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Can you recommend a book about big bang nucleosynthesis and chemical abundances?
I am interested in learning about big bang nucleosynthesis, nuclear fusion up to iron in stellar cores and beyond iron in supernovas, and into the lithium problem (galactic abundance anomoly for atomic numbers 3,4,5). As these topics are interlinked, I have been hoping to find good resources on the subject. Can anyone recommend a good source for self-study? In case it's relevant (regarding difficulty level), I'm a physics grad student lacking exposure to related topics.

I highly recommend Nucleosynthesis and Chemical Evolution of Galaxies by Bernard Pagel. It contains the basics of nuclear reactions andstellar evolution, chapters on big bang nucleosynthesis and light element production, as well as covering the broad swathe of stellar nucleoynthesis and how these link together into predicting the chemical evolution of galaxies.

Checked out a preview and I like it. Thanks for the suggestion.

My copy of "Principles of Stellar Evolution and Nucleosynthesis" by Donald D. Clayton is from 1968, but it covers what you're interested in and is intended for a graduate-level course.  It's also well-written.

Thanks for the rec, I will look into it. Just a thought, but hasn't our knowledge of nuclear reactions progressed a lot since 1968?

Yes, however most of what the author presents as needing more study turned out to be true.  A newer (1996) and also good & at the right level is David Arnett's "Supernovae and Nucleosynthesis."  I don't have the most modern books because ~$150 each is a bit much when I only read them recreationally.


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Umbra, penumbra and antumbra
I'm trying to understand following schema of the wikipedia page here.

Image subtitle is "Umbra, penumbra, and antumbra, for a lunar eclipse".
Distances and sizes are not to scale.
If distance Earth-Sun is about 150 Gm (gigameters) and Sun diameter is 1.4 Gm, $\arctan 1.4/150 = 0.57º$ (degrees). Thus, the Sun is seen from Earth with an angle of approx. 0.57º (0.01 radians). The Blue and red lines in schema should be near to parallel.
That means the point where umbra and antumbra touch is (being 12 Mm the approx diameter of Earth) at a distance of about $\frac{150}{1.4\times0.012}=1.3 \mathrm{Gm}$ from Earth.
As distance Earth-Moon is only 0.400 Gm, it seems impossible that Moon was never in the antumbra zone of the Earth.
I'm doing a basic error somewhere, but I do not know where.

As you say, it’s not to scale. Not at all. What is your question exactly?

As distance Earth-Moon is only 0.400 Gm, it seems impossible that Moon was never in the antumbra zone of the Earth.

That is correct. You were reading yet another poorly written wikipedia page. The information on that page isn't quite incorrect, but it isn't quite correct, either. There is no such thing as an annular lunar eclipse.
That page would have been much more meaningful had it used the Moon's shadow as the example. The Moon's umbra trails from 368000 km to 381000 km behind the Moon, depending on how close the Moon is to the Sun. This is very similar to the distance between the center of the Moon and the closest point on the surface of the Earth to the Moon, which varies from 362000 km to 400000 km.
When the Moon is directly between the Earth and the Sun, the resulting solar eclipse can be total or annular. When the Earth is directly between the Moon and the Sun, the resulting lunar eclipse is always total.

The calculations seem reasonable. The conclusion is wrong.
The Earths antumbra starts 1.3 Gm from Earth. At less than 1.3 Gm there is umbra, penumbra and non-shadow. The moon is 0.4 Gm from Earth.
$$0.4 < 1.3$$
Hence it is impossible for the moon to be in the Earth's antumbra.
However, the Moon's antumbra forms at almost exactly one lunar distance. We notice that the moon and the sun are almost exactly the same size in the sky. When the moon is further from the Earth, an annular eclipse can form, as a small part of the Earth will then be in the moon's antumbra.
I've made a scale model of the sun-earth-moon shadows, at https://ggbm.at/CbbvmBFq

Thanks for your answer, and very nice model. The scheme subtitle in wiki page is "... for a lunar eclipse" (and position of astros agrees, "Sun-Earth-Moon"). Thus, it seems we are talking about Earth umbra-antumbra and Moon traveling on them. As the second case, Moon in the antumbra of Earth is not possible, I do not understand the meaning of the wiki schema.

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The Sun or Our Galaxy revolves around what?
Like every planet they are in orbit of something or another planet like our solar system every planet revolves around the sun, my question is: what's the name of the planet or object which the sun revolves around; if the sun is stationary or if our galaxy is in motion, around what it is revolving.

The short answer is probably "it's complicated" - although one might say that we're orbiting around the centre of the galaxy. A good question, but as Tbb says it would probably be better asked (and may already be answered) on Astronomy.se.

See also https://astronomy.stackexchange.com/questions/20614/revolution-of-the-sun

As a previous comment states; it is complicated. But in principle the Sun is just one object in a large collection of objects which we call the Milky Way Galaxy. All the objects in this Galaxy orbit the barycentre (or barycenter, if you prefer the American spelling). The barycentre is, essentially, the common centre of mass of the whole galaxy. In the same way the planets in our solar system don't actually orbit the Sun, they orbit the Solar System Barycentre.
As for the Milky Way Galaxy, that too is a member of a larger group of galaxies, and they all orbit their common barycentre.

The Sun, like all other stars in the Milky Way Galaxy, is rotating around the Galactic Center. The Galactic Center is around 25,000 light years away, and it may contain a supermassive black hole. There is a very intense radio source in the center of the galaxy, called Saggitarius A*, which is thought to be the same as the black hole.

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Should we expect clues from a gamma ray burst causing massive extinctions on Earth?
I remember reading an article on the subject highlighting the periodicity of this event throughout history supposing it could be caused by the rotation of the sun around the galaxy.
My question is: Do we have any clues that can confirm this hypothesis? What should we expect to see? Shouldn't it leave any remains?

Do you mean "Gamma ray burst".

I think you're right and that's what I was thinking when using the term 'cosmic rays'.

A gamma ray burst could cause an extinction event. There is no evidence that this has ever happened to the Earth.
Gamma ray bursts are rare events, we don't know exactly how rare, but to cause a mass extinction a GRB would need to be near (6000 light years or less is near in this context) and pointed right at Earth. GRBs are rare enough for it to be possible that this has never happened. It is possible that one or two of the extinctions in the last billion years were due to GRBs
The mass extinction that is most likely to have been caused by a GRB is the Orodvician-Silurian Event, 444 million years ago. There is no positive evidence of a GRB, but it can't be excluded. The best that can be said is that the pattern of extinction is consistent with a GRB.
A gamma ray burst would leave few signs that could be used to prove it occurred. The radiation would strip the ozone layer, leaving the surface vulnerable to UV radiation. But it wouldn't leave a crater, or a geological layer.
Gamma ray bursts can occur when a hypernova occurs. There is no reason to think that the Earth is more vulnerable to a hypernova when it is in the galatic plane. The evidence of a 26 million year periodicity to mass extinctions is weak, and there is no evidence that closely ties this to the sun's orbit around the milky way. There is good reason to tie other extinctions to particular impacts, for example there is no need to hypothesise that a GRB occurred 65 million years ago, at the K-T mass extinction
Gamma ray bursts are one of the less likely causes of a mass extinction. Researcher David Thompson likens them to the danger of finding a polar bear in his closet. It could happen; it would be dangerous, but the chance of it happening is so low that it is not a risk worth worrying about.


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How accurate are NASA's eclipse predictions?
NASA's predictions for the August 2017 solar eclipse show that about half of Kansas City will see a total eclipse and the other half won't (they'll just see a "near" total eclipse, perhaps even Bailey's Beads?).
I thought this would be a great way of testing how accurate NASA's eclipse path prediction is: have people in Kansas City report, with location, whether they saw a total eclipse or not (of course, many people in Kansas City will probably head into the total eclipse zone just to see it, but hopefully few will remain outside it).
It then occurred to me that someone must've already done this for previous eclipses (I know it's done for lunar occultations of other stars).
However, googling only tells me that NASA's 2017 path predictions are more accurate than ever before, but not actually HOW accurate they are.
Where can I find data about accuracy of eclipse predictions?

I'd expect NASA is pretty darn accurate. They're very good at their job. Besides, how can you be sure people can be accurate enough to verify NASA? I'm sure NASA can specify the border of where a full eclipse occurs to a much much better accuracy than an individual can distinguish between a full eclipse or a 99.99% full eclipse.

@zephyr I disagree. The Sun is really really bright. There should be a very noticeable difference between 100% eclipse and 99.999999% eclipse. That's why we have effects like Bailey's Beads

Can you be confident the average person would know that difference though? I'm certainly not.

You're confusing accuracy and precision. You see that the predictions are precise, but you want to know if they're accurate.

@zephyr Even if only one billionth of the Sun is showing, it's brighter than Sirius. I'm thinking this is do-able... and, more importantly, it's been done.

Where can I find data about accuracy of [NASA's] eclipse predictions?  Have you considered asking NASA ?  They'd certainly know.

You haven't even defined the accuracy you're after.  Further,  "the sun is really really bright"  indicates a lack of understanding of the spatial distribution of the radiance, not to mention the effect of the atmosphere on those photons which manage to miss the moon.

The accuracy here would be in meters or something (the accuracy of the edge of the predicted path). Why would photons that miss the moon be different from any other photons arriving on Earth. I can't imagine the moon's gravitational attraction is enough to make much of a change.

Solar eclipse predictions are really good.  You aren't going to be able to discern error based on crowd-sourcing individual reports-- your uncertainty will be much greater then the error in the prediction.  But something like the method you suggest is used to check things that are harder to get right, like asteroid stellar occultations.

