Tangent-Impulse Transfer from Elliptic Orbit to an Excess Velocity Vector

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provided by Elsevier - Publisher Connector Chinese Journal of Aeronautics, (2014),27(3): 577–583

Chinese Society of Aeronautics and Astronautics & Beihang University Chinese Journal of Aeronautics

Tangent-impulse transfer from elliptic orbit to an excess velocity vector

Zhang Gang a,b,*, Zhang Xiangyu a, Cao Xibin a,b

a Research Center of Satellite Technology, Harbin Institute of Technology, Harbin 150080, China b National and Local United Engineering Research Center of Small Satellite Technology, Changchun 130033, China

Received 6 May 2013; revised 3 July 2013; accepted 14 October 2013 Available online 22 April 2014

KEYWORDS Abstract The two-body orbital transfer problem from an elliptic parking orbit to an excess veloc- Elliptic orbit; ity vector with the tangent impulse is studied. The direction of the impulse is constrained to be Escape trajectory; aligned with the velocity vector, then speed changes are enough to nullify the relative velocity. First, Excess velocity vector; if one tangent impulse is used, the transfer orbit is obtained by solving a single-variable function Orbital transfer; about the true anomaly of the initial orbit. For the initial circular orbit, the closed-form solution Tangent impulse is derived. For the initial elliptic orbit, the discontinuous point is solved, then the initial true anomaly is obtained by a numerical iterative approach; moreover, an alternative method is proposed to avoid the singularity. There is only one solution for one-tangent-impulse escape trajectory. Then, based on the one-tangent-impulse solution, the minimum-energy multi- tangent-impulse escape trajectory is obtained by a numerical optimization algorithm, e.g., the genetic method. Finally, several examples are provided to validate the proposed method. The numerical results show that the minimum-energy multi-tangent-impulse escape trajectory is the same as the one-tangent-impulse trajectory. ª 2014 Production and hosting by Elsevier Ltd. on behalf of CSAA & BUAA. Open access under CC BY-NC-ND license.

1. Introduction circular orbit, an approximate analytical solution was obtained for the minimum-energy three- and four-impulse transfers The two-body orbital transfer problem from a parking orbit to between a given circular orbit and a given hyperbolic velocity 1 a given excess velocity vector is a fundamental one in space vector at inﬁnity. For the transfer from the initial circular exploration. The minimum-energy trajectory optimization for orbit to an excess velocity vector, the two-impulse escape is this problem has been studied for many years. For the initial never simultaneously of lower cost than either the one- or three-impulse.2 Recently, Ocampo et al.3,4 studied the one- and three-impulse escape trajectories, which can be con- * Corresponding author. Tel.: +86 451 86413440 8308. structed to serve as initial guesses for determining constrained E-mail addresses: [email protected] (G. Zhang), zxyuhit@ optimal multi-impulsive escape trajectories. For the initial gmail.com (X. Zhang), [email protected] (X. Cao). elliptic orbit, the optimal three-impulse transfer between an Peer review under responsibility of Editorial Committee of CJA. elliptic orbit and an escape asymptote was solved.5 Moreover, a simple numerical technique was proposed to minimize the time-of-ﬂight for the multi-impulse transfer from an arbitrary Production and hosting by Elsevier elliptic orbit to a hyperbolic escape asymptote.6

1000-9361 ª 2014 Production and hosting by Elsevier Ltd. on behalf of CSAA & BUAA. Open access under CC BY-NC-ND license. http://dx.doi.org/10.1016/j.cja.2014.04.006 578 G. Zhang et al.

