# Dynamical Aspects of an Equilateral Restricted Four-Body Problem

Authors: Martha Álvarez-Ramírez; Claudio Vidal
Journal: Mathematical Problems in Engineering
 (2009)

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## Abstract

The spatial equilateral restricted four-body problem (ERFBP) is a four
body problem where a mass point of negligible mass is moving under the
Newtonian gravitational attraction of three positive masses (called the primaries) which move on circular periodic orbits around their center of mass
fixed at the origin of the coordinate system such that their configuration is
always an equilateral triangle. Since fourth mass is small, it does not affect
the motion of the three primaries. In our model we assume
that the two masses of the primariesm2 and m3 are equal to μ and the mass m1 is 1−2μ. The Hamiltonian function that governs the motion of the fourth mass is derived and it has three degrees of freedom depending periodically on time. Using a synodical system, we fixed the primaries in order to eliminate the time dependence. Similarly to the circular restricted three-body problem, we obtain a first integral of motion. With the help of the Hamiltonian structure, we characterize the region of the possible motions and the surface of fixed level in the spatial as well as in the planar case. Among other things, we verify that the number of equilibrium solutions depends upon the masses, also we show the existence of periodic solutions by different methods in the planar case.


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## Body

## 1. Introduction

Dynamical systems with few bodies (three) have been extensively studied in the past, and various models have been proposed for research aiming to approximate the behavior of real celestial systems. There are many reasons for studying the four-body problem besides the historical ones, since it is known that approximately two-thirds of the stars in our Galaxy exist as part of multistellar systems. Around one-fifth of these is a part of triple systems, while a rough estimate suggests that a further one-fifth of these triples belongs to quadruple or higher systems, which can be modeled by the four-body problem. Among these models, the configuration used by Maranhão [1] and Maranhão and Llibre [2], where three point masses form at any time a collinear central configuration (Euler configuration, see [3]), is of particular interest not only for its simplicity but mainly because in the last 10 years, an increasing number of extrasolar systems have been detected, most of them consisting of a “sun" and a planet or of a “sun" and two planets.We study the motion of a mass point of negligible mass under the Newtonian gravitational attraction of three mass points of massesm1, m2, and m3 (called primaries) moving in circular periodic orbits around their center of mass fixed at the origin of the coordinate system. At any instant of time, the primaries form an equilateral equilibrium configuration of the three-body problem which is a particular solution of the three-body problem given by Lagrange (see [4] or [3]). Two of these primaries have equal masses and are located symmetrically with respect to the third primary.We choose the unity of mass in such a way thatm1=1-2μ and m2=m3=μ are the masses of the primaries, where μ∈(0,1/2). Units of length and time are chosen in such a way that the distance between the primaries is one.For studying the position of the infinitesimal mass,m4, in the plane of motion of the primaries, we use either the sideral system of coordinates, or the synodical system of coordinates (see [5] for details). In the synodical coordinates, the three point masses m1, m2, and m3 are fixed at (3μ,0,0), (-(3/2)(1-2μ),1/2,0), and (-(3/2)(1-2μ),-1/2,0), respectively. In this paper, the equilateral restricted four-body problem (shortly, ERFBP) consists in describing the motion of the infinitesimal mass, m4, under the gravitational attraction of the three primaries m1, m2, and m3. Maranhão's PhD thesis [1] and the paper [2] by Maranhão and Llibre studied a restricted four body problem, where three primaries rotating in a fixed circular orbit define a collinear central configuration.In the ERFBP, the equations of motion ofm4 in synodical coordinates (x,y,z) are(1.1)ẍ-2ẏ=Ωx,ÿ+2ẋ=Ωy,z̈=Ωz,
where(1.2)Ω=Ω(x,y,z)=12(x2+y2)+1-2μρ1+μρ2+μρ3,ρ1=(x-3μ)2+y2+z2,ρ2=(x+32(1-2μ))2+(y-12)2+z2,ρ3=(x+32(1-2μ))2+(y+12)2+z2.We remark that the ERFBP becomesthe central force problem when μ=0, and m1=1 is situated in the origin of the system, while μ=1/2 results in the restricted three-body problem with the bodies m2 and m3 of mass 1/2.Our paper is organized as follows: Section2 is devoted to describing the most important dynamical phenomena that governs the evolution of asteroid movement and states the problem under consideration in the present study. In Section 3 reductions of the problem are discussed and a comprehensive treatment of streamline analogies is given. Section 4 is devoted to the principal qualitative aspect of the restricted problem—the surfaces and curves of zero velocity, several uses of which are discussed. The regions of allowed motion and the location and properties of the equilibrium points are established. We describe the Hill region. The description of the number of equilibrium points is given in Section 5, and in the symmetrical case (i.e., μ=1/3), we describe the kind of stability of each equilibrium. In Section 6, the planar case is considered. There, we prove the existence of periodic solutions as a continuation of periodic Keplerian orbits, and also when the parameter μ is small and when it is close to 1/2. Finally, in Section 8 we present the conclusions of the present work.Next, we will enunciate some four-body problem that has been considered in the literature. Cronin et al. in [6, 7] considered the models of four bodies where two massive bodies move in circular orbits about their center of mass or barycenter. In addition, this barycenter moves in a circular orbit about the center of mass of a system consisting of these two bodies and a third massive body. It is assumed that this third body lies in the same plane as the orbits of the first two bodies. The authors studied the motion of a fourth body of small mass which moves under the combined attractions of these three massive bodies. This model is called bicircular four-body problem. Considering this restricted four-body problem consisting of Earth, Moon, Sun, and a massless particle, this problem can be used as a model for the motion of a space vehicle in the Sun-Earth-Moon system. Several other authors have considered the study of this problem, for example, [8–11] and references therein. The quasi-bicircular problem is a restricted four body problem where three
masses, Earth-Moon-Sun, are revolving in a quasi-bicircular motion (i.e.
a coherent motion close to bicircular) also has been studied, see [12] and
references therein. The restricted four-body problem with radiation pressure was considered in [13], while the photogravitational restricted four body problem was considered in [14].


