Happy new year! While we’re thinking about years, why don’t we think about one of the first guys to explore the physical reason behind the year?

This nattily dressed gentleman is Johannes Kepler, who worked out three basic mathematical laws of planetary motion around the year 1605. He did so from scratch, purely based on phenomenological examination of astronomical data. Newton wouldn’t figure out the theory behind these laws for about another century, but with Newton’s classical mechanics we can show that his basic laws of motion imply that Kepler’s laws must be true.

The third of Kepler’s laws tells us how the length of a year varies among the different planets. Since a year on a particular planet is just the length of time required to make an orbit, this law relates the time per orbit – the period – to the distance between the planet and the sun. In mathematical language:

*The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. *

Planets move in ellipses, and the semi-major axis is just the distance to the center of the ellipse at the farthest point. For a period T and a semi-major axis r, we can write this law more compactly as:

Double the distance to the sun r and you’ll have increased the year T by a factor of 2^1.5 = 2.83 or so. For instance, the semi-major axis of Mars is about 1.524 times larger than that of earth, and so the Martian year ought to be 1.524^1.5 = 1.881 times larger than the earth’s 1 year. As indeed is exactly is – the Martian year is just under 687 earth days.

So can we prove this relationship from Newton’s laws? Of course! We can save ourselves a lot of trouble with the simplifying assumptions that the orbits are exactly circular. This is not far from true for the planets, and the intrepid are welcome to try the more challenging but perfectly doable proof for the general elliptical case. We’ll just do the circular case. We know by Newton’s law of gravitation, the gravitational force on a planet of mass m by the sun (mass M) is given by:

Where G is the universal gravitational constant and r is the radius of the orbit. Newton’s laws of motion also required that the force needed to keep an object moving in a uniform circle is equal to:

Both the above expressions equal F, and so they equal each other. Before writing that down, let’s figure out how to get rid of the v (which isn’t in Kepler’s laws explicitly) and replace it with T the orbital period. We know that by definition the velocity is distance/time. The distance is the circumference of the orbit which is just 2*pi*r, and the time is the period T. Thus v = 2*pi*r/T. Swap that into the above:

Fiddle with the algebra a bit and simplify, and we can write that as:

The m’s cancel, and everything remaining but r and T are constants. And they’re related in exactly the way Kepler wrote. He *was* a pretty smart guy.