A satellite is seen after every `8` hours over the equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of earth and satellite is same will be:

Options

A. 8 hours
B. 12 hours
C. 24 hours
D. 6 hours

Explanation

To solve the problem, we need to determine the time interval after which a satellite can be seen at the same place on Earth when its sense of rotation is the same as that of the Earth.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The satellite is seen every 8 hours when it moves in the opposite direction to the Earth's rotation.
- We need to find the time interval when the satellite moves in the same direction as the Earth's rotation.

2. **Angular Velocity of the Earth**:
- The Earth completes one full rotation (2π radians) in 24 hours.
- Therefore, the angular velocity (ω) of the Earth is:
\[ \omega_{\text{Earth}} = \frac{2\pi \text{ radians}}{24 \text{ hours}} = \frac{\pi}{12} \text{ radians/hour} \]

3. **Relative Angular Velocity (Opposite Direction)**:
- When the satellite moves in the opposite direction to the Earth, the relative angular velocity (ω_relative) is the sum of the angular velocities of the satellite and the Earth.
- Given that the satellite is seen every 8 hours, we can express the relative angular velocity as:
\[ \omega_{\text{relative}} = \frac{2\pi \text{ radians}}{8 \text{ hours}} = \frac{\pi}{4} \text{ radians/hour} \]

4. **Setting Up the Equation**:
- When the satellite moves in the opposite direction, we have:
\[ \omega_{\text{relative}} = \omega_{\text{satellite}} + \omega_{\text{Earth}} \]
- Substituting the values:
\[ \frac{\pi}{4} = \omega_{\text{satellite}} + \frac{\pi}{12} \]

5. **Solving for ω_satellite**:
- Rearranging the equation gives:
\[ \omega_{\text{satellite}} = \frac{\pi}{4} - \frac{\pi}{12} \]
- To combine the fractions, we find a common denominator (which is 12):
\[ \omega_{\text{satellite}} = \frac{3\pi}{12} - \frac{\pi}{12} = \frac{2\pi}{12} = \frac{\pi}{6} \text{ radians/hour} \]

6. **Relative Angular Velocity (Same Direction)**:
- When the satellite moves in the same direction as the Earth, the relative angular velocity is:
\[ \omega_{\text{relative}} = \omega_{\text{satellite}} - \omega_{\text{Earth}} \]
- Substituting the values:
\[ \omega_{\text{relative}} = \frac{\pi}{6} - \frac{\pi}{12} \]
- Again, finding a common denominator:
\[ \omega_{\text{relative}} = \frac{2\pi}{12} - \frac{\pi}{12} = \frac{\pi}{12} \text{ radians/hour} \]

7. **Finding the Time Period**:
- The time period (T) for the satellite to be seen at the same place when moving in the same direction is given by:
\[ T = \frac{2\pi}{\omega_{\text{relative}}} = \frac{2\pi}{\frac{\pi}{12}} = 2 \times 12 = 24 \text{ hours} \]

### Final Answer:
The time interval after which the satellite can be seen at the same place when its sense of rotation is the same as that of the Earth is **24 hours**.