Comet's speed

1. Mar 14, 2005

UrbanXrisis

A comet moves in an elliptical orbit around the sun. It's closest approach to the sun is 0.59 AU and its greatest distance form the sun is 35AU. If the comet's speed at its closest approach is 54 km/s what is the speed when it is farthest away? Angular momentum is conserved and the gravitational forec eserted by the Sun has a moment arm of zero.

Here's what I did...
$$I_{initial} \omega_{initial}=I_{final} \omega_{final}$$
moment of inertia is always the same...
$$\frac{v_i}{r_i}=\frac{v_f}{r_f}$$
$$\frac{54000m/s}{88262020000m}=\frac{v_f}{5.23593E12m}$$
$$v_f=3203419m/s$$

did I do this correctly?

2. Mar 15, 2005

tony873004

It's a bad question. A comet whose closest approach is 0.59AU, and has a velocity of 54 km/s at that point will only venture out 18.76 AU from the Sun before beginning its fall back towards the Sun. So it will be travelling from approximately the orbit of Venus to the orbit of Uranus. It will take 15 years for it to complete this half orbit, at which point it will be travelling 1.7 km/s.

Are you sure the question said the comet is orbiting the Sun, and not another star?