The path of totality, which eclipse maps show to be 70 miles wide, is actually narrower than that, by up to 1 mile, according to some experts.
An article by the Kansas City Star discusses this:

Those maps, provided by NASA and others, show a crisply defined, 70-mile-wide path of totality where the moon will block 100 percent of the sun. But they are not as precise as they appear, at least on their edges.
The southern edge of the path as shown on the maps could be off by as little as the length of a football field or as much as a half-mile, eclipse mapping experts say. Likewise for the northern edge, meaning the path of totality might be just 69 miles wide.
“This is an issue. This is really an issue, but it’s not advertised. … Yeah, all the maps are wrong,” Mike Kentrianakis, who is the solar eclipse project manager for the American Astronomical Society and who routinely consults with NASA, told The Star.
Xavier Jubier, a French engineer whose calculations have been used to create the interactive Google maps of the eclipse, confirmed to The Star by email that the actual path of the totality is slightly narrower than the 70 miles shown on current maps.
Ernest Wright, who created maps and other multimedia presentations on the eclipse for NASA, said he thought the map might be narrower by about 100 meters, slightly longer than a football field. He also said it’s possible that Kentrianakis is correct in his estimation that the path is narrower by a half-mile or more.

The article explains the reason the path of totality is narrower than the eclipse maps indicate is because they use a 41-year old value for the radius of the Sun, which is now known to be too small a value:

Wright explained that eclipse maps are made based on what is known about the relative sizes and positions of the moon and the sun. “We have really good information about the orbit of the moon, the positions of the sun, the positions of the Earth. All of that is really well nailed down,” Wright said. “In order to get more accuracy, we need to take into account the mountains and valleys on the moon, and the elevations on the Earth. And we’re starting to do that, as well.”
The size of the moon, in fact, has been measured to within a meter, and its position in the heavens has been measured to within a centimeter. “But the last sort of uncertainty might surprise you,” Wright said. “It’s the size of the sun.”
Jubier said that the current maps are accurate using the the 696,000-kilometer radius and other standards agreed upon in 1976 at a meeting of the International Astronomical Union. “This is perfectly accurate but we know it does use a solar diameter that is not large enough. Why don’t we change the value(?)” Jubier wrote. “Well simply because the IAU (International Astronomical Union) has not yet approved a new value. This is part of the research we’re doing and for which we’re looking for funding.”
He continued, “So technically speaking if the Sun is larger than the adopted IAU value, and we know it is, the eclipse path is necessarily narrower and our tools can simulate this, yet the standard maps for the public will still retain the currently adopted solar radius until a new value has been accepted."

Thanks! https://svs.gsfc.nasa.gov/4518 notes that the most precise maps they have "are high-resolution shapes with roughly 100-meter precision", which matches nicely with what you found. If you download data from HORIZONS for the Sun, it says both "Radius (photosphere)  = 6.963(10^5) km" AND "Solar Radius (IAU)    = 6.955(10^5) km", two slightly different values.


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Moon, center of mass & geometric center, tidal locking
Since the moon is tidally locked to earth, what about center of mass and geometric center? How far from each other are they? Can a celestial body be tidally locked to an other one if its mass distribution is perfectly homogeneous (center of mass at the very same place of geometric center)

Since the moon is tidally locked to earth, what about center of mass and geometric center? How far from each other are they?

They differ by about two kilometers.

Can a celestial body be tidally locked to an other one if its mass distribution is perfectly homogeneous (center of mass at the very same place of geometric center)?

Yes, but that condition is not what I would call "perfectly homogenous". What's needed is a non-spherical mass distribution. For example, a uniform density ellipsoidal body would have its center of mass and center of figure of coinciding, but would still be subject to torques. It's the moment of inertia that is key rather than the offset between center of mass and center of figure. An object whose center of mass and center of figure do not coincide necessarily has a non-spherical moment of inertia tensor.

To add to this, it's the tidal bulge, not the moon's gravitational irregularity that lead to the tidal locking.   A gravitational irregularity rotates around in circles.  A tidal bulge stays in one place ahead of or behind the straight line between the Planet and Moon.   That's not to say that the gravitational lumpiness has no effect.  It has some.  But the tidal bulge has a much greater effect in most tidal locking situations.

@userLTK -- If you are writing about the Earth's tidal bulge, this (a) would have zero impact on a spherically symmetric Moon and (b) doesn't exist. The Moon, however, does have a frozen tidal bulge.

I was talking about how the tidal bulge in general is the feature that leads to tidal locking, not gravitational asymmetry.    I thought that was relevant to the homogeneous question.

@DavidHammen I don't believe that tides don't exist for a second. There are multiple issues I have with the answer you linked. Besides, if you just look around the web, you'll see extremely varying answers and opinions about tides so its a really hard thing to get a straight, correct answer on. Kind of like the question about what makes planes fly.

@zephyr - I never wrote tides don't exist. I wrote that the tidal bulge doesn't exist. Laplace wrote pretty much the same thing when he came up with his dynamic theory of the tides.

Okay, tidal bulge instead of tides then. My claim still stands. I don't believe that it doesn't exist.

@zephyr - Humanity has been measuring the tides for centuries, including recently with extremely accurate orbiting radar altimeters. They don't see those bulges. They instead see Laplace's amphidromic systems. Newton's tidal bulges are a bit of a lie-to-children. While the concept explains the tidal forcing function nicely, it does not explain what happens on a rotating planet partially covered with a thin veneer of water. Laplace's theory does, and it doesn't have a bulge.

@DavidHammen Now I have to call you out. Are you talking about tides or tidal bulges! Tidal bulges 100% exist. That's just plain math and application of physics. I can agree that the tidal bulge may not always correspond to tides in the ocean for a variety of reasons like ocean currents, topology, etc. but the tidal bulge most certainly does exist. How could you explain the Moon moving away from the Earth, for example, if the tidal bulge did not exist? Furthermore, your linked question stating tides can't travel through water fast enough is faulty for the reasons one comment pointed out.

@zephyr - Yes, tidal bulges exist. The Moon has them. It has a mostly frozen set of bulge that wobble around a bit due to the eccentricity of the Moon's orbit. the solid Earth has them, but diminished because of elasticity in the solid Earth tides. These smallish tidal bulges and their time variations are hard to see, but they're their. The oceans on the other hand do not have the bulges you were taught about in high school or college. They cannot exist and they are not observed.

Let us continue this discussion in chat.


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What are reliable resources for weather and all other atmospheric obstructions?
Everyone knows you need clear skies to see what the stars or the Sun is doing. Weather is not too particularly hard to determine. There are many resources out there that will tell us the weather in any city. However, they very often lack other atmospheric obstructions. Naturally, light pollution is one of the biggest concerns. But another major concern is erratic atmospheric changes due to things like forest fires.
I'm looking for good resources that will tell me not only weather, but also light pollution and other obstructions such as smoke from forest fires.
(There's a solar eclipse coming up and I'm driving 500 miles to see it.)

Weather is not too particularly hard to determine You need to be made aware that forecasting weather accurately is exceptionally difficult, even on short time scales.  Don't underestimate this.

If you live in San Diego,  writing down "warm and sunny" is a sufficient weather forecast.  If you live in New England, don't trust any forecast more than one hour into the future.

@Stephen I mean so say there's plenty of resources.

You might try https://clearoutside.com/forecast/50.7/-3.52  It gives weather and visibility among other things.
They have an app for iPhone and android. Advertises site was developed by astronomers.


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Counter weight rod
Someone gave me an Orion StarBlast 4.5 EQ, and with it everything but the counterweight shaft. Because of this, the telescope is hard to use. It flops around quite a bit making viewing of objects difficult. Where can I buy just that part? I looked it up and all I could find is an "upgrade" to the rod for $70.

"what does the rod to gold metal mean?" Have you contacted customer services at orion telescopes? What did they say?

I have that same telescope in my collection! It served me well and I hope it will do the same for you.
This counterweight is not the one which comes with the telescope, but it should function much the same$-$it is $30.00, and you can find it here.
Clear skies!


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Who first observed another planet rotating?
We knew the earth rotates, but the moon is tidally locked.
To tell that another planet is rotating, you have to be able to distinguish its surface (I guess?). Who first did this?

Good question!!

Well, ... the moon is rotating, at one rotation per revolution.  But that's a whole 'nother flame war.

Mercury's surface was virtually unknown until Mariner 10 flew by and Venus is covered in clouds. Of the terrestrial planets, that leaves Mars.
It was likely Christiaan Huygens in November 1659 who first realised Mars was rotating when he tried to map it. His estimate of roughly 24 hours was fairly close.
I believe Galileo's earlier observations on Jupiter were with a telescope not powerful enough to resolve smaller 'surface' features (like the Great Red Spot), which may have either proven that Jupiter rotates, or at least that something on the surface is moving across it.

Nice answer! It might be better to link to both the Wikipedia article and the paper The Mapping of Mars separately, rather than linking to the footnote. (PDF also available). Wikipedia articles are constantly changing so a link to a footnote could disappear at any moment.

Wikipedia supports your view on Galileo, suggesting that the first sighting was either by Hooke in 1664 or Cassini in 1665 - https://en.wikipedia.org/wiki/Great_Red_Spot . Galileo died in 1642.

True, I imagine the Great Red Spot was identified before anything else of note (apart from the cloud bands), which was after Galileo's time. Interestingly, Galileo was among the first to infer that the Sun was rotating, by watching the movements of sunspots.


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Why doesn't the Sun explode?
I know stars explode because of the fuel causing a fusion compound pushing it apart and the fuel runs out and it 'bounces' for lack of a better term.
Given the fact that more massive stars are supposed to explode more quickly: why doesn't the Sun explode considering its massive size?

The sun isn't very massive as far as stars go. Astronomers consider a star to be "massive" when it has a mass of at least 5 times our own Sun, but many of the more massive stars can easily reach up to 100 solar masses or more.

Even the most massive stars we know burns for millions of years so it wouldn't explode on a human timescale.

it is exploding even right now so what's the matter?

@user6760  "exploding" ..  You keep using that word. I do not think it means what you think it means.

A star is just a sustained nuclear explosion.

@A.C.A.C. It really isn't a sustained nuclear explosion. The energy release in the Sun's core is 250 W per cubic metre.

@Sans Fusion inside the sun does not end in a bang.

Possible duplicate of The reason why the Sun doesn't explode?

Our sun is a particularly average sized star on the main sequence. It is not going to ever go "supernova" but instead will slowly swell and darken towards red, eventually swallowing Mercury and Venus.

(from http://www.oswego.edu/)
Very boring in the grand scheme of things. Which is good for us :-)

ack! yours is better than mine.

Can you say approximately when mercury and Venus will be swallowed?

