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? Radius of earth is 6400 km.

Explanation

Acceleration due to gravity at height h,
`g'=(g)/((1+(h)/(R_(e)))^(2))=(g)/(9)`
`rArr(1+(h)/(R_(e)))=3 rArrh=2R_(e)=2xx6400=12800` km