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Design of Small Circular Halo around L2

Kohta Tarao* and Jun’ichiro Kawaguchi** *Graduate Student, Department of Aeronautics and Astronautics, School of Engineering, The University of Tokyo ** The Institute of Space and Astronautics Science / Japan Aerospace Exploration Agency (ISAS/JAXA) 3-1-1 Yoshinodai, Sagamihara, Kanagawa 229-8510, JAPAN.

Abstract A point of - system, L1 or L2 is conceived one of the best point for astronomy and connection ports for outer . This paper describes a new useful control method for realizing and keeping small circular Halo around the L2 point. First, a control law is mathematically derived from the restricted three body problem formulation and the associated properties are discussed. And next, some useful cases for the space mission are verified numerically using real . The resulted is compact and circular realized with satisfactorily small thrust. The paper concludes the strategy is applied to many kinds of missions.

L2 点周りの小円ハロー軌道設計 多羅尾康太(東大・工・院)、川口淳一郎(ISAS/JAXA) 摘要 太陽―月・地球系におけるラグランジュ点の中で、L2 点は太陽と地球との距離がほぼ一定で、太 陽、地球との幾何学的関係が保たれ、またダウンリンク通信が太陽の影響を受けない点で L1 点よ りも有利である。そのため L2 点周辺の軌道は、赤外線天文衛星や外惑星探査のための起・終点と しての宇宙港の軌道として利用が考えられている。本論文では、新しい制御法により L2 点回りに 小さな円形の Halo 軌道が作れることを示す。まず初めに、円制限三体問題における線形化された 方程式から理論的に制御則を導く。その後、実軌道要素を用いて軌道制御方法を検証する。軌道制 御に必要な推力は十分小さく大変実用的な制御方法である。

INTRODUCTION million kilometers and also due to its highly alternating geometry. A also exists A is the point where the of around the L2 point naturally, however, the orbital two planets such as Sun and Earth is equal to plane rotates and the solar power availability is by their circular motions. L1 and often restricted due to the shadow by the Earth. L2 have comparatively low pseudo gravity potential Therefore, very small circular Halo orbits are among five Lagrange points, so can be needed for avoiding the shadow and maintaining reached to these points easily. Especially, in the the geometry. Such orbits do not exist naturally and Sun-Earth system, L1 or L2 is conceived the best they must rely on certain propulsion system. point for infrastructure that bridges the Earth There have been a variety of investigations about domain and interplanetary space, since it maintains the orbit control around the L2 point. However not only constant distance from the Sun and the those on small circular Halo orbits have never been Earth but also geometry with them. The L2 point well discussed so far. If the can provide any might be better than the L1 point, since the acceleration vector, it’s easy to realize small downlink communication is not affected by the circular Halo orbits with some control method such solar plasma. Therefore, there have been proposed a as LQR. Figure 1 shows a with LQR lot of orbit plans around the L2 point for infrared method. In upper-right figure, inner circle shows astronomy missions and a space port departing and the area of sun’s shadow and outer circle is the orbit. returning for outer planets cruise. Around the L2 So the satellite can avoid the shadow area easily. point, a natural Halo orbit exists. However, it is not Figure 2 shows thrust profile to maintain this orbit, practically useful due to its large size up to a which indicates that a continuous and variable SMALL CIRCULAR HALO ORBIT thrust is required. Concerning the propulsion system, chemical propulsion is not available to This section describes the formulation of a control strategy which realizes a small circular Halo maintain this orbit because of its large fuel orbit around L2. This strategy is mathematically consumption in continuous maneuvers. Electric obtained from linearized equations of motion in the propulsion is suitable for continuous maneuvers in circular restricted three body problem. This strategy terms of fuel consumption. However it’s difficult to provide any thrust vector, because its thrust is requires the following conditions: the thrust is practically small and nearly constant. Therefore, we constant and its direction rotates at a constant angular rate. In the following explanation, the could not realize this kind of small circular Halo primary body is the Sun and the secondary is the orbit with existing propulsion system. Earth. This paper discusses the way to realize circular closed loop orbits around L1 and L2 points with a constant thrust and constraining the thrust direction Derivation of control law In the circular restricted three body problem, in a certain plane. The paper assumes the thrusters nonlinear equations of motion around L2 point in are ion engines which have higher specific impulse nondimensional form are than others taking the fuel consumption into 1−mm account. The control law developed in the paper x−21yx −+−+() mg =−() x ++− 1 g() xgax + + LLLrr33 results in a uniform rotation around the Earth to L2 12 1−mm (1) line axis and is really useful for spacecraft design. yxy+−=−2 y − yay + rr33 The strategy obtained is a new and very simple one 12 1−mm and proprietary to Japan Aerospace Exploration zzzaz=− − + 1 33 Agency (JAXA) . rr12

