Lecture 4: The Initial-Value Problem

Lecture 4 The Initial-Value Problem Polar Coordinates and Orbital Plane Coordinates From Lecture 2 ⎤ ⎡ ⎡ 1 r ⎢ r ⎥ ⎣ 1 ⎣h ⎦= e sin f v µ ⎤⎡ 0 ⎦⎣ 1 + e cos f ir ⎤ ⎡ ⎦ ⎣ iθ ir iθ ⎤ ⎡ ⎦=⎣ cos f sin f − sin f cos f ⎤⎡ ⎦⎣ ie ⎤ ⎦ ip Representation in the Hodograph Plane Fig. 3.5 (a) and (b) from An Introduction to the Mathematics and Methods of Astrodynamics. Courtesy of AIAA. Used with permission. Physical Plane Hodograph Plane Flight-Direction Angle From the position and velocity equations in polar coordinates at the top of this page: µre sin f h In terms of the flight-direction angle γ (shown in the figures above), r·v= r = r ir v = v cos γ ir + v sin γ iθ =⇒ r · v = rv cos γ Also h = |r × v| = rv sin γ Therefore: def r · v rv cos γ h cot γ σ = √ = √ = √ µ µ µ σ= 16.346 Astrodynamics √ or, since p = p cot γ Lecture 4 h2 µ we have The Initial-Value Problem using Lagrange Coefficients F , G, Ft , Gt ⎡ ⎤ ⎡ r r cos f ⎣ ⎦=⎣ µ v − sin f h ⎤⎡ r sin f ie ⎤ ⎦⎣ ⎦ µ (e + cos f ) ip h r = F r0 + G v0 v = Ft r0 + Gt v0 or ⎡ ⎡ µ (e + cos f0 ) ⎣ ⎦ = ⎣ h2 µ ip sin f0 h2 ie     r r =Φ 0 v v0 ⎤  F with Φ = Ft G Gt #3.6 ⎤⎡ ⎤ r r0 − 0 sin f0 h ⎦ ⎣ ⎦ r0 cos f0 v0 h  and |Φ| = 1 The value of the determinant |Φ| = F Gt − GFt = 1 follows from the conservation of angular momentum r × v = (F Gt − GFt ) r0 × v0 = r0 × v0 Lagrange Coefficients in Terms of the True Anomaly Difference rr G = √ 0 sin θ µp r Gt = 1 − 0 (1 − cos θ) p r F = 1 − (1 − cos θ) p √ µ √ σ0 (1 − cos θ) − p sin θ Ft = r0 p where p r = √ r0 r0 + (p − r0 ) cos θ − p σ0 sin θ with θ = f − f0 and σ0 = Lagrange Coefficients in Terms of the Eccentric Anomaly Difference Define r0 · v0 √ µ Page 162 ϕ = E − E0 . Then: a F = 1 − (1 − cos ϕ) r √ 0 µa sin ϕ Ft = − rr0 where (3.42) √ √ µ G = aσ0 (1 − cos ϕ) + r0 a sin ϕ (4.41) a Gt = 1 − (1 − cos ϕ) r √ r = a + (r0 − a) cos ϕ + σ0 a sin ϕ and σ0 = r0 · v0 √ ≡ p cot γ0 √ µ Kepler’s Equation is then M − M0 = µ (t − t0 ) = (E − e sin E) − (E0 − e sin E0 ) a3 or, in terms of ϕ = E − E0 σ r µ (t − t0 ) = ϕ + √0 (1 − cos ϕ) − 1 − 0 sin ϕ 3 a a a 16.346 Astrodynamics Lecture 4 (4.43) Lagrangian Coefficients for Parabolic Orbits Pages 155–156 Since tan 12 f is obtained directly as the solution of Barker’s equation, it is more convenient to express all trigonometric functions in terms of this function of the true anomaly. Thus, the position and velocity vectors for the parabola in orbital plane coordinates are p (1 − tan2 12 f ) ie + p tan 12 f ip 2√ √ µp µp 1 i v=− tan 2 f ie + r r p r= r·v √ σ = √ = p tan µ Define Therefore, with χ = σ − σ0 , 1 2 f= √ p cot γ we have χ2 F =1− 2r √ 0 µχ Ft = − rr0 χ G = √ (2r0 + σ0 χ) 2 µ Gt = 1 − χ2 2r together with Barker’s equation and the equation of orbit: √ 6 µ(t − t0 ) = 6r0 χ + 3σ0 χ2 + χ3 r = r0 + σ0 χ + 21 χ2 σ = σ0 + χ Solving the Generalized Form of Barker’s Equation The solution is χ= √ p z − σ0 where z is obtained by solving the cubic equation z 3 + 3z = 2B with B= 1 p 3 2 √ [σ0 (r0 + p) + 3 µ(t − t0 )] Therefore, all the solution methods developed for Barker’s equation are applicable without modification provided that B > 0 . 16.346 Astrodynamics Lecture 4 Construction of a Parabola and its Tangents Pages 156–157 Fig. 4.7 from An Introduction to the Mathematics and Methods of Astrodynamics. Courtesy of AIAA. Used with permission. 16.346 Astrodynamics Lecture 4