Prove that at points near the surface of the Earth, the gravitational potential energy of the object is U = mgh.

Explanation

(i) Consider the Earth and mass system, with r, the distance between the mass m, and the Earths centre. Then the gravitational potential energy,
`U = -(GM_E M)/r" ".....(1)`
(ii) Here, `r = R_(E) + h`, where `R_(E)` is the radius of the Earth, h is the height above the Earth's surface.
`U = -G(M_e m)/((R_(E) + h))" "......(2)`
`If `h < < R_(e)`, equation (2) can be modified as
`U = -G(M_Em)/(R_(E)(1+h//R_E))`
`U = -G (M_E m)/(R_E) (1 + h//R_E)^(-1)" ".....(3)`
`[ :'` Binomial expression) `(1+x)^(-n) = 1 - nx`
`(1 + h/(R_E))^(-1) = (1 - h/(R_E))]`
(iii) By using Binomial expansion and neglecting the higher order terms, we get
`U = -G(M_Em)/(R_E) (1-h/(R_E))" "........(4)`
We know that, for a mass m on the Earths surface,
`G(M_Em)/(R_E) = mgR_E" "......(5)`
Substituting equation (5) in (4) we get,
`U = -mgR_(E) + mgh`
It is clear that the first term in the above expression is independent of the height h. for example, if the object is taken from height `h_(1)` to `h_(2)` then the potential energy at `h_(1)` is `U(h_1) = -mg R_(E) + mgh_(1) --(1)` and the potential energy at `h_(2)` is `U(h_2) = -mg R_(E)+ mgh_(2) ---(2)`. The potential energy difference between `h_(1) and h_(2)` is `U H_2) - U(h_1) = mg (h_2 - h_1)`. omitting the first term in (1) and (2), we get U = mgh.