3. Acceleration Due to Gravity

Derivation

Proof $g$ is directly proportional to $r$ below the surface

using

$$\rho = \dfrac{m}{V}, \quad \quad V = \dfrac{4}{3}\pi r^2, \quad \quad g = \dfrac{GM}{r^2}$$

we can solve for $m$

$$m = \dfrac{4}{3}\pi r^3 \rho$$

subbing $m$ into $g$

$$g = \dfrac{4}{3}G \pi \rho r$$

since

$$\dfrac{4}{3}G \pi \rho = \text{constant}$$

then

$$g \propto r$$