A satellite of earth of mass 'm' is taken from orbital radius 2R to 3R, then minimum work done is :-

Options

A. `(GMm)/(6R)`
B. `(GMm)/(12R)`
C. `(GMm)/(24 R)`
D. `(GMm)/(3R)`

Answer

To find the minimum work done when a satellite of mass 'm' is moved from an orbital radius of 2R to 3R, we can use the concept of gravitational potential energy. The work done is equal to the change in gravitational potential energy (ΔU) of the satellite.

### Step-by-Step Solution:

1. **Understand the Formula for Gravitational Potential Energy:**
The gravitational potential energy (U) of a satellite at a distance 'r' from the center of the Earth is given by the formula:
\[ U = -\frac{G M m}{r} \]
where:
- \(G\) is the gravitational constant,
- \(M\) is the mass of the Earth,
- \(m\) is the mass of the satellite,
- \(r\) is the distance from the center of the Earth.

2. **Calculate the Initial Potential Energy (U_initial) at 2R:**
For the satellite at an orbital radius of 2R:
\[ U_{initial} = -\frac{G M m}{2R} \]

3. **Calculate the Final Potential Energy (U_final) at 3R:**
For the satellite at an orbital radius of 3R:
\[ U_{final} = -\frac{G M m}{3R} \]

4. **Determine the Change in Potential Energy (ΔU):**
The change in potential energy as the satellite moves from 2R to 3R is:
\[ \Delta U = U_{final} - U_{initial} \]
Substituting the values we calculated:
\[ \Delta U = \left(-\frac{G M m}{3R}\right) - \left(-\frac{G M m}{2R}\right) \]
Simplifying this expression:
\[ \Delta U = -\frac{G M m}{3R} + \frac{G M m}{2R} \]

5. **Finding a Common Denominator:**
The common denominator for 3R and 2R is 6R. Rewriting the terms:
\[ \Delta U = -\frac{2G M m}{6R} + \frac{3G M m}{6R} \]
\[ \Delta U = \frac{(3 - 2)G M m}{6R} = \frac{G M m}{6R} \]

6. **Final Result for Work Done:**
The work done (W) is equal to the change in potential energy:
\[ W = \Delta U = \frac{G M m}{6R} \]

### Conclusion:
The minimum work done in moving the satellite from an orbital radius of 2R to 3R is:
\[ \frac{G M m}{6R} \]