Stability of the Lagrange Points,L4 and L5 Thomas Greenspan January 7,2014 Abstract A proof of the stability of the non collinear Lagrange Points,L4 and Ls.We will start by covering the basics of stability,stating a theorem (without proof)with a few corollaries, and then turn to the Lagrange points,proving first the stability of all Lagrange points in the z-direction and then restricting our attention to thethe points,L4 and Ls. Introduction The three body problem is one that has been studied for many centuries.It consists of considering 3 bodies,subject to only to mutually attracting forces (determined by the inverse square force, gravitation)and to solve for the motions of the three bodies.This task can be extremely complicated and,following Poincare's example,most mathematicians and physicists have interested themselves in the finding periodic solutions to the problem,that is,where x(t)=x(t+T)where T is the period [3.Indeed not only does this offer a useful framework from within which to work,but also describes states that are most applicable and useful to real world situations.Finding these periodic solutions is by no means trivial.It is a problem that has been worked on for centuries and still is in modern times,one of the latest publications dealing with it as recent as March of 2013 [9]. Among the first to interest themselves in and find solutions to the 3 body problem were Leonhard Euler in 1765 and Joseph Lagrange in 1772 [3].In his publication,Essai sur le probleme des trois corps (Essay on the 3-Body Problem),Lagrange proposed a method that had never been used until then,that of considering only the distances between the three bodies rather than their absolute positions [2].Through this method,he found that there are exactly 5 different configurations the three bodies can be arranged in so that their movement is both circular and periodic.Given the initial position of two of the masses (usually the largest masses),the 5 different locations that the third body can be in such that the solution is circular and periodic are called the 5 Lagrange or Libration points. Lagrange himself did not actually believe that instances of the 5 Libration points existed in the "real world",as he states at the beginning of his derivation [2].Indeed,it was not until 1906 that the first example,an asteroid sharing Jupiter's orbit but ahead of it by close to 60 with respect to the sun,was discovered by professor Max Wolf at Heidelberg [8].Because the asteroid and those that were subsequently discovered at the L4 and Ls points of the Jupter-sun system were named after characters in the Illiad,asteroids in L4 and Ls of any system are generally referred to as Trojan asteroids
Stability of the Lagrange Points, L4 and L5 Thomas Greenspan January 7, 2014 Abstract A proof of the stability of the non collinear Lagrange Points, L4 and L5. We will start by covering the basics of stability, stating a theorem (without proof) with a few corollaries, and then turn to the Lagrange points, proving first the stability of all Lagrange points in the z-direction and then restricting our attention to the the points, L4 and L5. Introduction The three body problem is one that has been studied for many centuries. It consists of considering 3 bodies, subject to only to mutually attracting forces (determined by the inverse square force, gravitation) and to solve for the motions of the three bodies. This task can be extremely complicated and, following Poincar´e’s example, most mathematicians and physicists have interested themselves in the finding periodic solutions to the problem, that is, where x(t) = x(t + T) where T is the period [3]. Indeed not only does this offer a useful framework from within which to work, but also describes states that are most applicable and useful to real world situations. Finding these periodic solutions is by no means trivial. It is a problem that has been worked on for centuries and still is in modern times, one of the latest publications dealing with it as recent as March of 2013 [9]. Among the first to interest themselves in and find solutions to the 3 body problem were Leonhard Euler in 1765 and Joseph Lagrange in 1772 [3]. In his publication, Essai sur le probl`eme des trois corps (Essay on the 3-Body Problem), Lagrange proposed a method that had never been used