If the acceleration due to gravity at a height h from the surface of the earth is the same as that at a depth h below the surface of the earth, prove that `h=(sqrt(5)-1)/2R` (R=radius of the earth).

Explanation

Acceleration due to gravity on the earth.s surface ,
`g=(GM)/(R^2)`
Acceleration due to gravity at a height h from the earth.s surface ,`g_1=(GM)/((R+h)^2)`
Hence, `(g_1)/(g)=(R^2)/((R+h)^2)`
Also, taking the earth as a uniform sphere of radius R, acceleration due to gravity at a depth h
`g_2=g(R-h)/R`
`therefore (g_1)/(g_2)=(R^3)/((R+h)^2(R-h))`
If `g_1=g_2`
`R^3=(R^2-h^2)(R+h)=R^3+R^2h-h^2R-h^3`
`or, h^2+Rh-R^2=0`
`therefore h=(-Rpmsqrt(R^2+4R^2))/2=((-1pmsqrt(5))R)/2`
As negative value of h is not acceptable,
`h=(sqrt(5)-1)/2R`