# sln9 - HW#9 Problem 1 For hyperbolic orbit 2 V 42 e = 1 2 =...

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HW#9 Problem 1 For hyperbolic orbit 42 22 2 2 11 1 (1 ) ,0 ) p p V ee V ae ra e aa a re VV V µ µµ ∞∞ =+ →∆= =− →= <→= Problem 2 Given 1 2 3 2 2 1 1.524 4 E M sun RR A U R RA AU yr µπ == = U U Non-Hohmann Transfer 1 2.048 1.524 T T p A T rA U aA U = = Heliocentric Transfer Orbit 2 2 2 ) 1 ) 2 ) T T p pT T T sun T T T T TT T T s u n T sun r ra e e e a p pa e h ph p ξ =−→ = =→ = At point 2 From energy equation - 2 2 2 sun H R T V   speed at pt.2 in the transfer orbit 2 sun M V R = speed of mars wrt sun 1 2 2 cos cos T TH H h hR V RV φφ −− = angle between 2 H V and M V

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() 2 2 /2 2 2c o s MH M H M VV V V V 2 φ −− =+ Flyby 22 2 / 1 / 1.1 , 1.5 1 11 sin 2sin 2 2s i n 2 PP PM M flyby M rR rV e ee µ δ ==  =→=   ∆= Larger P r is farther away from the planet, less gravity assistance.
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## This note was uploaded on 08/26/2008 for the course EAS 4510 taught by Professor Fitz-coy during the Spring '05 term at University of Florida.

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sln9 - HW#9 Problem 1 For hyperbolic orbit 2 V 42 e = 1 2 =...

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