until then, that of considering only the distances between the three bodies rather than their absolute positions [2]. Through this method, he found that there are exactly 5 different configurations the three bodies can be arranged in so that their movement is both circular and periodic. Given the initial position of two of the masses (usually the largest masses), the 5 different locations that the third body can be in such that the solution is circular and periodic are called the 5 Lagrange or Libration points. Lagrange himself did not actually believe that instances of the 5 Libration points existed in the “real world”, as he states at the beginning of his derivation [2]. Indeed, it was not until 1906 that the first example, an asteroid sharing Jupiter’s orbit but ahead of it by close to 60◦ with respect to the sun, was discovered by professor Max Wolf at Heidelberg [8]. Because the asteroid and those that were subsequently discovered at the L4 and L5 points of the Jupter-sun system were named after characters in the Illiad, asteroids in L4 and L5 of any system are generally referred to as Trojan asteroids. 1
Figure 1:Lagrange Points The three first points,L1,L2, L3 are all aligned with the earth-sun vector.L4 and Ls form an equilateral triangle with the sun and earth. The arrows indicate potentials and show the potential peaks, 5 wells and saddles. To find these points,Lagrange needed to make some approximations (indeed,the system is known today to be chaotic otherwise).The results nevertheless are often quite good at accurately describing dynamics of our solar system.In our case we consider the circular restricted 3-body problem as described below. The circular restricted 3-body problem The assumptions are as follows: Two of the masses,mi and m2,are much heavier than the third one which we thus consider to be negligible.The center of mass is thus on a line between mi and m2. The masses follow a circular orbit around the center of mass. We will describe our system by putting it in the form: 正X因=AX阳 1 Stability and its consequences In any dynamical system with one or more equilibrium points,it is important to know whether the equilibrium points are stable,that is,whether a point at equilibrium can be subjected to small perturbations and still stay close to the equilibrium point.We use the following definition: Definition An equilibrium solution,zo,to a dynamical system is considered: stable if for every small >0,there exists a 6>0 such that every solution,(t),with initial conditions l(to)-zoll 6 is such that ll(t)-zoll Vt to. asymptotically stable if,in addition to being stable,there exists 6o >0 such that every solution, x(t),with initial conditions (to)-zoll 6o is such that x(t)>zo as too. In words,an equilibrium point is stable if,given a small distance e,there always exists a distance, 6,such that any solution with initial conditions within 6 of the equilibrium point will always stay within e of the equilibrium for any time t.As we will see,Lagrange Points do not fulfill the much stronger condition of asymptotically stability but they are stable in some cases. 2
Figure 1: Lagrange Points The three first points, L1, L2, L3 are all aligned with the earth-sun vector. L4 and L5 form an equilateral triangle with the sun and earth. The arrows indicate potentials and show the potential peaks, wells and saddles. To find these points, Lagrange needed to make some approximations (indeed, the system is known today to be chaotic otherwise). The results nevertheless are often quite good at accurately describing dynamics of our solar system. In our case we consider the circular restricted 3-body problem as described below. The circular restricted 3-body problem The assumptions are as follows: • Two of the masses, m1 and m2, are much heavier than the third one which we thus consider to be negligible. The center of mass is thus on a line between m1 and m2. • The masses follow a circular orbit around the center of mass. We will describe our system by putting it in the form: d dtX(t) = AX(t) 1 Stability and its consequences In any dynamical system with one or more equilibrium points, it is important to know whether the equilibrium points are stable, that is, whether a point at equilibrium can be subjected to small perturbations and still stay close to the equilibrium point. We