1. bookVolume 6 (2021): Issue 1 (January 2021)
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Periodic orbits in the restricted problem of three bodies in a three-dimensional coordinate system when the smaller primary is a triaxial rigid body

Published Online: 31 Dec 2020
Page range: 429 - 438
Received: 06 Apr 2020
Accepted: 17 Jul 2020
Journal Details
License
Format
Journal
First Published
01 Jan 2016
Publication timeframe
2 times per year
Languages
English
Abstract

In this paper, we have studied the equations of motion for the problem, which are regularised in the neighbourhood of one of the finite masses and the existence of periodic orbits in a three-dimensional coordinate system when μ = 0. Finally, it establishes the canonical set (l, L, g, G, h, H) and forms the basic general perturbation theory for the problem.

Keywords

MSC 2010

Introduction

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.

Equations of Motion

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 X˙i=Hpi;p˙i=Hxi(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 H=12(p12+p22+p32)+n(p1x2p2x1)(1μ)r1μr2μ2r23+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=(x1μ)2+x22+x32r22=(x1μ+1)2+x22+x32p1=x1x2p2=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(2σ1σ2) n = 1 + {3 \over 4}\left( {2{\sigma _1} - {\sigma _2}} \right)

where σ1=a2c25R2 {\sigma _1} = {{{a^2} - {c^2}} \over {5{R^2}}} , σ2=b2c25R2 {\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.

Regularisation of the Solution

We regularise the solution by Levi-Civita's (18) transformation generated by S=(μ+ξ12ξ22)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 xi=spi;πi=si(i=1,2,3) {x_i} = {{\partial s} \over {\partial {p_i}}};{\pi _i} = {{\partial s} \over {{\partial _i}}}(i\; = \;1,\;2,\;3) where πi is the momenta associated with the new coordinate ξi.

We have from Eqs (3) and (4) π1=sξ1=2ξ1p1+2ξ2p2,π2=sξ2=2ξ2p1+2ξ1p2,π3=sξ3=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=π1ξ1π2ξ22(ξ12+ξ22),p2=π1ξ2π2ξ12(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+ξ12ξ22 {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(ξ2π1ξ1π2)2nμ2ξ2(ξ1π2+ξ2π1)1μr1μr2μ2r23(2σ1σ2)+6μr25(σ1σ2)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=12(ξ12ξ22)+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 = Co + μC1

Now we introduce a new independent variable τ instead of t defined by dt=r1dτ(t=0atτ=0) dt = {r_1}\;d\tau \;(t = 0{\rm{at}}\tau = 0)

The equations of motion (1) will be transformed into didτ=kπi,dπidτ=ki(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(HC)=π2r18ξ2+12π32r1+r1n2(ξ2π1π2ξ12cn)nμr12ξ2(ξ1π2+ξ2π1)(1μ)r1μr2r1μ2r23(2σ1σ2)+6r1μr25(σ1σ2)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 Ko + μK1, where Ko=π2r18ξ2+12π32r1+r1n2(ξ2π1π2ξ12co)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 K1=nr12ξ2(ξ1π2+ξ2π1)r1r2r12r23(2σ1σ2)+6r1μr25(σ1σ2)ξ12ξ22(cco)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.

