Worked example 12.4: Halley's comet

Question: The distance of closest approach of Halley's comet to the Sun is $0.57 {\rm AU}$. (1 AU is the mean Earth-Sun distance.) The greatest distance of the comet from the Sun is 35AU. The comet's speed at closest approach is $54 {\rm km/s}$. What is its speed when it is furthest from the Sun?

Answer: At perihelion and aphelion, the comet's velocity is perpendicular to its position vector from the Sun. Hence, at these two special points, the comet's angular momentum (around the Sun) takes the particularly simple form

$$l = m r u.$$

Here, $m$ is the comet's mass, $r$ is its distance from the Sun, and $u$ is its speed. According to Kepler's second law, the comet orbits the Sun with constant angular momentum. Hence, we can write

$$r_0 u_0 = r_1 u_1,$$

where $r_0$ and $u_0$ are the perihelion distance and speed, respectively, and $r_1$ and $u_1$ are the corresponding quantities at aphelion. We are told that $r_0=0.57 {\rm AU}$, $r_1=35 {\rm AU}$, and $u_0 = 54 {\rm km/s}$. It follows that

$$u_1 = \frac{u_0 r_0}{r_1} = \frac{54\times 0.57}{35} = 0.879 {\rm km/s}.$$