At a place, value of g is less by 1% than its value on the surface of the Earth (Radius of Earth = 6400 km). The place is

Let g be the acceleration due to gravity at the surface of the Earth. At a certain place, acceleration due to gravity becomes, g′=g−1 of g=g−g100=0.99g At height h, acceleration due to gravity becomes, g′=g(1−2hR)⇒0.99g=g(1−2hR) 0.99=1−2hR;2hR=1−0.99=0.01 h=0.01×R2=0.01×64002km=32km At depth d, acceleration due to gravity is given by g′=g(1−dR)⇒0.99g=g(1−dR) dR=1−0.99=0.01 d=0.01×R=0.01×6400km=64km