Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/118

on aura attention de faire évanouir auparavant (Article XXXIII) le terme constant $f_{1}\mathrm {H} _{1}\,;$ ce qui donnera $\mathrm {H} _{1}=0.$

Soit, pour abréger,

${\begin{aligned}&\Phi _{1}(a_{1},a_{2})={\frac {2\mu _{1}}{\mu _{2}-\mu _{1}}}\Xi _{1}(a_{1},a_{2})+{\frac {f_{1}}{(\mu _{2}-\mu _{1})^{2}}}{\widehat {\Gamma }}_{1}(a_{1},a_{2}),\\&\Phi _{2}(a_{1},a_{2})={\frac {2\mu _{1}}{2(\mu _{2}-\mu _{1})}}\Xi _{2}(a_{1},a_{2})+{\frac {f_{1}}{4(\mu _{2}-\mu _{1})^{2}}}{\widehat {\Gamma }}_{2}(a_{1},a_{2}),\\&\Phi _{3}(a_{1},a_{2})={\frac {2\mu _{1}}{3(\mu _{2}-\mu _{1})}}\Xi _{3}(a_{1},a_{2})+{\frac {f_{1}}{9(\mu _{2}-\mu _{1})^{2}}}{\widehat {\Gamma }}_{3}(a_{1},a_{2}),\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

et pareillement

${\begin{aligned}&\Phi _{1}(a_{1},a_{3})={\frac {2\mu _{1}}{\mu _{3}-\mu _{1}}}\Xi _{1}(a_{1},a_{3})+{\frac {f_{1}}{(\mu _{3}-\mu _{1})^{2}}}{\widehat {\Gamma }}_{1}(a_{1},a_{3}),\\&\Phi _{2}(a_{1},a_{3})={\frac {2\mu _{1}}{2(\mu _{3}-\mu _{1})}}\Xi _{2}(a_{1},a_{3})+{\frac {f_{1}}{4(\mu _{3}-\mu _{1})^{2}}}{\widehat {\Gamma }}_{2}(a_{1},a_{3}),\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

et ainsi de suite.

Soit de plus

$\gamma _{1}={\frac {\mu _{1}}{m-\mu _{1}}}\beta _{1}-{\frac {3f_{1}}{8(m-\mu _{1})^{2}}}\,;$

on trouvera

${\begin{aligned}y_{1}=&-{\frac {2\mu _{1}\varepsilon _{1}}{\mathrm {M} _{1}}}\sin(\mathrm {M} _{1}t+\omega _{1})\\&+\chi _{2}\left[\Phi _{1}(a_{1},a_{2})\sin(\mu _{2}-\mu _{1})t+\Phi _{2}(a_{1},a_{2})\sin 2(\mu _{2}-\mu _{1})t+\ldots \right]\\&+\chi _{3}\left[\Phi _{1}(a_{1},a_{3})\sin(\mu _{3}-\mu _{1})t+\Phi _{2}(a_{1},a_{3})\sin 2(\mu _{3}-\mu _{1})t+\ldots \right]\\&+\chi _{4}\left[\Phi _{1}(a_{1},a_{4})\sin(\mu _{4}-\mu _{1})t+\Phi _{2}(a_{1},a_{4})\sin 2(\mu _{4}-\mu _{1})t+\ldots \right]\\&+\mathrm {K} _{1}\gamma _{1}\sin 2(m-\mu _{1})t.\end{aligned}}$

XXXIX.

Puisque $\varphi _{1}=\mu _{1}t+ny_{1}$ (Article IV), on aura, en connaissant $\gamma _{1},$ l’expression du mouvement vrai du premier satellite par son mouvement moyen.