The motion of the infinitesimal mass in the restricted three–body problem is considered in the vicinity of the triangular point L₄, when the more massive primary is considered as a source of radiation. General coordinates are taken as polar coordinates ($r,\theta$) centered at the triangular point L₄. A time–independent nonlinear second–order ordinary differential equation for r as a function of θ is derived. Approximations to periodic solutions of finite size are obtained following the geometrical dynamics approach of Rand and Podgorski.¹

The simplicity and elusiveness of the three–body problem have attracted a number of mathematicians for centuries. There are names of many great mathematicians (Euler, Lagrange, Jacobi, Hill, Hamilton, Poincaré, Birkhoff etc.), who have worked on this problem and made important contributions. The book of Szebehely² provides systematic coverage of the literature on the subject as well as derivations of some of the important results. Even today the problem of three–body is as enigmatic as ever. If two of the finite bodies move in circular coplanar orbits about their common center of mass and the third body is too small to affect the motion of the two bodies, then the problem is called circular restricted three body problem (RTBP). In the circular problem, two finite masses are fixed in a co–ordinate system rotating with the orbital angular velocity and origin is at the center of mass of the two bodies. It resembles an important dynamical system for the study of new investigations regarding motions not only in the solar system but also in other planetary systems. Motion of small space objects (asteroid, comet, ring, spacecraft, satellite etc.) in the solar system as well as Sun–planet systems (Sun–Earth system, Sun–Jupiter system etc.) are the best examples of RTBP. In 1772, the famous mathematician Lagrange discovered that in a rotating frame, there are five stationary or equilibrium points at which the restricted mass would remain fixed if placed there. Three of them lie on the line connecting the two finite masses, called collinear equilibrium points and remaining two are located at equidistant from the two finite masses, called triangular equilibrium points. That is, the two masses and the triangular points are thus located at the vertices of the equilateral triangle in the plane of the circular orbits. The problem becomes more interesting when it also includes the other type of space structures such as belt, disk, ring etc., which are present in the solar system.³ Different aspects of this problem such as conditions for existence of equilibrium points, stability property (linear and nonlinear), periodicity of the orbits etc., with perturbation factors in the form of radiation pressure, oblateness etc. have been studied by many authors, Some of the studies.^4–9 Periodic orbits of finite size around the Lagrangian point L₄ had been the subject of investigations.^10,11 Geometrical dynamics is the study of the geometry of the orbits in configuration space of a dynamical system without reference to the system’s motion in time. It is an alternative approach to study the motion around the Lagrangian points. Rand & Podogorski¹ were the first to introduce this approach to planar RTBP it terms of the polar coordinates ($r,\theta$ ) centered at L₄ to study the motion around it in the RTBP. Sharma and Subba Rao¹² employed their method to study the motion around L₄ in the planar RTBP when the more massive primary is an oblate spheroid with its equatorial plane coincident with the plane of motion. In this paper we have utilized the same approach in the planar RTBP when the more massive primary is a source of radiation. We have used the polar coordinates ($r,\theta$ ) centered at the triangular liberation point L₄. A time–independent nonlinear second–order ordinary differential equation for r as a function of is derived. Approximations to periodic solutions are obtained by perturbations and Fourier series. These solutions represent periodic orbits around L₄.

The equations of motion for the circular photogravitational planar RTBP in the dimensionless barycentric synodic coordinates x, y are:^2,13

$\ddot{x}-\text{2}\dot{y}-x=-V_x$   (1)

$\ddot{y}+2\dot{x}-y=-V_y$   (2)

where

$\text{V}=-\text{q}\left(1-\text{µ}\right)/{\text{r}}_1-\text{µ}/{\text{r}}_2$

$r_1{}^2={\left(x-\mu \right)}^2+y^2; r_2{}^2={\left(x+1-\mu \right)}^2+y^2.$   (3)

q is the mass reduction factor for a given particle. In order to obtain the equations of motion in terms of polar coordinates ($r,\theta$);

$x=a+rcos\left(\theta +\alpha \right),$

$y=b+rsin\left(\theta +\alpha \right),$   (4)

$\alpha$ is the angle which the major axis of the ellipse makes with x–axis. (a,b) are the coordinates of L₄. Differentiating, we get

