Periodic orbits in the restricted problem of three bodies in a three-dimensional coordinate system when the smaller primary is a triaxial rigid body

In this paper, we wish to study the three-dimensional generalisation of the problem studied by Bhatnagar (12,13,14) for the circular case. Since the Hamilton-Jacobi equation for generating a solution takes an unmanageable form for any solution, we have assumed that the third coordinate (₃ of the infinitesimal mass is of the 0(μ). It will be interesting to observe that various equations and results worked out by Bhatnagar can be deduced from our results. In Section 2 we have determined the canonical form of the equations of motion, and in Section 3 these equations are regularised by the generalised Levi-Civita's transformation for three dimensions. Eqs (20)–(22) establish the canonical set (l, L, g, G, h, H) and Eq (32) form the basis of the general perturbation theory for the problem under consideration. During the last few years, many mathematician and astronomers have studied different types of periodic orbits in the restricted problem. Some of them are Giacaglia (7), Mayer and Schmidt (17), Markellos (19), Hadjidemetriou (10,11), Bhatnagar and Taqvi (15), Gomez and Noguera (8), Kadrnoska and Hadrava (9), Peridios et al. (21), Ahmad (1), Elipe and Lara (4), Mathlouthi (23), Scuflaire (22), Caranicolas (20), Poddar et al. (5, 6), Abouelmagd and Guirao (2) and Abouelmagd et al. (3). In this work, we have presented an analytical study of the existence of periodic orbits for μ = 0 in the restricted problem of three bodies in a three-dimensional coordinate system when the smaller primary is a triaxial rigid body.

The equations of motion in the canonical form of an infinitesimal mass under the gravitational field of two finite and unequal masses and moving in circles are given by (1) X˙i = ∂H∂pi;p˙i = −∂H∂xi(i = 1, 2, 3) {\dot X_i}\; = \;{{\partial H} \over {\partial pi}};{\dot p_i}\; = \; - {{\partial H} \over {\partial xi}}(i\; = \;1,\;2,\;3) where the Hamiltonian function H and consequently the energy integral is given by (2) H = 12(p12 + p22 + p32) + n(p1x2 − p2x1) − (1−μ )r1 − μr2 − μ2r23 + 3μ2r25x22 = C H\; = \;{1 \over 2}({p_1}^2\; + \;{p_2}^2\; + \;{p_3}^2)\; + \;n\left( {{p_1}{x_2}{\;_ - }\;{p_2}{x_1}} \right)\; - \;{{\left( {1 - \mu \;} \right)} \over {{r_1}}}\; - \;{\mu \over {{r_2}}}\; - \;{\mu \over {2r_2^3}}\; + \;{{3\mu } \over {2r_2^5}}x_2^2\; = \;C and C is a function of μ = C(μ) = C₀ + μ(C₁). r12 = (x1 − μ)2 +x22 + x32r22 = (x1−μ+1)2 + x22+x32p1 = x1−x2 p2=x˙2+x1p3= x˙3 \matrix{ {r_1^2\; = \;{{({x_1}\; - \;\mu )}^2}\; + x_2^2\; + \;x_3^2} \cr {r_2^2\; = \;{{\left( {{x_1} - \mu + 1} \right)}^2}\; + \;x_2^2 + x_3^2} \cr {{p_1}\; = \;{x_1} - {x_2}\;} \cr {{p_2} = {{\dot x}_2} + {x_1}} \cr {{p_3} = \;{{\dot x}_3}} \cr }

Mean motion n=1+34(2σ1−σ2) n = 1 + {3 \over 4}\left( {2{\sigma _1} - {\sigma _2}} \right)

where σ1=a2−c25R2 {\sigma _1} = {{{a^2} - {c^2}} \over {5{R^2}}} , σ2=b2−c25R2 {\sigma _2} = {{{b^2} - {c^2}} \over {5{R^2}}} , a, b, c = semi-axes of the triaxial rigid body, R = the dimensional distance between the primaries and (x₁, x₂, x₃) are equal to the synodic rectangular dimensionless coordinates of the infinitesimal mass in a uniformly rotating system.

We regularise the solution by Levi-Civita's (18) transformation generated by (3) S=(μ+ξ12−ξ22)p1+2ξ1ξ2p2+ξ3p3 S = (\mu + \xi _1^2 - \xi _2^2){p_1} + 2{\xi _1}{\xi _2}{p_2} + {\xi _3}{p_3} Such that (4) xi=∂s∂pi;πi=∂s∂i(i = 1, 2, 3) {x_i} = {{\partial s} \over {\partial {p_i}}};{\pi _i} = {{\partial s} \over {{\partial _i}}}(i\; = \;1,\;2,\;3) where π_i is the momenta associated with the new coordinate ξi.

