IntroductionEquation of MotionThe Jacobi IntegralLagrangian Equilibrium PointsLocation of Equilibrium PointsStability of Equilibrium PointsConclusionIntroductionEquation of MotionThe Jacobi IntegralLagrangian Equilibrium . . .Location of Equilibrium . . .Stability of Equilibrium . . .ConclusionHome PageTitle PageJJ IIJ IPage 1 of 28Go BackFull ScreenCloseQuitThe Restricted Three-Body ProblemShinichi TokoroMay, 22 2002AbstractThe method to find gravitational equilibrium point and stability of them willbe demonstrated.1. IntroductionWe will discuss about the gravitational interaction among three bodies in the caseof that third body has negligible mass compared with the other two bodies. Manyscientist has been analyzing three body problems and they are Euler, Lagrange, Laplace,Jacobi, Le Verrier, Hamiltion, Poincare, and Birkhoff. The general three body problemsare still not discovered thoroughly in stead of contribution of those scientists. If thetwo bodies move around in circular, coplanar orbits about their common center ofmass and the third mass is negligible so it doesn’t affect the motion of the other twobodies, and the problem of the third body is called the circular, restricted, three-bodyproblems. Although, real planetary orbital and movem ent are not coplanar and circular,the restricted three-body problem give us reasonable approximation for certain systems;also, the qualitative behaviour of the motion can be analyzed without too complicatingIntroductionEquation of MotionThe Jacobi IntegralLagrangian Equilibrium . . .Location of Equilibrium . . .Stability of Equilibrium . . .ConclusionHome PageTitle PageJJ IIJ IPage 2 of 28Go BackFull ScreenCloseQuitmethod. Mainly, we will discus the equation of the three-body problem, the locationand stability of equilibrium points in the case of circular restriction.2. Equation of MotionWe consider a tiny particle move around gravitational field of two mass m1and m2,and also two masses are creating circular orbit with their common center of mass. Weuse a set of axes a, b, c as a frame referred to the center of the system. Let the a axislie along the line between m1and m2at time t = 0 and the c axis perpendicular to thea − b plane; also, let the coordinates of the two masses in this frame be (a1, b1, c1) and(a2, b2, c2). Also, the angular velocity and a constant separation are fixed about theircommon center of mass. Let the unit of mass be chosen such that u = G(m1+m2) = 1,and we assume that m1> m2and define¯u =m2m1+ m2(1)then, in this system of units, the two masses are¯u = Gm1Gm1=1m1+ m2m1=m1+ m2m1+ m2−m2m1+ m2= 1 −m2m1+ m2u1= Gm1= 1 − ¯u (2)IntroductionEquation of MotionThe Jacobi IntegralLagrangian Equilibrium . . .Location of Equilibrium . . .Stability of Equilibrium . . .ConclusionHome PageTitle PageJJ IIJ IPage 3 of 28Go BackFull ScreenCloseQuitu2= Gm2=1m1+ m2m2u2= Gm2= ¯u (3)where ˆu < 1/2. Let the coordinates of the particle in the sidereal system, be (a, b, c)and the equation of motion of the particle arem¨r = ΣF=Gmm1r21~u +Gmm2r22~u¨r =u1r31a − 1 − a, b1− b, c1− c+u2r32a2− a, b2− b, c2− c< ¨a,¨b, ¨c > =u1r31a1− a, b1− b, c1− c+u2r32a2− a, b2− b, c2− c¨a = u1a1− ar31+ u2a2− ar32, (4)¨b = u1b1− br31+ u2b2− br32, (5)¨c = u1c1− cr31+ u2c2− cr32, (6)IntroductionEquation of MotionThe Jacobi IntegralLagrangian Equilibrium . . .Location of Equilibrium . . .Stability of Equilibrium . . .ConclusionHome PageTitle PageJJ IIJ IPage 4 of 28Go BackFull ScreenCloseQuitrr2r1xbyaOPu2u1Figure 1: A planar view of the relationship between the sidereal coordinates (a, b, c) ofthe particle at the point P . O is the center of mass of two bodieswhere, from Figure 1,r21= (a1− a)2+ (b1− b)2+ (c1− c)2, (7)r22= (a2− a)2+ (b2− b)2+ (c2− c)2. (8)In the general three-body problem, these equations still work because they don’trequire any assumptions about the shape of two mass’s orbital. We can rewrite theseIntroductionEquation of MotionThe Jacobi IntegralLagrangian Equilibrium . . .Location of Equilibrium . . .Stability of Equilibrium . . .ConclusionHome PageTitle PageJJ IIJ IPage 5 of 28Go BackFull ScreenCloseQuitequationsr21= (x + u2)2+ y2+ z2, (9)r22= (x − u2)2+ y2+ z2, (10)where (x, y, z) are the coordinates of the particle with respect to the rotating, or synodicsystem. Their coordinates are related to the coordinates in the sidereal system by thesimple rotation, so we write these in matrix formabc=cos nt −sin nt 0sin nt cos nt 00 0 1xyz(11)to obtain the equation of accelerations. We need to take the second derivertives of thes eequations by using the chain-rule.˙a˙b˙c=cos nt −sin nt 0sin nt cos nt 00 0 1˙x˙y˙z+−n sin nt −n cos nt 0n cos nt −n sin nt 00 0 0xyz=˙x cos nt − ˙y sin nt − xn sin nt − yn cos nt˙x sin nt + ˙y cos nt + xn cos nt − ny sin nt˙z˙a˙b˙c=cos nt −sin nt 0sin nt cos nt 00 0 1˙x − ny˙y + nx˙z. (12)IntroductionEquation of MotionThe Jacobi IntegralLagrangian Equilibrium . . .Location of Equilibrium . . .Stability of Equilibrium . . .ConclusionHome PageTitle PageJJ IIJ IPage 6 of 28Go BackFull ScreenCloseQuitSimilarly,¨a¨b¨c=cos nt −sin nt 0sin nt cos nt 00 0 1¨x − 2n ˙y − n2x¨y + 2n ˙x − n2y¨z. (13)In the above matrix equation n ˙x and n ˙y are called Corioli’s acceleration and n2xand n2y are called the centrifugal acceleration. Substitute for a, b, c, ¨a,¨b, and ¨c, in Eqs.(4),(5), and (6) become(¨x−2n ˙y−n2x) cos nt−(¨y+2n ˙x−n2y) sin nt =u1x1− xr31+ u2x2− xr32cos nt+u1r31+u2r32y sin nt,(14)(¨x−2n ˙y−n2x) sin nt+(¨y+2n ˙x−n2y) cos nt =u1x1− xr31+ u2x2− xr32sin nt−u1r31+u2r32y cos nt,(15)¨z = −u1r31+u2r32z. (16)After the calculation, we can write these acceleration as the gradient of a scalar functionU:¨x − 2n ˙y =∂U∂x, (17)¨y − 2n ˙x =∂U∂y, (18)¨z =∂U∂z, (19)IntroductionEquation of MotionThe Jacobi IntegralLagrangian Equilibrium . . .Location of Equilibrium . . .Stability of Equilibrium . . .ConclusionHome PageTitle PageJJ IIJ IPage 7 of 28Go BackFull ScreenCloseQuitwhere U = U(x, y, z) is given byU =n22(x2+ y2) +u1r1+u2r2. (20)3. The Jacobi IntegralIf we multiply Eq. (17) by ˙x, and Eq. (18) by ˙y, and Eq.