The period of revolution of planet A round from the sun is 8 times that of B. The distance of A from the sun is how many times greater then tht of B from the sun ?

Options

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

Explanation

To solve the problem, we can use Kepler's Third Law of planetary motion, which states that the square of the period of revolution of a planet (T) is directly proportional to the cube of the semi-major axis of its orbit (R). Mathematically, this can be expressed as:

\[ T^2 \propto R^3 \]

### Step-by-Step Solution:

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

2. **Given information**: We know that the period of planet A is 8 times that of planet B:
\[ T_A = 8 T_B \]

3. **Substituting the values**: Substitute \( T_A \) in the Kepler's law 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:
\[ \frac{64 T_B^2}{T_B^2} = 64 \]
So we have:
\[ 64 = \frac{R_A^3}{R_B^3} \]

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

6. **Taking the cube root**: To find the ratio of the distances, take the cube root of both sides:
\[ R_A = \sqrt[3]{64} R_B \]
Since \( \sqrt[3]{64} = 4 \):
\[ R_A = 4 R_B \]

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

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