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Kepler's third law

Gravity and orbits · level 6, lesson 4 of 6

A question to start

In 1619 Johannes Kepler, working from years of careful naked-eye observations of Mars by Tycho Brahe, found a strange rule: for every planet, the square of its year is proportional to the cube of its distance from the Sun. Why squares and cubes?

The idea

Take the two results from lesson 3, \(v = \sqrt{GM/r}\) and \(T = 2\pi r/v\). Square the second and put in the first:

\[T^2 = \frac{4\pi^2 r^2}{v^2} = \frac{4\pi^2 r^2}{GM/r} = \frac{4\pi^2}{GM}\, r^3.\]

That is Kepler's third law: \(T^2 \propto r^3\), with the same constant \(4\pi^2/GM\) for everything orbiting the same body. Newton showed that it holds for ellipses too, with \(r\) replaced by the semi-major axis \(a\), half the longest width of the ellipse.

In handy units. For planets round the Sun, measure \(T\) in years and \(a\) in astronomical units (1 AU is the Earth's mean distance). The Earth gives \(1^2 = 1^3\), so the constant is 1, and

\[T^2 = a^3 \qquad (T \text{ in years, } a \text{ in AU}).\]
Planet \(a\) (AU) \(a^3\) \(T = \sqrt{a^3}\) (years) Measured
Venus 0.723 0.378 0.615 0.615
Earth 1.000 1.000 1.000 1.000
Mars 1.524 3.537 1.881 1.881 (687 days)
Jupiter 5.203 140.8 11.87 11.86

Round Mars. Phobos at 9,376 km and Deimos at 23,463 km. The distances are in the ratio 2.502, so the periods should be in the ratio \(2.502^{1.5} = 3.96\). Measured: \(30.31 / 7.65 = 3.96\). ✓

Worked examples

1. Mars's year. \(T = \sqrt{1.524^3} = \sqrt{3.537} = 1.881\) years \(= 687\) days.

2. An asteroid. An asteroid takes 4.6 years to go round the Sun. Its distance: \(a = T^{2/3} = 4.6^{2/3} = \sqrt[3]{21.2} = 2.77\) AU, in the asteroid belt between Mars and Jupiter. (That is Ceres.)

3. Deimos from Phobos. \(T_D = 7.65 \times (23463/9376)^{1.5} = 7.65 \times 3.96 = 30.3\) h. ✓

Try it

  1. Saturn is 9.54 AU from the Sun. Find its year.
  2. A comet has a year of 76 years (Halley's). Find its semi-major axis.
  3. A satellite round Mars has a period of 12 hours. Use Phobos's orbit (9,376 km, 7.65 h) to find its distance.
  4. Derive \(T^2 = 4\pi^2 r^3 / GM\) from \(v = \sqrt{GM/r}\).
  5. The ship to Mars follows an ellipse with \(a = 1.262\) AU. How long would a full lap take? (You will use half of it in level 7.)
Answers
  1. \(T = \sqrt{9.54^3} = \sqrt{868} = 29.5\) years.
  2. \(a = 76^{2/3} = \sqrt[3]{5776} = 17.9\) AU.
  3. \(r = 9376 \times (12/7.65)^{2/3} = 9376 \times 1.350 = 12{,}660\) km.
  4. \(T = 2\pi r/v\), so \(T^2 = 4\pi^2 r^2 / v^2 = 4\pi^2 r^2 \cdot r / GM\).
  5. \(T = \sqrt{1.262^3} = \sqrt{2.010} = 1.418\) years \(= 518\) days. Half is 259 days.

Common mistake

Using the constant from one centre for another. \(T^2 = a^3\) in years and AU is only for things going round the Sun. Round Mars, the constant is \(4\pi^2/GM_\text{Mars}\), or compare with a known moon of Mars.

Where it leads