The acceleration due to gravity at a height h above the surface of the Earth is given by
gp=(R+h)2gR2...(1)
The acceleration due to gravity at a depth h below the surface of the Earth is given by
gq=g(1−Rh)...(2)
According to the given problem,
gp=gq...(3)
Equations (1), (2) and (3) imply that
(1+Rh)2g=g(1−Rh)⇒(1−R2h2)(1+Rh)=1...(4)
Take Rh=x
So, from equation (4), it follows that
x3−x+x2=0⇒x2+x−1=0⇒x=25−1⇒h=2R(5−1)