02677-T-2 OPTIMAL LOW THRUST ESCAPE VIEWED AS A RESONANCE PHENOMENON* by Robert A. Jacobson and William F, Powers The University of Michigan Ann Arbor, Michigan May 1970 *Performed under NASA Grant NGR 23-005-329.

Abstract In this study a complete second order perturbation solution to a modified optimal low thrust escape problem is presented. The optimal thrust direction is shown to be tangential to first order, and oscillatory to second order with period equal to that of the initial circular reference orbit. The improvement of the optimal trajectory over a tangential thrust escape trajectory is shown to be a second order resonance type effect.

I. Introduction The problem of low thrust escape from an initial circular orbit has been studied by many researchers using a wide variety of methods. Escape using a specified thrust program such as tangential, circumferential, or radial has been studied from both a numerical and an approximate analytical viewpoint (References 1-12) Escape using an optimal control, determined by the calculus of variations, has also been solved numerically13-15 but little analytical work has been done in this area16 On the other hand, numerous studies of the optimal close-orbit transfer problem, both analytical and numerical, have been reported (References 12,17-23). In this study a modification of the problem of minimum time escape from an initial near circular orbit under low constant thrust acceleration will be considered. As a result of previous numerical studies, it is known that tangential thrust is near optimal, and that the optimal control angle exhibits an oscillatory behavior with a period near that of the osculating orbital period, and with a mean value near tangential. The main purpose of this analysis is to explain: (1) the relationship between the optimal and the tangential controls, and (2) the physical significance of the oscillatory behavior of the optimal steering angle. II. Formal Problem Definition The specific problem to be studied is as follows: given a space vehicle in an initial near-circular orbit with energy E0, find the control angle program which will take the vehicle to a specified energy level Ef in minimum time. The vehicle is assumed to be subject to a low constant thrust acceleration engine, the gravity field is inverse square, and all motion is confined to the initial orbital plane. The equations of motion are: = V siny 9 = - cosy r V = - siny + a cos c = (V- _ -I) cosy + - sincf r rV V 1

2 where r = radial distance o = polar angle V = total velocity magnitude y = flight path angle measured from the local horizontal ( = control angle measured from the velocity direction a = thrust acceleration as shown in Figure 1. - T/m = a reference direction Figure 1 The optimal control program is obtained by application of the calculus of variations, where the performance index and the Hamiltonian are J - tf - to 11 = \V siny + kz r cosy - \3 siny + 4 -Zv cos + a(\3 cos4 + -+ sin), and the optimal control is defined by sin4 = X4+ 4.V C O.S + /= Vcos, -- X3V I\X + X2 V2 4 3

3 The multiplier equations are x X = x2 s COSy - 2X37 siny + X(rf - r /2 =0 \3 = -\ siny - Xz cos -X4 r + cosy+ a V sinq~ \4 = -1Vcos y + X2-sinNy +V3 cos +\4(y - - siny The boundary conditions are to - 0, V (tf) - = E 2 f r (tf) f r(0) = Ro X (tf) = 0 0 (0) =, 4 (tf) = 0 V(0) =,t = Ro0o, H(tf) -= 1 2a t 3 (tf) (' of r (tf)V(tf) 0 where the last four terminal conditions are given by the transversality conditions, and the particular choice of the initial flight path angle will be explained later. III. Approximate Analytical Solution Since 0 does not influence the problem and X2 (t) 0, only the sixth order system defined by (r,v,y, X, X3, 4 ) will be considered. A solution in the form of an expansion in powers of a, the thrust acceleration, will be assumed as follows: r = ro + ar + a2rz + * * V = Vo + aV1 + azV2 +''. Y = o + ayl + a22 +''' X = ak11 + aZX12 +\3 = a31 + a 3 2 + ~* X4 = aX41 + a2X42 + * sinc = po + apl + a2Pz +''' Cos c = ao + aal + a2a2 +'''

4 Ilia. Zero order state variable solutions After substituting the above expansions into the system of equations and grouping terms in powers of a, the zero order state equations are ro = Vo sinyo VO = -2 sinyo ro 1 = V^ IL ) O Vo ro rZ%/su The solution of these equations subject to the given initial conditions is ro(t) = Ro Vo(t) = Rolo Yo(t) = o That is, the zero order solution is a circular orbit of radius Ro. IIIb. First order state and multiplier solutions The first order system of equations obtained from the expansions is rl = V1 sinyo + Voy1 cos yo pv = EfL sinyo - ylcosYo) + a0 Y= = ( - Vor 2 ) cosNo 1 V (m) (Y_ sin( o coso) ro2 ro r V 1 X. 1 Z- V( --- + V_) \o \41 = - Vo-II + o31 + Po V0 roo - - + 2- )4 L KIvo j A3 Vo L o Vl + / J Since the multiplier equations are independent of the first order state equations, they may be solved easily when the zero order state solution is known. The general solution is

