# Existence and Linear Stability of Equilibrium Points in the Robe’s Restricted Three-Body Problem with Oblateness

Authors: Jagadish Singh; Abubakar Umar Sandah
Journal: Advances in Mathematical Physics
 (2012)

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## Abstract

This paper investigates the positions and linear stability of an infinitesimal body around the equilibrium points in the framework of the Robe’s circular restricted three-body problem, with assumptions that the hydrostatic equilibrium figure of the first primary is an oblate spheroid and the second primary is an oblate body as well. It is found that equilibrium point exists near the centre of the first primary. Further, there can be one more equilibrium point on the line joining the centers of both primaries. Points on the circle within the first primary are also equilibrium points under certain conditions and the existence of two out-of-plane points is also observed. The linear stability of this configuration is examined and it is found that points near the center of the first primary are conditionally stable, while the circular and out of plane equilibrium points are unstable.


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## Body

## 1. Introduction

Robe [1] considered a new kind of restricted three-body problem in which, one of the primaries of mass m1 is a rigid spherical shell, filled with homogenous, incompressible fluid of density ρ1; the second one is a point mass m2 located outside the shell and moving around the mass m1 in a Keplerian orbit; the infinitesimal mass m3 is a small sphere of density ρ3, moving inside the shell and is subject to the attraction of m2 and the buoyancy force due to the fluid of the first primary. Further, he discussed the linear stability of an equilibrium point obtained in two cases. In the first case, the orbit of m2 around m1 is circular and in the second case, the orbit is elliptic, but the shell is empty (there is no fluid inside it) or densities of m1 and m3 are equal. Since then various studies (e.g., [2–4]) under different assumptions have been carried out.In his study, Robe [1] assumed that the pressure field of the fluid ρ1 has a spherical symmetry around the center of the shell and he took into account only one out of the three components of the pressure field which is due to the own gravitational field of the fluid ρ1. He did not consider the other two components arising from the attraction of m2 and the centrifugal force. Taking care of all these three components of the pressure field, A. R. Plastino and A. Plastino [5] reanalyzed the Robe’s. But in their study, they assumed the hydrostatic equilibrium figure of the first primary as Roche’s ellipsoid (see Figure 1). They found that when the density parameter D is taken as zero, every point inside the fluid is an equilibrium point; otherwise the center of the ellipsoid is the only equilibrium point and it is linearly stable.Figure 1
The Robe’s CRTBP with oblate primaries.Hallan and Rana [3] investigated the existence of all equilibrium point and their stability in the Robe’s [1] restricted three-body problem. It was seen that the Robe’s elliptic restricted three-body problem has only one equilibrium point for all values of the density parameter K and the mass parameter μ, while the  Robe’s circular restricted three-body problem can have two, three, or infinite numbers of equilibrium points. As regards to the stability of these equilibria, they confirmed the stability result given by Robe [1] of the equilibrium point (-μ,0,0), whereas triangular and circular points are always unstable. The equilibrium point collinear with the center of the shell and the second primary was found to be stable under some conditions.Hallan and Mangang [4] studied the Robe’s [1] restricted three-body problem by considering the full buoyancy force as in A. R. Plastino and A. Plastino [5] and assuming the hydrostatic equilibrium figure of the first primary as an oblate spheroid. They derived the pertinent equations of motion and discussed the existence of equilibrium point and their linear stability.The participating bodies in the classical restricted three-body problem are strictly spherical in shape, but in actual situations several heavenly bodies, such as Saturn and Jupiter, are sufficiently oblate. The minor planets and meteoroids have irregular shape. The lack of sphericity, or the oblateness, of the planet causes large perturbations from a two-body orbit. The motions of artificial Earth satellites are examples of this. Global studies of problems with oblateness have been carried out by many researchers (e.g., [6–9]).Therefore, our effort in this paper aims at investigating the equilibrium points and their stability in the Robe’s circular restricted three-body problem when the hydrostatic equilibrium figure of the fluid of the first primary is an oblate spheroid and the second one is an oblate spheroid as well. The model of this study can be used to study the small oscillation of the Earth’s inner core taking into account the Moon’s attraction.This paper is organized as follows; Section2 represents the equations of motion; the existence of the equilibrium points is mentioned in Section 3, while Section 4 investigates their linear stability; Section 5 discusses the results obtained; the conclusion is drawn in Section 6.


## 2. Equation of Motion

Let the first primarym1 be a fluid of density ρ1 in the shape of an oblate spheroid as assumed by Hallan and Mangang [4]; let the second primary m2 be an oblate body too as Sharma and Subba Rao [6] assumed, which describes a circular orbit around m1.We adopt a uniformly rotating coordinate systemOx1x2x3 with origin at the center of mass m1, Ox1 pointing towards m2, with Ox1x2 being the orbital plane of m2 coinciding with the equatorial plane of m1. Then, the equations of motion of the infinitesimal body of density ρ3 in the coordinate system take the form [4, 6]:
(2.1)x¨1-2nx˙2=∂U∂x1,x¨2+2nx˙1=∂U∂x2,x¨3=∂U∂x3,
where
(2.2)U=V+n2{(x1-(m2/(m1+m2))R)2+x22}2,V=B+B′-ρ1ρ3[B+B′+n2{(x1-(m2/(m1+m2))R)2+x22}2],B=πGρ1[I-A1x12-A1x22-A2x32],B′=Gm2[(R-x1)2+x22+x32]1/2+Gm2α22[(R-x1)2+x22+x32]3/2-3Gm2α2x322[(R-x1)2+x22+x32]5/2,I=2a12A1+a22A2,A1=a12a2∫0∞duΔ(a12+u),A2=a12a2∫0∞duΔ(a22+u),Δ2=(a12+u)2(a22+u),n2=G(m1+m2)R2(1+32α1+32α2);α1=a12-a325R2,α2=a22-a425R2,D=1-ρ1ρ3.
Here V is the potential that explains the combined forces upon the infinitesimal mass, B denotes the potential due to the fluid mass of the first primary, B′ stands for the potential due to the second primary, R is the distance between the primaries, and G is the gravitational constant. n is the mean motion. a1,a2 and a3,a4 are the equatorial and polar radii of the first and second primary, respectively. I stands for the polar moment of inertia, while Ai  (i=1,2) are the index symbols. α1 and α2 are the oblateness coefficients of the first and second primaries, respectively.We choose the unit of mass such that the sum of the masses of the primaries is taken as unity, thus we takem2=μ, 0<μ=m2/(m1+m2)<1. For the unit of length, we take the distance between the primaries as unity, that is, R=1 and the unit of time is also selected such that G=1. With these units and substituting the expression for the potential B due to the fluid in the first primary and the potential B′ due to the second oblate primary, the equations of motion (2.1) are recast to the form:
(2.3)x¨1-2nx˙2=∂U∂x1,x¨2+2nx˙1=∂U∂x2,x¨3=∂U∂x3,
where
(2.4)U=D[πρ1{I-A1(x12+x22)-A2x32}+μ[(1-x1)2+x22+x32]1/2DDDD+μα2[(1-x1)2+x22+x32]3/2-3μα2x322[(1-x1)2+x22+x32]5/2+n2{(x1-μ)2+x22}2],n2=(1+32α1+32α2).These above equations of motion of the infinitesimal massm3 under the framework of the Robe’s circular restricted three-body problem have been obtained by taking into account the shapes of the primaries, the full buoyancy force, the forces due to the gravitational attraction of the second primary, and the gravitational force exerted by the fluid of density ρ1. In the case when the second primary is not an oblate spheroid (i.e., α2=0), the equations are the same as those of Hallan and Mangang [4].


