# Tangent Orbital Rendezvous Using Linear Relative Motion withJ2 Perturbations

Authors: Gang Zhang; Dongzhe Wang; Xibin Cao; Zhaowei Sun
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

The tangent-impulse coplanar orbit rendezvous problem is studied based on the linear relative motion forJ2-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 J2 perturbations is given to verify the efficiency of these proposed techniques.


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## Body

## 1. 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 [1–4]. 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 two-impulse 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 solution-existence 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 theJ2 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 J2 perturbation, Gim and Alfriend [16] used the geometric method to obtain an STM for the elliptic target orbit with the influence of the J2 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 J2 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 theJ2 perturbation. Based on the STM in the elliptic target orbit with the influence of the J2 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.


## 2. Relative Motion with theJ2 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, theXℋ axis is taken to be the radius vector from the center of the Earth to the center of the target, the Zℋ axis is aligned with the angular momentum vector of the target orbit, and the Yℋ axis completes the right-hand 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 J2 perturbation (other external accelerations are neglected) in the LVLH frame of target can be written as [18]
(1)ddt[xyzx˙y˙z˙]=[000100000010000001a41a42a4302ωz0a51a52a53-2ωz02ωxa61a62a630-2ωx0][xyzx˙y˙z˙],
where X(t)=[x,y,z,x˙,y˙,z˙]T denotes the state vector and the elements in the coefficient matrix are  (2a)a41=ωz2+2μRt3+G0(1-3sin2i-tsin2u-t),(2b)a42=ω˙z+G0sin2i-tsin(2u-t),(2c)a43=-ωxωz+G0sin(2i-t)sinu-t,(2d)a51=-ω˙z+G0sin2i-tsin(2u-t),(2e)a52=ωx2+ωz2-μRt3+14G0[-1+sin2i-t(7sin2u-t-2)],(2f)a53=ω˙x-14G0sin(2i-t)cosu-t,(2g)a61=-ωxωz+G0sin(2i-t)sinu-t,(2h)a62=-ω˙x-14G0sin(2i-t)cosu-t,(2i)a63=ωx2-μRt3+14G0[-3+sin2i-t(5sin2u-t+2)],where G0=6nt2J2(a-t/Rt)3, J2=1.0826269×10-3, nt is the mean motion, R is the radius, a is the semimajor axis, i is the orbit inclination, and u=ϖ+θ is the argument of latitude, where ϖ is the argument of perigee and θ is the true anomaly. The symbol (·)¯ denotes the mean orbit elements. The subscript t denotes the target orbit.The angular velocity vector of the LVLH frame of the target is(3)ω=[ωx,ωy,ωz]T,
whose components are (4a)ωx=Ω˙tsinitsinut+i˙tcosut,(4b)ωy=Ω˙tsinitcosut-i˙tsinut=0,(4c)ωz=Ω˙tcosit+ϖ˙t+θ˙t,where Ω denotes the right ascension of ascending node. Considering the J2 perturbation, the derivatives of the orbital elements with respect to time are [16]
(5)i˙=-3J2RE24μ(1+ecosθ)3[a(1-e2)]7/2sin(2i)sin(2u),Ω˙=-3J2RE2μ(1+ecosθ)3[a(1-e2)]7/2cosisin2u,ϖ˙=-3J2RE22μ(1+ecosθ)4e[a(1-e2)]7/2×[(2+ecosθ1+ecosθ)(3sin2isinu-1)cosθ000+(2+ecosθ1+ecosθ)sinθsin2isin(2u)]-Ω˙cosi,θ˙=3μJ2RE2(1+ecosθ)32e[a(1-e2)]7/2×[(3sin2isin2u-1)(1+ecosθ)cosθ00+(2+ecosθ)sinθsin2isin(2u)]+μ(1+ecosθ)2[a(1-e2)]3/2,
where RE is the radius of the Earth, e is the eccentricity, and  μ is the gravity parameter. With (4a)–(5), the second-order derivatives of it,Ωt,ϖt,θt can be obtained; and then it is not difficult to obtain the derivatives of ω˙x and ω˙z in (2a)–(2i).


