Coaching Centre
Kepler’s laws of planetary motion
Kepler's three laws, elliptical orbits and eccentricity, why planets speed up near the Sun, deriving Kepler's third law from Newton's law of gravitation, and using it to find the mass of a planet or star.
For eight years Johannes Kepler tried to fit the orbit of Mars to circles, the shape everyone since the ancient Greeks had assumed. His best circle-based model was out by 8 arcminutes, about a quarter of the Moon's width in the sky. Most astronomers would have shrugged. Kepler trusted the measurements of his old employer, Tycho Brahe, threw the circles away, and found the ellipse.
Kepler published his three laws between 1609 and 1619, purely from data. He had no idea why they were true. Seventy years later, Newton showed that all three follow from his law of gravitation. In the HSC you need to state the laws, explain them in terms of force and energy, and use the third law quantitatively.
The first law: orbits are ellipses
Each planet moves in an ellipse with the Sun at one focus. An ellipse has two foci; the Sun sits at one and the other is empty. The closest point to the Sun is the perihelion and the furthest is the aphelion. (For orbits around Earth, the words are perigee and apogee.)
How stretched an ellipse is is measured by its eccentricity $e$, from 0 (a circle) up to nearly 1 (very long and thin). Most planets have nearly circular orbits:
| Body | Eccentricity | What it means |
|---|---|---|
| Venus | 0.007 | almost a perfect circle |
| Earth | 0.017 | 3% closer to the Sun in early January than in early July |
| Mars | 0.093 | noticeably off-centre, which is why Kepler could find the ellipse |
| Halley's Comet | 0.967 | long, thin orbit, from inside Venus's orbit out beyond Neptune's |
Earth is closest to the Sun in early January, during the Australian summer. Seasons are caused by the tilt of Earth's axis, not the distance to the Sun.
The second law: equal areas in equal times
A line from the Sun to a planet sweeps out equal areas in equal times. Close to the Sun, the line is short, so to sweep the same area the planet must move further along its orbit: it speeds up. Far from the Sun, it slows down.
In terms of forces: in an ellipse, gravity is not perpendicular to the velocity (except at perihelion and aphelion). As the planet falls inward, gravity has a component along the motion and does positive work, so the planet speeds up. As it moves outward, gravity does negative work, slowing it.
In terms of energy: the total energy $E = K + U$ is constant. Near perihelion, gravitational potential energy is lowest and kinetic energy highest; near aphelion the reverse. In a circular orbit, the distance never changes, so the speed doesn't either. The second law is also a statement of conservation of angular momentum, and that gives a handy result: the speed is inversely proportional to the distance at perihelion and aphelion, $v_p r_p = v_a r_a$.
The third law: the period–radius relation
The square of a planet's period is proportional to the cube of its average distance from the Sun: $T^2 \propto r^3$. Kepler found this by trying combinations of numbers. Newton's gravity explains it. For a circular orbit, gravity provides the centripetal force:
$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}, \qquad v = \dfrac{2\pi r}{T}$
$\dfrac{GM}{r} = \dfrac{4\pi^2 r^2}{T^2} \quad\Rightarrow\quad \dfrac{r^3}{T^2} = \dfrac{GM}{4\pi^2}$
This is on the HSC formulae sheet. The right side depends only on the mass $M$ of the central body, so every object orbiting the same body has the same value of $r^3/T^2$. For an ellipse, $r$ is the semi-major axis (the average of the perihelion and aphelion distances).
Mars orbits at 1.524 AU from the Sun (1 AU is Earth's mean distance). How long is a Martian year?
Working in ratios like this avoids $G$ and $M$ entirely. It's the fastest method whenever two bodies orbit the same object.
Jupiter's moon Io orbits at a radius of $4.22 \times 10^8$ m with a period of 1.77 days. Find the mass of Jupiter.
This is how we know the mass of every planet with a moon, the mass of the Sun, and even the mass of the black hole at the centre of our galaxy (about 4 million Suns, from the orbits of stars close to it). It's also how astronomers find the masses of stars with exoplanets.
- With eccentricity 0.5, the planet is 3 times closer at perihelion than at aphelion, and moves 3 times faster. Watch the shaded slices: they look different but have equal areas.
- Set the eccentricity to 0, then 0.2, then 0.8. How does the speed ratio change?
- Switch to T² against r³. For the planets, is the graph a straight line through the origin? The gradient gives the mass of the Sun: $1.99 \times 10^{30}$ kg.
- Now choose Jupiter's moons. What mass does the gradient give for Jupiter? Compare it with the worked example.
Test yourself
A comet in an elliptical orbit has the greatest kinetic energy
Potential energy is lowest at perihelion, so kinetic energy is highest there.
A planet orbits 4 times further from its star than another planet of the same star. Its period is
$T \propto r^{3/2}$ and $4^{3/2} = 8$.
Which quantity can be found from the radius and period of a moon's orbit?
$r^3/T^2 = GM/4\pi^2$, where $M$ is the central (planet's) mass.
In an elliptical orbit, gravity does positive work on a planet
Moving inward, gravity has a component along the velocity, so the planet speeds up.
Label the parts of an elliptical orbit.
Which of Kepler's laws does each statement relate to?
Kepler's first law states that planets move in with the Sun at one . The point of closest approach to the Sun is called . The second law says a line from the Sun to a planet sweeps out equal in equal times. The third law says $T^2$ is proportional to the of the orbital radius.
(a) Starting from Newton's law of universal gravitation, derive Kepler's third law for a circular orbit. (b) Saturn's moon Titan orbits at $1.22 \times 10^9$ m with a period of 15.9 days. Calculate the mass of Saturn.
(a) For a circular orbit, gravity provides the centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$, so $v^2 = \dfrac{GM}{r}$. With $v = \dfrac{2\pi r}{T}$: $\dfrac{4\pi^2 r^2}{T^2} = \dfrac{GM}{r}$, so $\dfrac{r^3}{T^2} = \dfrac{GM}{4\pi^2}$, a constant for a given central mass.
(b) $T = 15.9 \times 86\,400 = 1.374 \times 10^6$ s. $M = \dfrac{4\pi^2 r^3}{GT^2} = \dfrac{4\pi^2(1.22 \times 10^9)^3}{(6.67 \times 10^{-11})(1.374 \times 10^6)^2} = 5.69 \times 10^{26}$ kg.