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How long could a world like the Moon survive with an Earth mass black hole in the center?
I asked that question at the worldbuilding stack exchange.
https://worldbuilding.stackexchange.com/questions/202273/how-long-could-a-planet-or-moon-survive-if-it-had-an-earth-mass-black-hole-withi
Obviously if a small world had a primordial black hole with a greater mass inside, it would have a higher surface gravity and escape velocity than a world of that size made out of even the densest known materials.  Thus the possibility of a fictional habitable world much smaller than otherwise would be the case.
The answers to this question What would happen if a small black hole fell into a star? indicate that I might get a good answer here.

The surface gravity is ~133.7 m/s^2 (~13.63 g). That's not habitable by humans. For 1 g gravity you need a BH of Schwarzschild radius 0.54963 mm, just under 0.062 Earth masses. That gives an escape speed of ~52% that of Earth.


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Why the ancient Greeks associated planets with gods?
They associated the Sun with Helios and the Moon with Selene, with the names of these gods also being used for these astronomical bodies. For other astronomical bodies, a name and an associated god were present: Phainon (Saturn) was associated with Cronus, Phaethon (Jupiter) with Zeus, Pyroeis (Mars) with Ares, Phosporos (Venus) with Aphrodite, and Stilbon (Mercury) with Hermes. (And today these planets are named after the Roman equivalents of these gods).
Why they did associate planets with gods?

The Greeks were influenced by the Babylonians https://en.wikipedia.org/wiki/Babylonian_astrology#Planets_and_gods It seems to be common for ancient civilisations to associate planets with gods. https://en.wikipedia.org/wiki/List_of_solar_deities

In same vein, Romans named the planets

There is a good explanation here as well: https://mythology.stackexchange.com/questions/6151/why-did-the-romans-name-the-planets-after-their-gods

Naming planets with the name associated with their god is true not only for ancient Greeks but also for other cultures, such as Vedic. Religion and scriptures were the major means through which people were united at that time. Motion and the presence of the Sun and moon were affecting human lifestyle directly. Ancient philosophers associated them with their gods. There is always a trend to associate things that we do not understand clearly with some superpowers.
Moreover, They explained celestials in accordance to their luminosity, color, shape, position, etc. in skies. There are numerous constellations named after their shapes. Personification of objects could make general people understand what the objects are. Ancient people must have tried to associate celestials with their gods having similar characteristics. Or ancient people wrote scriptures mentioning these celestials as some super powerful creatures and named them in accordance to their culture.
People may have thought those bright objects in the night sky must have a direct effect on our lives. They were personified as gods. I am not sure about Greek culture but in Vedic culture, philosophers have associated different planets with different aspects of natural phenomena such as rain, eclipse, etc. The attraction of humans toward astrology till this time shows how people still think.

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Is the distance covered always equal to the time taken?
Is the distance covered by light from an extremely distant source in light years equal to the time taken in years? Surely it would have to be, for the speed of light is absolute, after all.

There are a few different definitions of "distance" in cosmology https://en.wikipedia.org/wiki/Distance_measure

There is no "actual distance".

The tricky aspect of cosmology is that in an expanding universe, distances are not absolute. If the light takes a sufficiently long time to get to us, the intervening space may have expanded significantly and the number of light years traversed by the light is not equal to the number of years it had been moving.
For example, JADES-GS-z13-0 has a redshift of 13.2. That means light from it has traveled for 13.4 billion years. But its present proper distance is 33.6 billion light-years. But when the light emitted it was just 2.4 billion light-years away from us.

33.6 - 2.4 = 31.2 / 2 = 15.6 + 2.4 = 18 - 13.4 = 4.6

@CuriousCat - Remember that the expansion has changed nonlinearly, you need to use integrals dependent on the scale factor.

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Why are distant type Ia supernovae dimmer than expected?
Why are the (extremely) distant type Ia supernovae dimmer than expected?
Relative to what? Their cosmological red shift? Well, what else.
Please don't just say "Because the expansion is accelerating".
Of course I accept that but what's the actual mechanism?
Is it simply because the CRS\z increases linearly and
the intensity decreases as the inverse-square with distance,
so it gets affected more, by the increased distance?

Very closely related if not a duplicate https://astronomy.stackexchange.com/questions/35847/why-dimmer-high-redshift-supernovae-means-the-expansion-is-accelerating-if-the

The supernovae are observed at some fixed redshift. Their peak brightness will depend on how far away they are. Changing the cosmological parameters changes the relationship between redshift and distance. In this case, introducing the cosmological constant, means the supernovae are further away, compared to the expectations of a decelerating expansion, for a given redshift and are hence dimmer.
A point of confusion here is that redshift is only directly related to proper distance now. The Hubble "constant" in fact changes with time in a complex way (it was bigger in the past) depending on the values of the matter and dark energy densities.
If we estimate the distance to the supernovae using Hubble's law and the vale of $H_0$ today then we overestimate the light-travel distance to the supernovae if the expansion has been slowing down. On the other hand, if the expansion has accelerated over the last few billion years then that brings the estimated light travel time back up towards the naive Hubble's law value and even exceeds it for redshifts of $z\sim 0.5$. Thus when we say the distant supernovae are dimmer than expected, we mean they are further away than expected for a decelerating universe containing only matter.

U are right. The C/RS\z and I/ntensity\d do not disagree. It is simply relative to what was expected, which was a decelerating expansion.


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What is Hubble's parameter/constant H0 supposed to mean, exactly?
Surely it must mean that every second each Mpc expands by that much. Well what else, could it be.
So if the expansion is accelerating, then how can H0 be decreasing!?
Let's look at Hubble's law: v=H0D, so

D
v:

1
70,

2
140,

3
210;

Mpc
km/s.

How does accelerating expansion change this?
(Yes I know, that our local group, let alone galaxy, is not actually expanding).
Am I missing something?

The Hubble parameter is not a direct measure of how fast the universe is expanding; it is the ratio of the rate of expansion of the scale factor $a$ (can think of this as the distance between two galaxies) to the scale factor itself, $$H = \frac{\dot{a}}{a}\ .$$
$H$ is not a constant (in cosmic time) at all. The label $H_0$ means the value of $H$ now.
An accelerating expansion means $\dot{a}$ is increasing. But of course $a$ is increasing at the same time as the universe expands. Thus it is quite possible (as indeed seems to be the case in our universe) for the Hubble parameter to be decreasing with time whilst $\dot{a}$ is becoming larger. It just means $a$ is increasing faster than $\dot{a}$.
If the Hubble parameter were constant then this would means that the expansion was exponential and the scale factor would grow exponentially with time - an extreme case of an accelerating expansion, which our universe may approach asymptotically in the distant future. See here and here.
An accelerating universe does not change Hubble's law at all. $v = H_0 d$ is perfectly correct as long as you correctly interpret the meanings of $d$ and $v$. They are the "proper distance" (the length of a ruler stretched between us and the distant object now) and the rate of change of that distance. In particular, $d$ is not the light travel time distance and $v$ is not linearly proportional to redshift. For these reasons, a plot (or table) of redshift against light travel time distance would not be a straight line (except at smallish distances, where it is approximately linear).

U are right: 1) If H0 was 0, then vs would be 0, expansion would stop and CRSs would disappear. 2) If vs stayed constant, then H0 would have to decrease, significantly, progressively, and anything in/between produces an accelerating expansion.

Accelerated expansion means that a given galaxy recedes from us faster in the future than it does today.
A decreasing Hubble rate $H$ means that for a set distance $d$, the recession rate of objects at distance $d$ in the future is lower than the recession rate of objects at distance $d$ today.
In the former case, we fix the object that we are thinking about. Since it recedes, this means that we are considering an ever-increasing distance as time goes on. In the latter case, we fix the distance, which means that we think about different objects at different times.


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Can we measure how fast space is expanding from how quickly gravitationally lensed galaxies or stars are blurring or getting sharper?
A galaxy that lenses another celestial body should as it moves away from Earth alter the sharpness of what it lenses by gravity. Could this be used to determine how quickly space is expanding?

Gravitational lenses don't bend light like glass lenses.  They don't focus light to a point, and so they don't form a real image in the way that a glass lens does.  So there is no "sharpness", and only virtual and distorted images of the lensed object.

