Tangent Orbital Rendezvous Using Linear Relative Motion with J 2 Perturbations

The tangent-impulse coplanar orbit rendezvous problem is studied based on the linear relative motion for J 2 -perturbed elliptic orbits. There are three cases: (1) only the first impulse is tangent; (2) only the second impulse is tangent; (3) both impulses are tangent. For a given initial impulse point, the first two problems can be transformed into finding all roots of a single variable function about the transfer time, which can be done by the secant method. The bitangent rendezvous problem requires the same solution for the first two problems. By considering the initial coasting time, the bitangent rendezvous solution is obtained with a difference function. A numerical example for two coplanar elliptic orbits with J 2 perturbations is given to verify the efficiency of these proposed techniques.


Introduction
The orbital rendezvous problem is a fundamental one in aerospace engineering for human space activities.If the impulse maneuver is adopted, a single impulse is enough to finish the orbital interception mission.But for the orbital rendezvous mission, it needs at least two impulses, in which the second impulse is to correct the terminal velocity.If the transfer time is assigned, the required initial velocity for the chaser can be obtained by solving Lambert's problem [1][2][3][4].However, there is no constraint on the direction of the required initial velocity vector.If the required initial velocity vector is aligned with the current velocity of the chaser, the impulse is a tangent one, and then only the speed change will null the relative velocity.
Actually, the tangent orbit problem has existed for many years.The well-known Hohmann transfer is a classical twoimpulse bitangent transfer between two coplanar circular orbits [5].Because of its simple tangent-impulse direction and the minimum-energy cost among all the two-impulse transfers, the Hohmann transfer has been widely used in engineering applications.However, the Hohmann transfer is only available for coplanar circular orbits and coplanar coaxial elliptic orbits.Recently, for coplanar noncoaxial elliptic orbits, Adamyan et al. [6] used its geometric characteristics to obtain the analytical expression for the bitangent transfer orbital parameters.Based on the orbital hodograph theory, Thompson et al. [7] solved the coplanar bitangent orbit by a numerical iterative approach, that is, the regula falsi method, which is like the secant method.However, instead of retaining the last two points, the regula falsi method makes sure to keep one point on either side of the root.Furthermore, Zhang et al. [8] provided a simple and closed-form solution for the tangent transfer orbit problem associated with its solutionexistence conditions.In addition, Zhang and Zhou [9] studied a noncoplanar tangent orbit technique based on a new definition of orbit "tangency" condition.However, for the orbit rendezvous problem, the flight time of the chaser and that of the target are required to be equal.For this purpose, Zhang et al. [10,11] solved the tangent orbit rendezvous problem with the same terminal velocity direction and the two-impulse bitangent rendezvous problem, respectively.
All the previous methods with tangent orbits are only for the two-body orbit problem, but not taking the  2 perturbation into account.If the relative range between the chaser and the target is short, the linear relative motion equations can be used to solve the required velocity vector and the tangent orbit problem.The linear Hill-Clohessy-Wiltshire (HCW) equations [12,13] and the Tschauner-Hempel (TH) equations [14] are used for the circular and elliptic target orbits, respectively.When the transfer time is fixed, the required initial relative velocity can be derived from the state transition matrix (STM).Because the coefficient matrices are constant for the HCW equations, the STM can be easily obtained.For the TH equations, by using the true anomaly of the target instead of the transfer time as the independent variable, simpler equations can be derived from the TH equations then the STM can also be obtained [15].For the STM with the  2 perturbation, Gim and Alfriend [16] used the geometric method to obtain an STM for the elliptic target orbit with the influence of the  2 perturbation.In their method, the closed-form STMs for both mean elements and osculating elements were obtained.Moreover, Yamada et al. [17] derived an STM in eccentric orbits with  2 by using the osculating orbital elements of the target in the initial state as the nominal orbital elements.
This paper studies the tangent-impulse orbital rendezvous problem between two coplanar elliptic orbits with the  2 perturbation.Based on the STM in the elliptic target orbit with the influence of the  2 perturbation [16], the required initial relative velocity vector can be obtained.Then two functions are defined for the tangent impulse velocity.By using the secant method, the tangent to initial/final orbit can be obtained.The bitangent orbit indicates that it is the same solution for the tangent to initial orbit problem and the tangent to final orbit problem.

