1 minute read Introductory

From the archive (2024). Corrected in October 2026; see the note at the end.

The two-body problem can be solved exactly. This post shows the key step: separating the motion into the motion of the centre of mass and the relative motion of the two bodies. It is the starting point for my 2024 research project on the three-body problem.

Centre of mass

Define the centre-of-mass position

\[\mathbf{R}=\frac{m_1 \mathbf{r}_1+m_2 \mathbf{r}_2}{m_1+m_2}.\]

Since the masses are constant, the total momentum is

\[\mathbf{P}=m_1\dot{\mathbf{r}}_1 + m_2\dot{\mathbf{r}}_2 = (m_1+m_2)\,\dot{\mathbf{R}}.\]

How does the centre of mass move?

With no external forces, the total momentum is conserved, so \(\dot{\mathbf{R}}=\mathbf{P}/(m_1+m_2)\) is constant and

\[\mathbf{R}(t)=\mathbf{R}(0)+\frac{\mathbf{P}}{m_1+m_2}\, t .\]

The centre of mass moves in a straight line at constant velocity.

Relative motion

Now suppose the bodies interact through a conservative force that depends only on the relative position \(\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2\), so the potential is \(V = V(\mathbf{r}_1 - \mathbf{r}_2)\).

Solving the definition of \(\mathbf{R}\) for \(\mathbf{r}_1\),

\[\mathbf{r}_1=\mathbf{R}+\frac{m_2}{m_1+m_2}\, \mathbf{r},\]

and differentiating twice, using \(\ddot{\mathbf{R}} = \mathbf{0}\),

\[\ddot{\mathbf{r}}_1=\frac{m_2}{m_1+m_2}\,\ddot{\mathbf{r}} .\]

Hence

\[\frac{m_1 m_2}{m_1+m_2}\, \ddot{\mathbf{r}}=m_1 \ddot{\mathbf{r}}_1=-\nabla_1 V\left(\mathbf{r}_1-\mathbf{r}_2\right)=-\nabla V(\mathbf{r}),\]

that is,

\[\mu\, \ddot{\mathbf{r}}=-\nabla V(\mathbf{r}), \qquad \mu \equiv \frac{m_1 m_2}{m_1+m_2} .\]

The quantity \(\mu\) is the reduced mass, and it can also be written as

\[\frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}.\]

The two-body problem is therefore equivalent to a single body of mass \(\mu\) moving in the fixed potential \(V(\mathbf{r})\), which can be solved exactly. With three bodies no such reduction is available, and that is where the real difficulty begins.

Exercises

  • From \(\dot{\mathbf{R}}=\mathbf{P}/(m_1+m_2)\), show that \(\mathbf{R}(t)=\mathbf{R}(0)+\mathbf{P}t/(m_1+m_2)\). (Secondary school)
  • Define \(\nabla\). What is the difference between \(\nabla\) and \(\nabla_1\)? (Undergraduate, years 1–2)
  • What does the formula for the reduced mass remind you of? (Hint: electric circuits.) (Secondary school)
  • Show that two-body motion is planar. (Undergraduate, years 1–2)

Changes on 25 October 2026

What was wrong. Nothing in the derivation. The total momentum was used without being defined, and a link to my supervisor’s old staff page no longer worked.

What changed. The momentum is defined, vectors are in bold and the dead link is removed.

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