Graph reading: g(r) inside and outside the Earth

Problem

Graph reading: $g(r)$ inside and outside the Earth. The graph shows $g(r)$ as a function of $r$ starting from the Earth’s centre: it rises linearly up to $r = R_T$, then decreases as $1/r^2$. Explain the physical origin of the two behaviours.

Linear growth inside, decrease $\propto 1/r^2$ outside; the maximum is at the surface ($r = R_T$).

Solution

Inside the Earth ($r < R_T$). By Gauss’s theorem (or Newton’s shell theorem), a spherical shell exerts no gravitational force inside itself: at a point at distance $r$ from the centre, only the mass contained within radius $r$ matters. For an Earth of uniform density $\rho$: $M(r) = \frac{4}{3}\pi r^3 \rho$ hence $g(r) = \frac{G\,M(r)}{r^2} = \frac{G}{r^2}\cdot \frac{4}{3}\pi r^3 \rho = \frac{4}{3}\pi G \rho\, r$ which is linear in $r$: at the centre $g(0) = 0$ (the surrounding mass pulls in all directions and cancels out), and it grows up to the surface.

Outside the Earth ($r > R_T$). Now the whole mass $M_T$ is enclosed within $r$ and acts as if concentrated at the centre: $g(r) = \frac{GM_T}{r^2}$ which decreases as $1/r^2$.

Maximum at the surface. The two branches meet at $r = R_T$, where $g$ reaches its maximum value $g_0 = GM_T/R_T^2 \approx 9{,}8$ m/s$^2$: this is the point where all of the Earth’s mass is at the smallest possible distance from the surface.

$\ev{r<R_T:\ g \propto r \ \text{(linear)};\quad r>R_T:\ g \propto 1/r^2;\quad g_{\max} \text{ at the surface}}$

Links

Topics: Gravitation Concepts: Law of universal gravitation · Weight force Skills: Graph reading