo- sphere and at the base of the corona.

This is not saying that exactly 10 cm wavelength radiation (3.0 GHz) is optimal. Ten centimeters is being used in this context as a nice round number. 10.7 cm radiation is definitely in the "region of 10 cm". Schmahl and Kundu reported in "Microwave proxies for sunspot blocking and total irradiance" that anything in the range of 1000 MHz to 9400 MHz works quite nicely. 3.0 GHz (3000 MHz) is more or less in the middle of that range, logarithmically.
Measurements of the F10.7 radio flux go back to 1946. This was a serendipitous choice as a proxy for measuring variations in the Sun's output, not just in the microwave / radio frequencies, but also in very high frequencies (ultraviolet and X-rays).

References:
Tapping, K. F. "The 10.7 cm solar radio flux ($F_{10.7}$)." Space Weather 11.7 (2013): 394-406
Schmahl, E. J., and M. R. Kundu. "Microwave proxies for sunspot blocking and total irradiance." Journal of Geophysical Research: Space Physics 100.A10 (1995): 19851-19864


What causes the abundance of dust to 'float' in the thin lunar atmosphere? Several missions to the Moon have revealed a lunar horizon glow, such as the ones seen below by the spacecraft Surveyor 7 in 1968:

Image source: NASA

According to the NASA article "Model Helps Search for Moon Dust Fountains" written in 2010 suggests the cause of the phenomenon is due to sunlight shining through a dust layer suspended in the very thin atmosphere (exosphere) of the moon.  From the NASA webpage:

  "It has been suggested that electrostatic forces play a role in the ejection of dust from the lunar surface, and its dynamics in the atmosphere, but we really don’t understand how it gets there in such high abundances."

Given that there is no wind or precipitation in the lunar sky, my question is:

What causes the abundance of dust to 'float' in the thin lunar atmosphere?

A recent study in Geophysical Research Letters — "Dust charging and transport on airless planetary bodies" (Wang et al. 2016) and reported in the ScienceAlert article "NASA just explained why Moon dust is 'levitating' above the lunar surface" — seems to explain the phenomenon through laboratory simulation experiments which observe what occurs when micron-sized dust is exposed to ultraviolet (UV) radiation or electrically-charged gases known as plasmas.

From these experiments, the researchers observed that:

  In both cases, the dust particles would leap several centimetres above the surface.

The experiment observations can be seen below:

Image source: Science Alert article linked above

Thus, one can conclude that on the Moon light would scatter through these levitating particles to create the horizon glow.

The mechanism behind the 'levitation' is stated as being:

  This is due to the reaction [with the plasma and UV light] causing an emission and re-absorption of electrons inside 'micro-cavities' formed between neighbouring particles, which can generate unexpectedly large electrical charges and intense repulsive forces.

  These forces cause the dust particles or clusters to lift off the
  surface, or 'levitate'.

Although the effects are observed with the lunar horizon glow, the researchers expect that effect would cause a suspension of dust across the entire lunar surface to some degree.

A significant implication of this research is stated in the NASA article "Dust “Floats” Above Lunar Surface: Electrostatic Dust Transport Reshapes Surfaces of Airless Planetary Bodies" as:

  The laboratory observations also showed dusty surfaces becoming smooth as a consequence of dust mobilization. These electrostatic dust processes could help to explain the formation of the “dust ponds” on asteroid Eros and comet 67P, and the unexpectedly smooth surface on Saturn’s icy satellite Atlas.


Could Hot Jupiter systems sustain Trojans? My understanding is that there is a gravitationally stable region between the Sun and Mercury where asteroids could - in theory - orbit for long periods without being perturbed. However these hypothetical vulcanoids are seemingly non-existent; it being likely that their orbits decay through other means besides the influence of a planet, such as the Yarkovsky effect. We also have trojan asteroids, which share the orbits of planets such as Jupiter but spend their time in the leading and trailing Lagrange Points, keeping their orbits stable.

If a Hot Jupiter such as 51 Pegasi b were to host trojan asteroids, would their proximity to the planet allow their orbits to remain stable for longer, or would their orbits decay regardless? If a Hot Jupiter migrated inwards towards its star, is it more likely to have swept away any possible trojans?

Yes, hot Jupiters can have stable Trojans.

A great analysis of this very problem is given by Kislyakova et al. (2015). They suggest that super-hot outgassing Trojans may be an explanation for the ultraviolet anomalies seen in the WASP-12 and HD 189733 systems. As part of their analysis, they explore the photogravitational stability of Trojans of various size, and while they show that meter-sized bodies can be ejected on timescales of around a million years, Io-sized objects can be stable over the lifetime of the system. It may also be worth noting that although they found many stable tadpole orbits, they did not find any stable horse-shoe orbits (although they do not show that such an orbit cannot exist).

As for the actual formation and capture of such a Trojan, Beauge' et al. (2007) modeled the inward migration of a massive planet, with its semi-major axis decaying over a time scale of $10^5 - 10^7~\mathrm{yr}$. Their results suggest that a Trojan satellite with a tadpole orbit close to $L_4$ will be maintained during the migration.


What instruments are used to measure the distance to the Sun? In the 18th century, the distance to the Sun was measured using the transits of Venus. Nowadays, the timing of radar pulses is used to measure the distance of things in our solar system. What instruments are used to perform these measurement? What do they look like (bonus points for pictures)? Are there observatories with "radar guns" dedicated to measuring the distance of the Sun?

They look like radio telescopes, because that is what they are.
Some radio telescopes have been outfitted to transmit rather than receive, requiring using a different radio telescope to receive the reflected signal. Other radio telescopes have been outfitted to transmit as well as receive in a pulsed manner. These can be used for radar astronomy without the help of a separate transmitter.


Information on the star at the centre of the Helix "Eye Of God" Nebula Can someone please tell me what the mass (or size) of the star at the center of the Helix Nebula is, and how much larger it is than our own star (we call the Sun or Sol).  Also, what is the name of this star?  I cannot find a website that addresses these questions.  I believe I have heard, however, that the star is roughly 100 times larger than our sun.

The central star is a hot DAO white dwarf. As far as I can tell, the numbers that are the most reliable are those from Napiwotzki et al. (1999). They use a direct spectral fitting technique to estimate the surface gravity and effective temperature of the central white dwarf star. These are $g = 10^{(7.00 \pm 0.22)}$ cm/s$^2$ and $T_{\rm eff} = 103600 \pm 5500$ K.

They use these numbers and theoretical white dwarf evolutionary models (there is little alternative for a star that isn't part of a binary system) to estimate a mass of $0.57 \pm 0.02 M_{\odot}$.