use the following definition: Definition An equilibrium solution, x0, to a dynamical system is considered: • stable if for every small ε > 0, there exists a δ > 0 such that every solution, x(t), with initial conditions ||x(t0) − x0|| 0 such that every solution, x(t), with initial conditions ||x(t0) − x0|| < δ0 is such that x(t) → x0 as t → ∞. In words, an equilibrium point is stable if, given a small distance ε, there always exists a distance, δ, such that any solution with initial conditions within δ of the equilibrium point will always stay within ε of the equilibrium for any time t. As we will see, Lagrange Points do not fulfill the much stronger condition of asymptotically stability but they are stable in some cases. 2
Beyond characterizing the dynamical system in question,considering stability is extremely impor- tant for any real-world applications.Indeed it is virtually impossible for any object to be precisely at an actual equilibrium point.Thus it is crucial to know what an object will do when close to an equilibrium for these are the actual dynamics that the object will necessarily follow. Before starting,we need a theorem and a few corollaries. Theorem 1.1 Given a system X'(t)=AX(t)where A has distinct paired complex eigenvalues, a1+iB1,a1-iB1...,ak+iBk:ak-iBx.Let T be the matrix such that T-AT is in canonical form: B T where Then the general solution of X'(t)=AX(t)is TY(t)where aleat cos Bit+breait sin Bit -aleait sin Bit+bieait cos Bit Y(t)= (1.1) akeaxt cos Bkt+bkeakt sin Bkt -akeakt sin Brt bkeakt cos Bxt The proof of this theorem is beyond the scope of this paper,but can be found in chp.6 of Differential Equations,Dynamical Systems,and an Introduction to Chaos 4]. Corollary 1.2 A point in a dynamical system whose matrir of equations of motion,A,has an eigenvalue with positive real part is unstable. Proof Let m be the index s.t.am >0.To show the point is stable,it suffices to show that the point is unstable in only one direction.To do so we let ai=bi=0Vim and assume that either am0 or bm0(or both).This is effectively considering small perturbations of the equilibrium point in only one direction.We denote ci;=(T)i.i(where T is the matrix described in theorem 1.1). We begin by noting that T is invertible so every column must have at least one non-zero entry and no column is the multiple of another column.This means that,for every 1 0.For any row,n,we get: E>am(Cn.2m-1-Cn.2m) -cmt[Cn.2m-1(am cos Bmt+om sin Bmt)+cn,2m (-am cos Bmt+om sin Bmt)] (1.2) There are two cases to consider: 3
Beyond characterizing the dynamical system in question, considering stability is extremely important for any real-world applications. Indeed it is virtually impossible for any object to be precisely at an actual equilibrium point. Thus it is crucial to know what an object will do when close to an equilibrium for these are the actual dynamics that the object will necessarily follow. Before starting, we need a theorem and a few corollaries. Theorem 1.1 Given a system X0 (t) = AX(t) where A has distinct paired complex eigenvalues, α1 +iβ1, α1 −iβ1 . . . , αk +iβk, αk −iβk. Let T be the matrix such that T −1AT is in canonical form: T −1AT = B1 . . . Bk where Bi = αi βi −βi αi Then the general solution of X0 (t) = AX(t) is T Y (t) where Y (t) = a1e α1t cos β1t + b1e α1t sin β1t −a1e α1t sin β1t + b1e α1t cos β1t . . . ake αkt cos βkt + bke αkt sin βkt −ake αkt sin βkt + bke αkt cos βkt (1.1) The proof of this theorem is beyond the scope of this paper, but can be found in chp. 6 of Differential Equations, Dynamical Systems, and an Introduction to Chaos [4]. Corollary 1.2 A point in a dynamical system whose matrix of equations of motion, A, has an eigenvalue with positive real part is unstable. Proof Let m be the index s.t. αm > 0. To show the point is stable, it suffices to show that the point is unstable in only one direction. To do so we let ai = bi = 0 ∀i 6= m and assume that either am 6= 0 or bm 6= 0 (or both). This is effectively considering small perturbations of the equilibrium point in only one direction. We denote ci,j = (T)i,j (where T is the matrix described in theorem 1.1). We begin by noting that T is invertible so every column must have at least one non-zero entry and no column is the multiple of another column. This means that, for every 1 ≤ j ≤ k, there exists an i such that ci,2j−1 6= 0 or ci,2j 6= 0 and ci,2j−1 6= ±ci,2j (if this second condition were not the case then we would have (T)2j−1 = ±(T)2j which is impossible). Suppose the point is stable. Then we have ||T Y (0) − T Y (t)|| 0. For any row, n, we get: ε > am(cn,2m−1 − cn,2m) − e αmt [cn,2m−1 (am cos βmt + bm sin βmt) + cn,2m (−am cos βmt + bm sin βmt)] (1.2) There are two cases to consider: 3