5 X 11 (t) = C2. + e, Cs (Woit + IP) X31=(t) 0 + 2cco(tf3 X 41 (t) = Ci1 sin (wot + f3) From the expansion first order at tfth at of the terminal conditions in powers of a we find to XI I(tf) - UjoX3 I(t f) = 0 41tf = 0 X31 (t )=I? which implies XI, (t)::- a2 a2 X 41 (t = 0 It follows immediately that a0O. 1 Po = 0 005- cs = =1 s in c = 0 Therefore., the optimal "escape" control gential thrust. program to first order is tan The first order state equations may now be solved using the tangential control program. The general first order state solution is r, (t) = 3+ - t + Roc, sinwot - Rocz cos ujot V1 (t) =- 1 -C3- t + Rowoc2 COS WO~t - Row0c 1 sin wot 2 I~' (t) = R002 + c 1 COS wt + c2 sin w0t, The initial conditions for this system are r, (0) =0 V1 (0) =0 2 -Y (0)

6 which implies rl(t) = 2 t V1 (t) = -t 2 Y 1 (t) R= T 0 It should be noted that the above set of initial conditions, which correspond to an orbit of low eccentricity, was first suggested by Lawden'6 in an effort to simplify the higher order solutions by elimination of the oscillatory first order motion of the escape spiral. IIIc. Second order state and multiplier solutions After substitution of the zero order solution into the second order equations, we have =i2 = Ro0oY2 + YV1 V2 = -Rooz 2 + 2woz r 1 21 3 2wo 2 1 = O - r2 + 1 2 -3 - 2o- + 1 RO RO R 0o ro R0r Rowo 12 = ---- X42 - 2w02Y 131 Ro )32 = - X42 - XllYl Ro )[42 = - Rowo 12 + Roo2X3 2 - V X1 - 2o2 X31 r 1 4= _42 x31Vo The multiplier equations are independent of the state equations as in the first order case, and may be solved easily when the first order solutions are known. The general solution is X 2 (t) = woA2 - zt + osot+ P) X32(t) = A + Cos (ot + X42 (t) - — o + A1 sin (Wot + p) The second order terminal conditions obtained from the expansion of the

7 boundary conditions in powers of a are k12 (tf) = Co(X3z - 31 ro - X3I ) =V X32 (tf) - 0 X42 (tf) = 0 which implies A1 sin (cotf + p) = > 0 A2 + A cos (oot + ) = 0 Aj cos (wootf + P) = 0 The second order multiplier solution is then () 3 1 sinO(t-tf) X32 (t) = - Row2a si (t- tf) X42 (t) - -c a2 [l-osCOo(t-tf)] From the control angle expansion given in the Appendix, it follows that tan4 - sin = - R [1- cos o (t - tf) ] The optimal control angle is, therefore, oscillatory with frequency o and amplitude of order a. The second order state equations may now be solved using the oscillatory control program. The complete solution with zero initial conditions on the second order state is 18 - cos w0tf r2 (t) - R z tZ 4 (1 - cos 0t) R 0a)0 2Rwoo4 + 4 [cos oo (t - tf) + wot sin wco(t - tf) - cos Ootf] 18 - cos oOotf V2 (t)= 3 1 - Cos Wt) - 1 [cos o(t- tf) + ot sincoo(t-tf) - cos otf] R~owo 3IIf

8 18 t 18-cos wotf yz 2t) = R-23t + t cosO(t-t) - ( ZR2 4o sin Ot In the above expressions it can be seen that "resonance" type terms of the form t sint and t cost have been introduced by the control input which oscillates at the natural frequency of the second order solution. It will be shown that it is precisely these terms which make the optimal "better" than the tangential escape trajectory. We now have a complete second order expansion for the state variables and multipliers of the optimal control problem as defined in section II. The only remaining unknown is the final time which may be found by application of the terminal energy condition. Clearly, the solution as given will not hold to escape, i. e., Ef 0, since the various terms in the expansion will become large and invalidate the assumed solution form. But if Ef is near Eo, then the solution should be accurate. Furthermore, the control angle for this energy increase problem should behave like that in the initial portion of the escape trajectory. IV. Energy Increase Comparison The rate of change of energy is E = aV cosc The small parameter expansion form of this equation is E = a[VOaO + a(Voal + Vlao) + a2 (V2 ao + a2 Vo + aVl) ] since V = Vo + aV1 + a2V2 cos 4 = a0 + aac + a2a2 From the solution in Section III and the angle expansion given in the Appendix, it follows that oQ = 1 a1 = 0 (optimal) 2 2 1 [1- coso(t- tf)] no -) 2

