Two-body problem

The two-body problem determines the motion of two particles interacting only with one another. For a central force it separates into uniform centre-of-mass motion and one relative-coordinate problem with the reduced mass.

For two particles of equal mass $m$ and opposite charges $\pm q$ separated by $s$, the centre of mass lies halfway between them. In a nonradiating classical model, each radius is $s/2$ and $k_e{q}^2/s^2=m{\omega }^{2}s/2$. Hence ${\omega }^2=2k_e{q}^2/(m{s}^3)$ and $T=2\pi \sqrt{m{s}^3/(2k_e{q}^2)}$. Equivalently, relative motion has reduced mass $m/2$. Radiation and quantum binding invalidate an indefinitely stable classical orbit for actual microscopic charges.

For equal masses and relative position $\mathbf{r}$, the individual positions are ${\mathbf{r}}_A=\mathbf{R}-\mathbf{r}/2$ and ${\mathbf{r}}_B=\mathbf{R}+\mathbf{r}/2$, where the center of mass has uniform velocity in an isolated system. Bound inverse-square force motion makes $\mathbf{r}$ an ellipse; each mass follows a half-size ellipse in the center-of-mass frame. Adding a nonzero center-of-mass drift generally makes their laboratory trajectories nonclosed. The relative orbital period is not necessarily a period of either laboratory position.

Two equal point masses can move around the same fixed circle of radius $a$ in diametrically opposite positions. Their stationary centre of mass is the circle's center. A separation $2a$ gives gravitational force $G{m}^2/(4a^2)$ on each mass, and Newton's second law equates it to $ma{\omega }^2$. The common angular speed is therefore fixed by the displayed relation. Opposite tangential velocities produce the same sense of angular rotation.

Two point masses released from rest at separation $a$ have negative mechanical energy $-G{m}_1{m}_2/a$ but collide in time $\pi \sqrt{a^3/[8G(m_1+m_2)]}$. Their separation approaches zero along a regular trajectory before collision. The energy sign alone therefore does not supply a positive lower separation bound.