About 10 billion years from next Tuesday. Approximately.

There are two things to discuss here: (a) why the Sun does not explode; and (b) why the Sun will not explode.
(a) An explosion occurs when the timescale for the energy release by some process is much shorter than the timescale on which a system can adjust to damp the energy release process. In the present day Sun, nuclear fusion is a very slow process: on average it takes many billion years for a proton to fuse with another. This timescale is quite temperature dependent, so you might have thought the centre of the Sun might heat up quickly, leading to a runaway "explosion". However, an increase in temperature leads to an increase in pressure that would expand the Sun, reducing the core density and temperature and decreasing the rate of nuclear fusion again. The timescale for the Sun to react in this way is just millions of years, so this acts like a thermostat that keeps the reactions under control.
(b) Stars more massive than the Sun burn through their hydrogen and other heavier fuels, and end up with an inert iron core from which no further energy can be extracted. The core subsequently collapses and the gravitational potential energy released by the core collapse powers a supernova explosion. The reason that the core does not quietly collapse into a black hole (at least in the supernovae we see!) is that the core "bounces" when neutrons in the core are squeezed very close together. This provides a quantum mechanical effect called degeneracy pressure that resists a complete collapse.
Ironically, it is degeneracy pressure in the solar core that prevents our Sun reaching the supernova stage. After burning through its hydrogen and helium the core of the Sun would consist of carbon and oxygen; but it will never become hot enough to commence fusion of these elements, because electron degeneracy pressure will support it from contracting and getting any hotter. Only more massive stars will ever attain the core temperatures required to ensure progress towards a terminal iron core.

Finally! You might also add in your answer, for the sake of clarity, that the graviational energy you mention in point (b) refers to a bouncing of the outer layers over the dense core. This should help visualize the explosion mechanism for profanes.

Whether a star explodes or not is given by how quickly nuclear fusion happens inside it. If fusion takes place at a steady pace, the star does not explode. If lots of material fuse at once, the star explodes.
Our Sun, like most other stars, just keeps fusing material at a slow and steady pace.
It takes a different kind of star for the explosion to happen. Those stars that do explode are called supernovas, and there are a few different kinds of them. They could simply be much bigger than the Sun, and the conditions within these giant stars eventually trigger an explosion. Or they could be a white dwarf with a companion star nearby that keeps depositing material on the dwarf - when the pile is big enough, an explosion occurs.
None of that applies to our Sun.

Although, if the sun will eventually expand to the point of swallowing Mercury and Venus, their material has the chance of fusing. Probably not enough to explode, but dang, it'd be hot here!

@Sans Not really. If that happens, the planets will simply melt off and vaporize. The heavier the elements, the harder it is for them to fuse, and planets are made of much heavier stuff than stars.

Massive star supernovae do not happen because of rapid nuclear fusion.

Our sun (I'm assuming you're from Earth or at least the solar system) is actually not all that big, compared to other stars. The gravitational pull of the mass of the sun is kept in check by the fusion that this pull provides. Thus the Sun is at exactly the equilibrium of these two forces.
In other words, the Sun doesn't explode because its forces are balanced.
It also won't explode in the future because the mass of the Sun is not enough to trigger a supernova. It will more "swell up and blow away".

If I weren't from this solar system would I be speaking English?...  otherwise I'd probably be trying to communicate in Morse code right now

It won't explode in the future because it's collapse is halted by electron degeneracy pressure.

Someone else asked a similar question that was linked here as a duplicate. It seems that question isn't quite the same, but similar enough that it makes sense to put it here.
That person asked:
I know that the Sun is made up of hydrogen and helium.
Also, hydrogen explodes when it touches a flame.
But why does the Sun not explode?
The kind of explosion that happens when hydrogen touches flame is an exothermic chemical reaction in which hydrogen is rapidly combined with oxygen (oxidation).  When heat is applied (flame) to hydrogen (H2) in the presence of oxygen (O2), it causes a molecule of O2 to separate and each oxygen atom to bind with a molecule of H2.  When this happens, it releases a large amount of energy ( believe this is caused by the breaking of the O2 bonds), and you end up with water (probably steam) and heat (fire).
The kind of "buring" that happens in stars is thermonuclear fission. When two atoms of hydrogen are brought together by immense heat and pressure, they fuse into an atom of helium.  In this process, enormous amounts of energy are released.  This is the same nuclear reaction that takes place in a thermonuclear fusion bomb.  It does not involve a chemical change, but a nuclear one.
Let me take a half-step back on that.  A chemical change or reaction is when the atoms rearrange themselves in molecules.  In most such reactions, energy is either absorbed (endothermic) or released (exothermic).   Sometimes fairly large amounts are released.  Explosives such as gunpowder, TNT, C4, and the like are all chemical explosives and mostly their reactions involve very rapid oxidation of the substance being burned. The burning of a match, candle or campfire is the same type of chemical oxidation reaction, but at a much slower pace.
Nuclear reactions come in two main types: fission and fusion.  Fission is the splitting of larger atoms into smaller ones.  The isotope of uranium with an atomic mass of 235 (92 protons and 143 neutrons) has a half-life of about 700 million years (which is far shorter than the 4.5 billion year half-life of the more common U238).  Atoms naturally decay over time, and in a span of 700 million years, a mass of U235 will naturally split on its own and the number of atoms will reduce by half (hence the term half-life).  When an atom of U235 splits, it releases a bunch of energy, produces an atom of krypton and one of barium, and releases a few extra neutrons. When one of these neutrons meets another atom of U235, it joins that atom, which destabilizes it, and causes the reaction to happen again.  Note: because we are talking about reactions in the nucleus of the atom and resulting changes to the atoms themselves, we are talking about nuclear reactions, not chemical ones.
The reaction I mentioned above takes place in U235 naturally. If you don't have much of it and it's not in a dense enough state, it just sits there like a lump of rock and seems to do nothing.  Most of the neutrons it releases don't find another atom to play with, and end up just flying off into space.  But if you squeeze it down enough or get enough of it together, more neutrons are meeting up with other atoms than are flying off into nowhere.  This creates a chain reaction, which is the basis for nuclear power and nuclear weapons.
A reactor works by controlling the rate of this reaction, keeping it from going too fast.  Lumps of uranium are held just close enough together to react,but not rapidly, and rods of neutron-absorbing material (usually graphite) are slid between the lumps to help absorb the neutrons and prevent them from causing more reactions. A nuclear meltdown occurs when the reaction gets out of control and heats up so much that it literally melts the framework holding it all together.
This isn't enough, however, to create a nuclear explosion.  That occurs when you apply pressure and squeeze the material down, increasing its density.   U235 isn't actually suited for this, but element 94, plutonium, is. When you squeeze plutonium down to a high enough density, this reaction happens extremely rapidly.  First one atom of Pu239 splits, releasing its energy in the form of light and heat along with additional neutrons.  Then those extra neutrons find other atoms of Pu239 and cause them to split.  Every split creates at least twice as many extra neutrons as splitting atoms.  The first atom splits to cause 2 to split, which split and cause 2 each for four, then eight, sixteen, 32, 64, 128, 256, 512, 1024, 2048, 4096... each "generation" of these splits is double the previous. This all happens at a time scale measured in tiny fractions of a second, the time it takes light to go only a few millimeters.  This all builds up tremendous amounts of heat and pressure, which, unconstrained, erupt out of the core of the bomb.
As it turns out, only a small percentage of the original mass is actually lost to fission. The heat and pressure build up so great so fast that it forces the atoms away from each other in the explosion, thus ending the reaction.  In fact, in a fission bomb, before anything other than high-energy x- and gamma rays leaves the bomb case, the reaction is complete, all that's left is for it to expand outward due to the immense pressure.
The other kind of nuclear reaction, fusion, is also used in bombs.  In fusion, as I mentioned above, huge amounts of pressure are used to force atoms of lighter elements, typically hydrogen, together.  Every time two atoms of hydrogen fuse into one, a huge amount of heat and energy are released (to be honest, I've never understood why energy is released here and not absorbed, but I'm not a nuclear physicist, just a guy who likes to read a lot).  In bombs, the 'trigger' for the fusion is actually a fission bomb.  The fission bomb creates the heat and pressure necessary, and when hydrogen atoms are added to the mix (usually in the form of tritium or deuterium), the fusion reaction begins and starts adding its heat and pressure to the pot, which helps sustain the reaction long enough to put out an amount of energy several times that which the fusion bomb started.
If you remember back to your high school chemistry lessons, Boyle's Law states $P_1V_1 = P_2V_2$. P is pressure, V is volume.  This formula shows that pressure and volume are inextricably interconnected.  Charles' law states that $V_1T_2 = V_2T_1$, showing the interrelation between volume and temperature.  Gay-Lussac's law, $P_1T_2 = P_2T_1$ shows the relation between pressure and temperature. These all come together into the combined gas law and Avogadro's law.  They show that if you increase the pressure, volume, or temperature of a mass, the others are affected as well. An increase in pressure results in an increase of temperature.  A decrease of volume creates an increase in pressure which create an increase in temperature.  If you increase the volume, the temperature and pressure decrease.  Etc...
Ok, so, now here's where it all applies to stars.
In the result of the Big Bang, most of the matter created was hydrogen. There was some helium and a little lithium, but the overwhelming majority was hydrogen.
When the hydrogen molecules floating through space started to clump together through gravity, they started to condense down into smaller and smaller volumes.  Separately, the gravity of a single atom of hydrogen is minuscule.  But as more and more of them interacted, the combined gravity began to build up.  The clouds became more and more compact - denser, and their gravity brought in more and more mass, which added their gravity to the total.  Things began to heat up, due to the pressure caused by the volume.  The gravity started to get to be enough to continue to pull things toward the center of mass, into the "gravity well" caused by the cloud.  As the heat increased due the the pressure caused by gravity, it caused outward pressure.  The forces continued to balance each other.  As more mass fell in, gravity increased, causing more heat, which caused more pressure, which pushed outward... and so on.
At a certain point, the gravity gets so great, that it exerts enough pressure on the center of the mass to cause nuclear fusion.  This, in turn, causes more pressure outward, but the force of gravity of the entire mass is great enough to keep it all together.  This is when a star is born.  The pressure of gravity has caused a gravitational collapse to ignite into a star.  The balance of the forces of gravity and heat and pressure are what keep it going.  If there wasn't enough pressure outward, it would collapse further into a singularity (a black hole), if there's too much pressure outward, it would either explode (type II supernova) or at least expand outward until balance, what we call hydrostatic equilibrium, is achieved again.
As a star ages and burns through its supply of hydrogen, the equilibrium changes and at some point it begins to burn helium. After that, carbon. Depending on the original mass, it may fuse other materials up to iron.
In some stars, the mass is enough that when the fusion can no longer be supported, there is no longer enough outward pressure, and the core collapses in on itself.  The gravity increases as it shrinks, and there's not enough heat or pressure to overcome it.  It collapses to a point of infinite density and becomes a gravitational singularity and a black hole is created.  In other stars, the mass isn't quite enough to keep collapsing, and the pressure overcomes it in a massive explosion (type II supernova).  In other stars, the star belches out wave after wave of gas as it expands and contracts, finally belching enough away to no longer have enough reactive mass.  It shrinks down into a white dwarf - a hot ball of atomic particles that slowly cools over billions and billions of years until they are no longer luminous, becoming a black dwarf (the time for this to occur is believed to be so long that there are, as yet, no black dwarfs in the universe... but there someday should be, long, long in the future).
So, to circle back to the original questions... why doesn't the sun explode, and why doesn't the hydrogen in the sun explode... the main reason is gravity. It reaches a balance.  The heat and pressure pushing  are balanced against the gravity pulling in.  There is no chemical oxidation going on, because the atoms of hydrogen are being fused into helium and there's little oxygen to be consumed in the oxidation process to begin with.  Someday we expect the sun will run out of hydrogen to burn.  As it starts burning helium, it will expand out past the orbit of Mars, then at some point it will shed its cooler, outer layers and contract into a white dwarf which will slowly cool down to a dying ember in the ever darkening universe.

