B The singularities of gravitational collapse and cosmology

  • Thread starter Cerenkov
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Hello.

I'd just like to check a some points concerning the two kinds of singularities that Penrose and Hawking describe in this paper. https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1970.0021 The Singularities of Gravitational Collapse and Cosmology.

1.
According to the Cosmic Censorship Hypothesis, we should not expect to see any naked singularities. I take this to mean that all black hole singularities are expected to be concealed behind event horizons.

2.
But, to date, have we detected the presence of an event horizon around any black hole?

3.
If not, is this site the best place to await news of observations of an event horizon? https://eventhorizontelescope.org/

4.
When it comes to the initial singularity, is this also hidden from us, behind the 'fog' of the CMBR?

5.
By what means (if any) could we see further back than the 380,000 years limit set by the CMBR?

6.
Am I right to conclude that nearly fifty years on from the publishing of this paper, we are still waiting upon observational evidence to confirm or rule out Hawking and Penrose's findings about the initial singularity?

Any help given at a basic level would be appreciated.

Thank you.

Cerenkov.






 
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According to the Cosmic Censorship Hypothesis, we should not expect to see any naked singularities. I take this to mean that all black hole singularities are expected to be concealed behind event horizons.
Yes.

to date, have we detected the presence of an event horizon around any black hole?
You can't. An event horizon is not something that is directly detectable from the outside. You have to infer its presence from indirect evidence. We have strong indirect evidence for event horizons.

Also, this question seems backwards given your initial question. The question you should be asking if you're interested in the cosmic censorship hypothesis is not whether we've detected any event horizons, but whether we've detected any naked singularities. (And the answer to that is that we haven't.) We don't know whether the actual black holes we believe we see have actual singularities inside them.

is this site the best place to await news of observations of an event horizon?
I don't know if it's the "best" place but it's certainly a good place to look.

When it comes to the initial singularity, is this also hidden from us, behind the 'fog' of the CMBR?
First, we don't know if there even was an initial singularity; our best current model of the universe does not require there to have been one.

Second, cosmic censorship has nothing to do with initial singularities; it doesn't say that the initial singularity, if there is one, has to be hidden from us by anything. So while it's true that we can't visually see further back than the time the CMBR was created, that has nothing to do with cosmic censorship.

By what means (if any) could we see further back than the 380,000 years limit set by the CMBR?
By using a method of "seeing" that does not involve light. There are at least two that might work: neutrinos and gravitational waves. Unfortunately our ability to detect those things is not sufficient (yet) for us to "see" usefully that far back.

Am I right to conclude that nearly fifty years on from the publishing of this paper, we are still waiting upon observational evidence to confirm or rule out Hawking and Penrose's findings about the initial singularity?
Yes. But, as noted above, this has nothing to do with cosmic censorship.
 
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Thank you for this helpful reply, PeterDonis.

Yes, sorry for not making it clear, but I do realize that cosmic censorship applies only to black holes and not to the initial singularity - should such a thing exist. The idea I was trying to get across was that both kinds of singularity described in the Hawking-Penrose paper seem to be hidden from direct observation. Black hole singularities (should they exist) by their event horizons and the initial singularity (ditto) by the obscuring effects of the CMBR.

A few more questions, if I may.

Firstly, about those two avenues of investigation that don't employ electromagnetic radiation - neutrinos and gravitational waves. Are you referring to possible measurements of the Cosmic Neutrino Background? And when it comes to gravitational waves are you referring to efforts like this project, https://lisa.nasa.gov/?

Secondly, I'm curious as to how 'strong' a gravitational field GR has been tested in. Ok, I've used the word 'strong' when that's probably not appropriate. So, could you please indicate a better way of expressing and asking this question? I ask because the Sun's gravity well seems, to my naive understanding, to be the 'strongest' gravitational field we have observed, to date.

Thank you.

Cerenkov.
 
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both kinds of singularity described in the Hawking-Penrose paper seem to be hidden from direct observation
But in different ways. We can imagine a universe with an initial singularity that could be seen; just imagine one with a clear optical path all the way back. That might be unlikely, but it's not against the laws of physics.

By contrast, a singularity inside an event horizon is hidden by the laws of physics: it's not just unlikely but impossible to see it from the outside.