The existing methods for the impulse escape trajectory opti- vf ¼ v0 þ Dv ¼ðv0 þ kÞv0=v0 ð1Þ mization problem do not include any constraint on the impulse 7,8 where rj = krjk, vj = kvjk, j = 0, f, and 0 and f denote the ini- direction. If the impulse direction is aligned with the velocity tial and final orbits, respectively. A relationship between the vector, the impulse is called ‘‘tangent impulse’’ and a speed magnitudes of velocity and position vectors is change will finish the impulse maneuver to nullify the relative sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi velocity. The tangent orbit problem has also existed for many 2 1 years. The classical Hohmann transfer is a cotangent transfer vj ¼ l ð2Þ rj aj and is the minimum-energy one among all the two-impulse transfers between coplanar circular orbits and between copla- where l is the standard gravitational parameter, and a the nar coaxial elliptic orbits.9 The cotangent transfer problem semimajor axis. Then, the magnitude of the velocity can be between coplanar noncoaxial elliptic orbits has aroused con- written as a function of the initial true anomaly, siderable interest in recent years. The numerical solution was rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 10 l 2 obtained based on the orbital hodograph theory. Moreover, v0 ¼ ð1 þ e þ 2e0 cos u Þ ð3Þ p 0 0 the closed-form solution was obtained by using the geometric 0 11 12 þ characteristics and by the flight-direction angle, respec- where u is the true anomaly, u1 the true anomaly at infinity of tively. The latter reference also gave the closed-form solution the excess hyperbolic orbit, e the eccentricity, and p the semil- for the solution-existence condition. In addition, Zhang and atus rectum. Zhou13 studied the tangent orbit technique in 3D based on a From the energy equation, the semimajor axis of the final new definition of orbit ‘‘tangency’’ condition for noncoplanar hyperbolic orbit is orbits. Different from the orbital transfer problem, the orbital l ÀÁ rendezvous problem requires the same time-of-flight for both af ¼ 2 ð4Þ vþ the chaser and the target. Zhang et al. solved the two-impulse 1 rendezvous problem between two coplanar elliptic orbits with þ where v1 is the excess velocity vector. Substituting Eq. (4) into only the second impulse14 and both impulses being tangent,15 Eq. (2) gives respectively. Furthermore, Zhang et al.16 solved the two- sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁ ÀÁ 2 2l 2 2l impulse rendezvous problem between coplanar elliptic and þ þ vf ¼ v1 þ ¼ v1 þ ð1 þ e0 cos u0Þ ð5Þ hyperbolic orbits. r0 p0 This paper studies the coplanar orbit escape problem from an elliptic orbit to an excess velocity vector only with the tan- Thus, for a given initial true anomaly u0 of the impulse gent impulse. For a given initial orbit and a given excess veloc- point, the magnitude of the tangent impulse is ity vector, the one-tangent-impulse transfer trajectory is k ¼ vf v0 obtained by solving a single-variable piecewise function. Then sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁ 2 2l the optimal multi-tangent-impulse escape trajectory is þ ¼ v1 þ ð1 þ e0 cos u0Þ obtained by the genetic method. p0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l 2 2. Orbital elements of transfer orbit ðÞ1 þ e0 þ 2e0 cos u0 ð6Þ p0

Assume that the spacecraft moves in a given initial orbit, a tan- The eccentricity vector of the final orbit is gent impulse Dv with magnitude k is imposed at P1 (see Fig. 1), 1 2 l where the position vector relative to the Earth’s center F1 is r0 ef ¼ ðvf Þr0 ðr0 vfÞvf ð7Þ l r0 and the velocity vector is v0, then the velocity vector of the transfer orbit (or the final hyperbolic orbit) at P1 is By using the following expression k k l r v ¼ 1 þ ðr v Þ¼ 1 þ r e sin u 0 f v 0 0 v h 0 0 0 0 0 0 ffiffiffiffiffiffiffi p k e0 sin u0 ¼ lp0 1 þ ð8Þ v0 1 þ e0 cos u0 the eccentricity vector in Eq. (7) can be written as ÀÁ 1 2 l e ¼ vþ þ ð1 þ e cos u Þ r f l 1 p 0 0 0 0 ) ffiffiffiffiffiffiffi 2 p k e0 sin u0 lp0 1 þ v0 ð9Þ v0 1 þ e0 cos u0