## 2. Statement of the Problem

It is known that equilateral configurations of three-bodies with arbitrary massesm1, m2, and m3 on the same plane, moving with the same angular velocity, form a relative equilibrium solution of the three-body problem (see e.g., [4] or [3]). More precisely, we consider three particles of masses m1, m2, and m3 (called primaries) each describing, at any instant, a circle around their center of masses (which is fixed at the origin), with the same angular velocity ω and such that its configuration at any instant is an equilateral triangle (see Figure 1). Now, we consider an infinitesimal particle m4 attracted by the primaries m1, m2, and m3 according to Newton's gravitational law. Let r be the position vector of m4.Figure 1
The equilateral restricted four body problem in inertial coordinates.Figure 2
The equilateral restricted four body problem in a rotating frame.The equations of motion can be written as(2.1)r''=∇U,
where ()′ denotes derivative with respect to t and(2.2)U=U(r;t,m1,m2,m3)=m1∥r-r1(t)∥+m2∥r-r2(t)∥+m3∥r-r3(t)∥,
with r1(t),r2(t), and r3(t) representing the position of each primary, respectively. To remove the time dependence of the system (2.1), we consider the orthonormal moving frame in ℝ3, given by {e1,e2,e3} where(2.3)e1=e1(t)=eiωt,e2=e2(t)=ie1,e3=e3(t)=(0,0,1)
with i2=-1. This orthonormal moving frame corresponds to the synodical system. Then, (2.1) can be written as
(2.4)x1′′-2ωx2′-ω2x1=∂U∂x1,x2′′+2ωx1′-ω2x2=∂U∂x2,x3′′=∂U∂x3,
where(2.5)U=U(x1,x2,x3)=m1d1+m2d2+m3d3,d1=(x1-α1)2+(x2-β1)2+x32,d2=(x1-α2)2+(x2-β2)2+x32,d3=(x1-α3)2+(x2-β3)2+x32,
where rj(t)=eiωtζj, with ζj=αj+iβj for j=1,2,3. Applying the above notation, we can write r=(x1+ix2)e1+x3e3, r1=ζ1e1, r2=ζ2e1, r3=ζ3e1, and so ∥r-rj∥=∥(x1+ix2)+x3e3-ζj∥ for j=1,2,3.We perform the reparametrization of timedτ=ωdt, then the system (2.4) is transformed into(2.6)ẍ1-2ẋ2-x1=1ω2∂W∂x1,ẍ2+2ẋ1-x2=1ω2∂W∂x2,ẍ3=1ω2∂W∂x3,
where the dot denotes the derivative with respect to τ, and the potential W is given by(2.7)W=W(x1,x2,x3)=m1ρ1+m2ρ2+m3ρ3
with(2.8)ρ1=(x1-α1)2+(x2-β1)2+x32,ρ2=(x1-α2)2+(x2-β2)2+x32,ρ3=(x1-α3)2+(x2-β3)2+x32.
If we define μ1=m1/M, μ2=m2/M, and μ3=m3/M, where M=m1+m2+m3, the equations of motions (2.4) become(2.9)ẍ1-2ẋ2-x1=Mω2∂W∂x1,ẍ2+2ẋ1-x2=Mω2∂W∂x2,ẍ3=Mω2∂W∂x3,
where(2.10)W=W(x1,x2,x3)=Mμ1ρ1+Mμ2ρ2+Mμ3ρ3.For simplicity, we will consider an equilateral triangle of side 1 and so we obtain thatM/ω2=1.