where γ L denotes the distance from Earth to L2, x 104 x-y(LQR) x 104 y-z(LQR) 2 2 and µ =Me/(Ms+Me). Representative time here is 1 1 normalized with the Earth’s revolution period, x,y,z 0 0

y(km) z(km) are normalized with the distance between Sun and -1 -1

-2 -2 Earth and ax,ay,az are the elements of the thrust

-2 -1 0 1 2 -2 -1 0 1 2 4 4 acceleration respectively. Figure 3 shows the x(km) x 10 y(km) x 10 4 4 x 10 x 10 x-z(LQR) Orbit(LQR) coordinate system for these equations. The 2 2 1 equations of motion are linearized as 1 0

0 z(km) -1 xy− 221  −+=( BxaxL )

z(km) -2 -1 2 4   (2) -2 x 10 0 y ++21xB()L −y =ay 2 y(km) -2 0 1 4 -2 -1 0 1 2 -2 -1 x 10 4 x(km) x(km) x 10  zBzaz+=L Figure 1 Controlled Orbit with LQR method where,    1− µµ (3) BL =+  3 γ 3 x 10-6 Required Maneuver(LQR) ()1+ γ L L  2 

0 In Eqs.(2), if we choose an initial condition as:

-2 w xy (4) -4 x(0)==zx (0) 0, (0) = yzwy (0), (0) =z (0) )

2 ay k -6 z az Spacecraft -8 y ax Sun Earth -10 r Maneuver(m/s x -12

-14 L2 µ 1-µ γL2 -16 -18 0 100 200 300 400 500 600 700 800 Figure 3 Circular Restricted Three Body Time(day) Problem Figure 2 Thrust Profile z z the solution for the linearized equations of motion z’ y z’ y is θ control vector φ θ φ y(0) Sun Earth Sun Earth x()tt= sinω x’ x’ k xy x x (5) yt()= y (0)cosωxy t projection plane control plane zt()= y (0)sinω t z Figure 4 Projection Plane and Control Plane where ω 2 ++21B k = xy L . (6) 1cos(+ θ −φ ) (14) 2ωxy If θ ≠ φπ+−(2n 1) (n is integer), fourth order This is known for Lissajous solution. To obtain the coefficient is positive. Zero-th order coefficient is circular orbit, we assume the following relations. transformed to zy'=== sinωtxbzyy , ' ', cosωt (7) 22 00−−−{(1BBLL + 2 )(1cos( +θ +φθφ ))12 ++ B L +−− (12 BB LL + )sinsin} These equations are similar to Lissajous solution (sinθφ sin≥ 0) with the constraint that the trajectory projected on a (15) T plane perpendicular to ()cosθ ,0,sinθ axis should 22 −−−{(1BBLL + 2 )(1cos( +θ −φθφ ))12 ++ B L + (13 − B L )sinsin} be circular. This requirement is illustrated in figure (sinθφ sin< 0) 4. Then we obtain (16) x =−yt0 ()β cosθθω sin sin Since BL is nearly equal to 4, zero-th order . (8) yy= 0 cosω t coefficient is negative. If θ =+φπ(2n − 1) , the