whose magnitude is the eccentricity ef of the final orbit. The normalized eccentricity vector is 2 3 cos xf cos Xf sin xf sin Xf cos If ef 6 7 êf ¼ ¼ 4 cos xf sin Xf þ sin xf cos Xf cos If 5 ð10Þ Fig. 1 Transfer to an excess velocity vector by a tangent impulse, ef sin xf sin If example 1. Tangent-impulse transfer from elliptic orbit to an excess velocity vector 579 where I denotes the orbit inclination, x the argument of peri- Define an XYZ frame, which is obtained by rotating the gee, and X the right ascension of ascending node. With a ECI frame with 3–1 sequence for angles X and i, respectively. coplanar tangent impulse, the orbit plane will not change, The origin of the XYZ frame is also the center of the Earth. which indicates that If = I0, and Xf = X0. In addition, the Then the X and Y axes are in the orbit plane, and Z axis is argument of perigee can be obtained as aligned with the angular momentum vector of the initial (or – final) orbit. atan2ðÞe^fk3 = sin I0; e^fk1 cos X0 þ e^fk2 sin X0 sin I0 0 xf ¼ The eccentricity vector Eq. (9) in the ECI frame can be atan2ðÞe^ ; e^ sin I ¼ 0 fk2 fk1 0 transformed into that in the XYZ frame. Note that the ð11Þ flight-direction angle c0 is the angle from the position vector where e^ ; e^ ; and e^ are three components of ê in the r0 to the velocity vector v0, then from Eq. (9) the component fk1 fk2 fk3 f of the eccentricity vector in the Y axis is Earth centered inertial (ECI) frame, where the origin is the ÀÁ center of the Earth, the k1 axis is in the vernal equinox direc- p0 þ 2 e0 sinu0 efy ¼ v þl sinðx0 þu Þpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0 2 tion, the k3 axis is the Earth’s rotation axis and perpendicular 1þe0 cosu 1 e 2e cosu 0 þ 0 þ 0 0 to equatorial plane, and the k2 axis is in the equatorial plane ÀÁ p0 þ 2 and completes the right-hand system. The four-quadrant v1 þ2l sinðx0 þu0 þc0Þð19Þ 1þe0 cosu0 inverse tangent function a = atan2(c1, c2) is defined as the angle satisfying both sin a = c1 and cos a = c2. The value which includes two variables, u0 and c0. Since the flight-direc- range of a is (p, p]. tion angle is a function only of the initial true anomaly The angular momentum of the initial/final orbit is p e0 sin u0 hj ¼ rj vj ð12Þ c0 ¼ arctan ð20Þ 2 1 þ e0 cos u0 and hj = khjk, then the true anomaly of P1 on the final orbit it is known that can be obtained from 8 8 > e sin u > > sin c ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 0 > hf < 0 2 < sin u ¼ r0 vf 1 þ e þ 2e cos u f lr e 0 0 0 0 f ð13Þ > ð21Þ > > 1 þ e0 cos u0 :> r0 ef :> cos c ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cos uf ¼ 0 2 r0ef 1 þ e0 þ 2e0 cos u0

With the above expressions, all six classical orbital elements Then, Eq. (19) can be rewritten as a function only of u0, of the final orbit are obtained. ÀÁ p =l 2 e sinu e ¼ 0 vþ þ 1 sinðx þ u Þ 0 0 fy 1 e cosu 1 0 0 1 e2 2e cosu þ 0 0 þ 0 þ 0 0 3. Analysis of the problem ÀÁ p0=l þ 2 v1 þ 2 ½e0 cosx0 þ cosðx0 þ u0Þ ð22Þ 1 þ e0 cosu0 The eccentricity ef of the transfer orbit is obtained by Eq. (9). However, a simple explicit expression for ef only about the ini- Similarly, the component in the X axis is tial true anomaly will be derived in the following paragraphs. ÀÁ p =l 2 e sinu Note that the flight-direction angles at the impulse point P1 for e ¼ 0 vþ þ 1 cosðx þ u Þ 0 10 fx 1 e cosu 1 0 0 1 e2 2e cosu initial and final orbits are the same. From Eq. (12) it is known þ 0 0 þ 0 þ 0 0 p =l ÀÁ r2v2 h2 0 þ 2 0 f 2 f v1 þ 2 ½ðe0 sinx0 sinðx0 þ u0Þ 23Þ sin c0 ¼ ¼ pf ð14Þ 1 þ e cosu l l 0 0 Z Since the flight-direction angle is For the eccentricity vector, there is no component in the axis,qthusffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the eccentricity can also be obtained as 2 2 e0 sin u0 e e e , which is the same as that via Eq. (9). cot c ¼ ð15Þ f ¼ fx þ fy 0 1 e cos u þ þ 0 0 Moreover, the normalized excess velocity vector ^v1 in the ECI frame can be transformed into that in the XYZ frame as we have 2 3 2 32 3 1 ðÞ1 þ e cos u 2 v^þ 10 0 cosX sinX 0 sin2 c ¼ ¼ 0 0 ð16Þ 6 1x 7 6 76 7 0 2 2 4 þ 5 4 54 5 þ 1 þ cot c0 1 þ e0 þ 2e0 cos u0 v^1y ¼ 0 cosi sini sinX cosX 0 ^v1 0 0 sini cosi 001 Substituting Eq. (16) into Eq. (14) yields 2 3 ð24Þ 2 ÀÁ cosX sinX 0 p0=l 2 2l 6 7 þ þ v1 þ ð1 þ e0 cos u0Þ 4 cosisinX cosicosX sini 5^v 1 þ e2 þ 2e cos u p ¼ 1 0 ÀÁ0 0 0 2 sinisinX sinicosX cosi ¼ af 1 ef ð17Þ where v^þ ; and v^þ are the components in the X and Y axes, By using Eq. (4) the eccentricity can be obtained as 1x 1y respectively. The components in the Z axis is v^þ ¼ 0. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1z ÀÁ 2 The true anomaly uþ associated with ^vþ on the final orbit 1 2 p 1 1 2 þ 0 ef ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞe0 sinu þ v þð1 þ e0 cosu Þ 2 0 1 l 0 is then 1 þ e0 þ 2e0 cosu0 ð18Þ þ 1 u1 ¼ arccos ð25Þ ef which is a function only of the initial true anomaly u0. 580 G. Zhang et al.

