Two masses m_1 and m_2 are initially at rest | vedclass

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(C) By the law of conservation of energy,the initial potential energy at infinity is $0$ and the initial kinetic energy is $0$. At a separation distan

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$[2G(m_1 + m_2)/d]^{1/2}$

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(C) By the law of conservation of energy,the initial potential energy at infinity is $0$ and the initial kinetic energy is $0$. At a separation distance $d$,the potential energy is $-G m_1 m_2 / d$. The total energy is conserved:$0 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 - \frac{G m_1 m_2}{d}$From the conservation of linear momentum,since the system is initially at rest,the center of mass remains at rest:$m_1 v_1 = m_2 v_2 \implies v_2 = \frac{m_1 v_1}{m_2}$Substituting $v_2$ into the energy equation:$\frac{G m_1 m_2}{d} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 (\frac{m_1 v_1}{m_2})^2 = \frac{1}{2} m_1 v_1^2 (1 + \frac{m_1}{m_2}) = \frac{1}{2} m_1 v_1^2 (\frac{m_1 + m_2}{m_2})$Solving for $v_1$: $v_1 = \sqrt{\frac{2 G m_2^2}{d(m_1 + m_2)}}$. Similarly,$v_2 = \sqrt{\frac{2 G m_1^2}{d(m_1 + m_2)}}$.The relative velocity of approach is $v_{rel} = v_1 + v_2 = \sqrt{\frac{2 G}{d(m_1 + m_2)}} (m_2 + m_1) = \sqrt{\frac{2 G (m_1 + m_2)}{d}}$.

Two masses $m_1$ and $m_2$ are initially at rest and are separated by a very large distance. If the masses approach each other subsequently due to gravitational attraction between them,their relative velocity of approach at a separation distance of $d$ is:

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