zy=+0 ()cosθ βθ sin sin ω t equation is 22 2 2 (17) Similarly, we impose the control thrust to be {3(3+ BBBLLL−−++−= 2)sinφω} (12 (3 1)sin)0 φ constant and to rotate at a constant angular rate on This equation has positive solution of ω . T the plane perpendicular to ()cosφ ,0,sinφ axis as If sinφ = 0 , existence ofω is proved in the same illustrated in figure4. Then the acceleration vector is way. written as Thus it is proved that, in this strategy, we can aT =−()αφsin sinωαtt , cos ω , α cosφ sinω t. (9) choose any projection plane including y axis and Substituting Eq. (8) and Eq. (9) into Eq (2), then we any control rotation axis in x-z plane, depending on obtain following equations. our needs. yyBy(sinθ −+++−=βθωω cos) 2 2( 2 1) ( sin θβθα cos) sinφ 000L Application for space mission (10) Considering about application for space mission, 2 −−yy00ω 2sincos()θβ − θω +−=( ByL 1) 0 α (11) the orbit should not pass through shadow area. 2 Therefore we choose y-z plane as a projection plane, −+yBy00()cossinθ βθω +L ( cossincos θβθ +=−) αφ that is,θ = nπ . Then we can easily design as the (12) satellite avoids the shadow area as illustrated in Eliminatingα and β using Eq (10), Eq (11), and Eq figure 5. In this case, the relation between β and (12). (sinφ ≠ 0) ω is 43 (1++ cosθφ cos sin θφω sin ) ++ 2(sin θ sin φω ) ωω42+ 42BB−++ 212 βωβω 2∓ 122210 BB ++−= 2 (18) { ()()LL} LL ++{1BB + ( − 2)cosθ cosφθφωθφ − 2 B sin sin }2 − 2 B (sin + sin )ω LL L L Upper sign is forθ =2nπ and lower forθ =−(2n 1)π . 22 2 −+{2BBLL +−−+ (12)coscos( BB LLθφ −+ BB LL )sinsin}0 θφ = For β to exist, the following equation has to be (13) formed. This is a fourth order equation in terms ofω . In this 22 64221221ωωBBB− 42+−++ ω ω 2 +−> B0 (19) equation, the fourth order coefficient is opposite ()LLL{ ( )( )}( L) sign of zero-th order coefficient. Therefore So,ω is in the range: ω exists for anyθ andφ . In fact, in Eq (13), fourth 1.6581< ω < 2.0091 , (20) order coefficient can be transformed as where ω is dimensionless. Since ω =1 is x 10-4 Elements of Control Thrust corresponding to the angular rate of 360deg/year, 1.5 the period of resulted Halo orbit is from 0.5 to 0.7 ay(β(+)) year. Ifθ = 2nπ , β is 1 ay(β(-)) 24 4 22 6BBLLωωω±++−+++− (6 ) ( 2(2 B L 1) ω (2 B L 1) )(2 ω 2 B L 1) 0.5 . az(θ=(2n-1)π) β= 442 ωω+−++2(2BBLL 1) (2 1) 0 ax(β(-))

(21) ax,ay,az ax(β(+)) If θ =−(2n 1)π , β is opposite sign of θ = 2nπ . -0.5

Figure 6 shows the relation between ω and β . -1 Then elements of thrust are written as az(θ=2nπ) 2 -1.5 1.6 1.7 1.8 1.9 2 2.1 aBxL=−2(ω ωβ + 2 + 1) ω 2 . (22) aByL=−ω + −12 + ωβ Figure 7 Elements of Control Thrust

