A space vehicle is traveling at 4800 km/h relative to Earth when the exhausted rocket motor (mass 4m) is disengaged and sent backward. The relative speed between motor and command module (mass m) is then 82 km/h. What is the speed of the command module relative to Earth just after the separation ?

Answer

To solve the problem, we will use the principle of conservation of momentum and the information provided about the velocities involved.

### Step-by-Step Solution:

1. **Identify the masses and initial velocity:**
- Mass of the command module (CM) = \( m \)
- Mass of the rocket motor (RM) = \( 4m \)
- Total mass of the space vehicle = \( m + 4m = 5m \)
- Initial speed of the space vehicle relative to Earth, \( V_s = 4800 \) km/h.

2. **Define the velocities after separation:**
- Let the speed of the command module after separation be \( V_{CM} \).
- The rocket motor is sent backward with a relative speed of \( 82 \) km/h with respect to the command module. Therefore, the speed of the rocket motor after separation can be expressed as:
\[ V_R = V_{CM} - 82 \text{ km/h} \]

3. **Apply conservation of momentum:**
- Before separation, the total momentum of the system is:
\[ P_{initial} = (5m) \cdot V_s = 5m \cdot 4800 \]
- After separation, the total momentum is:
\[ P_{final} = (m \cdot V_{CM}) + (4m \cdot V_R) \]
- Substituting \( V_R \):
\[ P_{final} = m \cdot V_{CM} + 4m \cdot (V_{CM} - 82) \]
- Simplifying this gives:
\[ P_{final} = m \cdot V_{CM} + 4m \cdot V_{CM} - 328m = 5m \cdot V_{CM} - 328m \]

4. **Set initial momentum equal to final momentum:**
\[ 5m \cdot 4800 = 5m \cdot V_{CM} - 328m \]
- Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ 5 \cdot 4800 = 5 \cdot V_{CM} - 328 \]

5. **Solve for \( V_{CM} \):**
\[ 24000 + 328 = 5 \cdot V_{CM} \]
\[ 5 \cdot V_{CM} = 24328 \]
\[ V_{CM} = \frac{24328}{5} = 4865.6 \text{ km/h} \]

### Conclusion:
The speed of the command module relative to Earth just after the separation is approximately \( 4865.6 \) km/h.