At a point above the surface of Earth, the gravitational potential is `-5.12xx10^(7)Jkg^(-1)` and the acceleration due to gravity is `6.4ms^(-2)`. Assuming the mean radius of the Earth to be 6400 km, calculate the height of this point above the Earth's surface.

Explanation

Let r be the distance of the given point from the centre of the Earth.
Then,
Gravitational potential `=-(GM)/(r)`
`=-5.12xx10^(7)" "...(1)`
and acceleration due to gravity,
`g=(GM)/(r^2)=6.4" "...(2)`
Dividing (1) by (2) we get
`r=(5.12xx10^7)/(6.4)`
`=8xx10^(6)m=8000km`
`therefore` Height of the point from Earth's surface
`=r-R=8000-6400`
`=1600 km`