## 3. Position of Equilibrium Points

The equilibrium points are the solutions of the equations:(3.1)Ux1=Ux2=Ux3=0.
That is,
(3.2)Ux1=D[-2πρ1x1A1+μ(1-x1)[(1-x1)2+x22+x32]3/2DDDD+3μα2(1-x1)2[(1-x1)2+x22+x32]5/2-15μα2(1-x1)x322[(1-x1)2+x22+x32]7/2+n2(x1-μ)]=0,(3.3)Ux2=Dx2[-2πρ1A1-μ[(1-x1)2+x22+x32]3/2DDDDD-3μα22[(1-x1)2+x22+x32]5/2+15μα2x322[(1-x1)2+x22+x32]7/2+n2]=0,(3.4)Ux3=Dx3[-2πρ1A2-μ[(1-x1)2+x22+x32]3/2-3μα22[(1-x1)2+x22+x32]5/2DDDDD-3μα2[(1-x1)2+x22+x32]5/2+15μα2x322[(1-x1)2+x22+x32]7/2]=0.


### 3.1. Equilibrium Points Near the Centre of the First Primary

The positions of the equilibrium points near the first primary are the solutions of (3.2) when Ux1=0, x1≠0, x2=x3=0, D≠0, and n2=1+(3/2)(α1+α2). The x1 coordinate of the equilibrium points are then the roots of the equation:
(3.5)-2πρ1x1A1+μ(1-x1)|1-x1|3+3μα2(1-x1)2|1-x1|5+(1+32α1+32α2)(x1-μ)=0.We first determine the roots of (3.5) in the absence of oblateness, that is, the case when the primaries are spherical. In this case, the roots are [4]
(3.6)x11=1+μ+μ2+8μπρ1A1-4μ2(1-2πρ1A1),x12=1+μ-μ2+8μπρ1A1-4μ2(1-2πρ1A1).
The term A1 which appears in (3.2) and is due to the fluid mass affects these roots. Therefore, these roots will be real if the discriminant is nonnegative, that is if
(3.7)μ+8πρ1A1-4≥0.
When (1/4)μ≥1-2πρ1A1>0, both roots are greater than unity and we reject them because they lie outside the first primary. Now, if 1-2πρ1A1<0, we have x12>1 and x11<1. Further, we see that x11>-1 when 1-2πρ1A1<-(3/4)μ. Thus, in the case when 1-2πρ1A1<-(3/4)μ, the point (x11,0,0) lies within the first primary if |x11|<a1. When 1-2πρ1A1<-(3/4)μ, |x11|<a1; x11 is a root of (3.5). Hence for 1-2πρ1A1=0, the only root is x11=2 which lies outside the first primary and we neglect it. Hence, for α1=0, α2=0, x11=0 is always a root of (3.5) and x1=x11 is also a root provided 1-2πρ1A1<-(3/4)μ, |x11|<a1.Now, we find the roots of (3.5) when oblateness of both primaries is considered (i.e., α1≠0, α2≠0).Let the roots be such that(3.8)x1=0+p1,|p1|≪1,x1=x11+p2,|p2|≪1.
Putting these values in (3.5), multiplying throughout by (1-p1)4, expanding and neglecting second and higher powers of p1,α1,α2, as they are very small quantities, we have
(3.9)p1≅-3α12μ2πρ1A1-(1+2μ).
Similarly, putting x1=x11+p2 in (3.5) and then simplifying it, we get
(3.10)(1-x11)4[(x11-μ)(1+32α1+32α2)-2πρ1A1x11]+p2(1-x11)3[(1-3x11)(1-2πρ1A1)+2μ]+μ(1-x11)2-2p2(1-x11)[(1-x11)2{x11-2πρ1A1x11-μ}+μ]=-3μα22.
Multiplying (3.8) by (1-x11)2, simplifying and then using it in (3.10), we get
(3.11)p2≅-(1-x11)[(x11-μ)((3/2)α1+(3/2)α2)][(1-3x11)(1-2πρ1A1)+2μ]-3μα22(1-x11)3[(1-3x11)(1-2πρ1A1)+2μ].
A substitution of (3.11) in the second equation of (3.8) at once gives the position of the other equilibrium point near the center of the first primary.


#### 3.1.1. Positions of Circular Points

The positions of the circular points are sought using the first two equations of system (3.1) with the conditions x1≠0, x2≠0, x3=0; that is, they are the solutions of
(3.12)x1[-2πρ1A1-μ{(1-x1)2+x22}3/2-3μα22{(1-x1)2+x22}5/2+1+32α1+32α2]+μ{(1-x1)2+x22}3/2+3μα22{(1-x1)2+x22}5/2-μ(1+32α1+32α2)=0,(3.13)-2πρ1A1-μ{(1-x1)2+x22}3/2-3μα22{(1-x1)2+x22}5/2+1+32α1+32α2=0.
Solving the above equations and knowing that μ≠0, we get
(3.14)1{(1-x1)2+x22}3/2+3α22{(1-x1)2+x22}5/2-n2=0.
We let
(3.15)(1-x1)2+x22=r2.
Substituting (3.15) in (3.14), and simplifying, we get
(3.16)n2r5-r2-32α2=0.
Now, we let
(3.17)r=1+ε,ε≪1.
Substituting (3.16) in (3.15), neglecting second and higher powers of ε, we get
(3.18)ε=-12α1.
Therefore, (3.17) is now expressed as
(3.19)r≅1-12α1.
A substitution of (3.14) in (3.13) yields
(3.20)2πρ1A1=n2(1-μ).
Therefore, when 2πρ1A1=n2(1-μ), the points on the circle given by (3.15) with x3=0 and r=1-(1/2)α1 lying within the first primary are also equilibrium points. The general coordinates of these circular points are given by (1+rcosθ,rsinθ,0), where θ is a parameter. When y=0, the circular points coalesce to those lying on the line joining the primaries.