Gravitational lenses don't form images that may be sharp or not.
In the diagram, the light from the point of the arrow is brought together at a point, forming an image.  Notice that the light that passes through the middle of the lens is bend least.  The "sharpness" results from whether you place your film (or CCD etc) exactly at the point that the image forms (in which case you'll capture a sharp picture) or not quite at that position (in which case the image will be less sharp.
But in a gravitational lens there is no image formed.  The light that passes close to the lens is bent most. This is more complicated when the lens is not a single point, but the effect is that there is no image, so there is nothing that can be "sharp".  What we see are multiple virtual images.  For example  we might see light that has come along paths "b" and "c", so we would see the object in multiple locations. The actual effect is to produce multiple, distorted virtual images, all of which are "sharp".

I didn't realise how far removed gravitational lenses are from glass lensing!


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How can Hubble's original data show that the universe is expanding?
The diagram from Hubble's original data only goes from 0 to 2 Mpc (see here), and yet our Local Group is not supposed to be expanding and it is 1.5 Mpc in radius. How can the data show that the universe is expanding when it mostly just includes the Local Group?

The original diagram is I think referring to this one (with mislabelled y-axis units):

If you estmate the Hubble constant ($H_0$) from the solid straight line, you get $H_0 \simeq 500$ km/s per Mpc. This is about 7 times as big as the best estimates of $H_0$ today.
There are two reasons for this:
(1) Hubble was slightly fortunate; the galaxies he looked at were relatively nearby and not fully immersed in the "Hubble flow". In other words they would have been quite badly affected by the scatter caused by local peculiar motions, leading to a very noisy estimate of $H_0$. The scatter around the straight line is about 150 km/s.
(2) More imprtantly, Hubble got the distances wrong. The most distant 4 galaxies are in the Virgo cluster; a present-day distance estimate is 16.5 Mpc, i.e. 8 times further away than Hubble estimated. This means his Hubble parameter was about 8 times too big. Note that the distances, estimated in a few cases from Cepeid variables, but mostly using the apparent magnitudes of the brightest star(s) in each galaxy, would be uniformly wrong. i.e. It is just a calibration/scaling error on the x-axis, which should be about 7-8 times bigger.
A horizon of 16 Mpc for a Hubble diagram is still not very big and the peculiar motions of galaxies around the general Hubble flow would still play a big role here. Ideally, you would want measurements at $>50$ Mpc and Hubble diagrams today do cover a much larger volume of space. See for example Figure 3 in Kirshner (2003), showing a Hubble diagram for galaxies with type Ia supernovae going out to around 700 Mpc. More recent work expands this to several Gpc.

Ah, yes, I remember now! I know that our local group is not expanding but is the space within it still expanding and the gravity holds everything back, and can we still use the cosmological redshift measurements, within it, after allowing for any peculiar velocity?

This editor redlights REDSHIFT. Should we use 1 word or 2?

@CuriousCat: The Stack Exchange editor has no proofreading features, it's your browser (and/or OS and/or third-party software) doing that. Presumably its built-in dictionary was not compiled by cosmologists (but is at least still comprehensive enough to include the common English words "red" and "shift").

I don't know which Hubble diagram you're looking at but modern Hubble diagrams go up to hundreds of Mpc (see e.g. https://www.pnas.org/doi/10.1073/pnas.2536799100 Fig 3), well in the range where dark energy dominates.
If you're looking at Hubble's original diagram, do remember that the distance ladder has been much more accurately calibrated than in his day; we are able to estimate astronomical distances much better than Hubble could.

Yes I did mean the original one. Yes but my question is still valid.


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Does a cosmological redshift occur within our local group, of galaxies?
I know that it is not supposed to be expanding but space is space and it is expanding everywhere else outside it so why not inside it too, and gravity just holds everything back.
I suppose if it is, it would be insignificant, on a cosmic scale, for what is 3 MPcs, out of 14.2K, but then again our instruments are so precise now, but then again there are all the other Doppler and gravitational shifts.
The kinematic/velocity time dilation is caused by the motion thru space, or it moving thru U. It's the same, thing. So the out-flowing space would cause time to run a bit slower in our outer regions. And, if space flows out from our local group, wouldn't it also flow out from a black hole!?
And I mean/t the redshift caused by the expansion of space and not the Doppler shift caused by the peculiar velocities, of course.

I guess that the correct answer is yes there is a cosmological redshif within the local group of galaxies, but the various motions of the galaxies in the local group involve far greater redshift or blueshift that masks the cosmological redshift.   Trying to detect the cosmological redshift in the local group would be like trying to detect the breathing of someone running straight at you.

@MA.Golding: Ye's, but then again the COBE, SCP and High-z SST was also very hard.

The large scale motion is simply the net sum of all the small scale motions.   Otherwise, you get tears in the fabric of space.

No. The cosmological redshift is due to recessional motion. There is no additional redshift due to "expansion of space" (which is not a physical phenomenon).
The only reasons "expansion of space" is often emphasized as the source of the redshift are:

It's the most convenient way to calculate the cosmological redshift for cosmologically distant objects. Over cosmological distances, recession speeds are not uniquely defined, and it is tricky to construct the speed that correctly corresponds to the cosmological redshift. (By "cosmological distances" I mean comparable to the Hubble scale, so billions of light years or more.)

Often, the author holds the misconceived notion that "expansion of space" is physical. That's a common misconception that has even made its way into published scientific articles.

A useful reference on this topic: https://arxiv.org/abs/0808.1081


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What would happen if a huge planet like Jupiter or Saturn was (or could be) terraformed?
Even if these planets could be terraformed, it is very unlikely for this to be done in the future, if planets like Mars or Venus were terraformed. What would happen if Jupiter or Saturn could be terraformed?

Even if you were to define what terraforming means for a gas giant planet, "What would happen" is a question too broad for a Stack Exchange answere

Sorry, but this question doesn't make sense.

They are made of gases and thus do not have solid base.

@JamesK It's not so much that it doesn't make sense, as it's very vague and off-topic. There was for example idea of making a shell around Saturn, but that was sci-fi not astronomy.

The gas giants cannot be terraformed unless they cease to be gas giants.
Let's assume, they have a solid core it is maybe 10 or 20 times the mass of Earth. However Jupiter has a mass of around 300 earth masses, so there is about 280 or 290 earth masses of gas on Jupiter. At the fictious surface of Jupiter's core, the gas is supercritical, thus so compressed that it's in the state where you cannot distinguish liquid from gas anymore due to the high pressure regime. And when you want to remove the excess gas, you remove most of Jupiter's mass and convert it into a so-called super-earth.
The ratio is not as big in favour of gas for Saturn (around 100 earth masses), but its solid core is assumed to be of similar size (here we actually know it has one). But still the atmosphere is at the core's boundary at the supercritical state, and removing that would remove the essence of Saturn from it along with most of its mass.
Settling on the moons of the gas giants sounds like a much more probably thing to do.


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Why is v lowercase and D uppercase in $v=H_0D$?
Why is v almost always written in lowercase and D in uppercase in $v=H_0D$?
OK, v is in lowercase, as usual, but then why is D in uppercase? What's so different/special about it?
In my physics school text/book and Wikipedia it is written as v=H0D, and I always assumed that was the correct form. Yes I have seen it written as v=H0d, but I always thought that was a mistake. I think I assumed that it was D because both my sources pointed out that it was a proper distance, as opposed to a co-moving distance, I suppose. I cannot believe it is just arbitrary. Physics symbols are case sensitive. Was v=Hod considered too confusing!?

It's only that way becasue that if how you typed it.  If you typed V= H0d then the V is uppercase and the d is lowercase.

@James K: No it isn't! It is almost always written as v=H0D. Just Google it.

However your (English) words are written like this and not "Hauewwer juhr wörds ahr not written laik sis" - it's all a matter of convention and making communication easy. A decent source ALWAYS tells the reader what each symbol in an equation is supposed to mean. Just google it. v as in velocity, D as in distance, H as in Hubble. capitalization to not confuse with V for volume or d for ... also distance. why ever.

related: https://astronomy.stackexchange.com/questions/36298/where-do-the-symbols-lambda-and-beta-come-from-for-celestial-latitude-and/37063#37063

It's moderately common to use d for diameter and D for distance, but this is merely a common (not universal) convention. Someone using d for distance is not making a "mistake".