Relative Motion with the 𝐽 2 Perturbation
2.1.Linear Equations of Relative Motion.Define an orthogonal coordinate system, where the origin of the coordinate system is the center of the target, the  H axis is taken to be the radius vector from the center of the Earth to the center of the target, the  H axis is aligned with the angular momentum vector of the target orbit, and the  H axis completes the righthand system.This orthogonal coordinate system is called the local-vertical-local-horizontal (LVLH) frame of the target.Similarly, the LVLH frame of the chaser can also be defined.It should be notified that the LVLH coordinate system is a rotation frame.Since there is no maneuver on the target in orbit rendezvous, the linear relative motion equations with only the  2 perturbation (other external accelerations are neglected) in the LVLH frame of target can be written as [18] d d where X() = [, , , ẋ , ẏ , ż ] T denotes the state vector and the elements in the coefficient matrix are where  0 = 6 2   2 (  /  ) 3 ,  2 = 1.0826269 × 10 −3 ,   is the mean motion,  is the radius,  is the semimajor axis,  is the orbit inclination, and  =  +  is the argument of latitude, where  is the argument of perigee and  is the true anomaly.The symbol (⋅) denotes the mean orbit elements.The subscript  denotes the target orbit.
The angular velocity vector of the LVLH frame of the target is whose components are where Ω denotes the right ascension of ascending node.
Considering the  2 perturbation, the derivatives of the orbital elements with respect to time are [16] where   is the radius of the Earth,  is the eccentricity, and  is the gravity parameter.With (4a)-( 5), the secondorder derivatives of   , Ω  ,   ,   can be obtained; and then it is not difficult to obtain the derivatives of ω  and ω  in (2a)-(2i).

State Transition Matrix (STM).
To obtain the STM for the linear relative motion equations with the  2 perturbation, the nonsingular orbital elements are defined as  = [, , ,  1 ,  2 , Ω] T , where  1 =  cos  and  2 =  sin .
When the nonsingular orbital elements for the chaser and the target are both obtained, the differential orbital elements between the chaser and the target are denoted by  =   −  .The subscripts  and  denote the chaser orbit and the target orbit, respectively.The geometric transformation between the state vector X and the differential nonsingular osculating orbital elements  osc is [16] where () is the matrix for the two-body motion and () is the matrix perturbed by  2 .
(7) All the elements of the matrices (), (),   (,  0 ), () can be obtained in [16].Finally, the STM for the relative states with  2 effects is and the state vector is 2.3.Relative State in the LVLH Frame of the Target.Note that the LVLH frame of the target is a rotation frame.The time derivative of an arbitrary vector  in the earth-centeredinertial (ECI) frame can be written in terms of the derivative with respect to the LVLH frame of the target as where  is the angular velocity vector of the rotating frame which is obtained in (3).Assume that the position vector and the velocity vector of the chaser in the ECI frame are R  and V  , respectively, and those of the target are R  and V  , respectively.The transformation matrix  HI from the ECI to the LVLH frames is Using (11), the transformation matrices  HI  for the LVLH frame of the target and  HI  for the LVLH frame of the chaser are both obtained.
Then the relative position vector in the LVLH frame of the target is Leting  = r in (10), the relative velocity vector in the LVLH frame of the target is

Problem Analysis of Tangent Rendezvous
3.1.Tangent to the Initial Orbit.For the orbital interception and rendezvous problem, the required initial relative velocity can be obtained from the STM if the transfer time is fixed.
The final state vector X(  ) = [r T  , k T  ] T can be obtained from the initial state vector as If the final time   is fixed and the final relative position r  is assigned, for example, r  = [0, 0, 0] T for the interception and rendezvous problem, the required initial relative velocity in the LVLH frame of the target is Then, the first velocity impulse in the LVLH frame of the target is where k 0 is the initial relative velocity.By transforming the first impulse velocity vector in the LVLH frame of the target to that in the LVLH frame of the chaser, we can obtain The normalized vector of Δk 0 is For the coplanar case, ΔV 0 = 0 is satisfied.The flight-direction angle  is defined as the angle from the position vector to the velocity vector in the ECI frame (see Figure 1).From the definition it is known that the range of the flight-direction angle is  ∈ (0, ).At the initial time, the flight-direction angle of the chaser  0 can be written as a function only of the initial true anomaly If the first velocity impulse is required to be tangent to the initial orbit of the chaser, the first impulse velocity vector in the ECI frame must be aligned with the initial velocity vector of the chaser in the ECI frame.Thus, the following expression should be satisfied: Note that  0 is a function only of  0 , and Δk 0 can be obtained by (18) if the final time   is fixed.Therefore,  1 is a function only of the final time   or the transfer time Δ tran ≜   −  0 if the first impulse is at the initial time.Thus, the value of Δ tran for  1 (Δ tran ) = 0 is required to be solved.