#### 3.1.2. Positions of Out-of-Plane Equilibrium Points

The out-of-plane points have no analogy in the classical restricted three-body problem. However the investigation concerning these points in the photogravitational restricted three-body problem was first carried out by Radzievskii [10]. Afterwards, other researchers, for instance Douskos and Markellos [8], Singh and Leke [11], and so forth, have worked on the out-of-plane points. In this section, we locate these points for our study, as it has remained an open problem to date.The positions of the out-of-plane equilibrium points of the Robe’s problem with oblate primaries are the solutions of the first and last equations of (3.1) with x2=0, D≠0; that is,
(3.21)x1[-2πρ1A1-μ[(1-x1)2+x32]3/2-3μα22[(1-x1)2+x32]5/2+15μα2x322[(1-x1)2+x32]7/2+n2]+μ[(1-x1)2+x32]3/2+3μα22[(1-x1)2+x32]5/2-15μα2x322[(1-x1)2+x32]7/2-n2μ=0,(3.22)x3[-2πρ1A2-μ[(1-x1)2+x32]3/2-9μα22[(1-x1)2+x32]5/2+15μα2x322[(1-x1)2+x32]7/2]=0.
From (3.22), since x3≠0, we have
(3.23)-2πρ1A2=μ[(1-x1)2+x32]3/2+9μα22[(1-x1)2+x32]5/2-15μα2x322[(1-x1)2+x32]7/2.
Let
(3.24)l2=(1-x1)2+x32.
Then, (3.23) and (3.21) may be written respectively:
(3.25)15μα2x322l7=2πρ1A2+μl3+9μα22l5,x1[-2πρ1A1-μl3-3μα22l5+15μα2x322l7+n2]+μl3+3μα22l5-15μα2x322l7-n2μ=0.
Now, from first equations (3.25), we get
(3.26)x3=±l15μα2[μ(9α2+2l2)+4πρ1A2l5]1/2.
The use of (3.24) in second equation of (3.25) yields
(3.27)x1=(2πρ1A2+n2μ)l5+3μα2[2πρ1(A2-A1)+n2]l5+3μα2.
We use the software package Mathematica(Wolfram 2004) to compute the coordinates of the out-of-plane equilibrium points denoted by L6 and L7 starting with the initial values x1=1-μ and x3=3α2 in the case where we have kept up to first order terms in both the numerator and the denominator; we then get(3.28)x1=2μ+3μα1+4A2πρ12+3α1-4A1πρ1+4A2πρ1-3{μA1+(1-μ)A2}α2πρ1[1+3α1(1+2A2πρ1-2A1πρ1)+4πρ1(A2-A1+A12πρ1+A22πρ1-2A1A2πρ1)],x3=2/15μμ3(1+2πA2μ2ρ1)μα2+73μ/10(μ+2πA2μ3ρ1)α22μ3(1+2πA2μ2ρ1).
The location of the out-of-plane equilibrium points can be obtained by solving numerically equations (3.26) and (3.27) using (3.24).Now, from the expression for the density parameter(3.29)D=(1-ρ1ρ3).
We assume that ρ1≠ρ3, then D>or<0. In the case when the density parameter is positive, we have
(3.30)ρ1<ρ3.
Hence, numerically we choose
(3.31)a12=0.94,a22=0.9,a32=0.82,a42=0.8,μ=0.01,π=3.14.
Then,
(3.32)α1=0.024,α2=0.02,ρ1=0.236.
Now, we perform a numerical exploration of computing the out-of-plane points in the case of the Earth-Moon system. To do this, we arbitrarily choose values for the Ai  (i=1,2). We found that when A1=2.5076 and A2=2.555, the positions of the out-of-plane points (x1,0,±x3):
(3.33)x1=3.3527,x3=0.271418.
The abscissae of the out-of-plane point is outside the possible region of motion of the infinitesimal mass and so we neglect it. However, in the case when the Ai  (i=1,2) are chosen such that
(3.34)|A1-A2|≪1,A1=0.7,A2=0.68(say),Ai∈(0,0.7], the point L6, and L7 are, respectively,
(3.35)x1=0.98611,x3=0.271381
and lies within the fluid.


## 3.1. Equilibrium Points Near the Centre of the First Primary

The positions of the equilibrium points near the first primary are the solutions of (3.2) when Ux1=0, x1≠0, x2=x3=0, D≠0, and n2=1+(3/2)(α1+α2). The x1 coordinate of the equilibrium points are then the roots of the equation:
(3.5)-2πρ1x1A1+μ(1-x1)|1-x1|3+3μα2(1-x1)2|1-x1|5+(1+32α1+32α2)(x1-μ)=0.We first determine the roots of (3.5) in the absence of oblateness, that is, the case when the primaries are spherical. In this case, the roots are [4]
(3.6)x11=1+μ+μ2+8μπρ1A1-4μ2(1-2πρ1A1),x12=1+μ-μ2+8μπρ1A1-4μ2(1-2πρ1A1).
The term A1 which appears in (3.2) and is due to the fluid mass affects these roots. Therefore, these roots will be real if the discriminant is nonnegative, that is if
(3.7)μ+8πρ1A1-4≥0.
When (1/4)μ≥1-2πρ1A1>0, both roots are greater than unity and we reject them because they lie outside the first primary. Now, if 1-2πρ1A1<0, we have x12>1 and x11<1. Further, we see that x11>-1 when 1-2πρ1A1<-(3/4)μ. Thus, in the case when 1-2πρ1A1<-(3/4)μ, the point (x11,0,0) lies within the first primary if |x11|<a1. When 1-2πρ1A1<-(3/4)μ, |x11|<a1; x11 is a root of (3.5). Hence for 1-2πρ1A1=0, the only root is x11=2 which lies outside the first primary and we neglect it. Hence, for α1=0, α2=0, x11=0 is always a root of (3.5) and x1=x11 is also a root provided 1-2πρ1A1<-(3/4)μ, |x11|<a1.Now, we find the roots of (3.5) when oblateness of both primaries is considered (i.e., α1≠0, α2≠0).Let the roots be such that(3.8)x1=0+p1,|p1|≪1,x1=x11+p2,|p2|≪1.
Putting these values in (3.5), multiplying throughout by (1-p1)4, expanding and neglecting second and higher powers of p1,α1,α2, as they are very small quantities, we have
(3.9)p1≅-3α12μ2πρ1A1-(1+2μ).
Similarly, putting x1=x11+p2 in (3.5) and then simplifying it, we get
(3.10)(1-x11)4[(x11-μ)(1+32α1+32α2)-2πρ1A1x11]+p2(1-x11)3[(1-3x11)(1-2πρ1A1)+2μ]+μ(1-x11)2-2p2(1-x11)[(1-x11)2{x11-2πρ1A1x11-μ}+μ]=-3μα22.
Multiplying (3.8) by (1-x11)2, simplifying and then using it in (3.10), we get
(3.11)p2≅-(1-x11)[(x11-μ)((3/2)α1+(3/2)α2)][(1-3x11)(1-2πρ1A1)+2μ]-3μα22(1-x11)3[(1-3x11)(1-2πρ1A1)+2μ].
A substitution of (3.11) in the second equation of (3.8) at once gives the position of the other equilibrium point near the center of the first primary.


### 3.1.1. Positions of Circular Points

The positions of the circular points are sought using the first two equations of system (3.1) with the conditions x1≠0, x2≠0, x3=0; that is, they are the solutions of
(3.12)x1[-2πρ1A1-μ{(1-x1)2+x22}3/2-3μα22{(1-x1)2+x22}5/2+1+32α1+32α2]+μ{(1-x1)2+x22}3/2+3μα22{(1-x1)2+x22}5/2-μ(1+32α1+32α2)=0,(3.13)-2πρ1A1-μ{(1-x1)2+x22}3/2-3μα22{(1-x1)2+x22}5/2+1+32α1+32α2=0.
Solving the above equations and knowing that μ≠0, we get
(3.14)1{(1-x1)2+x22}3/2+3α22{(1-x1)2+x22}5/2-n2=0.
We let
(3.15)(1-x1)2+x22=r2.
Substituting (3.15) in (3.14), and simplifying, we get
(3.16)n2r5-r2-32α2=0.
Now, we let
(3.17)r=1+ε,ε≪1.
Substituting (3.16) in (3.15), neglecting second and higher powers of ε, we get
(3.18)ε=-12α1.
Therefore, (3.17) is now expressed as
(3.19)r≅1-12α1.
A substitution of (3.14) in (3.13) yields
(3.20)2πρ1A1=n2(1-μ).
Therefore, when 2πρ1A1=n2(1-μ), the points on the circle given by (3.15) with x3=0 and r=1-(1/2)α1 lying within the first primary are also equilibrium points. The general coordinates of these circular points are given by (1+rcosθ,rsinθ,0), where θ is a parameter. When y=0, the circular points coalesce to those lying on the line joining the primaries.