We have from Eqs (3) and (4) π1=∂s∂ξ1=2ξ1p1+2ξ2p2,π2=∂s∂ξ2=−2ξ2p1+2ξ1p2,π3=∂s∂ξ3=p3 {\pi _1} = {{\partial s} \over {\partial {\xi _1}}} = 2{\xi _1}{p_1} + 2{\xi _2}{p_2},{\pi _2} = {{\partial s} \over {\partial {\xi _2}}} = - 2{\xi _2}{p_1} + 2{\xi _1}{p_2},{\pi _3} = {{\partial s} \over {\partial {\xi _3}}} = {p_3} From these equations, we have p1=π1ξ1−π2 ξ22(ξ12+ξ22),p2=π1ξ2−π2 ξ12(ξ12+ξ22) {p_1} = {{{\pi _1}{\xi _1} - {\pi _{2\;}}{\xi _2}} \over {2\left( {\xi _1^2 + \xi _2^2} \right)}},{p_2} = {{{\pi _1}{\xi _2} - {\pi _{2\;}}{\xi _1}} \over {2\left( {\xi _1^2 + \xi _2^2} \right)}} Further p3=π3,x1=μ+ξ12−ξ22 {p_3} = {\pi _3},{x_1} = \mu + \xi _1^2 - \xi _2^2 x2=2ξ1ξ2,x3=ξ3 {x_2} = 2{\xi _1}{\xi _2},{x_3} = {\xi _3} The Hamiltonian Eq. (2) given in terms of these new variables is H=π28ξ2+π322+n(ξ2π1−ξ1π2)2−nμ2ξ2(ξ1π2+ξ2π1)−1−μr1−μr2−μ2r23(2σ1−σ2)+6μr25(σ1−σ2)ξ12ξ22 =C=const. \matrix{ {H = {{{\pi ^2}} \over {8{\xi ^2}}} + {{\pi _3^2} \over 2} + {{n({\xi _2}{\pi _1} - {\xi _1}{\pi _2})} \over 2} - {{n\mu } \over {2{\xi ^2}}}\left( {{\xi _1}{\pi _2} + {\xi _2}{\pi _1}} \right) - {{1 - \mu } \over {{r_1}}} - {\mu \over {{r_2}}} - {\mu \over {2r_2^3}}(2{\sigma _1} - {\sigma _2}) + 6{\mu \over {r_2^5}}({\sigma _1} - {\sigma _2})\xi _1^2\xi _2^2} \hfill \cr {\;\;\; = C = {\rm{const}}{\rm{.}}} \hfill \cr } where r12=ξ4+ξ32 r_1^2 = {\xi ^4} + \xi _3^2 , r22=1−2(ξ12−ξ22)+ξ4+ξ22 r_2^2 = 1 - 2(\xi _1^2 - \xi _2^2) + {\xi ^4} + \xi _2^2 , π2=π12+π22 {\pi ^2} = \pi _1^2 + \pi _2^2 , ξ2=ξ12ξ22 {\xi ^2} = \xi _1^2\xi _2^2 , C = C_(o) +C₁ (μ) and = C_o + μC₁

Now we introduce a new independent variable τ instead of t defined by (5) dt=r1 dτ (t=0atτ=0) dt = {r_1}\;d\tau \;(t = 0{\rm{at}}\tau = 0)

The equations of motion (1) will be transformed into (6) didτ=∂k∂πi,dπidτ=−∂k∂i(i=1,2.3) {{{d_i}} \over {d\tau }} = {{\partial k} \over {\partial {\pi _i}}},{{d{\pi _i}} \over {d\tau }} = - {{\partial k} \over {{\partial _i}}}(i = 1,2.3) Where K is the new Hamiltonian given by K=r1(H−C)=π2r18ξ2+12π32r1+r1n2(ξ2π1−π2ξ1−2cn) −nμr12ξ2(ξ1π2+ξ2π1)−(1−μ)−r1μr2−r1μ2r23(2σ1−σ2)+6r1μr25(σ1−σ2)ξ12ξ22. \matrix{ {K = {r_1}\left( {H - C} \right) = {{{\pi ^2}{r_1}} \over {8{\xi ^2}}} + {1 \over 2}\pi _3^2{r_1} + {{{r_1}n} \over 2}({\xi _2}{\pi _1} - {\pi _2}{\xi _1} - {{2c} \over n})} \hfill \cr {\;\;\;\;\; - {{n\mu {r_1}} \over {2{\xi ^2}}}({\xi _1}{\pi _2} + {\xi _2}{\pi _1}) - (1 - \mu ) - {{{r_{1\mu }}} \over {{r_2}}} - {{{r_1}\mu } \over {2r_2^3}}(2{\sigma _1} - {\sigma _2}) + {{6{r_1}\mu } \over {r_2^5}}({\sigma _1} - {\sigma _2})\xi _1^2\xi _2^2.} \hfill \cr }