At a distance now known to be around  $219 \pm 24$ pc (Harris et al. 2007), this makes the luminosity of the white dwarf around 100 times that of the Sun. A bit of calculation shows that for this gravity and mass, then although the star is small (about $0.04 R_{\odot}$), it is much bigger than most white dwarfs. That is because it is recently formed (planetary nebulae are recently formed objects) and hot; the white dwarf is still contracting towards its final equilibrium size of around $0.015 R_{\odot}$.

The CDS SIMBAD entry for the star is http://simbad.u-strasbg.fr/simbad/sim-id?Ident=WD+2226-210&NbIdent=1&Radius=2&Radius.unit=arcmin&submit=submit+id

It is listed in several catalogues, so goes by various names e.g. 2MASS J22293854-2050136, WD 2226-210.

The most recent (and probably most accurate) distance measurement is from the Gaia DR2 parallax survey, giving $201 \pm 4$ pc ($655 \pm 13$ light years).


Client-side JavaScript astronomy libraries which are not based on node.js? Are there any client-side JavaScript astronomy libraries? Almost every library I see appears to be server side for node.js. I realize some node.js libraries can be adapted for client-side use, so I'd also be happy with a server-side library that happens to work client-side as well.
EDIT: I did find https://github.com/mourner/suncalc but it appears to be abandoned, has 35+ open issues, many related to accuracy, and, despite its description, appears to give only solar and lunar positions, not planetary positions.

I found "Astronomy" from "cosinekitty" on Github:
https://github.com/cosinekitty/astronomy
It supports both node.js and browser.
Browser demos:
https://github.com/cosinekitty/astronomy/tree/master/demo/browser
Testing the library without downloading whole repository:
Save this file locally:
https://github.com/cosinekitty/astronomy/blob/master/demo/browser/positions.html
Into the file, replace line
<script src="astronomy.browser.js"></script>

by
<script src="https://cdn.jsdelivr.net/gh/cosinekitty/astronomy@1.0.1/source/js/astronomy.js"></script>

Reference: https://cdn.jsdelivr.net/gh/cosinekitty/astronomy/
Change this line:
let observer = new Astronomy.Observer(latitude, longitude, elevation);

to
let observer = new Astronomy.MakeObserver(latitude, longitude, elevation);

Remove line:
 <link rel="stylesheet" href="astro_demo.css" />

You should get the table in the file filled up with data calculated by the library.

Another one is astronomy-bundle-js:
https://cdn.jsdelivr.net/gh/andrmoel/astronomy-bundle-js@6.0.4/src


Is Io a magic energy machine? Io gets its volcanic heat from being flexed by Jupiter's gravity, and this seems to go on forever.  That sounds like a magic energy source, since Jovian gravity is perpetual and Io seems to remain in orbit perpetually.  Have we discovered magic?

The energy driving Io's volcanism is derived from its orbital energy and (to a lesser extent by the orbital energy of the other moons and the energy of Jupiter's rotation).  Similar mechanisms power the cryo-volcanism on Enceladus and the other icy moons.  The available orbital energy reservoir is huge compared to that required to drive the volcanic activity.

This does imply that the gravitational interactions on all the moons in the system are changing their orbits and rotations, albeit slowly.  Given unlimited time, the orbit of Io might decay to a point where it falls within Jupiter's Roche limit and is torn apart.

(Note: the answer to How can tidal heating lower Io's orbit describes this more precisely, and points out that tidal forces are transferring energy from Jupiter's rotation to the orbits of its moons, so Io's orbit in the long term gets larger (to the point of it leaving Jupiter orbit) rather than decaying.)


What’s the closest earth like planet we have discovered in our galaxy? What’s the closest earth like planet or moon to earth we have discovered in our galaxy?

Well, this really does depend on what is considered an Earth-like planet. For instance, one could conclude that Venus is an Earth-like planet based off of them being similar in mass and also in regards to the similarities in radius. As for Earth-like moon, Titan would be a great candidate if located inside the so-called Goldilocks zone.

Counting planets outside our solar system, the nearest planet I could find was about 4.243 light-years away. Astronomers found it on August 24, 2016. It is a rocky planet in the habitable zone of its parent star Proxima Centauri, the closest star to Earth. The name of the planet is Proxima b, this planet is approximately  1.3 times the mass of Earth and possesses an orbital period of roughly 11.2 Earth days.

However, since Proxima Centauri is classified as a red dwarf this casts reasonable doubts on whether Proxima b is habitable. To be habitable Proxima b would have to deal with low stellar flux, which causes a high probability of tidal locking, and high stellar variation.

List of nearest planets

List of possible habitable moons


L2 point ephemeris (celestial mechanics) I'm a master student and I am trying to get the L2 ephemeris for some calculations in my master's project. It was kinda hard to find a file with the L2 ephemeris, but once I found it, I tried the next syntax.

from jplephem.spk import SPK
kernel = SPK.open('L2_de431.bsp')
position = kernel[3,392].compute(2457061.5)

but I'm getting an exception: "Only SPK data types 2 and 3 are supported".

I tried the same syntax with a planetary ephemeris file (de432s.bsp) and it works fine.

Can someone help me with that or knows any other ephemeris file for L2?

L2 file is here: https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/lagrange_point/

Bad news, this type of SPK file has a different sort of interpolation that is not supported by  the jplephem package (Hermite interpolation vs Chebyshev polynomials). You can find this out by doing:

In [1]: print(len(kernel.segments))
1
In [2]: print(kernel.segments[0].describe())
2415020.50..2506696.50  Earth Barycenter (3) -> Unknown Target (392)
  frame=1 data_type=12 source=Sun-EarthMoon L2

which shows that the segment is of data type 12. Looking this up in the NAIF SPK documentation shows that this is Type 12: Hermite Interpolation --- Equal Time Steps; looking in the jplephem code in the spk.Segment._load() routine, shows the code that produces the error you are seeing and the supported types (2 and 3). According to the NAIF SPK docs these are Type 2: Chebyshev (position only) and Type 3: Chebyshev (position and velocity).

I think your options are either:

add support for this type of interpolation to jplephem (the math is at least documented in NAIF docs),
see if SpiceyPy has support for this more unusual type of SPK/BSP,
find an alternative source for an L2 ephemeris.


What is the temperature 55 km beneath the surface of Mars? What is the temperature 55 km (34.18 miles) beneath the surface of Mars?  The reason I ask is that I want to know if it might be habitable for a possible future colony if they could dig that deep (which might be easier in the lower gravity environment).  If the answer to my previous question is correct, this is the depth that the atmospheric pressure would be equivalent to that on Earth. At what depth on Mars would the atmosphere have equal pressure of that on Earth?

