Kepler's Second Law

Multiplying our planet's tangential equation of motion, (304), by $r$, we obtain

$$r^2\,\ddot{\theta} + 2\,r\,\dot{r}\,\dot{\theta} = 0.$$ (320)

However, the above equation can be also written

$$\frac{d(r^2\,\dot{\theta})}{dt} = 0,$$ (321)

which implies that

$$h = r^2\,\dot{\theta}$$ (322)

is constant in time. It is easily demonstrated that $h$ is the magnitude of the vector ${\bf h}$ defined in Eq. (291). Thus, the fact that $h$ is constant in time is equivalent to the statement that the angular momentum of our planet is a constant of its motion. As we have already mentioned, this is the case because gravity is a central force.

Figure 32: Kepler's second law.

Suppose that the radius vector connecting our planet to the origin sweeps out an angle $\delta\theta$ between times $t$ and $t+\delta t$ (see Fig. 32). The approximately triangular region swept out by the radius vector has the area

$$\delta A \simeq \frac{1}{2}\,r^2\,\delta\theta,$$ (323)

since the area of a triangle is half its base ( $r\,\delta\theta$) times its height ($r$). Hence, the rate at which the radius vector sweeps out area is

$$\frac{dA}{dt} =\frac{1}{2} \lim_{\delta t\rightarrow 0}\frac... ...}{\delta{t}}= \frac{r^2}{2}\,\frac{d\theta}{dt} = \frac{h}{2}.$$ (324)

Thus, the radius vector sweeps out area at a constant rate (since $h$ is constant in time)--this is Kepler's second law. We conclude that Kepler's second law of planetary motion is a direct consequence of angular momentum conservation.