A geostationary satellite is orbiting the earth at height 11R above surface of the earth. R being radius of earth. If the time period of this satellite is 24hr find out time period of another satellite which is revolving at 2R from surface of earth

Options

A. `6 sqrt(2)`
B. 5
C. 3
D. 8

Answer

To solve the problem of finding the time period of another satellite revolving at a height of 2R from the surface of the Earth, we can use Kepler's third law of planetary motion, which relates the time period of a satellite to its orbital radius.

### Step-by-Step Solution:

1. **Understanding the Geostationary Satellite:**
- A geostationary satellite has a time period (T1) of 24 hours and is located at a height of 11R above the Earth's surface.
- The radius of the Earth (R) is given, so the total distance from the center of the Earth (r1) for the geostationary satellite is:
\[ r_1 = R + 11R = 12R \]

2. **Using Kepler's Third Law:**
- Kepler's third law states that the square of the time period (T) of a satellite is proportional to the cube of the semi-major axis (r) of its orbit:
\[ T^2 \propto r^3 \]
- For two satellites, we can write:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

3. **Finding the Orbital Radius of the Second Satellite:**
- The second satellite is at a height of 2R above the Earth's surface, so its distance from the center of the Earth (r2) is:
\[ r_2 = R + 2R = 3R \]

4. **Substituting the Values:**
- We know \( T_1 = 24 \) hours and we need to find \( T_2 \):
\[ \frac{(24 \text{ hours})^2}{T_2^2} = \frac{(12R)^3}{(3R)^3} \]

5. **Calculating the Ratio:**
- Simplifying the right side:
\[ \frac{(12R)^3}{(3R)^3} = \frac{12^3}{3^3} = \frac{1728}{27} = 64 \]
- Therefore:
\[ \frac{(24)^2}{T_2^2} = 64 \]

6. **Solving for T2:**
- Rearranging gives:
\[ T_2^2 = \frac{(24)^2}{64} \]
\[ T_2^2 = \frac{576}{64} = 9 \]
\[ T_2 = 3 \text{ hours} \]

### Final Answer:
The time period of the satellite revolving at a height of 2R from the surface of the Earth is **3 hours**.