= Solution
For two bodies interacting through a <potential energy> that depends only on their separation, introduce the <centre of mass> coordinate and the <two-particle relative coordinate>:
$$
\mathbf R=\frac{m_1\mathbf r_1+m_2\mathbf r_2}{M},\qquad\mathbf r=\mathbf r_1-\mathbf r_2,\qquad M=m_1+m_2.
$$
Their <kinetic energy> separates exactly:
$$
T=\frac12M\dot{\mathbf R}^2+\frac12\mu\dot{\mathbf r}^2,\qquad \boxed{\mu=\frac{m_1m_2}{m_1+m_2}.}
$$
The <centre of mass> moves uniformly if there is no external <force>, while the internal <two-body problem> reduces to one effective particle of <reduced mass> $\mu$:
$$
M\ddot{\mathbf R}=0,\qquad \mu\ddot{\mathbf r}=-\nabla U(r).
$$
The original trajectories follow from $\mathbf r_1=\mathbf R+(m_2/M)\mathbf r$ and $\mathbf r_2=\mathbf R-(m_1/M)\mathbf r$.
For <Newtonian gravity>, $U=-Gm_1m_2/r$, so the relative <acceleration> is $\ddot{\mathbf r}=-GM\mathbf r/r^3$. Conserved relative <angular momentum> and <energy> then give the usual <conic sections>. For a bound orbit of relative semimajor axis $a$,
$$
\boxed{E=-\frac{Gm_1m_2}{2a},\qquad T_{\rm orb}=2\pi\sqrt{\frac{a^3}{G(m_1+m_2)}}.}
$$
The two bodies share this <orbital period> and orbit their common <centre of mass> with semimajor axes in inverse proportion to their <masses>. If $m_2\gg m_1$, then $\mu\simeq m_1$, recovering the fixed-heavy-centre approximation. Using $M$ as the internal inertial <mass> would confuse centre-of-mass translation with relative motion.
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