6.522, Space Propulsion Prof. manuel martinez-sanchez Lecture 3: Approximate Av for Low-Thrust Spiral climb Assume initial circular orbit atV=V Thrust is applied tangentially. By conservation of energy, assuming the orbit remains near-circular v 2r2 dt dr= a dt When we integrate =△V The result appears to be trivial, but it is not. Notice that the"velocity increment"Av 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 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 Av would be l(-a)dt, so the result in general is 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 1 of 9
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 ⎜ ⎟ ⎝ ⎠ , d - a dt 2r r ⎛ ⎞ µ µ ⎜ ⎟ ⎝ ⎠ 2 dr a 2r dt r µ µ -3 1 2 2 r dr a dt 2 µ When we integrate, bt 0 a dt = V, ∆ ∫ and so 1 -1 tb 2 2 0 - r =V µ ∆ ( ) 0 b V= - r rt µ µ ∆ or final ∆V=v -v co c (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 bt 0 (-a) dt ∫ , so the result in general is ∆ ∆ V= vc (2)
F For simplicity, assume now a=M constant, Which is actually optimal for many situations. Equation(1)can be recast as and solving for r (3) I、at This shows how the radial distance spirals out"in time. In principle this says r,oo at t=-co, 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 Av as follows The radial velocity r can be calculated from( 3) by differentiation. Notice that this is in the nature of an iteration since r was implicitly ignored in the energy balance which led to(3. We obtain (4) The tangential component v= r e is still approximated as the orbital velocity, i. e re= at =v 1-at The overall kinetic energy is therefore V =r+re The escape point is defined by its having zero total energy, i.e. 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 2 of 9
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 r r µ µ and solving for r, 0 0 2 2 co 0 r r 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 vco t = a , a crude indication of “escape”. But of course, the orbit is no longer “nearcircular” 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 = 1 - at v ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ i (4) The tangential component v = r θ θ i is still approximated as the orbital velocity, i.e., co 0 co at r = = - at = v 1 - rr v µ µ ⎛ ⎞ θ ⎜ ⎟ ⎝ ⎠ i (5) The overall kinetic energy is therefore 2 0 2 2 2 2 2 2 co co 6 co co ar 4 v at v v =r + r = 1- + 2 2v at 1 - v ⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎜ ⎟ ⎛ ⎞ ⎛ ⎞ ⎝ ⎠ ⎜⎟ ⎢ ⎥ θ ⎜ ⎟ ⎝ ⎠ ⎢ ⎥ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎣ ⎦ ⎝ ⎠ i i (6) The escape point is defined by its having zero total energy, i.e
μ 2 r Substituting at at aro /v20 1-a (7) where % is the ratio of thrust to gravity, a small number for us. Since at=AV, Equation(7)reads (2 In a more rigorous analysis the factor 2 4= 1. 19 turns out to be actually 0.79 ∧-sa 1-0.79v (more exact) 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 3 of 9
16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 3 of 9 2 e e v = 2 r µ , or 2 e 0 co e 1 v r - = 0. 2v r ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Substituting, 2 0 2 2 2 co 6 co co co ar 2 1 at at v 1 10 2v v at 1 v ⎛ ⎞ ⎜ ⎟ ⎛⎞ ⎛⎞ ⎝ ⎠ ⎜⎟ ⎜⎟ − + −− = ⎝⎠ ⎝⎠ ⎛ ⎞ ⎜ ⎟ − ⎝ ⎠ or ( )2 2 2 0 co 6 co co 4 ar v at = 1- v at 1 - v ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ ( ) 1 4 1 4 1 0 4 2 co co 2 0 at 2a 2ar 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 V v 1- 2 esc. co ⎡ ⎤ ∆≅ ν ⎢ ⎥ ⎣ ⎦ (9) In a more rigorous analysis, the factor 21/4 = 1.19 turns out to be actually 0.79: 1 4 V = v 1 - 0.79 esc. co ⎡ ⎤ ∆ ⎢ ⎥ ⎣ ⎦ ν (10) (more exact)
