A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $$\frac{R}{2}$$, and the other mass, in a circular orbit of radius $$\frac{3R}{2}$$. The difference between the final and initial total energies is:

For a body of mass $$m$$ moving around a planet of mass $$M$$ in a circular orbit of radius $$R$$, the total mechanical energy of the system is the sum of its kinetic and potential energies. For every circular orbit we use the well-known result

$$E \;=\; -\,\frac{G M m}{2 r},$$

where $$r$$ is the orbital radius and $$G$$ is the universal gravitational constant.

Initially the whole mass $$m$$ is in the orbit of radius $$R$$, so the initial total energy is

$$E_i \;=\; -\,\frac{G M m}{2R}.$$

At the given instant, the body splits into two equal parts, each of mass $$\dfrac{m}{2}$$. We neglect the gravitational interaction between the two fragments themselves and consider only their interaction with the planet.

The first fragment of mass $$\dfrac{m}{2}$$ now moves in a circular orbit of radius $$\dfrac{R}{2}$$. Using the same energy formula with $$r=\dfrac{R}{2}$$ we get

$$E_1 \;=\; -\,\frac{G M \left(\dfrac{m}{2}\right)}{2\left(\dfrac{R}{2}\right)} \;=\; -\,\frac{G M \left(\dfrac{m}{2}\right)}{R} \;=\; -\,\frac{G M m}{2R}.$$

The second fragment of mass $$\dfrac{m}{2}$$ moves in a circular orbit of radius $$\dfrac{3R}{2}$$. Inserting $$r=\dfrac{3R}{2}$$ gives

$$E_2 \;=\; -\,\frac{G M \left(\dfrac{m}{2}\right)}{2\left(\dfrac{3R}{2}\right)} \;=\; -\,\frac{G M \left(\dfrac{m}{2}\right)}{3R} \;=\; -\,\frac{G M m}{6R}.$$

Adding the energies of both fragments yields the final total energy of the system:

$$E_f \;=\; E_1 + E_2 \;=\; \left(-\,\frac{G M m}{2R}\right) + \left(-\,\frac{G M m}{6R}\right) \;=\; -\,\frac{3G M m}{6R} - \frac{G M m}{6R} \;=\; -\,\frac{4G M m}{6R} \;=\; -\,\frac{2G M m}{3R}.$$

Now we calculate the change in energy, i.e. the difference between the final and the initial energies:

$$\Delta E \;=\; E_f - E_i \;=\; \left(-\,\frac{2G M m}{3R}\right) - \left(-\,\frac{G M m}{2R}\right).$$

Combining the two terms by bringing them to a common denominator $$6R$$, we obtain

$$\Delta E \;=\; -\,\frac{4G M m}{6R} + \frac{3G M m}{6R} \;=\; -\,\frac{G M m}{6R}.$$

The change is negative, indicating that the total mechanical energy of the system decreases in magnitude by $$\dfrac{G M m}{6R}$$.

Hence, the correct answer is Option C.