$\dot{x}=\dot{r}\text{cos}\left(\theta +\alpha \right)-r\dot{\theta }sin\left(\theta +\alpha \right),$

$\dot{y}=\dot{r}sin\left(\theta +\alpha \right)+r\dot{\theta }cos\left(\theta +\alpha \right).$

Differentiating again, we get

$\ddot{x}=\ddot{r}cos\left(\theta +\alpha \right)-2\dot{\theta }\dot{r}sin\left(\theta +\alpha \right)-r{\dot{\theta }}^{2}cos\left(\theta +\alpha \right)-r\ddot{\theta }sin\left(\theta +\alpha \right),$

$\ddot{y}=\ddot{r}sin\left(\theta +\alpha \right)+2\dot{\theta }\dot{r}sin\left(\theta +\alpha \right)-r{\dot{\theta }}^{2}sin\left(\theta +\alpha \right)+r\ddot{\theta }cos\left(\theta +\alpha \right).$

Substituting these expressions in the equations of motion (1 and 2) and simplifying, we get

$\text{r-r}{\dot{\theta }}^2\text{-2r}\dot{\theta }{\text{-V}}_{\text{x}}{\text{cos(θ+α)-V}}_{\text{y}}\text{sin}\left(\text{θ+α}\right)\text{+acos}\left(\text{θ+α}\right)\text{+bsin}\left(\text{θ+α}\right)\text{+r,}$  (5)

$r^2\ddot{\theta }+2r\dot{r}\left(\text{θ+1}\right){\text{=V}}_{\text{x}}{\text{sin(θ+α)-V}}_{\text{y}}\text{cos(θ+α)-asin}\left(\text{θ+α}\right)\text{+bcos(θ+α),}$  (6)

where

${\text{V}}_{\text{x}}\text{=q}\left(\text{1-μ}\right)\left(\text{x-μ}\right)/{\text{r}}_{\text{1}}{}^{\text{3}}\text{+μ}\left(\text{x+1-μ}\right)/{\text{r}}_{\text{2}}{}^{\text{3}},$

${\text{V}}_{\text{y}}\text{=y}\left[\text{(1-μ)}/{\text{r}}_{\text{1}}{}^{\text{3}}\text{+μ}/{\text{r}}_{\text{2}}{}^{\text{3}}\right],$

${\text{V}}_{\text{r}}\text{=q}\left(\text{1-μ}\right)/{\text{r}}_{\text{1}}{}^{\text{3}}\left[\left(\text{x-μ}\right)\text{cos}\left(\text{θ+α}\right)\text{+ysin}\left(\text{θ+α}\right)\right]\text{+μ}/{\text{r}}_{\text{2}}{}^{\text{3}}\left[\left(\text{x+1-μ}\right)\text{cos}\left(\text{θ+α}\right)\text{+ysin(θ+α)}\right],$

${\text{V}}_{\text{r}}\text{=q}\left(\text{1-μ}\right)/{\text{r}}_{\text{1}}{}^{\text{3}}\left[\left(\text{x-μ}\right)\text{cos}\left(\text{θ+α}\right)\text{+ysin}\left(\text{θ+α}\right)\right]\text{+μ}/{\text{r}}_{\text{2}}{}^{\text{3}}\left[\left(\text{x+1-μ}\right)\text{cos}\left(\text{θ+α}\right)\text{+ysin(θ+α)}\right],$

${\text{V}}_{\text{θ}}\text{=q}\left(\text{1-μ}\right)/{\text{r}}_{\text{1}}{}^{\text{3}}\left[\text{-r}\left(\text{x-μ}\right)\text{sin}\left(\text{θ+α}\right)\text{+yrcos}\left(\text{θ+α}\right)\right]\text{+ μ}/{\text{r}}_{\text{2}}{}^{\text{3}}\left[\text{-r}\left(\text{x+1-μ}\right)\text{sin}\left(\text{θ+α}\right)\text{+yrcos(θ+α)}\right],$$V_{x }\text{cos}(\theta +\alpha )+V_y\sin \left(\theta +\alpha \right)=V_r,$

$V_x\sin \left(\theta +\alpha \right)-V_y\text{cos}(\theta +\alpha )=-V_{\theta }/r.$

