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. `8h`
B. `12h`
C. `24h`
D. `6h`

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 the sense of rotation of the Earth and the satellite is the same.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
The satellite is seen every 8 hours when it moves in the opposite direction to the Earth's rotation. This means that the satellite's relative motion with respect to the Earth's surface results in it being visible at the same point every 8 hours.

2. **Identifying the Time Period of Earth**:
The time period of the Earth's rotation (T1) is 24 hours. This is the time it takes for the Earth to make one complete rotation.

3. **Using the Relative Motion Concept**:
When the satellite moves in the opposite direction to the Earth's rotation, we can use the formula for the effective time period (T2) of the satellite:
\[ \frac{1}{T} = \frac{1}{T1} + \frac{1}{T2} \]
Here, T is the time interval at which the satellite is seen, T1 is the Earth's period (24 hours), and T2 is the satellite's period (which we know is 8 hours).

4. **Substituting Known Values**:
We know T = 8 hours and T1 = 24 hours. We can rearrange the formula:
\[ \frac{1}{8} = \frac{1}{24} + \frac{1}{T2} \]

5. **Solving for T2**:
Rearranging gives:
\[ \frac{1}{T2} = \frac{1}{8} - \frac{1}{24} \]
To solve this, we find a common denominator (24):
\[ \frac{1}{8} = \frac{3}{24} \]
Thus,
\[ \frac{1}{T2} = \frac{3}{24} - \frac{1}{24} = \frac{2}{24} = \frac{1}{12} \]
Therefore, T2 = 12 hours.

6. **Considering the Same Direction**:
Now, if the satellite moves in the same direction as the Earth's rotation, we again use the formula:
\[ \frac{1}{T} = \frac{1}{T1} - \frac{1}{T2} \]
Substituting T1 = 24 hours and T2 = 12 hours:
\[ \frac{1}{T} = \frac{1}{24} - \frac{1}{12} \]
Finding a common denominator (24):
\[ \frac{1}{12} = \frac{2}{24} \]
Thus,
\[ \frac{1}{T} = \frac{1}{24} - \frac{2}{24} = -\frac{1}{24} \]
This indicates that the satellite will be seen at the same point after 24 hours.

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