Tangent to the Final Orbit.
After the first impulse, the relative velocity vector at the initial time reaches k ℎ0 ; then the relative velocity vector at the final time   can be obtained from (14) as For the rendezvous problem, the required final relative velocity is zero.Thus, the second impulse in the LVLH frame of the target is The normalized vector of Δk  is For the coplanar case, ΔV  = 0 is also satisfied.
If the second impulse is required to be a tangent one, the second impulse vector in the ECI frame must be aligned with the final velocity vector of the chaser in the ECI frame; in other words, the direction of the final velocity vector of the chaser and that of the target are the same in the ECI frame.Thus, the second impulse in the ECI frame is aligned with the final velocity vector of the target in the ECI frame, which is equivalent to the condition that the second impulse in the LVLH frame of the target is aligned with the final relative velocity vector in the LVLH frame of the target, since r  = [0, 0, 0] T is satisfied for the rendezvous problem.At the final time, the flight-direction angle of the target is Therefore, the second impulse is tangent if and only if It is known that  2 is also a function only of the transfer time   or Δ tran if the first impulse is at the initial time.Thus, the value of Δ tran for  2 (Δ tran ) = 0 will be solved for the tangent to final orbit problem.
Mathematical Problems in Engineering 5

Bitangent Orbit.
The two-impulse method is the simplest and the most fundamental one to fulfill the rendezvous mission.For the tangent to initial orbit problem, only the first impulse is tangent, whereas for the tangent to final orbit problem, only the second impulse is tangent.Moreover, the bitangent rendezvous requires the transfer orbit to be tangent to the initial orbit and also tangent to the final orbit; that is, both impulses are tangent.Since both impulses are imposed on the chaser, then each impulse is required to be aligned with the velocity vector of chaser in the ECI frame.Thus, the same value of Δ tran is required to simultaneously satisfy two conditions, that is,  1 = 0 and  2 = 0.However, if the first impulse is at the initial time, there may not exist this value of Δ tran .When considering the initial coasting time Δ 1 ≜  1 −  0 , the first impulse is not at the initial time  0 but at the time  1 .
Then  1 and  2 are functions of two variables, that is,  1 and   or Δ 1 and Δ tran .Therefore, two variables Δ 1 and Δ tran are to be solved for the following equations: The total energy cost for the two-impulse rendezvous is

Solution Procedure
4.1.Tangent to the Initial/Final Orbit Problem.Assume that the first impulse is at the initial time  0 .Then for the tangent to initial orbit problem, the value of Δ tran is to be solved for  1 = 0. Since the linear relative motion equations cause large errors for the long-time case, the transfer time is assumed to be not greater than one mean orbit period of the target, that is, Δ tran ∈ (0,   ], where   denotes the mean orbit period of the target.For different values of Δ tran , the value of  1 (Δ tran ) will change.Then, the extreme points of  1 can be obtained by numerical methods.Assume that in the range (0,   ] there are  extreme points satisfying Δ tran,1 < Δ tran,2 < ⋅⋅⋅ < Δ tran, .Then there are altogether  + 1 piecewise ranges in (0,   ], that is, (0, Δ tran,1 ], (Δ tran,1 , Δ tran,2 ], . . ., (Δ tran, ,   ].If there is no extreme point in (0,   ], then  1 (Δ tran ) will decrease or increase monotonically.For each piecewise range (Δ tran,1 , Δ tran,2 ], there are two cases.

Bitangent Orbit Rendezvous Problem.
For the bitangent orbit rendezvous problem, two variables Δ 1 ∈ (0,   ]; and Δ tran ∈ (0,   ] in (26) are solved.Although the sequential two-point secant method and the sequential  + 1 point secant method can be used to solve the nonlinear vector equation [19], the initial guesses are not easy to select.Herein, we will use an alternative method to solve the bitangent orbit rendezvous problem.
For a given initial coasting time Δ 1 ∈ (0,   ], the first impulse is at  1 =  0 + Δ 1 ∈ ( 0 ,  0 +   ]; the transfer time Δ tran in one mean orbital period of the target for the tangent to initial/final orbit problem can be obtained by the proposed method as Δ tran, 1 and Δ tran, 2 .There may exist many solutions of Δ tran, 1 and Δ tran, 2 for a given initial coasting time Δ 1 .Define a new function where  1 is the  1 th solution of Δ tran, 1 and  2 is the  2 th solution of Δ tran, 2 such that the expression [Δ tran, 1 − Δ tran, 2 ] 2 is minimized.Then  = 0 should be satisfied for the bitangent rendezvous problem.It should be notified that  is a function only of Δ 1 .Then, by using a numerical iterative algorithm, for example, the secant method, all solutions of Δ 1 can be obtained in the range (0,   ].

