Engineering Mechanics - DynamicsChapter 13va1rp1b⎛⎝⎞⎠vp1=va11.181103×ms=When the rocket travels along the second elliptical orbit , from Eq. , we have a1e'−1e'+⎛⎝⎞⎠b=e'a−b+ba+=e'0.8462=Substitute r0rp2=a=rp2a=vp21e'+()GMerp2=vp28.58103×ms=Applying Eq. 13-20, we have va2rp2bvp2=va2715.021ms=For the rocket to enter into orbit two from orbit one at A,its speed must be decreased byΔvva1va2−=v466ms=If the rocket travels in a circular free - flight trajectory , its speed is given by Eq. 13-25vcGMea=vc6.315103×ms=The speed for which the rocket must be decreased in order to have a circular orbit is p2vc−=v2.27kms=*Problem 13-124An asteroid is in an elliptical orbit about the sun such that its perihelion distance is d. If theeccentricity of the orbit is
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