The depth below the earth's surface at which the acceleration due to gravity ' g ' becomes g n is ( R = radius of the earth, n is an integer, n >1)

Step-by-step solution

Acceleration due to gravity at surface, g= ( G M R^2 ) where, G is the universal gravitational constant, R is the radius of earth. At a depth h below the surface, considering force balance: aligned & G M ( R-h R )^3 m (R-h)^2 =m g^ & g^ = ( G M R^2 ) (R-h) R =g ( R-h R )= g n & R-h R = 1 n & -h R = 1-n n aligned h = ( n -1) R n