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 `R/2`. And the other mass, in a circular orbit of radius `(3R)/(2)`. The difference between the final and initial total energies is :

Options

A. (a)`-(GMm)/(2R)`
B. (b)`+(GMm)/(6R)`
C. (c)`(GMm)/(2R)`
D. (d)`-(GMm)/(6R)`

Explanation

To solve the problem, we need to calculate the initial and final gravitational potential energies of the system and then find the difference between them.

### Step 1: Calculate the Initial Energy
The initial energy of the body of mass \( m \) moving in a circular orbit of radius \( R \) around a planet of mass \( M \) can be calculated using the formula for gravitational potential energy (U) and kinetic energy (K).

The gravitational potential energy (U) is given by:
\[ U = -\frac{G M m}{R} \]

The kinetic energy (K) of the mass \( m \) in circular motion is given by:
\[ K = \frac{1}{2} m v^2 \]
where \( v \) is the orbital velocity. The orbital velocity can be calculated using the formula:
\[ v = \sqrt{\frac{G M}{R}} \]
Thus, the kinetic energy becomes:
\[ K = \frac{1}{2} m \left(\frac{G M}{R}\right) = \frac{G M m}{2R} \]

So, the total initial energy (E_initial) is:
\[ E_{\text{initial}} = K + U = \frac{G M m}{2R} - \frac{G M m}{R} = -\frac{G M m}{2R} \]

### Step 2: Calculate the Final Energy
After the body splits into two equal masses (\( \frac{m}{2} \)), we have:
- The first mass (\( \frac{m}{2} \)) moves in a circular orbit of radius \( \frac{R}{2} \).
- The second mass (\( \frac{m}{2} \)) moves in a circular orbit of radius \( \frac{3R}{2} \).

**For the first mass (\( \frac{m}{2} \)) at radius \( \frac{R}{2} \):**
- Gravitational potential energy:
\[ U_1 = -\frac{G M \left(\frac{m}{2}\right)}{\frac{R}{2}} = -\frac{G M m}{R} \]
- Kinetic energy:
\[ v_1 = \sqrt{\frac{G M}{\frac{R}{2}}} = \sqrt{\frac{2 G M}{R}} \]
\[ K_1 = \frac{1}{2} \left(\frac{m}{2}\right) v_1^2 = \frac{1}{2} \left(\frac{m}{2}\right) \left(\frac{2 G M}{R}\right) = \frac{G M m}{4R} \]

Total energy for the first mass:
\[ E_1 = K_1 + U_1 = \frac{G M m}{4R} - \frac{G M m}{R} = -\frac{G M m}{4R} \]

**For the second mass (\( \frac{m}{2} \)) at radius \( \frac{3R}{2} \):**
- Gravitational potential energy:
\[ U_2 = -\frac{G M \left(\frac{m}{2}\right)}{\frac{3R}{2}} = -\frac{G M m}{3R} \]
- Kinetic energy:
\[ v_2 = \sqrt{\frac{G M}{\frac{3R}{2}}} = \sqrt{\frac{2 G M}{3R}} \]
\[ K_2 = \frac{1}{2} \left(\frac{m}{2}\right) v_2^2 = \frac{1}{2} \left(\frac{m}{2}\right) \left(\frac{2 G M}{3R}\right) = \frac{G M m}{6R} \]

Total energy for the second mass:
\[ E_2 = K_2 + U_2 = \frac{G M m}{6R} - \frac{G M m}{3R} = -\frac{G M m}{6R} \]

### Step 3: Calculate the Final Total Energy
The final total energy (E_final) is:
\[ E_{\text{final}} = E_1 + E_2 = -\frac{G M m}{4R} - \frac{G M m}{6R} \]

To combine these fractions, we find a common denominator (12R):
\[ E_{\text{final}} = -\left(\frac{3G M m}{12R} + \frac{2G M m}{12R}\right) = -\frac{5G M m}{12R} \]

### Step 4: Calculate the Difference in Energy
Now, we find the difference between the final and initial total energies:
\[ \Delta E = E_{\text{final}} - E_{\text{initial}} = -\frac{5G M m}{12R} - \left(-\frac{G M m}{2R}\right) \]
Convert \(-\frac{G M m}{2R}\) to a fraction with a denominator of 12R:
\[ -\frac{G M m}{2R} = -\frac{6G M m}{12R} \]
Thus,
\[ \Delta E = -\frac{5G M m}{12R} + \frac{6G M m}{12R} = \frac{G M m}{12R} \]

### Final Answer
The difference between the final and initial total energies is:
\[ \Delta E = \frac{G M m}{12R} \]