The question is built on a false premise and $v = H_0 D$ is not a universal convention.
If a child asked you, you could point them first to the BBC bitesize revision notes on Hubble's law, which uses $v = H_0 d$.
Then hyperphysics notes, which uses $v = H_0 r$.
(These were the two of the first three non-pdf hits I got when I googled "Hubble's law")
I would then explain to the child that any algebraic symbol can have any meaning you choose to attach to them.

Ye's and no. Ye's I just Googled it and I was surprised by how many hits had it as v=H0d but I always assumed that was a mistake, coz my school physics text/book and Wikipedia have it as v=H0D. I think I assumed it was D because it was a proper distance, as opposed to a comoving distance, which should be d? I cannot believe it's just arbitrary.

@CuriousCat - well, convention is not the same as arbitrary. And there is no entity that dictates symbol usage and holds people to it.

@Jon Custer: Really!? Oh, come on. Physics symbols are case sensitive, and I have only ever seen F=ma, for example, written as such. Never as f=MA.


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Is REDSHIFT one word or two?
I (myself) prefer one but an editor I use redlines it as a misspelling, so which one should we use here?

You don't need quora, only a dictionary https://dictionary.cambridge.org/dictionary/english/redshift

@James K: Which one? Many spell it as 2 words, as does this editor and Quora's too, as does the European Space Agency/ESA.

I think it's rather normal for the English language, when two nouns are combined to form a new noun, to start out separated, then hyphenated, and finally conjoined. In addition to "red shift → red-shift → redshift", examples include "light year → light-year → lightyear", and also "minutes of arc → arc minute → arc-minute → arcmin", I think. "Light-seconds" are still hyphenated though, perhaps because it's not as popular a word.

It's your web browser that's doing the spell check, not Stack Exchange or Quora or the ESA website.

I think the question is OK. From time to time we have questions that ask to clarify certain items necessary to write or publish about Astronomy. See for example How is the star 'Spica' pronounced? and When we say a variable star is "fainting" does it mean something more or different than "dimming" or "fading"? At only 3.6 questions per day, it's not like we're so overwhelmed that we need to split hairs on topicality. After all, this has a well-defined, and it turns out interesting answer!

And that answer is fact-based, so closing for "This question is likely to be answered with opinions rather than facts " seems pointless. voting to leave open

I've rolled back the edits.  Please don't use edit to ask more questions.  If you want to know how to spell "Black hole" please use a dictionary.

I’m voting to close this question because the spelling of a word is best answered with a dictionary.  This is utterly trivial to answer.  There is nothing "interesting" about it at all.

Perhaps, as this asks "which spelling should be used here, this should be moved to meta.

It’s a single word. Used to be double, but quickly became single.
https://en.wiktionary.org/wiki/redshift
https://en.wikipedia.org/wiki/Redshift
https://www.merriam-webster.com/dictionary/redshift
https://www.oed.com/dictionary/redshift_n?tab=factsheet#26403329
https://books.google.com/ngrams/graph?content=redshift%2C+red-shift&year_start=1800&year_end=2019&corpus=en-2019&smoothing=3
(I’d curious about the diminishing number of mentions of “redshift” since 2000—not that “red-shift” has increased during the same period, on the contrary)

But then why does this editor and Quora's too redline it, as a misspelling? Wot, it's/they're out of date?

Scientific terms very often get redlined by non-scientific spell checkers. I've also had spell checkers ping on "email" because they think it should have a hyphen. Language evolves and programs react slowly.


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How do we work out the light travel time on a cosmic scale?
I just read this article in the AUSTRALIAN SKY & TELESCOPE magazine, Nov/Dec 2022 Issue 140, on P16, KEEP YOUR DISTANCE: How far away are the objects we see in the universe? And on P23:
"And here is where our story takes a mind-bending turn:"

What does distance even mean in the expanding universe? On scales of the Solar System, we can understand it fairly easily. But for really remote galaxies, cosmic expansion makes the concept of distance quite tricky. In fact, many cosmologists protest that giving distances for anything farther than a couple of billion light-years should be avoided.
Suppose you measure a galaxy's redshift to be z=1.5, meaning that visible light emitted with a wavelength of 500 nm by the galaxy has been shifted by 1.5 * 500 = 750 nm to an observed infrared wavelength of 1250 nm. The Hubble-Lemaitre Law tells you that the light has been travelling through expanding space for some 9.5 billion years. Intuitively, you'd conclude that the galaxy is 9.5 billion light-years away.
However, you can't simply convert the light-travel time into a distance. When the light was emitted 9.5 billion years ago, the universe was smaller and the galaxy was a "mere" 5.8 billion light-years away. Because space is expanding, it took the light 9.5 billion years to reach the Earth. But by the time the light finally arrives here, the galaxy's "true" (or proper) distance has increased to 14.6 billion light-years.

How do they know that the light has travelled for 9.5 billion years? How do you work/figure it out? And, is it now really 14.6 B light-years to the other galaxy or the original/starting point? The other galaxy has also moved away!

Does this answer your question? Is the distance covered always equal to the time taken?

@ProfRob: The stuff in [square brackets] is what I put in, for my own benefit, and as a matter of fact, 21 B LYs sounds to me to be more like what the proper distance should be now! The other galaxy has also moved away. Are my questions, at the end, really that unclear!?

The light travel time is calculated as the cosmic time interval between when the light was emitted and when it was received at the Earth.
Calculation of this quantity depends on the cosmological parameters that describe the expansion of the universe because, for a given initial separation between the distant galaxy and the Earth (or where the Earth would eventually form), a faster-expanding universe would mean the light took longer to reach us and vice-versa.
There is also a connection between redshift and the cosmic time at which the light was emitted and between the cosmological parameters and the age of the universe now. As a result, there is a non-straightforward relationship between the light travel distance and the redshift, $z$ of a galaxy.
$$d_{LT} = \frac{c}{H_0}\int^{z}_{0} \frac{dz}{(1+z)E(z)^{1/2}}\ , $$
where
$$E(z) = \Omega_m (1+z)^3 + \Omega_\Lambda +\Omega_k (1+z)^2 + \Omega_r (1+z)^4$$
with $H_0$ equal to the present Hubble constant (about 70 km/s/Mpc), $\Omega_m$ the current ratio of gravitating matter density to the critical density (about 0.3), $\Omega_\Lambda$ the current ratio of dark energy density to the critical density (about 0.7) and $\Omega_k$ and $\Omega_r$ are the equivalent density ratios for "curvature" and radiation, which are small in the current universe.
The integral needs to be (in general) calculated numerically. The derivation of this formula can be found in most cosmology textbooks or here. A convenient "cosmology calculator" that implements these equations can be found here. For $z=1.5$ in a flat universe with the parameters estimated from the Planck data, then I get $d_{LT} \simeq 9.5$ billion light years and the light travel time is thus 9.5 billion years.
The distance to the galaxy was smaller than this when the light was emitted and, because the expansion has continued since then, the galaxy is further away than this now. The figure of 14.6 billion light years is the approximate proper distance to the emitting galaxy (if it still exists?) now.
The proper distance now (also called the co-moving distance) is calculated as
$$d_{\rm prop} = \frac{c}{H_0}\int^{z}_{0} \frac{dz}{E(z)^{1/2}}\ . $$
The proper distance when the light was emitted (9.6 billion years ago) is equal to the proper distance now multiplied by the scale factor then (the scale factor now being defined as unity), which was $(1+z)^{-1}$. We thus get a proper distance when the light was emitted of $14.6/2.5 =  5.8$ billion light years.

Minor comment: the integral can be done analytically for the flat matter+$\Lambda$ case.

Let's first be clear that there is no unique way to identify the time or distance between two events. This is true in every relativistic context; just think about relativistic time dilation and length contraction. In fact, the proper time along light's path through spacetime (in vacuum) is always zero. So what does light travel time actually mean?
Fortunately, there is stuff everywhere in the universe, and that stuff is easy to track, in the sense that you can uniquely associate a flow velocity with each point, at least at scales much larger than galaxies (such that the matter is effectively a fluid, in the technical sense). This means you can consider times and distances as measured by the stuff.
The idea is that you can define the time to be that measured by clocks that were synchronized at the beginning of the universe, when all the stuff would have been in the same place, and those clocks simply move along with the rest of the stuff. This is likely how the article is defining the light travel time: it's the time difference between the source's clock at the emission time and the target's clock at the receiving time.
With this array of clocks, you can also define distances to be those measured "at constant time". Here's a conceptual illustration of the relevant distances. The important feature of this illustration is that I am depicting the source ("them") and the target ("us") in a symmetrical way, moving apart at the same speeds. This is demanded by how I defined the clocks, above. We need the speeds to be the same so that the clocks of the source and the target run at the same rate.