## 3. Equations of Motion and Preliminary Results

From (2.9), we deduce that the equations of motion of the ERFBP in synodical coordinates are given by the system of differential equations(3.1)ẍ1-2ẋ2=Ωx1,ẍ2+2ẋ1=Ωx2,ẍ3=Ωx3,
where(3.2)Ω=Ω(x1,x2,x3)=12(x12+x22)+W(x1,x2,x3),W=W(x1,x2,x3)=1-2μρ1+μρ2+μρ3,
with
(3.3)ρ1=(x1-3μ)2+x22+x32,ρ2=(x1+32(1-2μ))2+(x2-12)2+x32,ρ3=(x1+32(1-2μ))2+(x2+12)2+x32.Analogously to the circular three-body problem, we can verify that the system (3.1) possesses a first Jacobi type integral given by(3.4)C=12(ẋ12+ẋ22+ẋ32)-Ω(x1,x2,x3).
Thus we have the following result.Proposition.
The Jacobi-type function (3.4) is a first integral of the ERFBP for any value of μ.Proof.
Differentiating (3.4) with respect to the time, we get
(3.5)dCdt=ẋ1ẍ1+ẋ2ẍ2+ẋ3ẍ3-x1ẋ1-x2ẋ2-∂W∂x1ẋ1-∂W∂x2ẋ2-∂W∂x3ẋ3,
and using (2.9) we can reduce the obtained expression to
(3.6)dCdt=ẋ1(x1+∂W∂x1+2ẋ2)+ẋ2(x2+∂W∂x2-2ẋ1)+ẋ3∂W∂x3-x1ẋ1-x2ẋ2-∂W∂x1ẋ1-∂W∂x2ẋ2-∂W∂x3ẋ3=0.
Hence C is a constant of motion.In order to write the Hamiltonian formulation of the ERFBP we introduce the new variables(3.7)x=x1,y=x2,z=x3,X=ẋ-y,Y=ẏ+x,Z=ż.
Hence, it is verified that system (3.1) is equivalent to an autonomous Hamiltonian system with three degrees of freedom with Hamiltonian function given by(3.8)H=H(x,y,z,X,Y,Z)=12(X2+Y2+Z2)+(yX-xY)-W.
Therefore, the Hamiltonian system associated is(3.9)ẋ=y+X,Ẋ=Y+Wxẏ=-x+Y,Ẏ=-X+Wyż=Z,Ż=Wz.Of course, the phase space where the equations of motion are well defined is(3.10)ℳ={(x,y,z,X,Y,Z)∈(ℝ3∖{(3μ,0,0),(-32(1-2μ),12,0),(-32(1-2μ),-12,0)})×ℝ3},
where the points that have been removed correspond to binary collisions between the massless particle and one of the primaries.Additionally, the spatial ERFBP admits the planar case as a subproblem, that is,z=Z=0 is invariant under the flow defined by (3.9).On the other hand, we see that there are two limiting cases in the ERFBP, which we described below.(a)
Ifμ=0, we obtain a central force problem, with the body of mass m1=1 at the origin of the coordinates.(b)
Ifμ=1/2, we obtain the circular restricted three-body problem, with masses m2=m3=1/2.Note thatμ=1/3 corresponds to the symmetric case, that is, where the masses of the primaries are all equal to 1/3.It is easily seen that the equations of motion (3.9) are invariant by the symmetry(3.11)S:(x,y,z,X,Y,Z,τ)→(x,-y,z,-X,Y,-Z,-τ).
This means that if ψ(τ)=(x(τ),y(τ),z(τ),X(τ),Y(τ),Z(τ)) is a solution of the system (3.9), then φ(t)=(x(-τ),-y(-τ),z(-τ),-X(-τ),Y(-τ),-Z(-τ)) is also a solution. We note that this symmetry corresponds to a symmetry with respect to the xz-plane. In the planar case, the symmetry corresponds to symmetry with respect to the x-axis.


## 4. Permitted Regions of Motion

In this section, we will see that the functionΩ(x,y,z) allows us to establish regions in the (x,y,z) space, where the motion of the infinitesimal particle could take place. We will use similar ideas to those developing in [15, 16].By using (3.4), the surface of zero velocity is defined by the set(4.1)ℛC:(x,y,z)∈ℝ3suchthatΩ(x,y,z)=-C,foranylevelC.
This set corresponds to the so-called Hill region. We note that C≥0 implies ℛC=ℝ3∖{(3μ,0,0),(-3/2(1-2μ),1/2,0),(-3/2(1-2μ),-1/2,0)}. That is, the region of all possible motions is given by the whole phase space and so the infinitesimal particle is free to move; in particular escape solutions are permitted.In the spatial case, the surfaces that separate allowed and nonallowed motions are called zero-velocity surfaces, and for the planar case the set that separates the allowed and nonallowed motions is called zero-velocity curve. The shape and size of zero velocity sets-C=Ω(x,y,z) depend on C and μ. They correspond to the boundary of the Hill regions. The zero-velocity set (∂ℛC) is defined by the equation(4.2)Ω=12(x2+y2)+1-2μ(x-3μ)2+y2+z2+μ(x+(3/2)(1-2μ))2+(y-1/2)2+z2+μ(x+(3/2)(1-2μ))2+(y+1/2)2+z2=-C,
only for C<0 and any value of μ. Next, we give a list of all possible situations that may appear when this condition is fulfilled.(1)
z→±∞ on the ∂ℛC in which case x2+y2→-2C, this means, that around the z-axis the variables (x,y) must be asymptotic to a circle of radius -2C.(2)
x→∞or-∞ (resp., y→∞or-∞) on the ∂ℛC, when C→-∞.(3)
For|C| very large this implies that (x,y) can be sufficiently close to one of the primaries, or the infinitesimal mass is close to infinity.(4)
Sincex2+y2 is a factor of Ω on ∂ℛC, then small values for -C are not allowed.By simplicity, we will only show zero-velocity surfaces for the caseμ=1/3 and different values of the integral of motion C. Figures 3, 4, 5, and 6 show evolution of zero-velocity surfaces for several C values.Evolution of zero-velocity surface in the three-dimensional ERFBP forμ=1/3. (a) C=-1/4. (b) C=-1/2. (c) C=-3/4.
(a)(b)(c)Evolution of zero-velocity surface in the three dimensional ERFBP forμ=1/3. (a) C=-1. (b) and (c) C=-1.6 under different points of view.
(a)(b)(c)Evolution of zero-velocity surface in the three dimensional ERFBP forμ=1/3. All cases correspond to C=-1.7 under different points of view.
(a)(b)(c)Evolution of zero-velocity surface in the three dimensional ERFBP forμ=1/3. (a) C=-2. (b) C=-3. (c) C=-5.
(a)(b)(c)


