At a point above the surface of the earth, the gravitational potential is −5.12×107Jkg−1 and the acceleration due to gravity is 6.4 ms−2. Assuming the mean radius of the earth to be 6400 km, the height of this point above the earth's surface is.

Let $r$ be the distance of the given point from the centre of the earth. Then Gravitational potential

$= - \frac{GM}{r} = - 5 .12 \times 10^{7} \dots (i)$

And acceleration due to gravity,

$g = \frac{GM}{r^{2}} = 6 .4 \dots (ii)$

Dividing Eq. (i) by Eq. (ii), we get

$r = \frac{5.12 \times 10^{7}}{6.4}$

$= 8 \times 10^{6} m$

= 8000 km

$∴$ Height of the point from earth's surface $= r - R = 8000 - 6400$ = 1600 km

At a point above the surface of the earth, the gravitational potential is $- 5.12 \times 10^{7} J {kg}^{- 1}$ and the acceleration due to gravity is $6.4 {ms}^{- 2} .$ Assuming the mean radius of the earth to be 6400 km, the height of this point above the earth's surface is.