The particles A and B of mass m each are separated by a distance r . Another particle C of mass M is placed r . Another particle C of mass M is placed at the midpoint of A and B . Find the work done in taking C to a point equidistant r form A and B without acceleration (G = Gravitational constant and only gravitational constant and only gravitational intereaction between A, B and C is considered )

Options

A. `(GMm)/r`
B. `(2GMm)/r`
C. `(3GMm)/r`
D. `(4GMm)/r`

Answer

To solve the problem of finding the work done in moving particle C from the midpoint between particles A and B to a point equidistant from both A and B, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Configuration:**
- Particles A and B each have mass \( m \) and are separated by a distance \( r \).
- Particle C of mass \( M \) is initially placed at the midpoint of A and B, which is at a distance \( \frac{r}{2} \) from both A and B.

2. **Calculate the Initial Gravitational Potential Energy (U_initial):**
- The gravitational potential \( V \) at a point due to a mass \( m \) is given by:
\[ V = -\frac{Gm}{d} \]
where \( G \) is the gravitational constant and \( d \) is the distance from the mass to the point.
- The potential at the midpoint due to both A and B:
\[ V_{initial} = V_A + V_B = -\frac{Gm}{\frac{r}{2}} - \frac{Gm}{\frac{r}{2}} = -\frac{2Gm}{\frac{r}{2}} = -\frac{4Gm}{r} \]

3. **Identify the Final Configuration:**
- The final position of particle C is at a distance \( r \) from both A and B.

4. **Calculate the Final Gravitational Potential Energy (U_final):**
- At the final position, the potential due to A and B is:
\[ V_{final} = V_A + V_B = -\frac{Gm}{r} - \frac{Gm}{r} = -\frac{2Gm}{r} \]

5. **Calculate the Change in Gravitational Potential Energy (ΔU):**
- The change in potential energy when moving from the initial to the final position is:
\[ \Delta U = U_{final} - U_{initial} = \left(-\frac{2Gm}{r}\right) - \left(-\frac{4Gm}{r}\right) = -\frac{2Gm}{r} + \frac{4Gm}{r} = \frac{2Gm}{r} \]

6. **Calculate the Work Done (W):**
- The work done by an external agent in moving the mass C is equal to the change in gravitational potential energy:
\[ W = M \cdot \Delta U = M \cdot \frac{2Gm}{r} = \frac{2GMm}{r} \]

### Final Answer:
The work done in taking particle C to the new position is:
\[ W = \frac{2GMm}{r} \]