Sun-Jupiter-Hektor-Skamandrios

# Hill four-body problem with oblate tertiary: an application to the Sun-Jupiter-Hektor-Skamandrios system

Jaime Burgos-García Autonomous University of Coahuila, C.P. 25020, Saltillo, Mexico Alessandra Celletti Department of Mathematics, University of Roma Tor Vergata, Via della Ricerca Scientifica 1, 00133 Roma (Italy) Catalin Gales Department of Mathematics, Al. I. Cuza University, Bd. Carol I 11, 700506 Iasi (Romania) Marian Gidea Department of Mathematical Sciences, Yeshiva University, New York, NY 10016 (USA)  and  Wai-Ting Lam Department of Mathematical Sciences, Yeshiva University, New York, NY 10016 (USA)
###### Abstract.

We consider a restricted four-body problem with a precise hierarchy between the bodies: two point-mass bigger bodies, a smaller one with oblate shape, and an infinitesimal body in the neighborhood of the oblate body. The three heavy bodies are assumed to move in a plane under their mutual gravity, and the fourth body moves under the gravitational influence of the three heavy bodies, but without affecting them.

We start by finding the triangular central configurations of the three heavy bodies; since one body is oblate, the triangle is isosceles, rather than equilateral as in the point mass case. We assume that the three heavy bodies are in such a central configuration and we perform a Hill’s approximation of the equations of motion describing the dynamics of the infinitesimal body in a neighborhood of the oblate body. Through the use of Hill’s variables and a limiting procedure, this approximation amounts to sending the two other bodies to infinity. Finally, for the Hill approximation, we find the equilibrium points of the infinitesimal body and determine their stability. As a motivating example, we consider the dynamics of the moonlet Skamandrios of Jupiter’s Trojan asteroid Hektor.

## 1. Introduction

The discovery of binary asteroids has led to considering dynamical models formed by four bodies, two of them being typically the Sun and Jupiter. Among possible four-body models (see also [HS86, Sch98, GJ03, SB05, ARV09, BGD13b, BGD13a, BG16, KMJ17]), a relevant role is played by the models in which three bodies lie on a triangular central configuration. Given that asteroids have often a (very) irregular shape, it is useful to start by investigating the case in which one body has an oblate shape. Among the different questions that this model may rise, we concentrate on the existence of equilibrium points and the corresponding linear stability analysis. Within such framework, we consider a four-body simplified model and we concentrate on the specific example given by the Trojan asteroid 624 Hektor, which is located close to the Lagrangian point of the Sun-Jupiter system, and its small moonlet.

In our model Sun, Jupiter and Hektor form an isosceles triangle (nearly equilateral) whose shape remains unchanged over time. The small body represents Hektor’s moonlet Skamandrios. Obviously, we could also replace the moonlet by a spacecraft orbiting Hektor. The system Sun-Jupiter-Hektor-Skamandrios plays a relevant role for several reasons. Indeed, Hektor is the largest Jupiter Trojan, it has one of the most elongated shapes among the bodies of its size in the Solar system, and it is the only known Trojan to possess a moonlet (see, e.g., [DLZ12] for stability regions of Trojans around the Earth and [LC15] for dissipative effects around the triangular Lagrangian points).

The study of asteroids with satellites presents a special interest for planetary dynamics, as they provide information about constraints on the formation and evolution of the Solar system.

Another motivation to study the dynamics of a small body near a Trojan asteroid comes from astrodynamics, as NASA prepares the first mission, Lucy, to the Jupiter’s Trojans, which is planned to be launched in October 2021 and visit seven different asteroids: a Main Belt asteroid and at least five Trojans.

As a model for the Sun-Jupiter-Hektor dynamics, we consider a system of three bodies of masses , which move in circular orbits under mutual gravity, and form a triangular central configuration. We refer to these bodies as the primary, the secondary, and the tertiary, respectively. We assume that the first two bodies of masses are spherical and homogeneous, so they can be treated as point masses, while the third body of mass is oblate. We describe the gravitational potential of in terms of spherical harmonics, and we only retain the most significant ones. We show the existence of a corresponding triangular central configuration, which turns out to be an isosceles triangle; if the oblateness of the mass is made to be zero, the central configuration becomes the well-known equilateral triangle Lagrangian central configuration. We stress that when is oblate, the central configuration is not the same as in the non-oblate case, since the overall gravitational field of is no longer Newtonian; it is well known that central configurations depend on the nature of the gravitational field (see, e.g., [CLPC04, APC13, DSZ, MS17]). We note that there exist papers in the literature (e.g., [APHS16]), which consider systems of three bodies, with one of the bodies non-spherical, which are assumed to form an equilateral triangle central configuration. Such assumption, while it may lead to very good approximations, is not physically correct.

