Question 12.109: Based on observations made during the 1996 sighting of comet...

\begin{aligned}\frac{1}{r_{0}} & =\frac{G M}{h^{2}}=(1+\varepsilon), \quad \frac{1}{r_{1}}=\frac{G M}{h^{2}}(1+\varepsilon), \quad r_{1}=\frac{1+\varepsilon}{1-\varepsilon} r_{0} \\a & =\frac{1}{2}\left(r_{0}+r_{1}\right)=\frac{r_{0}}{1-\varepsilon}, \quad b=\sqrt{r_{0} r_{1}}=\sqrt{\frac{1+\varepsilon}{1-\varepsilon}} r_{0} \\h & =\sqrt{G M r_{0}(1+\varepsilon)} \\\tau & =\frac{2 \pi a b}{h}=\frac{2 \pi r_{0}^{2}(1+\varepsilon)^{1 / 2}}{(1-\varepsilon)^{3 / 2} \sqrt{G M r_{0}(1+\varepsilon)}} \\& =\frac{2 \pi r_{0}^{3 / 2}}{\sqrt{G M(1-\varepsilon)^{3 / 2}}}=\frac{2 \pi r_{0}^{3 / 2} \tau_{0}}{2 \pi R_{E}^{3}(1-\varepsilon)^{3 / 2}} \\& =\left\lgroup\frac{r_{0}}{R_{E}}\right\rgroup^{3 / 2} \frac{1}{(1-\varepsilon)^{3 / 2}} \tau_{0} \\& =(0.230)^{3 / 2} \frac{1}{(1-0.999887)^{3 / 2}} \tau_{0}=91.8 \times 10^{3} \tau_{0}\end{aligned}