K can be put in the form K_o + μK₁, where (7) Ko=π2r18ξ2+12π32r1+r1n2(ξ2π1−π2ξ1−2co′)−1=−<o(say) {K_o} = {{{\pi ^2}{r_1}} \over {8{\xi ^2}}} + {1 \over 2}\pi _3^2{r_1} + {{{r_1}n} \over 2}({\xi _2}{\pi _1} - {\pi _2}{\xi _1} - 2c_o^\prime) - 1 = - < o\left( {{\rm{say}}} \right) where c0′=c0n c_0^\prime = {{{c_0}} \over n} and (8) K1=nr12ξ2(ξ1π2+ξ2π1)−r1r2−r12r23(2σ1−σ2)+6r1μr25(σ1−σ2)ξ12ξ22−(c−co)μr1+1 {K_1} = {{n{r_1}} \over {2{\xi ^2}}}({\xi _1}{\pi _2} + {\xi _2}{\pi _1}) - {{{r_1}} \over {{r_2}}} - {{{r_1}} \over {2r_2^3}}(2{\sigma _1} - {\sigma _2}) + {{6{r_1}\mu } \over {r_2^5}}({\sigma _1} - {\sigma _2})\xi _1^2\xi _2^2 - {{(c - {c_o})} \over \mu }{r_1} + 1

The form given to k₀ ensures that the orbits which are analytically continued from the two-body orbits will belong to the K = 0 manifold. These are the solution to the regularised equation of the restricted problem. Here we have assumed that k₀ is negative (5). Thus, the corresponding two-body problem will admit bounded orbits as a solution in rotating coordinates. We can easily show that ||<1.

To write the Hamilton-Jacobi equation corresponding to the Hamilton k₀, we take πi=∂w∂i(i=1,2,3) {\pi _i} = {{\partial w} \over {{\partial _i}}}(i = 1,2,3)

For generating a solution, we shall choose k₀ for our Hamiltonian function. Since τ is not involved in k explicitly, the Hamilton-Jacobi equation corresponding to k₀ may be written as (9) 18[(∂w∂ξ1)2+(∂w∂ξ2)2]r1ξ2+12(∂w∂ξ3)2r1+nr12{ξ2∂w∂ξ1−ξ1∂w∂ξ2−2co′}=α. {1 \over 8}\left[ {{{\left( {{{\partial w} \over {\partial {\xi _1}}}} \right)}^2} + {{\left( {{{\partial w} \over {\partial {\xi _2}}}} \right)}^2}} \right]{{{r_1}} \over {{\xi ^2}}} + {1 \over 2}{\left( {{{\partial w} \over {\partial {\xi _3}}}} \right)^2}{r_1} + {{n{r_1}} \over 2}\left\{ {{\xi _2}{{\partial w} \over {\partial {\xi _1}}} - {\xi _1}{{{{\partial w} \over {\partial \xi }}}_2} - 2c_o^\prime} \right\} = \alpha . where α = 1 − ɛ.

