Question

Find the height from the earth's surface where g will be $25\%$ of its value on the surface of earth (R$=6400$km). (b) Find the percentage increase in the value of g at a depth h from the surface of earth.

Answer 1

a. At a height h above the earth's surface, we have
$g^{\prime }=g{\left(\frac{R}{R+h}\right)}^2\Rightarrow \frac{g}{4}=g{\left(\frac{R}{R+h}\right)}^2\Rightarrow \frac{R}{R+h}=\frac{1}{2}$
$\Rightarrow h=R=6400$km
b. At a depth h below the earth's surface, we have
$g^{\prime }=g(1-\frac{h}{R})\Rightarrow \frac{g^{\prime }}{g}=1-\frac{4}{6400}$
$\frac{g-g^{\prime }}{g}=\frac{4}{6400}\Rightarrow \frac{\Delta g}{g}=\frac{!}{16}$
$\therefore \%$ age$=\frac{\Delta g}{g}\times 100=\frac{1}{16}\%$ increase.