The value of the acceleration due to gravity at a height h from the surface of the earth is `g_1` and that at a depth h below the earth's surface is `g_2`. Show that (radius of earth R `gt gt`h) `(g_2)/(g_1) =1+h/R`

Explanation

It is known `g_1=1-(2h)/R and g_2=1-h/R`
`therefore (g_2)/(g_1)=(1-h/R)/(1-(2h)/R)=(1-h/R)(1-(2h)/R)^(-1)`
`=(1-h/R)(1+(2h)/R)[because Rgt gth]`
`=1-h/R+(2h)/R[because Rgt gt h]`
`=1+h/R`