Answer
See answer.
Work Step by Step
a. Use Kepler’s third law to find the mean orbital radius of Hale-Bopp, using Earth’s data.
$$(\frac{r_1}{r_2})^3=(\frac{T_1}{T_2})^2$$
$$r_1=r_2 (\frac{T_1}{T_2})^{2/3}$$
$$r_{HB}=r_{Earth} (\frac{T_{HB}}{T_{Earth}})^{2/3}$$
$$r_{HB}=(1AU) (\frac{2400y}{1y})^{2/3}=180AU$$
b. This value is the average of the nearest and farthest distances of the comet from the sun. Assuming that the nearest distance is 1 AU, the farthest distance is twice the value calculated, or 360AU (rounding off).
c. Refer to figure 5-29. When the comet is 360 times farther away, it is moving at 1/360 of the speed in order to sweep out equal areas in equal time.
$$\frac{v_{closest}}{v_{farthest}}=\frac{360}{1}$$
