Question

A planet revolves around the sun with a time period 27 times that of planet B. Planet A is at \( x \) times the distance of planet B from the sun. Find the value of \( x \).

• A. \( 13 \)
• B. \( 12 \)
• C. \( 10 \)
• D. \( 15 \)

Hint:

Kepler's third law is very useful when solving problems involving the relationship between the time period and the distance of a planet from the sun.

Answer 1

The Correct Option is A

We are given that the time period of planet A is 27 times that of planet B. We need to find the value of \( x \), the ratio of the distances of planet A and planet B from the sun.

1. Step 1: Use Kepler's third law of planetary motion: Kepler's third law states that the square of the time period \( T \) of a planet is directly proportional to the cube of its distance \( r \) from the sun: \[ T^2 \propto r^3 \]

2. Step 2: Set up the equation using the proportionality: Let the time period of planet A be \( T_A \) and the distance of planet A from the sun be \( r_A \), and similarly, for planet B, let the time period be \( T_B \) and the distance be \( r_B \). From Kepler's third law, we have: \[ \frac{T_A^2}{T_B^2} = \left( \frac{r_A}{r_B} \right)^3 \] We are told that \( T_A = 27T_B \), so: \[ \frac{(27T_B)^2}{T_B^2} = \left( \frac{r_A}{r_B} \right)^3 \] Simplifying: \[ 27^2 = \left( \frac{r_A}{r_B} \right)^3 \] \[ 729 = \left( \frac{r_A}{r_B} \right)^3 \] Taking the cube root of both sides: \[ \frac{r_A}{r_B} = 9 \] Thus, the value of \( x = 9 \).