For the coplanar orbit escape problem with the tangent 2 p). Finally, the numerical value of us can be obtained impulse, we have by the golden search with the boundary values as the , þ þ þ initial guesses. F atan2ðefy; efxÞþu1 atan2ðv^1y; v^1xÞ¼0 ð26Þ where atan2(efy, efx) denotes the argument of perigee for the þ þ 4.2. Solution for initial circular orbit final orbit, u1 the true anomaly associated with ^v1 in the final þ þ orbit, and atan2ðv^1y; v^1xÞ the angle determined by the excess velocity components in the X and Y axes (see Fig. 1). Note that For the initial circular orbit, e0 = 0 is satisfied. From Eq. (9) it F is a function only of the initial true anomaly u0. In Section 4, is known that the eccentricity vector of the final orbit is Eq. (26) will be solved. ÀÁ 1 l 1 2 l e ¼ v2 r ¼ vþ þ r ð30Þ f l f r 0 l 1 r 0 4. Solution of the problem 0 0 whose magnitude is þ þ þ ÀÁ ÀÁ For a given v1, the value of atan2ðv^1y; v^1xÞ in Eq. (26) is 1 2 l r 2 e ¼ vþ þ r ¼ 1 þ 0 vþ ð31Þ invariable. In addition, ef > 1 is satisfied for the final hyper- f 1 0 1 l r0 l bolic orbit, then from Eq. (25) it is known that þ For the initial circular orbit with a tangent impulse, the u1 2ðp=2; pÞ. In the following subsections, two methods are proposed to obtain the solutions for Eq. (26). eccentricity vector of the final orbit is aligned with the radius vector of the impulse point. Then the argument of the impulse 4.1. Discontinuous point u point is s þ þ þ u0 ¼ atan2 v^1y; v^1x u1 In Eq. (26), there is a discontinuous point for atan2(efy, efx)at us which satisfies efy(us) = 0 and efy(us d) > 0 (or efy(us + þ þ 1 ¼ atan2 v^1y; v^1x arccos ð32Þ d) < 0), where d is a constant small enough. ef Multiplying both sides of the equation efy =0 by 2 which is a closed-form solution. lð1 þ e0 cos u0Þð1 þ e0 þ 2e0 cos u0Þ gives hiÀÁ 2 þ 2 4.3. Solution for initial elliptic orbit ð1 þ e þ 2e0 cos u0Þ p0 v þ lð1 þ e0 cos u0Þ sinðx0 þ u0Þ 0 hi1 þ 2 e0 sin u0 p0ðv1Þ þ 2lð1 þ e0 cos u0Þ Once the discontinuous point of F is obtained, the equation F=0 can be solved by the secant method. In the whole range ½¼e0 cos x0 þ cosðx0 þ u0Þ 0 ð27Þ [0, 2p), there are two piecewise ranges, e.g., [0, us) and [us,2p). Simplifying it yields In addition, it is known that F(0) = F(2p) and F(u d) n hihi s ÀÁ = F(us)+2p. Extensive numerical examples show that F g u , 1 e2 p vþ 2 l 1 e2 le e p vþ 2 ð 0Þ ð þ 0Þ 0 1 þ þðþ 0Þ 0 þ 0 0ð 1Þ changes monotonously. Then there is only one solution in hiÀÁ þ 2 the range [0, 2p). If F(0)ÆF(us d) < 0, the solution is in cos u0g sinðx0 þ u0Þþe0 p0 v1 þ 2l hiÀÁ [0, us) and it can be obtained by the secant method, 2 2 þ 2 sin x0 þ 2le0 sin x0 cos u0 e0 p0 v1 þ 2l ÀÁ Fðu0;nÞ 3 u0;nþ1 ¼ u0;n u0;n u0;n1 cos x0 sin u0 le0 cos x0 sinð2u0Þ¼0 ð28Þ Fðu0;nÞFðu0;n1Þ n 1; 2; 3; ... The derivative of g(u0) with respect to u0 is ð ¼ Þ hi dg u ÀÁ ð33Þ 0 , ð 0Þ 2 þ 2 g ðu0Þ ¼ð1 þ e0Þ p0 v1 þ l cosðx0 þ u0Þ du0 hiÀÁ where the initial guesses are u0,0 = 0, and u0,1 = us d. 2 þ 2 Otherwise, the solution is in [u ,2p) and it can be obtained þð1 þ e0Þle0 þ e0p0 v1 cosð2u0 þ x0Þ s hiÀÁ by Eq. (33) with the initial guesses u0,0 = us, and 2 2 þ 2 2le0 sin x0 sin u0 e0 p0 v1 þ 2l cos x0 cos u0 u0,1 =2p d. Next, some numerical examples are given to show the 2le3 cos x cosð2u Þð29Þ 0 0 0 monotonicity of function F. The perigee height of the initial