aBzL=−∓()ω ω=1.7136 ω=2.0082 Figure 7 shows the elements of control thrust. If ax z z is equal to zero, ω is 1.7136 or 2.0082, which means the thrust rotation axis is x axis (the Earth to θ=2nπ y y L2 line axis). It is very useful for astronomy satellite because the satellite can expose the same plane toward Sun by spinning around x-axis. If az is equal to zero, ω is 1.976. It’s also useful. z z Especially, ifω is 1.7136, the direction of control θ=(2n-1)π y y vector is same as radial direction of circular orbit. This is very useful for the space port because we can orient the direction of the entrance to the radial Figure 8 Relation between Orbit and Control direction of circular orbit and then we can easily Vector control the traffic system. Figure 8 shows the relation between orbit and control thrust vector. There is another orbit to be useful for space mission. If the trajectory is completely circular and z Shadow the control thrust vector is parallel to the orbital Projection Spacecraft plane in the x-y-z frame defined in figure 3, it’s also y Plane useful for the space mission. This condition is met x by choosing β = 0 and θ =+φπn . In this case, Sun barycenter Earth L2 θ and ω are obtained uniquely as

−1 22()BBBLLL−−+ 1()() 2 1 3 1 Figure 5 Application for Space Missions θ =−sin ,ωθφπ = ( = + 2n ) ()()3131BB−+() 6 B − 2 LL L Existing Curve of Control Thrust (23) 0.4 −1 (24) θ =+=+=+−sin( 1/(3BBLL 1),) ωθφπ (3 1)/2 ( (2 n 1)) 0.3 The inclination of orbit plane is 0.2 θ=2nπ ±71.57deg (θ =+φπ 2n ) and ±73.17( θ =+φπ (2n − 1) ) . This 0.1 orbit can be also applied for deep space ports.

0

β

-0.1 Numerical Simulation θ=(2n-1)π -0.2 Numerical simulations are conducted to verify

-0.3 the derived small circular Halo orbit control in case of projection plane is y-z plane and ax=0. First, the -0.4 1.6 1.65 1.7 1.75 1.8 1.85 1.9 1.95 2 2.05 2.1 ω control law derived is applied to the nonlinear equation of motion of the circular restricted three Figure 6 Existing Curve of β body problem. Secondly we verify it using real ephemeris of Sun, Earth and . After several x 104x-y(Nonlinear) x 104y-z(Nonlinear) trial-and-error iterations, we select the initial 2 2 conditions as Table 1. Figure 9 and figure 10 show 1 1 0 0 the orbit motion of the satellite in nonlinear y(km) z(km) equation of motion. In figure 9, L2 shifts for -1 -1 -2 -2 approximately 6186km to barycenter. Period is -2 0 2 -2 0 2 -7 4 4 213.15days, acceleration is 5.7571×10 m/sec and x(km) x 10 x 104 y(km)y(km) x 10 x 104x-z(Nonlinear) Trajectory(Nonlinear) the ∆V are 18.156 m/sec/year. In figure 10, L2 2 2 1 shifts for approximately 6190km to barycenter. 1 -8 0 0 Period is 181.88days, acceleration is 7.6245×10 z(km)

z(km) -1 m/sec and ∆V is 2.4045 m/sec/year. Figure 11 and -1 -2 -2 2 figure 12 show this strategy works out even in real 0 x 104 -2 -2 0 2 -2 0 2 4 y(km) x(km) x 10 ephemeris in spite of the complicated motion. x(km) x 104 Acceleration and ∆V are same as these in figure 9 Figure 10 Nonlinear Simulation and figure 10 respectively and period is about (ω=2.0082,θ=(2n-1)π) 210days. In any case, acceleration and ∆V are very small. Therefore, this strategy can be used for space mission with existing propulsion system such as x 104 x-y x 104 y-z electric propulsion or solar sails. 2 2