### 2.2. State Transition Matrix (STM)

To obtain the STM for the linear relative motion equations with theJ2 perturbation, the nonsingular orbital elements are defined as κ=[a,u,i,q1,q2,Ω]T, where q1=ecosϖ    and    q2=esinϖ. 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 δκ=κc-κt. The subscripts c and t denote the chaser orbit and the target orbit, respectively.The geometric transformation between the state vectorX and the differential nonsingular osculating orbital elements δκosc is [16]
(6)X(t)={A(t)+3J2RE2B(t)}δκosc(t),
where A(t) is the matrix for the two-body motion and B(t) is the matrix perturbed by J2.Let the matrixD(t) denote the transformation matrix from the relative mean elements to the relative osculating elements, and let the state transition ϕ-κ be for the relative mean elements; then we have [16]
(7)δκosc(t)=D(t)δκmean(t)=D(t)ϕ-κ(t,t0)δκmean(t0)=D(t)ϕ-κ(t,t0)D-1(t0)δκosc(t0),D(t)=∂κosc(t)∂κmean(t).
All the elements of the matrices A(t),B(t),ϕ-κ(t,t0),D(t) can be obtained in [16]. Finally, the STM for the relative states with J2 effects is
(8)ΦJ2(t,t0)={A(t)+3J2RE2B(t)}D(t)ϕ-κ(t,t0)D-1(t0)×{A(t0)+3J2RE2B(t0)}-1
and the state vector is
(9)X(t)=ΦJ2(t,t0)X(t0).


### 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 vectorρ in the earth-centered-inertial (ECI) frame can be written in terms of the derivative with respect to the LVLH frame of the target as
(10)dρdt=dρdt|rot+ω×ρ,
where ω 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 Rc and Vc, respectively, and those of the target are Rt and Vt, respectively. The transformation matrix Cℋℐ from the ECI to the LVLH frames is(11)Cℋℐ=[cosucosΩ-sinucosisinΩcosusinΩ+sinucosicosΩsinusini-sinucosΩ-cosucosisinΩ-sinusinΩ+cosucosicosΩcosusinisinisinΩ-sinicosΩcosi]. Using (11), the transformation matrices Ctℋℐ for the LVLH frame of the target and Ccℋℐ for the LVLH frame of the chaser are both obtained.Then the relative position vector in the LVLH frame of the target is(12)r=Ctℋℐ·(Rc-Rt).
Leting  ρ=r in (10), the relative velocity vector in the LVLH frame of the target is
(13)v=Ctℋℐ·(Rc-Rt)-ω×r.


