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 solve the problem of finding 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 Given Information
We are given:
- Time period of the satellite, \( T_s = 6830 \) seconds.
- The satellite is moving in the same direction as the Earth's rotation (west to east).

### Step 2: Calculate the Time Period of the Earth
The time period of the Earth, \( T_e \), can be calculated as follows:
\[ T_e = 24 \text{ hours} = 24 \times 3600 \text{ seconds} = 86400 \text{ seconds} \]

### Step 3: Determine the Angular Velocities
The angular velocity of the satellite (\( \omega_s \)) and the angular velocity of the Earth (\( \omega_e \)) can be expressed as:
\[ \omega_s = \frac{2\pi}{T_s} \quad \text{and} \quad \omega_e = \frac{2\pi}{T_e} \]

### Step 4: Calculate the Relative Angular Velocity
Since the satellite is moving in the same direction as the Earth's rotation, the relative angular velocity (\( \omega_{relative} \)) is given by:
\[ \omega_{relative} = \omega_s - \omega_e = \frac{2\pi}{T_s} - \frac{2\pi}{T_e} \]

### Step 5: Find the Time Period for the Satellite Relative to the Earth
The time period for the satellite to complete one revolution relative to the Earth can be calculated using the formula:
\[ T_{relative} = \frac{2\pi}{\omega_{relative}} \]
Substituting the expression for \( \omega_{relative} \):
\[ T_{relative} = \frac{2\pi}{\left(\frac{2\pi}{T_s} - \frac{2\pi}{T_e}\right)} = \frac{1}{\left(\frac{1}{T_s} - \frac{1}{T_e}\right)} \]

### Step 6: Substitute the Values
Now, substituting the values of \( T_s \) and \( T_e \):
\[ T_{relative} = \frac{1}{\left(\frac{1}{6830} - \frac{1}{86400}\right)} \]

### Step 7: Calculate the Result
Calculating the above expression:
1. Find \( \frac{1}{6830} \) and \( \frac{1}{86400} \):
\[ \frac{1}{6830} \approx 0.000146 \quad \text{and} \quad \frac{1}{86400} \approx 0.00001157 \]
2. Subtract the two:
\[ \frac{1}{6830} - \frac{1}{86400} \approx 0.000146 - 0.00001157 \approx 0.00013443 \]
3. Take the reciprocal:
\[ T_{relative} \approx \frac{1}{0.00013443} \approx 7436 \text{ seconds} \]

### Final Answer
The time interval between two successive times at which the satellite appears vertically overhead to an observer at a fixed point on the equator is approximately \( 7416 \) seconds.

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