Generating Solution

To write the Hamilton-Jacobi equation corresponding to the Hamilton k0, we take πi=wi(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 18[(wξ1)2+(wξ2)2]r1ξ2+12(wξ3)2r1+nr12{ξ2wξ1ξ1wξ22co}=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=ξcosϕ,ξ2=sinϕ {\xi _1} = \;\xi \;{\rm{cos}}\phi ,\;\;\;\;\;{\xi _{2\;\;\;\;}} = {\rm{sin}}\phi Equation (9) may be written as 18[(wξ)2+1ξ2(wϕ)2]+12ξ2(wξ3)2+12nξ2[wϕ2c0]=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 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 (uz)2=H¯22n(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 f(z)=G22n(G+c0)H¯22αz2n(G+c0)H¯2z2 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 u(Z,G,α)[H¯22n(G+c0,)]1/2Z1Z2f(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 Z1=a(1e),Z2=a(1+e)Z=Z1cos2l2+Z1sin2l2=a(1ecosl) \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 ξi=K0πi,(i=1,2,3)ξ1=K0π1=π14r1ξ2+12nr1ξ2ξ2=K0π2=π14r1ξ2+12nr1ξ2ξ3=K03=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 Z1Zdzf(z)=(ττ0)[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}}}} e=[1G2L2]1/21f(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 } l=(ττ0)[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 wL=uL=l {{\partial w} \over {\partial L\;}}\;\;\; = \;\;{{\partial u\;\;} \over {\partial L}} = l wG=2+uG=2+nL2+G2[H¯2+2n(G+c0)]1/2sinlf=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=1e20ldl1ecosl f = \sqrt {1 - {e^2}} \mathop {\int_0^l } {{dl} \over {1 - e\cos l}} wH=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, ) since k0 = α − 1. It follows that K0=L[H¯2+2n(G+c0)]1/21>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 dLdτ=k0l=0,L=constant=L0(say)dGdτ=k0g=0,G=constant=G0(say)dH¯dt=k0h=0,H¯=constant=H0(say)dld=k0L=[H22N(G+c0)]12=const=nll=nlτ+lodgd=k0G=nL[H¯2+2n(G+c0)]1/2=const=ngg=nlgτ+godhd=k0H=LH[H¯2+2n(G+c0)]1/2=const=nhh=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 ϕ=12g+1n[L2+G2]2[H¯22n(G+c0)]sinl,whene1ϕ=12g+1LN2[H¯22nc0]sinl,whene=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 ξi, πi (i = 1, 2, 3) can be expressed by the canonical elements we have ξ1=±zcosϕ=±a(1ecosl)cosϕξ2=±zsinϕ=±a(1ecosl)sinϕξ3=hH¯(L2G2)12[H¯22n(G+c0]12sinlπ1=wξcosϕwϕsinϕξπ2=wξsinϕwϕcosϕξπ3=H¯w=dud=dudzdzd=2ξdudz={H¯22n(G+c0)f(z)}z212=±2eLsinla(1ecosl) \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=2elsinlcos2Gsin±[a1(1ecosl]12,π2=2elsinlsin+2Gcos±[a1(1ecosl]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), ξ1=±2asinl2cosϕξ2=±2asinl2sinϕξ3=hH¯LH¯22c0sinlπ1=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. x1=ξ12ξ22x2=2ξ1ξ2x3=ξ3p1=12z{π1ξ1π2ξ2}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 X1=x1costx2sint,X2=x1sint+x2costX3=x3X˙1=p1costp2sintX˙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 dt=r1dτ dt = \;{r_1}d\tau or t=0τξ2dτ+o(μ) t = \;\mathop {\int_0^\tau } {\xi ^2}\;d\tau + o\left( \mu \right)

Therefore tt0=0τzdτdzdz=a[H¯22n(G+c0]120l(1ecosl)=a(1esinl)[H¯22n(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 K=K0+μK1=L[H¯22n(G+c0]1/21+μ[12{r12(ξ1π1ξ2π2)}r1r2r12r23(2σ1σ2)+6r1r25(σ1σ2)ξ12ξ22(cc0)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(ξ12ξ22)+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 dldτ=KL=[H¯22n(G+c0]12+μRLdgdτ=KG=nL[H¯22n(G+c0]1/2+μRLdhdτ=KH=LH[H¯22n(G+c0]1/2+μRHdLdτ=Kl=μRldGdτ=Kg=μRgdH¯dτ=Kh=μRh \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 ξiπi 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}}}

Conclusion

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 ξiπi 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}}} .

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