## 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, theXℋ axis is taken to be the radius vector from the center of the Earth to the center of the target, the Zℋ axis is aligned with the angular momentum vector of the target orbit, and the Yℋ axis completes the right-hand 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 J2 perturbation (other external accelerations are neglected) in the LVLH frame of target can be written as [18]
(1)ddt[xyzx˙y˙z˙]=[000100000010000001a41a42a4302ωz0a51a52a53-2ωz02ωxa61a62a630-2ωx0][xyzx˙y˙z˙],
where X(t)=[x,y,z,x˙,y˙,z˙]T denotes the state vector and the elements in the coefficient matrix are  (2a)a41=ωz2+2μRt3+G0(1-3sin2i-tsin2u-t),(2b)a42=ω˙z+G0sin2i-tsin(2u-t),(2c)a43=-ωxωz+G0sin(2i-t)sinu-t,(2d)a51=-ω˙z+G0sin2i-tsin(2u-t),(2e)a52=ωx2+ωz2-μRt3+14G0[-1+sin2i-t(7sin2u-t-2)],(2f)a53=ω˙x-14G0sin(2i-t)cosu-t,(2g)a61=-ωxωz+G0sin(2i-t)sinu-t,(2h)a62=-ω˙x-14G0sin(2i-t)cosu-t,(2i)a63=ωx2-μRt3+14G0[-3+sin2i-t(5sin2u-t+2)],where G0=6nt2J2(a-t/Rt)3, J2=1.0826269×10-3, nt is the mean motion, R is the radius, a is the semimajor axis, i is the orbit inclination, and u=ϖ+θ is the argument of latitude, where ϖ is the argument of perigee and θ is the true anomaly. The symbol (·)¯ denotes the mean orbit elements. The subscript t denotes the target orbit.The angular velocity vector of the LVLH frame of the target is(3)ω=[ωx,ωy,ωz]T,
whose components are (4a)ωx=Ω˙tsinitsinut+i˙tcosut,(4b)ωy=Ω˙tsinitcosut-i˙tsinut=0,(4c)ωz=Ω˙tcosit+ϖ˙t+θ˙t,where Ω denotes the right ascension of ascending node. Considering the J2 perturbation, the derivatives of the orbital elements with respect to time are [16]
(5)i˙=-3J2RE24μ(1+ecosθ)3[a(1-e2)]7/2sin(2i)sin(2u),Ω˙=-3J2RE2μ(1+ecosθ)3[a(1-e2)]7/2cosisin2u,ϖ˙=-3J2RE22μ(1+ecosθ)4e[a(1-e2)]7/2×[(2+ecosθ1+ecosθ)(3sin2isinu-1)cosθ000+(2+ecosθ1+ecosθ)sinθsin2isin(2u)]-Ω˙cosi,θ˙=3μJ2RE2(1+ecosθ)32e[a(1-e2)]7/2×[(3sin2isin2u-1)(1+ecosθ)cosθ00+(2+ecosθ)sinθsin2isin(2u)]+μ(1+ecosθ)2[a(1-e2)]3/2,
where RE is the radius of the Earth, e is the eccentricity, and  μ is the gravity parameter. With (4a)–(5), the second-order derivatives of it,Ωt,ϖt,θt can be obtained; and then it is not difficult to obtain the derivatives of ω˙x and ω˙z in (2a)–(2i).


## 2.2. State Transition Matrix (STM)

To obtain the STM for the linear relative motion equations with theJ2 perturbation, the nonsingular orbital elements are defined as κ=[a,u,i,q1,q2,Ω]T, where q1=ecosϖ    and    q2=esinϖ. 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 δκ=κc-κt. The subscripts c and t denote the chaser orbit and the target orbit, respectively.The geometric transformation between the state vectorX and the differential nonsingular osculating orbital elements δκosc is [16]
(6)X(t)={A(t)+3J2RE2B(t)}δκosc(t),
where A(t) is the matrix for the two-body motion and B(t) is the matrix perturbed by J2.Let the matrixD(t) denote the transformation matrix from the relative mean elements to the relative osculating elements, and let the state transition ϕ-κ be for the relative mean elements; then we have [16]
(7)δκosc(t)=D(t)δκmean(t)=D(t)ϕ-κ(t,t0)δκmean(t0)=D(t)ϕ-κ(t,t0)D-1(t0)δκosc(t0),D(t)=∂κosc(t)∂κmean(t).
All the elements of the matrices A(t),B(t),ϕ-κ(t,t0),D(t) can be obtained in [16]. Finally, the STM for the relative states with J2 effects is
(8)ΦJ2(t,t0)={A(t)+3J2RE2B(t)}D(t)ϕ-κ(t,t0)D-1(t0)×{A(t0)+3J2RE2B(t0)}-1
and the state vector is
(9)X(t)=ΦJ2(t,t0)X(t0).


