Increase in Gravitational Potential Energy of Mass Raised Above Earth Surface

To find the increase in gravitational potential energy of a body of mass $m$ when it is raised from the surface of the Earth to a height $h = 2R_e$, we use the expression for gravitational potential energy.

The gravitational potential energy $U(r)$ of a mass $m$ at a distance $r$ from the center of the Earth is given by: $U(r) = -\frac{GMm}{r}$

where $G$ is the gravitational constant and $M$ is the mass of the Earth.

1. Initial Potential Energy at the Surface of the Earth ($r_1 = R_e$): $U_1 = -\frac{GMm}{R_e}$

2. Final Potential Energy at a Height $h = 2R_e$ ($r_2 = R_e + h = R_e + 2R_e = 3R_e$): $U_2 = -\frac{GMm}{3R_e}$

3. Increase in Gravitational Potential Energy ($\Delta U$): $\Delta U = U_2 - U_1 = -\frac{GMm}{3R_e} - \left(-\frac{GMm}{R_e}\right)$ $\Delta U = \frac{GMm}{R_e} \left(1 - \frac{1}{3}\right) = \frac{2}{3} \frac{GMm}{R_e}$

4. Relating to Acceleration Due to Gravity ($g$): The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R_e^2}$, which gives $GM = g R_e^2$.

   Substituting $GM = g R_e^2$ into the expression for $\Delta U$: $\Delta U = \frac{2}{3} \frac{(g R_e^2) m}{R_e} = \frac{2}{3} m g R_e$

Thus, the increase in potential energy is $\frac{2}{3} m g R_e$.

Correct Option: D