·Case1:bm≠0. Let n be the index s.t.cn,2m-1≠0orcm,2m≠0 and cn,2m-1≠-cn,2m·For simplicity we consider only the values t=(k2+/2)/B;for k EZ.Our equation then becomes: eamtbm[cn.2m-1+cn.2m]am(cn.2m-1-Cn.2m) As temm[cn.2m-1+cn.2m]dominates and the value goes to oo depending on the sign of bm Cn.2m-1+cn.2m.Thus the solution is clearly not stable. ·Case2:bm=0. Similarly to the first case,let n be the index s.t.cn.2m-10or cn.2m0and cn.2m-1Cn.2m. Again for simplicity,consider only the values t =(k2m)/B;for k EZ.Our equation then becomes: e>eamtam[cn.2m-1-cn.2m]+am(cn.2m-1-cn.2m) Astt[Cn.2m-1-cn.2m]dominates and the value goes to depending on the sign of am Cn.2m-1-cn.2m.Thus the solution is clearly not stable. Thus in either case,the point is not stable. ◆ Corollary 1.3 A point in a dynamical system whose matrix of equations of motion,A,has purely imaginary (non-zero)eigenvalues is stable. Proof Suppose that ai=0 Vi in(1.1).As seen in(1.2)the components of the solution becomes: lcn2i-1(a:cos月t+b:sin,t)+cn.2i(-a:cos3,t+bsin3,t切 i=1 Since time,t,only appears in sines and cosines,it is clear that these values are completely bounded. Thus by choosing a1,...,ak and bi,...,bk appropriately (i.e.small enough),we can make the solutions stay within e of the initial state for any s>0 and we thus conclude that the points are stable. ◆ 2 The Lagrange Points Recall from lecture that the first step we used to finding the Lagrange Points is to consider the bodies in a rotating reference frame such that the two heavier masses mi and m2 do not move. Furthermore,since the movement of the three planets is planar,we can define the plane they move in to be the z-y plane with both heavier masses on the z-axis(with the origin at the center of mass).Let R be the distance between m and m2,we then have the positions of the two heavier masses as: r1= m2R ,0,0 miR r2= -,0,0 (2.1) m1+m2 m1+m2 Proposition 2.1 The angular frequency of the rotating reference frame,S,is given by: 22R3=G(M1+M2) (2.2) This only holds because we are considering the circular restricted 3-body problem as discussed above and follows directly from Kepler's third law(for more information see Landau Livshits p. 231])
• Case 1: bm 6= 0. Let n be the index s.t. cn,2m−1 6= 0 or cn,2m 6= 0 and cn,2m−1 6= −cn,2m. For simplicity we consider only the values t = (k2π + π/2)/βi for k ∈ Z. Our equation then becomes: ε > e αmt bm cn,2m−1 + cn,2m + am(cn,2m−1 − cn,2m) As t → ∞, e αmt bm cn,2m−1 + cn,2m dominates and the value goes to ±∞ depending on the sign of bm cn,2m−1 + cn,2m . Thus the solution is clearly not stable. • Case 2: bm = 0. Similarly to the first case, let n be the index s.t. cn,2m−1 6= 0 or cn,2m 6= 0 and cn,2m−1 6= cn,2m. Again for simplicity, consider only the values t = (k2π)/βi for k ∈ Z. Our equation then becomes: ε > e αmt am cn,2m−1 − cn,2m + am(cn,2m−1 − cn,2m) As t → ∞, e αmtam cn,2m−1 − cn,2m dominates and the value goes to ±∞ depending on the sign of am cn,2m−1 − cn,2m . Thus the solution is clearly not stable. Thus in either case, the point is not stable. . Corollary 1.3 A point in a dynamical system whose matrix of equations of motion, A, has purely imaginary (non-zero) eigenvalues is stable. Proof Suppose that αi = 0 ∀i in (1.1). As seen in (1.2) the components of the solution becomes: X k i=1 [cn,2i−1 (ai cos βit + bi sin βit) + cn,2i (−ai cos βit + bi sin βit)] Since time, t, only appears in sines and cosines, it is clear that these values are completely bounded. Thus by choosing a1, . . . , ak and b1, . . . , bk appropriately (i.e. small enough), we can make the solutions stay within ε of the initial state for any ε > 0 and we thus conclude that the points are stable. . 2 The Lagrange Points Recall from lecture that the first step we used to finding the Lagrange Points is to consider the bodies in a rotating reference frame such that the two heavier masses m1 and m2 do not move. Furthermore, since the movement of the three planets is planar, we can define the plane they move in to be the x-y plane with both heavier masses on the x-axis (with the origin at the center of mass). Let R be the distance between m1 and m2, we then have the positions of the two heavier masses as: r1 = − m2R m1 + m2 , 0, 0 r2 = m1R m1 + m2 , 0, 0 (2.1) Proposition 2.1 The angular frequency of the rotating reference frame, Ω, is given by: Ω 2R 3 = G(M1 + M2) (2.2) This only holds because we are considering the circular restricted 3-body problem as discussed above and follows directly from Kepler’s third law (for more information see Landau & Livshits p. 23 [1]). 4
