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)