9 For the tangential thrust program cos ) -1, which implies ao = 1 al - (tangential) a2 = 0 The rates of change of energy on the optimal and the tangential trajectories are then E [pt = a[Vo + aV1 + a2 (V2 + Voaz)]opt E ltan = a[Vo + aVl + a2z ] tan Since there is no difference between the optimal and the tangential trajectories in the zero and first order solutions, Vlopt t Itan 1 lopt 1 ltan The second order expression for the velocity along the tangential trajectory can be determined easily. Since sin c = 0 along the tangential trajectory, then Pi -0. After substituting pl = 0 into the second order equations of motion above, the second order tangential state solution can be obtained. The second order tangential velocity is 8 Vz [ _0 —0- (1- cos w0t) Upon integration of the E-equations, the energy changes along the two trajectories at any time are AEo a2 oEIoptt -- atRw + 4Roo4 [(33 + 2 cos wo(t - tf))ot - 36 sin cot + 2 cos ootf sin ot - - sin 2oo(t - tf) - - sin Zootf]} AEItan = a{Root- at2 + a[ 32 ot- 32sinwot]}

10 The energy difference between the two trajectories at any time is a3 r A(AE) = 4R 1 + 2cos O (t - tf)] wot - 4sin wot + 2cos cotf sina ot - sin 2o (t - tf) - s in 2cootf At the final time A(AE) f 4 1 [3cotf +.sin 2cootf - 4sinootf] It can be shown that A(AE)f is positive for all wootf. In fact for small oootf the series expansion of the trigonometric expressions can be used to show that A (AE) f a (ot )5 Ef =40R0004 f Therefore, at the time the optimal trajectory reaches the specified terminal energy, the energy level is higher than the energy level at the corresponding time on the tangential trajectory. This implies that the optimal trajectory will reach a specified terminal energy level faster than the tangential trajectory. It must be noted, however, that at intermediate times the energy on the tangential may be greater than on the optimal. In other words, the tangential trajectory may reach intermediate energy levels sooner than the optimal. This phenomenon should not be entirely unexpected since the minimum time trajectory to a given energy level is not the minimum time trajectory to all lower energy levels. It is only the minimum time trajectory from the initial state to all of the states occurring along the optimal. It is well known that the tangential thrust program at each instant maximizes the rate of energy increase along a trajectory5. Therefore, it is reasonable to ask how the optimal manages to improve on the tangential. Since it has been shown that the optimal is tangential to first order, higher-order terms must produce the difference. In looking at the second order solution, we find that the difference between the tangential and

11 optimal velocities may or may not be positive at any given time. However, the difference between their mean values taken over a revolution-to-go, from t = tf - (2N+ 2)Tr/o to t = tf - 2Nir/o, is greater than zero, i. e., V2 AV = 20 — > 0. V opt tan 2Rvw 0 We conclude then that on the average the velocity is higher along the optimal trajectory. If the tangential velocity is compared to the component of the optimal velocity in the optimal thrust direction, the mean value of the optimal velocity component is also found to exceed the mean value of the tangential velocity, i.e., V cos - -V = 3a > 0 opt tan 4Ro > Therefore, not only does the optimal velocity exceed the tangential velocity on the average, but also its component in the direction of thrust exceeds the tangential velocity. Recalling that the rate of energy increase depends only upon the thrust acceleration and the velocity component along the thrust vector, we see that the optimal improves on the tangential by maintaining a higher velocity component in the direction of thrust. The key to the higher velocity on the optimal is the existence of the "resonance" type terms in the optimal velocity expression which have been introduced by the control angle oscillations. Considering the motion from a physical viewpoint, on the escape spiral the low thrust engine does work on the spacecraft causing its energy to increase. The vehicle spirals outward increasing its potential energy and decreasing its kinetic energy. The rate of energy increase depends highly upon the vehicle's velocity, i. e., its kinetic energy, and therefore decreases as the vehicle moves out. The tangential thrust program maximizes the rate of energy increase at each point along the trajectory but makes no direct effort to control the vehicle's velocity. The optimal thrust program, on the other hand, causes the energy to increase in such a way that the rate of increase of potential energy and rate of decrease of