The equations of motion then become

$\ddot{r}-r{\dot{\theta }}^2-2r\dot{\theta }=-V_r+a\cos \left(\theta +\alpha \right)+b\sin \left(\theta +\alpha \right)+r=-U_r,$  (7)

$r^2\ddot{\theta }+2r\dot{r}\left(\theta +1\right)=-V_{\theta }+\left[\text{b}\cos \left(\theta +\alpha \right)-a\sin \left(\theta +\alpha \right)\right]r=-U_{\theta },$  (8)

where

$U=V-r^2/2-r\left[\text{a}\cos \left(\theta +\alpha \right)+b \sin \left(\theta +\alpha \right)\right].$  (9)

The Jacobian Integral is

${\dot{r}}^2/2+{\dot{r}}^2{\dot{\theta }}^2/2+U=h=constant$   (10)

Taking r as a function of $\theta (r^{\prime })$

$\dot{r}=r^{\prime }\dot{\theta }\text{,}$

$\ddot{r}=r^{″}{\dot{\theta }}^2+r^{\prime }\ddot{\theta }.$ Also

$\ddot{r}=r{\dot{\theta }}^2+2r\dot{\theta }-U_r,$

$\ddot{\theta }=-U_{\theta }/r^2-2\dot{r}(\theta +1)/r$

${\dot{\theta }}^2=2\left(h-U\right) /\left(r^2+{r^{\prime }}^2\right)$

$r{\dot{\theta }}^2+2r\dot{\theta }-U_r=r^{″}{\dot{\theta }}^2+r^{\prime }\left[-U_{\theta }/r^2-2\dot{r}(\theta +1)/r\right]$

Substituting to above equation, we get a time–independent second–order ordinary differential equation

$\text{2 }\left(\text{h-U}\right)(r^2-r{r}^{″}+2{r^{\prime }}^2)-2{[2(h-u)]}^{\frac{1}{2}}{[r^2+{r^{\prime }}^2]}^{\frac{3}{2}}+(r^2+{r^{\prime }}^2)\left((r^{\prime }/r)U_e-r{U}_r\right)=0.$

Here it has been assumed that $\dot{\theta }<0$ for periodic orbits around L₄.²

Location of the triangular liberation point L₄,

$a=-\frac{1}{2}+\frac{ϵ}{3}+\mu$(13)

$b=\sqrt{3}\left(\frac{1}{2}-\frac{ϵ}{9}\right)$(14)

$tan2\alpha =\sqrt{3}\left(1-\frac{8ϵ}{9}-2\mu \right)$ (15)

$q=1- \epsilon ,$  (16)

where $\epsilon$ is a small quantity. Transforming of potential equation and substituting it in the equation below, we get

$U=V-r^2/2-r\left[\text{a}\cos \left(\theta +\alpha \right)+b \text{sin}\left(\theta +\alpha \right)\right]$     (17)

$U=U_0+r^{2}g\left(\theta \right)+r^{3}f\left(\theta \right)+0\left(r^4\right)$ (18)

where

$U_0=-1+ϵ$

$g\left(\theta \right)=\frac{3}{4}(-1+ϵ+\left(-\frac{5ϵ}{3\lambda }+\lambda \right)\text{Cos}\left[2\theta \right])$

$f\left(\theta \right)=\left(-\frac{3}{16}+\frac{15ϵ}{16}+\frac{3\mu }{8}\right)\text{Cos}\left[\alpha +\theta \right]+\left(\frac{5}{8}-\frac{5ϵ}{4}-\frac{5\mu }{4}\right)\text{Cos}\left[3\alpha +3\theta \right]+\left(\frac{3\sqrt{3}}{16}-\frac{41ϵ}{16\sqrt{3}}\right)\text{Sin}\left[\alpha +\theta \right]-\frac{5ϵ\text{Sin}\left[3\alpha +3\theta \right]}{8\sqrt{3}}$