There are nevertheless ways in which the picture is inexact. It's on a flat screen, while the Universe's spacetime is curved. Also, even in flat spacetime, the constant-clock-time surfaces would be hyperbolas (due to time dilation), not the horizontal lines that the picture suggests.

This doesn't tell the OP how the 9.5 billion years is calculated. I believe the issue is not with the concept you have drawn but in the actual calculation. The quoted article says Hubble's law is used in the form $cz/H_0$, but this in fact gives 21 billion light years.

@ProfRob: The stuff in [square brackets] is what I put in, for my own benefit, and as a matter of fact, 21 B LYs sounds to me to be more like what the proper distance should be now! The other galaxy has also moved away.

I second @ProfRob . This answer does not fully answers the question. A more detailed explanation is required, especially for users that can't see the image for any reason. Without the visual aid, the worth of this question drops to zero. I downvoted it.

Updated with more detail. Still a conceptual answer; thanks @ProfRob for the numerical answer.

Bright objects have emission spectra, within them are very narrow clearly defined frequencies that are due to the arrangements of electrons in that type of element. By looking at these frequencies they give a "fingerprint" that is unique for each element, complex objects like stars and clouds of dust are made up of multiple different elements but each element's frequencies are still visible.
Because the universe is growing the travelling light has its wavelength stretched out, longer wavelength light is towards the red end of the spectrum which is why the phenomenon is called "redshift". These fingerprints are redder than they would be for a nearby object. The amount of this shift tells us about how long the light has been travelling and how much the universe has grown since it was emitted.
Now that sounds a little circular in that we use redshift to work out the travel time, as the redshift amount depends on the travel time... However redshift isn't the only way to measure distance there are ways of measuring distance against types of objects with a known brightness and nearer objects with parallax, by using these different techniques astronomers have built up a catalogue of how redshift varies with distance/time.
TLDR: We know the light is that old because we knew what colour it was when it started and light doesn't just change colour by itself, but from being stretched which takes time.

I don;t think the OP is confused about how redshift is measured or that higher redshift means the light is "older". The questions asks about the origin of the numbers in the quoted article and how they are calculated.


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One analytic solution of Schwarzschild metrics and Maxwell's equations in this curved space-time
One analytic solution (ref. 1) in projected local orthonormal tetrad e_lambda^, (^ is not power mark in this case)
B_theta^ = (1/r^3)musin(theta)[ 6x^3(1-x^(-1))^(1/2)ln(1-x^(-1))+6x^2*(1-0.5*x^(-1))/(1-x^(-1))^(1/2) ],
B_r^ = (1/r^3)2mucos(theta)[ -3x^3ln(1-x^(-1))-3x^2(1+0.5x^(-1)) ], x = r/(2*M).
(Conversion to SI units M -> MG/c^2, mu -> mu0/(4Pi)mu, mu now Am^2 in SI units, mu0 = 4Pi10^(-7)mkgs^(-2)*A^(-2) in SI units)
Maple 9 plot would give for neutron star (near surface, max limit about neutron star (magnetar here) calculation?)
max B_theta^ = about 1.210^13T (equator, theta=Pi/2), max B_r^ = about 5.210^12T (pole, theta=0),
Maple 9 plot would give for neutron star (near surface, min limit calculation ?)
max B_theta^ = about 7.4510^9T (equator; theta=Pi/2), max B_r^ = about 3.2510^9T (pole, theta=0).
I have used following in my calculation: M = 1.5M_Sun = 2.9827510^30kg,(I don't remember reference of this), c = 2.9979248810^8ms^(-1), G = 6.674310^(-11)**m^3kg^(-1)s^(-2), mu = 1.610^32gausscm^3 = 1.610^29A*m^2, (= magnetar magnetic moment for max neutron star magnetic moment, I don't remember reference of this),
(observed values (Ref.2) for magnetic moments are mu = about range (10^29->10^32)gausscm^3 = (10^26->10^29)Am^2).
mu0 = 4Pi10^(-7)mkg*s^(-2)A^(-2), SI-units of B are [B] = kgs^(-2)*A^(-1) = T = Tesla.
If I calculate with mu = 10^29Am^2 as observed max for neutron stars I get from Maple 9 plots (near surface):
max B_theta^ = about 7.4510^12T (equator, theta = Pi/2), max B_r^ = about 3.2510^12T (pole, theta = 0).
Observed values of magnetic moments of neutron stars are from Ref.2
Please take a look these analytic formulae and the Ref.1,
(this article was 22.2.2024 in sci.physics.relativity, but Google Groups ended and stopped 22.2.2024, all old postings are still readable after 22.2.2024, for example mine use search term "Hannu Poropudas".)
Q1. Please give me comments about how usable these analytic formulae are ?
Q2. Maple 9 plots given about B_theta^, B_r^ numbers: What is possible range of M ? and what is possible range of magnetic moment of neutron star ?
Q3. Max ranges observed for neutron stars (M=1.5M_Sun): max B_theta^ = about 7.4510^9T -> 7.4510^12T (equator, theta=Pi/2), max B_r^ = about 3.2510^9T -> 3.2510^12T (pole, theta=0) ?
Reference 1.
Wasserman Ira and Shapiro Stuart L., 1983. Masses, radii, and magnetic fields of pulsating X-ray sources: Is the "Standard" Model self-consistent ? The Astrophysical Journal, 265: 1036-1046, February 15. (G=1, c=1 units, c.g.s units, electromagnetic units not SI units).
Reference 2.
Ghosh P. and Lamb F. K. 1979. Accretion by rotating magnetic neutron stars. II. Radial and vertical structure of the transition zone in disk accretion. The Astrophysical Journal, 232:259-276, 1979 August 15, page 273.

Hello Hannu.  Please not that this site has TeX syntax enabled so you can write maths like  $B_\theta$ with dollar signs

I'am sorry but I have not TeX available at the moment.

No, this site has TeX available.  You just need to retype your maths in TeX format.  At the moment it is almost illegible

I'am sorry, but it has been about 18 years when I used LaTex last time. So I cannot edit so at the moment.

This question could probably do with some editing. Of your three questions, Q1 seems to be overly broad and general with no possible "best" or "correct" answer.  Q2 seems answerable,  Q3 doesn't appear to be a question

"18 years..."  Time to learn again.  It's up to you.  But at the moment the notation is almost illegible.  So do you want answers or not?

Yes, but it will take little longer time than at the moment?

The question does not start with a sentence I understand. It's unclear to me where single formula start and end. The plots referred to are not shown. Symbols not explained. Links to references work on the internet. Why should we search on google groups for your name instead of you linking to whatever further information might be required to understand your question or improving the legibility in the first place?

Please look "edit" version where I have put multiplication mark (*)  in correct places.

I found with Google two partial answers to Q2 question: I. Typically, astronomers consider the mass of a neutron star to range from 1.4 to 2.9 solar masses. Then, if the collapsed core has more than three solar masses it becomes a black hole. II. The mass of a neutron star can vary, but most neutron stas have a mass ranging from about 1.4 to 2.1 times the mass of the Sun. This range is known as the "Chandrasekhar limit" and represents the maximum mass at which a white dwarf star can remain stable before collapsing into a neutron star.

Putting in the * has just made things worse, since the * is used in markdown for italics.  There is a right way to format maths, and you are not doing it the right way. You say ^ is not power mark.... so what is it?  I've voted to close, as I don't think this question can be answered in this state.

It’s very inconvenient to read, I suggest replacing the text of the formulas with screenshots from Maple, what do you think?

I don't want to waste my time to this my trying to convert my ASCII text to LaTEX form.