oW THRUST TO ESCAPE: RESULTS OF NUMERICAL COMPUTATIONS DEFINITIONS: Sesc =S(E=0 (linear distance travelled) △V ds丿 0 06 0.89079 28 063 0.86 0.8850.79 LO 0.63 0.80 10 78 0.6 09650,0000.88071 to a good approximation 1-0.79 0.88 =0.63 Edelbaum's Sub-Optimal Climb and Plane Change Instead of optimizing the tilt profile a(0), Edelbaum(1961, 1973)just kept J a constant during each orbit, then optimized |a I(R) 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 4 of 9
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 =s E=0 = = 2r F 2a 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 ∆ ν esc 0 r 0.88 r ν esc dr 0.63 ds ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 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 α (R )
<0< We still have(Equation 1) d i sin a cos e but now(using a=1) (sin a cos e)=+sin a 2 cos ede==sin a s de Similarly we still have dR 2FR cos夏 which needs no averaging Dividing(33) by(32)and dropping the averaging sign d i tan a We also have d△V_F/M dt F cOS d△V1√/R dr 2 cos a 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 5 of 9
16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 5 of 9 So, ( ) + for - < < 2 2 = 3 - for < < 2 2 ⎧ π π α θ ⎪⎪ α θ ⎨ π π ⎪ α θ ⎪⎩ (31) We still have (Equation 1) 2 di F R = sin cos , d M α θ θ µ but now (using α ≡ 1 ) 2 -2 1 2 sin cos = sin cos d = sin π α θ α θ θ α θ π π π ∫ 2 di 2 R F = sin d M α θ πµ ∴ (32) Similarly, we still have 3 dR 2FR = cos d M α θ µ (33) which needs no averaging. Dividing (33) by (32) and dropping the averaging sign, di tan = dR R α π (34) We also have d V FM = , dR dR dt ∆ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ and now 3 2 1 2 dR F R 2 cos , dt M = α µ 3 dV 1 R = dR 2 cos ∆ µ α (35)
To optimize a(R), we again look for min dv for a given d For the hamiltonian H √R di tan a(r) 2 cos a(r)'dE The control"variable is a, the stable variable"is i and the independent variable (replacing time) is R The optimality conditions are dRai」 and the "transversality condition is satisfied automatically, because i is prescribed at both ends. From (37 λ= constant and from(37a), using-=1+tan a, ta 2VR五+tFaR=0sma (39) Pending determination of Ai,(39)indicates that the thrust tilt amplitude a 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 a use the constraint on the total a From (34), tana From(39), 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 6 of 9
16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 6 of 9 To optimize α ( ) R , we again look for d V min. dR ∆ for a given di dR . For the Hamiltonian ( ) ( ) 3 i 1 di R tan R H= + - 2 cos R dR R µ ⎛ ⎞ α λ ⎜ ⎟ α π ⎝ ⎠ (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” i 2 1 R i R = 0 δ ⎡ ⎤ λ ⎣ ⎦ is satisfied automatically, because i is prescribed at both ends. From (37b), d i = 0 dR λ λi = constant (38) and from (37a), using 1 2 = 1+ tan , cos α α i i 3 2 1 tan R - = 0 sin = 2 R R 1+ tan µ α λ 2λ ⇒ α α π πµ (39) Pending determination of λi , (39) indicates that the thrust tilt amplitude α 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 λi , use the constraint on the total ∆i . From (34), 2 1 R i R tan = dR R α ∆ π ∫ (40) From (39)
