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 $2R$?

Step 1: Determine the radius of the circular orbit.

The orbit is at an altitude of $2R$ above the surface of the planet of radius $R$.

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

Step 2: Find the total energy of the satellite at the planet's surface.

Initially, the satellite is at rest on the surface, so its kinetic energy is zero.

$$E_{\text{surface}} = KE + PE$$ $$E_{\text{surface}} = 0 - \frac{GmM}{R}$$ $$E_{\text{surface}} = -\frac{GmM}{R}$$

Step 3: Find the total energy of the satellite in the circular orbit.

The gravitational force provides the necessary centripetal force for the circular orbit.

$$\frac{mv^2}{r} = \frac{GmM}{r^2}$$ $$v = \sqrt{\frac{GM}{r}}$$

The kinetic energy in orbit is:

$$KE_{\text{orbit}} = \frac{GmM}{2r}$$

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

$$E_{\text{orbit}} = \frac{GmM}{2r} - \frac{GmM}{r}$$ $$E_{\text{orbit}} = -\frac{GmM}{2r}$$

Substitute $r = 3R$ into the total energy expression:

$$E_{\text{orbit}} = -\frac{GmM}{6R}$$

Step 4: Calculate the minimum energy required for the launch.

The minimum energy required is the difference between the final orbital energy and the initial surface energy.

$$\Delta E = E_{\text{orbit}} - E_{\text{surface}}$$ $$\Delta E = -\frac{GmM}{6R} - \left(-\frac{GmM}{R}\right)$$ $$\Delta E = \frac{GmM}{R}\left(1 - \frac{1}{6}\right)$$ $$\Delta E = \frac{5GmM}{6R}$$

Answer: The minimum energy required is $\dfrac{5GmM}{6R}$.