Problem 25 - A Complete Resource Book in Physics for JEE Main

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

Step-by-Step Solution

Answer

The distance of planet A from the sun is 4 times greater than that of planet B from the sun. The correct answer is (C) 4.

Step 1: Understand and apply Kepler's Third Law of Planetary Motion

Kepler's Third Law states that the square of the period of revolution of a planet around the sun is proportional to the cube of the average distance between the planet and the sun. Mathematically, it can be expressed as: \[T^2 \propto r^3\] where \(T\) is the period of revolution and \(r\) is the average distance between the planet and the sun.

Step 2: Set up the ratio of periods of revolution

The problem states that the period of revolution of planet A is 8 times that of planet B. We can express this as: \[T_A = 8 \cdot T_B\]

Step 3: Use Kepler's Third Law to find the ratio of distances

Since \(T^2 \propto r^3\), we can write the relationship between the periods of revolution and the average distances for planets A and B as: \[(T_A)^2 \propto (r_A)^3\] \[(T_B)^2 \propto (r_B)^3\] Now we can find the ratio of their distances: \[\frac{(r_A)^3}{(r_B)^3} = \frac{(T_A)^2}{(T_B)^2}\] By substituting the given ratio of periods from Step 2: \[\frac{(r_A)^3}{(r_B)^3} = \frac{(8 \cdot T_B)^2}{(T_B)^2}\]

Step 4: Simplify and find the ratio of distances

Now observe that the right-hand side of the equation simplifies as: \[\frac{(8 \cdot T_B)^2}{(T_B)^2} = 8^2 = 64\] So, \[\frac{(r_A)^3}{(r_B)^3} = 64\] Taking the cube root of both sides: \[\frac{r_A}{r_B} = \sqrt[3]{64} = 4\] Therefore, the distance of planet A from the sun is 4 times greater than that of planet B from the sun. The correct answer is (C) 4.