Solution of the differential equation will be of the form

$r^2=\frac{M}{N+\text{cos}2\text{θ}}.$

Substituting this equation into (12) and

at $\theta =0°$

$\begin{array}{l}-2\sqrt{2}{\left(\frac{M}{1+N}\right)}^{3/2}\sqrt{1+h-\in -\frac{3M\left(-1+\in -\frac{5ϵ}{3\lambda }+\lambda \right)}{4\left(1+N\right)}}+\frac{2M\left(-1+N\right)\left(1+h-ϵ-\frac{3M\left(-1+\in -\frac{5\in }{3\lambda }+\lambda \right)}{4\left(1+N\right)}\right)}{{\left(1+N\right)}^2}-\\ \frac{M^2\left(3\left(-1+\lambda \right)\lambda +\in \left(-5+3\lambda \right)\right)}{2{\left(1+N\right)}^2\lambda }=0\end{array}$  (19)

$\theta =90°$

$\begin{array}{l}-2\sqrt{2}{\left(\frac{M}{-1+N}\right)}^{3/2}\sqrt{1+h-\in -\frac{3M\left(-1+\in +\frac{5\in }{3\lambda }-\lambda \right)}{4\left(-1+N\right)}}+\frac{2M\left(1+N\right)\left(1+h-\in -\frac{3M\left(-1+\in +\frac{5\in }{3\lambda }-\lambda \right)}{4\left(-1+N\right)}\right)}{{\left(-1+N\right)}^2}+\\ \frac{M^2\left(3\lambda \left(1+\lambda \right)-\in \left(5+3\lambda \right)\right)}{2{\left(-1+N\right)}^2\lambda }=0\end{array}$ (20)

After equating both the equations and simplifying, we get

$M=-\frac{\left(-1+h+ϵ\right)\left(4\left(1+3N^2\right)+6N\left(-1+N^2\right)\left(-\frac{5ϵ}{3\lambda }+\lambda \right)\right)}{-6N\left(-1+ϵ\right)+\frac{5ϵ}{\lambda }-3\lambda +N^2\left(-\frac{10ϵ}{\lambda }+6\lambda \right)}$

Where

$N=-\frac{4+\sqrt{-8+9{\lambda }^2} \text{Sign}\left[-1+h+ϵ\right]}{3\lambda }$

Values of $h> U_0$ correspond to the short–period orbits and $h< U_0$ corresponds to the long–period orbits.
For M and N to be real, it is necessary that

$-8+9{\lambda }^2>0,$

which is equivalent to the usual stability criterion,$\mu \left(1-\mu \right)<1/27.$

To obtain the periodic solution when r3 term is considered for inclusion in U, we adopt the perturbation scheme:

$r=X_1ϵ+X_2{ϵ}^2+\dots$

where

$X_1=M/\left(N+\text{cos}2\text{θ}\right),$

$ϵ={\left|h+U_0\right|}^{1/2}.$

Substituting it in (12) and (18) and equating to zero the coefficients of like powers of ϵ, we obtain a set of linearized differential equations $X_n(\theta )$. After some algebra, the equation $X_2(\theta )$becomes

$F_1{X}_2{}^{’’}+F_2{X}_2{}^{’}+F_3{X}_2+F_4=0,$(21)

where $F_1=-4k_1^2{k}_2$

$F_2=\frac{2k_1\left(-3k+k_2\left(4k_1+2k_3+\frac{G^{\prime }{k}_2{X}_1}{X_1{}^{\text{'}}}\right)\right)X_1{}^{\text{'}}}{X_1}$

$F_3=2Gk{k}_2+2k_1\left(-3k+k_2\left(2k_3+k_2\left(-4G+\frac{G^{\prime }{X}_1{}^{\text{'}}}{X_1}\right)+2k_1\left(2-\frac{X_1{}^{\text{'}\text{'}}}{X_1}\right)-4G\left(k_2+X_1^{\text{'}2}-X_1{X}_1{}^{\text{'}\text{'}}\right)\right)\right)$

$F_4=2F{k}_1{k}_2^2{X}_1{X}_1{}^{\text{'}}-F{X}_1^2{k}_2(k_2\left(6-k_3-6G{X}_1^2\right)+2k_1\left(k_2+X_1^{\text{'}2}-X_1{X}_1{}^{\text{'}\text{'}}\right))$

