16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 1 of 9 16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 3: Approximate ?V for Low-Thrust Spiral Climb Assume initial circular orbit, at co 0 v=v = r μ . Thrust is applied tangentially. Call F a= M . By conservation of energy, assuming the orbit remains near-circular v r ?? μ ?? ?? null , d -a dt 2r r μμ?? ?? ?? null 2 dr a dt r2r μμ null -3 2 1 2 r dr a dt 2 μ null When we integrate, b t 0 a dt = V,? ∫ and so b t 1-1 22 0 -r =Vμ? () 0b V= - rrt μμ ? or final co c V=v -v? (1) The result appears to be trivial, but it is not. Notice that the “velocity increment” V? is actually equal to the decrease in orbital velocity. The rocket is pushing forward, but the velocity is decreasing. This is because in a r -2 force field, the kinetic energy is equal in magnitude but of the opposite sign as the total energy (potential = - 2× kinetic). If thrust were applied opposite the velocity (negative a), the definition of V? would be b t 0 (-a) dt ∫ , so the result in general is c V= v?? (2) 16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 2 of 9 For simplicity, assume now F a = M = constant, which is actually optimal for many situations. Equation (1) can be recast as 0 -=at rr μμ and solving for r, 00 22 co 0 rr r= = at at 1- 1- v r ?? ?? ???? ???? ???? μ ???? (3) This shows how the radial distance “spirals out” in time. In principle, this says r → ∞ at co v t= a , a crude indication of “escape”. But of course, the orbit is no longer “near- circular” when approaching escape, so this result is not precise. One can get some improvement for the estimation of escape V? as follows. The radial velocity r i can be calculated from (3) by differentiation. Notice that this is in the nature of an iteration, since r i was implicitly ignored in the energy balance which led to (3). We obtain 0 co 3 co 2a r v r= at 1- v ?? ?? ?? i (4) The tangential component v= r θ θ i is still approximated as the orbital velocity, i.e., co 0co at r= = -at=v 1- rr v ?? μμ θ ?? ?? i (5) The overall kinetic energy is therefore 2 0 2 2 22 22 coco 6 co co ar 4 vv vat =r + r = 1- + 22v at 1- v ?? ?? ?? ?? ?? ?? ?? θ ?? ?? ? ? ?? ???? ?? ?? ?? ?? ii (6) The escape point is defined by its having zero total energy, i.e., 16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 3 of 9 2 e e v = 2r μ , or 2 e0 co e vr1 -=0. 2v r ?? ?? ?? Substituting, 2 0 22 2 co 6 co co co ar 2 v 1at at 110 2v v at 1 v ?? ?? ?? ?? ?? ?+ ??= ?? ?? ???? ?? ? ?? ?? or () 2 22 0co 6 co co 4ar v at =1- v at 1- v ?? ?? ???? ?? ?? () 1 4 1 4 1 0 4 2 co co 2 0 2arat 2a 1- = = = 2 v v r ?? ?? ?? ν ???? μ ???? ?? (7) where 2 0 a = r ν μ (8) is the ratio of thrust to gravity, a small number for us. Since at= V? , Equation (7) reads () 1 4 esc. co Vv1-2 ?? ?? ν ?? ?? (9) In a more rigorous analysis, the factor 2 1/4 = 1.19 turns out to be actually 0.79: 1 4 esc. co V=v1-0.79 ?? ? ?? ?? ν (10) (more exact) 16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 4 of 9 LOW THRUST TO ESCAPE: RESULTS OF NUMERICAL COMPUTATIONS DEFINITIONS: () 2 0 F m = r ν ?? μ ?? ?? co 0 v= r μ () () 2 co esc. 0 v s=sE=0= = F 2a 2r M μ (linear distance travelled) F a= M ?? ?? ?? ν esc 0 T r esc. dr ds ?? ?? ?? co V v ? esc. 0 s r esc. 0 r r ν co 1 4 V 1- v ? ν 10 -2 8.9 0.63 0.75 50 0.89 0.79 10 -3 28 0.63 0.86 500 0.885 0.79 10 -4 88 0.63 0.92 5,000 0.88 0.80 10 -5 278 0.64 0.96 50,000 0.88 0.71 So, to a good approximation, () 1 esc 4 co V 1-0.79 v ? νnull esc 0 r 0.88 r ν null esc dr 0.63 ds ?? ?? ?? null Edelbaum’s Sub-Optimal Climb and Plane Change Instead of optimizing the tilt profile ( )α θ , Edelbaum (1961, 1973) just kept α constant during each orbit, then optimized ( )Rα . 