Posted by Lonely on Saturday, July 30, 2011 at 12:15pm.
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
= __?
11.23 m/s
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