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

Explanation

Let us consider the Earth and mass system, with r, the distance between the mass 'm' and the Earth's centre. Then the gravitational potential energy,
`U=-(GM_(e)m)/(r)" "...(1)`
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 `hltltR_(e)`, equation (2) can be modified as
`U=-G(M_(e)m)/(R_(e)(1+h//R_(e)))`
`U=-G(M_(e)m)/(R_e)(1+h//R_(e))^(-1)" "...(3)`
By using Binomial expansion and neglecting the higher order terms, we get
`U=-G(M_(e)m)/(R_(e))(1-(h)/(R_e))" "...(4)`
We know that, for a mass m on the Earth's surface,
`G(M_(e)m)/(R_e)=mgR_(e)" "...(5)`
Substituting equation (5) in (4) we get,
`U=-mgR_(e)+mgh" "...(6)`
In the equation (6) the first term can be omitted or taken to zero. Thus it can be stated that the gravitational potential energy stored in the paritcle of mass m at a height h from the surfacd of the Earth is U = mgh.