Are you referring to possible measurements of the Cosmic Neutrino Background? And when it comes to gravitational waves are you referring to efforts like this project, https://lisa.nasa.gov/?
As general ways of observing those things (neutrinos and gravitational waves), yes. But that doesn't mean we will necessarily get information from them that will help us to "see" further back into the early universe than the time of the formation of the CMBR.

the Sun's gravity well seems, to my naive understanding, to be the 'strongest' gravitational field we have observed, to date.
No, not even close. GR has been tested for binary pulsars, which are neutron stars and have significantly stronger gravity than anywhere in the solar system, and now that LIGO has detected gravitational waves, we have tests of GR for black hole mergers, where gravity is stronger still.

The simplest quick estimate of how "strong" gravity is in a reasonably isolated system is the ratio ##M / R##, where ##M## is the mass of the system and ##R## is its radius. Here "mass" is in geometric units, i.e., ##G m / c^2##, where ##m## is the mass in conventional units. So, for example, ##M## for the Sun is about 1.477 km, and ##R## is about 696,000 km, so the ratio ##M / R## for the Sun is about ##2.1 \times 10^{-6}##. But for neutron stars, ##M## is roughly the same as the Sun but ##R## is about 1000 times smaller, so the ratio ##M / R## is about ##10^{-3}##, For a black hole, by definition, ##M / R = 1/2##, since ##R## is the horizon radius, which is ##2M##. Gravitational waves from black hole mergers come from the region just outside the horizon, so that's basically the ##M / R## ratio in the region where they are emitted.

Another way of quantifying how "strong" gravity is is to look at spacetime curvatures. The simplest quick estimate of this for an isolated system is ##M / R^3##, since that's the order of magnitude of the Riemann tensor components at the surface. But this number has units (curvature units, i.e., inverse length squared), whereas ##M / R##, above, is dimensionless; a dimensionless number is a better parameter to use for what I think you're looking for.
 
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Many thanks Peter.

One last thing, please. Since my level here is rated at B for Basic, I'm probably going to put my foot in it - but I just have to ask. In the ratio M / R for the Sun, the radius is expressed in kilometers. So why is mass expressed in geometric units of 1.477 kilometers? Shouldn't it be 1.477 Gm/c2?

Thanks,

Cerenkov.
 

PeroK

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Many thanks Peter.

One last thing, please. Since my level here is rated at B for Basic, I'm probably going to put my foot in it - but I just have to ask. In the ratio M / R for the Sun, the radius is expressed in kilometers. So why is mass expressed in geometric units of 1.477 kilometers? Shouldn't it be 1.477 Gm/c2?

Thanks,

Cerenkov.
You should check yourself that ##GM/c^2## has the dimension of length.

The ##1.477 km## is what you get if you use SI units - or any units where length is in metres.
 
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Thanks PeroK.

That's helpful, but a certain degree of B-level confusion remains. I seem to see PeterDonis dividing 1.477 kilometers by 696,000 kilometers. Ok, I can agree the result, 2.1. But where do cubic centimeters come into it?

Help!

Cerenkov.
 

PeroK

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Thanks PeroK.

That's helpful, but a certain degree of B-level confusion remains. I seem to see PeterDonis dividing 1.477 kilometers by 696,000 kilometers. Ok, I can agree the result, 2.1. But where do cubic centimeters come into it?

Help!

Cerenkov.
I haven't looked at your link, if the question relates to that!
 
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I just have to ask. In the ratio M / R for the Sun, the radius is expressed in kilometers. So why is mass expressed in geometric units of 1.477 kilometers?
So that the ratio ##M / R## is a dimensionless number.

Shouldn't it be 1.477 Gm/c2?
I'm not sure what you mean by this. ##Gm / c^2## isn't a unit, it's a formula. But if you take the mass ##m## of the Sun in conventional units (kilograms), and calculate ##Gm / c^2##, what do you get? And what units is that result in?
 
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Hello again Peter.

Thank you for your patience.

But please understand that I do not properly understand the language you are using. Not modern English, but the language of mathematics. If you wanted a parallel example to illustrate what I mean, please consider the following. As a teenager I read books on astronomy and taught myself to read the Greek symbols used in designating the magnitude of the stars - Alpha, Beta, Gamma, Delta, etc. Nowadays I can look at a Greek Interlinear New Testament, which is written in Koine (1st century Greek symbols) and see words like Pneuma and Prophetu and Ekklesiai and Morphothe leaping out at me. I can relate these words to the English that I use and understand that Spirit (breath) and Prophecy and Church and Shape (form) are what is meant. But please don't ask me anything about the syntax or grammar that should be used in Koine! I simply don't know it or understand it.

And there is a kind of parallel with my understanding of the language of mathematics. I know the basics; addition, subtraction, division and multiplication, but beyond that I'm just guessing. So, if you couch a question to me in terms beyond that, then I'm simply going to struggle. Likewise with your assuming that I can read the symbols you use and know what they are and how they work together.

