Reduced mass

Remarks

Let the bodies be at $x_1, x_2 \in \R^3$, let $M = m_1 + m_2$, let $R = (m_1 x_1 + m_2 x_2)/M$ be the centre of mass and let $r = x_2 - x_1$. Then the total kinetic energy splits as $$\frac{m_1}{2} \norm{x_1'}^2 + \frac{m_2}{2} \norm{x_2'}^2 = \frac{M}{2} \norm{R'}^2 + \frac{\mu}{2} \norm{r'}^2.$$ Check: $x_1 = R - \frac{m_2}{M} r$ and $x_2 = R + \frac{m_1}{M} r$; the cross terms cancel because $m_1 \frac{m_2}{M} - m_2 \frac{m_1}{M} = 0$, and the $r$ terms give $\frac{m_1 m_2^2 + m_2 m_1^2}{2 M^2} \norm{r'}^2 = \frac{\mu}{2} \norm{r'}^2$. So if the force between the bodies depends only on $r$, the relative motion obeys $\mu r'' = F(r)$, the equation of one body of mass $\mu$.

Example: for the Earth and the Moon, with mass ratio about $81$, the reduced mass is about $\frac{81}{82}$ of the mass of the Moon. The formula needs both masses finite and positive; if $m_1 \to \infty$, then $\mu \to m_2$, which is the case of a fixed centre. $\mu$ is less than $\min(m_1, m_2)$, and $\mu = m/2$ when $m_1 = m_2 = m$.