We take ξ₃ of the order of μ, then we have r1=ξ2 +0(μ) {r_1} = {\xi ^2}\; + 0(\mu ) Putting ξ1= ξ cosϕ, ξ2 =sinϕ {\xi _1} = \;\xi \;{\rm{cos}}\phi ,\;\;\;\;\;{\xi _{2\;\;\;\;}} = {\rm{sin}}\phi Equation (9) may be written as (10) 18[(∂w∂ξ)2+1ξ2(∂w∂ϕ)2]+12ξ2(∂w∂ξ3)2+12nξ2[−∂w∂ϕ−2c0′]=α {1 \over 8}\left[ {{{\left( {{{\partial w} \over {\partial \xi }}} \right)}^2} + {1 \over {{\xi ^2}}}{{\left( {{{\partial w} \over {\partial \phi }}} \right)}^2}} \right] + {1 \over 2}{\xi ^2}{\left( {{{\partial w} \over {\partial {\xi _3}}}} \right)^2} + {1 \over 2}n{\xi ^2}\left[ { - {{\partial w} \over {\partial \phi }} - 2c_0^\prime} \right] = \alpha Whose solution of Eq. (10) may be written as (11) W=u(ξ)+Gϕ+H¯ξ3 W = u\left( \xi \right) + G\phi + \bar H{\xi _3} where G is an arbitrary parameter and taking ξ² = z we have (12) (∂u∂z)2=H¯2−2n(G+c0′)z2f(z) {\left( {{{\partial u} \over {\partial z}}} \right)^2} = {{{{\bar H}^2} - 2n\left( {G + c_0^\prime} \right)} \over {{z^2}}}f\left( z \right) where (13) f(z)=G22n (G+c0′)−H¯2−2αz2n(G+c0′)−H¯2−z2 f(z) = {{{G^2}} \over {2n\;\left( {G + c_0^\prime} \right) - {{\bar H}^2}}} - {{2\alpha z} \over {2n\left( {G + c_0^\prime} \right) - {{\bar H}^2}}} - {z^2} We suppose that G+c0'<0 G + c_0^\prime < 0 then the equation f (z) = 0 has two positive roots z₁ and z₂ and is positive between them. Also z1+z2=−2α2n(G+c0′)−H¯2>0z1z2=−G22n (G+c0′)−H¯2>0 \matrix{ {{z_1} + {z_2} = - {{2\alpha } \over {2n\left( {G + c_0^\prime} \right) - {{\bar H}^2}}} > 0} \cr {{z_1}{z_2} = - {{{G^2}} \over {2n\;\left( {G + c_0^\prime} \right) - {{\bar H}^2}}} > 0} \cr } The solution of Eq. (12) is (14) u(Z,G,α)[H¯2−2n(G+c0,)]1/2∫Z1Z2f(z)Zdz u(Z,G,\alpha ){\left[ {{{\bar H}^2} - 2n\left( {G + c_0^,} \right)} \right]^{{\kern 1pt} 1{\kern 1pt} /{\kern 1pt} 2{\kern 1pt} }}\mathop {\int_{{Z_1}}^{{Z_2}} }{{\sqrt {f\left( z \right)} } \over Z}dz

Let us introduce the parameter a, e, l using the relation (15) Z1=a(1−e),Z2=a(1+e)Z=Z1cos2l2+Z1sin2l2=a(1−ecosl) \matrix{ {{Z_1} = a\left( {1 - e} \right),{Z_2} = a\left( {1 + e} \right)} \cr {Z = {Z_1}\mathop {\cos }\nolimits^2 {l \over 2} + {Z_1}\mathop {\sin }\nolimits^2 {l \over 2} = a(1 - e\cos l)} \cr } where 0≤e≤1. It may be noted that Z = Z₁ when l = 0.

The equations of motion to K₀ are (16) ξi′=∂K0∂πi,(i=1,2,3)ξ1′=∂K0∂π1=π14r1ξ2+12nr1ξ2ξ2′=∂K0∂π2=π14r1ξ2+12nr1ξ2ξ3′=∂K0∂3=π3r1 \matrix{ {\xi _i^\prime = {{\partial {K_0}} \over {\partial {\pi _i}}},(i = 1,2,3)} \cr {\xi _1^\prime = {{\partial {K_0}} \over {\partial {\pi _1}}} = {{{\pi _1}} \over 4}{{{r_1}} \over {{\xi ^2}}} + {1 \over 2}n{r_1}{\xi _2}} \cr {\xi _2^\prime = {{\partial {K_0}} \over {\partial {\pi _2}}} = {{{\pi _1}} \over 4}{{{r_1}} \over {{\xi ^2}}} + {1 \over 2}n{r_1}{\xi _2}} \cr {\xi _3^\prime = {{\partial {K_0}} \over {\partial 3}} = {\pi _3}{r_1}} \cr } Here ′ denotes differentiation with respect to τ

Now 14(ξ1π1+ ξ2π2)=ξξ′ {1 \over 4}\left( {{\xi _1}{\pi _1} + \;{\xi _2}{\pi _2}} \right) = \xi \xi '

Therefore dzdτ=H¯2+2(G+c0′)⋅f(z) {{dz} \over {d\tau }} = \sqrt {{{\bar H}^2} + 2\left( {G + c_0^\prime} \right)} \cdot \sqrt {f\left( z \right)}

Integrating, we have (17) ∫Z1Zdzf(z)=(τ−τ0)[H¯2+2n(G+c0′)]1/2 \mathop {\int_{{Z_1}}^Z } {{dz} \over {\sqrt {f\left( z \right)} }} = (\tau - {\tau _0}){\left[ {{{\bar H}^2} + 2n\left( {G + c_0^\prime} \right)} \right]^{1/2}} where z = z₁ at τ = τ₀.

