A body is orbiting around earth at a mean radius which is two times as greater as the parking orbit of a satellite, the period of body is

Options

A. 4 days
B. 16 days
C. `2 sqrt(2)` days
D. 64 days

Explanation

To solve the problem, we will use Kepler's Third Law of planetary motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. This can be mathematically expressed as:

\[ T^2 \propto r^3 \]

### Step-by-Step Solution:

1. **Identify the Mean Radius of the Orbits**:
- Let the mean radius of the parking orbit of the satellite be \( r \).
- The mean radius of the body orbiting around the Earth is given as \( 2r \).

2. **Apply Kepler's Third Law**:
- According to Kepler's Third Law, we can write the relationship for two different orbits:
\[ \frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} \]
where:
- \( T_1 \) is the period of the satellite in the parking orbit,
- \( T_2 \) is the period of the body in the orbit at radius \( 2r \),
- \( r_1 = r \) (the radius of the parking orbit),
- \( r_2 = 2r \) (the radius of the body’s orbit).

3. **Substitute the Values**:
- We know \( r_2 = 2r \) and \( r_1 = r \), so:
\[ \frac{T_2^2}{T_1^2} = \frac{(2r)^3}{r^3} \]
- Simplifying the right side:
\[ \frac{T_2^2}{T_1^2} = \frac{8r^3}{r^3} = 8 \]

4. **Solve for \( T_2 \)**:
- Rearranging gives:
\[ T_2^2 = 8T_1^2 \]
- Taking the square root of both sides:
\[ T_2 = \sqrt{8}T_1 = 2\sqrt{2}T_1 \]

5. **Determine \( T_1 \)**:
- If we assume \( T_1 \) (the period of the satellite in the parking orbit) is 1 day (as is common for geostationary satellites), then:
\[ T_2 = 2\sqrt{2} \times 1 \text{ day} \]

6. **Final Calculation**:
- Calculating \( 2\sqrt{2} \):
\[ 2\sqrt{2} \approx 2 \times 1.414 \approx 2.828 \text{ days} \]

### Conclusion:
The period of the body orbiting around the Earth at a mean radius which is two times greater than the parking orbit of a satellite is approximately \( 2\sqrt{2} \) days.