The Newtonian Two-Body Problem

An isolated Newtonian binary is exactly reducible to two independent motions: uniform translation of the center of mass and a one-body Kepler problem for the separation vector. The reduction is exact because the interaction depends only on the relative position and internal forces satisfy action–reaction; no small-mass approximation is required.

Required background. Classical and Variational Mechanics provides action principles, canonical variables, and mechanical conservation laws; Newtonian Gravity provides the gravitational potential, Poisson equation, and inverse-square force law. Helpful background. Calculus and Linear Algebra provides multivariable differentiation, linear maps, inner products, and basis changes; Ordinary Differential Equations provides initial-value problems, linearization, and asymptotic solution methods.

Exact separation of the Lagrangian

Let the masses be $m_1,m_2\gt0$, positions $\mathbf r_1,\mathbf r_2$, total mass $M=m_1+m_2$, and separation $\mathbf r=\mathbf r_1-\mathbf r_2$. Define

$$\mathbf R=\frac{m_1\mathbf r_1+m_2\mathbf r_2}{M}, \qquad \mu=\frac{m_1m_2}{M}.$$

Here $\mu$ is the reduced mass, not the frequently used gravitational parameter $GM$.

Invert the definitions:

$$\mathbf r_1=\mathbf R+\frac{m_2}{M}\mathbf r, \qquad \mathbf r_2=\mathbf R-\frac{m_1}{M}\mathbf r.$$

For the Newtonian gravitational interaction, the two-particle Lagrangian is

$$L=\frac{m_1}{2}\dot{\mathbf r}_1^2 +\frac{m_2}{2}\dot{\mathbf r}_2^2 +\frac{Gm_1m_2}{r}.$$

Substitution cancels the cross terms and gives

$$L=\frac{M}{2}\dot{\mathbf R}^{2} +\left(\frac{\mu}{2}\dot{\mathbf r}^{2}+\frac{Gm_1m_2}{r}\right).$$

The coordinate $\mathbf R$ is cyclic, so $M\dot{\mathbf R}=\mathbf P$ is constant. The relative equation is

$$\mu\ddot{\mathbf r} =-\frac{Gm_1m_2}{r^3}\mathbf r, \qquad \ddot{\mathbf r}=-\frac{GM}{r^3}\mathbf r.$$

The first form displays the effective particle of mass $\mu$; the second shows that the relative acceleration depends on the total mass $M$.

Conserved quantities and reconstruction

The total energy separates as

$$E=\frac{\mathbf P^2}{2M} +\left(\frac{\mu}{2}\dot{\mathbf r}^{2}-\frac{Gm_1m_2}{r}\right) \equiv E_{\rm CM}+E_{\rm rel}.$$

Likewise, the angular momentum about the chosen origin is

$$\mathbf J=\mathbf R\times\mathbf P+\mu\mathbf r\times\dot{\mathbf r}.$$

In the center-of-momentum frame, $\mathbf P=0$ and $\mathbf R$ is constant. After solving the Kepler problem for $\mathbf r(t)$, the individual trajectories are recovered algebraically:

$$\mathbf r_1-\mathbf R=\frac{m_2}{M}\mathbf r, \qquad \mathbf r_2-\mathbf R=-\frac{m_1}{M}\mathbf r.$$

Both bodies trace similar conics about the common barycenter, with size ratio $a_1/a_2=m_2/m_1$. Their periods and eccentricities agree because their position vectors differ only by constant factors and a sign.

Worked example: equal and unequal binaries

For $m_1=m_2=m$, the reduced mass is $m/2$ and each body lies at distance $r/2$ from the barycenter. A circular separation $r$ obeys

$$\Omega^2=\frac{G(2m)}{r^3}.$$

Each body’s speed is $\Omega r/2$. Summing both kinetic energies and the potential gives

$$E_{\rm rel}=-\frac{Gm^2}{2r},$$

consistent with the Kepler circular-orbit result.

If instead $m_2\ll m_1$, then $\mu\simeq m_2$, $M\simeq m_1$, and $\mathbf r_1-\mathbf R\simeq(m_2/m_1)\mathbf r$. The heavier body is nearly fixed at the barycenter, but “nearly” is an approximation made only after the exact reduction.

Checks, limits, and failures

• Translational invariance is essential. An external potential depending separately on $\mathbf r_1$ and $\mathbf r_2$ generally couples center-of-mass and relative motion.
• The cancellation of kinetic cross terms uses the mass-weighted center $\mathbf R$. The geometric midpoint works only for equal masses.
• Newton’s third law gives $\ddot{\mathbf R}=0$ only for the closed two-particle model. An external force or a net momentum flux can accelerate the material barycenter: isotropic mass or radiation loss need not do so, whereas anisotropic emission, external tides, or propulsion can. When the emitted field or exhaust is included, total momentum remains conserved for the enlarged isolated system.
• Point masses have no collision scale. A physical merger or tidal-disruption criterion requires radii, internal structure, dissipation, or relativistic dynamics.
• In general relativity, global center-of-mass separation and an instantaneous potential are not fundamental in this form. The reduction is a weak-field benchmark, not a covariant binary model.

The exact center-of-mass separation and reduced-mass reconstruction above follow Tong 2013, §§5.1–5.1.5, pp. 67–72, PDF.

Transfer task

Show that the kinetic energy is minimized, for fixed relative velocity $\dot{\mathbf r}$, in the center-of-momentum frame.

Solution

The separated kinetic energy is

$$T=\frac{M}{2}\dot{\mathbf R}^{2}+\frac{\mu}{2}\dot{\mathbf r}^{2}.$$

At fixed $\dot{\mathbf r}$, the second term is fixed and the first is nonnegative. Its unique minimum occurs at $\dot{\mathbf R}=0$, equivalently $\mathbf P=0$. This is the Newtonian center-of-momentum frame.

Continue to Parabolic and Hyperbolic Encounters to express unbound relative trajectories in terms of asymptotic speed and impact parameter.

References

• Tong, David. Dynamics and Relativity. University of Cambridge lecture notes, Lent 2013. §§5.1–5.1.5, pp. 67–72, PDF.

Further reading

• MIT OpenCourseWare 16.07, Lecture L16, “Central Force Motion: Orbits”, Fall 2009.