If we look at Mars' possible geothermal gradient (see Earth's) which is about 25 °C per km.   Using the low estimate of Mars's gradient to be 1/4 that of Earth's Source, that's a bit over 6° C per km. so 55 km, 330° C.  Added that to Mars' average surface temperature of -55 C, you're talking 275° C or 527° F at 55 km underground, and that's a low estimate.

Estimates of the thickness of Mars' crust are often less than 55 km and if you're talking about digging into mantle or even, half way to the mantle, that's both an unsustainable dig and unlivable temperatures.

Source

  The average thickness of the planet's crust is about 50 km (31 mi),
  with a maximum thickness of 125 km (78 mi)

and

Source

Also, small point, but the 55km estimate assumes uniform temperature.  The simpler pressure calculations don't take into account temperature fluctuation.   With warmer temps and warmer lighter air at the bottom of the dig, which would be inevitable, you'd probably need 60 km, maybe 65 km of depth to achieve 1 atm.  There's also circulation issues when you have cold heavier atmosphere above warm lighter atmosphere and there's drainage/flooding problems for the lowest part of a valley, so to avoid flooding, you might want to keep the surface temperature below freezing, to avoid runoff and silt build-up, or dig even deeper so there's a lake of sorts that catches the melting ground-ice around the dig and the living area is above the lake.

This doesn't mean that deep valley living on mars will never be possible, but it's likely impossible with it's current atmosphere.  With extensive terraforming, tapping gases from underground and/or from it's ice-caps or by crashing comets into Mars (that would take a lot of comets), over time, enough of an atmosphere could be created where deep valley living might work.    A simulated atmospheric pressure can also be created with a bubble-dome of sorts, or one of my favorites, a stretch-wrap around the entire planet, but for now that's way beyond our means.

For the near term, living underground on Mars is probably the way to go.   Dig 50 or 100 feet underground and you'd have the means to create a near vacuum seal, which on Mars, you'd want to do, you could create earth like air pressure and you'd be protected from solar and cosmic rays and access to plenty of underground water.   underground on Mars is likely the way to go, at least, in the near future.   There's no shortages of articles on this.  Here's one 6.

Humans can also live at less than 1 atm.  People have settled as high as 16,700 feet, which is somewhere in the neighborhood of 55% of 1 atm.


Why do different instrumental filters use different magnitude systems (Vega vs AB)? Suppose I wanted to construct the spectral energy distribution (SED) of an object. Further suppose I observe this object through the 9 broad-band filters using the Subaru Suprimecam (just for an example) here: http://www.naoj.org/Observing/Instruments/SCam/sensitivity.html

As the statement below the table of broad-band filters says, the Johnson-Cousins filters (capital letters) are on the Vega magnitude system whereas the SDSS filters (lowercase letters) are on the AB magnitude system. I understand what these two magnitude systems are and how to convert between them.

My question is: what is the point of the two sets of filters being on different magnitude systems, one Vega and the other AB? It's not as if the Johnson-Cousins filter hardware physically has the Vega spectrum built into it, right...?

If I do aperture photometry on an I-band image and an i'-band image, will the first one return Vega magnitude of the object and the second return AB magnitude of the object, by default? And if I wanted to do SED fitting, then I need to be consistent such that all of my data points (even from other telescopes) need to be on the same magnitude system (e.g., convert the I-band Vega magnitude into an I-band AB magnitude)? (Or do that conversion first, and only then convert to flux units via $\nu F_{\nu}$ or $\lambda F_{\lambda}$).

I'd guess it's largely a matter of historical accident and inertia, mixed with a certain amount of taste.

Vega magnitudes are the traditional system, based on how bright the star Vega is within the filter (+ detector response, etc.). The Johnson and Cousins filter systems were defined back in the early 1950s when Vega magnitudes were the default, and were to some extent meant to reproduce earlier photographic magnitudes. The AB system was proposed a bit later (1960s) and was originally used for narrow-band photometry and spectrophotometry, not broad-band photometry like the Johnson-Cousins and similar filters. Some more modern filter systems -- such as the SDSS filter system -- have been defined using AB instead, because the creators preferred it.

Some filter systems have both definitions. For example, the near/mid-IR filters of the IRAC camera on the Spitzer Space Telescope have both Vega and AB definitions; some papers use one, some the other. (Sometimes papers that are part of the same overall project.)

If I do aperture photometry on an I-band image and an i'-band image, will the first one return Vega magnitude of the object and the second return AB magnitude of the object, by default?

By default, no. Doing aperture photometry implies choosing a calibration: that is, a conversion between observed counts and magnitudes. (Such a calibration might already be provided in an image header, e.g., from SDSS.) It's certainly possible to do aperture photometry on an I-band image calibrated to i-band, and vice-versa. I have on occasion done things like aperture photometry on R-band images calibrated to K-band, though I certainly wasn't expecting things to be very accurate.

if I wanted to do SED fitting ...

I think the answer depends on what form your model SED is in. If for some reason your SED is in AB magnitudes, then, yes, you'd want to convert all your observed data to AB magnitudes. If, on the other hand, your model is in flux units, then you'd want to convert your data to flux units (either Vega mag to flux units or AB mag to flux units).


What is the farthest reported distance from observer from which a solar system body has occulted a star? I originally wrote "...farthest distance from Earth..." but changed to "observer" in case the occultation was observed from a space telescope. It may not matter much but I didn't want to over-constrain or burden the answer.

The occultation of a roughly 13th magnitude star by a roughly 40 km diameter object in the Kuiper Belt; 2014 MU69 has been successfully observed! I'm hoping to be able to look up the star's name direct. On 2017-Jun-03 03:07 UTC it was about 43.294 AU from the Sun, and 42.435 AU from Earth, (using the JPL Horizons ephemeris.)

I'm wondering if this is the farthest distance from which a solar system body would have been seen to occult a star?

Wikipedia lists some of the most farthest objects from the sun in the solar system. The third farthest object "Eris" was the subject of one of the most distant stellar occultations observed from Earth. From this occultation, its diameter, density and albedo was calculated.
On 6 November 2010, Eris occulted a faint star (V $\sim$ 17.1) which was observed from Chile. The following data was recorded from the occultation:

Radius, RE = 1163 $\pm$ 6 km
Density, $\rho$ = 2.5 $\pm$ 0.05 g/cm3
Albedo, $p_V$ = $0.96 ^{+0.09} _{-0.04}$
Distance, ~98 AU

Eris is now near its aphelion (at 97.6 AU), and is the most remote body presently observed in the solar system. With an orbital period of almost 560 years, it will take more than 240 years from now to reach its perihelion, at 37.8 AU.

