Example application of Kepler's third law: The Moon's orbit

On this page, we will show how to calculate the semimajor axis of the Moon's orbit using Kepler's third law.

The Moon completes one orbit around Earth approximately every 27 days. Thus, the Moon’s orbital period T is approximately 27 days.

The Kepler constant K for a body orbiting Earth is 9.91·10-14 s2/m3, or, equivalently, 1.33·10-23 days2/m3.

Using this information and Kepler’s third law, we can determine the semimajor axis of the Moon’s orbit. Kepler’s third law can be expressed mathematically as

T2 = K·a3.

We know the period of the Moon’s orbit T and Kepler’s constant K. We want to calculate the semimajor axis a of the Moon’s orbit, so we isolate a by dividing both sides of the previous equation by K. We get

a3 = T2/K.

Then, we take the cubic root of both sides of the equation,

a =(T2/K) 1/ 3.

Lastly, we substitute the values of T and K in their correct units to find the value of a:

a= ((27 days)2 /(1.33·10-23 days2/m3))1/ 3= 3.8 · 108 m.

Therefore, the semimajor axis of the Moon’s orbit is

a= 3.8 · 108 m= 3.8 · 105 km= 2.4 · 105 miles.