The moonlet Skamandrios is represented by a fourth body, of infinitesimal mass, which moves in a vicinity of under the gravitational influence of , but without affecting their motion. We consider the motion of the infinitesimal mass taking place in the three-dimensional space; it is not restricted to the plane of motion of the three heavy bodies. This situation is referred to as the spatial circular restricted four-body problem, and can be described by an autonomous Hamiltonian system of -degrees of freedom.

We ‘zoom-in’ onto the dynamics in a small neighborhood of by performing a Hill’s approximation of the restricted four-body problem. This is done by a rescaling of the coordinates in terms of , writing the associated Hamiltonian in the rescaled coordinates as a power series in , and neglecting all the terms of order in the expansion, since such terms are small when is small. This yields an approximation of the motion of the massless particle in an -neighborhood of , while and are ‘sent to infinity’ through the rescaling. This model is an extension of the classical lunar Hill problem [Hil78]. Since the tertiary is assumed to be oblate, and the corresponding central configuration formed by the three heavy bodies is not an equilateral triangle anymore, this model also extends Hill’s approximation of the restricted four-body problem developed in [BGG15].

The Hill approximation is more advantageous to utilize for this system than the restricted four-body problem, since it allows for an analytical treatment, and yields more accurate numerical implementations when realistic parameters are used. The main numerical difficulty in the restricted four-body problem is the large differences of scales among the relevant parameters, i.e. the mass of Hektor is much smaller than the masses of the other two heavy bodies. The rescaling of the coordinates involved in the Hill approximation reduces the difference of scales of the parameters to more manageable quantities; more precisely, in normalized units the oblateness effect in the restricted four-body problem is of the order , while in the Hill approximation is of the order (see Section 3.2 for details).

Once we have established the model for the Hill four-body problem with oblate tertiary, we study the equilibrium points and their linear stability. We find that there are pairs of symmetric equilibrium points on each of the -, -, and -coordinate axes, respectively. The equilibrium points on the - and -coordinate axes are just a continuation of the corresponding ones for the Hill four-body problem with non-oblate tertiary [BGG15]. The equilibrium points on the -coordinate axis constitute a new feature of the model. In the case of Hektor, these equilibrium points turn out to be outside of the body of the asteroid but very close to the surface, so they are of potential interest for low altitude orbit space missions, such as the one of NASA/JPL’s Dawn mission around Vesta ([Del11]).

This work is organized as follows. In Section 2 we describe in full details the restricted four-body model in which the tertiary is oblate; in particular, we describe the isosceles triangle central configuration of three bodies in which two bodies are point masses and the third is oblate. Hill’s approximation is introduced in Section 3. The determination of the equilibria and their stability is given in Section 4.

## 2. Restricted four-body problem with oblate tertiary

In this section we develop a model for a restricted four-body problem, which consists of two bigger bodies (e.g., the Sun and Jupiter), a smaller body – called tertiary – with oblate shape (e.g. an asteroid), and an infinitesimal mass (e.g., moonlet) around the tertiary.

As mentioned in Section 1, we consider the three masses as moving under the mutual gravitational attraction; the bodies with masses and are considered as point masses, while is the oblate body. We normalize the units of mass so that .

We assume that the bodies with masses , , move on a triangular central configuration, which will be determined in Section 2.3, once the gravitational field of the oblate body has been discussed in Section 2.2. We will concentrate on the specific example given by the asteroid Hektor and its moon Skamandrios, where Hektor moves on a central configuration with Jupiter and the Sun. Orbital and physical values are given in Section 2.1. The positions of the three man bodies in the triangular central configuration is computed in Section 2.4, while the equations of motion of the moonlet – with infinitesimal mass moving in the vicinity of – are given in Section 2.5.

### 2.1. Data on the Sun-Jupiter-Hektor-Skamandrios system

The models which we will develop below will be applied to the case of the Sun-Jupiter-Hektor-Skamandrios system. We extract the data for this system from [JPL, MDCR14, Des15].