### 4.1. The Planar Case

As we mentioned in last section, the set{z=Z=0} is invariant under the flow, and so the motion of the infinitesimal body lies on the xy plane that contains the primaries. In Figure 7, we show the evolution of the function Ω in the planar case for different values of the parameter μ.Evolution of the graph ofΩ(x,y) on the xy plane for different values of μ.
(a)(b)(c)(d)(e)(f)Next we show the evolution of the Hill's regions as well as the zero velocity curves, forμ=1/3 and many values of the Jacobian constant C; the permissible areas are shown on Figures 8, 9, 10, and 11 shading.Evolution of zero-velocity curves and Hill's region in the planar ERFBP forμ=1/3, where shading represents permissible areas. (a) C=-1.6, (b) C=-1.6775, (c) C=-1.6795.
(a)(b)(c)Evolution of zero-velocity curves and Hill's region in the planar ERFBP forμ=1/3, where shading are permissible areas. (a) C=-1.7, (b) C=-1.75, (c) C=-1.765.
(a)(b)(c)Evolution of zero-velocity curves and Hill's region in the planar ERFBP forμ=1/3, where shading are permissible areas. (a) C=-1.775. (b) C=-1.8. (c) C=-2.
(a)(b)(c)Evolution of zero-velocity curves and Hill's region in the planar ERFBP forμ=1/3, where shading are permissible areas. (a) C=-3, (b) and (c) both figures correspond to C=-5 with different window size.
(a)(b)(c)In Figure12, we show the behavior of level curves in the planar case for some values of μ and for different energy levels.Energy level curves for some values of the parameterμ in the planar case.
(a)(b)(c)(d)(e)(f)


## 4.1. The Planar Case

As we mentioned in last section, the set{z=Z=0} is invariant under the flow, and so the motion of the infinitesimal body lies on the xy plane that contains the primaries. In Figure 7, we show the evolution of the function Ω in the planar case for different values of the parameter μ.Evolution of the graph ofΩ(x,y) on the xy plane for different values of μ.
(a)(b)(c)(d)(e)(f)Next we show the evolution of the Hill's regions as well as the zero velocity curves, forμ=1/3 and many values of the Jacobian constant C; the permissible areas are shown on Figures 8, 9, 10, and 11 shading.Evolution of zero-velocity curves and Hill's region in the planar ERFBP forμ=1/3, where shading represents permissible areas. (a) C=-1.6, (b) C=-1.6775, (c) C=-1.6795.
(a)(b)(c)Evolution of zero-velocity curves and Hill's region in the planar ERFBP forμ=1/3, where shading are permissible areas. (a) C=-1.7, (b) C=-1.75, (c) C=-1.765.
(a)(b)(c)Evolution of zero-velocity curves and Hill's region in the planar ERFBP forμ=1/3, where shading are permissible areas. (a) C=-1.775. (b) C=-1.8. (c) C=-2.
(a)(b)(c)Evolution of zero-velocity curves and Hill's region in the planar ERFBP forμ=1/3, where shading are permissible areas. (a) C=-3, (b) and (c) both figures correspond to C=-5 with different window size.
(a)(b)(c)In Figure12, we show the behavior of level curves in the planar case for some values of μ and for different energy levels.Energy level curves for some values of the parameterμ in the planar case.
(a)(b)(c)(d)(e)(f)