Reference:

Sicardy, B., Ortiz, J., Assafin, M. et al. A Pluto-like radius and a
high albedo for the dwarf planet Eris from an occultation. Nature
478, 493–496 (2011). https://doi.org/10.1038/nature10550
(PDF)


Is it possible to calculate the focal length of a refactor I built? I am trying to build a very low cost refractor telescope. It's for a project, so I need to be able to provide "data" about the telescope. But it seems I can't calculate anything (magnification, to be more specific) without knowing the focal length of my telescope. However, because I built it myself, I do not know what my focal length is. How can I find that out, or how can I calculate the magnification?

On another note, what is the best combination for the best magnification? I am not looking for anything great, just the best magnification possible.

I can have a 50mm lens with a fl of 50cm, 30cm, and 20cm and a 38mm lens with each of the same focal lengths. Which is the best pair?

The focal length of a telescope is the focal length of the lens at the front of the scope. In your case it's either 50, 30, or 20 cm.

To magnify the image formed by one of those objective lenses, you need an eyepiece. None of the lenses you listed will work as an eyepiece unless you want a magnification of 2.5 (50/20).

Eye pieces generally have a focal length of 25 to 30 mm or less. To get a quality image, you should buy a used eyepiece.


Is there a comfortable part of Venus's upper atmosphere? I know the surface of Venus is very hot, but that it gets cooler the higher you go in the atmosphere.  I have heard that there is a section in the atmosphere of Venus that might be somewhat habitable even, that where it is about 1 atmosphere, the temperature is slightly over 100 F.  Not really comfortable, but possibly livable temperature wise (likely with some cooling mechanism).  However, I recently heard that near the poles, the atmosphere is extremely cold: http://www.iflscience.com/space/death-plunge-venus-spacecraft-reveals-hottest-planet-not-so-hot/

Does this mean that at the height where the atmospheric pressure is 1 atmosphere, and somewhere between the pole and the equator would lie a comfortable zone of 72 F?  "Extra credit" for suggesting / providing reasoning for a likely latitude where this condition would exist (if it does).  If this does exist, it might make an ideal place for a blimp like sky colony.  I am considering adding this to my Planetary Settlers video game.

There is a layer of Venus's atmosphere that has bright white clouds, a white sky, no acid rain, 1 atm pressure, a warm tropical climate and 0.9 gravities...

It is the cloud layers between 50 and 70km altitude above the ground.


Can someone in a deep well see the stars in daytime? I am wondering whether it is possible for a guy in a deep well to see the stars in the daytime.

Now the background is dark. So might the stars be visible?

No, because being in a deep well doesn't make the background dark.

The reason we can't see stars in the daytime is the atmosphere, which scatters sunlight in all directions such that along every sightline from the ground toward the sky you receive much more photons from the Sun than from a star.

That is, the photons that makes it impossible to see Vega, for instance, are the photons from the Sun that make their last scattering event at a molecule lying between you and Vega. Being in a well blocks the star- and sunlight from other directions, but lets through both star- and sunlight from Vega's direction.

Like this:


What are the positions of the Earth, Moon, and Sun at a time when there is the least difference between high and low tides? Does this happen when there is a full moon?

'Neap' tides when there is the least difference between high and low tide, generally occur around the time of the first and third quarter moons.
'Spring' tides when there is the largest difference between high and low tide, generally occur around the time of the New and Full moons.
The exact timing is complicated by your position on earth, the shape / size and depth of the ocean, coastlines also the moons distance from the earth and earths distance from the sun.


In a binary star system, what relation determines the eccentricity of the three orbits (for $m_1$, $m_2$, and the reduced mass)? In a binary star system, why does $$e_1 = e_2 = e$$

where $e_1$,$e_2$, and $e$ are the eccentricities of the three orbits of $m_1$,$m_2$, and the reduced mass, respectively.

Isn't it just conservation of momentum? Without any external forces, the centre of mass of the binary system must stay in the same place.

For example, imagine that you had a binary system, where one star had a circular orbit and the other, with equal mass, was eccentric. Clearly, as the orbits proceeded, the centre of mass, which is half way between the stars, would have a position which oscillates with time. But with no external forces applied to the binary system, this is forbidden by the conservation of linear momentum.


Where do stellar classes names come from? Under Morgan-Keenan classification stellar classes are called O, B, A, F, G, K, M, but what is the logic behind these letters? I can't see any pattern and I doubt they have just been chosen randomly. I also wasn't able to find an answer on the Internet, so is there any reason stellar classes are called this way?

It seems like these letters are inherited from Draper classification system, which used letters to subdivide classes defined by Secchi. Secchi classification uses numbers I to V, while Draper did use letters in alphabet in order. Some of the letters got dropped, and classes reordered later to match physical features (temperature / spectrum) better.

I believe Secchi and Draper added categories as new types of stars were discovered (eg carbon stars and Wolff-Rayet stars seem to be added later), and thus the categorization did not remain as logical as it could've been had it been designed with full knowledge from scratch. Modern system wanted to retain (some of the) existing labels but order them based on some physical feature.

Wikipedia article seems to have pretty decent details on this too: https://en.wikipedia.org/wiki/Stellar_classification


Accessing SDSS server SQL server via python Is there a way to directly access any of the Sloan Digital Sky Survey servers via SQL via Python?
Currently I have been able to run SQL queries from the website they provide (for example https://skyserver.sdss.org/dr12/en/tools/search/sql.aspx), but it's not clear to me how I can access them directly from a Python interface.
Any suggestions what connection details I would need to use are highly appreciated.

You can do it via the astroquery SDSS module; there is a function called query_sql.


How galactic density falls using Oort constants I am given the observations that Oort constants $A$ and $B$ are, respectively: $14.5 \text{km s}^{-1} \text{kpc}^{-1}$ and $-12 \text{km s}^{-1} \text{kpc}^{-1}$. From these, I am supposed to conclude that the galaxy density falls off as $\sim R^{-2}$. (Source of problem: Paper 2 Question 7 of 2016, Cambridge Astrophysics Tripos - see page 17.)
I don't see how I can pull this off, would appreciate any help.

In the earlier part of the question, I proved that at the solar location, it is true that
$$\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}+2\left(A^{2}-B^{2}\right)$$
which might help somehow, but I don't see how.

