A body of mass is taken from earth surface to the height $$h$$ equal to twice the radius of earth ($$R_e$$), the increase in potential energy will be: ($$g$$ = acceleration due to gravity on the surface of earth)

Potential energy at a distance r from the centre of earth is given by :

$$U = -\frac{GMm}{r}$$

Initial potential energy at earth surface :

$$U_1 = -\frac{GMm}{R_e}$$

Given height,

$$h = 2R_e$$

Therefore, final distance from centre of earth :

$$r = R_e + 2R_e = 3R_e$$

Final potential energy :

$$U_2 = -\frac{GMm}{3R_e}$$

Increase in potential energy :

$$\Delta U = U_2 - U_1$$

$$= -\frac{GMm}{3R_e} + \frac{GMm}{R_e}$$

$$= \frac{2GMm}{3R_e}$$

Using,

$$g = \frac{GM}{R_e^2}$$

⇒

$$GM = gR_e^2$$

Substituting,

$$\Delta U = \frac{2mgR_e^2}{3R_e}$$

$$\Delta U = \frac{2}{3}mgR_e$$

Final Answer :

$$\frac{2}{3}mgR_e$$