A particular comet has an elliptical orbit. When closest to the Sun (perihelion) it is at a distance of 8.823 1010 m and moves with a speed of 54.6 km/s. The greatest distance between this comet and the Sun (aphelion) is 6.067 1012 m.

Question

a. Calculate its speed at aphelion.

Answer 1

At perihelion and aphelion, the comet's velocity is perpendicular to the radial direction between sun and comet.

At these times, velocity * distance = = orbital angular momentum, which is constant

Therefore
6.067*10^2 Vaph = 8.823*10^10*54.6 km/s

Solve for Vaph.

This is a special case of Kepler's second law.