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.2 a3 2 1 a E ψ Identify ⇐⇒ µq q r(t⇐⇒a3 − τ)= E − sin E + sin E 0at − τ ⇐⇒ 1(t 22 − t1)since t2 − t1 =(t2 − τ)+(τ − t1)=2(t2 − τ). From the classical relation between the true and eccentric anomalies: 21 1+ e 21 2a − q 21 q 2 tan2 1 E tanf = tan E = tan E = = 22x 2 =1 − e 2 q 2 ⇒a tan2 1f 2 + tan2 1 E + x 2where we have defined x = tan21 E and = tan21 f2 2 . Since 1F0S √r r cos θ 1 cos f FP F S cos f = = 1 2 2 then =−= 0 2 −0F0P1 2 2(r1 + r2) 1 + cos f F0P2 + F0S Lecture 16 Non-Singular ``Gauss-Like'' Method for the BVP Derivation of the Time Equation Start with the Lagrange time equation and the equation for the mean point radius 3 1 2√ µ(t2 − t1)= a2 (ψ − sin ψ cos φ) (1) r2 10 = a(1 − cos φ)= r0p(1 + tan ψ2) (2) 1 r = 1 (r10p 2 21 + r2)+ √r1r2 cos θ 2 (3)FS = √ r1r2 cos 1θ 2 (4) Eliminate cos φ and compare with the elementary form of Kepler’s equation 1 µr(t− t )= ψ − sin ψ + 0 sin ψ16.346 Astrodynamics Lecture 16 Fig. 7.4 from An Introduction to the Mathematics and Methods of Astrodynamics. Courtesy of AIAA. Used with permission.Relation to Gauss’ Classical Method The time equation 1 (t2 µ q3 q q 2 tanq32 − t1) × = E × a3 − sin E + sin E = E a − sin E + a 1 + ta = sibecomes 1 µ 8 4 tan31 E(t2 − t31 2 2 q3 1) × tan sin × E = E − E + ( + x)3 2 ( + x)(1 + Then, since q = r0 = r2 0p(1 + tan1E2 )= r0p(1 + x) we have an expression for the transfer time as a function only of E ≡ ψ : µ 4 tan31E 4 tan31E3 (t2 − t1) 2= E sin E + 28r× [( +3 x)(1 + x)]20p−( + x)(1 + x) m3 E − sin E m = m +[( + x)(1 + x)]3 4 tan3 1E ( + x)(1 + x2Following Gauss, we can define y as 2 m y = so that y 3− y 2E − sin E= m( + x)(1 + x) 4 tan31 E2 where µ(t2 2 − t1)r1 + r1 m = and = 2 √− 2 r1r2 cos θ 2 8r3 r1 + r10p2 +2√r1r2 cos θ2 Note: The new y does not have the same geometric significance as Gauss’ yParameter and Semimajor Axis q 2x r2 = = 0p(1 + x)1 2x 2xy= = =a + x a ⇒ a r0p(1 + x)( + x) mr0p p sin φ c c(1 + x)2 cy2(1 + x)2 y2√ (1 + x)= = = = = p sin ψ 2a sin2 mψ 8ax 4mr0p m 12E n2 12En E x) ) . 2 where we have used the equation c =2a sin ψ sin φ from Lecture 9, Page 1 16.346 Astrodynamics Lecture 16Comparing the Structure of Gauss’ Method and the New Method Gauss’ Method New Method 1D √≡ r1r2 cos 11θD √≡ 4(r1 + r2 +2 r1r2 cos 22r1 + r2 2√1 − r cos = 1r2 θ r+ r2 = 1 2 − 2√r1r2 cos 1 θ 24D 4D µ(tm = 2 − t1)2 µ(t2 m = 2 D3 − t1)88D3 x ≡ sin2 1 ψ x 2≡ tan2 1 ψ 2y 2 m = y 2 m = + x ( + x)(1 + x) y 3 − y 2 2ψ = m − sin 2ψ ψ ψ y 3 sin − 3 y 2 = m −sin ψ 4 tan3 1 ψ 2p cy2 p cy2 = = (1 + x)2 pm 4mD pm 4mD 1 2y2x1 2y2x = (1 − x) = a mD a mD Universal Form The equations are universal when x is extended to include the other conics: ⎧ ⎨ tan2 14(E2 − E1) ellipse x = 0 parabola ⎩ − tanh2 14(H2 − H1) hyperbola θ)16.346 Astrodynamics Lecture
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