<p>Properly, your formula should read</p>
<p><span class="math-container">$g(r) = \frac{4\pi G}{(r+r_c)^2}[ \int_{0}^{r}{\rho}(r)r^{2}.dr +F ]$</span></p>
<p>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 <span class="math-container">$r_c$</span>. Also you do not use the <span class="math-container">$3$</span>. 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 <span class="math-container">$1/3$</span>. Of course when you add the core mass, that is computed over a volume and thus does include the <span class="math-container">$1/3$</span> factor if you are calling the core a homogemeous ball of radius <span class="math-container">$r_c$</span> and density <span class="math-container">$\rho_c$</span>. Thus <span class="math-container">$F=\rho_cr_c^3/3$</span>.</p>
<p>Otherwise your approach seems correct.</p>

CC BY-SA 4.0
2023-08-11T07:07:44.663Z
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!
co323
CC BY-SA 4.0
2023-08-11T09:31:39.940Z
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.
co323
CC BY-SA 4.0
2023-08-10T23:41:37.490Z
2023-08-10T23:51:14.823Z
2023-08-10T23:51:14.823Z
Oscar Lanzi
<p>I'm trying to derive the formula for gravitational acceleration as a function of Earth radius <span class="math-container">$g(r)$</span>, given a spherical shell model where each shell has a constant density <span class="math-container">$\rho(r)$</span>. If I set <span class="math-container">$r=0$</span> to be the core-mantle boundary,
I think it's something like:</p>
<p><span class="math-container">$g(r) = \frac{4\pi G}{3r^{2}}[ \int_{0}^{r}{\rho}(r)r^{2}.dr +F ]$</span></p>
<p>Where <span class="math-container">$F$</span> accounts for core mass. But I would like confirmation/references if possible!
Thanks!</p>
<p>Edit:
Here's my thinking thus far: at a radius <span class="math-container">$r$</span>,</p>
<p><span class="math-container">$g(r) = \frac{G}{r^{2}}\int_{V}\rho.dV$</span></p>
<p>where <span class="math-container">$V$</span> is the volume below <span class="math-container">$r$</span>. Since <span class="math-container">$V = \frac{4\pi}{3} r^{3}$</span>, <span class="math-container">$dV = 4\pi r^{2}.dr$</span>. So</p>
<p><span class="math-container">$g(r) = \frac{4\pi G}{r^{2}} \int_{0}^{r}\rho(r)r^{2}.dr$</span>, plus a constant for the mass of the core.</p>

CC BY-SA 4.0
2023-08-10T09:39:05.967Z
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)
JeopardyTempest
CC BY-SA 4.0
2023-08-10T11:02:26.003Z
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.
co323
CC BY-SA 4.0
2023-08-10T11:19:17.953Z
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.
David Hammen
CC BY-SA 4.0
2023-08-10T12:15:54.733Z
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.
co323
CC BY-SA 4.0
2023-08-11T03:49:35.240Z
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
JeopardyTempest
<geophysics><gravity>
What is a general formula for Earth gravity as a function of radius, given a spherical shell model?
co323
# What is a general formula for Earth gravity as a function of radius, given a spherical shell model?

(Asked by: username_0 on 2023-08-10)

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.

**Tags:** geophysics, gravity

> **Comments:**
> **username_1**: 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)  
> **username_0**: 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.  
> **username_2**: 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.  
> **username_0**: 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.  
> **username_1**: 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  

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## 1 answers

### Answer by username_3 on 2023-08-10. Score: 1 | Accepted answer

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.

> **Comments:**
> **username_0**: 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!  
> **username_0**: 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.  

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