Question

At a point above the surface of earth, the gravitational potential is $-5.12\times {10}^{7}J/kg$ and the acceleration due to gravity is $6.4m/s^2$. Assuming the mean radius of the earth to be $6400km$, calculate the height of this point above the earth's surface.

Answer 1

Let $r$ be the distance of the given point from the centre of the earth. Then,
Gravitational potential $=-\frac{GM}{r}=-5.12\times {10}^{7}J/kg....\left(1\right)$
and acceleration due to gravity,
$g=\frac{GM}{r^2}=6.4m/s^2.....\left(2\right)$
Dividing $\left(1\right)$ by $\left(2\right)$, we get
$r=\frac{5.12\times {10}^7}{6.4}=8\times {10}^{6}m=8000km$
$\therefore$ Height of the point from earth's surface $=r-R=8000-6400=1600km$.