## 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 vectorρ in the earth-centered-inertial (ECI) frame can be written in terms of the derivative with respect to the LVLH frame of the target as
(10)dρdt=dρdt|rot+ω×ρ,
where ω 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 Rc and Vc, respectively, and those of the target are Rt and Vt, respectively. The transformation matrix Cℋℐ from the ECI to the LVLH frames is(11)Cℋℐ=[cosucosΩ-sinucosisinΩcosusinΩ+sinucosicosΩsinusini-sinucosΩ-cosucosisinΩ-sinusinΩ+cosucosicosΩcosusinisinisinΩ-sinicosΩcosi]. Using (11), the transformation matrices Ctℋℐ for the LVLH frame of the target and Ccℋℐ for the LVLH frame of the chaser are both obtained.Then the relative position vector in the LVLH frame of the target is(12)r=Ctℋℐ·(Rc-Rt).
Leting  ρ=r in (10), the relative velocity vector in the LVLH frame of the target is
(13)v=Ctℋℐ·(Rc-Rt)-ω×r.


## 3. 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 vectorX(tf)=[rfT,vfT]T can be obtained from the initial state vector as
(14)[rfvf]=[Φ11(tf,t0)Φ12(tf,t0)Φ21(tf,t0)Φ22(tf,t0)][r0v0].
If the final time tf is fixed and the final relative position rf is assigned, for example, rf=[0,0,0]T for the interception and rendezvous problem, the required initial relative velocity in the LVLH frame of the target is
(15)vh0=-Φ12-1Φ11r0+Φ12-1rf.Then, the first velocity impulse in the LVLH frame of the target is(16)Δvt0=vh0-v0,
where v0 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
(17)Δvc0=Ccℋℐ(Ctℋℐ)T·Δvt0.
The normalized vector of Δvc0 is
(18)Δv^c0=Δvc0∥Δvc0∥=[Δv^c0xΔv^c0yΔv^c0z].
For the coplanar case, Δv^c0z=0 is satisfied.The flight-direction angleγ 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 γ∈(0,π). At the initial time, the flight-direction angle of the chaser γc0 can be written as a function only of the initial true anomaly θc0(19)γc0=π2-arctan(ecsinθc01+eccosθc0).Figure 1
The flight-direction angle and the LVLH frame.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:(20)F1≜-Δv^c0x+Δv^c0ytanγc0=0.
Note that γc0 is a function only of t0, and Δv^c0 can be obtained by (18) if the final time tf is fixed. Therefore, F1 is a function only of the final time tf or the transfer time Δttran≜tf-t0 if the first impulse is at the initial time. Thus, the value of Δttran for F1(Δttran)=0 is required to be solved.


### 3.2. Tangent to the Final Orbit

After the first impulse, the relative velocity vector at the initial time reachesvh0; then the relative velocity vector at the final time tf can be obtained from (14) as
(21)vf=Φ21r0+Φ22vh0=(Φ21-Φ22Φ12-1Φ11)r0+Φ22Φ12-1rf.For the rendezvous problem, the required final relative velocity is zero. Thus, the second impulse in the LVLH frame of the target is(22)Δvtf=0-vtf=-vtf.
The normalized vector of Δvtf is
(23)Δv^tf=Δvtf∥Δvtf∥≡[Δv^tfxΔv^tfyΔv^tfz].
For the coplanar case, Δv^tfz=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, sincerf=[0,0,0]T is satisfied for the rendezvous problem. At the final time, the flight-direction angle of the target is
(24)γtf=π2-arctan(etsinθtf1+etcosθtf).
Therefore, the second impulse is tangent if and only if
(25)F2≜-Δv^tfx+Δv^tfytanγtf=0.
It is known that F2 is also a function only of the transfer time tf or Δttran if the first impulse is at the initial time. Thus, the value of Δttran for F2(Δttran)=0 will be solved for the tangent to final orbit problem.


### 3.3. 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 ofΔttran is required to simultaneously satisfy two conditions, that is, F1=0 and F2=0. However, if the first impulse is at the initial time, there may not exist this value of Δttran.When considering the initial coasting timeΔt1≜t1-t0, the first impulse is not at the initial time t0 but at the time t1. Then F1 and F2 are functions of two variables, that is, t1 and tf or Δt1 and Δttran. Therefore, two variables Δt1 and Δttran are to be solved for the following equations:
(26)F1(Δt1,Δttran)=0,F2(Δt1,Δttran)=0.The total energy cost for the two-impulse rendezvous is(27)Δvtotal=∥Δvc0∥+∥Δvtf∥=∥Δvt0∥+∥Δvtf∥.