Introducing L by relation α=L [H¯2+2n(G+c0′)]1/2>0 , L>0 \alpha = L\;{\left[ {{{\bar H}^2} + 2n\left( {G + c_0^\prime} \right)} \right]^{1/2}} > 0\;,\;L > 0 We have a=L[H¯2+2n(G+c0′)]1/2 a = {L \over {{{\left[ {{{\bar H}^2} + 2n\left( {G + c_0^\prime} \right)} \right]}^{1/2}}}} (18) e=[1−G2L2]1/2≤1f(z)= aesinl \matrix{ {e = {{\left[ {1 - {{{G^2}} \over {{L^2}}}} \right]}^{1/2}} \le 1} \cr {\sqrt {f\left( z \right)} = \;ae\sin l} \cr } (19) l=(τ−τ0)[H¯2+2n(G+c0′)]1/2 l = (\tau - {\tau _0}){\left[ {{{\bar H}^2} + 2n\left( {G + c_0^\prime} \right)} \right]^{1/2}} Now taking L and G for the arbitrary constants instead of α and G, the solution may be given by the relation (20) ∂w∂L = ∂u ∂L=l {{\partial w} \over {\partial L\;}}\;\;\; = \;\;{{\partial u\;\;} \over {\partial L}} = l (21) ∂w ∂G =2 + ∂u∂G =2 +nL2+G2[H¯2+2n(G+c0′)]1/2sinl−f=g (say). {{\partial w\;} \over {\partial G}}\; = 2\;\; + \;{{\partial u} \over {\partial G\;}}\; = 2\; + {{n\sqrt {{L^2} + {G^2}} } \over {{{\left[ {{{\bar H}^2} + 2n\left( {G + c_0^\prime} \right)} \right]}^{1/2}}}}{\rm{sin}}l - f = g\;\left( {{\rm{say}}} \right). where f=1−e2∫0ldl1−ecosl f = \sqrt {1 - {e^2}} \mathop {\int_0^l } {{dl} \over {1 - e\cos l}} (22) ∂w∂H = 3+nL2+G2[H¯2+2n(G+c0′)]1/2sinl=h (say) {{\partial w} \over {\partial H}}\; = \;{\;_3} + {{n\sqrt {{L^2} + {G^2}} } \over {{{\left[ {{{\bar H}^2} + 2n\left( {G + c_0^\prime} \right)} \right]}^{1/2}}}}{\rm{sin}}l = h\;\left( {{\rm{say}}} \right) and for e = 1, we have G = 0, f = 0. Eqs (20)–(22) establish the canonical set (l, L, g, G, h, H̅) since k₀ = α − 1. It follows that K0=L[H¯2+2n(G+c0′)]1/2−1>0 {K_0} = L{\left[ {{{\bar H}^2} + 2n\left( {G + c_0^\prime} \right)} \right]^{1/2}} - 1 > 0 and therefore, for the problem generated by this Hamiltonian (regularised two-body problems in Rotating coordinates), we have (23) dLdτ = − ∂k0∂l= 0 , L= constant= L0(say)dGdτ = − ∂k0∂g = 0 , G= constant= G0(say)dH¯dt = − ∂k0∂h =0,H¯=constant = H0(say)dld = ∂k0∂L=[H2−2N(G+c0′)]12=const=nl∴l=nlτ+lodgd = ∂k0∂G=−nL[H¯2+2n(G+c0′)]1/2=const=ng∴g=nlgτ+godhd = ∂k0∂H=LH[H¯2+2n(G+c0′)]1/2=const=nh∴h=nhτ+ho \matrix{{{{dL} \over {d\tau }}\; = \; - \;{{\partial {k_0}} \over {\partial l}} = \;\;0\;,\;\;L = \;\;{\rm{constant}} = \;{L_0}({\rm{say}})} \cr {{{dG} \over {d\tau }}\; = \; - \;{{\partial {k_0}} \over {\partial g}}\; = \;\;0\;\;,\;\;G = \;\;{\rm{constant}} = \;{G_0}({\rm{say}})} \cr {{{d\bar H} \over {dt}}\; = \; - \;{{\partial {k_0}} \over {\partial h}}\; = 0,\bar H = {\rm{constant}}\; = \;{H_0}({\rm{say}})} \cr {{{dl} \over d}\; = \;\;{{\partial {k_0}} \over {\partial L}} = {{\left[ {{H^2} - 2N(G + c_0^\prime)} \right]}^{{1 \over 2}}} = {\rm{const}} = {n_l}\therefore l = {n_l}\tau + {l_o}} \cr {{{dg} \over d}\; = \;\;{{\partial {k_0}} \over {\partial G}} = {{ - nL} \over {{{\left[ {{{\bar H}^2} + 2n\left( {G + c_0^\prime} \right)} \right]}^{1/2}}}} = {\rm{const}} = {n_g}\therefore g = {n_{\lg }}\tau + {g_o}} \cr {{{dh} \over d}\; = \;\;{{\partial {k_0}} \over {\partial H}} = {{LH} \over {{{\left[ {{{\bar H}^2} + 2n\left( {G + c_0^\prime} \right)} \right]}^{1/2}}}} = {\rm{const}} = {n_h}\therefore h = {n_h}\tau + {h_o}} \cr } where l₀, g₀, h₀ are the values of l, g, h respectively at τ = 0.