### 3.1.2. Positions of Out-of-Plane Equilibrium Points

The out-of-plane points have no analogy in the classical restricted three-body problem. However the investigation concerning these points in the photogravitational restricted three-body problem was first carried out by Radzievskii [10]. Afterwards, other researchers, for instance Douskos and Markellos [8], Singh and Leke [11], and so forth, have worked on the out-of-plane points. In this section, we locate these points for our study, as it has remained an open problem to date.The positions of the out-of-plane equilibrium points of the Robe’s problem with oblate primaries are the solutions of the first and last equations of (3.1) with x2=0, D≠0; that is,
(3.21)x1[-2πρ1A1-μ[(1-x1)2+x32]3/2-3μα22[(1-x1)2+x32]5/2+15μα2x322[(1-x1)2+x32]7/2+n2]+μ[(1-x1)2+x32]3/2+3μα22[(1-x1)2+x32]5/2-15μα2x322[(1-x1)2+x32]7/2-n2μ=0,(3.22)x3[-2πρ1A2-μ[(1-x1)2+x32]3/2-9μα22[(1-x1)2+x32]5/2+15μα2x322[(1-x1)2+x32]7/2]=0.
From (3.22), since x3≠0, we have
(3.23)-2πρ1A2=μ[(1-x1)2+x32]3/2+9μα22[(1-x1)2+x32]5/2-15μα2x322[(1-x1)2+x32]7/2.
Let
(3.24)l2=(1-x1)2+x32.
Then, (3.23) and (3.21) may be written respectively:
(3.25)15μα2x322l7=2πρ1A2+μl3+9μα22l5,x1[-2πρ1A1-μl3-3μα22l5+15μα2x322l7+n2]+μl3+3μα22l5-15μα2x322l7-n2μ=0.
Now, from first equations (3.25), we get
(3.26)x3=±l15μα2[μ(9α2+2l2)+4πρ1A2l5]1/2.
The use of (3.24) in second equation of (3.25) yields
(3.27)x1=(2πρ1A2+n2μ)l5+3μα2[2πρ1(A2-A1)+n2]l5+3μα2.
We use the software package Mathematica(Wolfram 2004) to compute the coordinates of the out-of-plane equilibrium points denoted by L6 and L7 starting with the initial values x1=1-μ and x3=3α2 in the case where we have kept up to first order terms in both the numerator and the denominator; we then get(3.28)x1=2μ+3μα1+4A2πρ12+3α1-4A1πρ1+4A2πρ1-3{μA1+(1-μ)A2}α2πρ1[1+3α1(1+2A2πρ1-2A1πρ1)+4πρ1(A2-A1+A12πρ1+A22πρ1-2A1A2πρ1)],x3=2/15μμ3(1+2πA2μ2ρ1)μα2+73μ/10(μ+2πA2μ3ρ1)α22μ3(1+2πA2μ2ρ1).
The location of the out-of-plane equilibrium points can be obtained by solving numerically equations (3.26) and (3.27) using (3.24).Now, from the expression for the density parameter(3.29)D=(1-ρ1ρ3).
We assume that ρ1≠ρ3, then D>or<0. In the case when the density parameter is positive, we have
(3.30)ρ1<ρ3.
Hence, numerically we choose
(3.31)a12=0.94,a22=0.9,a32=0.82,a42=0.8,μ=0.01,π=3.14.
Then,
(3.32)α1=0.024,α2=0.02,ρ1=0.236.
Now, we perform a numerical exploration of computing the out-of-plane points in the case of the Earth-Moon system. To do this, we arbitrarily choose values for the Ai  (i=1,2). We found that when A1=2.5076 and A2=2.555, the positions of the out-of-plane points (x1,0,±x3):
(3.33)x1=3.3527,x3=0.271418.
The abscissae of the out-of-plane point is outside the possible region of motion of the infinitesimal mass and so we neglect it. However, in the case when the Ai  (i=1,2) are chosen such that
(3.34)|A1-A2|≪1,A1=0.7,A2=0.68(say),Ai∈(0,0.7], the point L6, and L7 are, respectively,
(3.35)x1=0.98611,x3=0.271381
and lies within the fluid.


## 3.1.1. Positions of Circular Points

The positions of the circular points are sought using the first two equations of system (3.1) with the conditions x1≠0, x2≠0, x3=0; that is, they are the solutions of
(3.12)x1[-2πρ1A1-μ{(1-x1)2+x22}3/2-3μα22{(1-x1)2+x22}5/2+1+32α1+32α2]+μ{(1-x1)2+x22}3/2+3μα22{(1-x1)2+x22}5/2-μ(1+32α1+32α2)=0,(3.13)-2πρ1A1-μ{(1-x1)2+x22}3/2-3μα22{(1-x1)2+x22}5/2+1+32α1+32α2=0.
Solving the above equations and knowing that μ≠0, we get
(3.14)1{(1-x1)2+x22}3/2+3α22{(1-x1)2+x22}5/2-n2=0.
We let
(3.15)(1-x1)2+x22=r2.
Substituting (3.15) in (3.14), and simplifying, we get
(3.16)n2r5-r2-32α2=0.
Now, we let
(3.17)r=1+ε,ε≪1.
Substituting (3.16) in (3.15), neglecting second and higher powers of ε, we get
(3.18)ε=-12α1.
Therefore, (3.17) is now expressed as
(3.19)r≅1-12α1.
A substitution of (3.14) in (3.13) yields
(3.20)2πρ1A1=n2(1-μ).
Therefore, when 2πρ1A1=n2(1-μ), the points on the circle given by (3.15) with x3=0 and r=1-(1/2)α1 lying within the first primary are also equilibrium points. The general coordinates of these circular points are given by (1+rcosθ,rsinθ,0), where θ is a parameter. When y=0, the circular points coalesce to those lying on the line joining the primaries.


