Radius of planet is R and time for rotation is 24hrs. A geostationary satellite is at an altitude 11 R of parallel. Find time period of a satellite which is at an altitude of 2R?

Options

A. 12hrs
B. 8hrs
C. 4hrs
D. 3hrs

Answer

To solve the problem, we will use Kepler's Third Law of planetary motion, which 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 around a planet.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Radius of the planet, \( R \)
- Time period of the geostationary satellite, \( T_1 = 24 \) hours
- Altitude of the geostationary satellite, \( h_1 = 11R \)
- Altitude of the satellite for which we need to find the time period, \( h_2 = 2R \)

2. **Calculate the Orbital Radius for the Geostationary Satellite:**
- The orbital radius \( r_1 \) for the geostationary satellite is the sum of the planet's radius and the altitude:
\[ r_1 = R + h_1 = R + 11R = 12R \]

3. **Calculate the Orbital Radius for the Satellite at Altitude \( 2R \):**
- The orbital radius \( r_2 \) for the satellite at altitude \( 2R \) is:
\[ r_2 = R + h_2 = R + 2R = 3R \]

4. **Apply Kepler's Third Law:**
- According to Kepler's Third Law:
\[ \frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} \]
- Substitute the values of \( r_1 \) and \( r_2 \):
\[ \frac{T_2^2}{(24 \text{ hours})^2} = \frac{(3R)^3}{(12R)^3} \]

5. **Simplify the Equation:**
- Calculate the ratio:
\[ \frac{T_2^2}{576} = \frac{27R^3}{1728R^3} \]
- This simplifies to:
\[ \frac{T_2^2}{576} = \frac{27}{1728} = \frac{1}{64} \]

6. **Solve for \( T_2^2 \):**
- Rearranging gives:
\[ T_2^2 = 576 \times \frac{1}{64} \]
- Calculate \( T_2^2 \):
\[ T_2^2 = 9 \times 576 = 81 \]

7. **Find \( T_2 \):**
- Taking the square root:
\[ T_2 = \sqrt{81} = 9 \text{ hours} \]

### Final Answer:
The time period of the satellite at an altitude of \( 2R \) is **9 hours**.