12 kinetic energy are reduced. The higher kinetic energy on the optimal then gives the vehicle more capability for increasing its energy as it moves out. The vehicle uses this additional capability in the latter portion of the trajectory to add more energy than it could on a tangential thrust trajectory, and in this way achieves escape in less time. The oscillation in the optimal control angle is a result of the trade-off between keeping the rate of energy increase high and the rate of kinetic energy decrease low. V. Numerical Results In order to test the accuracy of the approximate analytic solution, a comparison was made with an exact optimum energy increase trajectory. The exact solution was generated by numerically solving the two point boundary value problem using a secant iteration method. The initial values of the analytic multipliers were used as first guesses in the iteration scheme and seemed to work quite well. The initial values for the state variables were r(0) = 6.67817 X 106 meters V(0) - 7.72580 X 106 meters/sec y(0) = 2. 19518 X 102 radians These correspond to an initial orbit with the following eccentricity and energy -3 eo = 2.0 X 10 EO - -2. 98440 X 107 newton-meters/kg The specified terminal energy was E = -2.86218 X 107 newton-meters/kg which corresponds to otf - 6-rr in the analytic solution. Since W0 = 1.15687 X 10, the analytic tf = 1.629353 X 104 seconds. The terminal time found in the numerical solution was tf = 1. 629351 X 104 seconds. In Figures 2 and 3 a comparison between the analytical and numerical solutions is made. The state variables are in close agreement, as are the costate variables, with the primary differences appearing as a slight

13 mean value offset in the velocity costate, and as a small period discrepancy in both the velocity and flight path angle costates. The optimal control angles are also close with only slight differences in period and amplitude. A solution was also carried out using a terminal energy of Ef = -2. 58695 X 107 newton-meters/kg In the analytic solution this energy level occurs at 0otf = 20-rr, or tf = 5.43117 X 104 seconds. Numerical results give a final time, tf = 5.43116 X 104 seconds; and again the analytical and numerical state and costate variables differed only slightly. The determination of the full limitations of this approximation are currently under study. VI. Conclusions As a result of this analysis we reaffirm the well known fact that for low thrust spiral escape trajectories, tangential thrust is nearly time optimal, and in fact, is optimal to first order in the thrust-acceleration expansion solution. In addition, we now conclude that the observed oscillation in the optimal control angle is a second order resonance type phenomenon which reduces the velocity loss and therefore increases the rate of energy gain along the trajectory.

References 1. Tsien, H S., "Take-Off from Satellite Orbit, " Journal of the American Rocket Society, Vol. 23,No. 4, July-August 1953. 2. Benney, D. J., "Escape from a Circular Orbit Using Tangential Thrust, " Jet Propulsion, Vol.28, No.3, March 1958. 3. Dobrowolski, A., "Satellite Orbit Perturbations Under a Continuous Radial Thrust of Small Magnitude, " Jet Propulsion, Vol. 28, No. 10, October 1958. 4. Lass,H., J. Lorell," Low Acceleration Takeoff from a Satellite Orbit," ARS Journal, Vol. 31 No. 1, January 1961. 5. Cohen, M. J., "Low Thrust Spiral Trajectory of a Satellite of Variable Mass," AIAA Journal, Vol. 3, No. 10, October 1965. 6. Ting, L., S. Brofman, "On Take-off from Circular Orbit by Small Thrust," Z. Angew Math. Mech.,44, 1964. 7. Zee, Chong-Hung, "Low Constant Tangential Thrust Spiral Trajectories, AIAA Journal, Vol. I, No. 7, July 1963. 8. Okhotsimskii, D.E., "Investigation of Motion in a Central Field under the Influence of a Constant Tangential Acceleration," Cosmic Research, Vol. 2, No. 6, November-December 1964. 9. Shi, Y.Y., M. C. Eckstern, "Ascent or Descent from Satellite Orbit by Low Thrust, "AIAA Journal, Vol.4, No. 12, 1966. 10. Perkins, F. M., "Flight Mechanics of Low Thrust Spacecraft," Journal of the Aerospace Sciences, May 1959. 1 1. Moeckel, W. E., "Trajectories with Constant Tangential Thrust in Central Gravitational Fields," NASA Technical Report R- 53, 1960. 12. Melbourne, W.G., "Interplanetary Trajectories and Payload Capabilities of Advanced Propulsion Vehicles," JPL Technical Report 32-68, January 1961. 13. Efimov, G. B., D.E. Okhotsimskii, "Optimal Acceleration of a Spacecraft in a Central Field," Cosmic Research, Vol. 3, No. 6, NovemberDecember 1965. 14. Sherman, B., "Low Thrust Escape Trajectories," Proceedings IAS Symposium on Vehicle Systems Optimizations, New York, 1961.

