COMET HALLEY
Comet Halley moves in an elongated elliptical orbit around the sun (Fig. 13.21). Its distances from the sun at perihelion and aphelion are 8.75 × 10^7 km and 5.26 × 10^9 km, respectively. Find the orbital semi-major axis, eccentricity, and period.
IDENTIFY and SET UP:
We are to find the semi-major axis a, eccentricity e, and orbital period T. We can use Fig. 13.18 to find a and e from the given perihelion and aphelion distances. Knowing a, we can find T from Kepler’s third law, Eq. (13.17).
T=\frac{2 \pi a^{3 / 2}}{\sqrt{G m_{\mathrm{S}}}} (13.17)
EXECUTE:
From Fig. 13.18, the length 2a of the major axis equals the sum of the comet–sun distance at perihelion and the comet–sun distance at aphelion. Hence
a=\frac{\left(8.75 \times 10^{7} \mathrm{~km}\right)+\left(5.26 \times 10^{9} \mathrm{~km}\right)}{2}=2.67 \times 10^{9} \mathrm{~km}Figure 13.18 also shows that the comet–sun distance at perihelion is a – ea = a(1 – e). This distance is 8.75 × 10^7 km, so
e=1-\frac{8.75 \times 10^{7} \mathrm{~km}}{a}=1-\frac{8.75 \times 10^{7} \mathrm{~km}}{2.67 \times 10^{9} \mathrm{~km}}=0.967From Eq. (13.17), the period is
\begin{aligned}T &=\frac{2 \pi a^{3 / 2}}{\sqrt{G m_{\mathrm{S}}}}=\frac{2 \pi\left(2.67 \times10^{12} \mathrm{~m}\right)^{3 / 2}}{\sqrt{\left(6.67 \times 10^{-11} \mathrm{~N} \cdot\mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(1.99 \times 10^{30} \mathrm{~kg}\right)}} \\&=2.38 \times 10^{9} \mathrm{~s}=75.5 \text { years }\end{aligned}
EVALUATE: The eccentricity is close to 1, so the orbit is very elongated (see Fig. 13.21a). Comet Halley was at perihelion in early 1986 (Fig. 13.21b); it will next reach perihelion one period later, in 2061.
