Finding what Kepler devoted his life to in a couple of minutes (benefit of being in the future involved)

We are bounded in a nutshell of Infinite Space: Worksheet # 10, Problem #3: Finding what Kepler devoted his life to in a couple of minutes (benefit of being in the future involved)

3. For a planet of mass \(m\) orbiting a star of mass \(M_\star\), at \(a\) distance a in a circular orbit, start with the Virial Theorem and derive Kepler’s Third Law of motion. Assume that \(m \ll M_\star\) Remember that since m is so small, the semimajor axis, which is formally \( a = a_p + a_\star\) reduces to \(a = a_p\) (make sure you understand why).

First off, we’ll explain the concept of the axes of an ellipse. Technically, all orbits are ellipses, but many, especially in our solar system, are contained enough so they appear to be circles. However, the fact they are ellipses is what makes the solar system make sense, for the observations Tycho Brahe and Johannes Kepler made it clear ellipses were the best descriptions of these orbits, and off that Kepler created his equations. Here, the objects we are comparing are so different in size (\(m \ll M_\star\)), that the radius of either object is negligible and we can simply focus on the distance between them: \(a = a_p\).

Now, we will derive this equation a bit differently than how Kepler did it, we now being able to use the Virial theorem to describe the relationship of kinetic and potential energy. So we start with this equation: \[ K = -\frac{1}{2} U,\] and we can redefine these with the general definition of gravitational potential energy and kinetic energy, with \(a = a_p\),  \[ \frac{1}{2} mv^2 = -\frac{1}{2} \left(- \frac{GMm}{a} \right),\] which can be simplified into:  \[ v^2 = \frac{GM}{a},\] and we can now define our speed differently since we are assuming very circular orbits, we can use the definition of velocity for going around a circle once: \[ v = \frac{2\pi a}{P},\] where P is the period (time) for the rotation to occur.

Plugging that in, we have: \[\frac{4\pi^2 a^2}{P^2} = \frac{GM}{a},\] which is simplified and we find: \[ P^2=  \frac{4\pi^2 a^3}{GM},\] which is Kepler’s equation for the Period of planetary rotation around the Sun (or other central star in a solar system).