## 5. Equilibrium Solutions

It is verified that the equilibrium solutions of the system (3.9) or equivalently (3.1) are given by the critical points of the function Ω=Ω(x,y,z) or simply they are the solutions of the following system of equations:(5.1)(1-2μ)x-3μρ13+μ(x+32(1-2μ))(1ρ23+1ρ33)=x,(1-2μ)yρ13+μ(y-1/2ρ23+y+1/2ρ33)=y,-(1-2μρ13+μρ23+μρ33)z=0.
From the last equation we see that the coordinate z must be zero, so the critical points are restricted to the plane xy, and are given by the solutions of the first two equations.It is known (see [17]) that the number of equilibrium solutions of the system (5.1) is 8, 9 or 10 depending on the values of the masses, m1, m2 and m3 which must be positive. Six of them are out of the symmetry axis (i.e., out of the x-axis), therefore on the axis of symmetry we must have 2, 3 or 4. From the analysis done it follows that the number of the equilibrium solutions depends on the parameter μ. This implies that finding the critical points is a non-trivial problem, and this is one of the main differences with the problem studied by Maranhão in his doctoral thesis [1], because there, the number of critical points did not depend on the parameter μ.The critical points on the axisy=0 are the zeros of the function(5.2)Fμ(x)=(1-2μ)|x-3μ|(x-3μ)3+2μx+(3/2)(1-2μ)ρ23-x,
where(5.3)ρ1=|x-3μ|,ρ2=ρ3=(x+32(1-2μ))2+14.An explicit computation shows that in the limit case problems the number of equilibrium points corresponding to the system (5.1) is as follows.(a)
The function (5.2) with μ=0 results in
(5.4)F0(x)=x|x|3-x
whose zeros are x=-1 and x=1, and so there are two equilibrium points.(b)
Takingμ=1/2 in (5.2) becomes
(5.5)F1/2(x)=x(x2+(1/4))3/2-x
with zeros given by x=-(3/2), x=0 and x=(3/2). We conclude that there are three equilibrium points.From numerical simulations we get that the number of critical points along thex–axis is given in Table 1. Observe that μ*:=0.266318 is the bifurcation value.Table 1
Number of critical points on thex-axis.
μ=00<μ<μ*μ=μ*μ*<μ<1/2μ=1/222343In the symmetric case when all the masses are equals (i.e.,μ=1/3) we have that the graph of F1/3 is similar to the one shown in Figure 13. As a consequence, there are exactly 4 equilibrium solutions on the x-axis, and therefore there are exactly 10 equilibrium solutions. Of course, (0,0,0) is an equilibrium solution.Graph ofF1/3. (a) x<3/3. (b)x>3/3.
(a)(b)In general for any equilibrium solution of the form(x0,y0,0), the linearized system (3.9) in the planar case give us that the characteristic polynomial is(5.6)CA(λ)=(λ2-Wzz(x0,y0,0))λ4+(2-Wxx(x0,y0,0)-Wyy(x0,y0,0))λ2+(-Wxy2(x0,y0,0))1+Wxx(x0,y0,0)+Wyy(x0,y0,0)+Wxx(x0,y0,0)Wyy(x0,y0,0)-Wxy2(x0,y0,0)),
whose roots are(5.7)λ=±Wzz(x0,y0,0),λ=±12ρ±,
where ρ±is given by(5.8)ρ±=-4+2(a+c)±2(a-c)2+4b2-8(a+c)
with a=Wxx(x0,y0,0), b=Wxy(x0,y0,0) and c=Wyy(x0,y0,0). A very simple result is the following.Lemma.
The roots ofp(ρ)=ρ2+Aρ+B are real and negative if and only if A>0, B>0 and Δ=A2-4B≥0.Associating to our characteristic polynomial (5.6) we have(5.9)A=2-Wxx(x0,y0,0)-Wyy(x0,y0,0),B=1+Wxx(x0,y0,0)+Wyy(x0,y0,0)+Wxx(x0,y0,0)Wyy(x0,y0,0)-Wxy2(x0,y0,0).
Now, we remark that(5.10)Wzz(x0,y0,0)=-[1-2μρ13+μρ23+μρ33]<0.
Consequently we have the following result:Corollary.
In the spatial ERFBP for any equilibrium solution(x0,y0,0) we have at least two pure imaginary eigenvalues associated to the linear part, which are given by λ=±-Wzz(x0,y0,0)i.From this corollary we deduce that to study the nonlinear stability in the Lyapunov  sense of each equilibrium solution of the spatial ERFBP is not a simple problem, because we need to take into account the existence or not of resonance in each situation. Leandro in [17] studied the spectral stability in some situations (according to the localization of the equilibrium solution along the symmetry-axis). In a future work we intend to study the Lyapunov stability of each equilibrium.


### 5.1. Analysis of the Symmetrical Case,μ=1/3

As we have said previously in the symmetrical case (i.e.,μ=1/3) there are 10 equilibrium solutions and one of them is (0,0,0). Here we have ρ1=ρ2=ρ3=1/3, a=b=33/2 and c=0. Consequently, the characteristic roots are(5.11)λ1=-33i,λ2=33i,λ3=-1263-4+4-63,λ4=1263-4+4-63,λ5=-1263-4-4-63,λ6=1263-4-4-63.
Therefore, we have the following result.Corollary.
In the symmetrical spatial ERFBP the equilibrium solution(0,0,0) is unstable in the Lyapunov  sense.In general, it is possible to prove that the equilibrium solutions on thex–axis are x1=-0.9351859666722429, x2=-0.23895830919534947 and x3=1.1799984048894328, and by symmetry it follows:Corollary.
In the symmetrical spatial ERFBP all the equilibrium solutions are unstable in the Lyapunov sense.According to [17] we have the following corollary.Corollary.
In the symmetrical planar ERFBP all the equilibrium solutions are unstable in the Lyapunov sense.


## 5.1. Analysis of the Symmetrical Case,μ=1/3

As we have said previously in the symmetrical case (i.e.,μ=1/3) there are 10 equilibrium solutions and one of them is (0,0,0). Here we have ρ1=ρ2=ρ3=1/3, a=b=33/2 and c=0. Consequently, the characteristic roots are(5.11)λ1=-33i,λ2=33i,λ3=-1263-4+4-63,λ4=1263-4+4-63,λ5=-1263-4-4-63,λ6=1263-4-4-63.
Therefore, we have the following result.Corollary.
In the symmetrical spatial ERFBP the equilibrium solution(0,0,0) is unstable in the Lyapunov  sense.In general, it is possible to prove that the equilibrium solutions on thex–axis are x1=-0.9351859666722429, x2=-0.23895830919534947 and x3=1.1799984048894328, and by symmetry it follows:Corollary.
In the symmetrical spatial ERFBP all the equilibrium solutions are unstable in the Lyapunov sense.According to [17] we have the following corollary.Corollary.
In the symmetrical planar ERFBP all the equilibrium solutions are unstable in the Lyapunov sense.


