What is a general formula for Earth gravity as a function of radius, given a spherical shell model?

Question

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.

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.