Asked by Lonely
Halley comet which passes around the sun every 76 years has an elliptical orbit. when closest to the sun it is at a distance of 8.823 x 10^10 m and moves with a speed of 54.6 km/s. The greatest distance between hally's comet and the sun ( aphelion ) is 6.152 x 10^12 m. calculate its speed at aphelion. answer should be in m/s
Answers
Answered by
drwls
The easiest way to answer this question is to use Kepler's second law. At perihelion and aphelion, the rate of sweeping out area of the ellipse is the same. It is related to the law of conservation of angular momentum.
Thus,
(R^2*V)aphelion = (R^2*V)perihelion
Vaphelion = 54600 m/s*[(8.823*10^10)/6.152*10^12]^2
= __?
Thus,
(R^2*V)aphelion = (R^2*V)perihelion
Vaphelion = 54600 m/s*[(8.823*10^10)/6.152*10^12]^2
= __?
Answered by
Anonymous
11.23 m/s
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