Determine the ratio of (a) the gravitational field strengths…

Determine the ratio of (a) the gravitational field strengths and (b) the gravitational potentials at points at a height $h = 37\ \mathrm{km}$ above the Earth's surface and at points on the surface. The Earth's radius is $R = 6374\ \mathrm{km}$.

Step 1: Idea

Outside the Earth, its field and potential are the same as if its whole mass $M$ were concentrated at the centre. What matters is therefore the distance from the centre: $R$ on the surface and $R + h$ at height $h$. In a ratio, $\gamma$ and $M$ cancel, so we do not need the Earth's mass at all.

Step 2: (a) Field strength

The field falls with the square of the distance: $\frac{G_h}{G_0} = \frac{\gamma M/(R + h)^2}{\gamma M/R^2} = \left(\frac{R}{R + h}\right)^2$.

\[ \frac{G_h}{G_0} = \frac{6374^2}{6411^2} = \frac{40\,627\,876}{41\,100\,921}. \]

\[\frac{G_h}{G_0} \approx 0.9885\]

At 37 km the field is about 1.15% weaker than on the surface.

Step 3: (b) Potential

The potential falls only with the first power of the distance: \[\frac{\varphi_h}{\varphi_0} = \frac{-\gamma M/(R + h)}{-\gamma M/R} = \frac{R}{R + h} = \frac{6374}{6411}.\]

\[\frac{\varphi_h}{\varphi_0} \approx 0.9942\]

Both potentials are negative; the ratio is less than 1 because the potential at height is closer to zero (higher). It changes by only 0.58%, half as much as the field, because it depends on the first power of the distance and not the second.

Check

For small $h$: $ (1 + h/R)^{-2} $ $\approx 1 - 2h/R $ $= 1 - 0.0116 $ $= 0.9884\ \checkmark $

$ (1 + h/R)^{-1} $ $\approx 1 - h/R $ $= 0.9942\ \checkmark $