MIT OpenCourseWare http://ocw.mit.edu 16.346 Astrodynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 7 Optimum Orbital Transfer Velocity Requirements for Orbital Transfer Optimum Single-Impulse Transfer #11.2 We can apply Euler’s tangent to the hyperbola, derived in Lecture 6, to formulate the solution of the optimum velocity impulse problem. v0 = initial velocity of spacecraft at P ∆v1 = v1 − v0 = vρ1 + v c1 − v0 = required velocity impulse Then the necessary and sufficient condition for an optimum transfer is: ∆v1 ⊥ (vρ1 − v c1) = ⇒ (vρ1 − v c1) · (vρ1 + v c1 − v0)=0 To convert this to an algebraic equation, we have vρ1 2 1 vρ1  − 1 − (v0 · i r1) − (v0 · i c) =0 v v vc1 c1 c1 Define px 2 = vρ1 = m where pm =2r1r2 sin2 12 θ v c1 p c Then  1 r1r2 sin θ 2r1r2 v c1 = c√µp = µc x cos 1 θ where x> 02 16.346 Astrodynamics Lecture 7 Orbit of destination planetDeparture orbitP (launch)F (Sun)r1r2cv0Q (arrival)Orbit of spacecraftv1∆v1θFigure by MIT OpenCourseWare.Optimality condition: x 4 −  2r1r2 cos 1 θ ( · i ) x 3 + 2r1r2 2 v0 r1cosµc µc We have √µpv0 0 i r1 = v0 cos γ0 = cot γ0    1 θ (v0 · i ) x − 1=02 c· r1 √µp0 v0 · i c = v0 cos(φ1 − γ0)= (cos φ1 cot γ0 + sin φ1) r1 Define P =2r2p0 cos 12 θ cot γ0 Q =2r2p0 cos 12 θ (cos φ1 cot γ0 + sin φ1) r1c r1c Then x 4 − Px3 + Qx − 1=0 Optimum Single-Impulse Transfer from a Circular Orbit with Q =2r23 cos 12 θ sin θx 4 + Qx − 1=0 c3 The result of Problem 11–4 in the textbook is that the optimum single impulse transfer from a circular orbit can be expressed as 24 4(c/r2)3 sin ν − tan ν == Q sin θ√1 + cos θ where the angle ν , introduced in Section 6.5, is given by x 2 = pm = cot21 ν2p and 0 ≤ ν ≤ π . In this form, the equation can be solved for ν almost by inspection using a table of trigonometric functions. One of my past students, Peter Neirinckx, showed that this equation could be solved by successive substitutions. Peter’s argument was that since ν is in the second quadrant, the transformation ν = π + α would be convenient and the algorithm could be expressed recursively in the form: α = arctan[−(sin α +4/Q)]n+1 n His initial guess was α0 = −45◦ and convergence was very rapid indeed. The exact solution is developed on Pages 521–522. 16.346 Astrodynamics Lecture 7  Solving the General Optimization Problem by Successive Substitutions A more recent student, Phil Springmann, in 2003, showed that the general case is similar in form to the optimum transfer from a circular orbit. Phil’s development is x 4 − Px3 + Qx − 1=0 cot41ν − P cot31ν + Q cot 1ν − 1=02 2 2(cos41ν − sin41ν) − P cos31ν sin 1ν + Q cos 1ν sin31ν =02 2 2 2 2 2cos ν − 1P [sin ν(1 + cos ν)] + 1Q[sin ν(1 − cos ν)]=0 4 4(Q + P ) sin ν − (Q − P) tan ν =4 Then, as a recursive algorithm, with ν = π + α −[(Q + P ) sin α +4] α n+1 = arctan n Q − P 16.346 Astrodynamics Lecture 7Earth-to-Mars Departure Velocity for Orbital Transfer Introduce the dimensionless quantity 2 ∆v1 ∆E = v0 which is the amount of energy needed at point P1 to transfer from a circular orbit to an elliptical orbit for a voyage to P2 . We find that 1. ∆E is a double-valued function of a having an infinite slope at a = a m = 12 s where s = 12 (r1 + r2 + c) which is the smallest value of a for which an elliptical path from P1 to P2 is possible. 2. As a increases, the slope of the upper branch of ∆E is always positive, while the slope of the lower branch is negative for a near a m and has a minimum for a = aM . Both branches then approach horizontal asymptotes. 16.346 Astrodynamics Lecture 7 1.21.00.80.60.40.200123∆v1v0amaMa, astronomical unitsθ = 120oAsymptote of lower branchAsymptote of upper brancham : Minimum-energy trajectoryaM : Minimum-departure-velocity trajectoryFigure by MIT OpenCourseWare.Minimum Departure Velocity for Orbital Transfer from Earth to the Planets The figure below gives plots of ∆v1 v0  M as a function of the transfer angle θ for a voyage from earth to each of the other planets of the solar system. The curves for the two most remote planets Neptune and Pluto are not shown because, with the scale used, they would be indistinguishable from the curve for Uranus. The sections of the curves for Jupiter, Saturn and Uranus corresponding to parabolic trajectories as the minimum-velocity paths, are characterized by broken lines. For the special case in which θ = 180◦ , the trajectory is of the Hohmann type. (For transfer angles θ larger than 180◦ , the parabolic trajectories are ficitious optimums since they would not be closed in the counterclockwise direction.) 16.346 Astrodynamics Lecture 7 1.81.61.41.21.00.80.60.40.200 20 40 60 80 100 120 140 160 180180200220240260280300320340360θ, degreesSaturnUranusJupiterMarsMercuryVenus∆v1v0( )Elliptic trajectoryParabolic trajectoryFigure by MIT