## 3.1.2. Positions of Out-of-Plane Equilibrium Points

The out-of-plane points have no analogy in the classical restricted three-body problem. However the investigation concerning these points in the photogravitational restricted three-body problem was first carried out by Radzievskii [10]. Afterwards, other researchers, for instance Douskos and Markellos [8], Singh and Leke [11], and so forth, have worked on the out-of-plane points. In this section, we locate these points for our study, as it has remained an open problem to date.The positions of the out-of-plane equilibrium points of the Robe’s problem with oblate primaries are the solutions of the first and last equations of (3.1) with x2=0, D≠0; that is,
(3.21)x1[-2πρ1A1-μ[(1-x1)2+x32]3/2-3μα22[(1-x1)2+x32]5/2+15μα2x322[(1-x1)2+x32]7/2+n2]+μ[(1-x1)2+x32]3/2+3μα22[(1-x1)2+x32]5/2-15μα2x322[(1-x1)2+x32]7/2-n2μ=0,(3.22)x3[-2πρ1A2-μ[(1-x1)2+x32]3/2-9μα22[(1-x1)2+x32]5/2+15μα2x322[(1-x1)2+x32]7/2]=0.
From (3.22), since x3≠0, we have
(3.23)-2πρ1A2=μ[(1-x1)2+x32]3/2+9μα22[(1-x1)2+x32]5/2-15μα2x322[(1-x1)2+x32]7/2.
Let
(3.24)l2=(1-x1)2+x32.
Then, (3.23) and (3.21) may be written respectively:
(3.25)15μα2x322l7=2πρ1A2+μl3+9μα22l5,x1[-2πρ1A1-μl3-3μα22l5+15μα2x322l7+n2]+μl3+3μα22l5-15μα2x322l7-n2μ=0.
Now, from first equations (3.25), we get
(3.26)x3=±l15μα2[μ(9α2+2l2)+4πρ1A2l5]1/2.
The use of (3.24) in second equation of (3.25) yields
(3.27)x1=(2πρ1A2+n2μ)l5+3μα2[2πρ1(A2-A1)+n2]l5+3μα2.
We use the software package Mathematica(Wolfram 2004) to compute the coordinates of the out-of-plane equilibrium points denoted by L6 and L7 starting with the initial values x1=1-μ and x3=3α2 in the case where we have kept up to first order terms in both the numerator and the denominator; we then get(3.28)x1=2μ+3μα1+4A2πρ12+3α1-4A1πρ1+4A2πρ1-3{μA1+(1-μ)A2}α2πρ1[1+3α1(1+2A2πρ1-2A1πρ1)+4πρ1(A2-A1+A12πρ1+A22πρ1-2A1A2πρ1)],x3=2/15μμ3(1+2πA2μ2ρ1)μα2+73μ/10(μ+2πA2μ3ρ1)α22μ3(1+2πA2μ2ρ1).
The location of the out-of-plane equilibrium points can be obtained by solving numerically equations (3.26) and (3.27) using (3.24).Now, from the expression for the density parameter(3.29)D=(1-ρ1ρ3).
We assume that ρ1≠ρ3, then D>or<0. In the case when the density parameter is positive, we have
(3.30)ρ1<ρ3.
Hence, numerically we choose
(3.31)a12=0.94,a22=0.9,a32=0.82,a42=0.8,μ=0.01,π=3.14.
Then,
(3.32)α1=0.024,α2=0.02,ρ1=0.236.
Now, we perform a numerical exploration of computing the out-of-plane points in the case of the Earth-Moon system. To do this, we arbitrarily choose values for the Ai  (i=1,2). We found that when A1=2.5076 and A2=2.555, the positions of the out-of-plane points (x1,0,±x3):
(3.33)x1=3.3527,x3=0.271418.
The abscissae of the out-of-plane point is outside the possible region of motion of the infinitesimal mass and so we neglect it. However, in the case when the Ai  (i=1,2) are chosen such that
(3.34)|A1-A2|≪1,A1=0.7,A2=0.68(say),Ai∈(0,0.7], the point L6, and L7 are, respectively,
(3.35)x1=0.98611,x3=0.271381
and lies within the fluid.


## 4. Linear Stability of the Equilibrium Points

In order to study the linear stability of any equilibrium point(x10,x20,x30) of an infinitesimal body, we displace it to the position (x1,x2,x3) such that
(4.1)(x10+ξ,x20+η,x30+ζ),
where ξ,η,ζ are small displacements, and then linearize equation (2.3) to obtain the equations:
(4.2)ξ¨-2nη˙=(Ux1x10)ξ+(Ux1x20)η+(Ux1x30)ζ,η¨+2nξ˙=(Ux1x20)ξ+(Ux2x20)η+(Ux2x30)ζ,ζ¨=(Ux1x30)ξ+(Ux2x30)η+(Ux3x30)ζ,
where the partial derivatives are evaluated at the equilibrium points.


### 4.1. Equilibrium Points Near the Center of the First Primary

In order to consider the motion near any equilibrium point in thex1x2-plane, we let solutions of the first two equations of (4.2) be
(4.3)ξ=Aexp⁡(λt),η=Bexp⁡(λt),
where A, B, and λ are constants.Taking first and second derivatives of the above, substituting them into the first two equations of system (4.2) and has a non-zero solution when(4.4)|(λ2-Ux1x10)(2nλ+Ux1x20)(2nλ-Ux1x20)(λ2-Ux2x20)|=0.
Expanding the determinant, we have
(4.5)λ4-(Ux1x10+Ux2x20-4n2)λ2+Ux1x10Ux2x20-(Ux1x20)2=0.
Equation (4.5) is the characteristic equation corresponding to the variational equations (4.2) in the case when motion is considered in the x1,x2-plane.Now, the values of the second-order partial derivatives of the equilibrium point(xL,0,0), where xL=p1 stands for the first equilibrium and xL=x11+p2 for the second one, are given as
(4.6)Ux1x10=Dμ[-(1-xL)3-(3α2/2)(1-xL)+2xL(1-xL)2+6xLα2+n2(1-xL)5xL(1-xL)5],Ux2x20=Dμ[-(1-xL)3-(3/2)α2(1-xL)-xL(1-xL)2-(3/2)α2xL+n2(1-xL)5xL(1-xL)5],Ux3x30=-D[2πρ1A2+μ(1-xL)3+9μα22(1-xL)5],Ux1x20=0=Ux2x30=Ux1x30.
Substituting these in (4.2), we at once have the variational equations:
(4.7)ξ¨-2nη˙=Ux1x10ξ,η¨+2nξ˙=Ux2x20η,(4.8)ζ¨=-D[μ(1-xL)3+9μα22(1-xL)5+2πρ1A2]ζ,
where the partial derivatives have been computed at each equilibrium point xL.Now, (4.8) is independent of (4.7), the solution being a periodic function is bounded and therefore, the motion of the infinitesimal body in the x3 direction is stable.Now, the characteristic equation of the equilibrium points(xL,0,0) corresponding to the system (4.7) is
(4.9)λ4-(Ux1x10+Ux2x20-4n2)λ2+Ux1x10Ux2x20=0,
where
(4.10)Ux1x10=3Dμ(α12xL),(4.11)Ux2x20=3Dμ(-1+α12xL).
These equations have been obtained using binomial expansion and ignoring terms with second and higher power in p1,p2,α2, and their product.Now, letλ12 and λ22 be the roots of (4.9), then, the equilibrium point is stable if both the roots are real and negative. This means that their sum must be negative and their product must be positive. Hence, the points (xL,0,0) will be stable if the following two conditions hold:
(4.12)λ12+λ22=Ux1x10+Ux2x20-4n2<0,(4.13)λ12λ22=Ux1x10Ux2x20>0.
Now, in the case of the first equilibrium point xL=p1, if we suppose in (4.10) that p1<0 then, Ux1x10<0 since 0<μ<1, D>0, 0<α1≪1 and when p1>0, we have Ux1x10>0.Similarly, in (4.11), if we suppose p1<0 then Ux2x20<0.For the casep1>0, we will have Ux2x20>0 when α1>2|p1| which is not possible, hence Ux2x20<0.In the case0<p1<α1/2, Ux1x10>0, and Ux2x20>0.Also, if0<α1/2<p1, we see that Ux1x10>0 and Ux2x20<0. Thus, for the case p1<0, the equilibrium point is stable. For 0<p1<α1/2, the equilibrium point is stable if the condition (4.12) holds. When 0<α1/2<p1, the equilibrium point is unstable.Next, for the other equilibrium point positioned atxL=x11+p2, when x11>0,then x11′>0 since |p2|≪1 and the equilibrium point is stable if the conditions (4.12) and (4.13) are satisfied. If x11<0 then x11′<0; it makes Ux1x10<0, Ux2x20<0. Therefore, when x11<0, both the conditions (4.12) and (4.13) are fulfilled and the equilibrium point is stable.


