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.
a. Calculate its speed at aphelion.
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.