Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential ene

Question

Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass $m$ when taken to a height $h$ from the surface of the earth (of radius $R$ and mass $M$), is given by:

Accepted Answer

1. **Formula:** The gravitational potential energy of a mass $m$ at a distance $r$ from the center of the Earth is given by $U(r) = -\frac{GMm}{r}$ [Eq. 7.23, 7.24].
2. **Initial State:** At the surface of the Earth, the distance from the center is $R$. Thus, the initial potential energy is $U_i = -\frac{GMm}{R}$.
3. **Final State:** At a height $h$ above the surface, the distance from the center is $r = R + h$. Thus, the final potential energy is $U_f = -\frac{GMm}{R+h}$.
4. **Change in Potential Energy ($\Delta U$):**
   $$ \Delta U = U_f - U_i $$
   $$ \Delta U = \left( -\frac{GMm}{R+h} \right) - \left( -\frac{GMm}{R} \right) $$
   $$ \Delta U = \frac{GMm}{R} - \frac{GMm}{R+h} $$
   $$ \Delta U = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) $$
   $$ \Delta U = GMm \left( \frac{R+h-R}{R(R+h)} \right) $$
   $$ \Delta U = \frac{GMmh}{R(R+h)} $$
   (Note: This expression reduces to $mgh$ only when $h \ll R$).

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