Gravitation Near Earth's Surface — Question 1

Question 1

Half of surface gravity occurs well above a nearby mountaintop

Treat Earth as a nonrotating sphere of radius $R$. Outside it, $g(r)=GM/r^2$ and $g_0=GM/R^2$.

1. Find the altitude $h$ where $g=g_0/2$.

2. For $h=.01R$, find the exact gravity ratio and compare with the linear approximation.

Question 1 - Solution

Half of surface gravity occurs well above a nearby mountaintop

(a) Derivation and calculation

For $x=h/R$, the exterior inverse-square law gives $$\frac{g(h)}{g_0}=\left(\frac R{R+h}\right)^2=(1+x)^{-2}.$$ For half gravity, $$(1+x)^2=2,\qquad\boxed{h=(\sqrt2-1)R\simeq.414214R}.$$

(b) Result and interpretation

At $h=.01R$, $$\boxed{g/g_0=1/1.01^2=0.98029605}.$$ The small-altitude expansion is $$(1+x)^{-2}=1-2x+3x^2+O(x^3).$$ First order gives $0.980000$; its difference from the exact ratio is $2.9605\times10^{-4}$.

Physical check

The expansion parameter is $h/R$; the half-gravity altitude is too large for a first-order estimate.