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Step-by-Step Solution
Step 1: Identify the Relevant Concept
We want the change in gravitational potential energy when a mass $m$ is moved from the Earth’s surface to a height $2R$ above the surface. Here, $R$ is the Earth’s radius.
Step 2: Formula for Gravitational Potential Energy
The gravitational potential energy $U$ at a distance $r$ from the center of Earth is given by:
$U = - \frac{GMm}{r},$
where $G$ is the gravitational constant and $M$ is the mass of the Earth.
Step 3: Initial Potential Energy ($U_i$)
When the body is on the Earth’s surface, $r = R$. So,
$U_i = - \frac{GMm}{R}.$
Step 4: Final Potential Energy ($U_f$)
The mass is taken to a height $2R$ above the surface. Thus, the new distance from the Earth’s center is $r = R + 2R = 3R$. Hence,
$U_f = - \frac{GMm}{3R}.$
Step 5: Calculate the Change in Potential Energy ($\Delta U$)
We find the difference between final and initial potential energies:
$\Delta U = U_f - U_i
= \left(- \frac{GMm}{3R}\right) - \left(- \frac{GMm}{R}\right)$
$= - \frac{GMm}{3R} + \frac{GMm}{R}
= \frac{2 \, GMm}{3R}.$
Step 6: Use the Relation between $G M$ and $g$
At the Earth’s surface, $g = \frac{GM}{R^2}$, so $GM = g R^2$. Substitute this into the expression for $\Delta U$:
$\Delta U = \frac{2 \, \bigl(g R^2\bigr) \, m}{3R}
= \frac{2}{3} \, m \, g \, R.$
Step 7: Final Answer
Therefore, the change in potential energy is:
$\Delta U = \frac{2}{3} \, m \, g \, R.$