The angle θ is obtained from the equation (24) ϕ=12g+1n[L2+G2]2[H¯2−2n(G+c0′)]sinl , when e ≠ 1ϕ=12g+1 L N2[H¯2−2nc0′]sinl , when e =1. \matrix{{\phi = {1 \over 2}g + {{1n\left[ {{L^2} + {G^2}} \right]} \over {2\left[ {{{\bar H}^2} - 2n\left( {G + c_0^\prime} \right)} \right]}}\sin l\;,\;\;\;{\rm{when}}\;\;e\; \ne \;\;1} \cr {\phi = {1 \over 2}g + {{1\;L\;N} \over {2\left[ {{{\bar H}^2} - 2nc_0^\prime} \right]}}\sin l\;,\;\;\;\;\;\;\;\;{\rm{when}}\;\;e\; = 1.} \cr } The variables ξ_i, π_i (i = 1, 2, 3) can be expressed by the canonical elements we have (25) ξ1=± z cosϕ=± a(1−ecosl)cosϕξ2= ± zsinϕ = ± a(1−ecosl)sinϕξ3= h−H¯(L2−G2)12[H¯2−2n(G+c0′]12sinlπ1=∂w∂ξcosϕ − ∂w∂ϕsinϕξπ2= ∂w∂ξsin ϕ− ∂w∂ϕcosϕξπ3=H¯∂w∂ = dud= dudz⋅dzd=2ξdudz={H¯2−2n(G+c0′)f(z)}z212=±2eLsinla(1−ecosl) \matrix{ {{\xi _1} = \pm \;\sqrt {z\;\;} \cos \phi = \pm \;\sqrt {a(1 - e\cos l)} \cos \phi } \cr {{\xi _2} = \; \pm \;\sqrt z \sin \phi \;\; = \; \pm \;\sqrt {a(1 - e\cos l)} {\rm{sin}}\phi } \cr {{\xi _3} = \;\;{{h - \bar H{{\left( {{L^2} - {G^2}} \right)}^{{1 \over 2}}}} \over {{{\left[ {{{\bar H}^2} - 2n(G + c_0^\prime} \right]}^{{1 \over 2}}}}}\sin l} \cr {{\pi _1} = {{\partial w} \over {\partial \xi }}\cos \phi \; - \;{{\partial w} \over {\partial \phi }}{{\sin \phi } \over \xi }} \cr {{\pi _2} = \;{{\partial w} \over {\partial \xi }}\sin \;\phi - \;{{\partial w} \over {\partial \phi }}{{\cos \phi } \over \xi }} \cr {{\pi _3} = \bar H} \cr {{{\partial w} \over \partial }\; = \;{{du} \over d} = \;{{du} \over {dz}} \cdot {{dz} \over d} = 2\xi {{du} \over {dz}} = {{{{\left\{ {{{\bar H}^2} - 2n\left( {G + c_0^\prime} \right)f\left( z \right)} \right\}} \over {{z^2}}}}^{{1 \over 2}}} = \pm {{2eL\sin l} \over {\sqrt {a(1 - e\cos l)} }}} \cr } and ∂w∂=2G {{\partial w} \over \partial } = 2G