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text:
Is the eccentricity of a planets orbit relevant to finding whether the distance between said planet and its host star determines seasons?
The tilt of the Earth is the main factor that determines the seasons. It's also true that the eccentricity of Earth's orbit is very close to zero, implying that aphelion and perihelion distances are relatively close. If the eccentricity of Earth's orbit were close to but less than one, then would the distance between the Earth and the Sun become a dominant factor in determining the seasons?

Related question here with some good answers:  https://astronomy.stackexchange.com/questions/1184/why-dont-we-have-2-summers-and-2-winters?rq=1  And you wouldn't need to push the eccentricity close to 1 at all.   An eccentricity of 0.2 would give 2.25 times the solar energy at Perihelion than Aphelion which should be enough to overtake the tilt and govern the seasons.    An eccentricity of 0.3 would certainly overtake the tilt with the possible exception of perhaps the very Northern and Southern tips of the planet where the sun would still stay out for months then set for months.

Interesting question, but perhaps better worded as "is eccentricity or axial tilt the main factor in total irradiance?"    Then consider that any planet sans atmosphere doesn't care at all about axial tilt.

Yes it would. If the sun was much further away the season on earth would certainly be more winter-like. The reason distance is not the deciding factor for earth is the variation in distance is fairly small.


text:
Binaries consisting of a black hole and a non-black hole?
Are there any known binary systems in which one member is a black hole and the other is not a black hole (main sequence star, giant, neutron star, white dwarf, whatever)?
Googling seems to turn up only discussion of binaries in which both components are black holes. Is there some astrophysical reason why the kind of system I'm talking about cannot form? I think I remember that gravitational wave observatories have talked about black hole-neutron star mergers, but are such systems purely hypothetical so far?
I guess Sagittarius A* is sort of close to being an example, if you count it and the stars orbiting it (or orbiting their common barycenter) as a multiple star system.

Google is only showing hits for black-hole pairs because that's where the exciting research (gravity waves) is taking place right now.

@Mark I can detect Gravity waves at the beach. Perhaps you mean gravitational waves?

I made that post over a year ago and was cleaning all that up. I got suspended today for trying to improve my questions. You can fix that black whole question if you want but I no longer can make edits there.

different but related: Binary pairings that haven't been discovered yet?

LIGO/Virgo literally just published a paper about two BH-NS mergers yesterday! https://iopscience.iop.org/article/10.3847/2041-8213/ac082e

Black hole and main sequence star/giant star
We can observe binary systems containing a black hole by looking for emissions from accretion disks which may form when matter is transferred from the companion star. X-ray binaries and microquasars are particularly notable types.
The compact object doesn't have to be a black hole - neutron stars also sometimes show up - but in some cases, measurements indicate that the remnant is a black hole. In other cases, either the mass of the compact object is poorly constrained or could still be a massive neutron star, so for many systems, the nature of this object remains unknown.
Examples:

Cygnus X-1: The companion star here is HDE 226868, a massive O-type star.
V404 Cygni: The companion star is a K-type star (much less massive than HDE 226868); the system forms a microquasar, with occasional high-energy x-ray emission.

Black hole and white dwarf
It seems a logical assumption that black hole/white dwarf systems should be possible, and indeed they are. Mass loss through accretion by a companion object can change a star's evolutionary future (see the Algol paradox for a particularly weird case), but for many systems with a black hole, the "normal" star should still progress through the main sequence and post-main-sequence evolution, possibly becoming a white dwarf.
Matter transfer can still happen at this stage, of course, so emission may continue, although the nature of the radiation could be changed. It's true that there aren't many known systems of this type, but there almost certainly more.
Examples:

X9: The companion star is a white dwarf orbiting extremely close to the black hole. Bahramian et al. are confident that the companion is a white dwarf, and the main object is likely a black hole.

Black hole and neutron star
We now have two reasonable candidate neutron star-black hole systems thanks to a pair of LIGO detections observed in January 2020 and recently released. The primary components of both systems are certainly black holes, while the secondary components have masses that are fairly consistent with our understanding of neutron star mass limits. It's certainly possible that one or both could be an exotic compact object or a primordial black hole, but a neutron star nature seems far more likely.

Is the failure to detect a BH-NS system yet really puzzling?  The abstract on the article you linked says that they were only able to constrain the merger rate enough to eliminate the most optimistic models, and that it would take several more observing runs with no results before a continued failure became problematic.

@DanIsFiddlingByFirelight LIGO/Virgo just released yesterday a new paper about two BH-NS detections https://iopscience.iop.org/article/10.3847/2041-8213/ac082e

In general those binaries will be hard to detect, but there is no reason why they wouldn't form.
A whole class of objects, X-Ray binaries are in fact thought to be hosting a pair of BH and non-BH objects. The idea here is that the X-ray luminosity originates from mass accreted from the companion through Roche-Lobe-overflow.
The mass from the companion or 'donor' falls down the gravitational well of the BH, rams into its accretion disc and creates Bremsstrahlung as it deaccelerates violently. This radiation falls into the wavelength of X-Rays and escapes the binary system, to be picked up by our telescopes later.
In a small number of cases we've been successful in determining what the binary companion is, depicted in this famous graphic by Orosz:

See also here for more information
This graphic depicts the binary distance, approximate size of the accretion disc, and size of the companion star. The companion stars are very huge, because they reached their red giant stage, not because they have some other anomalous properties.
This stage is a normal ending of a star's life, just like what our sun will experience. Just in this case there's this black hole in the vicinity that has quite strong tides, which are able to strip away parts of the evolved star's outer shell.
This graphic is a few years old now, so I'm not sure what the recent state of affairs concerning the total number of those objects is. Also there are many more unconfirmed BH binary candidates amongst the X-ray binary population, and I don't even know anything about BH-exotic things pairings. But maybe someone else can comment on that.

Prior to the discovery of gravitational wave sources, the only confirmed stellar-sized black holes, were those detected in binary systems with more "normal" stars.
These were most detected via X-rays emitted by hot material that is accreted from the "normal" companion. The two main categories are high-mass and low-mass X-ray binaries: in the former, the black hole accreted the wind from a massive stellar companion; in the latter, the mass is transferred by Roche lobe overflow.
The black hole status is confirmed by measuring a dynamical mass for the unseen companion that exceeds any plausible upper limit for a stable neutron star. Read the review by McClintock (2003).

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text:
Why is Enceladus the only geologically active moon among its neighbours?
My understanding of why Enceladus is geologically active is that tidal forces from Saturn and - to a lesser extent - from the nearby larger moon Dione provide heat to the moon's interior, just like Jupiter does for Io.
Should the other mid-to-large-sized moons Mimas, Tethys, Dione and Rhea not also exhibit similar activity from the same forces? Granted, Dione and Rhea are further from Saturn and thus less susceptible to tidal stresses, but what about Tethys and Mimas? What special attributes does Enceladus possess that has allowed it to be geologically active in modern times? Is it purely down to the fact that the interior of Enceladus is more ice than rock compared to the other moons, or are there more factors to consider?

This puzzle is something called the Mimas paradox. The two bodies are nearly twins, but Mimas is substantially less active than Enceladus. There are currently several proposed solutions:

Mimas cooled quickly, before it entered its resonance with Tethys (which contributes to tidal heating, just as Dione does for Enceladus). A "hot start" from rapid radioactive decay (of aluminum-26 and iron-60) shortly after Enceladus' formation could have allowed its subsurface ocean to stay liquid.
Enceladus is rockier and therefore has a higher density. This would have led to quicker cooling, and would also mean that tidal dissipation could be higher for Enceladus than for Mimas. It does seem a bit paradoxical - one would imagine that more rock would mean less water ice and therefore a smaller subsurface ocean - but this isn't a problem.

It's worth noting that Tethys is thought to be made almost entirely of ice. If the rocky-Enceladus hypothesis is true, then that could be a major factor in the lack of geological activity on Tethys. However, Tethys likely once had a subsurface ocean. Dione currently does, and experiences tidal heating; however, its semi-major axis is large enough that the effects are not as strong as on Enceladus.

