A geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth's surface (R_("Earth") = 6400 " km")` will approximately be

Options

A. 1/2 hr
B. 1 hr
C. 2 hr
D. 4 hr

Answer

To solve the problem of finding the time period of a spy satellite orbiting a few hundred kilometers above the Earth's surface, we can use the relationship between the time period of a satellite and its orbital radius.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of the Earth, \( R_{\text{Earth}} = 6400 \, \text{km} \)
- Radius of the geostationary satellite, \( R_1 = 36000 \, \text{km} \)
- Time period of the geostationary satellite, \( T_1 = 24 \, \text{hours} \)

2. **Calculate the Radius of the Spy Satellite:**
- The spy satellite orbits a few hundred kilometers above the Earth's surface. Let's assume it orbits at a height of \( h = 400 \, \text{km} \).
- Therefore, the radius of the spy satellite, \( R_2 = R_{\text{Earth}} + h = 6400 \, \text{km} + 400 \, \text{km} = 6800 \, \text{km} \).

3. **Use the Formula for Time Period:**
- The time period \( T \) of a satellite is given by the formula:
\[ T \propto R^{3/2} \]
- This means that the ratio of the time periods of two satellites is related to the ratio of the cube of their radii:
\[ \frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2} \]

4. **Substitute the Known Values:**
- Substitute \( R_1 = 36000 \, \text{km} \) and \( R_2 = 6800 \, \text{km} \):
\[ \frac{T_2}{24} = \left( \frac{6800}{36000} \right)^{3/2} \]

5. **Calculate the Ratio:**
- Simplify the fraction:
\[ \frac{6800}{36000} = \frac{68}{360} = \frac{17}{90} \]
- Now calculate \( \left( \frac{17}{90} \right)^{3/2} \).

6. **Calculate \( \left( \frac{17}{90} \right)^{3/2} \):**
- First, find \( \sqrt{\frac{17}{90}} \):
\[ \sqrt{\frac{17}{90}} \approx 0.43 \quad (\text{approximately}) \]
- Now cube this value:
\[ \left(0.43\right)^3 \approx 0.079 \]

7. **Substitute Back to Find \( T_2 \):**
- Now substitute back to find \( T_2 \):
\[ \frac{T_2}{24} \approx 0.079 \implies T_2 \approx 24 \times 0.079 \approx 1.9 \, \text{hours} \]

8. **Final Result:**
- Therefore, the time period of the spy satellite is approximately \( 2 \, \text{hours} \).

### Conclusion:
The time period of the spy satellite orbiting a few hundred kilometers above the Earth's surface is approximately **2 hours**.