## 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 vectorX(tf)=[rfT,vfT]T can be obtained from the initial state vector as
(14)[rfvf]=[Φ11(tf,t0)Φ12(tf,t0)Φ21(tf,t0)Φ22(tf,t0)][r0v0].
If the final time tf is fixed and the final relative position rf is assigned, for example, rf=[0,0,0]T for the interception and rendezvous problem, the required initial relative velocity in the LVLH frame of the target is
(15)vh0=-Φ12-1Φ11r0+Φ12-1rf.Then, the first velocity impulse in the LVLH frame of the target is(16)Δvt0=vh0-v0,
where v0 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
(17)Δvc0=Ccℋℐ(Ctℋℐ)T·Δvt0.
The normalized vector of Δvc0 is
(18)Δv^c0=Δvc0∥Δvc0∥=[Δv^c0xΔv^c0yΔv^c0z].
For the coplanar case, Δv^c0z=0 is satisfied.The flight-direction angleγ 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 γ∈(0,π). At the initial time, the flight-direction angle of the chaser γc0 can be written as a function only of the initial true anomaly θc0(19)γc0=π2-arctan(ecsinθc01+eccosθc0).Figure 1
The flight-direction angle and the LVLH frame.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:(20)F1≜-Δv^c0x+Δv^c0ytanγc0=0.
Note that γc0 is a function only of t0, and Δv^c0 can be obtained by (18) if the final time tf is fixed. Therefore, F1 is a function only of the final time tf or the transfer time Δttran≜tf-t0 if the first impulse is at the initial time. Thus, the value of Δttran for F1(Δttran)=0 is required to be solved.


## 3.2. Tangent to the Final Orbit

After the first impulse, the relative velocity vector at the initial time reachesvh0; then the relative velocity vector at the final time tf can be obtained from (14) as
(21)vf=Φ21r0+Φ22vh0=(Φ21-Φ22Φ12-1Φ11)r0+Φ22Φ12-1rf.For the rendezvous problem, the required final relative velocity is zero. Thus, the second impulse in the LVLH frame of the target is(22)Δvtf=0-vtf=-vtf.
The normalized vector of Δvtf is
(23)Δv^tf=Δvtf∥Δvtf∥≡[Δv^tfxΔv^tfyΔv^tfz].
For the coplanar case, Δv^tfz=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, sincerf=[0,0,0]T is satisfied for the rendezvous problem. At the final time, the flight-direction angle of the target is
(24)γtf=π2-arctan(etsinθtf1+etcosθtf).
Therefore, the second impulse is tangent if and only if
(25)F2≜-Δv^tfx+Δv^tfytanγtf=0.
It is known that F2 is also a function only of the transfer time tf or Δttran if the first impulse is at the initial time. Thus, the value of Δttran for F2(Δttran)=0 will be solved for the tangent to final orbit problem.


## 3.3. 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 ofΔttran is required to simultaneously satisfy two conditions, that is, F1=0 and F2=0. However, if the first impulse is at the initial time, there may not exist this value of Δttran.When considering the initial coasting timeΔt1≜t1-t0, the first impulse is not at the initial time t0 but at the time t1. Then F1 and F2 are functions of two variables, that is, t1 and tf or Δt1 and Δttran. Therefore, two variables Δt1 and Δttran are to be solved for the following equations:
(26)F1(Δt1,Δttran)=0,F2(Δt1,Δttran)=0.The total energy cost for the two-impulse rendezvous is(27)Δvtotal=∥Δvc0∥+∥Δvtf∥=∥Δvt0∥+∥Δvtf∥.