Hence, to you Gm/c2 is a formula. But I did not know that. To me, you were mantergeistmanning. That is, speaking a language I could not recognize, could not parse and could not understand. Just as you cannot understand what it means to mantergeist. To me, since the subject in question is mass, then if upper case M is the mass of the system and upper case R is it's radius, then to my untrained eyes, the unexplained upper case G in Gm/c2 might be grams, which to me is a conventional unit of mass.

Now, I fully realize that you could ask me, 'If you can't do the math, then why are you asking questions that require it?'
Well, the only answer I can give you is that, like Salieri, I love the very thing that I have no talent for. Astronomy, astrophysics and cosmology fascinate and have done so five decades of my life. The key to understanding them well is math - but I have no key and therefore I cannot enter via the doorway of mathematical understanding. Sorry, but there it is.

One final point, Peter. Our relationship in this forum is totally asymmetric. You have all the power and I have none. Those like you have all the power and those like me have none. We rely on your goodwill and patience to inch forward slowly and painfully in our comprehension of these things. For your information, it's taken me over an hour to carefully consider the content of this post, to rethink it and reword it and to get it to say what I want. So please be considerate to me in your reply. It's a difficult thing for a man to publicly declare his lack of expertise, his lack of understanding and his lack of smarts. Especially in a forum like this one, where these things are so important.

Thanks again for your patience and for taking the time to read this.

Cerenkov.
 
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I seem to see PeterDonis dividing 1.477 kilometers by 696,000 kilometers.
Yes, that's what I'm doing, to get a dimensionless number.

where do cubic centimeters come into it?
Nowhere. There are no cubic centimeters anywhere in anything I've said. The formula ##G m / c^2## means Newton's gravitational constant ##G##, times the mass ##m## in conventional units (kilograms), divided by the speed of light ##c## squared.

to my untrained eyes, the unexplained upper case G in Gm/c2 might be grams, which to me is a conventional unit of mass.
This isn't a matter of not understanding math. It's a matter of not understanding the common symbols used in physics. There's nothing wrong with being that way to start with--none of us start out knowing all this stuff--but it really helps to acquire such an understanding, because, as you see, even a "B" level discussion about physics is going to end up using some standard symbols that will confuse you if you don't know what they mean.

For the correct meaning of the symbols I used, see above. You should be able to work out how the units come out given that information (if you have to, it's easy to look up the units of Newton's gravitational constant ##G##). Also note that, in standard symbols for SI units, grams are denoted by ##g## (note the lower case) and centimeters by ##cm##.

(Btw, even with your mistaken understanding of the symbols, there would be only square centimeters--reading "c" to mean "centimeters"--in the denominator, not "cubic" ones. :wink:)
 
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Peter, I am trying to reply to you by copying segments from your message of yesterday (#4). When I do so, I get this.

The simplest quick estimate of how "strong" gravity is in a reasonably isolated system is the ratio M/R" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: "PT Sans", san-serif; position: relative;">M/RM/R, where M" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: "PT Sans", san-serif; position: relative;">MM is the mass of the system and R" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: "PT Sans", san-serif; position: relative;">RR is its radius. Here "mass" is in geometric units, i.e., Gm/c2" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: "PT Sans", san-serif; position: relative;">Gm/c2Gm/c2, where m" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: "PT Sans", san-serif; position: relative;">mm is the mass in conventional units.

That was my attempt to copy-and-paste just two sentences. Any idea of how I can overcome this glitch, please? The Quote function, maybe?

Ok, I've Googled 'Dimensionless Numbers' and this was the first hit. https://en.wikipedia.org/wiki/Dimensionless_numbers_in_fluid_mechanics
Is this relevant (what have fluids got to do with mass or gravity?) or should I go with the second hit? https://en.wikipedia.org/wiki/Dimensionless_quantity

This link for G ? https://simple.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

This one for SI units? https://simple.wikipedia.org/wiki/International_System_of_Units

Btw, thanks for your latest response.

Cerenkov.
 

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That was my attempt to copy-and-paste just two sentences. Any idea of how I can overcome this glitch, please? The Quote function, maybe?
I believe that if you 'reply' to the entire post and then delete what you don't want to quote, that should fix the problem. Try it and see.
 
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The Quote function, maybe?
That's the best way to quote people's posts, yes.

The second link is probably better for dimensionless quantities in general.

The last two links look fine.
 
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My thanks and appreciation to PeterDonis and Drakkith for their helpful responses.

