Kepler's Second Law

Multiplying our planet's tangential equation of motion, (1.294), by $r$, we obtain

$$r^{2}\,\skew{5}\ddot{\theta} + 2\,r\,\skew{3}\dot{r}\,\skew{5}\dot{\theta} = 0.$$ (1.310)

However, the previous equation can be also written

$$\frac{d(r^{2}\,\skew{5}\dot{\theta})}{dt} = 0,$$ (1.311)

which implies that

$$h = r^{2}\,\skew{5}\dot{\theta}$$ (1.312)

is constant in time. It is easily demonstrated that $h$ is the magnitude of the vector ${\bf h}$ defined in Equation (1.281). 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 1.17: Kepler's second law.

Suppose that the radius vector connecting our planet to the origin (i.e., the Sun) sweeps out an angle $\delta\theta$ between times $t$ and $t+\delta t$. See Figure 1.17. The approximately triangular region swept out by the radius vector has the area

$$\delta A \simeq \frac{1}{2}\,r^{2}\,\delta\theta,$$ (1.313)

because 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

$$\frac{dA}{dt} = \lim_{\delta t\rightarrow 0}\frac{r^{\,2}\,\delta\theta}{2\,\delta{t}}= \frac{r^{2}}{2}\,\frac{d\theta}{dt} = \frac{h}{2}.$$ (1.314)

Thus, the radius vector sweeps out area at a constant rate (because $h$ 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.