Gravitation Inside Earth — Question 4

Question 4

An outward-increasing density produces a quadratic interior field

A spherical planet of radius $R$ has density $\rho(r)=\rho_0r/R$. Use $dM=4\pi r^2\rho(r)\,dr$ and the shell theorem.

1. Find enclosed mass, total mass and interior gravity.

2. Find $g(R/2)/g(R)$ and explain why the uniform-density rule would fail.

Question 4 - Solution

An outward-increasing density produces a quadratic interior field

(a) Derivation and calculation

Integrate shells using a dummy radius $s$: $$M(r)=\int_0^r4\pi s^2\rho(s)\,ds =\frac{4\pi\rho_0}{R}\int_0^rs^3\,ds =\frac{\pi\rho_0r^4}{R}.$$ Consequently, $$\boxed{M=\pi\rho_0R^3},\qquad \boxed{g(r)=\frac{GM(r)}{r^2}=\frac{\pi G\rho_0}{R}r^2}.$$

(b) Result and interpretation

Normalizing by the surface value, $$g(R)=\pi G\rho_0R,\quad \frac{g(r)}{g(R)}=(r/R)^2.$$ $$\boxed{g(R/2)/g(R)=1/4}.$$ The uniform-density rule would instead use $M(r)\propto r^3$ and give $1/2$; here $M(r)\propto r^4$.

Physical check

The total mean density is $3\rho_0/4$, below the surface density $\rho_0$.