$k_1=1-G{X}_1^2$

$k_2=X_1^2+X_1^{\text{'}2}$

$k_3=-2G{X}_1^2+G^{\prime }{X}_1{X}_1{}^{\text{'}}$

$k=k_2{k}_3+2k_1(k_2+X_1^{\text{'}2}-X_1{X}_1{}^{\text{'}\text{'}})$

We know that

$F={\beta }_1\text{Cos}\theta +{\beta }_2\text{Sin}\theta +{\beta }_3\text{Cos}3\theta +{\beta }_4\text{Sin}3\theta ,$

Where

${\beta }_1={\beta }_1{}^{\left(1\right)}\text{Cos}\alpha +{\beta }_1{}^{\left(2\right)}\text{Sin}\alpha$

${\beta }_2=-{\beta }_1{}^{\left(1\right)}\text{Sin}\alpha +{\beta }_1{}^{\left(2\right)}\text{Cos}\alpha$

${\beta }_3={\beta }_3{}^{\left(1\right)}\text{Cos}3\alpha +{\beta }_3{}^{\left(2\right)}\text{Sin}3\alpha$

${\beta }_4=-{\beta }_3{}^{\left(1\right)}\text{Sin}3\alpha +{\beta }_3{}^{\left(2\right)}\text{Cos}3\alpha$

with

${\beta }_1{}^{\left(1\right)}=(-\frac{3}{16}+\frac{15\in }{16}+\frac{3\mu }{8})$

${\beta }_1{}^{\left(2\right)}=(\frac{3\sqrt{3}}{16}-\frac{41\in }{16\sqrt{3}})$

${\beta }_3{}^{\left(1\right)}=(\frac{5}{8}-\frac{5\in }{4}-\frac{5\mu }{4})$

${\beta }_3{}^{\left(2\right)}=\frac{-5\in }{8\sqrt{3}}$

After some algebra, we obtain

$F_1(\theta )={\displaystyle \sum }_{j=0}^7{\alpha }_{2j}Cos[(2j)\theta ]$

${\text{F}}_2\text{(}\theta )={\displaystyle \sum }_{j=1}^7{z}_{2j}\text{Sin [(2j)}\theta \text{]}$

${\text{F}}_3\text{(}\theta )={\displaystyle \sum }_{j=0}^7{\tau }_{2j}\text{Cos [(2j) }\theta \text{]}$

${\text{F}}_4\text{(}\theta )={\displaystyle \sum }_{j=0}^4{o}_{2j+1}\text{Cos [(2j+1) }\theta \text{]+}{\displaystyle \sum }_{j=0}^4{\varphi }_{2j+1}\text{Sin [(2j+1) }\theta ]$

where $\alpha \text{'s}$ are known constants, which depend on$\lambda$ and N only. The coefficients of $\alpha \text{'s}$ occurring in the expressions for $F_i\left(i=1,2,3,4\right)$ are provided hereunder. Taking

$\sigma =3m+4n, \rho =4-3m\lambda , \gamma =-n+\lambda , \delta =-1+n\lambda ,$

the coefficients of $F_1$ are

${\alpha }_0=2a_0{b}_0+a_2{b}_2+a_4{b}_4+a_6{b}_6$ 

${\alpha }_2=a_6+2a_2{b}_0+2a_0{b}_2+a_4{b}_2+a_2{b}_4+a_6{b}_4+a_4{b}_6$ 

${\alpha }_4=a_4+2a_4{b}_0+a_2{b}_2+a_6{b}_2+2a_0{b}_4+a_2{b}_6$ 

${\alpha }_6=a_2+2a_6{b}_0+a_4{b}_2+a_2{b}_4+2a_0{b}_6$ 

${\alpha }_8=2a_0+a_6{b}_2+a_4{b}_4+a_2{b}_6$ 

${\alpha }_{10}=a_2+a_6{b}_4+a_4{b}_6$ 

${\alpha }_{12}=a_4+a_6{b}_6$ 

${\alpha }_{14}=a_6$  

where

$a_0=4n\rho \sigma +\left(1+n^2\right)({\rho }^2+2{\sigma }^2)$ 

$a_2=n{\rho }^2+4\left(1+n^2\right)\rho \sigma +2n({\rho }^2+2{\sigma }^2)$ 

$a_2=n{\rho }^2+4\left(1+n^2\right)\rho \sigma +2n({\rho }^2+2{\sigma }^2)$ 

$a_6=n{\rho }^2$ 

$b_0=3+24n^2+8n^4$ 

$b_2=8n(3+4n^2)$ 

$b_4=4(1+6n^2)$ 

$b_6=8n.$ 

If

${\zeta }_0=4(n^2\left(-4-15m\lambda +12mn+8n^2\right)+6mn-\left(4+3m\lambda \right))$ 

