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The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.