About those Maple 9 plots: I used in all plotting ranges from r = 0 to about 1000*(2MG/c^2). Range from 0 to (2MG/c^2) Maple 9 gives complex numbers, so

About those Maple 9 plots: I used in all plotting ranges from r = 0 to about 1000*(2MG/c^2). Range from r = 0 to (2MG/c^2) Maple 9 gives complex numbers, so this range is not plotted. Maple 9 plots then near singularity r = (2MG/c^2) to about 1000*(2MG/c^2)  deeply decreasing hyperbola like curve, which started about those maximum values which numbers I gave to B_theta^ and B_r^. About Q2. Range for mass of neutron stars is about 1.4M_Sun to about 2.9M_Sun and range for magnetic dipole moments for observed neutron stars is 10^26Am^2 to 10^29Am^2 in SI-units.


text:
Converting equatorial coordinates of a comet's pole into cometocentric orbital coordinates I and Phi
The comet's spin axis orientation is usually reported in equatorial coordinates RA, Dec; however sometimes these are shown in form of the Eulerian angles equatorial obliquity (I), i.e. the angle between the orbital and equatorial planes of the comet, and cometocentric solar longitude at perihelion (phi), measured from the vernal equinox of the comet in the sense of increasing true anomaly to the sub-solar meridian at perihelion. I am looking for the formulas to convert the equatorial coordinates of a comet's pole into I and Phi.
Description is given in: Sekanina, Z. 1981a. Rotation and precession of cometary nuclei. Ann. Rev. Earth Planet. Sci. 9, 113–145.


text:
Why wasn't the planet Uranus recognized by ancient cultures?
Mercury, Venus, Mars, Jupiter, and Saturn have been identified by ancient astronomers as they are visible with naked eye at night. The planet Uranus, despite being visible during very clear nights, wasn't recognized as such. Why?

It is very very faint.  It seems entirely unsurprising that it wasn't noticed as a planet.  Put yourself in their shoes. Would you notice it?

I'm guessing a combination of its low visibility and very long orbital period (~84 years, so a very long lifetime by ancient standards) means that nobody noticed it was "wandering" around the background of stars like the other planets did.

An astronomer in the ancient world stands a good chance of seeing Saturn complete an entire revolution, and maybe even two. It is very unlikely that even a single astronomer would ever have seen Uranus complete a cycle.

Add to that the fact that Saturn, the faintest ancient planet (comparing maximum magnitudes), is still brighter than all but two nighttime stars (Sirius is significantly brighter and Canopus is just a small bit brighter).  There's no missing Saturn in the night sky if you're looking around, and seeing that it has moved over the course of a year (~12 degrees per year) is very easy.  (Mercury is much more difficult to see but it moves so obviously even day to day and its brightness changes dramatically in such a way that anyone who was able to see it regularly would know it was weird.)

The source of light that is Uranus was observed as far back as 128 BC.  However it was misidentified as star as late as the 1760's.  Then it was observed by Herschel in 1781, who also misidentified it, but as a comet.  He had a telescope he built himself which would have been an impressive instrument for the day.  He contacted another astronomer, one Nevil Maskelyne, who first suggested it was not a comet, but a planet.  Further analysis by other astronomers made this more certain.  It was essentially the orbit (not very elliptical) which seems to have been the main reason for identifying it as a planet. That and the apparent lack of a tail.
It required advances in observational accuracy and mathematical analysis of the observations to work out the orbit to work out it was a planet.  These things simply were not possible to the required accuracy (or theoretical understanding) until Herschel's time.
From comments and by way of comparison, Saturn was considered a planet in ancient times. It has a shorter orbital period and is a great deal brighter than Uranus. It's relatively easy to spot this bright object whose (apparent) motion can be noticed and is easily distinguished from stars (unlike Uranus).  Saturn is the 8th brightest natural object in the night sky. Uranus is the 350th.  All the objects considered planets in ancient time were easier to see and notice the apparent motion of compared to stars.  Uranus was just beyond that "natural" threshold where this could be done without relatively advanced technologies and theory.

And then there was a big to-do about how the planet should be named, because in all recorded history, nobody had ever discovered a new planet.

...and he HAD to be accurate, because stating a body to be a planet requires absolute precision and truth and if misidentified, the person would have a become a joke especially during those times.

@A.R. You mean they could name it anything and they literally chose Uranus?

@pipe Yep. Many other names were considered, including "Herschel" (imagine if the planets were Mercury, Venus, Earth, Mars, Jupiter, Saturn, and Herschel; just doesn't feel right), "Neptune" (a fine name, but one that just punts the problem down the road a few decades), and "Georgium Sidis" (a completely ridiculous name which would have honored the British monarch at the time).

I feel like this answer doesn't actually respond to the question.  It provides interesting context for the discovery of Uranus, but it doesn't really explain why it wasn't identified as a planet earlier.  It didn't require advanced orbital calculations to determine that the other planets were planets, and it wouldn't have done for Uranus had it been readily visible and moved a significant amount night-to-night.  Uranus is near the edge of visibility, even in dark skies.  It's easily mistaken for any one of thousands of similar stars, and it moves very slowly.  That's why it wasn't recognized.

@JasonPatterson I think I did address the issue directly.  The fact is that it was not identified until the combination of technological precision, understanding and theory reached a point where t could be correctly and consistently identified.  Note that Herschel observed it was not a star but could not make the leap to "planet" alone.  For all history until that time there we a certain number of planets, now someone claimed there's another one ?  That's a huge conceptual leap for the time.  All done without photography, computers, electronics.

@StephenG-HelpUkraine Right, but why did it require that conceptual leap and Saturn didn't?

@JasonPatterson Saturn was recognized as planet very early on.  It has a shorter orbital period, is a great deal brighter than Uranus.  It's relatively easy to spot this bright object whose (apparent) motion can be noticed and is easily distinguished from stars (unlike Uranus).  In ancient times I doubt "planet" had the conceptual meaning we give it now.  By the time of Uranus's discovery, "planet" had a conceptual meaning we would recognize.  Saturn is the 8th brightest natural object in the night sky.  Uranus is the 350th.  There's no comparison in terms of identification as a planet.

@StephenG-HelpUkraine I think your comment about comparison to Saturn (or other planets) may give a helpful context, would you add it to your answer?

@AndrewT. Done.

@StephenG-HelpUkraine That was the intent of my earlier comment.  Apologies if it seemed terse or felt like an attack.  It originally began with, "I think this is an excellent answer and I have already upvoted it, but..."  A clear tone fell victim to character count.

@JasonPatterson Your comment led my editing more (hopefully) useful info into the answer I gave, which is always considered a good thing on SE.

@pipe Think about this way: They could choose any name they want so of course they would choose Uranus. Gotta make it count.

I like ScienceSnake's comment - a combination of faintness and its comparatively small apparent motion.  It was observed in 1690 by John Flamsteed on several occasions, but because it moved so little from night-to-night, he thought it a star and identified it as 34 Tauri.  Had it moved more he almost certainly would've recognised it as a local body. It's also possible he was observing around the time it went into retrograde, when it would move even less.
For the ancients to have identified Uranus as "wandering" they would have had to record the precise location of every naked eye object (4000+), and then done so again after some time, and identified Uranus as having moved.  This isn't easy.  It depends on where it is relative to the Earth, but Uranus moves at roughly 42 arc seconds per day.  It takes roughly 45 days to move the same distance in the sky as the apparent diameter of the moon.  If you made two star charts a month apart, you'd have to accurately identify that an object at the limit of visibility had moved less than the diameter of the moon - and to have done so without any instruments.
I believe had it been brighter - say as bright as Saturn - they would have discovered it.  As it was, it was just too dim and too slow to notice.

It would require a lucky coincidence
As noted by others, Uranus is faint and easy to misidentify as a fixed star. However, Bourtembourg 2013 claims that Uranus was observed already in 128 BC by Hipparchus and recorded in his star catalogue (that's the source for this claim on Wikipedia). They mention that the goal of the catalogue was to notice changes in the sky. So if somebody followed up on the (likely extremely tedious) observations by Hipparcus a couple times in the following years, they could have plausibly noticed the movement. Many later astronomers did notice that the star from the catalogue was missing, so it doesn't seem such a stretch that movement could have been noticed as well if the followup observation was made early enough.
Bourtembourg even argues that at the time of observation, Uranus must have been apparently almost motionless (due to just transitiong from apparent prograde to apparent retrograde motion), because otherwise Hipparchus would have noticed Uranus moving over consecutive observations. They then use this claim to support a particular dating of the observation time.
So taking Bourtembourg at their word, it seems that if either: a) Hipparchus did his observations at a different time or b) somebody actually tried to use the catalogue to check for motion in the following years, Uranus would at least rise suspicion and with a bit more luck it could have been recognized as a planet.
EDIT: But note that Hipparchus's original observation and his creation of a catalogue is already quite a lucky coincidence, so I'd say the ancients would need huge luck to classify Uranus as a planet.

