I'm trying to derive the formula for gravitational acceleration as a function of Earth radius $g(r)$, given a spherical shell model where each shell has a constant density $\rho(r)$. If I set $r=0$ to be the core-mantle boundary, I think it's something like:

$g(r) = \frac{4\pi G}{3r^{2}}[ \int_{0}^{r}{\rho}(r)r^{2}.dr +F ]$

Where $F$ accounts for core mass. But I would like confirmation/references if possible! Thanks!

Edit: Here's my thinking thus far: at a radius $r$,

$g(r) = \frac{G}{r^{2}}\int_{V}\rho.dV$

where $V$ is the volume below $r$. Since $V = \frac{4\pi}{3} r^{3}$, $dV = 4\pi r^{2}.dr$. So

$g(r) = \frac{4\pi G}{r^{2}} \int_{0}^{r}\rho(r)r^{2}.dr$, plus a constant for the mass of the core.

  • $\begingroup$ Perhaps you should further discuss your reasoning (sounds like a hw question, which people may help with, but usually when the poster is showing good faith in talking through what they've done) $\endgroup$ Aug 10, 2023 at 9:39
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    $\begingroup$ It's for a paper I'm writing... I genuinely can't find the answer after a literature trawl, only simplified solutions e.g. $\rho$ varies linearly with depth. $\endgroup$
    – co323
    Aug 10, 2023 at 11:02
  • $\begingroup$ Your equation is dimensionally incorrect. This is a start over situation. I also suggest you do a better job of researching literature. A good place to start is the 40+ year old paper on the Preliminary Reference Earth Model, the many papers cited by the PREM paper, and the 11000+ papers that cite the PREM paper as a reference. $\endgroup$ Aug 10, 2023 at 11:19
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    $\begingroup$ Thanks for your super-helpful comment; nowhere in the PREM paper do they explain how they calculated their gravity values. I am currently searching through the 11000+ papers, I just thought someone out there might know where to find the answer off the top of their head and be able to help me out. $\endgroup$
    – co323
    Aug 10, 2023 at 12:15
  • $\begingroup$ Apologies, didn't realize it was complex, thought integrating over a shell was usually fairly straightforward and a type of problem I thought I remembered from schools days, but sounds like it's more complex than I realized $\endgroup$ Aug 11, 2023 at 3:49

1 Answer 1


Properly, your formula should read

$g(r) = \frac{4\pi G}{(r+r_c)^2}[ \int_{0}^{r}{\rho}(r)r^{2}.dr +F ]$

where your radial distance (which is the argument in the inverse square law) is measured from the center of the shells and thus includes the core radius $r_c$. Also you do not use the $3$. The mass of each shell is the surface area times the density times the shell thickness, and the surface area does not have the factor of $1/3$. Of course when you add the core mass, that is computed over a volume and thus does include the $1/3$ factor if you are calling the core a homogemeous ball of radius $r_c$ and density $\rho_c$. Thus $F=\rho_cr_c^3/3$.

Otherwise your approach seems correct.

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    $\begingroup$ Great, that all makes sense! It would be good to have a literature reference for this specific case, but I think I can justify your formula for my use. Thanks! $\endgroup$
    – co323
    Aug 11, 2023 at 7:07
  • $\begingroup$ Edit: I will leave the denominator of the prefactor as $r^{2}$, and change the bottom limit of the integral to $R_{CMB}$, which is core-mantle boundary radius, and set $r=0$ to be the centre of the Earth - this makes more sense intuitively. $\endgroup$
    – co323
    Aug 11, 2023 at 9:31

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