A satellite of mass `5M` orbits the earth in a circular orbit. At one point in its orbit, the satellite explodes into two pieces, one of mass `M` and the other of masses `4M`. After the explosion the mass `M` ends up travelling in the same circular orbit, but in the opposite direction. After explosion the mass `4M` is

Options

A. In a circular orbit
B. unbound
C. elliptical orbit
D. data is insufficient to determine the nature of the orbit.

Explanation

To solve the problem, we will use the principle of conservation of momentum and analyze the motion of the two fragments after the explosion.

### Step-by-Step Solution:

1. **Initial Setup**:
- A satellite of mass \(5M\) is in a circular orbit around the Earth with velocity \(v_{5M}\).
- The satellite explodes into two pieces: one of mass \(M\) and the other of mass \(4M\).
- After the explosion, the mass \(M\) continues in the same circular orbit but in the opposite direction.

2. **Conservation of Momentum**:
- Before the explosion, the total momentum of the system is:
\[ P_{\text{initial}} = (5M) v_{5M} \]
- After the explosion, the momentum of the two pieces must equal the initial momentum:
\[ P_{\text{final}} = (4M) v_{4M} - (M) v_{5M} \]
- Here, \(v_{4M}\) is the velocity of the mass \(4M\) after the explosion, and we take the direction of \(v_{5M}\) as positive.

3. **Setting Up the Equation**:
- According to the conservation of momentum:
\[ 5M v_{5M} = 4M v_{4M} - M v_{5M} \]
- Rearranging gives:
\[ 5M v_{5M} + M v_{5M} = 4M v_{4M} \]
\[ 6M v_{5M} = 4M v_{4M} \]

4. **Solving for \(v_{4M}\)**:
- Dividing both sides by \(4M\):
\[ v_{4M} = \frac{6}{4} v_{5M} = \frac{3}{2} v_{5M} \]

5. **Relating Velocity to Gravitational Parameters**:
- The velocity \(v_{5M}\) in a circular orbit can be expressed as:
\[ v_{5M} = \sqrt{\frac{GM_e}{r}} \]
- Therefore, substituting this into the equation for \(v_{4M}\):
\[ v_{4M} = \frac{3}{2} \sqrt{\frac{GM_e}{r}} \]

6. **Comparing with Escape Velocity**:
- The escape velocity \(v_e\) from the Earth's gravitational field is given by:
\[ v_e = \sqrt{\frac{2GM_e}{r}} \]
- To determine if \(v_{4M}\) is greater than \(v_e\):
\[ v_{4M} = \frac{3}{2} \sqrt{\frac{GM_e}{r}} = \frac{3}{2} \cdot \frac{1}{\sqrt{2}} v_e = \frac{3}{2\sqrt{2}} v_e \approx 1.06 v_e \]
- Since \(v_{4M} > v_e\), the mass \(4M\) has enough kinetic energy to escape Earth's gravitational pull.

7. **Conclusion**:
- The mass \(4M\) is unbound and will escape from the Earth's gravitational field.

### Final Answer:
The mass \(4M\) is **unbound**.