Asked by Anonymous

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?
Answered by Scott
the angular momentum of the system is constant, so the velocity is inversely proportional the the radius of the orbit

Va = Vp * (Rp / Ra)
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