# lecture18 - v at each maneuver Triangle, v1, delta v1, vt...

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Δ v at each maneuver Triangle, v1, delta v1, vt with angle alpha1 Speeds for circular orbits are determined by radii, v 2 i = μ r i , v 2 f = μ r f , and the Hohmann analysis gives the transfer- orbit speeds, v 2 ta = μ ± 2 r i - 2 r i + r f ! v 2 tb = μ ± 2 r f - 2 r i + r f ! For the two maneuvers, use the law of cosines to include plane change to compute vector lengths, v a ) 2 = v 2 i + v 2 ta - 2 v i v ta cos α v b ) 2 = v 2 f + v 2 tb - 2 v f v tb cos(Δ i - α ) . L. Healy – ENAE404 – Spring 2007 – Lecture 18 (Apr. 3) 1

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Minimize over initial angle change The total velocity change “cost” is Δ v v a + Δ v b = q v 2 i + v 2 ta - 2 v i v ta cos α + r v 2 f + v 2 tb - 2 v f v tb cos(Δ i - α ) Optimal choice of angle α is found from mini- mizing Δ v ; Δ v ∂α = 0 or 0 = v i v ta sin α q v 2 i + v 2 ta - 2 v i v ta cos α - v f v tb sin(Δ i - α ) q v 2 f + v 2 tb - 2 v f v tb cos(Δ i - α ) . Use bisection method or Newton-Raphson iter- ation (if you calculate the derivative) will solve this for α . Solution depends on r i , r f and Δ i . L. Healy – ENAE404 – Spring 2007 – Lecture 18 (Apr. 3) 2
Worked example, Hohmann w/split plane change A geosynchronous transfer. Circular initial and ﬁnal orbits. Initial altitude 191km; r i = 6569km. Final altitude 35780km; r f = 42158km. Initial inclination 28 . 5 , ﬁnal inclination 0 . L. Healy – ENAE404 – Spring 2007 – Lecture 18 (Apr. 3) 3

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Speeds for split plane change The various speeds may be calculated from these values, v i = s μ r i = 7 . 790 km / sec v f = s μ r f = 3 . 075 km / sec v ta = v u u t μ ± 2 r i - 2 r i + r f ! = 10 . 25 km / sec v tb = v u u t μ ± 2 r f - 2 r i + r f ! = 1 . 597 km / sec . With Δ i = - 28 . 5 , the derivative Δ v/∂α is found to be zero at α = - 0 . 0378 = - 2 . 16 . L. Healy – ENAE404 – Spring 2007 – Lecture 18 (Apr. 3) 4
v for example With the optimized value of α = - 2 . 16 , we can compute the Δ v s for each burn. From the law of cosines formulas before, Δ v a = q v 2 i + v 2 ta - 2 v i v ta cos α =2 . 48 km / sec Δ v b = r v 2 f + v 2

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## lecture18 - v at each maneuver Triangle, v1, delta v1, vt...

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