Question

The period of revolution of the planet A around the sun is 27 times that of another planet B. If the distance of A from the sun is \( X \) times greater than that of B from the sun, then the value of \( X \) is

• A. \(8 \)
• B. \(4 \)
• C. \(9 \)
• D. \(3 \)
• E. \(12 \)

Hint:

For orbital motion, always use \( T^2 \propto r^3 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem is an application of Kepler's Third Law of planetary motion. The law states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis of its orbit, which for a nearly circular orbit can be taken as the mean distance (R) from the sun.
Step 2: Key Formula or Approach:
Kepler's Third Law can be written as: \[ T^2 \propto R^3 \quad \text{or} \quad \frac{T^2}{R^3} = \text{constant} \] For two planets, A and B, we can write the relationship as: \[ \left(\frac{T_A}{T_B}\right)^2 = \left(\frac{R_A}{R_B}\right)^3 \] Step 3: Detailed Explanation:
We are given the following information:
The period of planet A is 27 times the period of planet B: \(T_A = 27 T_B \implies \frac{T_A}{T_B} = 27\).
The distance of A is X times the distance of B: \(R_A = X R_B \implies \frac{R_A}{R_B} = X\).
Substitute these into Kepler's law equation: \[ (27)^2 = (X)^3 \] We can write 27 as \(3^3\): \[ (3^3)^2 = X^3 \] \[ 3^6 = X^3 \] To solve for X, we can take the cube root of both sides: \[ X = (3^6)^{1/3} = 3^{6/3} = 3^2 = 9 \] Step 4: Final Answer:
The value of X is 9.