16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 5 of 9 So, () + for - < < 22 = 3 - for < < 22 ππ? αθ ? ? αθ ? ππ ? αθ ? ? (31) We still have (Equation 1) 2 di F R =sin cos , dM α θ θμ but now (using 1α≡ ) 2 - 2 12 sin cos = sin cos d = sin π π θ αθ α θ θ α ππ ∫ 2 di 2 R F =sin dM α θπμ ∴ (32) Similarly, we still have 3 dR 2FR =cos dM α θμ (33) which needs no averaging. Dividing (33) by (32) and dropping the averaging sign, di tan = dR R α π (34) We also have dV FM = , dRdR dt ? ?? ?? ?? and now 3 2 1 2 dR F R 2cos, dt M = α μ 3 RdV 1 = dR 2 cos μ? α (35) 16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 6 of 9 To optimize ()R,α we again look for dV min. dR ? for a given di dR . For the Hamiltonian () () 3 i tan RR 1d H= + - 2cos R dR R ??α μ λ ?? απ ?? (36) The “control” variable is ,α the “stable variable” is i and the independent variable (replacing time) is R. The optimality conditions are i H =0 d H = dR i ? ? ? ??α ? λ ? ? ? ? ? (37) and the “transversality condition” 2 i 1 R i R =0 δ ?? λ ?? is satisfied automatically, because i is prescribed at both ends. From (37b), i d =0 dR λ i =λ constant (38) and from (37a), using 2 1 =1+tan , cos α α ii 3 2 1tan R - = 0 sin = 2RR 1+tan λ2λμα ?α ππμ α (39) Pending determination of λ i , (39) indicates that the thrust tilt amplitudeα increases over the mission, so that most of the plane-change activity is deferred to the last part of the climb, when the orbital velocity is lower. To find λ i , use the constraint on the total i ? . From (34), 2 1 R i R tan =dR R α ? π ∫ (40) From (39), 16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 7 of 9 2 2 i R= sin , 2 ??π μα ?? λ ?? so ()2d sindR 2d ==. Rsin tan α α αα Hence, 2 1 i 2 =d α α ?α π ∫ () i 2 = 21 ?αα π ? (41) A separate relationship between and 1 2 α αcomes from (39): 21 12 sin R v == sin R v 2 1 α α (42) where v 1 , v 2 denote the initial and final orbital velocities. Combining (41) and (42), 1 2 sin + i v = sin v 1 1 π?? α? ?? 2 ?? α ; 1 2 v cos i + cot sin i = v 1 ππ?? ?? ?α? ?? ?? 22 ?? ?? 1 2 v -cos i v cot = sin i 1 π ? 2 ?α π ? 2 2 11 22 sin i or sin = vv 1-cosi vv 1 π ? 2 ??α ?? π ? ?? 2 ?? ?? + (43) The most important quantity is the optimized V? . From (35), 22 11 RR 3 1dR1dR V= = 2cos2RcosR μμ ? αα ∫∫ () 2 2 1 1 R ii 2 R i 2211dd = = = cot - cot 2sincostan sin α 1 2 α λλλαα?????? α α ?????? πα α α π πα ?????? (44) 16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 8 of 9 Putting (from 39) i 1 2 vsin 1 λ =α π , and using, analog to (43), 2 1 v cos i- v cot = , sin i 2 π 2 α π ? 2 12 21 1 2 11 22 vv -cos i - cos i- sin i V=v vv sin i 1+ -2 cos i vv ????ππ π ?? ???? ? 22 ????2 ? π ?? ?π ? 2?? 2 ?? and simplifying, 22 12 12 V= v +v -2vvcos i π?? ?? ?? 2 ?? (45) Geometrically, ?V appears as the vector difference of 12v and v, except the angle between 12v and v is not the actual ?i, but i π ? 2 . The extra factor reflects the inefficiency NSSK ??1 ?? η ?? associated with thrusting through the full 180 null in each out-of-plane direction. Example Consider again LEO (400 Km) to GEO, with i=28.5? null ; 1 v = 7673 m/s, 2 v = 3072 m/s. From (46), 22 V = (7673) +(3072) - 2×7673×3072 cos 28.5 π?? ? ?? 2 ?? null 16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 9 of 9 V =5903 m/s? This is noticeably worse than the true optimum V = 5768 m/s? calculated for the case when ( )αθ is also modulated as tan cos α θ.～ The initial and final tilt angles are sin 28.5 ×3072 2 sin = = 0.3665 5903 1 π?? ?? ?? α null = 21.5 1 α null =+ i 2 21 π αα ? =66.3 2 α null These are smaller than the peak values MAX =30.5 1 α null , MAX 2 =72.2α null in this optimal case, but, of course, they are applied for the whole half-orbit.