Design of Small Circular Halo Orbit 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 Lagrange point of Sun-Earth system, L1 or L2 is conceived one of the best point for astronomy and connection ports for outer planets. This paper describes a new useful control method for realizing and keeping small circular Halo orbits 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 ephemeris. The resulted Halo orbit 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 Lissajous orbit also exists A Lagrange point is the point where the gravity 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 centrifugal force 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 satellites 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 satellite 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 circular orbit 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 barycenter Earth -10 r Maneuver(m/s x -12 -14 γ L2 -16 µ 1-µ L2 -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 .