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M.A. Sharaf, A.N. Saad, and M.I. Nouh where χ is to be considered as a new independent variable — a kind of generalized anomaly. It could be shown [12] that when χ is used as the independent variable instead of the time t, then the non-linear equations of motion can be converted into linear, constant coefficients, differential equations. For t0 and t, the variable χ can be related to the classical anomalies by: ( ) ⎧ a E−E0for α > 0 ⎪ ⎪ ⎛ 1 1 ⎞ χ= ⎨ a⎜tan f −tan f0for α= 0 2 2 ⎟ , (2.14) ⎪ ⎝ ⎠ ⎪ ⎪⎩ −a( H −H0) for α < 0 where E and H are respectively, the elliptic eccentric anomaly and the hyperbolic eccentric anomaly. Regarding the second point mentioned above, we shall consider for the family of transcendental functions, the generating functions: C n ∞ ( z) = ∑ ( −1) k = 0 k k z , n = 0,1,2,… (2.15) 2 ! ( k + n ) and are known as Stumpff functions [1]; they are well defined for a complex variable z, since the power series is convergent at any point in the complex plane. They are real values for real z. Note also that: 1 Cn(0) = . (2.16) n ! What concerns us in the subsequent analysis are the following relations between C2 and C3 of arguments ψ and 4ψ: C (4 ψ ) = [ C ( ψ ) + C ( ψ) −ψC ( ψ) C ( ψ)]/4 , (2.17) 3 2 3 3 2 1 2 2 3 2 C (4 ψ ) = {1 −ψC ( ψ)} . (2.18) • The universal form of the F, …, G functions which are valid for any type of conic motions are given as [11]: 2 χ 2 F = 1 − C2( α0χ) , (2.19) r 0 3 χ 2 G = t−t0 − C 3( α0χ) , (2.20) µ 0 2 2 { 0 3( 0 ) 1} F µ = χ α χ C α χ − rr 2.3. Universal Lambert Problem 2 , (2.21) G χ 2 = 1 − C2( α0χ) . (2.22) r The formulations of the universal Lambert problem could easy be obtained [11] using the two sets of Equations (2.8) to (2.11) and Equations (2.19) to (2.22), and summarized in stepwise form as follows: 90 The Arabian Journal for Science and Engineering, Volume 28, Number 1A. January 2003
• Equating Equations (2.8) and (2.19) and then solve for the universal variable χ we get 0 χ = − ∆ pC2 M.A. Sharaf, A.N. Saad, and M.I. Nouh rr (1 cos f ) , (2.23) where C2 = C2(α0χ 2 ). Also let C3 = C3(α0χ 2 ) below. • Substituting Equation (2.23) into Equation (2.21) and then equate it with F of Equation (2.10) we get where Also, A( ψC3−1) B = r0+ r+ , (2.24) C A= sin ∆f 2 r r0 1−cos∆f , (2.25) r r0 B = (1 −cos ∆f) , (2.26) p 0 2 ψ =α χ . (2.27) χ= B , (2.28) C 2 1 cos ∆ f = ( r.r 0 ) , (2.29) rr 0 1 sin ∆ f = r0× r . (2.30) rr 0 • Solve for ∆t = t–t0 from Equations (2.20) and (2.9) we get 3 µ∆ t = χ C + A B . (2.31) 3 • Finally, the unknown velocity vectors v0 and v are obtained from 1 v0 = r−r0F G ( ) , (2.32) 1 v = ( G r−r0) , (2.33) G where from Equations (2.25) = 1− B F , (2.34) r 0 B G = A , (2.35) µ G = 1− B . (2.36) r January 2003 The Arabian Journal for Science and Engineering, Volume 28, Number 1A. 91
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