Home Page | Astronomy Articles | Stars Page | Astronomy Links |

July 2005 Feature: Kepler's Third Law of Planetary Motion |

(With Modifications from Sir Isaac Newton) |

Kepler's Third Law of Planetary Motion, which correlates a planet's orbital period with a planet's distance from the Sun,
can be summed up by the simple equation: T^{2} = D^{3}. In this equation, T = the time of the planet's orbital period in Earth years and
D = the planet's mean distance from the Sun in astronomical units (AU). One AU = the mean distance between the Earth and the Sun.

Some texts use different letters, such as p^{2} = a^{3}
or p^{2} = r^{3}, but it's all the same and you're certainly at liberty to use whatever letters work best for you. I've always followed
the suggestion given by the *Golden Book SkyGuide*: "To remember which is squared and which is cubed, recall the area of
Broadway and 42nd Street in New York: Times Square!" For that reason, I have **T = time** of the planet's orbital period.

Let's give an example of how Kepler's Third Law works. Let's say that you know that Mars' mean distance from the Sun is 1.524 AU. We have all the information we need to figure out Mars' orbital period:

T^{2} = D^{3} |

T^{2} = 1.5424 x 1.5424 x 1.5424 |

T^{2} = 3.5396 |

T = 1.88 = Mars' orbital period in Earth years |

In the case of Mars, Kepler used data from Tycho Brahe to figure out the orbital period and distance of this planet -- not the Third Law of Planetary Motion, which had yet to be conceived. (For more on how Kepler determined the orbit of Mars, click here or here.) Kepler's stroke of genius was to derive his Laws of Planetary Motion from what he knew about Mars -- and then to apply them to the other planets as well. In fact, Kepler's Third Law governs all bodies orbiting the Sun, including asteroids and comets.

Let's take another example. Suppose we discovered an asteroid whose mean distance is 4 AU (4 times Earth's mean distance from the Sun). We can compute its orbital period:

T^{2} = D^{3} |

T^{2} = 4 x 4 x 4 |

T^{2} = 64 |

T = 8 = the asteroid's orbital period in Earth years |

Also, knowing the orbital period equals 8 Earth years, we can compute the asteroid's distance:

T^{2} = D^{3} |

8 x 8 = D^{3} |

64 = D^{3} |

4 = D = asteroid's distance from the Sun in astronomical units |

*Isaac Newton Upgrades Kepler's Third Law*

Yes, Kepler's Third Law works like a charm for figuring distances and orbital periods for bodies within our Solar System. But that's because the mass of any planet, asteroid or comet counts as insignificant, with the Sun claiming more than 99.8% of the total mass of the Solar System. If Solar System bodies were substantially more massive, Kepler's Third Law -- as stated -- would never pass the muster.

Let's presume that this body residing 4 AU from the Sun happens to have significant mass -- 1/10th the Sun's. Then how do we compute its orbital period? Fortunately, Isaac Newton's version of Kepler's Third Law equips us to find the answer. The upgraded equation has M = the mass of the Sun and m = the mass of the secondary body:

(M+m) x T^{2} = D^{3}

In our example, we have the secondary body 0.1 times as massive as the Sun. The solution to our problem is as shown:

(M+m) x T^{2} = D^{3} |

(1 + 0.1) x T^{2} = 4 x 4 x 4 |

1.1 T^{2} = 64 |

T^{2} = 64/1.1 |

T^{2} = 58.18 |

T = 7.63 = Orbital period in Earth years |

When you use this equation, remember to have the orbital period in Earth years, the distance in astronomical units and the mass in solar masses.

*What About A More Massive Sun?*

Newton's version of Kepler's Third Law allows us to figure out what Earth's orbital period would be -- given a different mass for the Sun. For our purposes, we'll say that the combined mass of the Sun and Earth equals the mass of the Sun (M+m = M), and not be in serious error, since the Sun is some 333,000 times more massive than Earth.

So let's have the mass of the Sun increase fourfold. Then what would our orbital period around the Sun be in this scenario? The answer:

M x T^{2} = D^{3} |

4 x T^{2} = 1 x 1 x 1 |

4 T^{2} = 1 |

T^{2} = 1/4 = 0.25 |

T = 1/2 = 0.5 = the Earth's orbital period in years |

So if the Sun were 4 times as massive, we'd orbit the Sun in half the time!

copyright 2005 by Bruce McClure

Super Cool: Kepler's Laws Calculator and Animation |