Reading the paper it seems to me to contain an lot of supposition and assumption rather than being a truly convincing demonstration that Uranus really was the observation. I wonder how this is viewed by more qualified scholars of astronomical history?

(the "more qualified scholars" there was meant to be in comparison me, reading it back I fear that wasn't clear)

@JackAidley I am not a scholar of history, neither of astronomy, but my impression is that whether the Uranus observation was really made or not, there is little doubt that Hipparchus and other ancient astronomers were in principle able to measure the position of objects as faint as Uranus with sufficient precision to notice the movement. However, there are many such objects and observations would have to be timed well and record keeping would have to be good, etc. so overall it is quite unlikely, but not completely inconceivable.


text:
Are spacecraft visits to Uranus and Neptune hard to plan?
Until today, Voyager 2 remains the only spacecraft to visit these planets. Are spacecraft visits to these planets hard to plan or develop?

Define "hard".  You have to squeeze $4 billion plus out of governments, and that's before you get off paper, but after you have at least a partial design and detailed plan.  Then you have to build stuff.  And it has to work, because you don't get second chances.

Voyager 2 was able to use multiple gravity assists.

https://space.stackexchange.com/a/45634/37747

Setting a goal and planning is difficult, but it is possible. But what is the purpose of such a journey?

Further to StephenG-HelpUkraine's excellent comment, can you  also define 'plan'?
Is that simply to plot the course, or does it include all the administration and logistics, as for instance what cargo is carried and even, how the cargo is distributed in the holds?

Uranus and Neptune are far - in orders of light-hours away. It took Voyager 2 nearly a decade to reach Uranus since its launch, costing almost $1B to build it in the first place.
It's not that we can't launch another mission, we're more than capable of doing it again, especially with the evolution of technology - but rather that it is ludicrously expensive and time-consuming, so NASA chose to allocate capital to other projects.
Even if we chose to launch another mission to the ice giants, expect up to a decade for development (depending on how fancy they make the probe(s)), then another decade for arrival. Not very timely.

In addition, if we were going to send a mission to one of the ice giants, we'd probably want to drop into orbit out there and get an extended data collection period, which is really difficult. You can either get there in a relatively reasonable time and be going so fast that you can't stop, or you can get there at a reasonable speed and take so long that the probe might fail from general wear and tear before it arrives.

I would expect the delta V requirement for either kind of mission could be greatly affected by the relative positions of Earth, Jupiter, and Saturn in their orbits.  The Voyager probes were launched when they were to exploit the particular alignment of the planets at that time; if you can readily find information about whether any future launch windows would be particularly advantageous, it might be useful to include it.

@supercat, most of the Voyager boost came from Jupiter.  An ideal Earth-Jupiter-Neptune or Earth-Jupiter-Uranus alignment happens about every twelve years; sub-ideal alignments happen almost every year.  (The Voyager Grand Tour alignment happens every 175 years, but you don't need that if you're just going to a single destination.)

@user253751 I'm not so sure about that. In a quick search it looks like the federal budget as a % of GDP has been relatively static, though overall trending slightly higher, for the last 50+ years. The definition of "public interest" has changed - NASA budget as % of federal budget peaked with Apollo and has been trending downward ever since (except for a slight bump early 1990s).

They're expensive for sure, and you will be giving something up if you plan and build for such a mission.  Since Voyager, the priorities haven't aligned for any such mission to be chosen over other alternatives.
However, the most recent Decadal Survey (which gives an idea to priorities that folks would like to see) has selected a Uranus orbiter and probe mission as the highest priority for flagship (big/expensive) missions.
So there is a reasonable chance of such a mission being funded this decade.

No, it's not. You shoot a probe past Jupiter for a slingshot assist, and you reach Uranus in six, and Neptune in eight years or so. You can build that probe from old blueprints of Galileo and Cassini, and more recent Juno etc.
The problem is to plan and design a probe and mission that will show something we haven't seen and don't already know or can guess very well. If Cassini hadn't been more than a somewhat improved version of Galileo, it would have sent back nice pictures of Saturn with little scientific value. That would've been, paraphrasing Ernest Rutherford, "stamp collecting".


text:
Area of Influence of Lagrange points?
By which I mean, the area around a Lagrange point where the linear-force approximation of the 3-body system defined at the Lagrange point itself is "good".
Background
Consider the restricted 3-body problem, where there are two masses that affect each other and a test one which feels the gravity (but does not effect) the other two. In this setting, the sphere(oid) of influence of a smaller of the two main masses (ie planet) is the region of space where the dynamics of the test mass is better approximated by considering only the gravitational force from the planet than only that from the star.
If we restrict the star and planet to moving around each other in a circle, we find that there are 5 Lagrange points where the test particle can be placed at rest with respect to the two principle masses (ie, in a co-rotating frame) and feel no net force. This naturally invites us to expand the total 3-body forces (including things the fictitious centrifugal Coriolis forces from being in a rotating frame) acting on that test mass as a Taylor series about a Lagrange point. If we keep the linear terms (or, equivalently, the quadratic terms in the series expansion of the pseudo-potential), we get the linearized approximation of the forces at that Lagrange point. This approximation is hugely useful and has been extensively studied for centuries because it can be solved exactly (unlike the full 3-body problem) and so can be used to determine the stability and general structure of the dynamics near the Lagrange point.
My question is, how far out is this linear approximation good? Specifically, how far away can we place the test mass before this approximation of the forces acting on it is less good than the 2-body approximation of only considering the star's or only the planet's gravity? By analogy with how the Sphere of Influence of the planet is defined, I'm calling this region of space where the linearized forces are a better approximation than the alternatives the area of influence of a Lagrange point.
Not stability. Note that I'm not asking about the region of stability. That's a completely different question. L1, L2, L3 have no stable orbits, yet it's a mathematical certainty that as you get close to these points the linear approximation of the forces become arbitrarily good. Indeed, this linearized approximation around L2 predicts unstable orbits that are very similar to (if not exactly the same as) the halo orbits we see at L2.
Is this a meaningful thing to define? I think it ought to be, because basic calculus tells us that as you approach the Lagrange point the linearized approximation becomes better, and matches exactly at the point itself. Very far away, this same approximation is obviously terrible. Hence, there must be a cross over point.
Has this been done already? I'd assume so, but I couldn't find anything on the matter online or in a textbook on the matter. It seems like it would be useful though, as it could be used to improve the patched conics approximation (moving between conic orbits around masses, and linear dynamics around Lagrange points; all of which can be solved exactly without numerical integration).
How could this range of influence be calculated? My intuition is that you could do it by taking the 3rd derivative of the total potential at the Lagrange point, and look for the distance where this contribution is of the same size as the 2nd derivative. Aka, look for where the neglected terms in the approximation start to matter. I'm not sure that this would be accurate in practice however. There's also the problem that the potential around the Lagrange point includes a velocity-dependent term (from the Coriolis fictitious force), which might potentially mean that the region of influence would be velocity dependent? Does this render the whole concept meaningless? Or does it just mean that we have to think of it as being in phase space?
EDIT:
Seeing some interest but a lack of answers, I did some preliminary numerical exploration. Below are examples of which types of approximation has the smallest error compared to the real dynamics (yellow is the heavier body, green the smaller one, and blue the linearized equations of motion around the respective Lagrange point):

The top one has a mass ratio of ~1/10 (similar to the Pluto-Charon system), the middle one ~1/82 (Earth-Moon) and the bottom one ~1/1000 (Jupiter-Sun). Note that because I'm ignoring the z axis and eccentricity completely, these shouldn't be considered accurate of those particular systems, I just wanted to show the dependency on mass ratio. All the forces were computed in a co-rotating frame with the principle two body dynamics and so including centrifugal forces. Coriolis forces are not included because they are linear in velocity and so represented exactly in the linear approximations at the Lagrange points. Note that this doesn't exactly match the definition of Sphere of Influence as defined in places like here, but it wasn't obvious to me how to map that definition exactly to this context. For "smallest error" I calculated the norm of the difference between the exact force vector and the force vector for each approximation. The source code can be found as a Mathematica notebook here.
A few observations: Firstly, the regions aren't tiny. They do shrink with mass ratio (and L4/L5 shrink much faster than the rest, to the point of being invisible at this resolution for a 1:1000 mass ratio) but L1/L2 remain pretty large, comparable with the SoI of the smaller mass. Secondly, they're not at all spheroid. L1 and L2 seem to eat into the smaller masses SoI in a Pacman like fashion, while L3 is weirdly rectangular. Thirdly, just because the linear approximation around a Lagrange point is better than ignoring one mass or the other entirely, it doesn't make the approximation good, it might be that a full 3-body treatment is needed in some places to get anything vaguely accurate or useful.
All this aside, I'm no closer to computing any of these areas analytically. Or even establishing how their general size scales with mass ratio.