Hektor is approximately located at the Lagrangian point of the Sun-Jupiter system. According to [Des15], Hektor is approximately km in size, and its shape can be approximated by a dumb-bell figure; the equivalent radius (i.e., the radius of a sphere with the same volume as the asteroid) is km111Note that [Des15] claims that there are some typos in the values reported in [MDCR14]..

Hektor spins very fast, with a rotation period of approximately hours (see the JPL Solar System Dynamics archive [JPL]).

The moonlet Skamandrios orbits around Hektor at a distance of approximately km, with an orbital period of days; see [Des15]. Its orbit is highly inclined, at approximately with respect to the orbit of Hektor, which justifies choosing as a model the spatial restricted four-body problem rather than the planar one; see [MDCR14].

We also note that the inclination of Hektor is approximately (see [JPL]). Although a more refined model should include a non-zero inclination, we will consider that Sun-Jupiter-Hektor move in the same plane, an assumption that is needed in order for the three bodies to form a central configuration. We will further assume that the axis of rotation of Hektor is perpendicular to the plane of motion.

For the masses of Sun, Jupiter and Hektor we use the values of kg, , and kg, respectively. For the average distance Sun-Jupiter we use the value km.

In Figure 1 we provide a comparison between the strength of the different forces acting on the moonlet: the Newtonian gravitational attraction of Hektor, Sun, Jupiter, and the effect of the non-spherical shape of the asteroid, limited to the the so-called coefficient, which will be introduced in Section 2.2.

### 2.2. The gravitational field of a non-spherical body

We first consider that the tertiary body, representing Hektor, has a general (non-spherical) shape. The gravitational potential, relative to a reference frame centered at the barycenter of the tertiary and rotating with the body, is given in spherical coordinates by (see, e.g., [CG18]):

 V(r,ϕ,λ)=GmHr ∞∑n=0(RHr)n n∑m=0Pnm(sinϕ) (Cnmcosmλ+Snmsinmλ) ,

where is the gravitational constant, is the mass of Hektor, is its average radius, are the Legendre polynomials defined as

 Pn(x)=12nn! dndxn(x2−1)nPnm(x)=(1−x2)m2 dmdxmPn(x) ,

and and are the spherical harmonics coefficients.

In the case of an ellipsoid of semi-axes , we have the following explicit formulas ([Boy97]):

 Sn,m = 0, C2p+1,2q = 0, C2p,2q+1 = 0, C2p,2q = 3R2pHp!(2p−2q)!22q(2p+3)(2p+1)!(2−δ0q) ⌊p−q2⌋∑i=0(a2−b2)q+2i[c2−12(a2+b2)]p−q−2i16i(p−q−2i)!(p+q)!i!.

In particular, and turn out to be given by simple expressions

 (2.1) C20 = c2−a22−b225R2H, (2.2) C22 = a24−b245R2H.

For the Sun-Jupiter-Hektor data, we take km, km, km, km, following [Des15] (see Section 2.1), we calculate the following coefficients:

 C20=−0.476775; C22=0.230232; C40=0.714275; C42=−0.078406; C44=0.009465; C60=−1.54769; C62=0.076832; C64=−0.002507; C66=0.000201.

Notice that each term is multiplied in (2.2) by the factor . For equal to the average distance from the moonlet to the asteroid, we have . Therefore, in the following we will ignore the effect of the coefficients , with , which are at least of .

Note that the value of computed above is significantly bigger in absolute value than the one reported in [MDCR14], which equals . The reason is that we use different estimates for the size of Hektor, following [Des15] (see Section 2.1).

If we consider a frame centered at the barycenter of the tertiary, and which rotates with the angular velocity of the tertiary about the primary, the time dependent gravitational potential takes of the form

 V(r,ϕ,λ)=GmHr 2∑n=0(RHr)n n∑m=0Pnm(sinϕ) Cnmcos(m(λ+Θt)),

where represents the frequency of the spin of Hektor.

For , the corresponding term in the summation (2.2) is equal to and is independent of time; for , the corresponding term is equal to , so is time-dependent. We do not consider the other terms in the sum (2.2).

Since the ratio of the rotation period of Hektor to the orbital period of the moonlet is relatively small, approximately , in this paper we will only consider the average effect of on the moonlet, which is zero.

In conclusion, in the model below we will only consider the effect of , which amounts to approximating Hektor as an oblate body (i.e., an ellipsoid of revolution obtained by rotating an ellipse about its minor axis); the dimensionless quantity is referred to as the zonal harmonic in the gravitational potential. The term corresponding to is the larger one, followed by that corresponding to ; however, since the term introduces a time dependence, thus further complicating the model, we start by disregarding it and plan to study its effect in a future work.

