newtonian mechanics - A simple explanation of Kepler's Third Law

Is there a simple way to explain how Kepler's third law follows from the inverse square law that of gravity (and laws of motion)

For example for Kepler's second law we can say it's because Gravity is a central force

What is it about the law of gravity which ultimately gives rise to Kepler's third law?

Answer

To show that the inverse square law of gravity leads to Kepler's 3rd law for an elliptical orbit which obeys the first and second laws, we make use of the conservation of mechanical energy: $$E_{orbit}=\frac{1}{2}mv^2-\frac{GMm}{r},$$ where $E_{orbit}$ is the constant energy, $r$ is the instantaneous distance of the planet from the sun, $v$ is the instantaneous speed, $G$ is the universal gravitational constant, $M$ is the sun mass, and $m$ is the planet mass.

Let us choose two convenient points along the ellipse:

• the perihelion ($r_1=a(1-e)$), where $e$ is the eccentricity and the $a$ is the semi-major axis. At this point the angular momentum $L=mv_1a(1-e)$.


• the aphelion ($r_2=a(1+e)$). At this point $L=mv_2a(1+e)$.

We equate the energies at these two point, and because of the 2nd Law, we know that the angular momentum magnitudes, $L$, are equal: $$\frac{L^2}{2ma^2(1+e)^2}-\frac{GMm}{a(1+e)}=\frac{L^2}{2ma^2(1-e)^2}-\frac{GMm}{a(1-e)}.$$

After some algebra we arrive at $$L^2=GMm^2a(1-e^2).$$

Next, we examine the geometry of the ellipse and we find the relationship between the semi-major axis and the semi-minor axis lengths are $$b^2=a^2(1-e^2).$$

Using this to eliminate the eccentricity gives us $$aL^2=GMm^2b^2.$$

Finally, according to Kepler's 2nd Law (and conservation of angular momentum) we know that the ratio of the area of the ellipse to the period of revolution, $T$, is a constant, proportional to the specific angular momentum, $L/m$.: $$\frac{\pi a b}{T}=\frac{k_0L}{m}.$$ Solving for the square of the period we get $$T^2={\pi^2a^2b^2}{k_0^2} \to T^2=\frac{\pi^2a^2\left(\frac{aL^2}{GMm^2}\right)}{\frac{k_0^2L^2}{m^2}}.$$ Resolving the algebra we get $$T^2=\frac{\pi^2}{k_0^2GM}a^3.$$