## 4. Solution Procedure

### 4.1. Tangent to the Initial/Final Orbit Problem

Assume that the first impulse is at the initial timet0. Then for the tangent to initial orbit problem, the value of Δttran is to be solved for F1=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, Δttran∈(0,T-t], where T-t denotes the mean orbit period of the target. For different values of Δttran, the value of F1(Δttran) will change. Then, the extreme points of F1 can be obtained by numerical methods. Assume that in the range (0,T-t] there are m extreme points satisfying Δttran,s1<Δttran,s2<⋯<Δttran,sm. Then there are altogether m+1 piecewise ranges in (0,T-t], that is, (0,Δttran,s1],(Δttran,s1,Δttran,s2],…,(Δttran,sm,T-t]. If there is no extreme point in (0,T-t], then F1(Δttran) will decrease or increase monotonically. For each piecewise range (Δttran,g1,Δttran,g2], there are two cases.(1)
IfF1(Δttran,g1+δ)·F1(Δttran,g2)>0, where δ is a positive constant small enough, there will be no solution in (Δttran,g1,Δttran,g2].(2)
IfF1(Δttran,g1+δ)·F1(Δttran,g2)<0, there will be a single solution in (Δttran,g1,Δttran,g2], which can be obtained by the secant method as
(28)Δttran(n+2)=Δttran(n+1)-(Δttran(n+1)-Δttran(n))×F1(Δttran(n+1))F1(Δttran(n+1))-F1(Δttran(n)),n=1,2,3,…,where Δttran(1)=Δttran,g1+δ,Δttran(2)=Δttran,g2.For the tangent to final orbit problem,F2(Δttran)=0 is required to be solved. By using the same method for F1(Δttran)=0, the tangent to final orbit problem can be solved with (28) by only replacing F2 with F1.


### 4.2. Bitangent Orbit Rendezvous Problem

For the bitangent orbit rendezvous problem, two variablesΔt1∈(0,T-t]; and Δttran∈(0,T-t] in (26) are solved. Although the sequential two-point secant method and the sequential N+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 timeΔt1∈(0,T-t], the first impulse is at t1=t0+Δt1∈(t0,t0+T-t]; the transfer time Δttran in one mean orbital period of the target for the tangent to initial/final orbit problem can be obtained by the proposed method as Δttran,F1 and Δttran,F2. There may exist many solutions of Δttran,F1 and Δttran,F2 for a given initial coasting time Δt1. Define a new function
(29)τ(Δt1)=Δttran,F1(j1)-Δttran,F2(j2),
where j1 is the j1th solution of Δttran,F1 and j2 is the j2th solution of Δttran,F2 such that the expression [Δttran,F1-Δttran,F2]2 is minimized. Then τ=0 should be satisfied for the bitangent rendezvous problem. It should be notified that τ is a function only of Δt1. Then, by using a numerical iterative algorithm, for example, the secant method, all solutions of Δt1 can be obtained in the range (0,T-t].


## 4.1. Tangent to the Initial/Final Orbit Problem

Assume that the first impulse is at the initial timet0. Then for the tangent to initial orbit problem, the value of Δttran is to be solved for F1=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, Δttran∈(0,T-t], where T-t denotes the mean orbit period of the target. For different values of Δttran, the value of F1(Δttran) will change. Then, the extreme points of F1 can be obtained by numerical methods. Assume that in the range (0,T-t] there are m extreme points satisfying Δttran,s1<Δttran,s2<⋯<Δttran,sm. Then there are altogether m+1 piecewise ranges in (0,T-t], that is, (0,Δttran,s1],(Δttran,s1,Δttran,s2],…,(Δttran,sm,T-t]. If there is no extreme point in (0,T-t], then F1(Δttran) will decrease or increase monotonically. For each piecewise range (Δttran,g1,Δttran,g2], there are two cases.(1)
IfF1(Δttran,g1+δ)·F1(Δttran,g2)>0, where δ is a positive constant small enough, there will be no solution in (Δttran,g1,Δttran,g2].(2)
IfF1(Δttran,g1+δ)·F1(Δttran,g2)<0, there will be a single solution in (Δttran,g1,Δttran,g2], which can be obtained by the secant method as
(28)Δttran(n+2)=Δttran(n+1)-(Δttran(n+1)-Δttran(n))×F1(Δttran(n+1))F1(Δttran(n+1))-F1(Δttran(n)),n=1,2,3,…,where Δttran(1)=Δttran,g1+δ,Δttran(2)=Δttran,g2.For the tangent to final orbit problem,F2(Δttran)=0 is required to be solved. By using the same method for F1(Δttran)=0, the tangent to final orbit problem can be solved with (28) by only replacing F2 with F1.


