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Step-by-Step Solution
Step 1: Write the formula for gravitational potential energy
The gravitational potential energy of a mass $m$ at a distance $r$ from the center of the Earth of mass $M$ is given by
$$
U = -\frac{GMm}{r}.
$$
Step 2: Calculate the initial potential energy at Earth's surface
At the surface of the Earth, $r = R$. Therefore, the initial potential energy $U_i$ is
$$
U_i = -\frac{GMm}{R}.
$$
Step 3: Calculate the final potential energy at height $2R$ above Earth's surface
The height from the surface is $h = 2R$. Hence, the distance from the center of the Earth is
$$
r = R + h = R + 2R = 3R.
$$
Therefore, the final potential energy $U_f$ is
$$
U_f = -\frac{GMm}{3R}.
$$
Step 4: Determine the change in potential energy
The change in potential energy $\Delta U$ is given by
$$
\Delta U = U_f - U_i.
$$
Substituting the expressions for $U_f$ and $U_i$, we get
$$
\Delta U = -\frac{GMm}{3R} - \left( -\frac{GMm}{R} \right).
$$
Simplifying,
$$
\Delta U = -\frac{GMm}{3R} + \frac{GMm}{R} = \frac{GMm}{R}\left(1 - \frac{1}{3}\right) = \frac{GMm}{R} \times \frac{2}{3}.
$$
Step 5: Express the result in terms of $mgR$
Since $g = \frac{GM}{R^2}$, we have $GM = gR^2$. Thus,
$$
\frac{GMm}{R} = \frac{gR^2 m}{R} = gRm.
$$
Hence,
$$
\Delta U = \frac{2}{3}gRm = \frac{2}{3}mgR.
$$
Final Answer
The change in potential energy of the body is
$$
\frac{2}{3} \, mgR.
$$