### 2.3. Central configurations for the three-body problem with one oblate body

We now consider only the three heavy bodies, of masses , with the body of mass being oblate, in which case we only take into account the term corresponding to in (2.2). We write the approximation of the gravitational potential of the tertiary (2.2) in both Cartesian and spherical coordinates (in the frame of the tertiary and rotating with the body):

 (2.3) V(x,y,z)=m3r−m3r(R3r)2(J22)(3(zr)2−1)=m3r+m3r(R3r)2(C202)(3sinϕ2−1),

where is the normalized mass of Hektor (the sum of the three masses is the unit of mass), is the average radius of Hektor in normalized units (the distance between Sun and Hektor is the unit of distance), the gravitational constant is normalized to , and .

We want to find the triangular central configurations formed by , , ; we will follow the approach in [APC13]. Since for a central configuration the three bodies lie in the same plane, in the gravitational field (2.3) of we set , obtaining

 (2.4) V(q)=m3r+Cm3r3,

where is the position vector of an arbitrary point in the plane, is the distance from , and we denote

 (2.5) C=R23J2/2>0.

Let be the position vector of the mass , for , in an inertial frame centered at the barycenter of the three bodies.

The equations of motion of the three bodies are

 (2.6) m1¨q1=m1m2(q2−q1)1∥q2−q1∥3+m1m3(q3−q1)[1∥q3−q1∥3+3C∥q3−q1∥5],m2¨q2=m2m1(q1−q2)1∥q1−q2∥3+m2m3(q3−q2)[1∥q3−q2∥3+3C∥q3−q2∥5],m3¨q3=m3m1(q1−q3)1∥q1−q3∥3+m3m2(q2−q3)1∥q2−q3∥3,

where the last terms in the first two equations are due to (2.4), and the gravitational constant is normalized to . Denote , for , , and the matrix with copies of each mass along the diagonal. Then (2.6) can be written as

 (2.7) M¨q=∇U(q),

where

 (2.8) U(q)=m1m21r12+m1m3(1r13+Cr313)+m2m3(1r23+Cr323)

is the potential for the three body problem with oblate .

Let us assume that the center of mass is fixed at the origin, i.e.,

 (2.9) Mq=3∑i=1miqi=0.

We are interested in relative equilibrium solutions for the motion of the three bodies, which are characterized by the fact they become equilibrium points in a uniformly rotation frame.

Denote by the block diagonal matrix consisting of diagonal blocks the form

 (cos(θ)−sin(θ)sin(θ)cos(θ))∈SO(2).

Substituting for some in (2.7), where , we obtain

 M(¨z+2ωJ˙z−ω2z)=∇U,

where is the block diagonal matrix consisting of diagonal blocks the form

 (2.10) (0−110).

The condition for an equilibrium point of (2.3) yields the algebraic equation

 (2.11) ∇U(z)+ω2Mz=0.

A solution of the three-body problem satisfying (2.11) is referred to as a central configuration. This is equivalent to , for , meaning that the accelerations of the masses are proportional to the corresponding position vectors, and all accelerations are pointing towards the center of mass. Thus, the solution is a relative equilibrium solution if and only if with being a central configuration solution, and the rotation being a circular solution of the Kepler problem.

Let be the moment of inertia. It is easy to see that this is a conserved quantity for the motion, that is, for some at all . Using Lagrange’s second identity (see, e.g., [GN12]), and that , normalizing the masses so that , the moment of inertia can be written as:

 (2.12) I(z)=∑1≤i

Thus, central configurations correspond to critical points of the potential on the sphere , which can be obtained by solving the Lagrange multiplier problem

 (2.13) ∇f(z)=0,I(z)−¯I=0,

where . In the above, we used the fact that .

We solve this problem in the variables for , since both and can be written in terms of these variables. This reduces the dimension of the system (2.13) from equations to equations. Denote , and let be the function expressed in the variable , that is . By the chain rule, . It is easy to see that the rank of the matrix is maximal provided that are not collinear (for details, see [CLPC04, APC13]). As we are looking for triangular central configurations, this condition is satisfied. Thus, if and only if . In other words, we can now solve the system (2.13) in the variable . We obtain

 (2.14) −1r212+ω2r12=0,−1r213−3Cr413+ω2r13=0,−1r223−3Cr423+ω2r23=0,m1m2r212+m1m3r213+m2m3r223=¯I.

