Two massive particles of mass m₁ and m₂ are released from rest from a very large distance. Find the speeds of the particles, when their distance of separation is r.

$m_1 \sqrt{\frac{2 G}{\left(m_1+m_2\right) r}}$

Since the net force acting on the system is equal to, $F=\vec{F}_1 + \vec{F}_2=0$, the momentum of the system at this instant, is equal to its initial momentum, that is zero because the system is released from rest.

$\Rightarrow\left|m_1 \vec{v}_1+m_2 \vec{v}_2\right|=0$

$\Rightarrow m_1 v_1=m_2 v_2$ (numerically)        ......(1)

Conservation of energy yields

$E_1=E_2$

$K_1+U_1=K_2+U_2$

Since the system is released from rest, K₁ = 0. Since the particles are released from very large distance, U₁ = 0

$\Rightarrow K_2+U_2=0$

$\Rightarrow\left(\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\right)+\left(-\frac{G m_1 m_2}{r}\right)=0$          .......(2)

Eliminating $N_2$ from (2) by putting $v_2=\frac{m_1 v_1}{m_2}$ from (1), we obtain,

$\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2\left(\frac{m_1 v_1}{m_2}\right)^2=\frac{G m_1 m_2}{r}$

$\Rightarrow \frac{m_1 v_1^2}{2}\left[1+\frac{m_1}{m_2}\right]=\frac{G m_1 m_2}{r}$

$\Rightarrow v_1=\sqrt{\frac{2 G m_2^2}{\left(m_1+m_2\right) r}}=m_2 \sqrt{\frac{2 G}{\left(m_1+m_2\right) r}}$

Similarly, $v_2=\frac{m_1 v_1}{m_2}=m_1 \sqrt{\frac{2 G}{\left(m_1+m_2\right) r}}$