Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 11.djvu/277

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SUR L’ÉQUATION SÉCULAIRE DE LA LUNE.

quent le produit de ${\frac {\delta a}{a}}$ par ces forces,

$2a\int \operatorname {d} \mathrm {R} ={\frac {2n'^{2}}{3n^{2}}}-{\frac {n'^{2}u}{n^{2}}}\left(1+{\frac {3}{2}}e'^{2}\right)+{\frac {3n'^{2}}{2n^{2}}}\int ue'de'\,;$

mais on a, à fort peu près,

$u=-{\frac {d^{2}u}{n^{2}dt^{2}}},$

ce qui donne, à cause de l’extrême lenteur avec laquelle $e'$ varie,

$\int ue'de'=-{\frac {du}{ndt}}{\frac {e'de'}{ndt}}\,;$

on aura donc

$2a\int \operatorname {d} \mathrm {R} ={\frac {2n'^{2}}{3n^{2}}}-{\frac {n'^{2}u}{n^{2}}}\left(1+{\frac {3}{2}}e'^{2}\right)-{\frac {3n'^{2}}{2n^{2}}}{\frac {du}{ndt}}{\frac {e'de'}{ndt}}.$

On trouvera pareillement

$a\left(x{\frac {\partial \mathrm {R} }{\partial x}}+y{\frac {\partial \mathrm {R} }{\partial y}}+z{\frac {\partial \mathrm {R} }{\partial z}}\right)=-{\frac {n'^{2}}{2n^{2}}}-{\frac {3n'^{2}e'^{2}}{4n^{2}}}-{\frac {n'^{2}u}{n^{2}}}\left(1+{\frac {3}{2}}e'^{2}\right).$

L’équation différentielle précédente deviendra ainsi, en y faisant $\left(1+{\frac {\delta a}{a}}\right)u=u',$

${\begin{aligned}0=&{\frac {d^{2}u'}{dt^{2}}}+n^{2}u'\left(1-{\frac {3\delta a}{a}}-{\frac {2n'^{2}}{n^{2}}}-{\frac {3n'^{2}e'^{2}}{n^{2}}}\right)\\&-{\frac {3n'^{2}}{2n^{2}}}{\frac {du'}{dt}}{\frac {e'de'}{dt}}+n^{2}{\frac {\delta a}{a}}+{\frac {1}{6}}n'^{2}-{\frac {3n'^{2}e'^{2}}{4}}\,;\end{aligned}}$

d’où l’on tire les deux équations suivantes ;

${\frac {\delta a}{a}}=-{\frac {n'^{2}}{6n^{2}}}+{\frac {3n'^{2}e'^{2}}{4n^{2}}},$

$0={\frac {d^{2}u'}{dt^{2}}}+n^{2}u'\left(1-{\frac {3n'^{2}}{2n^{2}}}-{\frac {21n'^{2}e'^{2}}{4n^{2}}}\right)-{\frac {3n'^{2}}{2n^{2}}}{\frac {du'}{dt}}{\frac {e'de'}{dt}}.$