R R Hence ,= A=2(a2-a1) (41) A separate relationship between a, and a, comes from (39): sIn where vi, v2 denote the initial and final orbital velocities Combining(41)and(42) al 1-C0s-Ai sn-△ In a, sin -Al VcOS I Ai The most important quantity is the optimized Av From(35), a。-假 dR/R 2入1 2dα (cot a, -cot a,) 4 rsna八(cos TaSnα 16.522, Space Propulsion Lecture 3 Prof. Manuel martinez-Sanchez Page 7 of 9
16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 7 of 9 2 2 i R = sin , 2 ⎛ ⎞ π µ α ⎜ ⎟ λ ⎝ ⎠ so dR 2d 2d sin ( ) = =. R sin tan α α α α Hence, 2 1 i 2 = d α α ∆ α π ∫ i ( ) 2 ∆ αα = 2 1 π − (41) A separate relationship between α1 and 2 α comes from (39): 2 1 1 2 sin R v = = sin R v 2 1 α α (42) where v1, v2 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 ⎛⎞ ⎛⎞ π π ∆ α∆ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ 2 2 1 2 v - cos i v cot = sin i 1 π ∆ 2 ⇒ α π ∆ 2 2 1 1 2 2 sin i or sin = v v 1 - 2 cos i v v 1 ⎛ ⎞ ⎜ ⎟ π ∆ 2 α ⎛ ⎞ π ⎜ ⎟ ∆ 2 ⎝ ⎠ ⎝ ⎠ + (43) The most important quantity is the optimized ∆V . From (35), 2 2 1 1 R R R R 3 1 dR 1 dR R V= = 2 cos 2 R cos R µ µ ∆ α α ∫ ∫ ( ) 2 2 1 1 R i i R 2 = = = cot - cot 1 1 2d d 2 22 i 2 sin cos tan sin α 1 2 α ⎛ ⎞ λ λλ ⎛ ⎞⎛ ⎞ α α ⎜ ⎟⎜ ⎟⎜ ⎟ α α ⎝ ⎠ πα α α π π ⎝ ⎠⎝ ⎠ α ∫ ∫ (44)
Putting(from 39) 2A1 =V;snα and using, analog to(43) COS sin-△i 1-cos△i-cos△i AV=V 2--1C0S=Ai and simplifying △V=2+v2-2 v, vcos△i (45) Geometrically AV appears as the vector difference of Vi and v2, except the angle between V, and v2 is not the actual Ai, but I AL. The extra factor reflects the associated with thrusting △i through the full 180 in each out-of-plane direction Example Consider again LEo (400 Km)to GEO, with Ai=28.5° v1=7673m/ V2=3072m/s. From(46), △V=(7673)2+(3072)-2×7673×3072c0s28.5 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 8 of 9
16.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 8 of 9 Putting (from 39) i 1 2 v sin 1 λ = α π , and using, analog to (43), 2 1 v cos i - v cot = , sin i 2 π 2 α π ∆ 2 1 2 2 1 1 2 1 1 2 2 v v - cos i - cos i - sin i v v V=v v v sin i 1+ - 2 cos i v v ⎛ ⎞⎛ ⎞ π π π ⎜ ⎟⎜ ⎟ ∆ ∆ ∆ 2 2 2 ⎝ ⎠⎝ ⎠ ∆ π ⎛ ⎞ π ∆ ⎜ ⎟ ∆ 2 2 ⎝ ⎠ and simplifying, 2 2 V = v + v - 2v v cos i 1 2 12 ⎛ ⎞ π ∆ ∆ ⎜ ⎟ ⎝ ⎠ 2 (45) Geometrically, ∆V appears as the vector difference of v and v 1 2 , except the angle between v and v 1 2 is not the actual ∆i, but i π ∆ 2 . The extra factor reflects the inefficiency NSSK ⎛ ⎞ 1 ⎜ ⎟ ⎝ ⎠ η associated with thrusting through the full 180D in each out-of-plane direction. Example Consider again LEO (400 Km) to GEO, with ∆i=28.5D ; v = 7673 m/s, 1 v = 3072 m/s. 2 From (46), 2 2 V = (7673) + (3072) - 2×7673×3072 cos 28.5 ⎛ ⎞ π ∆ ⎜ ⎟ ⎝ ⎠ 2 D
=5903m/s This is noticeably worse than the true optimum Av=5768 m/s calculated for the case when a(e)is also modulated as tan a -cos 0 The initial and final tilt angles are x285|×3072 =0.3665 5903 a1=215° a2 =a,+=Ai a2=66.3° These are smaller than the peak values a,Mx=30.5, arMax=72. 2 in this optimal case,but, of course, they are applied for the whole half-orbit 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 9 of 9
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 α θ( ) 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 D α1 = 21.5D =+ i 2 2 1 π αα ∆ α2 = 66.3D These are smaller than the peak values MAX α1 = 30.5D , 2MAX α = 72.2D in this optimal case, but, of course, they are applied for the whole half-orbit
