Two bodies of masses m_1 and m_2 initially at | vedclass

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(B) Initially,the bodies are at an infinite distance,so their total energy is $0$. When they are at a distance $r$,the total energy is the sum of kine

Vedclass · Accepted Answer

$\left[\frac{2 G\left(m_1+m_2\right)}{r}\right]^{1 / 2}$

Vedclass · Answer

(B) Initially,the bodies are at an infinite distance,so their total energy is $0$. When they are at a distance $r$,the total energy is the sum of kinetic energies and potential energy: $E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 - \frac{Gm_1m_2}{r} = 0$. Since the system is isolated,the center of mass remains at rest,so $m_1v_1 = m_2v_2$. Substituting $v_2 = \frac{m_1v_1}{m_2}$ into the energy equation,we get $\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2(\frac{m_1v_1}{m_2})^2 = \frac{Gm_1m_2}{r}$. Simplifying,$\frac{1}{2}m_1v_1^2(1 + \frac{m_1}{m_2}) = \frac{Gm_1m_2}{r}$,which gives $v_1 = \sqrt{\frac{2Gm_2^2}{r(m_1+m_2)}}$. Similarly,$v_2 = \sqrt{\frac{2Gm_1^2}{r(m_1+m_2)}}$. The relative velocity of approach is $v_{rel} = v_1 + v_2 = \sqrt{\frac{2G}{r(m_1+m_2)}} (m_1 + m_2) = \sqrt{\frac{2G(m_1+m_2)}{r}}$.

$\text{LIST-I}$	$\text{LIST-II}$
$A$. Gravitational constant	$I$. $[LT^{-2}]$
$B$. Gravitational potential energy	$II$. $[L^2 T^{-2}]$
$C$. Gravitational potential	$III$. $[ML^2 T^{-2}]$
$D$. Acceleration due to gravity	$IV$. $[M^{-1} L^3 T^{-2}]$

Two bodies of masses $m_1$ and $m_2$ initially at rest at infinite distance apart move towards each other under gravitational force of attraction. Their relative velocity of approach when they are separated by a distance $r$ is ($G=$ universal gravitational constant.)

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