### 4.2. Circular Points

At circular points(1+rcosθ,rsinθ,0), the values of the second partial derivatives with the use of (3.14) and neglecting the product α1α2 are
(4.14)Ux1x10=3Dμcos⁡2θ(n2+α2),Ux1x20=3Dμcos⁡θsin⁡θ(n2+α2),Ux2x20=3Dμsin2θ(n2+α2),Ux3x30=-D[2πρ1A2+μ(n2+3α2)].
Substituting these values in the variational equations (4.2), we get
(4.15)ξ¨-2nη˙=3Dμcos⁡2θ(n2+α2)ξ+3Dμcos⁡θsin⁡θ(n2+α2)η+(0)ζ,η¨+2nξ˙=3Dμcos⁡θsin⁡θ(n2+α2)ξ+3Dμsin2θ(n2+α2)η+(0)ζ,(4.16)ζ¨=-D[2πρ1A2+μ(n2+3α2)]ζ.
Equation (4.16) is independent of (4.15), it shows that the motion of the infinitesimal mass along the x3-direction is stable.Now, a substitution of these partial derivatives in the characteristic equation (4.5) yields
(4.17)λ4-[3Dμ(n2+α2)-4n2]λ2=0.
Let λ2=Λ in (4.17) then, we have
(4.18)Λ[Λ-{3Dμ(n2+α2)-4n2}]=0.
Hence, either
(4.19)Λ=0orΛ=3Dμ(n2+α2)-4n2,
which implies that
(4.20)λ=0twice,orλ=±[3Dμ(n2+α2)-4n2]1/2.
Therefore, (4.20) gives the roots of the characteristic equation (4.17). Hence we conclude that the circular points are unstable due to the presence of multiple roots.


### 4.3. Out-of-Plane Points

To determine the stability of the out-of-plane equilibrium points, we consider the following partial derivatives:(4.21)Ux1x1=D[-2πρ1A1+μ{[2(1-x1)2-x32]{(1-x1)2+x32}5/2}DD+32μα2{[4(1-x1)2-x32]{(1-x1)2+x32}7/2}-152μα2x32{[6(1-x1)2-x32]{(1-x1)2+x32}9/2}+n2],Ux1x3=D[-3μ(1-x1)x3{(1-x1)2+x32}5/2-152μα2(1-x1)x3{(1-x1)2+x32}7/2DD-15μα2(1-x1)22x3[(1-x1)2+x32]-7x33{(1-x1)2+x32}9/2],Ux2x2=D[-2πρ1A1-μ{(1-x1)2+x32{(1-x1)2+x32}5/2}DD-3μα22{(1-x1)2+x32{(1-x1)2+x32}7/2}+15μα2x322{(1-x1)2+x32{(1-x1)2+x32}9/2}+n2],Ux3x3=D[-2πρ1A2-μ{(1-x1)2-2x32{(1-x1)2+x32}5/2}DD-9μα22{(1-x1)2-4x32{(1-x1)2+x32}7/2}+15μα22{3x32[(1-x1)2+x32]-7x34{(1-x1)2+x32}9/2}].
Since x2=0, therefore the partial derivatives to be computed at the out-of-plane equilibrium points are
(4.22)Ux1x20=0=Ux2x10=Ux2x30=Ux3x20,Ux1x10=D[-2πρ1A1-2πρ1A2+μl3-3μx32l5+3μα22l5-9μα22l5-45μα2x32l7+105μα2x342l9+n2],Ux2x20=D[-2πρ1A1+2πρ1A2+3μα2l5+n2],Ux1x30=-3(1-x1)x3D[μl5+15μα22l7-35μα2x322l9],Ux3x30=D[3μx32l5+75μα2x322l7-105μα2x342l9].
Now, we let,
(4.23)Ux1x10=U11,Ux2x20=U22,Ux1x30=U13,Ux3x30=U33,Ux1x20=U12,Ux2x30=U23.
Using (4.23), the variational equation can be recast in the form:
(4.24)ξ¨-2nη˙=U11ξ+U13ζ,η¨+2nξ˙=U22η+U23ζ,ζ¨=U13ξ+U33ζ.
In order to consider the motion of the out-of-plane points, we let solution of the system (4.24) be
(4.25)ξ=Aexp⁡(λt),η=Bexp⁡(λt),ζ=Cexp⁡(λt),
where A,B,C, and λ are constants. ξ, η, and ζ are the small displacements in the coordinates of the infinitesimal body.Now, the characteristic equation corresponding to the variational equations (4.24) in the case of the out-of-plane point may be expressed as
(4.26)λ6-a1λ4+a2λ2+a3=0,
where the coefficients of the characteristic equation (4.26) are such that
(4.27)a1=D[-4πρ1A1-2πρ1A2+2n2],a2=14l16[D[-9μ2Dx32(-1+x1)2[2l4+5α2(3l2-7x32)]14l16DD+3l4μx32[2l4+5α2(5l2-7μx32)]14l16DD×[3Dμα2+2l5{(-4+D)n2-2DA1πρ1+2DA2πρ1}]14l16DD+D[6l4μx32+3α2{l4μ-25l2-35μx34}+2l9(n2-2A1πρ1+2A2πρ1)]14l16DD×[3α2{l4μ-2l2(15+l2)x32+35μx34}+2l6(l3n2+μ-2l3A1πρ1-2l3A2πρ1)]-9μ2Dx32(-1+x1)2[2l4+5α2(3l2-7x32)]]-9μ2Dx32(-1+x1)2[2l4+5α2(3l2-7x32)]]a3=3D34l18μx32[2l4+5α2(5l2-7μx32)][n2+3μα22l5-2A1πρ1+2A2πρ1]×[3α2{l4μ-2l2(15+l2)μx32+35μx34}+2l6(l3n2+μ-2l3A1πρ1-2l3A2πρ1)],
where l2=(1-x1)2+x32.These computations have been done using the software package Mathematica.For the stability analysis of the out-of-plane equilibrium point, we compute numerically the partial derivatives calculated at the out-of-plane points with the use of (3.28) and the following numerical values:
(4.28)μ=0.01,π=3.14,α1=0.024,α2=0.02A1=0.7,A2=0.68,ρ1=0.236,D=0.2133.Now, substituting the above values in the characteristic equation (4.27), we get
(4.29)λ6-0.192437λ4+0.248034λ2-0.0000197624=0.
Its roots are:
(4.30)λ1,2=-0.545066±0.448239i,λ3,4=±0.00892642,λ5,6=0.545066±0.448239i.
The positive root and the positive real part of the complex roots induce instability at the out-of-plane point. Hence, the motion of the infinitesimal mass around the out-of-plane equilibrium points is unstable for the specific numerical example given here. However, fuller discussion of their stability remains a theme for future research.


