Problem 18 - A Complete Resource Book in Physics for JEE Main

Question

Imagine a light planet revolving around a very massive star in a circular orbit of radius \(R\) with a period of revolution \(T\). If the gravitational force of attraction between the planet and the star is proportional to \(R^{-5 / 2}\), \(T^{2}\) is proportional to (A) \(R^{3}\) (B) \(R^{7 / 2}\) (C) \(R^{3 / 2}\) (D) \(R^{3.75}\)

Step-by-Step Solution

Answer

The relationship between \(T^2\) and R is given by: \[T^2 \propto R^{3/2}\] Hence, the correct answer is (C).

Step 1: Analyzing the Centripetal Force

The centripetal force acting on the planet must balance the gravitational force to keep it in a stable circular orbit. The centripetal force can be given as: \[F_c = \frac{m_p v^2}{R}\] where \(m_p\) is the mass of the planet, and v is the orbital velocity of the planet. The orbital velocity can be determined from the circumference of the orbit and the period of the revolution, as follows. \[v = \frac{2 \pi R}{T}\]

Step 2: Equating Gravitational Force and Centripetal Force

Now we will equate the centripetal force with the gravitational force: \[F_g = k R^{-5/2}\] where k is a proportionality constant. Therefore, \[\frac{m_p v^2}{R} = k R^{-5/2}\] Now, substitute the expression for velocity from step 1: \[\frac{m_p (2 \pi R / T)^2}{R} = k R^{-5/2}\]

Step 3: Simplifying the Equation and Solving for T^2

We will now simplify the equation and solve for \(T^2\): \[\frac{4m_p \pi^2 R^2}{T^2 R} = k R^{-5/2}\] \[4m_p \pi^2 R = k R^{1/2}T^2\] Now, rearrange the equation to find the proportionality between \(T^2\) and R: \[T^2 = \frac{4m_p \pi^2 R^{3/2}}{k}\] From this expression, we can clearly see that the relationship between \(T^2\) and R is given by: \[T^2 \propto R^{3/2}\] This corresponds to the answer choice (C) in the given options.