Kepler's Third Law

We have seen that the radius vector connecting our planet to the origin sweeps out area at the constant rate $dA/dt=h/2$ [see Equation (249)]. We have also seen that the planetary orbit is an ellipse. Suppose that the major and minor radii of the ellipse are $a$ and $b$, respectively. It follows that the area of the ellipse is $A=\pi\,a\,b$. Now, we expect the radius vector to sweep out the whole area of the ellipse in a single orbital period, $T$. Hence,

$$T = \frac{A}{(dA/dt)} = \frac{2\pi\,a\,b}{h}.$$ (257)

It follows from Equations (236), (237), and (256) that

$$T^2 = \frac{4\pi^2\,a^3}{G\,M}.$$ (258)

In other words, the square of the orbital period of our planet is proportional to the cube of its orbital major radius--this is Kepler's third law.

Note that for an elliptical orbit the closest distance to the Sun--the so-called perihelion distance--is [see Equations (236) and (255)]

$$r_p = \frac{r_c}{1+e} = a\,(1-e).$$ (259)

Likewise, the furthest distance from the Sun--the so-called aphelion distance--is

$$r_a = \frac{r_c}{1-e} = a\,(1+e).$$ (260)

It follows that the major radius, $a$, is simply the mean of the perihelion and aphelion distances,

$$a = \frac{r_p+r_a}{2}.$$ (261)

The parameter

$$e = \frac{r_a-r_p}{r_a+r_p}$$ (262)

is called the eccentricity, and measures the deviation of the orbit from circularity. Thus, $e=0$ corresponds to a circular orbit, whereas $e\rightarrow 1$ corresponds to an infinitely elongated elliptical orbit.

As is easily demonstrated from the above analysis, Kepler laws of planetary motion can be written in the convenient form

$$r = \frac{a\,(1-e^2)}{1-e\,\cos\theta},$$ (263)

$$r^2\,\dot{\theta} = (1-e^2)^{1/2}\,n\,a^2,$$ (264)

$$G\,M = n^2\,a^3,$$ (265)

where $a$ is the mean orbital radius (i.e., the major radius), $e$ the orbital eccentricity, and $n=2\pi/T$ the mean orbital angular velocity.