## 4.1. Equilibrium Points Near the Center of the First Primary

In order to consider the motion near any equilibrium point in thex1x2-plane, we let solutions of the first two equations of (4.2) be
(4.3)ξ=Aexp⁡(λt),η=Bexp⁡(λt),
where A, B, and λ are constants.Taking first and second derivatives of the above, substituting them into the first two equations of system (4.2) and has a non-zero solution when(4.4)|(λ2-Ux1x10)(2nλ+Ux1x20)(2nλ-Ux1x20)(λ2-Ux2x20)|=0.
Expanding the determinant, we have
(4.5)λ4-(Ux1x10+Ux2x20-4n2)λ2+Ux1x10Ux2x20-(Ux1x20)2=0.
Equation (4.5) is the characteristic equation corresponding to the variational equations (4.2) in the case when motion is considered in the x1,x2-plane.Now, the values of the second-order partial derivatives of the equilibrium point(xL,0,0), where xL=p1 stands for the first equilibrium and xL=x11+p2 for the second one, are given as
(4.6)Ux1x10=Dμ[-(1-xL)3-(3α2/2)(1-xL)+2xL(1-xL)2+6xLα2+n2(1-xL)5xL(1-xL)5],Ux2x20=Dμ[-(1-xL)3-(3/2)α2(1-xL)-xL(1-xL)2-(3/2)α2xL+n2(1-xL)5xL(1-xL)5],Ux3x30=-D[2πρ1A2+μ(1-xL)3+9μα22(1-xL)5],Ux1x20=0=Ux2x30=Ux1x30.
Substituting these in (4.2), we at once have the variational equations:
(4.7)ξ¨-2nη˙=Ux1x10ξ,η¨+2nξ˙=Ux2x20η,(4.8)ζ¨=-D[μ(1-xL)3+9μα22(1-xL)5+2πρ1A2]ζ,
where the partial derivatives have been computed at each equilibrium point xL.Now, (4.8) is independent of (4.7), the solution being a periodic function is bounded and therefore, the motion of the infinitesimal body in the x3 direction is stable.Now, the characteristic equation of the equilibrium points(xL,0,0) corresponding to the system (4.7) is
(4.9)λ4-(Ux1x10+Ux2x20-4n2)λ2+Ux1x10Ux2x20=0,
where
(4.10)Ux1x10=3Dμ(α12xL),(4.11)Ux2x20=3Dμ(-1+α12xL).
These equations have been obtained using binomial expansion and ignoring terms with second and higher power in p1,p2,α2, and their product.Now, letλ12 and λ22 be the roots of (4.9), then, the equilibrium point is stable if both the roots are real and negative. This means that their sum must be negative and their product must be positive. Hence, the points (xL,0,0) will be stable if the following two conditions hold:
(4.12)λ12+λ22=Ux1x10+Ux2x20-4n2<0,(4.13)λ12λ22=Ux1x10Ux2x20>0.
Now, in the case of the first equilibrium point xL=p1, if we suppose in (4.10) that p1<0 then, Ux1x10<0 since 0<μ<1, D>0, 0<α1≪1 and when p1>0, we have Ux1x10>0.Similarly, in (4.11), if we suppose p1<0 then Ux2x20<0.For the casep1>0, we will have Ux2x20>0 when α1>2|p1| which is not possible, hence Ux2x20<0.In the case0<p1<α1/2, Ux1x10>0, and Ux2x20>0.Also, if0<α1/2<p1, we see that Ux1x10>0 and Ux2x20<0. Thus, for the case p1<0, the equilibrium point is stable. For 0<p1<α1/2, the equilibrium point is stable if the condition (4.12) holds. When 0<α1/2<p1, the equilibrium point is unstable.Next, for the other equilibrium point positioned atxL=x11+p2, when x11>0,then x11′>0 since |p2|≪1 and the equilibrium point is stable if the conditions (4.12) and (4.13) are satisfied. If x11<0 then x11′<0; it makes Ux1x10<0, Ux2x20<0. Therefore, when x11<0, both the conditions (4.12) and (4.13) are fulfilled and the equilibrium point is stable.


## 4.2. Circular Points

At circular points(1+rcosθ,rsinθ,0), the values of the second partial derivatives with the use of (3.14) and neglecting the product α1α2 are
(4.14)Ux1x10=3Dμcos⁡2θ(n2+α2),Ux1x20=3Dμcos⁡θsin⁡θ(n2+α2),Ux2x20=3Dμsin2θ(n2+α2),Ux3x30=-D[2πρ1A2+μ(n2+3α2)].
Substituting these values in the variational equations (4.2), we get
(4.15)ξ¨-2nη˙=3Dμcos⁡2θ(n2+α2)ξ+3Dμcos⁡θsin⁡θ(n2+α2)η+(0)ζ,η¨+2nξ˙=3Dμcos⁡θsin⁡θ(n2+α2)ξ+3Dμsin2θ(n2+α2)η+(0)ζ,(4.16)ζ¨=-D[2πρ1A2+μ(n2+3α2)]ζ.
Equation (4.16) is independent of (4.15), it shows that the motion of the infinitesimal mass along the x3-direction is stable.Now, a substitution of these partial derivatives in the characteristic equation (4.5) yields
(4.17)λ4-[3Dμ(n2+α2)-4n2]λ2=0.
Let λ2=Λ in (4.17) then, we have
(4.18)Λ[Λ-{3Dμ(n2+α2)-4n2}]=0.
Hence, either
(4.19)Λ=0orΛ=3Dμ(n2+α2)-4n2,
which implies that
(4.20)λ=0twice,orλ=±[3Dμ(n2+α2)-4n2]1/2.
Therefore, (4.20) gives the roots of the characteristic equation (4.17). Hence we conclude that the circular points are unstable due to the presence of multiple roots.


