Halley's comet, which passed around the sun early in 1986, moves in a highly elliptical orbit with an eccentricity of 0.967 and a period of 76 years. Calculate its minimum  distances from the Sun.
Equation 8.49 relates the period of motion with the semimajor axes. Because $m$ (Halley's comet) $\ll m_{\text {Sun }}$
$$
\begin{aligned}
a & =\left(\frac{G m_{\text {Sun }} \tau^2}{4 \pi^2}\right)^{1 / 3} \\
& =\left[\frac{\left.\left(6.67 \times 10^{-11} \frac{\mathrm{Nm}^2}{\mathrm{~kg}^2}\right)\left(1.99 \times 10^{30} \mathrm{~kg}\right)\left(76 \mathrm{yr} \frac{365 \mathrm{day}}{\mathrm{yr}} \frac{24 \mathrm{hr}}{\mathrm{day}} \frac{3600 \mathrm{~s}}{\mathrm{hr}}\right)^2\right]}{4 \pi^2}\right]^{1 / 3} \\
a & =2.68 \times 10^{12} \mathrm{m}
\end{aligned}
$$
Using Equation 8.44 , we can determine $r_{\min }$ and $r_{\max }$
$$
\begin{aligned}
& r_{\min }=2.68 \times 10^{12} \mathrm{~m}(1-0.967)=8.8 \times 10^{10} \mathrm{~m} \\
\end{aligned}
$$