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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 gravitational attraction. Their relative velocity of approach at a separation distance $r$ between them is
${\left[ {2G\,\frac{{\left( {{m_1} - {m_2}} \right)}}{r}} \right]^{1/2}}$
${\left[ {\frac{{2G}}{r}\,\left( {{m_1} + {m_2}} \right)} \right]^{1/2}}$
${\left[ {\frac{r}{{2G\,\left( {{m_1}{m_2}} \right)}}} \right]^{1/2}}$
${\left[ {\frac{{2G}}{r}\,{m_1}{m_2}} \right]^{1/2}}$
Solution
By applying law of conservation of momentum,
${m_1}{v_1} – {m_2}{v_2} = 0 \Rightarrow {m_1}{v_1} = {m_2}{v_2}…\left( i \right)$
where $v_1$ and $v_2$ are the velocities of masses $m_1$ and $m_2$ at a distance $r$ from each other.
By conservation of energy,
$change\,in\,P.E=change\,in\,K.E.$
$\frac{{G{m_1}{m_2}}}{r} = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2\,\,\,…\left( {ii} \right)$
Solving eqn. $(i)$ and $(ii)$ we get
${v_1} = \sqrt {\frac{{2Gm_2^2}}{{r\left( {{m_1} + {m_2}} \right)}}} \,and\,{v_2} = \sqrt {\frac{{2Gm_1^2}}{{r\left( {{m_1} + {m_2}} \right)}}} $
Relative velocity of approach, $v_R$
$ = \left| {{v_1}} \right| + \left| {{v_2}} \right| = \sqrt {\frac{{2G}}{r}\left( {{m_1} + {m_2}} \right)} $