Unless you think otherwise Peter, I reckon it's best that I spend some time familiarizing myself with the basic and derived SI units before doing anything else. Oh, and reading up on dimensionless quantities, too. After that, I can return to this thread and see how well I can follow what you wrote (in #4) about estimating the 'strength' of gravity in a reasonably isolated system.

Small movements, first.

Thanks again for sticking with me.

Cerenkov.
 
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Hello again, Peter.

I've opted to use different colours to differentiate your words from mine.


The simplest quick estimate of how "strong" gravity is in a reasonably isolated system is the ratio ##M / R##, where ##M## is the mass of the system and ##R## is its radius. Here "mass" is in geometric units, i.e., ##G m / c^2##, where ##m## is the mass in conventional units.

M = mass of the system
R = radius of the system
G = Newton's gravitational constant
m = mass of the Sun
c = the speed of light

According to this page on SI units https://simple.wikipedia.org/wiki/International_System_of_Units
m is a base unit measure of length. So, if I read this right...

To derive the value of M I need to to take Newton's gravitational constant G, multiply it by m and then divide it by c, the speed of light, squared. (Yes, I know you told me this, but I'm working this out for myself, so please cut me a little slack.) Then M is divided by R.

According to this page https://en.wikipedia.org/wiki/Gravitational_constant
"The measured value of the constant is known with some certainty to four significant digits. In SI units its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2 "

So I need to take 6.674×10−11 m3⋅kg−1⋅s−2 and multiply it by m, the mass of the Sun expressed in kilograms. (1.9891 × 1030 kilograms ) Then I divide it by c ( 299,792,458 metres per second) squared. This gives me M, which I then divide by R, 696,000 km.

Ok, let's see. Using the calculator function on my laptop I get 89,875,517,873,681,764, when squaring c. That much I can do. The figure seems to agree with this... 8.98755179 × 1016 m2 / s2 ...which is what Google gave me when I looked up, 'speed of light squared'.

But I can't figure out how to multiply 6.674×10−11 m3⋅kg−1⋅s−2 and 1.9891 × 1030 kilograms. That kind of multiplication is beyond me because I don't know how to deal with the standard form.

Can you help, please?

Thanks,

Cerenkov.



So, for example, ##M## for the Sun is about 1.477 km, and ##R## is about 696,000 km, so the ratio ##M / R## for the Sun is about ##2.1 \times 10^{-6}##. But for neutron stars, ##M## is roughly the same as the Sun but ##R## is about 1000 times smaller, so the ratio ##M / R## is about ##10^{-3}##, For a black hole, by definition, ##M / R = 1/2##, since ##R## is the horizon radius, which is ##2M##. Gravitational waves from black hole mergers come from the region just outside the horizon, so that's basically the ##M / R## ratio in the region where they are emitted.

Another way of quantifying how "strong" gravity is is to look at spacetime curvatures. The simplest quick estimate of this for an isolated system is ##M / R^3##, since that's the order of magnitude of the Riemann tensor components at the surface. But this number has units (curvature units, i.e., inverse length squared), whereas ##M / R##, above, is dimensionless; a dimensionless number is a better parameter to use for what I think you're looking for.
 
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I've opted to use different colours to differentiate your words from mine.
All this does is make your post unusable since I can't quote and reply to the part you wrote. Please use the quote feature as it's intended to be used: quote what I wrote that you're responding to, and then respond to it in the body of your post. These tools are there to facilitate discussion; if you try to make up your own tools you're just making discussion harder.
 
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@Cerenkov, as far as the question you are trying to ask in your latest post (which, as I posted just now, I can't quote and respond to directly):

The mass ##m## of the Sun in conventional units is, as you found, ##1.9891 \times 10^{30}## kg. To get the mass in "geometric units", which I was using the symbol ##M## to denote, you calculate:

$$
M = \frac{G m}{c^2}
$$

We have ##G = 6.674 \times 10^{-11}##, ##m = 1.9891 \times 10^{30}##, and ##c = 299792458##. So that gives

$$
M = \frac{6.674 \times 10^{-11} \times 1.9891 \times 10^{30}}{\left( 299792458 \right)^2}
$$

Your calculator should be able to do this for you, since it should have a way to enter numbers in scientific notation; but if you want to try to do this by hand, the easiest way is to remember that multiplying exponential numbers like ##10^{-11}## and ##10^{30}## means adding the exponents. So you have

$$
M = \frac{6.674 \times 1.9891 \times 10^{-11} \times 10^{30}}{\left( 299792458 \right)^2} = \frac{6.674 \times 1.9891 \times 10^{19}}{\left( 299792458 \right)^2}
$$

We can simplify this further by plugging in the scientific notation value you got for ##c^2## and remembering that dividing exponential numbers means subtracting exponents:

$$
M = \frac{6.674 \times 1.9891 \times 10^{19}}{8.98755179 \times 10^{16}} = \frac{6.674 \times 1.9891 \times 10^{3}}{8.98755179}
$$

And now it's in a form that you should be able to punch into your calculator without having to use scientific notation (since ##10^3 = 1000##) to obtain an answer in meters.
 