We need to send some humans to these moons immediately if not sooner!  I'm joking, but without a doubt there's rather a lot we have yet to learn/verify about these moons.


text:
Metals and dust locked into planets
I wonder what is the contribution of metals and dust locked into planets, with respect to the amount in the interstellar medium (ISM). In other words, when we measure the metallicity and the dust mass of a galaxy we are sensitive to metals and dust in the various phases of the ISM, and thus the observed value is a lower limit. Is the "correction" a second-order variation?

The amount of metals locked up in planets is completely negligible compared to that of the interstellar medium (ISM).
First, in a typical stellar system, the mass fraction of planets is less than 1%, and of this by far the most mass will be in gaseous planets, rather than rocky planets. For instance, in the Solar System, the mass fraction of all planets is 0.1%, while rocky planets comprises only a fraction of $6\times10^{-6}$.
Second, typically only a small fraction of the gas in a galaxy is locked up in stars. For small galaxies — which are the most abundant — as little as ~5% of the gas is in stars (e.g. Schombert et al. 2001), although for large galaxies like the Milky Way, which are more efficient at converting gas into stars, the ISM-to-stellar gas fraction can be of order unity (e.g. Morokuma-Matsui & Baba 2015).
That is, the fraction of metals locked up in planets is of the order $10^{-5}\text{–}10^{-7}\!$, depending somewhat on the type of galaxy considered.
On the other hand, the fraction of metals locked up in dust is considerable, with typical values of roughly 1/3 of the metals in dust. Thus, if you want to know the metallicity of a galaxy by looking at emission or absorption lines from metals, you must either account for this factor (hoping that it's a universal factor which it may or may not be, see e.g. Zafar & Watson 2013), or measure metals that tend not to deplete to dust grains, e.g. zinc or sulfur.


text:
How do you convert an observed magnitude to a flux?
For a series of supernovae I have the observed magnitude and I want to convert these values to flux.
I understand that I need to apply:
$m - M_r = -2.5\log_{10} \frac{F}{F_r}$
where $M_r$ is a 'reference magnitude' with corresponding known 'reference flux'.
What exactly should I use as my reference values?

Very similar: https://astronomy.stackexchange.com/questions/13709/relationship-between-absolute-magnitude-of-a-star-and-its-luminosity/

Cheers buddy!!!!

@SirCumference Yet that original question does not use the term flux, which others may search for as a key term. Perhaps an answer is worthwhile here.

As I understand the question, it sounds backwards. Usually one takes the flux measurements and uses a reference to determine the magnitude.

The reference magnitude and flux really depends on what filter (i.e. range of wavelengths) the observation was made at (e.g., see the UBV system, or a more complete set of photometric passbands here). If you know the filter, and it is in one of the common ultraviolet, optical, or near infrared filters, then this website and this website provide useful reference fluxes. Using the common AB magnitude system then a source with a measured magnitude of zero in any band by definition has a reference flux of 3631 Jy (where 1 Jansky = 10-26 W Hz-1 m-2).


text:
Why are planetary orbits not all circular?
Why are the orbits of the planets elliptical ? The circular orbit also satisfies all the conditions of gravitational force. Then why did nature choose the elliptical orbits ?

Just wondering - but how is asking why something is NOT circular a duplicate of asking why something is circular?

A circle is an ellipse with major and minor axes equal.  equality is much rarer / harder to achieve than inequality (true both in algebraic systems and the universe)

Can anyone source that a circular orbit actually does satisfy the conditions of gravitational force? Don't Kepler's laws guarantee two distinct foci? If the the primary body is much more massive, the foci are closer together, but can they ever converge? You might be able to show both the primary and the orbiting body have a circular orbit around some third point, but that's not the same thing.

For a given amount of angular momentum, a circular orbit is the orbit with lowest energy.
Some orbits are elliptical (actually, all orbits approximate to ellipses) with a non-zero eccentricity, because they have yet to reach their lowest energy state, or because they have been excited from their minimum energy state circular orbit.

This! What Rob is alluding to is the principle of minimum energy, which dictates that a closed system (a system that can exchange energy but not mass, linear momentum, or angular  momentum with the external environment) will tend toward the system's minimum energy state if a path toward that minimum energy state exists. Tidal heating means that that pathway does exist.

Planets don't have elliptical or circular orbits. The elliptical orbit is the solution to the two body Newtonian gravity problem. As their are more than two particles in the solar system, the orbits aren't ellipses or circles.
A circle is an ellipse with eccentricity = 0. It's special in some ways, but for orbits it's just another elliptical orbit. The universe doesn't care what the eccentricity of the planets orbits is (providing it's not too big) There is no special reason why the eccentricity should be exactly 0. The eccentricity is essentially a random value. If you choose a number at random between 0 and 0.1 the probability of choosing exactly zero is 0.
The real question is Why are most planetary orbits nearly circular Since orbits of any eccentricty are possible it is (initially) surprising that all orbits are so close to circluar.

Good info, tho' probably a couple sig figs deeper than the OP wanted :-)

Don't laugh at anybody. Not good for your health.

While a circular orbit is possible and consistent with equations of gravitation, elliptical orbits are almost a necessity due to the surroundings of any orbit. If you take our solar system as an example, the center of mass will change as not all planets orbit at the same speed. The center of mass is found with the equation:
$$x_c=\frac{\sum\limits_{i=1}^{n}{m_{i}x_i}}{\sum\limits_{i=1}^{n}{m_{i}}}$$
As the planets mass distribution around the sun changes, the point which the planets orbit changes slightly, disrupting any circular orbits. All orbiting bodies will have some disruption, preventing orbits to always be perfectly circular. In addition, losing energy and angular momentum to other sources can cause bodies to lose their orbital shapes and degenerate into elliptical orbits.
In addition, please use this link also posted as a comment.


text:
Can one approximate the semi-major axis of an orbit as the average orbital distance for eccentric orbits?
Kepler's 3rd Law (regarding the relation between orbital period and the semi-major axis of an orbit) applies to all elliptical orbits. But as I understand it, the only reason it is safe to use the approximation that the semi-major axis of an orbit is approximately equal to the average orbital distance is because the planets of our solar system are generally not very eccentric.

As an example, the Earth is about 1 AU away from the Sun and has a very low orbital eccentricity, so one can say that the semi-major axis of the Earth's orbit is also approximately 1 AU in length. Since P^2 (yr) = a^3 (AU), one can deduce that the orbit of Earth should take about a year. But, what if the Earth had a higher eccentricity - If the orbital eccentricity of the Earth were higher, would it still be wise to approximate the orbital distance as the semi-major axis of an orbit?

The gist of this, is that your assumption is incorrect.   It's the semi-major axis that defines the period, not the average distance.   Newton worked this out when he invented calculus and derived Kepler's laws.   (Look up Derivation of Kepler's laws for some explanations).   Here's the Wikipedia version of the Math.
I should add that Kepler's three laws, and Newton's derivation of them still don't perfectly define orbits.  They assume that only the larger mass matters, when in reality, the two objects tug on Each other.    The Sun doesn't just tug on the Earth but the Earth tugs on the Sun, which speeds up the orbit a tiny bit (the mass ratio of 330,000 to 1 makes this close to negligible, but it's there).  More accurate Newtonian orbital calculations between two massive objects require more advanced mathematics.  Kepler's laws only work if the central object is much more massive than the 2nd object.
Ignoring the finer details and just looking at the three laws, Newton's derivation works out exactly to the semi-major axis.  The average distance doesn't enter into the equation.
Average distance of an orbiting object to it's Sun or central object is problematic anyway because the planet moves more slowly when it's more distant, so there's a few ways to calculate average distance.
By time the planet spends in each part of it's orbit (same as area as defined by Kepler's 3rd law)
Or, you can measure average distance by arc or angle, taking smaller and smaller angles and running an average.
Or you can take average distance by length of the ellipse to the specific foci.
That's three separate methods for measuring average distance.  Needless to say, this gets a little complicated.  The good news is, you don't need to do any calculation for average distance because the Semi-major axis is actually correct.
Does the semi-major axis equal the average distance, perhaps by arc or by length?   Not sure.   I'll try to work that out.  I know that it doesn't equal average by time.
Does that make sense.  I'm not sure I explained that as well as I should.
Edit
On average distance.  The standard average distance with orbits is by time or area.  Kepler's 3rd law says equal areas over equal time, so it's really two ways of saying the same thing, equal time and equal surface area.

I think your question is interesting and I'll try to work out a relation between semi major axis and average distance.  Off hand, it's a little complicated, but it seems to me that ratio between average distance and semi major axis changes with eccentricity but, I'd rather work that out before I say for certain.

If I understand correctly, I incorrectly assumed the law related average orbital distance with orbital period whereas the “simple math” - derivation yields the result that the period is related to the semi-major axis; The actual case is more complicated as smaller masses are no longer negligible. Is that correct? Slightly off-topic, but of the three methods used to calculate average orbital distance, would the most convenient be the first method since the swept area (or elapsed time in orbit) would be constant?

@mikey Smaller masses are closer to negligible.  It's when the two masses approach being more equal then Kepler's laws become less accurate.   As to your first point, yes, the derivation points to the semi-major axis.   As to your 3rd, I'll edit the answer above.

I’m guessing that you integrate along the orbital curve varying the angle for the second method. But I don’t yet have an idea for the third method.

Kepler's Third law applies just fine to binary stars. No need for one to be much more massive than the other.

The maths says that the semi-major axis is not a good measure of average distance for high eccentricity (elliptical) orbits.
There are basically two ways to measure this : (1) an average over the entire orbit on a purely geometric basis, and (2) the average over time.  These give quite different results - qualitatively different.
Average distance on geometric basis.
In a way this is the most obvious one.  The mathematics is :
$$R_g = \frac a {2\pi} \int_0^{2\pi}r\,d\theta$$
And we use for an ellipse :
$$r = \frac \rho {1+\epsilon\,cos\theta}$$
The result is (eventually :-) ) :
$$R_g = a\,\sqrt{1-\epsilon^2} \tag{1}$$
Average over time.
This is a tricker calculation, but in fact it's the most "human" average.  We want :
$$R_t = \frac 1 T \int_0^T r\,dt$$
The maths on this is tedious and not at all useful, but the result ( with a little help from Wolframalpha.com ) is :
$$R_t = a\,\frac 2 {\pi} \frac{2+\epsilon^2}{(1-\epsilon^2)^{\frac 5 2}}tan^{-1}\left( \sqrt{\frac {1-\epsilon}{1+\epsilon}} \right) \tag{2}$$
Now by any definition that's a messy equation, but the main point is to note the key difference between the time averaged expression (2) and the geometric one (1).
The geometric expression gets smaller as eccentricity gets larger.
The time averaged one gets larger as eccentricity gets larger (a slight simplification).
In fact for both $\frac R a$ starts off at $1.000$ but when $\epsilon\to 1$ then  $R_g\to 0$ and $R_t\to \infty$ !
Why ?
The geometric expression doesn't allow for the orbital speed of the object.  As you increase eccentricity an orbits ends up with having a faster close approach, so less time close up to it's primary.  As it moves away from the primary, the orbital speed also decreases, so it spends much more time at large distances than at short distances.