Some good posts on our sister sites on Lagrange points: https://physics.stackexchange.com/q/36092/123208 https://space.stackexchange.com/q/49002/38535 & links therein.

Thanks, I've had a skim through them, but I think they're asking a very different question. Stability is about whether the forces near the Lagrange point will tend to kick an object out or not. But that itself is working under the framework that the forces acting are the ones that come from a quadratic approximation of the true 3-body problem about that Lagrange point. This is a good approximation (arbitrarily good even) provided one is close enough to the Lagrange point. But I'm asking how far out this approximation can be made, regardless stability of orbits.

Although perhaps the Physics StackExchange would be a better place for this question. Are there rules/etiquettes about posting questions on multiple sites?

Cross-posting is discouraged. Your question is fine on this site, and all the celestial mechanics experts on Physics.SE also visit here.

Instead of editing answers into the question, you could self-answer.  If you have solved your problem yourself that should be celebrated! Your diagrams look interesting. One observation, the usual pictures of the pseudopotential in a rotating frame don't include the effect of the coriolis force,

@JamesK I considered it, but my edit is far from an complete answer. It's a numerical exploration of what an answer might look like. It doesn't answer whether: a SoI around Li is useful thing to consider (for example, as an improvement to patched conics); or if my way is the right way of defining it; or provide a general formula for its size. As for the Coriolis force, I didn't include it because it's treated exactly in the 2-body system (part of Keplerian dynamics) and in the linear approximation around Lagrange points (because it's already linear), so it doesn't contribute to the errors.

You may enjoy this closely related topic: Polar Orbits Around Binary Stars by Greg Egan. https://arxiv.org/abs/1510.05345

Right, so this is everything I've calculated on this topic, and as far as I'm able to go on this; at least, as far as I'm able without devoting serious mathematical machinery and computing power. This answer will serve as my own notes on the topic, so I'm being particularly thorough.
Problem Definition
Consider two masses in circular orbit around each other. We pick our units such that they have a separation of $1$ and with total mass $M_1 + M_2 = 1$ with $M_1 > M_2$. We define the mass parameter $\mu = M_2 / (M_1 + M_2)$, which is between $0$ and $1/2$, which acts as the sole parameter of the dynamical system. Throughout we'll be working in a co-rotating reference frame with the origin at the barycentre such that $M_1$ is always at $(-\mu, 0)$ and $M_2$ is at $(1-\mu, 0)$, and confining ourselves to the x-y plane. For reasons of mathematical simplicity (which also coincides with astronomical reality) we are mostly concerned with the situation where the two bodies have very different masses ($\mu << 1$) and will often expand our equations as a power series in $\mu$ in order to solve them.
In this set up, the total force vector (per unit mass, but this distinction will not be made from now on) acting on a test particle at point $\underline{r}$ is
$$
F_{exact}(\underline{r}) = -\frac{(1-\mu) \underline{r}_1} {r_1^3} -\frac{(-\mu)
 \underline{r}_2}{r_2^3} + \underline{r} - 2 \hat{\underline{z}}\times\underline{v}
$$
where $\underline{r}_{1,2}$ is the vector from $M_{1,2}$ to the point $\underline{r}$ respectively, and $\underline{v}$ is the velocity of the particle. The terms in that equation are the gravitational attraction from $M_1$, from $M_2$, the centrifugal pseudo force, and the Coriolis pseudo force respectively. We assume that this test particle is of negligible mass such that it does not perturb the orbits of the two main bodies.
It is well known that there are 5 points where $F_{exact} = 0$, known as the Lagrange points. Three are colinear: $L_1$ is between the two masses, $L_2$ beyond $M_2$, and $L_3$ on the opposite side of $M_1$. Finally, $L_{4,5}$ form an equilateral point on the plane with the two masses.
Approximate Forces
Because the three-body system (even in its simplified, restricted form) given above is non-integrable, it can be desirable to approximate the exact forces such that the problem becomes exactly solvable. A common way to do this is to ignore one or the other mass:
$$
F_{M_1}(\underline{r}) = -\frac{(1-\mu) \underline{r}_1} {r_1^3} + \underline{r} - 2 \hat{\underline{z}}\times\underline{v}\\
F_{M_2}(\underline{r}) = -\frac{(-\mu)
 \underline{r}_2}{r_2^3} - 2 \hat{\underline{z}}\times\underline{v}
$$
The first case corresponds to simply ignoring the smaller mass: transforming back to an inertial frame gives the standard Kepler problem for the test particle and $M_1$. For the second one we cancel out the attractive gravitational force with the centrifugal repulsion. Transforming from the rotating frame to one with fixed alignment but who's origin moves to stay centred on $M_2$ also reduces to the Kepler problem.
Both of these approximations are intuitively quite good if $r_i << r_j$, but become poor when the forces are close to cancelling out. This happens precisely at the Lagrange points. Hence, another approximation to consider is to expand $F_{exact}$ in the vicinity of the i'th Lagrange point (located at $\underline{R_i}$)
$$
F_{L_i}(\underline{r}) = 0 + \tilde{M} \underline{\Delta r} + O\big( \underline{\Delta r}^2 \big) - 2 \hat{\underline{z}}\times\underline{\Delta v}
$$
where $\underline{\Delta r} = \underline{r}-\underline{R_i}$ is the displacement from the Lagrange point, essentially shifting the centre of the co-ordinate system, and $\underline{\Delta v} = \underline{v}$ is the (unchanged) velocity in this translated frame. $\tilde{M}$ is a $2\times2$ matrix whose entries depend on $\mu$ and which Lagrange point is in question: its exact values can be found be differentiating $F_{exact}$. Note that, as the force is precisely $0$ at the Lagrange point itself, if we neglect quadratic and higher order terms in the displacement from $\underline{R_i}$, we have dynamics that are linear in $\underline{\Delta r}$. This makes them integrable and so completely solvable - this is precisely the method used in determining which Lagrange points are stable and which are not.
'Best' Approximation
Let us define the Area of Influence of the Lagrange point $L_i$ to be the region where
$$ |F_{L_i} - F_{exact}| < |F_{M_1} - F_{exact}| \text{    and    }\\
|F_{L_i} - F_{exact}| < |F_{M_2} - F_{exact}|$$
aka, where it is the 'best' approximation of the dynamics. Note that as the Coriolis force appears in exactly the same way in every expression for the force, it cancels out completely. The region defined this way thus depends only on position and not on general phase space. This definition is closely linked (though not quite identical) to both the definition for the Sphere of Influence and to the definition of the Hill Sphere
For illustrative purposes here, we'll define the Area of Influence of $M_2$ as the region where
$$ |F_{M_2} - F_{exact}| < |F_{M_1} - F_{exact}|$$
and similarly for $M_1$.
Solution
As shown in the (edited) question, it is possible to plot those areas via brute force numerical methods once the value of $\mu$ has been fixed.

where the arrows indicate the Lagrange points, the blue regions are their SoI, while the green one is the SoI of $M_2$ and yellow of $M_1$.
I have not been able to compute the exact boundary of the blue regions analytically. I'm even dubious whether an exact closed-form solution to this exist (just calculating the position of the colinear Lagrange points is already a quintic equation, and these are an input when it comes to calculating their SoI).
What I did do was find an approximation for the extent of the SoIs on the x-axis, as a series in $\mu$ that holds exactly as $\mu\to0$. The results for $L_1$ and $L_2$ are illustrated here

