Question

Two bodies of masses \( m_1 \) and \( m_2 \) are initially at rest at an infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. The relative velocity of approach at a separation distance \( r \) between them is:

• A. \( V = \sqrt{\frac{2G(m_1 - m_2)}{r}} \)
• B. \( V = \sqrt{\frac{2G(m_1 + m_2)}{r}} \)
• C. \( V = \sqrt{\frac{G(m_1 + m_2)}{r}} \)
• D. \( V = \sqrt{\frac{2G(m_1 m_2)}{r}} \)

Hint:

The conservation of energy and momentum principles are fundamental when dealing with two-body gravitational systems.

Answer 1

The Correct Option is B

The two bodies are initially at rest, so their initial relative velocity is zero. As they move towards each other due to gravitational attraction, the gravitational potential energy is converted into kinetic energy. The gravitational potential energy of the system when the bodies are at a distance \( r \) is given by: \[ U = -\frac{G m_1 m_2}{r} \] The total mechanical energy of the system is conserved. At any point during the motion, the kinetic energy of the system is equal to the change in potential energy. The total kinetic energy of the two bodies is given by: \[ K = \frac{1}{2} \mu V^2 \] where \( \mu \) is the reduced mass of the system, given by: \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \] Using the conservation of mechanical energy, the change in potential energy equals the total kinetic energy. Thus: \[ \Delta U = \Delta K \] \[ -\frac{G m_1 m_2}{r} = \frac{1}{2} \mu V^2 \] Substituting the value of \( \mu \): \[ -\frac{G m_1 m_2}{r} = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} V^2 \] Solving for \( V \): \[ V = \sqrt{\frac{2G (m_1 + m_2)}{r}} \] Thus, the relative velocity of approach at a separation distance \( r \) between them is: \[ V = \sqrt{\frac{2G (m_1 + m_2)}{r}} \] Conclusion: The correct answer is (2).