+n! and kudos for math! If you fix $a=1$, then the limit of $Rt$ is only 1.5, not $\infty$. That $(1-\epsilon^2)^{5/2}$ in the denominator seems to be the issue.

See this answer for a numerical verification of that.

I still believe that there is a mathematical error in your expression. With $a=1$ and eccentricity approaching unity ($\epsilon \rightarrow 1$), the apoapsis distance $a(1+\epsilon)$ approaches a value of 2, and the time-averaged mean distance 1.5. However your expression diverges to $\infty$. Are you planning on addressing this?

I can edit your answer for you if you like. Right now I think there is a mistake.

@uhoh Sure, go ahead.  To be honest I've lost track of this question mentally and if your maths is kosher, go for it.

OK I know exactly what you mean! I'll ping you when I do. I really admire the analytical approach, so it will be good practice for me.

Just to provide an analytical formula for @uhoh's correct time-averaged distance, here the derivation of $\langle r\rangle_t=1+\epsilon^2/2$:
$$a=1 \qquad c=e\qquad b=\sqrt{1-e^2}\\
\vec{r}=(\cos \beta-e,\sqrt{1-e^2}\sin \beta)\\
\vec{r'}=(-\sin \beta,\sqrt{1-e^2}\cos \beta)\\
|r|^2=\cos^2 \beta -2e\cos \beta+e^2+\sin^2\beta-e^2\sin^2\beta=1-2e\cos \beta+e^2\cos \beta=(1-e\cos \beta)^2\\
|\vec{r}\times\vec{r'}|=(\cos \beta-e)\sqrt{1-e^2}\cos \beta+\sin \beta\sqrt{1-e^2}\sin \beta=
\sqrt{1-e^2}(1-e\cos \beta)=2\,\mathrm{d}A/\mathrm{d}\beta=\text{const. }\mathrm{d}t/\mathrm{d}\beta\\\
\langle|r|\rangle_t=
\int_0^{T}|r|\mathrm{d}t=\int_0^{2\pi}{|r|\frac{\mathrm{d}t}{\mathrm{d}\beta}}\mathrm{d}\beta
=\frac{\sqrt{1-e^2}\int_0^{2\pi}{(1-e\cos\beta)^2}\mathrm{d}\beta}
{\sqrt{1-e^2}\int_0^{2\pi}({1-e\cos\beta})\mathrm{d}\beta}
=\frac{2\pi+e^2\pi}{2\pi}=1+e^2/2$$

This was intended to be a supplemental answer to StephenG's answer. However, there appears to be a problem with the expression for time-averaged distance in that answer. I think it's great to seek a mathematical expression, but it should be confirmed numerically.
I did a quick numerical double check and verified those general trends, but there may still be a problem with one of the expressions there.
Assuming a constant semi major axis of 1, the time-averaged distance rises from 1 at $\epsilon = 0$ to 1.5 at $\epsilon \rightarrow 1$, while for the $\theta$-averaged ("geometrical") distance drops from 1 down to zero.
I think both of us should now add the path-average for completeness by averaging over ds. :-)

Python script:
def deriv(X, t):
    x, v = X.reshape(2, -1)
    acc = -x * ((x**2).sum())**-1.5
    return np.hstack((v, acc))

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

halfpi, pi, twopi = [f*np.pi for f in 0.5, 1, 2]

time = np.linspace(0, twopi, 10001)[:-1]

a = 1.0
eps = np.hstack((0, 0.2, 0.5, 0.7, 0.9, 0.99))

orbits = []
for ep in eps:
    rperi = a * (1. - ep)
    vperi = np.sqrt(2./rperi - 1./a)
    X0 = np.array([rperi, 0, 0, vperi])

    answer, info = ODEint(deriv, X0, time, atol = 1E-12, full_output=True)
    xy = answer.T[:2]
    orbits.append(xy)

rs = [np.sqrt((xy**2).sum(axis=0)) for xy in orbits]
rmeans = [r.mean() for r in rs]

plt.figure()
plt.subplot(3, 2, 1)
for x, y in orbits:
    plt.plot(x, y)
plt.ylim(-1, 1)
plt.plot([0], [0], 'ok')
plt.subplot(3, 2, 3)
for r in rs:
    plt.plot(time, r)
plt.subplot(3, 2, 5)
plt.plot(eps, rmeans)
plt.plot(eps, rmeans, 'ok')
plt.plot(eps, np.ones_like(eps), '--k')
plt.ylim(0, 1.6)

theta = np.linspace(0, twopi, 10001)[:-1]
rs = [a * (1-ep**2)/(1 + ep*np.cos(theta)) for ep in eps]
rmeans = [r.mean() for r in rs]

plt.subplot(3, 2, 2)
for r in rs:
    x, y = [r*f(theta) for f in (np.cos, np.sin)]
    plt.plot(x, y)
plt.ylim(-1, 1)
plt.plot([0], [0], 'ok')
plt.subplot(3, 2, 4)
for r in rs:
    plt.plot(theta, r)
plt.subplot(3, 2, 6)
plt.plot(eps, rmeans)
plt.plot(eps, rmeans, 'ok')
plt.plot(eps, np.ones_like(eps), '--k')
plt.ylim(0, 1.6)
plt.suptitle("Time averages      Theta averages")
plt.show()

see my answer for derivation of $\langle r\rangle_t=1+\epsilon^2/2$

@MarkusSchmassmann I'll give it a look as soon as I get to a reasonable screen and some pencil&paper, seems great though!


text:
Why most planets don't spin retrograde?
Edit: just found duplicate
Since orbital velocity is greater on small orbits, and smaller on large orbits, why don't planets spin retrograde around their own axis?
One portion of planet's mass is closer to its star, and should move prograde relative to center of mass.
Part of the planet's mass far from the star orbits higher and should move retrograde relative to center of mass,
which should create a resulting retrograde spin, with rotation speed depending on planet's diameter.
What's wrong with this? ( since fact is : most planets spin prograde )

In this not to scale second illustration, planet is replaced by two masses linked by a beam. Kepler's law of equal surfaces swept per unit of time is used
to display the retrograde rotational force that should be applied relative to center of mass, which orbits black path.
EDIT : Ok I think I found where my mistake is(?) Differential gravitational pull (yellow arrows) counteracts retrograde rotational force (green arrow) Which results in no rotational force at all. Anyway I still don't get how conservation of momentum of everything orbiting the sun, also applies to spin direction of orbiting objects.

Tidal effects only matter for a nonhomogeneous body.

@CarlWitthoft at some point during their formation, planets were nonhomogeneous bodies, so what force made them spin around themselves in the first place, counterclockwise seen from above, same direction as around the sun?

Basically, go back to the pre-planet particle clouds, and as explained in
geek.com,

To answer the question we have to go back to the beginning, to the
  formation of a solar system like ours. Before a star and its planets
  exist, there’s just a cloud of disorganized gas and small molecules.
  This is often called a molecular cloud or “stellar nursery.” The Eagle
  and Orion Nebulae are some of the more famous stellar nurseries we
  have been able to observe with the Hubble Space Telescope. These
  clouds are composed mostly of molecular hydrogen, which wouldn’t be
  able to congregate outside the dense molecular clouds. These clouds
  can be of any size — not just massive structures like the Orion
  Nebula. Over time the energy of the molecules in the cloud pushing
  outward can be overcome by slower molecules collapsing together
  farther in. As long as there is sufficient mass in the molecular
  cloud, it continues collapsing in toward the center until it reaches a
  high enough mass to fuse hydrogen and become a new star. The spin that
  we see quite clearly now is related to the process of the molecular
  cloud collapsing. The original cloud was very, very large and made up
  of many individual molecules and small clumps of matter. On that
  scale, there is some small amount of rotation within the cloud. It
  could be caused by the gravity of nearby stellar objects, local
  differences in mass as the cloud churns, or even the impact of a
  distant supernova. The point is, most molecular clouds have at least a
  little rotation.
As the cloud collapses to form a star, it has what physicists call
  angular momentum. This is the movement an object has as it rotates
  around a central point. In a large system like a molecular cloud, each
  particle has some angular momentum, and it all adds together across a
  very wide area. That’s a lot of momentum, and it is conserved as the
  cloud continues to collapse. But how does that get us to objects that
  spin and orbit?
Imagine a figure skater spinning around with arms outstretched. That
  is a model of angular momentum just like a collapsing cloud of gas.
  When the arms are drawn inward, the rotational velocity goes up
  because the total angular momentum is conserved unless there is some
  external force acting on it. There are such forces acting on the
  figure skater, but less so on a collapsing molecular cloud. So if a
  molecular cloud was maybe a light year across, then collapsed down to
  be just a fraction of that, it would be a huge change in size. Just
  like the figure skater pulling her arms in, the velocity must increase
  to conserve angular momentum and therefore form a spinning
  protostellar disc. It is from this orbiting matter that all the
  planets form, and of course, they are also spinning and orbiting in
  the same direction because of the conservation of angular momentum.