Therefore, π1= 2elsinlcos −2Gsin±[a1(1−ecosl]12,π2= 2elsin lsin +2Gcos±[a1(1−ecosl]12 {\pi _1} = \;{{2el\sin l\cos \; - 2G\sin } \over { \pm {{\left[ {{a_1}(1 - e\cos l} \right]}^{{1 \over 2}}}}},{\pi _2} = \;{{2el{\rm{sin}}\;l\sin \; + 2G{\rm{cos}}} \over { \pm {{\left[ {{a_1}(1 - e\cos l} \right]}^{{1 \over 2}}}}} where ( is given by the first of Eq. (24). When e = 1 (G = 0), (26) ξ1=± 2asinl2cosϕξ2=± 2asinl2sinϕξ3=h− H¯LH¯2−2c0′sinlπ1 = 4L2acosl2cosϕπ2= 4L2acosl2sinϕπ3= H¯ \matrix{ {{\xi _1} = \pm \;\sqrt {2a} \sin {l \over 2}\cos \phi } \cr {{\xi _2} = \pm \;\sqrt {2a} \sin {l \over 2}\sin \phi } \cr {{\xi _3} = h - \;\;{{\bar HL} \over {{{\bar H}^2} - 2c_0^\prime}}\sin l} \cr {{\pi _1}\; = \;\;{{4L} \over {\sqrt {2a} }}\cos {l \over 2}\cos \phi } \cr {{\pi _2} = \;\;{{4L} \over {\sqrt {2a} }}\cos {l \over 2}\sin \phi } \cr {{\pi _3} = \;\;\bar H} \cr } where φ is given by the second of the Eq. (24)

The original synodic Cartesian coordinates are obtained from equations (μ = 0), i.e. (27) x1 = ξ12−ξ22x2=2 ξ1ξ2x3 = ξ3p1 = 12z{π1ξ1−π2ξ2}p2 = 12z{ξ1π2+ξ2π1}p3 = π3 \matrix{ {{x_1}\; = \;\;\xi _1^2 - \xi _2^2} \cr {{x_2} = 2\;{\xi _1}{\xi _2}} \cr {{x_3}\; = \;\;{\xi _3}} \cr {{p_{1\;}} = \;{1 \over {2z}}\left\{ {{\pi _1}{\xi _1} - {\pi _2}{\xi _2}} \right\}} \cr {{p_{2\;}} = \;{1 \over {2z}}\left\{ {{\xi _1}{\pi _2} + {\xi _2}{\pi _1}} \right\}} \cr {{p_3}\; = \;\;{\pi _3}} \cr } where z = a(1 − e cos l).

The sidereal Cartesian coordinates are given by (28) X1= x1cost− x2sint , X2= x1sint+ x2costX3= x3X˙1= p1cost−p2sintX˙2= p1sint+p2costX˙3= p3 \matrix{ {{X_1} = \;\;{x_1}\cos t - \;{x_2}\sin t\;\;\;,\;\;\;\;\;{X_2} = \;\;{x_1}\sin t + \;{x_2}\cos t} \cr {{X_3} = \;{x_3}} \cr {{{\dot X}_1} = \;{p_1}\cos t - {p_2}\sin t} \cr {{{\dot X}_2} = \;{p_1}\sin t + {p_2}\cos t} \cr {{{\dot X}_3} = \;{p_3}} \cr } where (29) dt= r1dτ dt = \;{r_1}d\tau or t= ∫0τξ2 dτ+o(μ) t = \;\mathop {\int_0^\tau } {\xi ^2}\;d\tau + o\left( \mu \right)

Therefore (30) t−t0= ∫0τzdτdzdz =a[H¯2−2n(G+c0′]12∫0l(1−ecosl) = a (1−esinl)[H¯2−2n(G+c0′]12 \matrix{ {t - {t_0} = \;\mathop {\int_0^\tau } z{{d\tau } \over {dz}}dz} \hfill \cr {\;\;\;\;\;\;\;\; = {a \over {{{\left[ {{{\bar H}^2} - 2n(G + c_0^\prime} \right]}^{{1 \over 2}}}}}\mathop {\int_0^l } \left( {1 - e\cos l} \right)} \hfill \cr {\;\;\;\;\;\;\;\; = \;{{a\;(1 - e\sin l)} \over {{{\left[ {{{\bar H}^2} - 2n(G + c_0^\prime} \right]}^{{1 \over 2}}}}}} \hfill \cr } where t₀ is a constant. It is seen that l is the eccentric anomaly of the problem of two-body.

