At what height above the earth's surface the acceleration due to gravity will be `(1)/(9)` th of its value at the earth's surface? Given radius of earth, `R = 6400km`

Explanation

We know `g= (GM)/(R^(2))` …(i)
Here M= Mass of earth
R= Radius of earth
Let `g_(h)` is acceleration due to gravity at height h from earth.s surface then `g_(h)= (GM)/((R+h)^(2))` …(ii)
Dividing equation (ii) with equation (i), we have
`(g_(h))/(g)= ((R )/(R+h))^(2)` ...(iii)
Putting `g_(h)= (g)/(9) and R= 6400km` is equation (iii), we have
`((6400)/(6400+ h))^(2) = (1)/(9) rArr (6400)/(6400+h) = (1)/(3) rArr 6400+ h= 19200` h= 12800km
So at height 12800 km from the surface of earth acceleration due to gravity becomes `(1)/(9)th` of its value at the surface of earth.