## 6. Continuation of Periodic Solutions in the Planar Case

In this section we prove the existence of periodic solutions in the ERFBP forμ sufficiently small in the planar case and by the use of the Lyapunov Center Theorem when μ is close to 1/2. In order to find periodic orbits of our problem we will use the continuation method developed by Poincaré which is one of the most frequently used methods to prove the existence of periodic orbits in the planar circular restricted three-body problem (see [15]). This method has been also used by other authors in different problems. In Meyer and Hall [5], we find a good discussion of the Poincaré continuation method to different n-body problem (see also [18]).In our approach we will continue circular and elliptic solutions of the Kepler problem with the body fixed in the origin of the system with mass1. We know that all the orbits of the Kepler problem with angular momentum zero are collision orbits with the origin. We assume that the angular momentum is not zero and we study the orbits that have positive distance of (-3/2,1/2) and (-3/2,-1/2). In the following lemma we resume the kind of orbits that we will consider.Lemma.
Fixeda>0 there exists a finite number of elliptic orbits with semi-major axis a, such that its trajectories are periodic in the rotating system and pass through the singularity of the other primaries (-3/2,1/2) or (-3/2,-1/2).The proof of this lemma can be found in [19].


### 6.1. Continuation of Circular Orbits

In this section we show that circular solutions of the unperturbed Kepler problem can be continued to periodic solutions of the ERTBP for small values ofμ. We introduce the polar coordinates given as x=rcosθ,y=rsinθ, thus ẋ=ṙcosθ-rθ̇sinθ and ẏ=ṙsinθ+rθ̇cosθ. So, Ẋ=ṙcosθ-r(θ̇+1)sinθ and Ẏ=ṙsinθ+r(θ̇+1)cosθ, consequently X2+Y2=ṙ2+r2(θ̇+1)2 and yX-xY=-r2(θ̇+1). Thus, the Hamiltonian (3.8) now is(6.1)H=ṙ2+r2(θ̇+1)22-r2(θ̇+1)-V(r,θ),
where(6.2)V(r,θ)=1-2μρ1+μ(1ρ2+1ρ3),
where(6.3)ρ1=(rcosθ-3μ)2+r2sin2θ,ρ2=(rcosθ+3/2(1-2μ))2+(rsinθ-1/2)2,ρ3=(rcosθ+3/2(1-2μ))2+(rsinθ+1/2)2.
The new coordinates are not symplectic. In order to obtain a set of symplectic coordinates (r,θ,R,Θ) we define R=ṙ (radial velocity in the sideral system) and Θ=r2(θ̇+1) (angular momentum in the sideral system), then H is(6.4)H=R2+Θ/r22-Θ-V(r,θ).
When μ=0 we have that(6.5)H=R2+Θ2/r22-Θ-1r,
is the Hamiltonian of the Kepler problem in polar coordinates. So, if μ is a small parameter, the Hamiltonian (3.8) assumes the form(6.6)H=R2+Θ2/r22-Θ-1r+𝒪(μ).
For μ=0, the Hamiltonian system associated is(6.7)ṙ=R,Ṙ=Θ2r3-1r2,θ̇=Θr2-1,Θ̇=0.LetΘ=c be a fixed constant. For c≠1, the circular orbit R=0, r=c2 is a periodic solution with period |2πc3/(1-c3)|. Linearizing the r and R equations about this solution gives(6.8)ṙ=R,Ṙ=-c-6r,
which has solutions of the form exp(±it/c3), and so the nontrivial multipliers of the circular orbits are exp(±i2π/(1-c3)) which are not +1, provided 1/(1-c3) is not an integer. Thus we have proved the following theorem (see details in [5]).Theorem.
Ifc≠1 and 1/(1-c3) is not an integer, then the circular orbits of the Kepler problem in rotating coordinates with angular momentum c can be continued into the equilateral restricted four body problem for small values of μ.


### 6.2. Continuation of Elliptic Orbits

In Section3, we saw that the ERFBP has the S-symmetry which when exploited properly proves that some elliptic orbits can be continued from the Kepler problem. The main idea is given in the following lemma, which is a consequence of the uniqueness of the solution of the differential equations and the symmetry of the problem.Lemma.
A solution of the equilateral restricted problem which crosses the line of syzygy (the x-axis) orthogonally at a timet=0 and later at a time t=T/2>0 is T-periodic and symmetric with respect to the line syzygy.That is, ifx(t) and y(t) is a solution of the equilateral restricted four body problem such that y(0)=ẋ(0)=y(T/2)=ẋ(T/2)=0, where T>0, then this solution is T-periodic and symmetric with respect to the x-axis.In Delaunay variables(l,g,L,G), an orthogonal crossing of the line of sizygy at a time t0 is(6.9)l(t0)=nπ,g(t0)=mπ,n,mintegers.
These equations will be solved using the Implicit Function theorem to yield the following theorem (see details in [5]).Theorem.
Letm,k be relatively prime integers and T=2πm. Then the elliptic T-periodic solution of the Kepler problem in rotating coordinates which satisfies
(6.10)l(0)=π,g(0)=π,L3(0)=mk
and does not go through (-3/2,1/2) and (-3/2,-1/2) can be continued into the equilateral restricted four body problem for small μ. This periodic solution is symmetric with respect to the line of syzygy.Proof.
The Hamiltonian of the ERFBP in Delaunay coordinates forμ sufficiently small is
(6.11)H=-12L2-G+𝒪(μ),
and the equations of motion are
(6.12)l̇=1L3+𝒪(μ),L̇=0+𝒪(μ),ġ=-1+𝒪(μ),Ġ=0+𝒪(μ).
Let L03=m/k, and let l(t,Λ,μ), g(t,Λ,μ), L(t,Λ,μ) and G(t,Λ,μ) be the solution which goes through l=π, g=π, L=Λ, G arbitrary at t=0; so, it is a solution with an orthogonal crossing of the line of syzygy at t=0.
From (6.12) l(t,Λ,0)=t/Λ3+π, g(t,Λ,0)=-t+π. Thus, l(T/2,L0,0)=(1+k)π and g(T/2,L0,0)=(1-m)π, and so when μ=0, this solution has another orthogonal crossing at time T/2=mπ. Also,
(6.13)det(∂l∂t∂l∂Λ∂g∂t∂g∂Λ)t=T/2,L=L0,μ=0=det(km-3π(k4m)1/3-10)≠0.
Thus, the theorem follows by the Implicit Function theorem.


