Gravitational potential energy formula

The gravitational potential energy of a mass $m$ at a distance $r$ from the center of the earth is $U(r) = -\frac{GMm}{r}$ where $G$ is the gravitational constant and $M$ is the mass of the earth.

Initial and final positions

• Initially, the object is on the earth's surface: $r_1 = R$
• It is raised to a height equal to the earth's radius $R$, so final distance from the center is: $r_2 = R + R = 2R$

Change in potential energy

Gain in potential energy is $\Delta U = U(2R) - U(R)$

Substitute: $\Delta U = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right)$

$\Delta U = -\frac{GMm}{2R} + \frac{GMm}{R}$

$\Delta U = \frac{GMm}{2R}$

Use relation between $g$ and $GM$

We know $g = \frac{GM}{R^2}$ so $GM = gR^2$

Substitute into $\Delta U$: $\Delta U = \frac{gR^2 m}{2R} = \frac{1}{2}mgR$

Match with options

$\boxed{\Delta U = \frac{1}{2}mgR}$

So the correct option is C.