Solving the results found in lecture,We find our Lagrange points,L4 and Ls: m1-m2 2 Ls= m1-m2 2R0 (2.3) m1+m2 m1+m2 To find out the dynamical stability of motion near the equilibrium points,we need to look at the generalized potential about the mass m3.Because we are in a rotating frame we must add both the coriolis acceleration and the centrifugal acceleration.Let r be the position vector of m3 and n the angular velocity such that =(0,0,)Furthermore let di and d2 be the distances between m3 and mi and m2 respectively: 2 =x+ m2R +y2+22, mR +y2+z2 (2.4) m1+m2 m1+m2 Proposition 2.2 The total acceleration,f,and the generalized potential,U are given by Gm1c-r_Gm2r-r2l-2n×i-×2×r d d (2.5) 0=-Gm1-Gm2-2mx-)-,(2+y) (2.6) d山d2 The second to last and last elements correspond to the coriolis and centrifugal accelerations respec- tively(for further information on these forces see Landau Livshits p.128 [1]). Note that,in addition to position,the potential is dependent on velocity and,although it has no effect on the position of the Lagrange points,it must be taken into account when looking at their stability.To do so we separate the components dependent on velocity: 0'=0+2(y-y)=-Gm-Gm2-2 2 2+y2) (2.7) d d Reducing (2.5)by components we get: Gm（c+品 Gm2（x-m1+m2 mIR +22g+22x= aU' 龙三 的 +22 (2.8) =- Gmy Gmay-20i+2y=- 8U' -22i d ay (2.9) 艺=、 Gm1z Gm22 0U' (2.10) d 8z Finally,it can be useful to rewrite the generalized potential as a sum of partial derivatives.Using the Taylor series expansion of U'around the Lagrange point (ro,y0,z0)we get (to second order): U'=U6+U(x-xo)+Ug(g-0)+U(z-20) +号【.e-oP+U-%P+U.e-2o月 +Uy(x-x0)(g-0)+U2(x-x0)(z-20)+Ug2(y-0)(z-20) 5
Solving the results found in lecture, We find our Lagrange points, L4 and L5: L4 = R 2 m1 − m2 m1 + m2 , √ 3 2 R, 0 ! , L5 = R 2 m1 − m2 m1 + m2 , − √ 3 2 R, 0 ! (2.3) To find out the dynamical stability of motion near the equilibrium points, we need to look at the generalized potential about the mass m3. Because we are in a rotating frame we must add both the coriolis acceleration and the centrifugal acceleration. Let r be the position vector of m3 and Ω the angular velocity such that Ω = (0, 0, Ω). Furthermore let d1 and d2 be the distances between m3 and m1 and m2 respectively: d 2 1 = x + m2R m1 + m2 2 + y 2 + z 2 , d2 2 = x − m1R m1 + m2 2 + y 2 + z 2 (2.4) Proposition 2.2 The total acceleration, ¨r, and the generalized potential, U are given by ¨r = − Gm1(r − r1) d 3 1 − Gm2(r − r2) d 3 2 − 2Ω × ˙r − Ω × Ω × r (2.5) U = − Gm1 d1 − Gm2 d2 − 2Ω(xy˙ − yx˙) − Ω 2 2 (x 2 + y 2 ) (2.6) The second to last and last elements correspond to the coriolis and centrifugal accelerations respectively (for further information on these forces see Landau & Livshits p. 128 [1]). Note that, in addition to position, the potential is dependent on velocity and, although it has no effect on the position of the Lagrange points, it must be taken into account when looking at their stability. To do so we separate the components dependent on velocity: U 0 = U + 2Ω(xy˙ − yx˙) = − Gm1 d1 − Gm2 d2 − Ω 2 2 (x 2 + y 2 ) (2.7) Reducing (2.5) by components we get: x¨ = − Gm1 x + m2R m1+m2 d 3 1 − Gm2 x − m1R m1+m2 d 3 2 + 2Ω ˙y + Ω2x = − ∂U0 ∂x + 2Ω ˙y (2.8) y¨ = − Gm1y d 3 1 − Gm2y d 3 2 − 2Ω ˙x + Ω2 y = − ∂U0 ∂y − 2Ω ˙x (2.9) z¨ = − Gm1z d 3 1 − Gm2z d 3 2 = − ∂U0 ∂z (2.10) Finally, it can be useful to rewrite the generalized potential as a sum of partial derivatives. Using the Taylor series expansion of U 0 around the Lagrange point (x0, y0, z0) we get (to second order): U 0 = U 0 0+U 0 x (x − x0) + U 0 y (y − y0) + U 0 z (z − z0) + 1 2 U 0 xx(x − x0) 2 + U 0 yy(y − y0) 2 + U 0 zz(z − z0) 2 +U 0 xy(x − x0)(y − y0) + U 0 xz(x − x0)(z − z0) + U 0 yz(y − y0)(z − z0) 5
