3.1 The laws of planetary motion  (Page 4/17)

Kepler’s third law

Kepler’s first two laws of planetary motion describe the shape of a planet’s orbit and allow us to calculate the speed of its motion at any point in the orbit. Kepler was pleased to have discovered such fundamental rules, but they did not satisfy his quest to fully understand planetary motions. He wanted to know why the orbits of the planets were spaced as they are and to find a mathematical pattern in their movements—a “harmony of the spheres” as he called it. For many years he worked to discover mathematical relationships governing planetary spacing and the time each planet took to go around the Sun.

In 1619, Kepler discovered a basic relationship to relate the planets’ orbits to their relative distances from the Sun. We define a planet’s orbital period , ( P ), as the time it takes a planet to travel once around the Sun. Also, recall that a planet’s semimajor axis, a, is equal to its average distance from the Sun. The relationship, now known as Kepler’s third law , says that a planet’s orbital period squared is proportional to the semimajor axis of its orbit cubed, or

When P (the orbital period) is measured in years, and a is expressed in a quantity known as an astronomical unit (AU)    , the two sides of the formula are not only proportional but equal. One AU is the average distance between Earth and the Sun and is approximately equal to 1.5 × 10 ⁸ kilometers. In these units,

Kepler’s third law applies to all objects orbiting the Sun, including Earth, and provides a means for calculating their relative distances from the Sun from the time they take to orbit. Let’s look at a specific example to illustrate how useful Kepler’s third law is.

For instance, suppose you time how long Mars takes to go around the Sun (in Earth years). Kepler’s third law can then be used to calculate Mars’ average distance from the Sun. Mars’ orbital period (1.88 Earth years) squared, or P ² , is 1.88 ² = 3.53, and according to the equation for Kepler’s third law, this equals the cube of its semimajor axis, or a ³ . So what number must be cubed to give 3.53? The answer is 1.52 (since 1.52 × 1.52 × 1.52 = 3.53). Thus, Mars’ semimajor axis in astronomical units must be 1.52 AU. In other words, to go around the Sun in a little less than two years, Mars must be about 50% (half again) as far from the Sun as Earth is.