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Two hypothetical planets of masses $m_1$ and $m_2$ are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres . What is their speed when their separation is $'d'$ ? (Speed of $m_1$ is $v_1$ and that of $m_2$ is $v_2$ )
$v_1 = v_2$
$\begin{array}{l}
{v_1}{\mkern 1mu} = {\mkern 1mu} {m_2}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} \\
{v_2}{\mkern 1mu} = {\mkern 1mu} {m_1}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}}
\end{array}$
$\begin{array}{l}
{v_1}{\mkern 1mu} = {\mkern 1mu} {m_1}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} \\
{v_2}{\mkern 1mu} = {\mkern 1mu} {m_2}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}}
\end{array}$
$\begin{array}{l}
{v_1}\, = \,{m_2}\,\sqrt {\frac{{2G}}{{{m_1}}}} \\
{v_2}\, = \,{m_2}\,\sqrt {\frac{{2G}}{{{m_2}}}}
\end{array}$
Solution
We choose reference point, infinity, where total energy of the system is zero.
So initial energy of the system $=0$
Final energy
$ = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2=\frac{{G{m_1}{m_2}}}{d}$
From conservation of energy,
$Initial\, energy=Final\, energy$
$\therefore 0 = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 = \frac{{G{m_1}{m_2}}}{d}$
$or\,\frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_1}v_2^2 = \frac{{G{m_1}{m_2}}}{d}\,\,…\left( i \right)$
By conservation of linear momentum
${m_1}{v_1} + {m_2}{v_2} = 0$
$or\,\frac{{{v_1}}}{{{v_2}}} = \frac{{{m_2}}}{{{m_1}}} \Rightarrow {v_2} = – \frac{{{m_1}}}{{{m_2}}}{v_1}$
Putting value of $v_2$ in equation $(1)$, we get
${m_1}v_1^2 + {m_2}{\left( { – \frac{{{m_1}{v_1}}}{{{m_2}}}} \right)^2} = \frac{{2G{m_1}{m_2}}}{d}$
$\frac{{{m_1}{m_2}v_1^2 + m_1^2v_1^2}}{{{m_2}}} = \frac{{2G{m_1}{m_2}}}{d}$
${v_1} = \sqrt {\frac{{2Gm_2^2}}{{d\left( {{m_1} + {m_2}} \right)}}} = {m_2}\sqrt {\frac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}} $
$Similarly\,{v_2} = {m_1}\sqrt {\frac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}} $
