Kepler's three laws

Four hundred years ago Johannes Kepler spent years working through careful measurements of where Mars had been seen in the sky. He found three rules that every planet obeys.

One. A planet’s path is not a circle. It is a slightly squashed circle called an ellipse, and the Sun is not in the middle but a little to one side.

Two. A planet speeds up as it comes nearer the Sun and slows down as it moves away.

Three. The further a planet is from the Sun, the longer its year, and not only because it has further to go. It also moves more slowly. Mercury goes round in 88 days. Neptune takes 165 years.

Kepler did not know why these rules worked. That took Isaac Newton, about seventy years later, who showed that all three follow from gravity.

First law: orbits are ellipses, with the Sun at one focus. How stretched an ellipse is gets measured by its eccentricity, from 0 for a circle up towards 1 for a long thin loop. The planets’ orbits are close to circles. The Earth’s eccentricity is 0.017 and you could not tell its orbit from a circle by eye. Mars is more lopsided at 0.094: it swings between 206.7 and 249.3 million km from the Sun. That difference was just big enough for Kepler to catch, which is why Mars gave the game away.

Second law: a line from the Sun to a planet sweeps out equal areas in equal times. Near the Sun the line is short, so the planet has to cover more of its path to sweep the same area. In other words it goes faster. The Earth moves at 30.3 km/s when nearest the Sun in January and 29.3 km/s when furthest in July. For a stretched orbit the change is huge. Halley’s Comet passes the Sun at about 54 km/s and creeps round the far end of its orbit, out beyond Neptune, at less than 1 km/s.

Third law: the square of the period goes as the cube of the orbit’s size. With the period P in years and the average distance a in au, it could not be simpler:

P² = a³

Planet	a (au)	a³	P (years)	P²
Mercury	0.387	0.058	0.241	0.058
Earth	1.000	1.000	1.000	1.000
Mars	1.524	3.54	1.881	3.54
Jupiter	5.203	140.9	11.86	140.7
Neptune	30.07	27,190	164.8	27,160

The two columns agree to about one part in a thousand. The small differences are real: the planets pull on one another, and their own mass counts for a little.

The same law holds for moons round a planet, but with a different constant, and the constant depends only on the mass of the thing in the middle. So timing a moon weighs its planet. That is how the masses of Jupiter, Saturn, Uranus and Neptune were first measured, and it is still how the mass of almost everything in astronomy is found.

Kepler found his laws by fitting curves to observations. Newton later showed that all three follow from one force that weakens with the square of the distance. This level writes each law as an equation and checks it against real numbers. The orbital figures are from NASA’s fact sheets, the constants are from JPL, and every result is worked out here.

The ellipse. Measure the planet’s distance $r$ from the Sun and its angle $\theta$ round the orbit, counted from the point of closest approach. With semi-major axis $a$ and eccentricity $e$, the first law is

$$r = \frac{a\,(1 - e^2)}{1 + e\cos\theta}$$

At $\theta = 0$ this gives the nearest point, and at $\theta = 180^\circ$ the furthest:

$$r_{\text{near}} = a\,(1 - e), \qquad r_{\text{far}} = a\,(1 + e)$$

For Mars, $a = 227.956$ million km and $e = 0.0935$. That gives 206.6 and 249.3 million km. The fact sheet lists 206.650 and 249.261.

Equal areas. In a short time the line from the Sun to the planet sweeps a thin triangle. Its area depends only on the distance $r$ and on the part of the velocity that is at right angles to the line, $v_\perp$:

$$\frac{dA}{dt} = \tfrac{1}{2}\, r\, v_\perp$$

The Sun’s pull acts along the line, so it cannot twist the planet round the Sun, and the planet’s angular momentum $L = m\,r\,v_\perp$ never changes. So $r\,v_\perp$ is constant, and the second law is the conservation of angular momentum in another form.

At the nearest and furthest points the whole velocity is at right angles to the line, so

$$r_{\text{near}}\, v_{\text{near}} = r_{\text{far}}\, v_{\text{far}}$$

For the Earth the distances are 147.095 and 152.100 million km, a ratio of 1.0340. The listed speeds are 30.29 and 29.29 km/s, a ratio of 1.0341.

The harmonic law. Take a small body in a circular orbit of radius $a$ round a mass $M$. Gravity supplies exactly the inward force needed to keep it on the circle:

$$\frac{G M m}{a^2} = \frac{m\, v^2}{a}$$

Its speed is one circumference in each period, $v = 2\pi a / P$. Put that in and the small mass $m$ cancels:

$$P^2 = \frac{4\pi^2}{G M}\, a^3$$

The same result holds for an ellipse, with $a$ the semi-major axis. For the Sun, $GM = 1.32712 \times 10^{20}$ m³/s². One astronomical unit is $1.495979 \times 10^{11}$ m. The formula then gives $P = 365.257$ days. The measured sidereal year is 365.256 days.

Measured in years and astronomical units, the constant $4\pi^2/GM$ is 1 for anything going round the Sun, which is why the law looked so simple to Kepler.

The planet’s own mass. Strictly, the two bodies go round their common centre of mass, and $M$ should be the sum of both masses:

$$P^2 = \frac{4\pi^2}{G\,(M + m)}\, a^3$$

The Sun has 1,047 times the mass of Jupiter and its moons, so for Jupiter $a^3/P^2$ comes out about one part in a thousand larger than for a body of no mass. That is the small mismatch in Jupiter’s row of the table in the level before this one.

Weighing Jupiter. Turn the law round and it gives the mass in the middle:

$$M = \frac{4\pi^2 a^3}{G\, P^2}$$

Io goes round Jupiter at 421,800 km in 1.769138 days. With $G = 6.67430 \times 10^{-11}$ m³ kg⁻¹ s⁻², this gives $1.90 \times 10^{27}$ kg. The listed mass of Jupiter is $1.898 \times 10^{27}$ kg.

Timing an orbit gives the product $GM$ far more exactly than either $G$ or $M$ is known alone. That is why JPL lists $GM$ for the Sun to 18 figures, while $G$ itself is uncertain by about 2 parts in 100,000.

Speed at any point. Energy is conserved as well as angular momentum. Together they give the speed at any distance $r$ on an orbit of semi-major axis $a$:

$$v^2 = G M \left( \frac{2}{r} - \frac{1}{a} \right)$$

For a circle $r = a$ and this is $v^2 = GM/a$, which for the Earth’s orbit is 29.78 km/s. At the Earth’s nearest and furthest points it gives 30.29 and 29.29 km/s, the listed values.

Halley’s Comet has $a = 17.94$ au and comes within 0.587 au of the Sun, so its furthest point is $2a - 0.587 = 35.3$ au. The formula gives 54.5 km/s at the near end and 0.91 km/s at the far end, a ratio of 60.

Leaving altogether. As $a$ grows without limit the orbit opens out and the body never returns. Setting $1/a = 0$ gives the speed needed to escape from distance $r$:

$$v_{\text{esc}} = \sqrt{\frac{2 G M}{r}}$$

It is $\sqrt{2}$ times the speed of a circular orbit at the same distance. At the Earth’s distance from the Sun that is 42.1 km/s.