In terms of the canonical variables, the complete Hamiltonian may be written as (31) K=K0+μK1=L[H¯2−2n(G+c0′]1/2−1+μ[ −12{r12(ξ1π1−ξ2π2)}− r1r2−r12r23(2σ1−σ2)+ 6r1r25(σ1−σ2)ξ12ξ22−(c−c0)r1μ+1] \matrix{ {K = {K_0} + \mu {K_1}} \cr { = L{{\left[ {{{\bar H}^2} - 2n(G + c_0^\prime} \right]}^{1/2}} - 1 + } \cr {\mu \left[ {\; - {1 \over 2}\left\{ {{{{r_1}} \over {^2}}\left( {{\xi _1}{\pi _1} - {\xi _2}{\pi _2}} \right)} \right\} - \;\;\;{{{r_1}} \over {{r_2}}} - {{{r_1}} \over {2r_2^3}}\left( {2{\sigma _1} - {\sigma _2}} \right) + \;6{{{r_1}} \over {r_2^5}}\left( {{\sigma _1} - {\sigma _2}} \right)\xi _1^2\xi _2^2 - {{\left( {c - {c_0}} \right){r_1}} \over \mu } + 1} \right]} \cr } where r12= ξ4+ξ32r22= 1+ξ4+2(ξ12−ξ22)+ξ32ξ2=ξ12+ξ22 \matrix{ {r_1^2 = \;{\xi ^4} + \xi _3^2} \cr {r_2^2 = \;1 + {\xi ^4} + 2\left( {\xi _1^2 - \xi _2^2} \right) + \xi _3^2} \cr {{\xi ^2} = \xi _1^2 + \xi _2^2} \cr } and ξ₁, ξ₂, ξ₃, π₁, π₂, π₃ are given by Eq. (25).

The equations of motion for the complete Hamiltonian are (32) dldτ= ∂K∂L= [H¯2−2n(G+c0′]12+μ∂R∂Ldgdτ= ∂K∂G=−nL[H¯2−2n(G+c0′]1/2+μ∂R∂Ldhdτ= ∂K∂H=LH[H¯2−2n(G+c0′]1/2+μ∂R∂HdLdτ= −∂K∂l=−μ∂R∂ldGdτ= −∂K∂g=−μ∂R∂gdH¯dτ= −∂K∂h=−μ∂R∂h \matrix{ {{{dl} \over {d\tau }} = \;{{\partial K} \over {\partial L}} = \;{{\left[ {{{\bar H}^2} - 2n(G + c_0^\prime} \right]}^{{1 \over 2}}} + \mu {{\partial R} \over {\partial L}}} \cr {{{dg} \over {d\tau }} = \;{{\partial K} \over {\partial G}} = {{ - nL} \over {{{\left[ {{{\bar H}^2} - 2n(G + c_0^\prime} \right]}^{1/2}}}} + \mu {{\partial R} \over {\partial L}}} \cr {{{dh} \over {d\tau }} = \;{{\partial K} \over {\partial H}} = {{LH} \over {{{\left[ {{{\bar H}^2} - 2n(G + c_0^\prime} \right]}^{1/2}}}} + \mu {{\partial R} \over {\partial H}}} \cr {{{dL} \over {d\tau }} = \; - {{\partial K} \over {\partial l}} = - \mu {{\partial R} \over {\partial l}}} \cr {{{dG} \over {d\tau }} = \; - {{\partial K} \over {\partial g}} = - \mu {{\partial R} \over {\partial g}}} \cr {{{d\bar H} \over {d\tau }} = \; - {{\partial K} \over {\partial h}} = - \mu {{\partial R} \over {\partial h}}\;} \cr }

These equations form the basis of the general perturbation theory for the problem in question.

The solution described by Eqs (25) or (26) is periodic if l and g have commensurable frequencies, i.e. if nlng=2n(G+Co/)L=pq {{{n_l}} \over {{n_g}}} = {{2n(G + C_o^/)} \over L} = {p \over q} where p and q are integers.

The period of ξ_iπ_i is 4πnl {{4\pi } \over {{n_l}}} and 4πng {{4\pi } \over {{n_g}}} , and therefore, in the case of commensurability the period of solution is 4πpnl {{4\pi p} \over {{n_l}}} or 4πqng {{4\pi q} \over {{n_g}}}

We have shown that the equations of motion for the problem are regularised by the generalised Levi-Civita's transformation for three dimensions in the neighbourhood of one of the finite masses and the existence of periodic orbits for μ = 0 in the three-dimensional coordinate systems.

Equations (20)–(22) establish the canonical set (l, L, g, G, h, H) and Eq. (32) form the basis of the general perturbation theory for the problem in question. The solution described by Eq. (25) or (26) is periodic if l and g have commensurable frequencies, that is, if nlng=2n(G+Co/)L=pq(say) {{{n_l}} \over {{n_g}}} = {{2n(G + C_o^/)} \over L} = {p \over q}({\rm{say}}) where p and q are integers.

The period of ξ_iπ_i is 4πnl {{4\pi } \over {{n_l}}} and 4πng {{4\pi } \over {{n_g}}} , so that in case of commensurability the period of solution is 4πpnl {{4\pi p} \over {{n_l}}} or 4πqng {{4\pi q} \over {{n_g}}} .