The exact force and it's various approximations in the vicinity of $M_2$. The black line is $F_{exact}$, the green one $F_{M_2}$, the orange one $F_{M_1}$, and the blue/purple ones $F_{L_{1,2}}$ respectively. Note that the latter lines are tangent to the exact force and touch it where it crosses the x-axis (the position of the Lagrange points) by construction. The vertical red lines are where I calculated (to lowest order in $\mu$) the boundaries of the SoI of the Lagrange points as defined above. That is, where either the green or orange line is as close to the black line as the blue/purple one.
Going outwards from $M_2$ we have the first boundary between the SoI of $M_2$ and of $L_{1,2}$ at a distance of $\mu^{1/2}$ from $M_2$. The position of the $L_1$ and $L_2$ themselves are a Hill radius away from $M_2$: $ (\mu/3)^{1/3}$. The boundary between the SoI of the Lagrange points and of $M_1$ are a further half Hill radius from themselves: $1/2 (\mu/3)^{1/3}$. For $L_3$ the results are simpler because they never interact with the SoI of $M_2$, here the boundary for the SoI of the Lagrange point is a symmetric $0.288... \mu^{1/2}$ along both direction.
I'll stress once again that these results are only for the x-axis and only to leading order in $\mu$. While the approximations for the boundary between the SoI of Lagrange points and $M_1$ holds pretty well even for largish $\mu \approx 1/100$, it converges much more slowly for the $M_2$ interface. Indeed, even for the $\mu=1/10000$ plotted we can see that the red line at the $M_2$-$L_2$ interface isn't quite where the purple and green lines cross. Also, because this only applies along the x-axis, it doesn't tell us anything about how the SoI extends in other directions. Nevertheless, based on the 2-d plot above, it's not an unreasonable approximation to treat them as roughly circular, with their extent in the y-direction being of a similar magnitude.
As the $L_{4,5}$ points are not on the x-axis, I could find no analytical results for them. Nonetheless, but playing around graphically with different values of $\mu$, I found what appears to be a decent heuristic for their SoI, as shown by:

The yellow background is the SoI of $M_1$ and the blue area the SoI of $L_4$. The dotted boundary is my heuristic for this: an ellipsoid with semi-major axis $5\mu/3$ and semi-minor axis $\mu/3$, rotated by $24.7...^{\circ}$. And symmetrically for $L_5$. Once again, this holds very well provided that $\mu$ is small enough, but discrepancies are clear once it gets above $1/100$.
To summarise the mixture of numerical and approximate analytical insights: the SoI of $L_{4,5}$ scales linearly with $\mu$, while for $L_3$ it scales as its square root; and for $L_{1,2}$ it scales with the cube root.
Application
The question remains of whether the sphere of influence of a Lagrange point, as defined here, is of any use. I propose that it might be a good way of 'upgrading' the patched conics approximation. In that method, orbits through a complicated system are approximated by only considering the dominant gravitational attraction and neglecting all others; resulting in a chain of Keplerian orbits as the test particle moves between the sphere of influence of various bodies of the solar system. The addition I suggest is to also consider the sphere of influence of the Lagrange points, chaining together the solutions of the linear dynamics in those regions with Keplerian orbits. This would be of negligible extra computational cost (because propagating a trajectory through a linear system comes down to a single matrix multiplication), and without the problems that numerically integrating the full 3-body dynamics brings. Importantly, it ought to represent the most important elements of 3-body dynamics, qualitatively if not precisely quantitatively. If I have the time, I'll put that claim to the test.

Is there a meaningful way of defining the area of influence of Lagrange points?

No.
For several reasons actually!
The basic one that you already recognize is that there's nothing there, there's no special gravity field that comes from these particular places in empty space.
And you don't experience gravity from the other two bodies any differently there than you do anywhere else.
This is a dynamical problem. It's not like you're on a "normal" 3-body orbit until you get close to a Lagrange point when suddenly a new kind of influence starts to kick in, you have to consider the full trajectory.
You should proceed with the simulations discussed in comments, they will provide a lot of insight.
That said
If you focus on the concept of libration points which are Lagrange points with some semblance of stability (e.g. L4 and L5), you could think about how large of an area objects can move and still stay associated with those points.
The problem is, even in 2D, it's a 4D problem! The potential plots we see for stability are usually zero-velocity potentials the trajectories and stability regions are different for each non-zero initial velocity you might choose.
See for example

Why shouldn't we illustrate spacecraft trajectories on top of static zero velocity pseudo-potential surfaces?

Definitely start simulating, it's really fun and instructive!
And feel free to post your result an an answer post!

Of course there aren't different forces around a Lagrange point, just like there aren't new or different forces that appear when you cross the boundary from the Sun's SoI to the Earth's. It's the 3-body problem everywhere. The whole concept of SoI is about which approximation is best: the 2-body problem with just the Sun or the one with just the Earth? My question is why not add the linearization of the forces around a Lagrange point to this list of approximations? I thought I explained that in the OP, can you say which bit you didn't understand so I can edit it to make it clearer to others?

@ScienceSnake I'll give it some more thought. btw there are more orbital mechanics folks in Space Exploration SE as well. You might consider asking a follow-up question there and linking back to your question here. We don't ask the same question in two sites in Stack Exchange, but it's quite common to ask related or follow-up questions in the same or a different site. Check it out :-)


text:
My question is about the warp of time and space near black holes
I have seen some recent articles indicating that there might not be a singularity at the center of black hole at all1. Have further heard commentary about black holes that spin incredibly fast.
I was wondering if this spin could actually provide enough centrifugal force to keep all of the mass in the Event Horizon of larger and faster spinning black holes. If this is the case can we use the rotation of the black hole to increase your mass to such an extent that you could Warp or Wormhole through the Event Horizon if any of this is true wouldn't a black hole have the thinnest Event Horizon near its poles as opposed to at its rotational Horizon.
What little research I've done because I'm not an astrophysicist has led me to the conclusion that warping through a solid object would probably be a very bad idea whereas a wormhole might actually provide an opportunity. Of course this is theoretical but we technically can't detect a singularity because the Event Horizon makes it impossible. Is this theoretically possible?

1 Science News Study finds that black hole inner horizons can be charged or discharged which links to Klein, Zahn and  Hollands (2021) PRL
Quantum (Dis)Charge of Black Hole Interiors

The theory of general relativity unites space and time into the concept of spacetime and describes gravity as a bending of that spacetime," Christiane Klein, one of the researchers who carried out the study, told Phys.org. "One of its most prominent predictions are black holes (i.e., regions of the spacetime from which even light cannot escape). If a black hole is electrically charged or rotating, its interior has an interesting feature: Inside the black hole, there is a surface with properties resembling those of the event horizon (i.e., the outer edge) of the black hole. It is therefore called the inner horizon.

"I have seen some recent articles"  You must link to these articles or your question is not clear.

Also do the words "Kerr metric" mean anything to you?

What does "thinnest Event Horizon" mean. An event horizon is a mathematical surface, it has zero thickness.

https://phys.org/news/2022-01-black-hole-horizons-discharged.html#:~:text=If%20a%20black%20hole%20is,therefore%20called%20the%20inner%20horizon.%22.     this is the link to the article to which I refer

https://phys.org/news/2022-01-black-hole-horizons-discharged.html#:~:text=If%20a%20black%20hole%20is,therefore%20called%20the%20inner%20horizon.%22

I don't think this question is focused enough to be answered on this site.  If you are interested in these topics, I can recommend the book "Black Holes and Time Warps" by Kip Thorne (theoretical physicist who was technical consultant on Interstellar)

The linked article is about charged black holes.  But your question is about rotating black holes.  The linked article doesn't say "there might not be a singularity", but is about interesting effects at the inner event horizon.  There is nothing in the article about "warping" or "wormholes".  Please give more details about what you have been reading concerning rotating black holes.

"Black Holes and Time Warps" by Kip Thorne I should look into that, thank you. James K you are correct. I can look up other articles and post them so that you can look into them as well if you like. As it turns out however wormholes apparently have a problem with the gravity that they themselves generate may cause them to collapse on themselves much less trying to send them through the tremendous gravity of a black hole but I thank you for your response anyway.