A satellite having time period of revolution `6830` sec is travelling in the same direction as the rotation of the earth, i.e. from west to east, the time interval between the two successive times at which it will appear vertically overhead to an observer at a fixed point on the equator is

Options

A. `86400` sec
B. `5000`sec
C. `7416` sec
D. `84` sec

Answer

To find the time interval between two successive times at which a satellite appears vertically overhead to an observer at a fixed point on the equator, we can follow these steps:

### Step 1: Understand the Problem
We have a satellite with a time period of revolution \( T_s = 6830 \) seconds, and we know that the Earth rotates once every \( T_e = 86400 \) seconds (which is 24 hours). The satellite is moving in the same direction as the Earth's rotation.

### Step 2: Calculate Angular Velocities
The angular velocity of the satellite \( \omega_s \) and the angular velocity of the Earth \( \omega_e \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Thus, we have:
\[ \omega_s = \frac{2\pi}{6830} \quad \text{and} \quad \omega_e = \frac{2\pi}{86400} \]

### Step 3: Find the Relative Angular Velocity
Since both the satellite and the Earth are rotating in the same direction, the relative angular velocity \( \omega_{rel} \) of the satellite with respect to the Earth is given by:
\[ \omega_{rel} = \omega_s - \omega_e \]

### Step 4: Set Up the Equation for the Time Interval
To find the time interval \( t \) between two successive times when the satellite is directly overhead, we can use the condition that the relative angular displacement must equal \( 2\pi \) (one complete revolution):
\[ \omega_{rel} \cdot t = 2\pi \]
Substituting for \( \omega_{rel} \):
\[ \left(\frac{2\pi}{6830} - \frac{2\pi}{86400}\right) t = 2\pi \]

### Step 5: Simplify the Equation
Dividing both sides by \( 2\pi \):
\[ \left(\frac{1}{6830} - \frac{1}{86400}\right) t = 1 \]
Now, we can find a common denominator to simplify:
\[ \frac{86400 - 6830}{6830 \times 86400} t = 1 \]

### Step 6: Solve for \( t \)
Rearranging gives:
\[ t = \frac{6830 \times 86400}{86400 - 6830} \]
Calculating the denominator:
\[ 86400 - 6830 = 79570 \]
Thus, we have:
\[ t = \frac{6830 \times 86400}{79570} \]

### Step 7: Calculate the Value
Now, performing the multiplication and division:
\[ t = \frac{588072000}{79570} \approx 7384.47 \text{ seconds} \]

### Conclusion
The time interval between two successive times at which the satellite will appear vertically overhead to an observer at a fixed point on the equator is approximately **7384.47 seconds**.