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 and R be the mass and radius of the Earth and g be the acceleration due to gravity at the Earth's surface.
On the surface of the Earth,
`g = (GM)/(R^(2)) " "…(1)`
where G is the universal gravitational constant.
Consider a body of mass m at an altitude h above the Earth's surface. Then, its distance from the centre of the Earth is R + h. If `g_(h)` is the acceleration due to gravity at the altitude h, the weight of the body there is `mg_(h)` and the gravitational force acting on it due to the Earth is `(GMm)/((R+h)^(2))`.
`therefore mg_(h)=(GMm)/((R+h)^(2))`
`therefore g_(h)=(GM)/((R+h)^(2))" "...(2)`
Dividing Eq. (2) by Eq. (1), we get, `(g_(h))/(g)=(R^(2))/((R+h)^(2))`
`therefore g_(h)=g((R)/(R+H))^(2)=g((R+h)/R)^(-2)=(1+(h)/(R))^(-2)" "...(3)`
Then, expanding the bracketed term on the right hand side and retaining only the first-order term in `h//R` for `h lt lt R`, we get,
`g_(h)~=g(1-(2h)/(R))" "...(4)`