### 6.3. Application of the Lyapunov Center Theorem

Forμ=1/2, we have three equilibrium solutions on the x-axis which are P1=(-3/2,0), P2=(0,0) and P3=(3/2,0). At the point P2, the associated eigenvalues are ±75+82 and ±75-82. Therefore, this equilibrium is unstable and by Lyapunov's Center Theorem (see [5]), we obtain the following theorem.Theorem 6.5.
There exists a one-parameter family of periodic orbits of the ERFBP emanating from the Euler equilibrium (forμ=1/2). Moreover, when approaching the equilibrium point along the family, the periods tend to 2π/-3+82.


## 6.1. Continuation of Circular Orbits

In this section we show that circular solutions of the unperturbed Kepler problem can be continued to periodic solutions of the ERTBP for small values ofμ. We introduce the polar coordinates given as x=rcosθ,y=rsinθ, thus ẋ=ṙcosθ-rθ̇sinθ and ẏ=ṙsinθ+rθ̇cosθ. So, Ẋ=ṙcosθ-r(θ̇+1)sinθ and Ẏ=ṙsinθ+r(θ̇+1)cosθ, consequently X2+Y2=ṙ2+r2(θ̇+1)2 and yX-xY=-r2(θ̇+1). Thus, the Hamiltonian (3.8) now is(6.1)H=ṙ2+r2(θ̇+1)22-r2(θ̇+1)-V(r,θ),
where(6.2)V(r,θ)=1-2μρ1+μ(1ρ2+1ρ3),
where(6.3)ρ1=(rcosθ-3μ)2+r2sin2θ,ρ2=(rcosθ+3/2(1-2μ))2+(rsinθ-1/2)2,ρ3=(rcosθ+3/2(1-2μ))2+(rsinθ+1/2)2.
The new coordinates are not symplectic. In order to obtain a set of symplectic coordinates (r,θ,R,Θ) we define R=ṙ (radial velocity in the sideral system) and Θ=r2(θ̇+1) (angular momentum in the sideral system), then H is(6.4)H=R2+Θ/r22-Θ-V(r,θ).
When μ=0 we have that(6.5)H=R2+Θ2/r22-Θ-1r,
is the Hamiltonian of the Kepler problem in polar coordinates. So, if μ is a small parameter, the Hamiltonian (3.8) assumes the form(6.6)H=R2+Θ2/r22-Θ-1r+𝒪(μ).
For μ=0, the Hamiltonian system associated is(6.7)ṙ=R,Ṙ=Θ2r3-1r2,θ̇=Θr2-1,Θ̇=0.LetΘ=c be a fixed constant. For c≠1, the circular orbit R=0, r=c2 is a periodic solution with period |2πc3/(1-c3)|. Linearizing the r and R equations about this solution gives(6.8)ṙ=R,Ṙ=-c-6r,
which has solutions of the form exp(±it/c3), and so the nontrivial multipliers of the circular orbits are exp(±i2π/(1-c3)) which are not +1, provided 1/(1-c3) is not an integer. Thus we have proved the following theorem (see details in [5]).Theorem.
Ifc≠1 and 1/(1-c3) is not an integer, then the circular orbits of the Kepler problem in rotating coordinates with angular momentum c can be continued into the equilateral restricted four body problem for small values of μ.


