= Solution
The <center of mass> is
$$
\mathbf R=\frac{m_1\mathbf x_1+m_2\mathbf x_2}{m_1+m_2}.
$$
By <Newton's third law>, $\mathbf F_{21}=-\mathbf F_{12}$, so
$$
(m_1+m_2)\ddot{\mathbf R}
=m_1\ddot{\mathbf x}_1+m_2\ddot{\mathbf x}_2
=\mathbf F_{12}+\mathbf F_{21}=0.
$$
Thus $\dot{\mathbf R}$ is constant. For $\mathbf r=\mathbf x_1-\mathbf x_2$,
$$
\ddot{\mathbf r}
=\left(\frac1{m_1}+\frac1{m_2}\right)\mathbf F_{12}.
$$
Therefore
$$
\boxed{\mu\ddot{\mathbf r}=\mathbf F_{12}},
\qquad
\boxed{\mu=\frac{m_1m_2}{m_1+m_2}},
$$
where $\mu$ is the <reduced mass>.
Back to article page
