Gravitation Near Earth's Surface — Question 8

Question 8

Two altitude readings can separate planetary radius from mass

For a spherical nonrotating planet, gravity is measured at the surface and at known altitude $h$. Their ratio is $q=g(h)/g(0)$, with $0<q<1$.

1. Infer $R$ and then $GM$ from these measurements.

2. For $q=100/121$ and $h=100\,\mathrm{km}$, find $R$ and explain the sensitivity when $q$ is near one.

Question 8 - Solution

Two altitude readings can separate planetary radius from mass

(a) Derivation and calculation

The field ratio eliminates $GM$: $$q=\frac{GM/(R+h)^2}{GM/R^2}=\left(\frac R{R+h}\right)^2.$$ Solving $\sqrt q(R+h)=R$ gives $$\boxed{R=\frac{h\sqrt q}{1-\sqrt q}},\qquad \boxed{GM=g(0)R^2}.$$

(b) Result and interpretation

For $q=100/121$, $\sqrt q=10/11$, so $$\boxed{R=10h=1000\ \mathrm{km}}.$$ At fixed $h$, differentiation quantifies sensitivity: $$\frac{d\ln R}{d\ln q}=\frac1{2(1-\sqrt q)},\qquad \frac{\delta R}{R}\simeq\frac{1}{2(1-\sqrt q)}\frac{\delta q}{q}.$$ The amplification is $5.5$ here and diverges as $q\to1$.

Physical check

A ratio close to one corresponds to a short baseline compared with the inferred radius.