Draw the graph of the Earth's gravitational potential…

Draw the graph of the Earth's gravitational potential against the distance $r$ of a point from the Earth's centre, $\varphi = \varphi(r)$.

Step 1: Idea

The potential is found from the field: going outward by $dr$ against a field of strength $G$ raises the potential by $G\,dr$, i.e. $d\varphi/dr = G(r)$. We know $G(r)$ inside and outside a uniform Earth (Problem 33), and we fix the constant by $\varphi(\infty) = 0$.

Step 2: Outside ($r \ge R$)

Outside, the Earth acts as a point mass: $\varphi(r) = -\frac{\gamma M}{r}$.

On the surface $\varphi(R) = -\gamma M/R$.

Step 3: Inside ($r \le R$)

Inside, $G = \gamma Mr/R^3$. Start from the surface value and integrate inward: $\varphi(r) = \varphi(R) + \int_R^r\gamma\frac{Mr'}{R^3}\,dr'$.

\[ \varphi(r) = -\frac{\gamma M}{R} + \frac{\gamma M(r^2 - R^2)}{2R^3}. \]

\[ \varphi(r) = -\frac{\gamma M}{2R}\left(3 - \frac{r^2}{R^2}\right). \]

\[\begin{gathered} \varphi(0) = -\frac{3\gamma M}{2R} \\ \varphi(R) = -\frac{\gamma M}{R} \end{gathered}\]

Step 4: Shape of the graph

Inside the Earth the graph is a parabola with its lowest point at the centre, where the potential is 1.5 times deeper than at the surface. At the surface it joins smoothly (with no kink) onto the hyperbola $-\gamma M/r$, which rises toward zero at large distances. The potential is negative everywhere and lowest at the centre, the bottom of the Earth's \"potential well\".

Check

Continuous at $r = R$, and the slopes match there too (both equal the surface field $\gamma M/R^2$) $\checkmark$

$d\varphi/dr = \gamma Mr/R^3$, equal to the field inside $\checkmark$