Question

show that value of g at a height h is same as the value of acceleration due to gravity at a depth d=2h

Answer 1

at height h,
g = GM/R^{2} 
g' = GM/ (R+h)^{2} 
g'/g = R^{2}/(R+h)^{2}
g'//g = (R+h/R)^{-2}
by binomial expansion
g'/g = (1- 2h/R)
g' = (1- 2h/R)g

at depth 2h,
density = d
volume = 4/3 \pi R^{3} 
volume at depth 2h = 4/3 \pi (R-2h)^{3}
g = G*d* 4/3 \pi R
g' = G*d* 4/3 \pi (R-d) 
g'/g = (R- 2h)/R
g'/g = 1- 2h/R) 
g' = (1 - 2h/R)g

Hence proved.. 

Answer 2

....1-2h/r................