What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

We have a planet of mass $$M$$ and radius $$R$$. A satellite of mass $$m$$ is to be placed in a circular orbit whose height above the planet’s surface is $$2R$$. Therefore the radius of the orbit, measured from the planet’s centre, is

$$r = R + 2R = 3R.$$

First, let us recall the expression for the gravitational potential energy of a mass $$m$$ at a distance $$r$$ from the centre of a planet of mass $$M$$:

$$U = -\dfrac{G M m}{r}.$$

Next, for a circular orbit we use the condition that the gravitational force provides the necessary centripetal force. Stating the formula,

$$\dfrac{G M m}{r^{2}} = \dfrac{m v^{2}}{r}.$$

Cancelling one factor of $$r$$ and solving for the orbital speed $$v$$ gives

$$v^{2} = \dfrac{G M}{r}, \qquad\text{so}\qquad v = \sqrt{\dfrac{G M}{r}}.$$

The kinetic energy of the satellite in this circular orbit is therefore

$$K = \dfrac{1}{2} m v^{2} = \dfrac{1}{2} m \left(\dfrac{G M}{r}\right) = \dfrac{G M m}{2 r}.$$

The total mechanical energy in the orbit is the sum of kinetic and potential energies:

$$E_{\text{orbit}} = K + U = \dfrac{G M m}{2 r} \;-\; \dfrac{G M m}{r} = -\dfrac{G M m}{2 r}.$$

Substituting $$r = 3R$$, we obtain

$$E_{\text{orbit}} = -\dfrac{G M m}{2 \times 3R} = -\dfrac{G M m}{6R}.$$

Now, let us look at the energy of the satellite initially, when it is resting on the planet’s surface. Because it starts from rest, its kinetic energy is zero, and its potential energy is

$$U_{\text{surface}} = -\dfrac{G M m}{R}.$$

Hence the initial total energy is simply

$$E_{\text{initial}} = -\dfrac{G M m}{R}.$$

The minimum energy that must be supplied by the launching mechanism equals the increase in the satellite’s total mechanical energy as it goes from the surface to the required orbit. Therefore,

$$\begin{aligned} \Delta E &= E_{\text{orbit}} - E_{\text{initial}} \\ &= \left(-\dfrac{G M m}{6R}\right) \;-\; \left(-\dfrac{G M m}{R}\right) \\ &= -\dfrac{G M m}{6R} + \dfrac{G M m}{R}. \end{aligned}$$

Bringing the terms to a common denominator of $$6R$$, we have

$$\Delta E = -\dfrac{G M m}{6R} + \dfrac{6 G M m}{6R} = \dfrac{5 G M m}{6R}.$$

This quantity is positive, meaning it represents the energy that must be provided. Thus the minimum launch energy is

$$\boxed{\dfrac{5 G M m}{6R}}.$$

Comparing with the given options, this matches Option C.

Hence, the correct answer is Option C.