A satellite is seen every `6 h` over the equator. It is known that it roates 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 a satellite that is seen every 6 hours over the equator and rotates in the opposite direction to that of the Earth, we can follow these steps:

### Step 1: Understand the Problem
The satellite is observed every 6 hours, which means it completes one full orbit relative to an observer on Earth in that time. The Earth rotates once every 24 hours.

### Step 2: Define Angular Velocities
Let:
- \( \omega_1 \) be the angular velocity of the Earth.
- \( \omega_2 \) be the angular velocity of the satellite.
- The angular velocity of the Earth \( \omega_1 = \frac{2\pi}{T_{Earth}} = \frac{2\pi}{24} \) radians per hour.

### Step 3: Calculate Angular Velocity of Earth
Calculating \( \omega_1 \):
\[ \omega_1 = \frac{2\pi}{24} = \frac{\pi}{12} \text{ radians per hour} \]

### Step 4: Relative Angular Velocity
Since the satellite rotates in the opposite direction to the Earth's rotation, the relative angular velocity \( \omega_{relative} \) is given by:
\[ \omega_{relative} = \omega_1 + \omega_2 \]
Given that the satellite is seen every 6 hours, we can express this as:
\[ \omega_{relative} = \frac{2\pi}{T_{relative}} = \frac{2\pi}{6} = \frac{\pi}{3} \text{ radians per hour} \]

### 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 Angular Velocity of Satellite
Rearranging the equation to solve for \( \omega_2 \):
\[ \omega_2 = \frac{\pi}{3} - \frac{\pi}{12} \]
To subtract these fractions, we need a common denominator:
\[ \frac{\pi}{3} = \frac{4\pi}{12} \]
Thus,
\[ \omega_2 = \frac{4\pi}{12} - \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4} \text{ radians per hour} \]

### Step 7: Conclusion
The angular velocity of the satellite about the center of the Earth is:
\[ \omega_2 = \frac{\pi}{4} \text{ radians per hour} \]