A 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 is `g'=g/((1+h/R_(e))^(2))=g/9 implies (1+h/R_(e))=3 implies h=2 R_(e)=128000` km.