Time period of revolution of polar satellite is around

Options

A. 6 minutes
B. 100 minutes
C. 8 hours
D. 24 hours

Answer

To determine the time period of revolution of a polar satellite, we can use the formula for the orbital period of a satellite. The time period \( T \) of a satellite in a circular orbit can be calculated using the formula:

\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]

Where:
- \( T \) is the time period of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite,
- \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the Earth (\( 5.972 \times 10^{24} \, \text{kg} \)).

### Step-by-Step Solution:

1. **Identify the parameters**:
- For a polar satellite, the altitude is typically around 700 km above the Earth's surface. The radius \( r \) from the center of the Earth will be:
\[ r = R_{Earth} + h \]
Where \( R_{Earth} \approx 6371 \, \text{km} \) and \( h \approx 700 \, \text{km} \).
\[ r = 6371 \, \text{km} + 700 \, \text{km} = 7071 \, \text{km} = 7071000 \, \text{m} \]

2. **Plug in the values into the formula**:
- Using the values for \( G \) and \( M \):
\[ T = 2\pi \sqrt{\frac{(7071000)^3}{(6.674 \times 10^{-11})(5.972 \times 10^{24})}} \]

3. **Calculate \( r^3 \)**:
- Calculate \( (7071000)^3 \):
\[ r^3 \approx 3.35 \times 10^{20} \, \text{m}^3 \]

4. **Calculate \( GM \)**:
- Calculate \( GM \):
\[ GM \approx (6.674 \times 10^{-11})(5.972 \times 10^{24}) \approx 3.986 \times 10^{14} \, \text{m}^3/\text{s}^2 \]

5. **Substitute and simplify**:
- Substitute \( r^3 \) and \( GM \) into the equation for \( T \):
\[ T = 2\pi \sqrt{\frac{3.35 \times 10^{20}}{3.986 \times 10^{14}}} \]

6. **Calculate the fraction**:
- Calculate the fraction:
\[ \frac{3.35 \times 10^{20}}{3.986 \times 10^{14}} \approx 8.39 \times 10^{5} \]

7. **Take the square root**:
- Take the square root:
\[ \sqrt{8.39 \times 10^{5}} \approx 915.5 \]

8. **Calculate \( T \)**:
- Finally, calculate \( T \):
\[ T \approx 2\pi \times 915.5 \approx 5757 \, \text{s} \approx 96 \, \text{minutes} \]

### Conclusion:
The time period of revolution of a polar satellite is approximately 100 minutes.