Numerical Examples
At the initial time, the nonsingular mean orbital elements of the target are The chaser orbit and the target orbit are coplanar.With the previous parameters, in the LVLH frame of the target, the initial relative position vector is r 0 = [2.4020,−4.2185, 0] T km, and the initial relative velocity vector is k 0 = [−0.7223,−2.6583, 0] T m/s.The mean orbit period of the target is   = 20859.684s.The range of the coasting time is Δ 1 ∈ (0,   ] and that of the transfer time is Δ tran ∈ (0,   ].

Tangent to Initial/Final
Orbit.If the impulse is imposed at the initial time, the curves of functions  1 and  2 for different transfer time Δ tran can be plotted in Figure 2. By using (28), the solution for  1 = 0 can be obtained as Δ tran, 1 = 0.704717  and that for  2 = 0 can be obtained as Δ tran, 2 = 0.059408  .
For the tangent to initial orbit problem, Δ tran, 1 = 0.704717  , the first impulse is aligned with the initial velocity vector in the ECI frame, that is, [−0.303824, 0, 0.952728] T .Thus, with a tangent impulse whose magnitude is 0.148890 m/s, the "true" curve of relative position is plotted in Figure 3, which is obtained by the numerical integral for nonlinear equations with the  2 perturbation.The final position error is [11.9539,−31.7167, 0] T m in the LVLH frame of the target.
For the tangent to final orbit problem, Δ tran, 2 = 0.059408  , the first impulse is [−5.09566,0, 2.56639] T m/s in the ECI frame and it is not a tangent impulse.The final velocity directions of the chaser and the target in the ECI frame are both [−0.744802, 0, 0.667285] T , which indicates that the second impulse is a tangent one.The magnitude of the second impulse is 4.13168 m/s.The curve of relative position is also plotted in Figure 3.The final position error is [−0.1616,0.1757, 0] T m in the LVLH frame of the target. versus Δ 1 can be plotted in Figure 4.The final solution for  = 0 can be obtained as Δ 1 =  , which is the coasting time for the bitangent rendezvous and the corresponding rendezvous time is Δ tran = 0.540606  , and the first impulse is a tangent impulse whose magnitude is 0.258068 m/s.The final velocity direction of the chaser in the ECI frame is [0.317860, 0, −0.948138] T and that of the target is [0.317854, 0, −0.948140] T ; thus they are almost the same.The magnitude of the second impulse is 0.258676 m/s.The curve of relative position is plotted in Figure 5, which includes the initial coasting arc and the rendezvous arc.The final position error is [61.0958,−105.1616,0] T m in the LVLH frame of the target.The total cost of two impulses for the bitangent rendezvous is 0.516744 m/s.

Conclusions
In this paper the short-range coplanar orbital rendezvous problem with tangent impulses on the chaser is studied using the linear relative motion with the  2 perturbation.Three cases are analyzed with either the first impulse or the second impulse or both impulses being tangent.For the first and the second tangent impulse problems, the numerical solutions are obtained by the secant method for a single variable function.The first tangent impulse can be used in the single-impulse interception problem.The bitangent rendezvous indicates that there exists the same solution for the first and the second tangent impulse problems, which can be obtained by considering the initial coasting time.This method is available for two coplanar elliptic orbits with the  2 perturbation.However, since this method is based on the linear relative motion equations, it is only suitable for the short-range case between two spacecraft.

Figure 1 :
Figure 1: The flight-direction angle and the LVLH frame.

Figure 3 : 8 𝜏Figure 4 :
Figure 3: Curves of relative position in the LVLH frame of the target for tangent to initial/final orbit.

Figure 5 :
Figure 5: Curves of relative position in the LVLH frame of the target for bitangent rendezvous.
Rendezvous.If the first impulse is not at the initial time, the solutions of Δ tran, 1 and Δ tran, 2 will change.For different initial coasting time Δ 1 , all solutions of Δ tran, 1 and Δ tran, 2 in one (mean) orbit period of the target and the corresponding costs are listed in Table1, which shows that there are many solutions of Δ tran, 1 when Δ 1 = 0.2  , Δ 1 = 0.8  , and Δ 1 = 0.9  .The plot of