Two bodies of masses \(m_1\) and \(m_2\) initially at rest at infinite distance apart move towards each…
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 ( \(\mathrm{G}=\) universal gravitational constant.)
Initially when the two masses are at an infinite distance from each other, their gravitational potential energy is zero. When they are at a distance \(r\) from each, the gravitational \(P E\) is
\(\mathrm{PE}=\frac{-G m_1 m_2}{r^2}\)
The minus sign-indicates that there is a decrease in PE. This gives rise to an increase in kinetic energy. If \(v_1\) and \(v_2\) are their respective velocities when they are a distance \(r\) apart then, from the law of conservation of energy, we have
\(\frac{1}{2} m_1 v_1^2=\frac{G m_1 m_2}{r} \text { or } v_1=\sqrt{\frac{2 G m_2}{r}}\)
and \(\frac{1}{2} m_2 v_2^2=\frac{G m_1 m_2}{r}\) or \(v_2=\sqrt{\frac{2 G m_1}{r}}\)
Therefore, their relative velocity of approach is
\(\begin{aligned}
v_1+v_2 & =\sqrt{\frac{2 G m_2}{r}}+\sqrt{\frac{2 G m_1}{r}}=\sqrt{\frac{2 G}{r}\left(m_1+m_2\right)} \\
& =\left(\frac{2 G\left(m_1 \times m_2\right)}{r}\right)^{\frac{1}{2}}
\end{aligned}\)