${\zeta }_2=16n(-3m{n}^2\lambda -3m\lambda +6mn+4n^2-4)$ 

${\zeta }_4=n(-36mn\lambda +24m+16n-9{\lambda }^{2}m)$ 

and

${\eta }_0=-48m\left(\left(1+n^2\right)\gamma +n\delta \right)-24m\left(1+4n^2+n^4\right)\lambda +16\left(2n\left(n\rho +\sigma \right)+\left(1+n^2\right)\left(\rho +2n\sigma \right)\right)-3{\zeta }_0$ 

${\eta }_4=-48mn\delta -48m{n}^2\lambda +16\left(\left(1+n^2\right)\rho +2n\left(n\rho +\sigma \right)\right)-3{\zeta }_4$ 

${\eta }_6=16n\rho .$ 

The coefficients of $F_2$ are

${\text{z}}_2=\left(1+4n^2\right){\Omega }_2+4n{\Omega }_4+{\Omega }_6$ 

$z_4=4n{\Omega }_2+2\left(1+2n^2\right){\Omega }_4+4n{\Omega }_6+{\Omega }_8$ 

$z_6={\Omega }_2+4n{\Omega }_4+2\left(1+2n^2\right){\Omega }_6+4n{\Omega }_8+{\Omega }_{10}$ 

$z_8={\Omega }_4+4n{\Omega }_6+2\left(1+2n^2\right){\Omega }_8+4n{\Omega }_{10}$ 

$z_{10}={\Omega }_6+4n{\Omega }_8+2\left(1+2n^2\right){\Omega }_{10}$ 

$z_{12}={\Omega }_8+4n{\Omega }_{10}$ 

$z_{14}={\Omega }_{10}$ 

where

${\Omega }_2=4\sigma {\eta }_0+\rho {\eta }_2-2\sigma {\eta }_4-\rho {\eta }_6$ 

${\Omega }_4=2\rho {\eta }_0+2\sigma {\eta }_2-2\sigma {\eta }_6$ 

${\Omega }_6=\rho {\eta }_2+2\sigma {\eta }_4$ 

${\Omega }_8=\rho {\eta }_4+2\sigma {\eta }_{6}a$ 

${\Omega }_{10}=\rho {\eta }_6.$ 

The coefficients of $F_3$ are given by

${\tau }_0=2n{\omega }_0+{\omega }_2$ 

${\tau }_2=2{\omega }_0+2n{\omega }_2+{\omega }_4$ 

${\tau }_4={\omega }_2+2n{\omega }_4+{\omega }_6$ 

${\tau }_6={\omega }_4+2n{\omega }_6+{\omega }_8$ 

${\tau }_8={\omega }_6+2n{\omega }_8+{\omega }_{10}$ 

${\tau }_{10}={\omega }_8+2n{\omega }_{10}+{\omega }_{12}$ 

${\tau }_{12}={\omega }_{10}+2n{\omega }_{12}$ 

${\tau }_{14}={\omega }_{12}$ 

where

${\omega }_0=2B_0{P}_0+B_2{P}_2+B_4{P}_4-2{\pi }_0{\zeta }_0-{\pi }_2{\zeta }_2-{\pi }_4{\zeta }_4,$ 

${\omega }_2=2B_2{P}_0+\left(2B_0+B_4\right)P_2+B_2{P}_4+B_4{P}_6-2{\pi }_0{\zeta }_2-{\pi }_4{\zeta }_2-{\pi }_6{\zeta }_4-{\pi }_2\left(2{\zeta }_0+{\zeta }_4\right),$ 

${\omega }_4=2B_0{P}_4+B_2\left(P_2+P_6\right)+B_4\left(2P_0+P_8\right)-2{\pi }_4{\zeta }_0-\left({\pi }_2+{\pi }_6\right){\zeta }_2-2{\pi }_0{\zeta }_4$ 