From the definitions of the Oort constants we know that
$$A + B = -\frac{dv}{dr} $$
at the solar radius.
But $A + B$ is small, implying that $dv/dr \simeq 0^{\dagger}$ and we are dealing with a flat(ish) rotation curve.
Assuming a spherically symmetric mass distribution with a density that is proportional to $r^\alpha$, we can write the centripetal acceleration at a radius $R$
$$\frac{v^2}{R} = \int^R_0 \frac{4\pi G r^2 \rho(r)}{R^2} \ dr\ . $$
Then, if $v$ is constant, we can just deal in proportionalities:
$$ R^{-1} \propto R^{-2}\int^R_0 r^2 r^\alpha \ dr\ , $$
$$ R^{-1} \propto R^{-2} R^{3+\alpha}\ ,$$
and so $\alpha =-2$.
Another variant is to tell you that $A=0$ in some galaxy. This means $dv/dr = v_0/r_0$, which means $v \propto r$ (i.e. solid body rotation). This in turn means $\alpha =0$.
$\dagger$ by small, what we mean is that the velocity gradient is small compared with $v_0/R_0$ where $v_0$ is the rotation speed where the Oort constants are defined. Since $A-B = v_0/R_0$ this means that
$$ |A+B| \ll |A-B| $$
which in this case is 2.5 vs 26.5 km/s per kpc.


Why the more-luminous star has larger surface area when two stars have the same temperature? If two stars have the same temperature but one is more luminous than
  the other, the more-luminous star must have a larger surface area, and
  therefore a larger radius than the dimmer star.

The light density of the two stars is obviously different. Must the brighter star have more lights?

I'd say your problem is that you are imagining the light rays emanating out from the center of the star.  Try it again with light emanating from each square area on the surface, and you will see that being at the same temperature means each square on the surface sends out the same number of rays, but there are just more such squares.  Then you will see a lot more light rays coming from the larger star, in proportion to its surface area.


What would be the outcome for life in our galaxy if the merger of the Milky Way and Andromeda creates a Quasar? If predictions are correct, the Milky Way and Andromeda are set to collide in around 4 billion years. If, when this occurs, a Quasar is formed by matter being accreted to the common galactic center (perhaps with the two supermassive black holes forming a binary), what would be the outcome for life in the new galaxy?

It's known that galactic collisions were much more common in the early universe; most Quasars are observed at high redshifts. It isn't clear if there are other conditions necessary to produce a quasar other than just the galactic collision itself, so obviously this is just speculation, but it is clear that Quasars are some of the most energetic systems in the Universe; if one were ignited so close to home, would every trace of life in the galaxy go out in a blaze of glory?

Quasars have very pointed jets perpendicular to the rotational axis of the accretion disc around the central black hole. As long as this jet doesn't happen to point into the galactic plane, the cosmic radiation of these jets won't be a too serious problem.

But a collision of two galaxies can become an eventual problem in a different way. A merger of galaxies triggers star formation and sweeps most of the galaxy free of interstellar dust and gas. So, after a short (hundreds of millions of years) period of time, we'll get lots of new stars. But they'll eventually exceed their main sequence lifetime. Due to lack of interstellar matter only very few new stars will form afterwards. That's a difficult settings for formation of planets and life.


How can the life time of a multiple star system, such as for example the trinary system PSR J0337+1715 be derived? As for example explained at the beginning of this blog post, the trinary system consists of a millisecond pulsar ($1.438$ times the mass of the sun) orbited by two white dwarfs. One of the white dwarfs ( $0.198$ solar masses) is very close to the pulsar and has an orbit period of $1.6$ d, whereas the other one ($0.410$ solar masses) is farther away and and needs about a year ($327$ d) to orbit the central pulsar.

Such a three-body system is in principle expected to show a chaotic behavior sooner or later, which means that collisions between these three celestial bodies can be expected and a finit life time of the system can be assumed.

Making some in my opinion way too hand-waving arguments, the blog post further explains that the collisions can not be expected too soon however, by taking into account that the distant white dwarf "sees" the inner white dwarf and the pulsar as a single central body and the relative motion of the inner white dwarf around the pulsar is rather stable and eliptic too.

Thinking about such multiple star systems as chaotic dynamical systems, another approach to estimate the liftime could be to make use of some chaos-theoretic methods which could for example involve the Lyapunov Exponent of the system, such that a large exponent would mean that the collisions happen soon and the stellar system has a rather short life time, whereas the converse would be true if the Lyapunov exponent is small (which is what I would expect for the system in my question).

So in short my question is: how can the liftime of a multiple star system be calculated in a not just hand-waving way?

This question is interestingly related to my issue, but it does not yet answer it...

Such a three-body system is in principle expected to show a chaotic behavior sooner or later. Nope. Hierchical multiple systems (such as this), where the semi-major axes differ by a factor ten or larger may well be stable for ever (never become chaotic), in particular if the eccentricities are low and if the most massive object is in a tight binary.

An unstable three-particle system will eventually result in (typically) the two most massive objects in a tight binary and the third particle ejected (unbound). The time scale for this to happen is of the order of several (10-100) dynamical times and is indeed a highly chaotic process.

The concept of the Lyapunov time scale is not too useful here. One issue is that as soon as one object is ejected (unbound), the system is no longer bounded, when the concept of Lyapunov becomes problematic. Another issue is that the Lyapunov time is defined in the limit of infinite time and does not necessary reflect the behaviour of the system over any finite time.

Finally, to answer your question. I think there is no stringent way. What one can do is to integrate numerically many realisations of the system, each equally in agreement with the data (and their uncertainties). Then one can see whether there are any stable configurations and how frequent they occur. Given that the system didn't form yesterday, it appears likely that it is indeed stable.


Meade ETX-90 with computerized mount slews too far during auto-alignment I've been trying to get my new ETX-90 to properly align itself these past few nights, but without success.  I've mostly been using the Easy align functionality, which has been choosing Arcturus and Altair as target stars.  Arcturus can become obscured by trees later in the night from my viewing location, but I can see it early enough in the night to use it for aligning the telescope.

The problem, as the title suggests, is that when the telescope starts the alignment process, it slews right past Arcturus (and to a lesser extent, above it).  The same thing happens if I do manage to get a fix on Arcturus and continue on to Altair.  If I try, instead, to perform a two star alignment, choosing Altair as the first star, the telescope will slew over 180°.  In both cases, after fixing on both stars Autostar complains that I didn't perform the alignment correctly.

It seems like the gear ratios are incorrect, however I've verified that they are the factory defaults, and have even performed a reset.  I've also tried to calibrate the motors and train the drives, using a landmark about 100m from my viewing location, although I don't own an eyepiece with a reticle.  The time is correct, and the location should be close enough (within a few arc-minutes), so nothing else seems wrong with my setup.

