The time of revolution of planet `A` round the sun is `8` times that of another planet `B`. The distance of planet `A` from the sun is how many times the distance of `B` from the sun

Options

A. 5
B. 4
C. 3
D. 2

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 revolution of a planet around the sun is directly proportional to the cube of the semi-major axis (average distance) of its orbit.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Let the time period of planet B be \( T_B \).
- Then, the time period of planet A is \( T_A = 8 T_B \).

2. **Applying Kepler's Third Law**:
- According to Kepler's Third Law, we have:
\[ \frac{T_A^2}{T_B^2} = \frac{R_A^3}{R_B^3} \]
- Here, \( R_A \) and \( R_B \) are the distances of planets A and B from the sun, respectively.

3. **Substituting the Values**:
- Substitute \( T_A = 8 T_B \) into the equation:
\[ \frac{(8 T_B)^2}{T_B^2} = \frac{R_A^3}{R_B^3} \]

4. **Simplifying the Left Side**:
- Calculate the left side:
\[ \frac{64 T_B^2}{T_B^2} = 64 \]
- Thus, we have:
\[ 64 = \frac{R_A^3}{R_B^3} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ R_A^3 = 64 R_B^3 \]

6. **Taking the Cube Root**:
- Taking the cube root of both sides:
\[ R_A = \sqrt[3]{64} R_B \]
- Since \( 64 = 4^3 \), we find:
\[ R_A = 4 R_B \]

7. **Conclusion**:
- Therefore, the distance of planet A from the sun is **4 times** the distance of planet B from the sun.

### Final Answer:
The distance of planet A from the sun is **4 times** the distance of planet B from the sun.

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