${\omega }_6=B_4{P}_2+2B_0{P}_6+B_2\left(P_4+P_8\right)-2{\pi }_6{\zeta }_0-{\pi }_4{\zeta }_2-{\pi }_2{\zeta }_4$ 

${\omega }_8=B_4{P}_4+B_2{P}_6+2B_0{P}_8-{\pi }_6{\zeta }_2-{\pi }_4{\zeta }_4$ 

${\omega }_{10}=B_4{P}_6+B_2{P}_8-{\pi }_6{\zeta }_4,$ 

${\omega }_{12}=B_4{P}_8$ 

with

$B_0=n\rho +(1+n^2)\sigma ,$ 

$B_2=\left(1+n^2\right)\rho +2n\sigma ,$ 

$B_4=n\rho .$ 

and

$P_0=-24m\left(-1+n^2\right)\left(-n+\gamma \right)-12m\left(2\left(1+n^2\right)\gamma +4n\delta \right)-4m(12n\delta +$ 

$\left(1+n^2\right)\left(-12n+3\lambda \right))+3\rho +8n\left(n\rho +\sigma \right)+2\left(-3+4n^2\right)(\rho +2n\sigma ),$ 

$P_2=-48m\left(-1+n^2\right)\delta -12m\left(8n\gamma +3\delta +4n^2\delta \right)-4m(6n\left(-3n+\gamma \right)+$ 

$12\left(1+n^2\right)\delta +9n\lambda )+10n\rho +6\sigma +4\left(-3+4n^2\right)\left(n\rho +\sigma \right)+8n\left(\rho +2n\sigma \right),a$ 

$\begin{array}{l}{P}_4=-24m\left(\gamma +2n\delta \right)-24m\left(-1+n^2\right)\lambda -4m\left(-12n+9\lambda +21n^2\lambda \right)\\ +2\left(-3+4n^2\right)\rho +8n\left(n\rho +\sigma \right)+6\left(\rho +2n\sigma \right),\end{array}$ 

$P_6=-12m\delta -36mn\lambda +10n\rho +6\sigma ,$ 

$P_8=3\rho .$  

${\pi }_0=6(2\left(m\left(1+n^2\right)-mn\lambda +n\rho \right)+\left(1+2n^2\right)\left(\rho +\sigma \right)),$ 

${\pi }_2=3\left(-4m\left(1+n^2\right)\lambda +\rho +8n\left(m+\sigma \right)\right),$ 

${\pi }_4=-12mn\lambda +12n\rho +6\sigma ,$ 

${\pi }_6=3\rho .$ 

Introducing

$q_2=4n\rho +8\left(1+n^2\right)\sigma ,$ 

$q_4=4\left(1+n^2\right)\rho +8n\sigma ,$ 

$q_6=4n\rho .$  

$s_1=-4\left(\left(1+5n^2\right)\rho +6n\sigma \right)-8n\left(3n\rho +\left(1+5n^2\right)\sigma \right)+{\zeta }_0,$ 

$s_2=-12n\rho -8n\left(\left(1+5n^2\right)\rho +6n\sigma \right)-8\left(3n\rho +\left(1+5n^2\right)\sigma \right)+{\zeta }_2,$ 

$s_4=-24n^2\rho -4\left(\left(1+5n^2\right)\rho +6n\sigma \right)+{\zeta }_4,$ 

$s_6=-12n\rho .$ 

The coefficients of $F_4$are

$o_1=\left(1+n^2\right)m_1+n(m_{-1}+m_3)$ 

$o_3=\left(1+n^2\right)m_3+n(m_1+o_3=\left(1+n^2\right)m_3+n(m_1+m_5)$ 

$o_5=\left(1+n^2\right)m_5+n(m_3+m_7)$ 

$o_7=\left(1+n^2\right)m_7+n(m_5+m_9)$ 

$o_9=\left(1+n^2\right)m_9+n(m_7+m_{11})$ 

${\varphi }_1=\left(1+n^2\right)n_1+n(n_{-1}+n_3)$ 

${\varphi }_3=\left(1+n^2\right)n_3+n(n_1+n_5)$ 

${\varphi }_5=\left(1+n^2\right)n_5+n(n_3+n_7)$ 

${\varphi }_7=\left(1+n^2\right)n_7+n(n_5+n_9)$ 

${\varphi }_9=\left(1+n^2\right)n_9+n(n_7+n_{11})$ 

where

$m_1=m_{-1}=\left(-q_2+2s_1+s_2\right){\beta }_1+\left(-3q_2-3q_4+s_2+s_4\right){\beta }_3$ 

