A geostationary satellite is orbiting the earth at a height of `5R` above the surface of the earth, `2R` being the radius of the earth. The time period of another satellite in hours at a height of `2R` form the surface of the earth is

Options

A. 5
B. 10
C. `6 sqrt(2)`
D. `6//sqrt(2)`

Explanation

To solve the problem, we need to find the time period of a satellite that is at a height of \(2R\) above the Earth's surface, where \(R\) is the radius of the Earth. We will use Kepler's Third Law of planetary motion, which states that the square of the time period of a satellite is directly proportional to the cube of the semi-major axis of its orbit.

### Step-by-Step Solution:

1. **Identify the Radius of the Orbits**:
- The radius of the Earth \(R\) is given.
- For the geostationary satellite, it is at a height of \(5R\) above the surface. Therefore, the total distance from the center of the Earth is:
\[ r_1 = R + 5R = 6R \]
- For the second satellite at a height of \(2R\) above the surface, the total distance from the center of the Earth is:
\[ r_2 = R + 2R = 3R \]

2. **Apply Kepler's Third Law**:
- According to Kepler's Third Law:
\[ T^2 \propto r^3 \]
- For the geostationary satellite:
\[ T_1^2 \propto (6R)^3 \]
- For the second satellite:
\[ T_2^2 \propto (3R)^3 \]

3. **Set Up the Proportionality**:
- We can write the ratio of the squares of the time periods:
\[ \frac{T_2^2}{T_1^2} = \frac{(3R)^3}{(6R)^3} \]

4. **Simplify the Ratio**:
- Simplifying the right-hand side:
\[ \frac{T_2^2}{T_1^2} = \frac{27R^3}{216R^3} = \frac{27}{216} = \frac{1}{8} \]
- Thus, we have:
\[ T_2^2 = \frac{1}{8} T_1^2 \]

5. **Find the Time Period of the Geostationary Satellite**:
- The time period \(T_1\) of a geostationary satellite is known to be 24 hours.

6. **Calculate \(T_2\)**:
- Substituting \(T_1 = 24\) hours into the equation:
\[ T_2^2 = \frac{1}{8} (24^2) \]
- Calculate \(24^2\):
\[ 24^2 = 576 \]
- Therefore:
\[ T_2^2 = \frac{576}{8} = 72 \]
- Taking the square root gives:
\[ T_2 = \sqrt{72} = 6\sqrt{2} \text{ hours} \]

### Final Answer:
The time period of the satellite at a height of \(2R\) from the surface of the Earth is \(6\sqrt{2}\) hours.