## 4.3. Out-of-Plane Points

To determine the stability of the out-of-plane equilibrium points, we consider the following partial derivatives:(4.21)Ux1x1=D[-2πρ1A1+μ{[2(1-x1)2-x32]{(1-x1)2+x32}5/2}DD+32μα2{[4(1-x1)2-x32]{(1-x1)2+x32}7/2}-152μα2x32{[6(1-x1)2-x32]{(1-x1)2+x32}9/2}+n2],Ux1x3=D[-3μ(1-x1)x3{(1-x1)2+x32}5/2-152μα2(1-x1)x3{(1-x1)2+x32}7/2DD-15μα2(1-x1)22x3[(1-x1)2+x32]-7x33{(1-x1)2+x32}9/2],Ux2x2=D[-2πρ1A1-μ{(1-x1)2+x32{(1-x1)2+x32}5/2}DD-3μα22{(1-x1)2+x32{(1-x1)2+x32}7/2}+15μα2x322{(1-x1)2+x32{(1-x1)2+x32}9/2}+n2],Ux3x3=D[-2πρ1A2-μ{(1-x1)2-2x32{(1-x1)2+x32}5/2}DD-9μα22{(1-x1)2-4x32{(1-x1)2+x32}7/2}+15μα22{3x32[(1-x1)2+x32]-7x34{(1-x1)2+x32}9/2}].
Since x2=0, therefore the partial derivatives to be computed at the out-of-plane equilibrium points are
(4.22)Ux1x20=0=Ux2x10=Ux2x30=Ux3x20,Ux1x10=D[-2πρ1A1-2πρ1A2+μl3-3μx32l5+3μα22l5-9μα22l5-45μα2x32l7+105μα2x342l9+n2],Ux2x20=D[-2πρ1A1+2πρ1A2+3μα2l5+n2],Ux1x30=-3(1-x1)x3D[μl5+15μα22l7-35μα2x322l9],Ux3x30=D[3μx32l5+75μα2x322l7-105μα2x342l9].
Now, we let,
(4.23)Ux1x10=U11,Ux2x20=U22,Ux1x30=U13,Ux3x30=U33,Ux1x20=U12,Ux2x30=U23.
Using (4.23), the variational equation can be recast in the form:
(4.24)ξ¨-2nη˙=U11ξ+U13ζ,η¨+2nξ˙=U22η+U23ζ,ζ¨=U13ξ+U33ζ.
In order to consider the motion of the out-of-plane points, we let solution of the system (4.24) be
(4.25)ξ=Aexp⁡(λt),η=Bexp⁡(λt),ζ=Cexp⁡(λt),
where A,B,C, and λ are constants. ξ, η, and ζ are the small displacements in the coordinates of the infinitesimal body.Now, the characteristic equation corresponding to the variational equations (4.24) in the case of the out-of-plane point may be expressed as
(4.26)λ6-a1λ4+a2λ2+a3=0,
where the coefficients of the characteristic equation (4.26) are such that
(4.27)a1=D[-4πρ1A1-2πρ1A2+2n2],a2=14l16[D[-9μ2Dx32(-1+x1)2[2l4+5α2(3l2-7x32)]14l16DD+3l4μx32[2l4+5α2(5l2-7μx32)]14l16DD×[3Dμα2+2l5{(-4+D)n2-2DA1πρ1+2DA2πρ1}]14l16DD+D[6l4μx32+3α2{l4μ-25l2-35μx34}+2l9(n2-2A1πρ1+2A2πρ1)]14l16DD×[3α2{l4μ-2l2(15+l2)x32+35μx34}+2l6(l3n2+μ-2l3A1πρ1-2l3A2πρ1)]-9μ2Dx32(-1+x1)2[2l4+5α2(3l2-7x32)]]-9μ2Dx32(-1+x1)2[2l4+5α2(3l2-7x32)]]a3=3D34l18μx32[2l4+5α2(5l2-7μx32)][n2+3μα22l5-2A1πρ1+2A2πρ1]×[3α2{l4μ-2l2(15+l2)μx32+35μx34}+2l6(l3n2+μ-2l3A1πρ1-2l3A2πρ1)],
where l2=(1-x1)2+x32.These computations have been done using the software package Mathematica.For the stability analysis of the out-of-plane equilibrium point, we compute numerically the partial derivatives calculated at the out-of-plane points with the use of (3.28) and the following numerical values:
(4.28)μ=0.01,π=3.14,α1=0.024,α2=0.02A1=0.7,A2=0.68,ρ1=0.236,D=0.2133.Now, substituting the above values in the characteristic equation (4.27), we get
(4.29)λ6-0.192437λ4+0.248034λ2-0.0000197624=0.
Its roots are:
(4.30)λ1,2=-0.545066±0.448239i,λ3,4=±0.00892642,λ5,6=0.545066±0.448239i.
The positive root and the positive real part of the complex roots induce instability at the out-of-plane point. Hence, the motion of the infinitesimal mass around the out-of-plane equilibrium points is unstable for the specific numerical example given here. However, fuller discussion of their stability remains a theme for future research.


## 5. Discussion

The equation of motion (2.3) is different from those of Hallan and Mangang [4] due to oblateness of the second primary. If we assume that the second primary is not oblate (i.e., α2=0), then these equations will fully coincide with those of Hallan and Mangang [4].Equation (3.9) gives the equilibrium position of the point (P1,0,0) near the center of the first primary and fully coincides with that of Hallan and Mangang [4]. It shows that the position of this equilibrium point does not depend on oblateness of the second primary, while the other equilibrium point (x11+P2,0,0) given by (3.11) is different from that of Hallan and Mangang [4] due to the appearance of oblateness of the second primary. When 2πρ1A1=n2(1-μ), points on the circle (1-x1)2+x22=r2, x3=0 lying within the first primary are also equilibrium points. These points are affected by oblateness of both primaries. Equations (3.28) give the positions of the out-of-plane points when only linear terms in oblateness of the second primary are retained. We have been able to show that the oblateness of the primaries allows the existence of the out-of-plane equilibrium points in the x1x3-plane within the first primary. These points have no analogy in the previous studies of the Robe’s restricted three-body problem.The linear stability analysis of the equilibrium solutions of the problem is investigated with the help of characteristic roots. The characteristic equation (4.9) in the case of the equilibrium point xL=p1 near the center of the first primary is the same as that of Hallan and Mangang [4], while that of the other point xL=x11+p2 near the center differs from that of Hallan and Mangang [4] due to oblateness of the second primary. The characteristic equation of the circular case (4.17) also differs from that of Hallan and Mangang [4] due to oblateness of the second primary. The stability in the first approximation of this configuration shows that points near the centre of the first primary are conditionally stable; the circular points are unstable. This confirms the earlier results of Hallan and Rana [3], Hallan and Mangang [4]. A numerical exploration shows that the out-of-plane equilibrium points are also unstable. This outcome validates the earlier results of Douskos and Markellos [8] and Singh and Leke [11] that the points are unstable.


## 6. Conclusion

We have derived the equations of motion and established the positions of the equilibrium points of the infinitesimal body in the Robe’s [1] restricted three-body problem with oblateness. The term “oblateness” is used in the sense that both primaries are considered as oblate spheroids under the effects of the full buoyancy force exerted by the fluid on the infinitesimal mass.We have obtained one equilibrium point(P1,0,0) near the centre of the first primary which will be on the left or right of the centre of the first primary accordingly as 2πρ1A1-2μ><1. This point is the same as that of Hallan and Mangang [4]. In addition to this, another equilibrium point (x11+P2,0,0) is found within the first primary on the line joining the center of the primaries when 1-2πρ1A1<-3μ/4 and |x11|<a1. When 2πρ1A1=n2(1-μ), points on the circle (1-x1)2+x22=r2, x3=0 lying within the first primary are also equilibrium points. We call them circular points. Finally, we have been able to show that the oblateness of the primaries allows the existence of the out-of-plane equilibrium points in the x1x3-plane within the first primary.The result of this paper can be summarized as follows. The restricted three-body problem under the framework of the Robe’s [1] problem with oblate primaries has the equilibrium points of the type: points near the center of the first primary, points on the circle (circular points), and two out-of-plane points L6,7. It is seen that points near the first primary are conditionally stable, the circular points are unstable, while the out-of-plane equilibrium points are unstable for the specific numerical example given here. The effect of drag forces as considered by Giordano et al. [2] under the present context, particularly as regards the analysis of the properties of the equilibrium points located inside the first primary, will be interesting.



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*Source: 679063-2012-08-16.xml*