This doesn't seem like a common problem, since none of my other research has turned up similar situations.  My most reasonable guess is that the gear ratios are in fact wrong, which would lead me to believe my new telescope is in fact not new, and has non-standard gears.  I haven't opened it up to verify that, though.

Is there a more plausible explanation and solution to my problem?

It seems unlikely that it would have non standard gears.  On the ETX forum that Bonnie refers to there is a thread about a problem similar to yours - its one of the earliest posts.  The problem turned out be grease on an encoder, which when cleaned off, fixed the problem.


Why do galaxies collide? If the universe is expanding outward, what is the processes for one galaxy to get off track enough to collide with another?

Say, the Andromeda Galaxy and the Milky Way.

The universe is expanding on a large scale. But locally things are always messy.

Locally, galaxies are not set in stone, they move relative to each other, and the directions are random. If they're moving towards each other fast enough, then they will collide.

Also, there's gravity. Some galaxies are bound to each other by gravity, and that will tend to pull them together.

As to why galaxies move at all, relative to each other - well, things in this universe have kinetic energy, and it's distributed randomly. Being distributed randomly, all kinds of scenarios are possible - things running away from each other, zooming past each other, bumping into each other, etc.

It's a messy and random universe, and the order of expansion becomes apparent only on the largest scale.


Reason for disqualifying Pluto as a Planet? Pluto is no longer a member of the planets in the solar system.

Why was it disqualified as a planet?
What is the criteria for classifying an Astronomical Object as a Planet?
Does this criteria apply to all Star Systems or just our Solar System?

Pluto:

It was disqualified as a planet because orbital dominance was not achieved in the case of Pluto. Orbital dominance means that a the planet candidate should have remove all the small bodies from its orbit, by impact, capture of gravitational disturbance.

Planet:

According to the IAU in its resolution B5 (IAU is the International Astronomical Union; in particular it is in charge of naming celestrial bodies), a planet is a celestial body:

in orbit around the Sun
with a sufficient mass so that self-gravity overcome rigid body forces (that assumes a hydrostatic equilibrium)
has cleared its neighbourhood.

Exoplanets:

As you see in the resolution B5, this definition is specified for "Planet in the Solar System". It is mainly because it does not really make sense now that it is already challenging to observe planets, so the distinction between a planet and a dwarf planet is not yet crucial for exoplanets. So the official definition of an exoplanet by IAU makes mainly a difference between a planet and a brown dwarf.

However, I guess the IAU definition of a planet should be kept for the other planetary systems if we are able to detect objects such as dwarf planets.


Why is the Lunar Node in a daily motion? I've read a lot about the Lunar Nodes online (on Wikipedia, astro.com and a bunch of other websites), but there is one thing I cannot figure out and the sources don't seem to clearly explain: why are the nodes moving on a daily basis? As the node is a point of intersection of Moon's orbit and the Ecliptic I understand that this point does not fall in the same place in space every time, but I would expect that once it's passed by the Moon it stays where the crossing occurred and its position gets updated on the next crossing of the Ecliptic, so in ephemeris I would expect to see the same values of the coords between the crossings, but instead I can see they're changing on the daily basis (ie. this ephemeris has 2 columns showing true and mean nodes' positions: http://www.astro.com/swisseph/ae/2000/ae_2016.pdf).
I'm sure the ephemeris cannot be wrong, so I must be missing some details. I thought that finding the Moon's nodes would be as easy as finding the dates/times/longitudes when the Moon's latitude equals 0, but apparently this way I can only find the times and static points of the crossings of the Ecliptic, not the moving nodes as they're shown in ephemeris tables.

The nodes are the two points on the orbit of the moon at which the plane of the orbit crosses the elliptic. The position of the nodes change in a predictable way due to the perturbation by the sun (and the other planets). Since the orbit is constantly changing, the position of the longitude. The node is not the position where the crossing last occurred. Rather it is the intersection of two imaginary lines in the sky, the ecliptic and the orbit of the moon.

The position of the nodes are measured as an angle, clockwise from the first point of Aries, thus it is a longitude, or right ascension. On the first of September 2016 the longitude of the ascending node of the moon is 162.484 degrees, and decreasing slowly.

The linked pdf indicates a value of 12♍41, apparently this means 12 degrees 41 minutes within Virgo. The notation is not used in astronomy. If you are interested in the scientific study of the sky you should be aware that astro.com is an astrology website. The presentation of ephemeris may not represent scientific practice. I consider modern astrology to be pseudoscience and fraudulent.


For any given meteor shower in a given year (e.g. Quadrantids, now) where can I see a rough prediction of how many days it will last? For a given shower the rate hitting the Earth's atmosphere versus time is usually a peak with a width of a few days, though there are some exceptional unicors.

Is there a place where I can view this envelope?

The context here would be to have some idea for a given year of a given shower how much less intense it would be one, two, or three days before or after maximum. I'm not looking for something precise or location dependent, but just to get an idea how critical it is seeing it on the "best" night versus a more opportune night.

Is there some place to go to see some predictions of many of these envelopes in order to look up a given one, or to see which ones are narrow and which ones are wide?

The International Meteor Organization publishes an annual Meteor Shower Calendar.
The PDF document for 2020 predicts a Quadrantid peak on January 4, 08:20 UT, and says:

  The peak is short-lived with an average duration (full width at half-maximum, FWHM – that is the period with ZHR above half of the peak level) of about four hours.

Unfortunately it only gives peak durations for a few showers.


If the Oort cloud has billions of objects larger than 20 km, then why do we have so few comets? According to Wikipedia:

“The Oort cloud may have billions of objects with absolute magnitude

brighter than 11 (corresponding to approximately 20-kilometre diameter)”

Most Oort cloud objects stay in the Oort cloud, and we never see them.

Comets don't last long. A comet is made of a mixture of ices and dust. A comet that enters the inner solar system has one of a few outcomes:

it could pass out of the solar system, and either never be seen again, or enter such a long orbit that it isn't seen again for an extremely long time (like Hale Bopp)
it could fall into the sun (or travel so close that it is completely broken up) (like comet ISON)
it could get caught into a short period and over time have all its ices boiled away by the heat of the sun. (asteroid 2015 TB145. Halley's comet is in the process of losing its ice with each orbit it grows smaller)
it could hit a planet. (Shoemaker-Levy 9)

In all cases, the comet is lost, destroyed or deactivated.

As comets have a short lifespan, they are rare.


Why are accreting objects portrayed with a white flash at the point where the gas stream from the star enters the accretion disk? When I see images of accreting objects, the gas from the donor has a white flash where it meets the accretion disk, like in the image below. Why does this happen if it is true, and why is it there?

