Problem 816 - JEE Physics

Question

Two bodies of masses \(m_{1}\) and \(m_{2}\) are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual a gravitational attraction Their relative velocity of approach at separation distance \(\mathrm{r}\) between them is (A) \(\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{-1 / 2}\) (B) \(\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}\) (C) \(\left[\mathrm{r} /\left\\{2 \mathrm{G}\left(\mathrm{m}_{1} \mathrm{~m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}\) (D) \(\left[\left(2 \mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}\right]^{1 / 2}\)

Step-by-Step Solution

Answer

The short answer to the question is: The relative velocity of approach between the two masses at separation distance \(r\) is given by the expression \(v_r = \left[\frac{2G(m_{1}+m_{2})}{r}\right]^{\frac{1}{2}}\), which corresponds to option (B).

Step 1: Expression for Gravitational Force

First, we need to recall the expression for gravitational force between the two masses \(m_1\) and \(m_2\) at distance \(r\) apart. It is given by: \[F = G \frac{m_{1}m_{2}}{r^2}\] where \(F\) represents the gravitational force, \(G\) is the gravitational constant, and \(r\) is the distance between the two masses.

Step 2: Expression for Gravitational Potential Energy

Next, let's find the expression for the gravitational potential energy (GPE) of the system as a function of distance \(r\). The GPE, represented as \(U(r)\), can be obtained by integrating the gravitational force with respect to distance: \[U(r) = -\int F dr\] \[U(r) = -\int G \frac{m_{1}m_{2}}{r^2} dr\] Upon integrating, we get: \[U(r) = -G \frac{m_{1}m_{2}}{r} + C\] where \(C\) is the constant of integration. Since the potential energy is zero at infinite separation, we get \(C = 0\). Thus, the expression for GPE becomes: \[U(r) = -G \frac{m_{1}m_{2}}{r}\]

Step 3: Conservation of Mechanical Energy

Now, let's apply the conservation of mechanical energy to the system. The total mechanical energy can be written as the sum of the kinetic energy (KE) and the potential energy (PE): \[E_{total} = KE + U(r)\] Initially, the masses are at rest and have zero kinetic energy. As the masses move, their kinetic energy increases while their gravitational potential energy decreases. Since mechanical energy is conserved, the final energy remains the same as the initial energy: \[0 = KE + (-G \frac{m_{1}m_{2}}{r})\]

Step 4: Derive the expression for Relative Velocity

We can represent the kinetic energy of the system in terms of the relative velocity between the two masses. The relative velocity, denoted as \(v_r\), can be written as: \[KE = \frac{1}{2} m_{1} \left(\frac{m_{2}}{m_{1}+m_{2}}\right) v_r^2+ \frac{1}{2} m_{2} \left(\frac{m_{1}}{m_{1}+m_{2}}\right) v_r^2\] Substituting the expression for KE into the conservation of mechanical energy equation, we get: \[0 = \frac{1}{2} m_{1} \left(\frac{m_{2}}{m_{1}+m_{2}}\right) v_r^2+ \frac{1}{2} m_{2} \left(\frac{m_{1}}{m_{1}+m_{2}}\right) v_r^2 - G \frac{m_{1}m_{2}}{r}\] Now, let's solve this equation for the relative velocity \(v_r\): \[v_r^2 = \frac{2G(m_{1}+m_{2})}{r}\] \[v_r = \sqrt{\frac{2G(m_{1}+m_{2})}{r}}\] The expression for the relative velocity of approach between the two masses at separation distance \(r\) is: \[v_r = \left[\frac{2G(m_{1}+m_{2})}{r}\right]^{\frac{1}{2}}\] Comparing the final expression with the given options, the correct answer is: (B) \(\left[\frac{2G(m_{1}+m_{2})}{r}\right]^{\frac{1}{2}}\)

Key Concepts

Gravitational Force

Gravitational force is a fundamental concept that explains how two masses attract each other due to their mass. This force is always attractive and acts along the line joining the centers of the two masses. The strength of the gravitational force depends on:

• The masses of the two objects
• The distance between their centers

The mathematical expression for gravitational force, formulated by Isaac Newton, is given by:\[F = G \frac{m_{1}m_{2}}{r^2}\]Here:

• \(F\) is the force of attraction between the two objects.
• \(G\) is the gravitational constant, a universal value that quantifies the strength of gravitational attraction.
• \(m_{1}\) and \(m_{2}\) are the masses of the two objects.
• \(r\) is the distance between the centers of the two masses.

The force decreases with an increase in distance, following an inverse square law. This means if the distance doubles, the gravitational force becomes one-fourth as strong.

Gravitational Potential Energy

Gravitational potential energy (GPE) represents the energy due to the position of an object in a gravitational field. It is the energy stored as a result of the gravitational force acting on a mass and can be thought of as the potential for converting to kinetic energy as the objects move toward each other.The equation for GPE is derived by integrating the expression for gravitational force with respect to distance:\[U(r) = -G \frac{m_{1}m_{2}}{r} + C\]Where \(C\) is a constant that is generally zero when considering large distances (infinite separation), simplifying the equation to:\[U(r) = -G \frac{m_{1}m_{2}}{r}\]A key aspect of GPE is that it becomes more negative as the objects get closer, indicating that more energy is stored. This relation shows that gravitational potential energy is inversely proportional to the distance \(r\), and grows stronger as objects move towards one another.

Conservation of Mechanical Energy

The principle of conservation of mechanical energy states that in a closed system with no external forces, the total mechanical energy remains constant. This total energy is the sum of kinetic energy (the energy of motion) and potential energy (the stored energy due to position).For two masses attracting each other gravitationally:

• Kinetic Energy (KE) initializes at zero since the masses start at rest.
• Gravitational Potential Energy (U) is maximized when they are at an infinite distance apart.

As the two masses draw closer, gravitational potential energy decreases while kinetic energy increases, keeping the total mechanical energy conserved:\[E_{total} = KE + U(r)\]Using this principle to determine the relative velocity of approach between the masses involves equating the change in potential energy to the increase in kinetic energy:\[0 = KE + (-G \frac{m_{1}m_{2}}{r})\]Through solving for kinetic energy in relation to velocity, we find that:\[v_r = \sqrt{\frac{2G(m_{1}+m_{2})}{r}}\]Therefore, the conservation of mechanical energy provides a powerful tool for predicting the behavior of systems under gravitational interactions.