## 4.2. Bitangent Orbit Rendezvous Problem

For the bitangent orbit rendezvous problem, two variablesΔt1∈(0,T-t]; and Δttran∈(0,T-t] in (26) are solved. Although the sequential two-point secant method and the sequential N+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 timeΔt1∈(0,T-t], the first impulse is at t1=t0+Δt1∈(t0,t0+T-t]; the transfer time Δttran in one mean orbital period of the target for the tangent to initial/final orbit problem can be obtained by the proposed method as Δttran,F1 and Δttran,F2. There may exist many solutions of Δttran,F1 and Δttran,F2 for a given initial coasting time Δt1. Define a new function
(29)τ(Δt1)=Δttran,F1(j1)-Δttran,F2(j2),
where j1 is the j1th solution of Δttran,F1 and j2 is the j2th solution of Δttran,F2 such that the expression [Δttran,F1-Δttran,F2]2 is minimized. Then τ=0 should be satisfied for the bitangent rendezvous problem. It should be notified that τ is a function only of Δt1. Then, by using a numerical iterative algorithm, for example, the secant method, all solutions of Δt1 can be obtained in the range (0,T-t].


## 5. Numerical Examples

At the initial time, the nonsingular mean orbital elements of the target are(30)a-t=RE+10000km,u-t=20deg,i-t=90deg,q-1t=0.295442,q-2t=0.0520945,Ω-t=0deg,
where RE=6378.13km is the radius of the Earth. Then, it is known that e-t=0.3, ϖ-t=10deg and θ-t0=10deg. The differential mean orbital elements between the chaser and the target are
(31)δa-=-1km,δu-=-0.021deg,δi-=0deg,δq-1=-1.96053×10-4,δq-2=-3.98826×10-5,δΩ-=0deg.
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 r0=[2.4020,-4.2185,0]Tkm, and the initial relative velocity vector is v0=[-0.7223,-2.6583,0]Tm/s. The mean orbit period of the target is T-t=20859.684s. The range of the coasting time is Δt1∈(0,T-t] and that of the transfer time is Δttran∈(0,T-t].


### 5.1. Tangent to Initial/Final Orbit

If the impulse is imposed at the initial time, the curves of functionsF1 and F2 for different transfer time Δttran can be plotted in Figure 2. By using (28), the solution for F1=0 can be obtained as Δttran,F1=0.704717T-t and that for F2=0 can be obtained as Δttran,F2=0.059408T-t.Figure 2
FunctionsF1 and F2 for different Δttran.For the tangent to initial orbit problem,Δttran,F1=0.704717T-t, 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 J2 perturbation. The final position error is [11.9539,-31.7167,0]T m in the LVLH frame of the target.Figure 3
Curves of relative position in the LVLH frame of the target for tangent to initial/final orbit.For the tangent to final orbit problem,Δttran,F2=0.059408T-t, 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.


