Answer
(a) Using Kepler's First Law and Eq. (3) we prove that
$A = \frac{1}{2}JT$
(b) Using the result from Exercise 23: $b = \sqrt {pa} $ and the area the ellipse: $A = \pi ab$ we show that
$A = \left( {\pi \sqrt p } \right){a^{3/2}}$
(c) Using the results from part (a) and part (b), we show that
${T^2} = \frac{{4{\pi ^2}}}{{GM}}{a^3}$
Work Step by Step
(a) Kepler's First Law states that the orbit of a planet is an ellipse with the sun at one focus. By Theorem 1 of Section 12.4, the area swept out by the planet's radial vector is
$A = \frac{1}{2}\mathop \smallint \limits_0^{2\pi } r{\left( t \right)^2}{\rm{d}}\theta $
Write
$A = \frac{1}{2}\mathop \smallint \limits_0^{2\pi } r{\left( t \right)^2}\theta '\left( t \right){\rm{d}}t$
By Eq. (3), we have $J = r{\left( t \right)^2}\theta '\left( t \right)$. So,
$A = \frac{1}{2}\mathop \smallint \limits_0^{2\pi } J{\rm{d}}t$
By Theorem 1, $J$ is constant. Therefore
$A = \frac{1}{2}J\mathop \smallint \limits_0^{2\pi } {\rm{d}}t$
Hence, $A = \frac{1}{2}JT$.
(b) From Exercise 23, we obtain $b = \sqrt {pa} $.
Since the area $A$ of the ellipse is $A = \pi ab$, so
$A = \pi a\sqrt {pa} $
Hence, $A = \left( {\pi \sqrt p } \right){a^{3/2}}$.
(c) From part (a) and part (b) we obtain $A = \frac{1}{2}JT$ and $A = \left( {\pi \sqrt p } \right){a^{3/2}}$.
So,
$\frac{1}{2}JT = \left( {\pi \sqrt p } \right){a^{3/2}}$
$T = \frac{2}{J}\left( {\pi \sqrt p } \right){a^{3/2}}$
We have $p = \frac{{{J^2}}}{k} = \frac{{{J^2}}}{{GM}}$. So, $\frac{{\sqrt p }}{J} = \frac{1}{{\sqrt {GM} }}$.
Thus, $T = \frac{{2\pi }}{{\sqrt {GM} }}{a^{3/2}}$
Hence, ${T^2} = \frac{{4{\pi ^2}}}{{GM}}{a^3}$.