When the stream of gas falls towards the disk it gains a significant fraction of the orbital kinetic energy (after all, it is falling from the top of the Roche lobe) which means that it is moving fast and then slows down sharply when it interacts with the disk. This produces a hot spot that in theory could reach $10^8$ K but in practice "merely" is very hot.


What Constellation Is This? (Photo Included) Observed on 9th of Jan 2021 in Newcastle Upon Tyne (UK) looking WSW at a little after Midnight.

If it was rather small, I suppose you were looking at the Pleiades, which is an open star cluster (not a constellation).
Take a look at this web application: https://stellarium-web.org/
You can set a time and location and find constellations and other objects really easily.


How long would a near-light-speed trip to proxima centauri take from the traveler's and the earth-observer's perspectives? Lets say the spaceship is traveling at .9c to Proxima Centauri. How long does this trip take from the observational standpoint of the passenger vs. the observer from Earth?

Proxima Centauri is about 4.25 light years away.

4.25 / 0.9 = 4.72 years, for the observer on earth. See fourth grade math for info on the calculation.

For the person on the ship, he is undergoing time dilation of about 2.29, which results in 4.72 / 2.29 = 2.06 years. See for example https://www.fourmilab.ch/cship/timedial.html for info the calculation of the dilation.


Can Earth be ejected out of the solar system this way? I found this image of what maybe a model of our solar system. Without a credible source I'm not sure I am looking at a star or black hole that comes close to our solar system and how it could throw Earth out of orbit from the sun and/or solar system?

I found it while Googling for how a solar system travels through space and did a picture search with no luck.

Is this possible? And what am I looking at?

It looks like the model is a simulation of what happens if a massive object flies by a very narrow ring system type of thing with a circular orbit.   The ring system setup seems improbable to me, and the massive object doesn't appear to fly in a straight line, so it's not very accurate for two reasons, but the gravitational effect seems accurate.    If a massive body was to fly into the solar system and pass near Earth it could send the Earth out of the solar system or in towards the sun.

Fortunately, bodies that massive almost never fly inside the solar system, so that scenario is unlikely to happen over the lifetime of the sun.


Classification of Active Galaxies I've read that there are 4 distinct although related forms of AGs : Seyferts, Quasars, Radio and Blazars. Are there any other forms possible? Also do these forms ever intermix to give something new?

Yes of course, there are other forms apart from the four stated above. Basically you have just two kinds:

the radio-quiet, and
the radio-loud.

But then again, there are two Seyferts

Seyfert I, and
Seyfert II

Apart from Quasar, Blazar and Radio galaxy, we also have presently,

BL Lac (Named after its prototype BL Lacarte (original member of Blazar type)), and
OVV (Optically Variable Violent Quasar or OVV Quasar; subtype of Blazar)

Best source to begin with is wikipedia.org of course.


Is it possible for a person to not see the new moon at different places on earth? It is known that theoretically we see the same moon phase everywhere on the earth, but practically I see that people are unable to observe the 0.8% waxing crescent at different places on earth. This has some grave consequences as in some cultures like Jewish,Muslim and hindu moon sighting can mark beginnings of festive months. This link claims that different countries have different eid beginnings:

  In India, Muslims will celebrate the festival on Monday or Tuesday, depending on the moon's sighting. In Pakistan, Eid will be celebrated on Tuesday as there are no chances of moon to be sighted on 27 July, The News Tribe reported.
  In the United Kingdom, Eid al-Fitr is likely to be celebrated on 29 July, according to timeanddate.com.
  In the United States, Muslims will mark the beginning of Shawwal on 28 July, the Fiqh Council of North America, an association of Muslims which is affiliated with the Islamic Society of North America (ISNA), stated in a press release.
  Muslims in Philippines will celebrate Eid on Tuesday as the official date of the end of Ramadan has been set as 29 July. The Philippine Government in their proclamation had declared 29 July as holiday throughout the country in observance of Eid al-Fitr.

So the question is , is it really possible for people living in different part of the world to see or not see the new moon (0.8% waxing crescent) to mark their festive seasons at different dates?

In other words do Automated calculations \ Lunar calendars give an exact picture of the moon sighting as viewed by a person on earth.

Note : These cultures put personal observation of the new moon a pre requisite to mark the months hence the motivation of this question.

+taupunkt basically answered this: the thin lunar crescent after New Moon is only visible for a short while after sunset.  While the sun is up, the moon is invisible and it becomes visible when, after sunset, the sky turns dark enough.  However, the moon will set typically 1-2 hours after sunset, and will become invisible some time before that due to atmospheric extinction.  Whether or not, and if so for how long, you can actually see the moon's crescent depends on the local cirumstances: the distance between the sun and moon (which correlates with the lunar phase and depends on your geographical longitude), the angle between the ecliptic and the horizon (depends on your latitude), whether you're in a mountainous area (and if so, whether you're on a mountain top or in a valley), atmospheric humidity, the weather (a few clouds at the western horizon is all it takes), whether you use your naked eye, binoculars or a telescope, your eyesight and experience, etc.

Because of this, it is difficult to predict when you can see the first/last moon crescent.  I use this paper to do that for the Netherlands (Google translation, should be ok since it's a table) and for the naked eye.  One thing that strikes me is that of the 25 cases of a first/last crescent sighting this year, the lunar phase ranges from 1.2-9.2%, with a mean of 4.0%, so 0.8% seems to be rather optimistic.  You may allow the use a telescope though.

The table doesn't list how long the moon is visible for (the prediction of whether the moon is visible or not is sufficiently uncertain by itself), but it will definitely not be more than the time difference between the rise/set times of the sun and moon (~1-2h).  If we use half of the average of those numbers, we get 45 minutes, which may well be optimistic but is not too unrealistic either (for a ~4% crescent).  The other ingredient is how long the lunar phase is at 0.8%.  Assuming that this means between 0.75% (below which it would be rounded down to 0.7%) and 0.85% (above which it would be 0.9%) this was the case for ~97 minutes for the last time this happened (July 27).  Together with the 45-minute observation window, this gives a ~142-minute window of opportunity, or about 10% of a day, indicating that the crescent would be visible from ~10% of the earth.

Remember that we assumed a 4% crescent here, not 0.8%, so this 10% is an overestimation.  My guess is that it is a huge overestimation, since in most places you will not be able to see the moon with the unaided eye at all.  Even if the moon were somehow visible for an instant everywhere, the fraction of the earth the 0.8% crescent is visible from would drop to ~6.7%.  If you want a more precise definition of 0.8%, e.g. 0.80%, or between 0.795% and 0.805%, the moon spends about 10 minutes at the desired phase and this would be visible from 0.7% of the earth.

