A geostationary satellite is orbiting the earth at a height of `5R` above the surface of the earth, `R` 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. `6sqrt(2)`
D. `6//sqrt(2)`

Answer

To find the time period of a satellite orbiting at a height of \(2R\) above the surface of the Earth, we can 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 height and radius:**
- The radius of the Earth, \(R\).
- The height of the geostationary satellite is \(5R\) above the surface.
- The height of the second satellite is \(2R\) above the surface.

2. **Calculate the total distance from the center of the Earth:**
- For the geostationary satellite:
\[ r_1 = R + 5R = 6R \]
- For the second satellite:
\[ r_2 = R + 2R = 3R \]

3. **Apply Kepler's Third Law:**
- According to Kepler's Third Law:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]
- Here, \(T_1\) is the time period of the geostationary satellite, which is \(24\) hours, and \(T_2\) is the time period of the satellite at height \(2R\).

4. **Substitute the known values:**
- Substitute \(r_1 = 6R\) and \(r_2 = 3R\):
\[ \frac{(24)^2}{T_2^2} = \frac{(6R)^3}{(3R)^3} \]

5. **Simplify the equation:**
- The \(R\) cancels out:
\[ \frac{(24)^2}{T_2^2} = \frac{6^3}{3^3} = \frac{216}{27} = 8 \]
- Therefore:
\[ \frac{(24)^2}{T_2^2} = 8 \]

6. **Cross-multiply to find \(T_2^2\):**
- Rearranging gives:
\[ T_2^2 = \frac{(24)^2}{8} \]
- Calculate \(24^2\):
\[ 24^2 = 576 \]
- Thus:
\[ T_2^2 = \frac{576}{8} = 72 \]

7. **Take the square root to find \(T_2\):**
- Therefore:
\[ T_2 = \sqrt{72} = 6\sqrt{2} \text{ hours} \]

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