$n_1=\left(q_2+2s_1-s_2\right){\beta }_2+(-3q_2+3q_4+s_2-s_4){\beta }_4$ 

$n_{-1}=-\left(q_2+2s_1-s_2\right){\beta }_2-(-3q_2+3q_4+s_2-s_4){\beta }_4$ 

$m_3=\left(q_2-q_4+s_2+s_4\right){\beta }_1+\left(-3q_6+2s_1+s_6\right){\beta }_{3}a\frac{}{}$ 

$n_3=\left(q_2+q_4+s_2-s_4\right){\beta }_2+\left(3q_6+2s_1-s_6\right){\beta }_4$ 

$m_5=\left(q_4-q_6+s_4+s_6\right){\beta }_1+\left(3q_2+s_2\right){\beta }_{3}a$ 

$n_5=\left(q_4+q_6+s_4-s_6\right){\beta }_2+\left(3q_2+s_2\right){\beta }_4$ 

$m_7=\left(q_6+s_6\right){\beta }_1+\left(3q_4+s_4\right){\beta }_3$ 

$n_7=\left(q_6+s_6\right){\beta }_2+\left(3q_4+s_4\right){\beta }_4$ 

$m_9=m_{11}=n_9=n_{11}=0$ 

For a periodic solution to (21), it is sufficient to set.¹²

$x_2(\theta )={\displaystyle \sum }_{\begin{array}{c}n=1\\ n odd\end{array}}^{ \infty }(a_n\text{Cos n}\theta +b_n\text{Sin n}\theta )$ 

Substituting this Fourier series in (21) and equating to zero coefficients of $\text{cos n }\theta$ and $\text{sin n }\theta$, we obtain a set of linear algebraic equations on $a_n$ and $b_n$, respectively $x_3(\theta )$. Periodic expression for and higher order terms can be obtained in a similar fashion.

There will be only one periodic orbit for given values of $h,ϵ$ and $\mu$. The value $h=-1$corresponds to the equilibrium solution at L₄. Value of $h>-1$ corresponds to short–periodic orbits while $h<-1$ correspond to long–periodic orbits.

Geometrical illustration of the foregoing analysis is provided in Figures 1–5 for some typical values of $\mu ,ϵ$ and $h$. The curves with dotted lines therein correspond to the linear analysis with X₁ term for r while the others correspond to the non–linear analysis with the inclusion of X₂ terms. Fourier series solution for X₂ has been attempted retaining terms up to $29\theta$ . Figure 1 & 2 illustrate the impact of the higher order terms in the analysis for the perturbed and unperturbed problems. However, it may be noted that the relative location of the origin with reference to the primaries in the two cases is not the same. Figure 3 refers to the short–period eigen frequency and is meant to illustrate the effects of radiation pressure. The curves corresponding to the long–period eigen frequency are indistinguishable and hence are omitted. It may thus be highlighted that the effect of higher order terms becomes significant for higher values of $\mu$ and surpasses the considerations of assumed radiation pressure effect.

Figure 1 $\mu =0.0369$,${\xi }^2=.0009$ and$ϵ=0.01$ .

Figure 2 $\mu =0.0369$ , ${\xi }^2=.0005$& $ϵ=0.01$ .

Figure 3 $\mu =0.0369$ ,${\xi }^2=.0005$ .

Figure 4 $\mu =0.0369$ ,${\xi }^2=.0005$ & $0.0009$ &$ϵ=0.01$ .

Figure 5 $\mu =0.012$ & 0.0369,${\xi }^2=.0005$ &$ϵ=0.001$ .

Finite periodic orbits have been generated at the triangular point L₄ in the photogravitational restricted three–body problem by considering the more massive primary as a source of radiation. The geometrical illustration of the periodic orbit shows the effects of $\mu ,ϵ$ and $h$. The effect of higher order terms becomes significant for higher values of $\mu$ .

Author declares there is no conflict of interest.

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