Quoting from wikipedia as to the mystery of  the misbehaving planets:

All eight planets in the Solar System orbit the Sun in the direction
  that the Sun is rotating, which is counterclockwise when viewed from
  above the Sun's north pole. Six of the planets also rotate about their
  axis in this same direction. The exceptions—the planets with
  retrograde rotation—are Venus and Uranus. Venus's axial tilt is 177
  degrees, which means it is spinning almost exactly in the opposite
  direction to its orbit. Uranus has an axial tilt of 97.77 degrees, so
  its axis of rotation is approximately parallel with the plane of the
  Solar System. The reason for Uranus's unusual axial tilt is not known
  with certainty, but the usual speculation is that during the formation
  of the Solar System, an Earth-sized protoplanet collided with Uranus,
  causing the skewed orientation.[6]
It is unlikely that Venus was formed with its present slow retrograde
  rotation, which takes 243 days. Venus probably began with a fast
  prograde rotation with a period of several hours much like most of the
  planets in the solar system. Venus is close enough to the Sun to
  experience significant gravitational tidal dissipation, and also has a
  thick enough atmosphere to create thermally driven atmospheric tides
  that create a retrograde torque.

Thanks for your input! I get why conservation of momentum makes planets orbit the sun in the same direction around it, I get why one 177 degrees tilt reverses Venus' spin direction, but there is something non intuitive and I still don't get why during their accretion, planets started spinning around their own axis in this specific direction in the first place. I mean it was particles orbiting the sun. I feel like if Saturn ring for some reason started agglomerating in one body, this body would start spinning retrograde, because particles orbiting closer to Saturn have greater velocity.

That is an intersting question as this is far from being obvious. Actually, for quite some time it was believed that planets should indeed spin retrograde (we can at least track it back to Laplace in his Exposition du Système du Monde published in 1976 where he developed his nebular hypothesis).
How to prove that planet should spin prograde? We'll have to go back to their formation history. Let's start with an ideal case: particles moving in eliptical orbits around a central object of mass $M$, with a proto-planet of mass $m_0$, and rotating in a eliptical orbit. Let's look at what happens at a given point at a distance $r$ from the central object. First, for a planetesimal in a tangential external orbit, then for a planetesimal in a tangential internal orbit.
Tangential external orbit
We can write Newton's Second Law ${\bf F} = m {\bf a}$ and project it on the current trajectory axis (normal and tangent), knowing that the acceleration is always something like $v^2/R$, with $R$ the curvature radius (that is different from $r$, the distance to the central object). We get for the normal component:
$$\frac{GM}{r^2} \cos \alpha = \frac{v_{ext}^2}{R_{ext}^2},$$
$R_{ext}$ being the curvature radius of the planetesimal in an external orbit. For the proto-planet:
$$\frac{GM}{r^2} \cos \alpha = \frac{v^2}{R^2},$$
$\alpha$ having the same value for both object, since it denotes the angle between the direction toward the curvature center and the central object and the orbits being tangential at this specific point, where $r$ the distance to the central object is also the same. Therefore:
$$\frac{v^2}{R^2} = \frac{v_{ext}^2}{R_{ext}^2}.$$
As $R < R_{ext}$, we have $v < v_{ext}$.
Tangetial internal orbit
With the same argument, we can show that $v > v_{int}$ (I let it as an exercice; extra bonus point if you do it with the energy conservation law).
Consequences
If you go in the reference frame of the protoplanet, you will see the internal planetesimal, rotating at a lower speed, coming backward, towards the proto-planet, and the external planetesimal, rotating at a faster speed, coming forward:

where $v'_{ext}$ and $v'_{int}$ are the velocity of the external and internal planetesimal in the reference fram of the proto-planet. That is why the rotation of the proto-planet tends to be in the same direction as the whole system.


text:
When would Oumuamua have entered the solar system?
So from what I've read, Oumuamua travels at around 26km/s when in interstellar space.
I also remember reading a while back that Voyager was travelling at around 18km/s when leaving the solar system.
Seeing as Voyager took around 35 years (I think) to exit the solar system, I'd guess that Oumuamua entered our solar system around 20–25 years or so ago?

One problem with answering this question is that it's not precisely clear where the solar-system ends or when (if) Voyager I crossed that border.
If we set the estimate at 19 billion km (per article) or 127 AU. (about 4 times the distance between Neptune and the Sun), then an estimate is possible.
Some conversions first.  Voyager left earth in 1977 and the announcement was in 2013, so it's usually listed as 36 years (it might be 35 and some months) but that's not all that relevant.
18 km/s x 31.5 million seconds in a year (close enough) is 567 million km per year.   It should have taken Voyager about 33.5 years at that velocity to reach 19 billion km.   However, Voyager's velocity wasn't consistent.  As it flies from the sun it slows down and it had several gravity assists along the way which sped it up and changed it's direction.   Mathematical calculations get a little tricky, but here's a chart of Voyager's velocity.

(Note: that's Voyager 2's velocity not Voyager 1.  I'm having trouble finding Voyager 1's velocity chart).
Here's it's direction, more of a fish-hook shaped path than a straight line, though it's been mostly straight since Saturn which is over 90% of it's journey.

Oumuamua's velocity at infinity, as it approached our solar-system was 26.33 km/s.   It speeds up as it approaches the sun and it will slow down as it leaves.   When it passed the 19 billion km "border", you have to add the escape velocity from our sun at that distance, which is about 3.7 km/s, so it entered the solar-system traveling at (close enough) 31 km/s or 976 million km/year.  At that velocity it would take 19.5 years to reach the sun (granted it will miss the sun), and in theory another 19.5 years to leave the solar system.   But that estimate doesn't take into account that it speeds up as it gets close to the sun, so it takes less time.   17-18 years (ballpark) might be a good estimate.  That's 17 (or so) years traveling in, another 17 (or so) to travel back out and reach the made-up border.
Unlike Voyager, nearly all of Oumuamua's journey is nearly straight, except for a quick curve around the sun.  That makes it's journey close to "V" shaped when viewed from a distance.  (I only mention that in relation to Voyager's more curvy fish-hook flight path, which was by design to pass by several planets).

There's probably someone here who can do the math (I'd prefer not to), and they can work out a more precise answer of when it entered the fuzzy 19 billion km border, when it will reach it's perihelion or closest to the sun, and when it will pass outside that border again.

I like the "science" answer posted by userLTK. Here I'll cheat and use a projection from the current ephemeris in JPL Horizons. Set up like this:

and using the script below to plot the results, I get the years 1995 and 2039 for the times that the comet passes 127 AU. That's about +/- 22 years from perihelion, nicely confirming userLTK's ballpark estimate. Outstanding is the question of the direction of the comet relative to any possible "tail" of the heliopause opposite the direction of the Sun. I think there is no hard, observational information about this currently, so sticking with an estimate of a radius seems acceptable.

According to the current ephemeris' header:
JPL/HORIZONS              1I/'Oumuamua (A/2017 U1)         2017-Dec-03 00:25:53
Rec #:838451 (+COV)   Soln.date: 2017-Nov-21_20:25:17      # obs: 115 (34 days)

Python script:
import numpy as np
import matplotlib.pyplot as plt

fname = 'Comet horizons_results.txt'

with open(fname, 'r') as infile:

    lines = infile.read().splitlines()

    iSOE = [i for i, line in enumerate(lines) if "$$SOE" in line][0]
    iEOE = [i for i, line in enumerate(lines) if "$$EOE" in line][0]

    lines = lines[iSOE+1:iEOE]

    lines = [line.split(',') for line in lines]
    JD  = np.array([float(line[0]) for line in lines])
    pos = np.array([[float(item) for item in line[2:5]] for line in lines])
    vel = np.array([[float(item) for item in line[5:8]] for line in lines])

r = np.sqrt((pos**2).sum(axis=1))
s = np.sqrt((vel**2).sum(axis=1))

years = 2017 + (JD - 2457754.5)/365.2564  # roughly

plt.figure()
plt.subplot(2, 1, 1)
plt.plot(years, r)
plt.plot(years, 127 + np.zeros_like(years), '-k')
plt.title("distance from solar system barycenter (AU)", fontsize=16)
plt.xlim(1980, 2060)
plt.ylim(0, None)
plt.subplot(2, 1, 2)
plt.plot(years, s)
plt.title("speed relative to solar system barycenter (AU/day)", fontsize=16)
plt.xlim(1980, 2060)
plt.ylim(0, None)
plt.suptitle("Asteroid 'Oumuamua (A/2017 U1) from JPL Horizons", fontsize=18)
plt.show()


text:
Rotation of the Earth about its own axis
I am wondering how the Earth got its original angular velocity. What was the cause that made the Earth rotate about its own axis?
What about its motion around the Sun? What gave the Earth the push to keep rotating around the Sun?

Origin of Earth rotation
There are two main phenomena that explain Earth rotation. First one is linked to the history of the solar system: there is angular momentum in the interstellar medium (ISM), so the collapsing cloud that formed our solar system had an initial angular momentum (rotation of the galaxy and turbulence in the ISM are typically a good candidate for that; see for example Mac Low & Klessen 2004). By conservation of the latter, particles in the collapsing cloud accelerate (the norm of the angular momentum $I$ is $mrv$, with $m$ the mass of the system, $r$ the radius and $v$ the velocity; as the cloud contracts, $r$ decreases and then $v$ has to increase) and due to centrifugal force, a disk will tend to form. By conservation of the angular momentum in the protoplanetary disk, you will also tend to form bodies that rotates.
The second phenomena is linked to the formation history of the Earth. Simply put, Earth formation history is due to collisions of planetesimals (let say they are big chunks of rock, formed by a collision history of smaller chunks of rock, formed by...); these planetesimals are orbiting the forming star in the protoplanetary disk, their rotation being linked to the disk formation history (see above). When planetesimals collide, two things can basically happen: either the collision is perfectly central (meaning that the momentum of one of the two colliding objects is pointed toward the center of mass of the second object) and there is no transfer of angular momentum, either the collision is lateral and there is a transfer of angular momentum. Depending on the configuration, you can either transfer momentum that will make the other body rotating on way or the other. That being said, you can then object that there should be a equirepartition of collisions pushing the proto-Earth to rotate in both directions. Which is true, but external planetesimals are rotating faster than internal ones. Therefore, from the proto-Earth perspective (meaning, in a physical sense, in its reference frame), internal planetesimals are seen as coming backward, from front, and external planetesimals are seen as coming forward, from behind. Long story short, there are both transfering momentum in the same "direction", forcing the proto-Earth to rotate in the same direction as the rotation of the protoplanetary disk (see for example Artem'ev & Radzievskii 1965).
Rotation around the Sun
This one's easy: gravity.
The Earth is constantly falling towards the Sun because of gravity, and the centripetal acceleration counteracts perfectly gravity, mainting the Earth on its orbit.


text:
If I look beyond (about same region, higher redshift) a galaxy, will I see its progenitors?
Is there a range of redhsift dz, such that if I look at z+dz from a massive object at redshift z, I will see its progenitors?
(in a LCDM scenario of halo assembling, with smaller halos evolving towards larger ones across the cosmic time)
I am looking for a detailed explanation of this (maybe naif) question.

