Question

The period of revolution of planet A around the sun is 8 times that of \( B \). The distance of \( A \) from the sun is how many times greater than that of \( B \) from the sun?

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Hint:

- Kepler’s Third Law: $T^2 \propto r^3$
- Larger time period $\Rightarrow$ planet is farther from the sun
- Always square the time ratio, then take cube root to find distance ratio

Answer 1

The Correct Option is C

Concept: Kepler’s Third Law of Planetary Motion
This law states that the square of the time period of revolution of a planet is directly proportional to the cube of its average distance from the sun: \[ T^2 \propto r^3 \] For two planets orbiting the same star, the ratio form is: \[ \left(\frac{T_A}{T_B}\right)^2 = \left(\frac{r_A}{r_B}\right)^3 \]

Step 1: Write the given information.
\[ T_A = 8T_B \;\;\Rightarrow\;\; \frac{T_A}{T_B} = 8 \]

Step 2: Substitute into Kepler’s law.
\[ \left(\frac{T_A}{T_B}\right)^2 = \left(\frac{r_A}{r_B}\right)^3 \] \[ 8^2 = \left(\frac{r_A}{r_B}\right)^3 \]

Step 3: Simplify the equation.
\[ 64 = \left(\frac{r_A}{r_B}\right)^3 \]

Step 4: Take cube root on both sides.
\[ \frac{r_A}{r_B} = \sqrt[3]{64} = 4 \] Final Result:
The distance of planet A from the sun is 4 times that of planet B. Answer: ${4}$