Then, there are two extreme points for g(u0) in the range orbit is hp = 1000 km, then the semimajor axis is [0, 2p): a0 =(RE + hp)/(1 e0) , where RE = 6378.13 km is the Earth radius, the initial orbital elements are I0 =40, x0 =30, and 0 þ (1) If g (0) > 0 , the maximum point umax and the minimum X0 =50. If the excess velocity vector is v1 ¼½2:4888; T point umin satisfy umax < umin, then us e (umax, umin), 0:1302; 1:6700 km/s in the ECI frame, then the normalized T thus us can be obtained by the golden search with the excess vector in the XYZ frame is [0.5000, 0.8660, 0] km/ initial guesses umax and umin. s. For different eccentricities e0 with 0.1, 0.3, 0.5, 0.7, 0.9, 0 (2) If g (0) < 0, the maximum point umax and the minimum the plots of F are given in Fig. 2, which shows that F monot- point umin satisfy umax > umin. It is known that us e (0, onously increases in both piecewise ranges, and the curve of F umin)ifg(0) > 0 and us e (umin,2p)ifg(0) < 0. Note is more linear for a smaller eccentricity. 2 þ 2 that gð0Þ¼ð1 þ e0Þ ½p0ðv1Þ þ lð1 þ e0Þ sin x0 , then In addition, for the eccentricity e0 = 0.5 and different g(0) > 0 when x0 e [0, p) and g(0) < 0 when x0 e (p, parameters kv, the ﬁnal excess velocity vector is Tangent-impulse transfer from elliptic orbit to an excess velocity vector 581

þ Fig. 4 Fs vs u0 for the same v1 and different eccentricities. þ Fig. 2 F vs u0 for the same v1 and different eccentricities.

þ Fig. 3 F vs u0 for the same eccentricity and different v1 . þ Fig. 5 Fc vs u0 for the same v1 and different eccentricities. þ T v1 ¼ kv½2:4888; 0:1302; 1:6700 =3 km/s in the ECI frame. range of u0 (but function F monotonously increases in both The plots of F are given in Fig. 3, which shows that F monot- piecewise ranges). The same result can be obtained for the cases onously increases in both piecewise ranges. Therefore, for dif- þ with the same eccentricity and different v1 . Thus, the extreme ferent eccentricities and excess velocity vectors, F always points need to be solved and then the solutions for Eqs. (34) changes monotonously in both piecewise ranges. and (35) are obtained. As a result, although the singularity problem is avoided, solving Eqs. (34) and (35) is not much simpler 4.4. Alterative method for avoiding the singularity than the proposed method with a discontinuous point.