1 1 Table 1 0 0 y(km) z(km) INITIAL CONDITIONS (position) -1 -1 xyz -2 -2 -2 0 2 -2 0 2 Figure8 -6185.9645184/AU 15000/AU 0 4 4 x(km) x 10 x 104 y(km)y(km) x 10 Figure9 -6190.0843209/AU 15000/AU 0 x 104 x-z Trajectory(Real Ephemeris) Figure10 -1564.72089/AU 15000/AU 0 2 2 Figure11 -1477.8948/AU 15000/AU 0 1 1 0 0 z(km) z(km) -1 -1 Table 2 -2 -2 20 INITIAL CONDITIONS (velocity) 4 -2 -2 0 2 x 10 -2 0 2 4 4 y(km) x(km) x 10 vx vy vz x(km) x 10 Figure8 ωβy(0) 0 -ωy(0) Figure 11 Real Ephemeris Simulation Figure9 ωβy(0) 0 -ωy(0) (ω=1.7136,θ=(2n-1)π) Figure10 ωβy(0) 0.00047 -ωy(0) Figure11 ωβy(0) 0.00047 -ωy(0)

x 104 x-y x 104 y-z x 104x-y(Nonlinear) x 104y-z(Nonlinear) 2 2

2 2 1 1 1 1 0 0 0 0 y(km) z(km) -1 -1 y(km) z(km) -1 -1 -2 -2 -2 -2 -2 0 2 -2 0 2 4 4 -2 0 2 -2 0 2 x(km) x 10 x 104 y(km)y(km) x 10 4 4 4 x(km) 4 y(km)y(km) x-z Trajectory(Real Ephmeris) x 10 x 10 x 10 x 10 4x-z(Nonlinear) x 10 Trajectory(Nonlinear) 2 2 2 2 1 1 1 1 0 0

0 z(km) -1 z(km) 0 z(km) -1 z(km) -1 -2 -1 -2 -2 2 -2 2 4 0 4 0 x 10 -2 -2 0 2 -2 0 2 4 x 10 -2 2 y(km) x(km) x 10 -2 0 2 y(km) -2 0 4 x(km) 4 4 x(km) x 10 x 10 x(km) x 10 Figure 12 Real Ephemeris Simulation Figure 9 Nonlinear Simulation (ω=2.0082,θ=(2n-1)π) (ω=1.7136,θ=(2n-1)π) CONCLUSION This paper shows the control strategy to realize small circular Halo orbit around L2. We can apply this strategy to any kind of system: Sun-Earth, Earth-Moon, etc. Resulted Halo orbit is well compact (about 15000km from L2) realized with satisfactory small thrust (about 0.1mN at 1000kg satellite).

This strategy allows choosing any projection plane including y axis (perpendicular to the Earth to L2 line axis and on the orbital plane of second ) and any rotation axis of thrust vector on the x-z plane (perpendicular to y axis). Therefore we can apply this strategy to many types of missions. If y-z plane is a projection plane, we can easily design the orbit to avoid the shadow area. Especially, if the rotation axis of thrust vector is x-axis (the Earth to L2 line axis), it’s useful for satellite and deep space port. If the orbit is circular and the thrust vector locks on the orbit plane, it’s especially also useful for space missions.

REFERENCES 1. Jun’ichiro Kawaguchi, Kohta Tarao, “The Way of Trajectory Planning for the Satellite”, Patent application number 2005-210266 2. Bong Wie, SPACE VEHICLE DYNAMICS AND CONTROL, AIAA EDUCATION SERIES, 1998, Chapter 3, pp. 240-256. 3. David Cielaszyk and Bong Wie, “New Approach to Halo Orbit Determination and Control”, Journal of Guidance, Control, and Dynamics, vol.19, No.2, March-April 1996, pp.266-273. 4. David L. Richardson, “Halo Orbit Formulation for the ISEE-3 Mission”, Journal of Guidance and Control, Vol.3, No.6, Nov-Dec. 1980, pp.543-548. 5. Keiko Kuroshima, Kohta Tarao, Jun’ichiro Kawaguchi, “Candidate Orbit for Infrared Astronomical Satellite”, 14th Workshop on Astrodynamics and Flight Mechanics, 2004, pp.318-323.