Two Bodies Become One

The hook

Two bodies attract or repel each other: the Earth and the Moon, two stars, a proton and an electron. Each body’s motion depends on where the other one is, so the equations are coupled, and it looks as if both motions must be found at once. Newton’s great simplification was to change variables. Instead of the two positions, use the centre of mass, which moves as if no forces acted at all, and the separation, which moves like a single particle with a modified mass. One hard problem becomes two easy ones.

This question carries out that separation for two particles on a line. Along the way there is a small miracle: the separation turns out to be the square root of a quadratic in \(t\), for a reason that has nothing to do with the question as set and everything to do with motion in a plane.

Before you start

The force is repulsive, so it pushes \(P\) (on the left) to the left and \(Q\) (on the right) to the right: \[2\ddot x_1 = -\frac{2}{z^3}, \qquad 1 \cdot \ddot x_2 = +\frac{2}{z^3}.\] These are coupled, because each involves \(z = x_2 - x_1\).

Working it through

Stage 1 The separation

Subtract: \(\ddot z = \ddot x_2 - \ddot x_1 = \frac{2}{z^3} + \frac{1}{z^3} = \frac{3}{z^3}\).■

The separation behaves like a single particle of mass \(\frac23\) pushed by the force \(\frac{2}{z^3}\). The number \(\frac23 = \frac{2 \times 1}{2 + 1}\) is the reduced mass \(\frac{m_1m_2}{m_1 + m_2}\).

Stage 2 Energy for the separation

From \(v\frac{\mathrm{d}v}{\mathrm{d}z} = \frac{3}{z^3}\), integrating gives \(\frac12v^2 = -\frac{3}{2z^2} + C\). Initially \(z = 1\) and \(\dot z = \dot x_2 - \dot x_1 = 0 - 1 = -1\), so \(\frac12 = -\frac32 + C\), giving \(C = 2\): \[v^2 = 4 - \frac{3}{z^2}.\] (This is conservation of energy for the separation: kinetic energy plus potential energy \(\frac{3}{2z^2}\) per unit reduced mass is constant.)

Stage 3 A surprisingly simple solution

With \(w = z^2\), \(\dot w = 2z\dot z\), so \[\dot w^2 = 4z^2\dot z^2 = 4z^2\Bigl(4 - \frac{3}{z^2}\Bigr) = 16w - 12.\] Differentiating, \(2\dot w\ddot w = 16\dot w\), so (whenever \(\dot w \neq 0\)) \(\ddot w = 8\). Hence \(w\) is a quadratic in \(t\): \(w = 4t^2 + \alpha t + \beta\). Initially \(w = 1\) and \(\dot w = 2z\dot z = -2\), so \(\beta = 1\) and \(\alpha = -2\): \[z = \sqrt{4t^2 - 2t + 1}. \qed\] (You can check directly: \(\dot w^2 = (8t - 2)^2 = 64t^2 - 32t + 4 = 16(4t^2 - 2t + 1) - 12\).) The particles come closest at \(t = \frac14\), when \(z = \frac{\sqrt3}{2}\), and then separate for ever.

Stage 4 The centre of mass

Adding the two equations of motion, \(2\ddot x_1 + \ddot x_2 = -\frac{2}{z^3} + \frac{2}{z^3} = 0\). The forces are equal and opposite (Newton’s third law), so the total momentum \(2\dot x_1 + \dot x_2\) is constant. It starts at \(2 \times 1 + 0 = 2\), and \(2x_1 + x_2\) starts at \(1\), so \[2x_1 + x_2 = 1 + 2t.\] The centre of mass, at \(\frac{2x_1 + x_2}{3}\), moves steadily at speed \(\frac23\). Combining with \(x_2 - x_1 = z\), \[x_1 = \frac{1 + 2t - \sqrt{4t^2 - 2t + 1}}{3}, \qquad x_2 = \frac{1 + 2t + 2\sqrt{4t^2 - 2t + 1}}{3}. \qed\]

Looking back

• Decouple by changing variables. The difference of the equations involved only \(z\); the weighted sum involved only the centre of mass.

• Energy integral. Writing \(\ddot z = v\frac{\mathrm{d}v}{\mathrm{d}z}\) turned a second-order equation into a first-order one: conservation of energy.

• Change the unknown. The equation for \(z\) was awkward; the equation for \(z^2\) was trivial.

What’s really going on

Why \(z^2\) is quadratic

For any force law of the form \(\ddot z = k/z^3\), the same calculation gives \(\frac{\mathrm{d}^2}{\mathrm{d}t^2}(z^2) = 2\dot z^2 + 2z\ddot z = 2\dot z^2 + \frac{2k}{z^2} =\) constant, since \(\frac12\dot z^2 + \frac{k}{2z^2}\) is the conserved energy. So \(z^2\) is always a quadratic in \(t\). Why?

Here is the hidden reason. Think of a particle moving freely, in a straight line at constant speed, in a plane. Its distance \(r\) from the origin satisfies \(r^2 = (\text{a quadratic in } t)\), since \(r^2 = (x_0 + at)^2 + (y_0 + bt)^2\). And in polar coordinates its radial equation of motion is \(\ddot r - r\dot\theta^2 = 0\); with the angular momentum \(h = r^2\dot\theta\) constant, that becomes \[\ddot r = \frac{h^2}{r^3}.\] The inverse-cube repulsion is exactly the centrifugal effect of the hidden angular motion. In our problem, \(z^2 = 4t^2 - 2t + 1 = (2t - \frac12)^2 + \frac34\) is the squared distance from the origin of a point moving at speed \(2\) along a straight line that passes the origin at distance \(\frac{\sqrt3}{2}\). The one-dimensional repelling particles are the shadow of free motion in two dimensions.

The two-body problem

The method of this question works for any force between two bodies that depends only on their separation and acts along the line between them, in three dimensions as well as one. The centre of mass moves uniformly, and the separation vector \(\mathbf{r}\) obeys \(\mu\ddot{\mathbf{r}} = \mathbf{F}(\mathbf{r})\), with \(\mu = \frac{m_1m_2}{m_1 + m_2}\). For gravity this is the Kepler problem, solved by Newton in the Principia: the separation traces an ellipse, parabola or hyperbola. The Earth and the Sun both orbit their common centre of mass, which lies inside the Sun; the Earth and the Moon orbit a point about \(4700\) km from the Earth’s centre, inside the Earth but well away from its centre, which is why the Earth wobbles monthly. Astronomers find planets round other stars by detecting exactly this wobble in the star.

The thinking skill

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Further reading

• John Taylor, Classical Mechanics (University Science Books), Chapter 8. The two-body problem, reduced mass and orbits.

• Daniel Kleppner & Robert Kolenkow, An Introduction to Mechanics (Cambridge University Press), Chapter 10, on central forces.

• Francesco Calogero, Classical Many-Body Problems Amenable to Exact Treatments (Springer). For the ambitious: the world of exactly solvable systems.