Although Section 4.3 gives a method to solve the one-tangent- 5. The minimum-fuel multi-tangent-impulse transfer impulse transfer problem from an elliptic parking orbit to a given excess velocity vector, there is a discontinuous point The previous section solves the one-tangent-impulse transfer when solving the equation F = 0. This subsection provides problem from a coplanar elliptic orbit to an excess velocity an alternative method to avoid the singularity. vector. However, if multiple tangent impulses are used to fulfill Eq. (26) can be rewritten as the mission, there are infinite solutions, and the minimum-fuel solution is the most interesting one. Assume that N tangent F , e þ sin uþ atan2 v^þ ; v^þ ¼ 0 ð34Þ s fy 1 1y 1x impulses are adopted, the orbital elements for each orbit arc are Ei =[ai, ei, Ii, xi, Xi, ui], then the initial orbital elements , þ þ þ Fc efx cos u1 atan2 v^1y; v^1x ¼ 0 ð35Þ are E0, the final orbital elements are EN = Ef. The position vector ri and the velocity vector vi for the orbital elements Ei Then solving Eq. (26) can be transformed into solving Eqs. are obtained by the transformation from orbital elements to (34) and (35). position and velocity vectors in the ECI frame. For the When Eq. (34) is considered, for the example in Section 4.3 of (i + 1)th tangent impulse, the impulse position is ui and the e F different eccentricities 0 with 0.1, 0.3, 0.5, 0.7, 0.9, the plots of s impulse magnitude is ki, then the velocity vector at ui reaches and Fc are given in Figs. 4 and 5, respectively. The figures show vi that there are no discontinuous points for F and F . However, vi 1 ¼ð1 þ kiÞ ð36Þ s c þ kv k we need to solve two related functions, i.e., i Eqs. (34) and (35); or we need to obtain all solutions for either With ri and vi+1, the orbital element Ei+1 is obtained by the of Eqs. (34) and (35) and delete the ones not satisfying the other. transformation from position and velocity vectors in the ECI In addition, Fs and Fc do not monotonously change in the whole frame to orbital elements, or directly by the proposed method 582 G. Zhang et al. in Section 2. Then, our purpose is to solve the true anomaly u i Table 2 Solutions of u for the same eccentricity and different of the impulse position and the corresponding impulse magni- 0 excess velocity vectors. tude ki(i = 1,2,...,N) such that the total cost kv Solution of u0 () Impulse magnitude k (km/s) XN Dvtotal ¼ ki ð37Þ 1 227.2309 2.4349 i¼1 3 252.8762 2.5256 5 279.9323 3.0033 is minimized. 7 302.2874 3.8274 When ui and ki of the tangent impulse for i = 1,2,...,N1 9 318.2679 4.9338 are fixed, the values of uN and kN for the final impulse can be obtained by using the proposed numerical method in Section 4. Thus, for the N tangent-impulse case, only 2(N1) variables s in the ECI frame. With a tangent impulse whose magnitude are needed in the numerical optimization approach. This paper is k = 2.5256 km/s, the orbit elements of the final hyperbolic will use the genetic method to obtain a numerical solution to orbit at P1 are Ef =[44288.9 km, 1.2294, 40, 335.5685, the minimum-fuel multi-tangent-impulse transfer problem. 