15. Irving, J. H., "Low Thrust Flight: Variable Exhaust Velocities in Gravitational Fields," Space Technology, 10-01-10-54, (H. Siefert, ed.) John Wiley and Sons, New York, 1959. 16. Lawden, D. F. "Optimal Escape from a Circular Orbit, " Astronautica Acta, 4, pp. 218-233, 1958. 17. Gobetz, Frank W., "Optimal Variable-Thrust Transfer of a PowerLimited Rocket between Neighboring Circular Orbits," AIAA Journal, Vol. 2, No. 2, February 1964. 18. McIntyre, John E., Luigi Crocco, "Linearized Treatment of the Optimal Transfer of a Thrust-Limited Vehicle between Coplanar Circular Orbits," Astronautica Acta, Vol. 12, No. 3, 1966. 19. McIntyre, John E., Luigi Crocco, "Higher Order Treatment of the Optimal Transfer of a Thrust-Limited Vehicle between Coplanar Circular Orbits," Astronautica Acta, Vol. 13, No. 1, 1967. 20. Edelbaum, T. N., "Optimum Power-Limited Orbit Transfer in Strong Gravity Fields," AIAA Journal, Vol. 3, No. 5, May 1965. 21. Edelbaum, T.N., "An Asymptotic Solution for Optimum PowerLimited Orbit Transfer, AIAA Journal, Vol.4, No. 8, August 1966. 22. Hinz, H.K., "Optimal Low-Thrust Near-Circular Orbit Transfer," AIAA Journal, Vol. 1, No. 6, June 1963. 23. Melbourne, W.G., Carl G. Sauer, "Optimum Thrust Programs for Power-Limited Propulsion Systems," JPL Technical Report 3 2-118, 1961.

Appendix In the solution of the systems of equations it was found easier to consider separate expansions of the control angle functions rather than to use the multiplier expressions directly. The optimal control program is defined by tan X4= X3V By assuming expansions of the form V = Vo + aV1 + a2V2 +''' \3 = aX31 + a2X32 + a3\33 +... \4 = a\41 + aX42z + a3\4 +... we obtain tan = 0 + arl + a2r2 + ** where " 41 X31 Vo 42 - X1 4 41 r 4V + --- X31 Vo X 31 1 r XA9( X32 X A32V 32 +V1 X33 V2^ T _1 V0 k43 - -42+- ----- +_41 -3 V o 2 J X 31V L Z O 31 ) 41 \3lVo +X3 V0 \31 VoJ1 Next consider sin _ = cos A =V which can be written as 1 cos + = [1 + tan2]-] a sin c = tanc cos c Substituting in the expression for tan 4 and expanding in a Taylor's series about a = 0: cos = a + aal + a a2 +.. where

a0 = (1 +n) 3 5 3 ~[3= Tj (i + o) 2 = M 1311 (1 + ) - + 2no2z ) (1 + ] Using the expansions for tan 4 and cos 4 we obtain sin c~ = Po + ap a + a *2 +' where 1 Po = no(1 + no) 2 Pi = q1 (1 +n) -2 _- no (1 + no) = (1 + - (- 3 + 2 ) (1 + + (1 + I

Figure 2. Radial Distance., Total Velocity, and Their Corresponding Costates 6. 0 x 0 Cd Cd cld 0 Cd 'H Cd Cd t12 0 U1 * 12( * 11! Exact -. Analytic -.-S 6. 6. Exact — Analytic 4-8-.111 4175 10,175 16,175 Time (Secs) 4175 10,175 16,175 Time (Secs) 0 C.) 0. 0 Exact-0 Analytic -8 - 7. 70 7. 7. 102. C.) 0 Cd o 102. 0 a) 0 (9 101. Exact -.9 -Analytica -8 - 4175 10.,175 16,1175 Time (Secs) 4175 10,175 Time (Secs) 16,175

Figure 3. Flight Path Angle, its Costate, and the Control Angle Time (Secs) 4175 10,175 16,175 N + 0 x r — b.CO r-4.24.23,4 0 '-I.) 0.r U -1 --4 --7 --10 --13 --16 Exact -GAnalytic -s-.22 4175 10,175 16,175 Time (Secs) Exact -- Analytic -a Time (Secs) 5 10,175.00 4-,i 0 4U -.03 -.06 -.09 -. 12 -.15