A satellite orbiting the circular orbit of radius `R` complete one revolution in `3h`. If orbital radius of geostationary satellite is `36000km`, then the orbital radius `R` of satellite is

Options

A. 6000 km
B. 9000 km
C. 12000 km
D. 15000 km

Explanation

To find the orbital radius \( R \) of the satellite that completes one revolution in 3 hours, we can use the relationship between the orbital period and the radius of the orbit. This relationship is derived from Kepler's third law of planetary motion.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Period of the satellite \( T_1 = 3 \) hours \( = 3 \times 3600 \) seconds \( = 10800 \) seconds.
- Orbital radius of the geostationary satellite \( R_2 = 36000 \) km \( = 36000 \times 1000 \) meters \( = 3.6 \times 10^7 \) meters.
- Period of the geostationary satellite \( T_2 = 24 \) hours \( = 24 \times 3600 \) seconds \( = 86400 \) seconds.

2. **Apply Kepler's Third Law:**
According to Kepler's third law, the square of the period of revolution of a satellite is directly proportional to the cube of the semi-major axis (orbital radius) of its orbit:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]
where \( R_1 \) is the radius of the satellite we want to find.

3. **Rearranging the Equation:**
Rearranging the equation gives:
\[ R_1^3 = R_2^3 \cdot \left(\frac{T_1^2}{T_2^2}\right) \]

4. **Substituting the Values:**
Substitute the known values into the equation:
\[ R_1^3 = (3.6 \times 10^7)^3 \cdot \left(\frac{(10800)^2}{(86400)^2}\right) \]

5. **Calculating the Ratios:**
Calculate the ratio:
\[ \frac{(10800)^2}{(86400)^2} = \left(\frac{10800}{86400}\right)^2 = \left(\frac{1}{8}\right)^2 = \frac{1}{64} \]

6. **Calculating \( R_1^3 \):**
Now substitute this back into the equation:
\[ R_1^3 = (3.6 \times 10^7)^3 \cdot \frac{1}{64} \]

7. **Calculating \( R_1 \):**
First, calculate \( (3.6 \times 10^7)^3 \):
\[ (3.6)^3 = 46.656 \quad \text{and} \quad (10^7)^3 = 10^{21} \]
Thus,
\[ (3.6 \times 10^7)^3 = 46.656 \times 10^{21} \]
Now divide by 64:
\[ R_1^3 = \frac{46.656 \times 10^{21}}{64} = 0.729 \times 10^{21} \text{ m}^3 \]

8. **Taking the Cube Root:**
Now take the cube root to find \( R_1 \):
\[ R_1 = (0.729 \times 10^{21})^{1/3} \approx 0.9 \times 10^{7} \text{ m} = 9 \times 10^6 \text{ m} = 9000 \text{ km} \]

### Final Answer:
The orbital radius \( R \) of the satellite is approximately **9000 km**.