## 6.2. Continuation of Elliptic Orbits

In Section3, we saw that the ERFBP has the S-symmetry which when exploited properly proves that some elliptic orbits can be continued from the Kepler problem. The main idea is given in the following lemma, which is a consequence of the uniqueness of the solution of the differential equations and the symmetry of the problem.Lemma.
A solution of the equilateral restricted problem which crosses the line of syzygy (the x-axis) orthogonally at a timet=0 and later at a time t=T/2>0 is T-periodic and symmetric with respect to the line syzygy.That is, ifx(t) and y(t) is a solution of the equilateral restricted four body problem such that y(0)=ẋ(0)=y(T/2)=ẋ(T/2)=0, where T>0, then this solution is T-periodic and symmetric with respect to the x-axis.In Delaunay variables(l,g,L,G), an orthogonal crossing of the line of sizygy at a time t0 is(6.9)l(t0)=nπ,g(t0)=mπ,n,mintegers.
These equations will be solved using the Implicit Function theorem to yield the following theorem (see details in [5]).Theorem.
Letm,k be relatively prime integers and T=2πm. Then the elliptic T-periodic solution of the Kepler problem in rotating coordinates which satisfies
(6.10)l(0)=π,g(0)=π,L3(0)=mk
and does not go through (-3/2,1/2) and (-3/2,-1/2) can be continued into the equilateral restricted four body problem for small μ. This periodic solution is symmetric with respect to the line of syzygy.Proof.
The Hamiltonian of the ERFBP in Delaunay coordinates forμ sufficiently small is
(6.11)H=-12L2-G+𝒪(μ),
and the equations of motion are
(6.12)l̇=1L3+𝒪(μ),L̇=0+𝒪(μ),ġ=-1+𝒪(μ),Ġ=0+𝒪(μ).
Let L03=m/k, and let l(t,Λ,μ), g(t,Λ,μ), L(t,Λ,μ) and G(t,Λ,μ) be the solution which goes through l=π, g=π, L=Λ, G arbitrary at t=0; so, it is a solution with an orthogonal crossing of the line of syzygy at t=0.
From (6.12) l(t,Λ,0)=t/Λ3+π, g(t,Λ,0)=-t+π. Thus, l(T/2,L0,0)=(1+k)π and g(T/2,L0,0)=(1-m)π, and so when μ=0, this solution has another orthogonal crossing at time T/2=mπ. Also,
(6.13)det(∂l∂t∂l∂Λ∂g∂t∂g∂Λ)t=T/2,L=L0,μ=0=det(km-3π(k4m)1/3-10)≠0.
Thus, the theorem follows by the Implicit Function theorem.


## 6.3. Application of the Lyapunov Center Theorem

Forμ=1/2, we have three equilibrium solutions on the x-axis which are P1=(-3/2,0), P2=(0,0) and P3=(3/2,0). At the point P2, the associated eigenvalues are ±75+82 and ±75-82. Therefore, this equilibrium is unstable and by Lyapunov's Center Theorem (see [5]), we obtain the following theorem.Theorem 6.5.
There exists a one-parameter family of periodic orbits of the ERFBP emanating from the Euler equilibrium (forμ=1/2). Moreover, when approaching the equilibrium point along the family, the periods tend to 2π/-3+82.


## 7. Numerical Results

In the Section8, we established theorems on the continuation of periodic solutions from the Kepler's problem in rotating coordinates to the ERFBP. In this section,  we present some numerical experiments that illustrates the thesis of Theorem 6.2.To find those circular orbits we first selected an angular momentumc such that c≠1 and 1/(1-c3)∉ℤ. By varying c we generated a set of initial conditions for Kepler problem in rotating coordinates given by the system (3.9) taking μ=0. We have chosen y0=0 and X0=0 for all orbits, ensuring that we were following a family of symmetric orbits; we have taken into account the fact that circular orbits satisfy r=c2.We have noticed that for values ofc=2,3,4,5,6,7,8,9,10 with μ=10-2 the orbit is close to the circular orbit, see Figure 14.Figure 14
Circular orbit forc=2 with μ=10-2.However, the circular orbits associated toc≈0,1 is close to the circular orbit if μ≤10-4, for instance c=9/10 can be continued for μ small and of the order 10-4 but not for higher values. The orbits obtained as a consequence of numerical simulations are shown in Figure 15.The circular orbit associated toc=9/10 for μ=0 continued to μ=10-4 which is not circular for μ=10-2.
(a)(b)(c)


## 8. Conclusions and Final Remarks

The spatial equilateral restricted four-body problem (ERFBP) is considered. This model of four-body problem, we have that three masses, moving in circular motion such that their configuration is always an equilateral triangle, the fourth mass being small and not influencing the motion of the three primaries. In our model we assume that two masses of the primariesm2 and m3 are equal to μ and the mass m1 is 1-2μ. In a synodical systems of coordinates the dynamics obeys to the system of differential equations(8.1)ẍ-2ẏ=Ωx,ÿ+2ẋ=Ωy,z̈=Ωz,
where(8.2)Ω=Ω(x,y,z)=12(x2+y2)+1-2μρ1+μρ2+μρ3,ρ1=(x-3μ)2+y2+z2,ρ2=(x+32(1-2μ))2+(y-12)2+z2ρ3=(x+32(1-2μ))2+(y+12)2+z2.In Section4 it is devoted to give the principal qualitative aspect of the restricted problem—the surfaces and curves of zero velocity, several uses of which are discussed. The regions of permitted motion and the location and properties of the equilibrium points are established. We describe the Hill region. The description of the number of equilibrium points is given in Section 5, and in the symmetrical case (i.e., μ=1/3) we are describing the kind of stability of each equilibrium. In Section 6 the planar case is considered. Here, we prove the existence of periodic solutions as continuation of periodic Keplerian orbits, when the parameter μ is small and when it is close to 1/2. Finally, in Section 7 we present some numerical experiments that illustrates the thesis of theorem concerning with the continuation of circular orbits of the Kepler problem to the ERFBP with μ small enough.In a work in progress we intend to continue the study of the ERFBP in different aspects of its dynamics. For example, the behavior of the flow near the singularities (collisions). The study of the escapes solutions (i.e., the unbounded solutions). Existence of chaos under the construction of a shift map. We desired to get periodic solutions under the use of numerical methods.



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*Source: 181360-2010-03-03.xml*