A satellite is seen every `6 h` over the equator. It is known that it rotates opposite to that of earth's direction. Then, the angular velocity (in radius per hour) of satellite about the centre of earth will be

Options

A. `(pi)/(2)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. `(pi)/(8)`

Explanation

To find the angular velocity of the satellite about the center of the Earth, we can follow these steps:

### Step 1: Understand the problem
The satellite is seen every 6 hours over the equator, which means it has a relative motion with respect to the Earth. The Earth rotates in one direction, and the satellite rotates in the opposite direction.

### Step 2: Define angular velocities
Let:
- \( \omega_1 \) = angular velocity of the Earth
- \( \omega_2 \) = angular velocity of the satellite

The Earth completes one rotation in 24 hours, so:
\[ \omega_1 = \frac{2\pi \text{ radians}}{24 \text{ hours}} = \frac{\pi}{12} \text{ radians/hour} \]

### Step 3: Determine the relative angular velocity
Since the satellite is moving in the opposite direction to the Earth's rotation, the relative angular velocity \( \omega_{relative} \) can be expressed as:
\[ \omega_{relative} = \omega_1 + \omega_2 \]

### Step 4: Use the given time period
The satellite is seen every 6 hours, which means its relative angular velocity can also be expressed as:
\[ \omega_{relative} = \frac{2\pi \text{ radians}}{6 \text{ hours}} = \frac{\pi}{3} \text{ radians/hour} \]

### Step 5: Set up the equation
Now we can set up the equation:
\[ \frac{\pi}{3} = \frac{\pi}{12} + \omega_2 \]

### Step 6: Solve for \( \omega_2 \)
To find \( \omega_2 \), rearrange the equation:
\[ \omega_2 = \frac{\pi}{3} - \frac{\pi}{12} \]

### Step 7: Find a common denominator
To subtract these fractions, find a common denominator, which is 12:
\[ \omega_2 = \frac{4\pi}{12} - \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4} \text{ radians/hour} \]

### Conclusion
The angular velocity of the satellite about the center of the Earth is:
\[ \omega_2 = \frac{\pi}{4} \text{ radians/hour} \]