A payload of mass m is resting on the surface of a spherical...

Initial energy of the payload on the surface of the planet is:

$E_i = K_i + U_i = 0 - \dfrac{GMm}{R} = -\dfrac{GMm}{R}$

The payload is established in a circular orbit at an altitude of $3R$. The orbital radius from the center of the planet is:

$r = R + 3R = 4R$

In a stable circular orbit, the necessary centripetal force is provided by the gravitational force:

$\dfrac{mv^2}{r} = \dfrac{GMm}{r^2} \Rightarrow mv^2 = \dfrac{GMm}{r}$

The final kinetic energy of the payload is:

$K_f = \dfrac{1}{2}mv^2 = \dfrac{GMm}{2r}$

The final potential energy is:

$U_f = -\dfrac{GMm}{r}$

The final total energy of the payload in the orbit is:

$E_f = K_f + U_f = \dfrac{GMm}{2r} - \dfrac{GMm}{r} = -\dfrac{GMm}{2r}$

Substituting $r = 4R$:

$E_f = -\dfrac{GMm}{2(4R)} = -\dfrac{GMm}{8R}$

The minimum work required is equal to the change in total mechanical energy:

$W = E_f - E_i = -\dfrac{GMm}{8R} - \left(-\dfrac{GMm}{R}\right)$

$W = \dfrac{GMm}{R} - \dfrac{GMm}{8R} = \dfrac{7GMm}{8R}$

Answer: $\dfrac{7 G M m}{8 R}$