I think I understand what you mean and the answer is: specifically no; but generally yes.
If we look out into space, we can divide up what we can see into redshift slices of arbitrary thickness. The light we see from galaxies in each slice has been travelling for a certain amount of time, so we are seeing that "piece of space" at a particular cosmic epoch corresponding to its redshift.
If we look past that galaxy to a higher redshift slice, then what we are seeing there is a different patch of space at a younger cosmic epoch. i.e You cannot see the same galaxy as it was at a different time. We see each and every galaxy only once.
But in general you will be seeing younger galaxies at higher redshift. In an LCDM model then smaller galaxies build up into larger ones, so there will be a higher density of galaxies at higher redshifts (even after accounting for cosmic expansion), but the galaxies they will produce by merging etc. are not yet visible to us. It is possible that we will see them in the future (depending on the exact values of the cosmological parameters and what redshift we are talking about) as that distant patch of space recedes from us and becomes older (in cosmic terms).

Rob, is this related to expansion? I mean in an hypothetical contracting universe could one observe the same object at two different epoch (z)?

@Alchimista No. Each galaxy is observed at a particular epoch.

That would mean a galaxy emitting and approaching us faster than light :) I reversed the universe too much :)

@Alchimista With gravitational lensing we can get 2 (or more) signals that have traveled over slightly different paths, through different gravity wells, so there is a small but significant time difference between them. There's some info at The Hubble Constant From Gravitational Lens Time Delays by Kochanek & Schechter.

PM 2Ring good point . The same could be with a close U. But I was thinking something different which implied a reverse Hubble flow with shrinking region at v > c. :) based on a reversal thinking done in a rush of enthusiasm.


text:
Does habitable zone take tidally locked bodies into account?
Imagine a tidally locked planet orbiting a red dwarf, where habitability is not situated close to terminator zone, but on a "small" cap normal to incident starlight (zenith) with permanent average +15 Celsius over the year, variations through year depending on orbit's eccentricity.
If there was in our solar system one (maybe not spherical) tidally locked body orbiting some close to circular orbit around the sun, how far from the sun would it be to meet those +15C average zenith cap temperature? Does it make habitable zone larger than it is?

How do you propose to deal with heat transfer?

@RobJeffries I thought everything in equilibrium would result in +15C at cap. With maybe one anticyclonic eye above cap, air there exchanging heat at slow rate.

I think you need to specify the temperature distribution and an albedo. In equilibrium the planet will re-radiate all energy incident upon it.

Thickness of the atmosphere is a factor.  Venus has almost no temperature variation, day to night or equator to pole.  The thinner the atmosphere, the greater the variation.    There's no straight forward answer to this because there's a few factors that need to be accounted for.

You might get some answers to this over at [Worldbuilding.SE].


text:
Is it fruitful to search for a correlation between rates of in-falling matter in black holes and the expansion rate of our universe?
To my understanding, white holes are hypothetical objects that are time-reversed black holes; black holes gravitationally "suck in" matter whereas white holes "spit" matter back out. As such, I think it's a nice idea that the Big Bang could be the "spitting of matter" via a white hole (assuming such an object exists).
This got me wondering how one could experimentally test for such an object. Since the expansion rate of the universe is increasing, could one possible explanation be that a white hole is continually "spitting out" matter such that an increase of flux of in-falling matter/space could be observed as an increasing rate of expansion of out-pouring matter/space? While I don't know of a way to detect this, I was wondering whether the following idea was fruitful or not.
Given that a supermassive black hole is believed to be at the center of every galaxy, given that most galaxies (sans Andromeda) are receding away at a speed proportional to their distance relative to us observers, and given that the spatial distribution of matter in our universe is isotropic and homogeneous, could one build a simulation of such a concept? To emulate isotropy and homogeneity, one could use a random number generator corresponding to a uniform distribution to place "galaxies" along a grid; accounting for gravitational forces, one can simulate the time-evolution of these galaxies and compare rates of in-falling matter to find a correlation between the rate of in-falling matter and the expansion rate of the universe (likely as a function of time-elapsed, and age and size per black hole). While the simulation does not definitely prove anything, providing a correlation could be interesting in my opinion.
Is this a stupid idea, or is there a potential to build upon this idea? If it is not fruitful, I am curious as to why.

I don’t think this is a duplicate. The post you linked to concerns the definition of a white hole, whereas my question is about the plausability of searching for a simulated correlation from observable quantities.

As a layman, I can tell you that there are some issues with your proposal.
The biggest one is that White holes probably don't exist.  They're purely theoretical and, even worse than that, they're mostly not even well defined theoretical objects.  For a physical hypothesis to work, every element of the theory needs to be precisely defined.  White holes aren't well defined.  (now, let me add that, perhaps White holes are well defined by Einstein's field equations, I'm not sure, but the specifics go above my paygrade).
And running time in reverse, a black hole really doesn't look anything like a big bang, and, granted, we don't know what the/a big bang looked like, but we know what a black hole swallowing matter looks like.   It absorbs matter by accretion disk and it shoots out jets along the accretion disks axis.   Running time backwards you can see streams of matter and the accretion disk reforming into stars and dust that flies away from the black hole but that gravitational consumption in reverse That looks NOTHING like the big bang is thought to have looked.
The moments after the big bang are believed to have been a very hot, dense and rapidly expanding quark/photon/fundamental particle soup kind of thing.   Nothing in today's universe, backwards or forwards in time, looks anything close to that, so running time backwards doesn't really help.
The very early universe and the ratio of hydrogen to helium and the presence and near perfect uniformity of the background radiation all imply a mostly uniform, very hot, very dense, and only the slightest bit lumpy rapidly expanding origin, that only works under modern physical laws if space expands faster than the speed of light.
We can re-create those conditions, but only for the tinyest fractions of a second, and with only a handful of particles, in particle accelerators. You might think that inside a black hole, similar conditions exist, but that's not entirely true.  Inside a black hole, gravity is dominant.   Due to faster than light expansion and uniformity, it's not believed that gravity was a dominant force in the moments after the big bang.  It was a force, but not a dominant force.   The space of space-time, for example, is very different inside a black hole than in the moments after the big bang.
A final issue with your question is whether the big bang was an explosion, or, more specifically that it originated from a source-point-origin at all, (as it's often presented), or whether it was an expansion that took place everywhere (everywhere stretch, so to speak).
A theoretical white hole ejecting matter would be a source point and a source point is subject to the inverse square rule and we'd expect to see a variation in density and evidence of a direction towards the origin.   We don't see that.  The observable universe appears remarkably homogeneous, which is only possible, from a source-point, if the size is enormously large.   The rate that distant galaxies move away from us is also very difficult to explain without some kind of dark energy expansion.
The variations in the image are exaggerated.  The cosmic background radiation is very homogeneous, implying we were inside the expansion, not part of some material that was ejected.

Source.
I realize that you didn't say explosion specifically, but asked about a source object ejecting material, but both have a similar effect.  Ejected material spreads out from it's source.   The universe doesn't appear to have been formed in that way.   Nice summary of the expansion theory here.
That's my layman's understanding anyway and I'm not sure I should have gone there.  Origin of the universe questions are kind of out there.


text:
Does anybody know how to get Sun-Jupiter Lagrangian points L4 and L5 RA and Dec?
I wanted to point at the L4 and L5 of Sun-Jupiter to find Trojan asteroids but I don't know how to calculate the RA and Dec of these.
Do you know of any software that does it for you or how to calculate them?

The L4 and L5 points are unstable orbits. The Trojans move in perturbed (ie slightly non-keplerian) orbits that move them around the L4 and L5 points. At any time you will find Trojans in a pair of lobes between about 30 and 90 degrees ahead of Jupiter.
You can find individual asteroids using NASA Horizons ephemeris
For example 624 Hektor, the largest Trojan (and placed in the Greek Camp around L4) will be found at RA 20 47 54.08, Dec -28 04 20.1 on Feb 10 2018. And 884 Piramus (in the Trojan Camp, L5) will be at RA 10 43 51.39 Dec +01 04 11.3.
(Jupiter is between them at RA 15 18 55.32 Dec -17 07 47.1)
Neither are easy to find, with a Magnitude of 15 or 16, due to their greater distance than main belt asteroids, and very dark colour. The image below shows the distribution of Trojans around Jupiter.

In order to find the rather large extent of space where Jupiter's Trojan asteroids can remain, one would use a map of the Celestial sphere, and look at the long stretch of points that happen to be +/- sixty degrees of wherever Jupiter happens to be at the moment. I show a screen shot from in-the-sky.org's Star Atlas page. The purple dot in the center next to the constellation name Libra is Jupiter, but because of some glitch it is not showing right now. Holding a cursor over the dot shows its name. The yellow diagonal line is the Ecliptic, and the +/- 60 degree regions will be at either end of the box. It's really really huge, there are no specific coordinates, and the only way to know an extremely faint dot would be an asteroid rather than a star would be an extremely detailed star map, or a series of exposures over hours or days showing the motion relative to the other objects.
EDIT: You can use any similar mapping program or website. You've asked about December, the programs can usually be walked backwards or forwards in time to whichever date you are interested in. The figure below is for mid-February, when I posted the answer.

You can also enjoy Scott Manleys YouTube Videos:

Asteroids In Resonance With Jupiter (First 3:2 resonances, then Trojans are shown separately)
Asteroid Discovery - 1980-2012 - UHDTV (several versions on this YouTube channel)