where U=U for any variable X.Note,however,that Urz= 1(0,0,20 andU收=毁on Uv:=0 and furthermore we have that by definition,U2=U=U:=0 for Lagrange points.We thus get: U'-U+U(-zo)+U(w-w)+U:(+U(-z0)(w-w) (2.11) 3 Linearization and stability in the z-direction To analyze the stability about the equilibrium points in a system such as the 3-body problem,we linearize the equations of motion and look at the small perturbations.Note that because of how we have set up the reference frame,the Lagrange points are fired points.We thus have the following equations: x=x0十6x 立=6 (3.1) y=0十dy 9=6y (3.2) 2=20+02 交=68 (3.3) Plugging in these values into(2.11)then gives us: U=6+号g(iajP+U6wP+U.(6a月+U%iai (3.4) Using(2.8)-(2.10)and(3.4)we get our equations: 6元=-UJz6x-Um6y+226i (3.5) =-Uwy-U2yi证-206i (3.6) 6龙=-U:26z (3.7) We thus get the following: 0 0 0 1 0 01 0 0 0 0 1 0 2 d 0 0 0 0 0 1 di dt -Uz 0 0 22 0 6元 (3.8) 9 60 -U' 0 -22 0 0 6问 6 0 0 -U:: 0 0 0 62 We start by considering only the z-direction. Theorem 3.1 All Lagrange points are stable in the z-direction. Proof Let 6r =y=0.By (3.5)-(3.7)we can see that 6z and 6i are independent of 6r,6i,6y, and y and vice-versa so we can reduce our matrix to 品()）=(8.)（周 6
where U 0 0 = U 0 (x0,y0,z0) and U 0 X = ∂U ∂X (x0,y0,z0) for any variable X. Note, however, that Uxz = Uyz = 0 and furthermore we have that by definition, U 0 x = U 0 y = U 0 z = 0 for Lagrange points. We thus get: U 0 = U 0 0 + 1 2 U 0 xx(x − x0) 2 + U 0 yy(y − y0) 2 + U 0 zz(z − z0) 2 + U 0 xy(x − x0)(y − y0) (2.11) 3 Linearization and stability in the z-direction To analyze the stability about the equilibrium points in a system such as the 3-body problem, we linearize the equations of motion and look at the small perturbations. Note that because of how we have set up the reference frame, the Lagrange points are fixed points. We thus have the following equations: x = x0 + δx x˙ = δx˙ (3.1) y = y0 + δy y˙ = δy˙ (3.2) z = z0 + δz z˙ = δz˙ (3.3) Plugging in these values into (2.11) then gives us: U 0 = U 0 0 + 1 2 U 0 xx(δx) 2 + U 0 yy(δy) 2 + U 0 zz(δz) 2 + U 0 xyδxδy (3.4) Using (2.8)-(2.10) and (3.4) we get our equations: δx¨ = −U 0 xxδx − U 0 xyδy + 2Ωδy˙ (3.5) δy¨ = −U 0 yyδy − U 0 xyδx − 2Ωδx˙ (3.6) δz¨ = −U 0 zzδz (3.7) We thus get the following: d dt x y z x˙ y˙ z˙ = d dt δx δy δz δx˙ δy˙ δz˙ = 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 −U 0 xx −U 0 xy 0 0 2Ω 0 −U 0 xy −U 0 yy 0 −2Ω 0 0 0 0 −U 0 zz 0 0 0 δx δy δz δx˙ δy˙ δz˙ (3.8) We start by considering only the z-direction. Theorem 3.1 All Lagrange points are stable in the z-direction. Proof Let δx = δy = 0. By (3.5)-(3.7) we can see that δz and δz˙ are independent of δx, δx˙, δy, and δy˙ and vice-versa so we can reduce our matrix to d dt δz δz˙ = 0 1 −U 0 zz 0 δz δz˙ 6
By (2.10)we have that U2= Gmi Gm2 d d (3.9) Furthermore,since di,d2 are distances we have di,d2>0 and thus U'>0. The eigenvalues of the above matrix are: ±iV0g (3.10) Since these are always imaginary,we have by corollary 1.3 that the point is stable and thus conclude that all Lagrange points are stable in the z-direction. ■ In the same way (and for the same reason)that we were able to consider only the z-direction,we can do the opposite and consider only the x and y directions.Our equation becomes: Ox 0 0 0 Ox d 0 0 0 1 dt -U -UJ 0 22 6 (3.11) oy Iy -22 0 67 It is this equation that we consider for the rest of the paper. 4 Stability of L4 and L5 The discovery of asteroids or other astrological bodies at the L4 and L5 points of almost every planet-sun system seems to indicate quite strongly that these points are indeed very stable.However, this is not always the case and requires that the heaviest mass,mI be significantly heavier than the second heaviest mass,m2.We have the following theorem: Theorem 4.1 The Lagrange points,La and Ls are stable in all directions if and only if m1≥ 25+3V69 ≈24.9599 m2 2 (4.1) Proof We start by using(2.7)to compute the partial derivatives found in(3.11).Note that for L4 and L5,d=d2=R.Evaluating our partial double derivatives at L4 and Ls gives: 2 Gm2 3Gm(c+品)】 3Gm2（x- miR m1+m2 -22 d d d =(品)》 2 G(m1+m2） 3Gm1 (m1+m2)R +m2 -(m1+m2)R 2(n1+m2) 2(m1+m2) R3 R6 1G(m1+m2-02 4 R 7