Then again, you may allow the use of a telecope, so that you can see the moon even during day time.  If you also don't worry about the weather, the window of observation would be 12 hours (since the moon is up for observers at about 50% of the earth's surface) and with the 97-minute time window for which the crescent is between 0.75% and 0.85%, the 0.8% crescent would be visible from about 57% of our planet's surface.  For 0.795%-0.805%, this would be 51% of the earth.

The conclusion is that a 0.8% crescent of the moon is visible from beteen 0% and 60% of the surface of the earth, depending on your exact definitions and your choice of accuracy and optical aid.


Where do rocks come from? Consider a loose pile of rocks in an early solar system, say a tiny asteroid. As it accretes more and more material, the pressure at its center eventually becomes great enough to collapse the gaps between the rocks, which further increases the density, which increases the gravitational force, ... which leads to a runaway process that compacts a loose pile of rubble into a near-spherical body of almost solid rock.  (And some ices and liquids.)

But where do the rocks themselves come from?  Surely at least that much pressure is needed to fuse dust into macroscopic rocks?  Why isn't this a chicken-and-egg problem?

In general, solids are made from from atoms that originate naturally in stellar nucleosynthesis,
see for example the composition of our sun's atmosphere from an older paper of
Asplund et al (2005):

Here you see for example, that silicon (Si), an important rock-forming element, has an abundance relative to hydrogen of $10^{-3}$, or for every 1000 hydrogen atoms there is one silicon atom. Also we find lots of oxygen, carbon, magnesium, iron,... all known rock-forming elements.

The above work has been performed using spectroscopy, and in a similar way we can detect solids in space mainly through three means:

Spectroscopy (again): Small particles of $\mu m$-sizes exist in vast numbers in the spectrae of red giants (I'll comment on the formation later), white Dwarves, in cold molecular clouds and even in far away Quasars. Those numbers are so huge that we can detect prominent features like the 10 $\mu m$-feature of the $Si-O$ bond that we find in many earthly minerals. See also the following example of a clear detection of silicate minerals in the envelope of a young star (the broad trough at 10 $\mu m$) (c)Uni Hawaii:

As mentioned, when solids grow, their numbers diminish. Thus big solids around the sizes of mm become invisible in spectrae, but sometimes can be detected through scattered light, as they undergo Mie-scattering with visible or infrared light when being very near to a stellar source.
Then one can take pictures like the one from Benisty et al. 2015 of the young stellar system MWC758, where we see dust spirals possibly caused by the existence of young planets:

Reflection of solar light: As solids grow to cm, meter, kilometer-sizes they become invisible for the above methods. Only at sizes of tens to hundreds of kilometers and only in our solar system can we then detect them again through the reflection of solar light.

Now that we have established that solids exist in space, let us review the theoretical perspective of growth mechanisms:

dust of up to $\mu m$ condenses out in the relatively cool atmospheres of red giants (AGB stars), where the ionized environment helps to grow solids through charge separation. Details e.e. in here.
this dust is blown away in the strong AGB star winds into the interstellar medium, where it helps to create new star systems (see also the second image I linked).
All successive growth to larger sizes has to happen in the interaction with gas, thus it usually happens in the relatively dense environment around young stars
The dust has now to overcome the bouncing barrier in order to grow into cm to meter-sized particles that we call 'pebbles'. How particles overcome this barrier is unclear in general, but the article I've linked mentions a few solutions.
Pebbles are defined by their property that they interact with the orbiting gas in protoplanetary discs through friction. This interaction allows them to trigger the streaming instability, a very efficient mechanism that is able to compress clumps of pebbles directly into km-sized solids.
Now small comets grow through direct collisions, up to planetary sizes. Only in this stage geological mechanisms can play a role, like the igneous separation of minerals that you've mentioned.

That's it for now. I think with this description it is clear that the whole problem is not chicken-and-egg, but if you think otherwise then please comment.


Hottest Possible Hydrogen-Fusing Stars I guess this is more a question about stellar models than anything else. I was wondering what is predicted to be the hottest possible stars that would still be hydrogen-burning.

What complicates this more is that the most massive stars will probably only sit close to the main sequence (as very early O-type) for several thousand years before gaining a WNh spectrum despite still being relatively early in the lifetime. So I'll be somewhat more precise and ask what is the hottest possible star predicted at ZAMS that won't just blow itself apart by radiation pressure.

Also tangential to the topic but have there been any stars found with an O1 spectrum, and is this even a spectral type that has models made for it?

An answer to your question is contained within What is the largest hydrogen-burning star? The hottest observed main sequence stars are of type O3V, with photospheric temperatures of about 50,000 K.

However, it is indeed possible that hotter main sequence stars may exist in the present-day universe, but have simply evolved into Wolf-Rayet stars (and lost a lot of mass). Indeed, Crowther et al. (2010) claim to see evidence for such objects in the cluster R136 in the Large Magellanic Cloud.

How hot could such objects be? A theoretical study by Bromm et al. (2001) suggests that very massive, metal-rich present-day stars might reach 65,000 K at a few hundred solar masses (see their Fig.1).

The hottest zero age main sequence stars ever would probably be very massive ($\sim 1000M_{\odot}$) ultra-metal poor, or even metal-free population III stars. These actually begin life as He burning stars until they produce enough carbon to commence the CNO H-burning cycle. Equation 6 and Fig.1 in Bromm et al. (2001), who provide a theoretical study of such objects, suggests effective temperatures of about $1.1\times 10^{5}$ K for such stars, with very little detailed dependence on mass and metallicity.

I guess such a star might be classified as O1, but according to Bromm et al. they are basically blackbodies once the mass exceeds about 300 solar masses.


How can pre-main sequence stars radiate more energy than main-sequence stars? How can a pre-main sequence star radiate more energy by gravitational contraction than a main-sequence star can by hydrogen fusion?

Although pre-main sequence stars have lower temperature, they are essentially huge clouds of gas, often as large as 1 pc wide. The luminousity being proportional to square of the radius is essentially large for pre-main sequence stars.

Plus, the problem with gravitational contraction is not the amount of energy that can be generated per second. That can actually be larger than that produced from fusion if you insist on really massive objects. The logic, when we ruled out gravitational contraction as a source of energy for Sun, was that gravitational contraction cannot sustain it for long. Geologists knew from their dating science that Sun had to be older than grav. contraction accounts for.

Thus, admit it. Gravitational contraction makes pre-main sequence stars radiate more energy than main sequence starts, but that's just luminousity and not really energy.
Luminousity is energy per second.