Problem 27 - A Complete Resource Book in Physics for JEE Main

Question

Given that mass of the earth is \(M\) and its radius is \(R\). A body is dropped from a height equal to the radius of the earth above the surface of earth. When it reaches the ground its velocity will be (A) \(\frac{G M}{R}\) (B) \(\left[\frac{G M}{R}\right]^{1 / 2}\) (C) \(\left[\frac{2 G M}{R}\right]^{1 / 2}\) (D) \(\left[\frac{2 G M}{R}\right]\)

Step-by-Step Solution

Answer

The velocity of the body when it reaches the ground will be \(v = \left[\frac{2 G M}{R}\right]^{1 / 2}\).

Step 1: Finding the initial potential energy and gravitational force

First, let's consider the potential energy of the body at its initial height equal to the Earth's radius (i.e., \(2R\) from the center of the Earth). The gravitational potential energy formula is: \[U = - \frac{G M m}{r}\] Where: - \(U\) is the potential energy, - \(G\) is the gravitational constant, - \(M\) is the mass of the Earth, - \(m\) is the mass of the body, and - \(r\) is the distance from the center of the Earth (which is equal to \(2R\)). The gravitational force acting on the body can be calculated using Newton's law of universal gravitation: \[F = \frac{G M m}{r^{2}}\]

Step 2: Applying the conservation of energy principle

Next, we will apply the conservation of energy principle. The initial potential energy of the body will convert into kinetic energy as it falls towards the Earth. The conservation of energy states that the sum of the potential and kinetic energies will be constant: \[U_{i} + K_{i} = U_{f} + K_{f}\] Where: - \(U_{i}\) and \(U_{f}\) are the initial and final potential energies, respectively, - \(K_{i}\) and \(K_{f}\) are the initial and final kinetic energies, respectively. At the initial point, the body is not moving, so its initial kinetic energy is zero: \(K_{i} = 0\). At the final point, when the body reaches the ground, its final potential energy is: \[U_{f} = - \frac{G M m}{R}\] So now the energy conservation equation becomes: \[- \frac{G M m}{2R} = - \frac{G M m}{R} + K_{f}\]

Step 3: Finding the final velocity of the body

The kinetic energy of the body can be calculated using the formula: \[K = \frac{1}{2} m v^{2}\] Where \(v\) is the velocity of the body. Now, replace \(K_{f}\) in the energy conservation equation with the kinetic energy formula: \[- \frac{G M m}{2R} = - \frac{G M m}{R} + \frac{1}{2} m v^{2}\] Solve for \(v\): \[v^{2} = 2\left(\frac{G M}{R} - \frac{G M}{2R}\right)\] \[v^{2} = 2\left(\frac{G M}{2R}\right)\] \[v = \sqrt{\frac{2 G M}{R}}\] Now, let's compare the calculated velocity with the given options: (A) \(\frac{G M}{R}\) (B) \(\left[\frac{G M}{R}\right]^{1 / 2}\) (C) \(\left[\frac{2 G M}{R}\right]^{1 / 2}\) (D) \(\left[\frac{2 G M}{R}\right]\) Our calculated velocity matches with option (C). Therefore, the correct answer is: \[v = \left[\frac{2 G M}{R}\right]^{1 / 2}\]