Calculating Speed of Comet at Greatest Distance from Sun

[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?

 

[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 traveling 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 traveling 1.7 km/s.

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

 

[thepatientmental]

Yes, you have correctly calculated the speed of the comet at its greatest distance from the sun. By using the conservation of angular momentum and the fact that the moment arm is zero at the greatest distance, you were able to equate the initial and final angular momenta and solve for the final speed. Your result of 3,203,419 m/s is consistent with the expected speed for a comet at that distance from the sun. Good job!