lecture16 - Phasing worked example Vallado Example 6–8,...

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Unformatted text preview: Phasing worked example Vallado Example 6–8, pp. 348–349 • Given: a i = a tg = 2ER = 12756km, Δ θ = 20 ◦ so the interceptor leads the target. • Find phasing time and Δ v . The angular velocity is n tg = 4 . 382 × 10- 4 rad / s Because the interceptor is not that far from the target, do at least a full revolution, k tg = 1, τ p = 2 πk tg + Δ θ n tg = 2 π + π/ 9 4 . 382 × 10- 4 rad / s = 1 . 513 × 10 4 s , and the phasing semimajor axis a p = 3 v u u t μ τ p 2 πk i ! 2 = 13220km . Notice that this is slightly larger than the orig- inal semimajor axis (radius). L. Healy – ENAE404 – Spring 2007 – Lecture 16 (Mar. 27) 1 Δ v , and taking longer Δ v = 2 v u u t 2 μ a tg- μ a p- s μ a tg = 196 m / s . Now take two orbits k i = k tg = 2, τ p = 2 πk tg + Δ θ n tg = 4 π + π/ 9 4 . 382 × 10- 4 rad / s = 2 . 947 × 10 4 s it takes longer, but a p = 3 v u u t μ τ p 2 πk i ! 2 = 12990km , the phasing semimajor axis is shorter and Δ v = 2 v u u t 2 μ a tg- μ a p- s μ a tg = 100 . 7 m / s , about half the total Δ v . L. Healy – ENAE404 – Spring 2007 – Lecture 16 (Mar. 27) 2 Rendezvous Phasing to a different orbit. • This is a Hohmann transfer. Solving this problem is a matter of timing. • Remember, the two orbits have different orbital periods and therefore the angle be- tween the interceptor and the target is con- tinuously changing. • The Hohmann tr ansfer time T t accounts for a phase delay called the lead angle α L , so we need to wait until the phase angle Δ θ is this minus the half orbit of the transfer. L. Healy – ENAE404 – Spring 2007 – Lecture 16 (Mar. 27) 3 Mathematically Recall the Hohmann transfer time T t is half an orbital period, T t = π v u u t a 3 tr μ and therefore the lead angle is determined from the t arget’s angular velocity n tg , α L = n tg T t . The Hohmann transfer involves an angular change of π , so we require a phase angle Δ θ = α L- π. L. Healy – ENAE404 – Spring 2007 – Lecture 16 (Mar. 27) 4 Wait time and synodic period If Δ θ i is the phase angle at the start, the amount of time one must w ait before making the first burn is τ w = Δ θ- Δ θ i + 2 πk n i- n tg , with k being a positive integer. The synodic period τ synodic = 2 π n i- n tg , is the time it takes for the relative angles to repeat; it is not the period of either orbit in general....
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lecture16 - Phasing worked example Vallado Example 6–8,...

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