Name: Gravitational Potential Energy
Uploaded: 2019-07-19
Duration: 9 min 10 s
Description: The energy possessed by a body because of its position in a gravitational field is known as Gravitational Potential Energy. Let’s learn about it in detail in...

Question

Prove that at points near the surface of the Earth, the gravitational potential energy of the object is

U = m g h

.

Accepted Answer

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,

Then the gravitational potential energy. $U = - \frac{G M m}{r}$
Here $r = R_{e} + h$ , where $R_{e}$ is the radius of the Earth, h is the height above the earth's surface.
$U = - \frac{G M_{e} m}{(R_{e} + h)}$ If $h < < R_{e}$ , this equation can be modified as
$U = - G \frac{M_{e} m}{R_{e} (1 + \frac{h}{R_{e}})} U = - G \frac{M_{e} m}{R_{e}} {(1 + \frac{h}{R_{e}})}^{- 1}$
By using Binomial expansion and neglecting the higher-order terms, we get
$U = - G \frac{M_{e} m}{R_{e}} (1 - \frac{h}{R_{e}}) . . . . . . . (i)$
We know that, for a mass m on the Earth’s surface,
$G \frac{M_{e} m}{R_{e}} = m g R_{e} . . . (i i)$
Substituting equation (ii) in (i) we get,
$U = - m g R_{e} + m g h$ ......(a)
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 $U = - m g R_{e} + m g h_{1}$ ............(iii)
and the potential energy at $h_{2}$ is $U = - m g R_{e} + m g h_{2}$ ............. (iv)
The potential energy difference between $h_{1}$ and $h_{2}$ is
$U (h_{2} - h_{1}) = m g (h_{2} - h_{1})$
The term $m g R_{e}$ in equations (iii) and (iv) plays no role in the result. Hence in the equation (a) the first term can be omitted or taken to zero. Thus it can be stated that The gravitational poten $U = m g h$ . On the surface of the Earth, U = 0, since h is zero.

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