Subsection 6.2.4 Derivation of Kepler’s III Law
The total area of an ellipse is given by
\begin{equation}
A = \pi a b\tag{6.2.32}
\end{equation}
Where a and b are semi-major and semi-mionor axes, respectively. Also from (6.2.31), we have -
\begin{align*}
or, \quad \frac{\,dA}{\,dt} \amp = \frac{L}{2m}
\end{align*}
Since the direction of areal velocity is same as direction of angular momentum.
\begin{align*}
or, \quad \,dt \amp = \frac{2m}{L}\,dA \\
or, \quad \int_0^T\,dt \amp = \frac{2m}{L}\int_0^A\,dA \\
or, \quad T \amp = \frac{2m}{L}A= \frac{2m}{L} (\pi a b) = \frac{2\pi m}{L} [a (a\sqrt{1-e^2})]
\end{align*}
using (6.2.23).
\begin{equation}
\therefore \quad T = \frac{2\pi m}{L} (a^2\sqrt{1-e^2})\tag{6.2.33}
\end{equation}
\begin{equation*}
or, \quad T^2 = \frac{4\pi^2 m^2}{L^2} [a^4 (1-e^2)] = \frac{4\pi^2 m^2}{L^2} [a^4 (\frac{l}{a})]
\end{equation*}
using (6.2.21)
\begin{equation*}
or, \quad T^2 = \frac{4\pi^2 m^2}{L^2} [l a^3]
\end{equation*}
on using (6.2.20)
\begin{equation*}
\therefore \quad T^2 = (\frac{4\pi^2}{GM}) a^3
\end{equation*}
\begin{equation}
\therefore \quad T^2 \propto a^3\tag{6.2.34}
\end{equation}