### 5.2. Bitangent Rendezvous

If the first impulse is not at the initial time, the solutions ofΔttran,F1 and Δttran,F2 will change. For different initial coasting time Δt1, all solutions of Δttran,F1 and Δttran,F2 in one (mean) orbit period of the target and the corresponding costs are listed in Table 1, which shows that there are many solutions of Δttran,F1 when Δt1=0.2T-t, Δt1=0.8T-t, and Δt1=0.9T-t. The plot of τ versus Δt1 can be plotted in Figure 4. The final solution for τ=0 can be obtained as Δt1=0.974968Tt-, which is the coasting time for the bitangent rendezvous and the corresponding rendezvous time is Δttran=0.540606T-t, 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.Table 1
Solutions ofΔttran,F1 and Δttran,F2 and their costs for different Δt1.
Tangent to initial orbit
Tangent to final orbit
Δ
t
1
Δ
t
tran
,
F
1
Δ
v
total
Δ
t
tran
,
F
2
Δ
v
total
(Orbit period)
(Orbit period)
(m/s)
(Orbit period)
(m/s)
0
0.704717
0.825201
0.059408
9.837243
0.1
0.502039
0.982136
0.033730
22.20027
0.2
0.004492
158.3517
0.918635
1.034115
0.260872
1.949606
0.3
0.897311
1.111503
0.561613
1.039989
0.4
0.801650
1.142299
0.342601
1.704567
0.5
0.707957
1.160767
0.193872
3.458322
0.6
0.615953
1.163889
0.097424
8.342175
0.7
0.526142
1.147815
0.034026
28.92826
0.8
0.006292
184.3432
0.875984
1.035060
0.440795
1.101123
0.983633
1.313181
0.9
0.022954
51.75954
0.644049
0.655018
0.370941
0.977307
0.837305
0.828427Figure 4
τ versus Δt1.Figure 5
Curves of relative position in the LVLH frame of the target for bitangent rendezvous.


## 5.1. Tangent to Initial/Final Orbit

If the impulse is imposed at the initial time, the curves of functionsF1 and F2 for different transfer time Δttran can be plotted in Figure 2. By using (28), the solution for F1=0 can be obtained as Δttran,F1=0.704717T-t and that for F2=0 can be obtained as Δttran,F2=0.059408T-t.Figure 2
FunctionsF1 and F2 for different Δttran.For the tangent to initial orbit problem,Δttran,F1=0.704717T-t, 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 J2 perturbation. The final position error is [11.9539,-31.7167,0]T m in the LVLH frame of the target.Figure 3
Curves of relative position in the LVLH frame of the target for tangent to initial/final orbit.For the tangent to final orbit problem,Δttran,F2=0.059408T-t, 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.


## 5.2. Bitangent Rendezvous

If the first impulse is not at the initial time, the solutions ofΔttran,F1 and Δttran,F2 will change. For different initial coasting time Δt1, all solutions of Δttran,F1 and Δttran,F2 in one (mean) orbit period of the target and the corresponding costs are listed in Table 1, which shows that there are many solutions of Δttran,F1 when Δt1=0.2T-t, Δt1=0.8T-t, and Δt1=0.9T-t. The plot of τ versus Δt1 can be plotted in Figure 4. The final solution for τ=0 can be obtained as Δt1=0.974968Tt-, which is the coasting time for the bitangent rendezvous and the corresponding rendezvous time is Δttran=0.540606T-t, 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.Table 1
Solutions ofΔttran,F1 and Δttran,F2 and their costs for different Δt1.
Tangent to initial orbit
Tangent to final orbit
Δ
t
1
Δ
t
tran
,
F
1
Δ
v
total
Δ
t
tran
,
F
2
Δ
v
total
(Orbit period)
(Orbit period)
(m/s)
(Orbit period)
(m/s)
0
0.704717
0.825201
0.059408
9.837243
0.1
0.502039
0.982136
0.033730
22.20027
0.2
0.004492
158.3517
0.918635
1.034115
0.260872
1.949606
0.3
0.897311
1.111503
0.561613
1.039989
0.4
0.801650
1.142299
0.342601
1.704567
0.5
0.707957
1.160767
0.193872
3.458322
0.6
0.615953
1.163889
0.097424
8.342175
0.7
0.526142
1.147815
0.034026
28.92826
0.8
0.006292
184.3432
0.875984
1.035060
0.440795
1.101123
0.983633
1.313181
0.9
0.022954
51.75954
0.644049
0.655018
0.370941
0.977307
0.837305
0.828427Figure 4
τ versus Δt1.Figure 5
Curves of relative position in the LVLH frame of the target for bitangent rendezvous.


## 6. 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 theJ2 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 J2 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.



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*Source: 531672-2013-04-23.xml*