# Kepler’s Third Law: the law of harmonies

- look at data about the motion of the planets
- explain Kepler’s third law
- investigate whether the law of harmonies also holds for satellites of the earth.

## Periods and average distances of planets

Most planets have elliptical orbits — this is*Kepler’s First Law*, with the added provision that in the case of a small planet revolving around a big sun, the sun will be at one of the foci of the elliptic orbit of the planet. In fact most orbits are quite close to being circles.

*Kepler’s Second Law*asserts that areas swept out by a planet in equal times are equal.

*Kepler’s Third Law*is the most complicated, and it relates the period \(\normalsize{T}\) of a planet, which is the time spent for one revolution around the sun, to the average distance \(\normalsize{R}\) to the sun.The units for both quantities are taken from the earth’s orbit: period is measured in years, which is the time taken for the earth to complete one of its orbits, and distance is taken in astronomical units (au), which is the average distance from the earth to the sun. This turns out to be about \(\normalsize{150}\) million kilometres.So for the earth \(\normalsize{T=1}\) and \(\normalsize{R=1}\) by definition.The question that Kepler answered was: what is the general relation between \(\normalsize{R}\) and \(\normalsize{T}\) for the other planets?

## Kepler’s third law

Kepler’s discovery was that the period \(\normalsize{T}\) and the average distance \(\normalsize{R}\) of a planet from the sun obeyed the relation \[\Large{\frac{T^2}{R^3}=1}.\]Or if we want to find \(\normalsize{T}\) from \(\normalsize{R}\), the expression is \[\Large{T=R^{\frac{3}{2}}}.\]This is a power law, but now with an exponent bigger than \(\normalsize{1}\). Here is the graph of \(\normalsize{y=x^\frac{3}{2}}\):## The data

Here is a table of periods and average distances for the planets, as well as the ratio \(\normalsize{T^2/R^3}\).Planet | Period (yr) | Average Distance (au) | \(\normalsize{T^2/R^3}\) |
---|---|---|---|

Mercury | 0.241 | 0.39 | 0.98 |

Venus | .615 | 0.72 | 1.01 |

Earth | 1.00 | 1.00 | 1.00 |

Mars | 1.88 | 1.52 | 1.01 |

Jupiter | 11.8 | 5.20 | 0.99 |

Saturn | 29.5 | 9.54 | 1.00 |

Uranus | 84.0 | 19.18 | 1.00 |

Neptune | 165 | 30.06 | 1.00 |

Pluto | 248 | 39.44 | 1.00 |

## The more general form of the Third Law

More generally, if several satellites orbit the same more massive object, then the period \(\normalsize{T}\) of an orbiting body will relate to the mean distance \(\normalsize{R}\) from the center of the massive object according to the power law \[\Large{T=A R^{3/2}}\]where \(\normalsize{A}\) is a constant that depends only on the massive object. In particular this law holds for other objects orbiting the sun, and also holds for satellites orbiting the earth. However in this latter case the constant \(\normalsize{A}\) will not be \(\normalsize{1}\).## The case of comets

Comets which are recurrent also obey Kepler’s law. The most famous one is Halley’s comet, with an orbital period of \(\normalsize{75.3}\) years and an average distance of \(\normalsize{17.55}\) au.You can compute that \[\Large{\frac{T^2}{R^3}=\frac{(75.3)^2}{(17.55)^3} \approx 1.048}\]which is pretty close to one.^{Lspn comet halley By NASA/W. Liller, Public domain, via Wikimedia Commons}

Q1(E): Jupiter has four moons. They also obey Kepler’s laws in their orbits around Jupiter. The distance of the moon called Io from Jupiter’s centre is \(\normalsize{4.2}\) units, and its period is \(\normalsize{1.8}\) Earth-days. Another moon is called Ganymede; it is \(\normalsize{10.7}\) units from Jupiter’s center. What is the period of Ganymede (pictured)?

^{Ganymede moon By NASA/JPL/DLR (Ganymede’s Trailing Hemisphere), Public domain, via Wikimedia Commons}

## The moon and other satellites of earth

## Discussion

## Answers

\[\Large \frac{(1.8)^2}{(4.2)^3}=\frac{T_G^2}{(10.7)^3}.\]A1.Kepler’s third law says that \(\normalsize T^2/R^3\) should be roughly the same for all bodies orbiting Jupiter. So if we call Ganymede’s orbital period \(\normalsize T_G\) then by comparing with Io’s orbit, we must haveNotice that whatever constant of proportionality occurs in Kepler’s law, it cancels here.Solving, we get \(\normalsize T_G\approx 7.3\) Earth-days. The actual value is \(\normalsize 7.155\) Earth-days.

#### Maths for Humans: Linear, Quadratic & Inverse Relations

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