50, 307.3077]. With these hyperbolic orbital parameters, after the time-of-flight 1.0 · 109 s for the two-body motion, 6. Numerical simulations the error of the velocity vector and the required excess velocity vector is 4.4282 · 105 km/s. When e = 0.5 and vþ 7 2:4888; 0:1302; 1:6700 T=3 Assume that a spacecraft moves in an initial elliptic orbit, 0 1 ¼ ½ km/s, the solution is u0 = 302.2874 and the corresponding where the perigee height is hp = 1000 km, the semimajor axis cost is k = 3.8274 km/s. The trajectories of the initial and final is a0 =(RE + hp)/(1 e0), where RE = 6378.13 km is the excess orbits in the orbit plane are plotted in Fig. 6. The initial Earth radius, the orbit inclination is I0 =40, the argument T position vector at P1 is r0 = [7353.9, 3923.4, 2610.9] km, of perigee is x0 =30, and the right ascension of ascending and the velocity vector of the initial orbit at P1 is node is X0 =50. T v0 =[3.8132, 4.8843, 5.0855] km/s in the ECI frame. With 6.1. One tangent impulse a tangent impulse whose magnitude is k = 3.8274 km/s, the orbit elements of the final hyperbolic orbit at P1 are Ef = [8134.7 km, 1.9924, 40, 359.8744,50, 332.4130]. With If the excess velocity vector is reached by a single tangent these hyperbolic orbital parameters, after the time-of-flight impulse, the solution is obtained by the secant method in Sec- 1.0 · 109 s for the two-body motion, the error of the velocity þ tion 4. If the excess velocity vector is v1 ¼½2:4888; T vector and the required excess velocity vector is 0:1302; 1:6700 km/s in the ECI frame, which is three times 8.1861 · 106 km/s. the velocity vector at u0 =0 with e0 = 0.5, then the normalized excess vector in the XYZ frame is [0.5000, 0.8660, T 6.2. Multiple tangent impulse 0] km/s; for different eccentricities e0 with 0.1, 0.3, 0.5, 0.7, 0.9, the plots of F are given in Fig. 2, and the plots of Fs If multiple tangent impulses are used for the orbit transfer to and Fc are given in Figs. 4 and 5, respectively. The solutions are listed in Table 1. the excess velocity vector problem, the minimum-energy solution of ½ui; ki; i ¼ 1; 2; ...; N 1 can be obtained by numerical If the eccentricity e0 = 0.5, for different kv, the final excess þ T optimization methods; the solution of ½uN; kN for the final velocity vector is v1 ¼ kv½2:4888; 0:1302; 1:6700 =3 km/s in the ECI frame, the plots of F are given in Fig. 3, and the solu- impulse is obtained by the method in Section 4. Herein, the tions are listed in Table 2, which shows that more energy cost genetic algorithm (‘‘ga’’ in MATLAB) is used for this optimi- is required for a larger final excess velocity. zation problem. The orbital elements of the initial orbit are the þ T same as that in Section 6.1 with e0 = 0.5 and When e0 = 0.5 and v ¼½2:4888; 0:1302; 1:6700 km/s, 1 vþ ¼½2:4888; 0:1302; 1:6700T km/s. Then the two, three, the solution is u0 = 252.8762 and the corresponding cost is 1 k = 2.5256 km/s. The trajectories of the initial and final excess orbits in the orbit plane (or the XYZ frame) are plotted in Fig. 1, where F1O is the direction of the final eccentricity vector in the X–Y plane. The initial position vector at P1 is T r0 = [9283.1, 4014.2, 8132.2] km, and the velocity vector T of the initial orbit at P1 is v0 = [0.4864, 5.2705, 2.5300] km/

