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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 Equation (4.24).] We have also seen that the planetary orbit is an ellipse. Now, the major and minor radii of such an ellipse are $ a=r_c/(1-e^{\,2})$ and $ b=(1-e^{\,2})^{1/2}\,a$ , respectively. [See Equations (A.107) and (A.108).] The area of the ellipse is $ A=\pi\,a\,b$ . We expect the radius vector to sweep out the whole area of the ellipse in a single orbital period, $ T$ . Hence,

$\displaystyle T = \frac{A}{dA/dt} = \frac{2\pi\,a\,b}{h} = \frac{2\pi\,a^{\,2}\,(1-e^{\,2})^{1/2}}{h}=\frac{2\pi\,a^{3/2}\,r_c^{\,1/2}}{h}.$ (4.32)

It follows from Equation (4.31) that

$\displaystyle T^{\,2} = \frac{4\pi^{\,2}\,a^{\,3}}{G\,M}.$ (4.33)

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 of planetary motion.

Richard Fitzpatrick 2016-03-31