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Ok, so the formula for calculating oblateness of a planet (in this instance, a gas or ice giant) goes like this:

$\text{Oblateness} = C\frac{R^3}{M P^2}$

I have seen C defined as a constant that depends on the mass distribution of the body in question. I would like to know how that is determined. One source just said using multiples of known values (Jupiter or Neptune). What would those values even be? I am wanting to know what the others are. I'm guessing it's the following:

R=radius

M=Mass

P=rotation period

I'm not really sure; this is my best guess, since the source doesn't explain it much further. If anyone knows (which I'm sure someone here does), please either confirm or inform me. Many thanks!

-M-

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  • $\begingroup$ Equatorial bulge? (The formula in the article is for the first flattening f.) $\endgroup$
    – AlexP
    Oct 4, 2022 at 0:13
  • $\begingroup$ Could you post a link to your source (if it is online)? $\endgroup$
    – sondre99v
    Oct 4, 2022 at 6:48
  • $\begingroup$ The source is a broken link and has not been saved on Internet Archive. I typed it down verbatim. The meanings behind R, M, & P are my guesses. I'm trying to find how oblate the gas giants I have in my systems are. To do that, I need to know what values to put in, and if this is even the correct formula for what I'm trying to do. $\endgroup$
    – Mike
    Oct 4, 2022 at 13:35
  • $\begingroup$ @Mike What is the oblateness? $\endgroup$
    – Daron
    Oct 4, 2022 at 16:50
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    $\begingroup$ @Mike Those formulae appear here too but they don't bother to say what the oblateness constant means. They just define $q = $ your stuff. $\endgroup$
    – Daron
    Oct 4, 2022 at 21:26

1 Answer 1

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The formula you want is:

$$\frac{a-b}{b} = \frac{5}{4}\frac{(2\pi/T)^2 R^3}{GM}$$

Where:

$a$ is the (larger) equatorial radius.

$b$ is the (smaller) polar radius.

$T$ is the time for one revolution.

$R = (b+2a)/3$ is the mean radius.

$M$ is the mass of the planet.

$G$ is Newton's constant.

The formula is derived in this thread where it seems to agree with the experimental numbers for Earth. This is good for you because it suggests we can indeed ignore how the planet is denser in the middle. See also Wikipedia and their source [8].

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  • $\begingroup$ So, what you're saying is my mean radius (which in this case, I'll say is 60421 km,) will be number crunched into a new set? Very nice. Now I have to figure out how to do maths with Newton's Constant...I thank you, sir! $\endgroup$
    – Mike
    Oct 4, 2022 at 23:03
  • $\begingroup$ Also, where can I enter this formula? I've tried a few calculators and Wolfram Alpha. It never lets itself get entered as we see presented here. $\endgroup$
    – Mike
    Oct 5, 2022 at 1:08
  • $\begingroup$ @Mike: Microsoft Excel or Google Sheets or LibreOffice Calc. $\endgroup$
    – AlexP
    Oct 5, 2022 at 8:49
  • $\begingroup$ Officially stumped. I could never figure out how to use the gravitational constant, so I avoided any calculations with it. I managed to get it entered into Wolfram Alpha, but it came back with some ridiculously high number that I couldn't make use of. So I guess I probably need to express the numbers correctly in SI units? $\endgroup$
    – Mike
    Oct 5, 2022 at 12:33
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    $\begingroup$ @Mike Yes yes yes. You need everything in SI units. kilos, seconds, metres, etc. Then Newton's constant is just a number. If I have input everything right the formula seems to give 0.42% rather than the real 0.3% ish number. Eh, close enough. $\endgroup$
    – Daron
    Oct 5, 2022 at 14:58

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