Prove that `g_(h)=g(1-(2h)/(R))`, where `g_(h)` is the acceleration due to gravity at altitude h and `hltltR` (R is the radikus of the Earth).

Explanation

Let, M be the mass of the earth, R be the radius of earth, h be the height at which acceleration due to gravity is to be found, g be acceleration due to gravity at the surface of the earth and `g_(n)` be acceleration due to gravity at height h.
On the surface of earth,
`g=(GM)/(r^(2))" ".......(i )`
At heigth h, from the earth's surface,
`g_(h)=(GM)/((R+h)^(2))" "......(ii)`
Dividing equation (ii) by equation (i) we get
`(g_(h))/(g)=(R^(2))/((R+h)^(2))`
`:.g_(h)=(gR^(2))/((R+h)^(2))`
`g_(h)=g""(R^(2))/(R^(2)[1+(h)/(R)])=g[1+(h)/(R)]^(-2)`
Expanding by using biknomial expansion and neglecting higher powers of `(h)/(R)" as "(h)/(R)ltlt1`
`:.g_(h)[1-(2h)/(R)]` represents the acceleration due to gravity at altitiude h.