A body of mass $m$ is taken from the Earth’s surface to the height equal to twice the radius ($R$) of the Eart

Question

A body of mass $m$ is taken from the Earth’s surface to the height equal to twice the radius ($R$) of the Earth. The change in potential energy of the body will be:

Accepted Answer

1. **Gravitational Potential Energy Formula:** The gravitational potential energy of a mass $m$ at a distance $r$ from the center of the Earth (mass $M$) is given by $U = -\frac{GMm}{r}$ [Eq. 7.23].
2. **Initial State:** At the Earth's surface, $r_1 = R$. Therefore, $U_i = -\frac{GMm}{R}$.
3. **Final State:** At a height $h = 2R$ above the surface, the distance from the center is $r_2 = R + h = R + 2R = 3R$. Therefore, $U_f = -\frac{GMm}{3R}$.
4. **Change in Potential Energy ($\Delta U$):**
   $$ \Delta U = U_f - U_i $$
   $$ \Delta U = \left( -\frac{GMm}{3R} \right) - \left( -\frac{GMm}{R} \right) $$
   $$ \Delta U = \frac{GMm}{R} - \frac{GMm}{3R} = \frac{GMm}{R} \left( 1 - \frac{1}{3} \right) = \frac{2GMm}{3R} $$
5. **Substitution using Gravity ($g$):** We know that $g = \frac{GM}{R^2}$, which implies $GM = gR^2$.
   Substituting this into the expression for $\Delta U$:
   $$ \Delta U = \frac{2(gR^2)m}{3R} = \frac{2}{3}mgR $$

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