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Kepler's second law

Multiplying our planet's tangential equation of motion, Equation (4.19), by $ r$ , we obtain

$\displaystyle r^{\,2}\,\skew{5}\ddot{\theta} + 2\,r\,\skew{3}\dot{r}\,\skew{5}\dot{\theta} = 0.$ (4.20)

However, this equation can be also written

$\displaystyle \frac{d(r^{\,2}\,\skew{5}\dot{\theta})}{dt} = 0,$ (4.21)

which implies that

$\displaystyle h = r^{\,2}\,\skew{5}\dot{\theta}$ (4.22)

is constant in time. It is easily demonstrated that $ h$ is the magnitude of the vector $ {\bf h}$ defined in Equation (4.6). 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 4.2: Kepler's second law.
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Suppose that the radius vector connecting our planet to the origin (i.e., the Sun) rotates through an angle $ \delta\theta$ between times $ t$ and $ t+\delta t$ . (See Figure 4.2.) The approximately triangular region swept out by the radius vector has the area

$\displaystyle \delta A \simeq \frac{1}{2}\,r^{\,2}\,\delta\theta,$ (4.23)

as 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

$\displaystyle \frac{dA}{dt} = \lim_{\delta t\rightarrow 0}\frac{r^{\,2}\,\delta\theta}{2\,\delta{t}}= \frac{r^{\,2}}{2}\,\frac{d\theta}{dt} = \frac{h}{2}.$ (4.24)

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


next up previous
Next: Kepler's first law Up: Keplerian orbits Previous: Plane polar coordinates
Richard Fitzpatrick 2016-03-31