Table 1 Solutions of u0 for different eccentricities and the same excess velocity vector.

e0 Solution of u0 () Impulse magnitude k (km/s) 0 300.9965 3.4688 0.1 292.2646 3.2307 0.3 273.3140 2.8359 0.5 252.8762 2.5256 0.7 231.7169 2.2595 0.9 210.2594 1.9600 Fig. 6 Transfer to an excess velocity vector by a tangent impulse, example 2. Tangent-impulse transfer from elliptic orbit to an excess velocity vector 583

Table 3 Solutions of ½ui; ki; i ¼ 1; 2; :::; N 1 and the total costs for different numbers of tangent impulses.

Impulse number N Solution of ½ui; ki ((), km/s) Solution of ½uN; kN ((), km/s) Total cost Dvtotal (km/s) 1 [252.8762, 2.525643] 2.525643 2 [0.1561, 4.3185 · 105] [252.8746, 2.525623] 2.525667 3 [34.4573, 0], [29.0005, 0] [252.8762, 2.525643] 2.525643 4 [5.9257, 0], [84.6521, 0], [12.9435, 4.8346 · 105] [252.8742, 2.525625] 2.525674 and four tangent impulses are used, respectively. The solutions 4. Jones DR, Ocampo C. Optimization of impulsive trajectories from are obtained as listed in Table 3, which shows that the costs of circular orbit to an excess velocity vector. J Guid Control Dyn optimal one-, two-, three- and four-impulse transfer are almost 2012;35(1):234–44. the same. For the one-tangent-impulse case, the solution is 5. Gerbracht RJ, Penzo PA. Optimum three-impulse transfer between an elliptic orbit and a non-coplanar escape asymptote. obtained directly by the method in Section 4, thus this optimi- In: AAS/AIAA astrodynamics specialist conference; 1968 Sep 3-5; zation method is not needed. It should be notified that the Jackson, WY. New York: American Astronautical Society; 1968. orbital elements Ei will not change when k = 0 even though 6. Robinson SB, Geller DK. A simple numerical procedure for the true anomalies are different. computing multi-impulse lunar escape sequences with minimal As a result, the optimal multi-tangent-impulse transfer time-of-flight. In: 20th AAS/AIAA space flight mechanics meeting; from coplanar elliptic orbit to an excess velocity vector is the 2010 Feb 14–17; San Diego, CA. also, Advances in the Astro- same as that with one tangent impulse. Because this result is nautical Sciences 2010; 136: 1015–24. only obtained by numerical methods, it is not easy to give a 7. Pontani M. Simple method to determine globally optimal orbital stick explanation. However, for different eccentricities and dif- transfers. J Guid Control Dyn 2009;32(3):899–914. ferent final excess velocity vectors, the result still holds by 8. Chen Y, Baoyin HX, Li JF. Design and optimization of a trajectory for moon departure near earth asteroid exploration. Sci using numerical optimization algorithms. China Phys Mech Astron 2011;54(4):748–55. 9. Vallado DA. Fundamentals of astrodynamics and applications. 2nd 7. Conclusions ed. El Segundo: Microcosm Publishing; 2001. p. 964–9. 10. Thompson BF, Choi KK, Piggott SW, Beaver SR. Orbital In this paper the coplanar orbit escape problem from an ellip- targeting based on hodograph theory for improved rendezvous tic orbit to an excess velocity vector is studied with the condi- safety. J Guid Control Dyn 2010;33(5):1566–76. 11. Adamyan VG, Adamyan LV, Zaimtsyan GM, Manandyan LT. tion that only tangent impulses are used. Since the impulse Double-pulse cotangential transfers between coplanar elliptic direction is aligned with the velocity vector, the magnitude orbits. J Appl Math Mech 2009;73(6):664–72. and the true anomaly of the position point of the tangent 12. Zhang G, Zhou D, Mortari D, Henderson TA. Analytical study of impulse are solved. For initial circular orbit, the solution can tangent orbit and conditions for its solution existence. J Guid be obtained in closed-form, whereas for initial elliptic orbit, Control Dyn 2012;35(1):186–94. the solution is only obtained by a numerical iterative algo- 13. Zhang G, Zhou D. Tangent orbit technique in three dimensions. J rithm. The one-tangent-impulse solution is the optimal one Guid Control Dyn 2012;35(6):1907–11. even though the multi-tangent-impulse can be used. The pro- 14. Zhang G, Zhou D, Bai XL. Tangent orbital rendezvous with the posed method for the coplanar tangent-orbit design of escape same direction of terminal velocities. J Guid Control Dyn trajectories can be used in space exploration, such as human 2012;35(1):335–9. 15. Zhang G, Zhou D, Sun ZW, Cao XB. Optimal two-impulse Lunar and Mars exploration missions. The tangent escape tra- cotangent rendezvous between coplanar elliptical orbits. J Guid jectories provide a new approach for orbit escape transfer with Control Dyn 2013;36(3):677–85. only speed changes. 16. Zhang G, Cao XB, Zhou D. Optimal two-impulse cotangent rendezvous between coplanar elliptic and hyperbolic orbits. J Guid Acknowledgements Control Dyn 2014;37(3):964–9.

This work was supported in part by the China Postdoctoral Zhang Gang received the Bachelor’s degree from School of Mathe- Science Foundation funded project (No. 2012M520753), the matics at Jilin University in 2007, and Ph.D. degree of aerospace Fundamental Research Funds for the Central Universities engineering at Harbin Institute of Technology in 2012, respectively. (No. HIT.NSRIF.2014307), and the Open Fund of National Now he is a Lecturer at Harbin Institute of Technology. His current research interests include orbit rendezvous, orbit transfer, relative Defense Key Discipline Laboratory of Micro-Spacecraft Tech- motion, and tangent orbit. nology (No. HIT.KLOF.MST.201303). Zhang Xiangyu is a Ph.D. student at School of Astronautics, Harbin References Institute of Technology. His area of research includes optimization of trajectory, relative motion dynamics and control and rendezvous and 1. Edelbaum TN. Optimal nonplanar escape from circular orbits. docking of spacecraft. AIAA J 1971;9(12):2432–6. 2. Gobetz FW, Doll JR. A survey of impulsive trajectories. AIAA J Cao Xibin is a professor and Ph.D. supervisor at School of Astro- 1969;7(5):801–34. nautics, Harbin Institute of Technology, Harbin, China. His current 3. Ocampo C, Saudemont RR. Initial trajectory model for a multi- research interests include spacecraft design and simulation, spacecraft maneuver moon-to-earth abort sequence. J Guid Control Dyn formation ﬂying dynamics and control, and application of micro- 2010;33(4):1184–94. satellite.