For a satellite at a distance 11R from the surface of a planet P of radius R. Its time period is 24hrs. Evaluate time period of another satellite at a distance 2R from surface of P

Answer

To solve the problem, 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 given data for the first satellite (S1)**:
- Distance from the surface of the planet \( h_1 = 11R \)
- Radius of the planet \( R \)
- Time period \( T_1 = 24 \) hours

2. **Calculate the orbital radius \( R_1 \) of the first satellite**:
\[ R_1 = R + h_1 = R + 11R = 12R \]

3. **Use Kepler's Third Law for the first satellite**:
\[ T_1^2 \propto R_1^3 \]
This can be expressed as:
\[ T_1^2 = k \cdot R_1^3 \]
where \( k \) is a constant.

4. **Identify the given data for the second satellite (S2)**:
- Distance from the surface of the planet \( h_2 = 2R \)

5. **Calculate the orbital radius \( R_2 \) of the second satellite**:
\[ R_2 = R + h_2 = R + 2R = 3R \]

6. **Use Kepler's Third Law for the second satellite**:
\[ T_2^2 \propto R_2^3 \]
This can be expressed as:
\[ T_2^2 = k \cdot R_2^3 \]

7. **Set up the ratio of the time periods using the proportionality from Kepler's Third Law**:
\[ \frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} \]

8. **Substitute the values of \( R_1 \) and \( R_2 \)**:
\[ \frac{T_2^2}{(24)^2} = \frac{(3R)^3}{(12R)^3} \]

9. **Simplify the right side**:
\[ \frac{T_2^2}{576} = \frac{27R^3}{1728R^3} \]
\[ \frac{T_2^2}{576} = \frac{27}{1728} = \frac{1}{64} \]

10. **Cross-multiply to find \( T_2^2 \)**:
\[ T_2^2 = 576 \cdot \frac{1}{64} \]
\[ T_2^2 = 9 \]

11. **Take the square root to find \( T_2 \)**:
\[ T_2 = 3 \text{ hours} \]

### Final Answer:
The time period of the second satellite at a distance of \( 2R \) from the surface of planet P is **3 hours**.