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All this does is make your post unusable since I can't quote and reply to the part you wrote. Please use the quote feature as it's intended to be used: quote what I wrote that you're responding to, and then respond to it in the body of your post. These tools are there to facilitate discussion; if you try to make up your own tools you're just making discussion harder.
Oops!
 
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@Cerenkov, as far as the question you are trying to ask in your latest post (which, as I posted just now, I can't quote and respond to directly):

The mass ##m## of the Sun in conventional units is, as you found, ##1.9891 \times 10^{30}## kg. To get the mass in "geometric units", which I was using the symbol ##M## to denote, you calculate:

$$
M = \frac{G m}{c^2}
$$

We have ##G = 6.674 \times 10^{-11}##, ##m = 1.9891 \times 10^{30}##, and ##c = 299792458##. So that gives

$$
M = \frac{6.674 \times 10^{-11} \times 1.9891 \times 10^{30}}{\left( 299792458 \right)^2}
$$

Your calculator should be able to do this for you, since it should have a way to enter numbers in scientific notation; but if you want to try to do this by hand, the easiest way is to remember that multiplying exponential numbers like ##10^{-11}## and ##10^{30}## means adding the exponents. So you have

$$
M = \frac{6.674 \times 1.9891 \times 10^{-11} \times 10^{30}}{\left( 299792458 \right)^2} = \frac{6.674 \times 1.9891 \times 10^{19}}{\left( 299792458 \right)^2}
$$

We can simplify this further by plugging in the scientific notation value you got for ##c^2## and remembering that dividing exponential numbers means subtracting exponents:

$$
M = \frac{6.674 \times 1.9891 \times 10^{19}}{8.98755179 \times 10^{16}} = \frac{6.674 \times 1.9891 \times 10^{3}}{8.98755179}
$$

And now it's in a form that you should be able to punch into your calculator without having to use scientific notation (since ##10^3 = 1000##) to obtain an answer in meters.
Thanks very much, Peter.

Here's how things went.

6.674 x 1.9891 x 10.e+19 = 1,327,525,340,000,000,000,000
8.98755179 x 10.e+16 = 898,755,179,000,000,000
The two values divided = 1,477.071143

6.674 x 1.9891 x 10.e+3 = 132,752.534
Divided by 8.98755179 = 14,770.71143

Now, these two results appear to agree with the value of M, which you gave earlier... 1,477 km.

So, dividing M (1,477) by R (696,000 km) we obtain... 0.002122126436

Which resembles the value 2.1 x 10-6, which you gave earlier, save for the decimal shift.

So far, so good?

Cerenkov.
 
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The two values divided = 1,477.071143
Yes, that's the correct value.

6.674 x 1.9891 x 10.e+3 = 132,752.534
Divided by 8.98755179 = 14,770.71143
You seem to be off by a factor of 10 here. I suspect that instead of entering ##10^3## in your calculator, i.e., ##1000##, you entered the equivalent of ##10 \times 10^3##, i.e., ##10,000##. In the first case that wouldn't matter since you entered two quantities of that form, one in the numerator and one in the denominator, so the factor of 10 error in each would cancel out. But the error shows up in this case since there is only one quantity that is in scientific notation.
 
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So, dividing M (1,477) by R (696,000 km) we obtain... 0.002122126436

Which resembles the value 2.1 x 10-6, which you gave earlier, save for the decimal shift.
You're off by a factor of 1000. The Sun's radius is 696,000 kilometers, but ##M## is in meters. So you need to multiply your ##R## by 1000 to make it meters (or divide ##M## by 1000 to make it kilometers). That will give you the answer I got earlier.
 
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Thanks again, Peter.

I'll re-run the exercise in the tomorrow, taking into account your two corrections. But please let me first say that it's such a relief (and a thrill) to get sequences of numbers that bear some resemblance to yours. My two errors can be reasonably put down to inexperience and can be corrected for.

Your patience is v.much appreciated.

Cerenkov.
 

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