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 Eq. (324)]. 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,

$$T = \frac{A}{(dA/dt)} = \frac{2\,\pi\,a\,b}{h}.$$ (332)

It follows from Eqs. (311), (312), and (331) that

$$T^2 = \frac{4\,\pi^2\,a^3}{G\,M}.$$ (333)

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 Eqs. (311) and (330)]

$$r_p = \frac{r_c}{1+e} = a\,(1-e).$$ (334)

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

$$r_a = \frac{r_c}{1-e} = a\,(1+e).$$ (335)

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

$$a = \frac{r_p+r_a}{2}.$$ (336)

The parameter

$$e = \frac{r_a-r_p}{r_a+r_p}$$ (337)

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.