Note that the function has positive derivative with , hence it is injective. Thus, the second and third equation in (2.14) yield . Solving for in the first and second equation we obtain:

 (2.15) ω=√1r312=√1r313+3Cr513.

Solving for yields:

 (2.16) r12:=v=(u5u2+3C)1/3.

Notice that implies .

The condition yields

 m1m2(u5u2+3C)2/3+(m1m3+m2m3)u2=¯I,

which is equivalent to

 m31m32u10(u2+3C)2=(¯I−(m1m3+m2m3)u2)3.

To simplify the notation, let , , , , obtaining

 az5(z+b)2=(¯I−cz)3.

The function has derivative

 k′(z)=3az6+8abz5+5ab2z4(z+b)4>0

and the function has derivative

 l′(z)=−3c(¯I−cz)2<0.

Since

 k(0)=0 and limz→+∞k(z)=+∞,

and

 l(0)=¯I3>0 and limz→+∞l(z)=−∞,

the equation has a unique solution with . We conclude that for each fixed , there is a unique solution of (2.14). Thus, we have proved the following result.

###### Proposition 2.1.

In the three-body problem with one oblate primary, for every fixed value of the moment of inertia there exists a unique central configuration, which is an isosceles triangle.

We note that, while [APC13] studies central configurations of three oblate bodies (as well as of three bodies under Schwarzschild metric), the isosceles central configuration found above is not explicitly shown there (see Theorem 4 in [APC13]).

To put this in quantitative perspective, when we use the data from Section 2.1 in (2.5), we obtain . If we set , from (2.16) we obtain . In terms of the Sun-Jupiter distance km, the distance differs from the corresponding distance in the equilateral central configuration by km. Practically, this isosceles triangle central configuration is almost an equilateral triangle.

### 2.4. Location of the bodies in the triangular central configuration

We now compute the expression of the location of the three bodies in the triangular central configuration, relative to a synodic frame that rotates together with the bodies, with the center of mass fixed at the origin, and the location of on the negative -semi-axis. We assume that the masses lie in the plane. Instead of fixing the value of the moment of inertia, we fix and have given by (2.16). Then, we obtain the following result.

###### Proposition 2.2.

In the synodic reference frame, the coordinates of the three bodies in the triangular central configuration, satisfying the constraints

 (2.17) (x2−x1)2+(y2−y1)2 = v2, (2.18) (x3−x1)2+(y3−y1)2 = 1, (2.19) (x3−x2)2+(y3−y2)2 = 1, (2.20) m1x1+m2x2+m3x3 = 0, (2.21) m1y1+m2y2+m3y3 = 0, (2.22) m1+m2+m3 = 1, (2.23) y1 = 0

are given by

 (2.24) x1=−√v2m22+v2m2m3+m23,y1=0,x2=2v2m2+v2m3−2v2m22−2v2m2m3−2m232√v2m22+v2m2m3+m23,y2=−v√4−v2m32√v2m22+v2m2m3+m23,x3=v2m2+2m3−2v2m22−2v2m2m3−2m232√v2m22+v2m2m3+m23,y3=v√4−v2m22√v2m22+v2m2m3+m23.
###### Proof.

Denote by and , so , , and . Substituting these in (2.20) we obtain . From (2.22) it follows , where we denoted . From (2.23) and (2.21) we have , and . From (2.17), we can solve for in terms of (see (2.25) below). From now on, the objective is to solve for and , which in turn will yield .

Equations (2.17), (2.18), (2.19) become:

 (2.25) A2+y22 = v2, (2.26) B2+¯μ2y22 = 1, (2.27) (B−A)2+(1+¯μ)2y22 = 1.

Adding (2.25) and (2.26), and subtracting (2.27) yields

 (2.28) 2BA−2¯μy22=v2.

From (2.25), , so (2.28) yields

 (2.29) B=v2+2¯μ(v2−A2)2A.

Substituting in (2.26) and solving for yields

 (2.30) A=±v2(2¯μ+1)2√v2¯μ2+v2¯μ+1.

Substituting in (2.28) and solving for yields

 (2.31) B=v24A(2¯μ+1)(v2¯μ+2)v2¯μ2+v2¯μ+1=±v2¯μ+22√v2¯μ2+v2¯μ+1.

with .

