Question

The acceleration due to gravity at a place is $0.2m/s^2.$ Find its height above the earth's surface.

Answer 1

Let height above earth surface be $hm$. Then

$g^{\prime }=g\left(1-\frac{2h}{R_E}\right)$

$\frac{g^{\prime }}{g}=\left(1-\frac{2h}{R_E}\right)$

$\frac{0.2}{9.8}=\left(1-\frac{2h}{R_E}\right)$

$\frac{2h}{R_E}=1-\frac{1}{49}$

$\frac{2h}{R_E}=\frac{48}{49}$

$h=\frac{48R_E}{98}$

$=\frac{24\times 6.4\times {10}^6}{49}$

$=3.1347\times {10}^{6}m$

$=3134.7km$

So, at height of $3134.7km$ value of $g$ is $0.2m/s^2$