240_Dynamics 11ed Manual

# 240_Dynamics 11ed Manual - Engineering Mechanics Dynamics...

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Engineering Mechanics - Dynamics Chapter 13 v a1 r p1 b v p1 = v a1 1.181 10 3 × m s = When the rocket travels along the second elliptical orbit , from Eq.[4] , we have a 1 e' 1 e' + b = e' a b + ba + = e' 0.8462 = Substitute r 0 r p2 = a = r p2 a = v p2 1 e' + () G M e r p2 = v p2 8.58 10 3 × m s = Applying Eq. 13-20, we have v a2 r p2 b v p2 = v a2 715.021 m s = For the rocket to enter into orbit two from orbit one at A, its speed must be decreased by Δ vv a1 v a2 = v 466 m s = If the rocket travels in a circular free - flight trajectory , its speed is given by Eq. 13-25 v c GM e a = v c 6.315 10 3 × m s = The speed for which the rocket must be decreased in order to have a circular orbit is p2 v c = v 2.27 km s = *Problem 13-124 An asteroid is in an elliptical orbit about the sun such that its perihelion distance is d . If the eccentricity of the orbit is
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