A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. then the time period of a spy satellite orbiting a frw hundred km (600 km) above the earth's surface (R=6400 km) will approximately be

Options

A. `(1)/(2)h`
B. `1h`
C. `2h`
D. `4h`

Explanation

To solve the problem of finding the time period of a spy satellite orbiting 600 km above the Earth's surface, we can use Kepler's Third Law of planetary motion. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- The radius of the geostationary satellite (R_g) = 36,000 km.
- The radius of the Earth (R_e) = 6,400 km.
- The altitude of the spy satellite = 600 km.
- Therefore, the radius of the spy satellite (R_s) = R_e + altitude = 6,400 km + 600 km = 7,000 km.

### Step 2: Apply Kepler's Third Law
Kepler's Third Law states that the square of the time period (T) of a satellite is directly proportional to the cube of the semi-major axis (R) of its orbit:
\[ \frac{T_g^2}{T_s^2} = \frac{R_g^3}{R_s^3} \]
Where:
- \(T_g\) = time period of the geostationary satellite = 24 hours.
- \(T_s\) = time period of the spy satellite (unknown).
- \(R_g\) = radius of the geostationary satellite = 36,000 km.
- \(R_s\) = radius of the spy satellite = 7,000 km.

### Step 3: Substitute the Known Values
Substituting the known values into the equation:
\[ \frac{(24 \text{ hours})^2}{T_s^2} = \frac{(36,000 \text{ km})^3}{(7,000 \text{ km})^3} \]

### Step 4: Simplify the Equation
Rearranging the equation to solve for \(T_s^2\):
\[ T_s^2 = T_g^2 \cdot \frac{R_s^3}{R_g^3} \]
Substituting \(T_g = 24\) hours:
\[ T_s^2 = (24 \text{ hours})^2 \cdot \frac{(7,000 \text{ km})^3}{(36,000 \text{ km})^3} \]

### Step 5: Calculate the Values
Calculating the right-hand side:
1. Calculate \( (24 \text{ hours})^2 = 576 \text{ hours}^2 \).
2. Calculate \( \frac{(7,000)^3}{(36,000)^3} = \left(\frac{7}{36}\right)^3 = \frac{343}{46,656} \approx 0.00735 \).

Now substituting back:
\[ T_s^2 = 576 \text{ hours}^2 \cdot 0.00735 \approx 4.24 \text{ hours}^2 \]

### Step 6: Find \(T_s\)
Taking the square root to find \(T_s\):
\[ T_s \approx \sqrt{4.24} \approx 2.06 \text{ hours} \]

### Step 7: Final Result
Thus, the time period of the spy satellite is approximately 2.06 hours, which can be rounded to about 2 hours.

### Conclusion
The final answer is:
**The time period of the spy satellite is approximately 2 hours.**