Substituting , and in , and choosing the negative sign for to agree with our initial choice that , after simplification, we first obtain the value of below. Then, substituting in , , , , we compute , , , , obtaining (2.24).

For future reference, we note that if we let in (2.24), we obtain

 (2.32) x1=−vm2,y1=0,x2=v(1−m2),y2=0,x3=v2(1−2m2),y3=√4−v22.

###### Remark 2.3.

In the case when the oblateness coefficient of is made equal to zero, then , and in (2.24) we obtain the Lagrangian equilateral triangle central configuration, with the position given by the following equivalent formulas (see, e.g., [BP13]):

 (2.33) x1=−|K|√m22+m2m3+m23K,y1=0,\vspace3mmx2=|K|[(m2−m3)m3+m1(2m2+m3)]2K√m22+m2m3+m23,y2=−√3m32m322 ⎷m32m22+m2m3+m23,\vspace3mmx3=|K|2√m22+m2m3+m23,y3=√32m122 ⎷m32m22+m2m3+m23,

where .

Notice that the equations (2.33) are expressed in terms of , while (2.24) are expressed in terms of ; we obtain corresponding expressions that are equivalent when we substitute in (2.33). One minor difference is that in (2.33) the position of is not constrained to be on the negative -semi-axis, as we assumed for (2.24); the position of in (2.33) depends on the quantity ; when we have , and the equations (2.24) become equivalent with the equation (2.33).

We remark that when , the limiting position of the three masses in (2.33) is given by:

 (2.34) x1=−m2,y1=0,z1=0,x2=1−m2,y2=0,z2=0,x3=1−2m22,y3=√32,z3=0,

with and representing the position of the masses and , respectively, and representing the position of the equilibrium point in the planar circular restricted three-body problem.

### 2.5. Equations of motion for the restricted four-body problem with oblate tertiary

Now we consider the dynamics of a fourth body in the neighborhood of the tertiary. This fourth body represents the moonlet Skamandrios orbiting around Hektor. We model the dynamics of the fourth body by the spatial, circular, restricted four-body problem, meaning that the moonlet is moving under the gravitational attraction of Hektor, Jupiter and the Sun, without affecting their motion which remains on circular orbits and forming a triangular central configuration as in Section 2.3. As before, we assume that Hektor has an oblate shape, with the gravitational potential given by (2.3).

The equations of motion of the infinitesimal body relative to a synodic frame of reference that rotates together with the three bodies is given by

 (2.35) ¨x−2ω˙y=∂~Ω∂x=~Ωx¨y+2ω˙x=∂~Ω∂x=~Ωy¨z=∂~Ω∂z=~Ωz,

where the effective potential is given by

 ~Ω(x,y,z)=12ω2(x2+y2)+(3∑i=1miri+m3r3(R3r3)2(C202)(3sin2ϕ−1))

with representing the -coordinates in the synodic reference frame of the body of mass , is the distance from the moonlet to the mass , for , and is the angular velocity of the system of three bodies around the center of mass given by (2.15). The perturbed mean motion of the primaries in the above equation depends on the oblateness parameter. The coordinates , , of the bodies , , , respectively, are given by (2.24), while , for .

For the choice that we have made and satisfying (2.16), we have

 (2.36) ω=1v3=√1+3R23J22=√1−3R23C202.

We remark that if we set we obtain the restricted three-body problem with one oblate body and agrees with the formulas in [McC63, SR76, AGST12]. If has no oblateness, i.e., , then .

We rescale the time so that in the new units the angular velocity is normalized to , obtaining

 (2.37) ¨x−2˙y=∂Ω∂x=Ωx,¨y+2˙x=∂Ω∂y=Ωy,¨z=∂Ω∂z=Ωz,

with

 Ω(x,y,z)=12(x2+y2)+1ω2(3∑i=1miri+m3r3(R3r3)2(C202)(3sin2ϕ−1)).

The equations of motion (2.37) have the total energy defined below as a conserved quantity:

We switch to the Hamiltonian setting by considering the system of symplectic coordinates with respect to the symplectic form , and making the transformation , and . We obtain:

 (2.38) H=12((px+y)2+(py−x)2+p2z)−12(x2+y2)−1ω2(3∑i=1miri+m3r3(R3r3)2(C202)(3sin2ϕ−1))=12(p2x+p2y+p2z)+ypx−xpy−1ω2(3∑i=1mir