Question

An asteroid in an elliptical orbit about the sun travels at 2.0 106 m/s at perihelion (the point of closest approach) at a distance of 3.8 108 km from the sun. How fast is it traveling at aphelion (the most distant point), which is 8.3 108 km from the sun?

Answer 1

the angular momentum of the system is constant, so the velocity is inversely proportional the the radius of the orbit

Va = Vp * (Rp / Ra)