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 \(v_2\) )

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

\({m_1}v_1^2 + {m_2}{\left( { - \frac{{{m_1}{v_1}}}{{{m_2}}}} \right)^2} = \frac{{2G{m_1}{m_2}}}{d}\)

\(\frac{{{m_1}{m_2}v_1^2 + m_1^2v_1^2}}{{{m_2}}} = \frac{{2G{m_1}{m_2}}}{d}\)

\({v_1} = \sqrt {\frac{{2Gm_2^2}}{{d\left( {{m_1} + {m_2}} \right)}}}  = {m_2}\sqrt {\frac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}} \)

\(Similarly\,{v_2} = {m_1}\sqrt {\frac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}} \)