By (2.10) we have that U 0 zz = Gm1 d 3 1 + Gm2 d 3 2 (3.9) Furthermore, since d1, d2 are distances we have d1, d2 > 0 and thus U 0 zz > 0. The eigenvalues of the above matrix are: ±i p U0 zz (3.10) Since these are always imaginary, we have by corollary 1.3 that the point is stable and thus conclude that all Lagrange points are stable in the z-direction. . In the same way (and for the same reason) that we were able to consider only the z-direction, we can do the opposite and consider only the x and y directions. Our equation becomes: d dt δx δy δx˙ δy˙ = 0 0 1 0 0 0 0 1 −U 0 xx −U 0 xy 0 2Ω −U 0 xy −U 0 yy −2Ω 0 δx δy δx˙ δy˙ (3.11) It is this equation that we consider for the rest of the paper. 4 Stability of L4 and L5 The discovery of asteroids or other astrological bodies at the L4 and L5 points of almost every planet-sun system seems to indicate quite strongly that these points are indeed very stable. However, this is not always the case and requires that the heaviest mass, m1 be significantly heavier than the second heaviest mass, m2. We have the following theorem: Theorem 4.1 The Lagrange points, L4 and L5 are stable in all directions if and only if m1 m2 ≥ 25 + 3√ 69 2 ≈ 24.9599 (4.1) Proof We start by using (2.7) to compute the partial derivatives found in (3.11). Note that for L4 and L5, d1 = d2 = R. Evaluating our partial double derivatives at L4 and L5 gives: U 0 xx = Gm1 d 3 1 + Gm2 d 3 2 − 3Gm1 x + m2R m1+m2 2 d 5 1 − 3Gm2 x − m1R m1+m2 2 d 5 2 − Ω 2 x= R 2 m1−m2 m1+m2 = G(m1 + m2) R3 − 3Gm1 (m1+m2)R 2(m1+m2) 2 + m2 −(m1+m2)R 2(m1+m2) 2 R5 − Ω 2 = 1 4 G(m1 + m2) R3 − Ω 2 7
Uw= Gm2 3Gm1y2 3Gm2y2 -n2 d的 d碣 = G(m1+m2)3Gm1R2+3Gm2R2 R3 R5 -2 5G(m1+m2）-2 心、 4R3 3Gm(e+品)y3Gm2(e-品 m1+m2 d 3Gm1(慢)9R+3Gm2(-号)岁R R5 3V3Gi(m-ma)3G(m ma) 4R3 where=+(m1-m2)/(m1 m2).Using (2.2)we get: %.- Uw=-4 =- v3 4±22 (4.2) Our matrix from (3.11)thus becomes 0 0 1 0 0 0 1 2 3￥24n2 0 22 (4.3) 3K02 0 -22 0 The matrix has the four following eigenvalues: .2 4=±2/2-V27x4-23 2V2+V27星-23 (4.4) By corollary 1.2 and corollary 1.3,we have that all the eigenvalues must all be imaginary (otherwise there will necessarily be at least one eigenvalue with Re(A)>0,making the point unstable)so V2-V27r星-23 must be completely real..Note that since±l≤l,V27k￥-23≤2 so theonly remaining condition is: 271-23≥0 (4.5)） After a little algebraic manipulation this gives us mi 1+√翳 25+3v69 m2 1-V景 2 (4.6) which gives us the expected condition for the stability of the Lagrange points,L4 and L5. ◆
U 0 yy = Gm1 d 3 1 + Gm2 d 3 2 − 3Gm1y 2 d 5 1 − 3Gm2y 2 d 5 2 − Ω 2 ! y=± √ 3 2 R = G(m1 + m2) R3 − 3Gm1 3 4R2 + 3Gm2 3 4R2 R5 − Ω 2 = − 5 4 G(m1 + m2) R3 − Ω 2 U 0 xy = − 3Gm1 x + m2R m1+m2 y d 5 1 − 3Gm2 x − m1R m1+m2 y d 5 2 x= R 2 m1−m2 m1+m2 ,y=± √ 3 2 R = ∓ 3Gm1 R 2 √ 3 2 R + 3Gm2 − R 2 √ 3 2 R R5 = ∓ 3 √ 3G(m1 − m2) 4R3 = − 3 √ 3 4 κ± G(m1 + m2) R3 where κ± = ±(m1 − m2)/(m1 + m2). Using (2.2) we get: U 0 xx = − 3 4 Ω 2 U 0 yy = − 9 4 Ω 2 U 0 xy = − 3 √ 3 4 κ±Ω 2 (4.2) Our matrix from (3.11) thus becomes 0 0 1 0 0 0 0 1 3 4Ω 2 3 √ 3 4 κ±Ω 2 0 2Ω 3 √ 3 4 κ±Ω 2 9 4Ω 2 −2Ω 0 (4.3) The matrix has the four following eigenvalues: λ± = ±i Ω 2 r 2 − q 27κ 2 ± − 23 σ± = ±i Ω 2 r 2 + q 27κ 2 ± − 23 (4.4) By corollary 1.2 and corollary 1.3, we have that all the eigenvalues must all be imaginary (otherwise there will necessarily be at least one eigenvalue with r Re(λ) > 0, making the point unstable) so 2 − q 27κ 2 ± − 23 must be completely real. Note that since |κ±| ≤ 1, q 27κ 2 ± − 23 ≤ 2 so the only remaining condition is: 27κ 2 ± − 23 ≥ 0 (4.5) After a little algebraic manipulation this gives us m1 m2 ≥ 1 + q 23 27 1 − q 23 27 = 25 + 3√ 69 2 (4.6) which gives us the expected condition for the stability of the Lagrange points, L4 and L5. . 8
5 Conclusion This result would in fact be completely unexpected and surprising were it not for the discovery of actual instances of astrological bodies at these points in our Solar system.Indeed,by looking at the second partial derivatives in (4.2)we find that U=U=-2<0 and U=U =-2<0. This would indicate that L4 and L5 are at peaks(local maxima)of the potential in the z-y plane and would thus imply that that these points are extremely unstable.What gives L4 and Ls their stability is simply the coriolis force discussed previously.Initially,a body at L4 and Ls start moving away from the equilibrium point but,as the body pick up speed,the coriolis force takes effect sending the body into an