Kepler's Laws and Binary Stars

Ellipses, not circles

For two thousand years astronomers assumed that planets move on circles, or on circles riding on circles. Johannes Kepler, working from Tycho Brahe's careful records of Mars, found that no set of circles fitted. The orbit of Mars is an ellipse, a stretched circle. The Sun is not at its centre but at a special point inside it, a focus.

An ellipse has two foci on its long axis. Half the length of that axis is the semi-major axis \(a\). The foci lie a distance \(ae\) on either side of the centre, where \(e\) is the eccentricity, a number from 0 to 1. With \(e = 0\) the two foci meet at the centre, and the ellipse is a circle. The nearest point to the Sun is the perihelion, at \(r_p = a(1 - e)\), and the farthest is the aphelion, at \(r_a = a(1 + e)\).

The Earth's orbit is nearly a circle, with \(e = 0.017\), so its distance from the Sun changes by only about \(3\,\%\) over a year. Halley's comet has \(e = 0.97\): its orbit is a long, thin ellipse that reaches from inside the orbit of Venus to beyond Neptune. Newton later showed that an inverse-square force gives exactly such ellipses, and the cannon of lesson 5 drew one.

Equal areas in equal times

Kepler's second discovery was about speed. Tycho's records showed Mars moving faster in some months than in others. Kepler found a simple rule that fixes the speed at every point of the ellipse, using the line drawn from the Sun to the planet.

In Fig 7.2 the dots crowd together far from the Sun and spread apart near it. The sector swept in one twelfth of a year is short and wide near the Sun, long and thin far away. Their areas are the same. So the planet must move fast when it is close and slowly when it is far. Here the velocity arrow is four times longer at the perihelion than at the aphelion.

The reason is the conservation of angular momentum. In the chapter on Angular Momentum you saw that the Sun's pull points straight at the Sun, so it has no torque about the Sun. The planet's angular momentum about the Sun, \(L = mrv_\perp\), is therefore constant. Here \(v_\perp\) is the part of the velocity at right angles to the line from the Sun.

At the perihelion and at the aphelion the velocity is at right angles to the line from the Sun, so there \(L = mrv\) exactly. Equal \(L\) gives \(v_p r_p = v_a r_a\). For the orbit in the figure, \(r_a = 4r_p\), so the planet moves four times faster at the perihelion. The Earth's own speed changes from \(30.3\) to \(29.3\,\text{km/s}\), as the chapter on Angular Momentum found.

The law of periods

Kepler searched for ten more years for a rule linking the planets to one another. In 1619 he found it: the square of the period is proportional to the cube of the orbit's size. For a circular orbit it follows in one line from lesson 5. Square \(T = 2\pi\sqrt{r^3/GM}\) to get \(T^2 = (4\pi^2/GM)\,r^3\). For an ellipse, \(r\) is replaced by the semi-major axis \(a\).

The constant \(4\pi^2/GM\) is the same for every planet, because it depends only on the Sun. So ratios are easy. Measure distances in astronomical units, where the Earth's orbit has \(a = 1\,\text{AU}\), and periods in years. Then \(T^2 = a^3\) exactly. Mars, at \(a = 1.52\,\text{AU}\), takes \(1.52^{3/2} = 1.87\) years, and Jupiter, at \(5.2\,\text{AU}\), takes \(11.9\) years.

The same law weighs the Sun. The Earth's orbit has \(r = 1.5 \times 10^{11}\,\text{m}\) and \(T = 3.16 \times 10^7\,\text{s}\). Then \(M = 4\pi^2 r^3/GT^2 = 2.0 \times 10^{30}\,\text{kg}\). The planet's own mass never appears, so the law weighs the central body alone. The Moon's month weighs the Earth in the same way.

Two stars, one centre

So far the Sun was held still. That is fair when one body is far heavier. But about half the stars in the sky are binary stars: two stars bound to each other, often of similar mass. Each pulls the other equally, so neither can stay still. Since no outside force acts, their centre of mass stays at rest, as in the chapter on Centre of Mass. Both stars circle that point.

The two stars must stay on opposite sides of the centre of mass, on one line through it. So they go round in the same time, with the same angular speed \(\omega\). The heavier star is nearer the centre. With \(d\) their separation, the centre of mass gives \(m_1r_1 = m_2r_2\), so \(r_1 = m_2d/(m_1 + m_2)\) and \(r_2 = m_1d/(m_1 + m_2)\).

Now write the force equation for star 1. The pull of star 2 is \(Gm_1m_2/d^2\), and it must supply the centripetal force \(m_1\omega^2 r_1\). Substitute \(r_1 = m_2d/(m_1 + m_2)\), and \(m_1m_2\) cancels to leave \(\omega^2 = G(m_1 + m_2)/d^3\). So the period of a binary is \(T = 2\pi\sqrt{d^3/G(m_1 + m_2)}\). It is Kepler's third law with the total mass in place of \(M\).

The reduced mass \(\mu = m_1m_2/(m_1 + m_2)\) says the same thing more compactly. The pull on star 1 is \(\mu\omega^2 d\). So the separation between the two stars changes exactly like the orbit of one body of mass \(\mu\). That body circles a fixed centre at the distance \(d\). The total energy of the pair is \(E = -Gm_1m_2/2d\), negative, so the pair is bound. When \(m_2 \ll m_1\), \(\mu \approx m_2\) and every result of lessons 5 and 6 returns.

The hook asked about Halley's comet. It races near the Sun and crawls far away because its angular momentum about the Sun is fixed: \(v_pr_p = v_ar_a\). Its period of about 76 years gives \(a = 76^{2/3} \approx 18\,\text{AU}\), which matches its nearest distance of \(0.59\,\text{AU}\) and its farthest of \(35\,\text{AU}\). Their mean is \(17.8\,\text{AU}\).