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

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When an object of mass m is raised to a height h, the potential energy stored in the object is mgh.

This can be derived using the general expression for gravitational potential energy.

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 =\(-\frac{GM_em}{r}\)  ...(1)

Here r = R+ h, where Re is the radius of the Earth, h is the height above the earth's surface

Substituting equation (4) and (5) we get, 

U = -mge + mgh …….. (6) 

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 h1 to h2 , then the potential energy at h1 is

U(h1) = – mgRe + mgh1 …(7) 

and the potential energy at h2 is 

U(h2) = – mgRe + mgh2 …(8)

The potential energy difference between h1 and h2 is

U(h2) – U(h1) = mg(h2 – h1) …(9)

The term mgRe in equation (7) and (8) plays no role in the result. Hence 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 particle of mass m at a height h from the surface of the Earth is U = mgh. On the surface of the Earth, U = 0, since h is zero.

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