effective orbit around the Lagrange point. Because of this effect,the areas around L4 and Ls that are effectively stable are in fact quite large as is illustrated by Figure 1.Indeed,this is the reason that so many trojan asteroids exist, some more than 5 off of 60,where the L4 and L5 are located.No satellites have been placed at these locations (unlike LI and L2 hosting the SOHO satellite [7]and the WMAP satellite [6] respectively despite their inherent instability),however,they were visited in 2009 by the STEREO satellites.As the only stable Lagrange points,L4 and L5 are unique phenomena in the solar system and the areas around them have been and are of great interest(for example as possible places of origin of the moon or locations from which to better observe solar storms [5])to the astrophysics community. References [1]E.M.Lifshitz L.D.Landau.Course of Theoretical Physics,volume 1.Elsevier Inc.,3rd edition, 2011. [2]Joseph Lagrange.Le Probleme des trois corps.1772. [3]Richard Montgomery.A new solution to the three-body problem.Technical report,Notices Amer.Math Soc,2001. [4]Robert L.Devaney Morris W.Hirsch,Stephen Smale.Differential Equations,Dynamical Sys- tems,and an Introduction to Chaos.Number 978-0-12-382010-5.Elsevier Inc.,3rd edition, 2013. [5]NASA.Stereo.http://www.nasa.gov/mission_pages/stereo/news/gravity_parking. html. [6]NASA.Wilkinson microwave anisotropy probe.http://map.gsfc.nasa.gov/. [7]NASA.Solar and heliospheric observatory.http://sohowww.nascom.nasa.gov/home.html, January 2014. [8 Seth B.Nicholson.The trojan asteroids.Leaflet of the Astronomical Society of the Pacific, 8:239.1961. [9]Milovan Dmitrasinovic.V.Suvakov.Three classes of newtonian three-body planar periodic orbits.Phys.Rev.Lett.,110:114301,Mar 2013. 9
5 Conclusion This result would in fact be completely unexpected and surprising were it not for the discovery of actual instances of astrological bodies at these points in our Solar system. Indeed, by looking at the second partial derivatives in (4.2) we find that Uxx = U 0 xx = − 3 4Ω 2 < 0 and Uyy = U 0 yy = − 9 4Ω 2 < 0. This would indicate that L4 and L5 are at peaks (local maxima) of the potential in the x-y plane and would thus imply that that these points are extremely unstable. What gives L4 and L5 their stability is simply the coriolis force discussed previously. Initially, a body at L4 and L5 start moving away from the equilibrium point but, as the body pick up speed, the coriolis force takes effect sending the body into an effective orbit around the Lagrange point. Because of this effect, the areas around L4 and L5 that are effectively stable are in fact quite large as is illustrated by Figure 1. Indeed, this is the reason that so many trojan asteroids exist, some more than 5◦ off of 60◦ , where the L4 and L5 are located. No satellites have been placed at these locations (unlike L1 and L2 hosting the SOHO satellite [7] and the WMAP satellite [6] respectively despite their inherent instability), however, they were visited in 2009 by the STEREO satellites. As the only stable Lagrange points, L4 and L5 are unique phenomena in the solar system and the areas around them have been and are of great interest (for example as possible places of origin of the moon or locations from which to better observe solar storms [5]) to the astrophysics community. References [1] E. M. Lifshitz L. D. Landau. Course of Theoretical Physics, volume 1. Elsevier Inc., 3rd edition, 2011. [2] Joseph Lagrange. Le Probl`eme des trois corps. 1772. [3] Richard Montgomery. A new solution to the three-body problem. Technical report, Notices Amer. Math Soc, 2001. [4] Robert L. Devaney Morris W. Hirsch, Stephen Smale. Differential Equations, Dynamical Systems, and an Introduction to Chaos. Number 978-0-12-382010-5. Elsevier Inc., 3rd edition, 2013. [5] NASA. Stereo. http://www.nasa.gov/mission_pages/stereo/news/gravity_parking. html. [6] NASA. Wilkinson microwave anisotropy probe. http://map.gsfc.nasa.gov/. [7] NASA. Solar and heliospheric observatory. http://sohowww.nascom.nasa.gov/home.html, January 2014. [8] Seth B. Nicholson. The trojan asteroids. Leaflet of the Astronomical Society of the Pacific, 8:239, 1961. [9] Milovan & Dmitra˘sinovi´c. V. Suvakov. Three classes of newtonian three-body planar periodic ˘ orbits. Phys. Rev. Lett., 110:114301, Mar 2013. 9