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Kepler's Second Law

Multiplying our planet's tangential equation of motion, (229), by $r$, we obtain
\begin{displaymath}
r^2\,\ddot{\theta} + 2\,r\,\dot{r}\,\dot{\theta} = 0.
\end{displaymath} (245)

However, the above equation can be also written
\begin{displaymath}
\frac{d(r^2\,\dot{\theta})}{dt} = 0,
\end{displaymath} (246)

which implies that
\begin{displaymath}
h = r^2\,\dot{\theta}
\end{displaymath} (247)

is constant in time. It is easily demonstrated that $h$ is the magnitude of the vector ${\bf h}$ defined in Equation (216). 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 18: Kepler's second law.
\begin{figure}
\epsfysize =1.5in
\centerline{\epsffile{Chapter05/fig5.06.eps}}
\end{figure}

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 18. The approximately triangular region swept out by the radius vector has the area

\begin{displaymath}
\delta A \simeq \frac{1}{2}\,r^2\,\delta\theta,
\end{displaymath} (248)

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
\begin{displaymath}
\frac{dA}{dt} = \lim_{\delta t\rightarrow 0}\frac{r^2\,\delt...
...\,\delta{t}}= \frac{r^2}{2}\,\frac{d\theta}{dt} = \frac{h}{2}.
\end{displaymath} (249)

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.


next up previous
Next: Kepler's First Law Up: Planetary Motion Previous: Conic Sections
Richard Fitzpatrick 2011-03-31