Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 12.djvu/244

petites relativement à $dx',dy'$ et $dz'.$ En faisant donc

$ds'={\sqrt {dx'^{2}+dy'^{2}+dz'^{2}}},$

et prenant pour le plan des $x$ et des $y$ celui de l’écliptique vraie, ce qui rend $dz'$ nul, on aura

${\begin{aligned}{\frac {\partial \mathrm {R} }{\partial x}}=&{\frac {(\mathrm {K'-K} )ds'dx'}{dt^{2}}}-{\frac {\mathrm {K} ds'dx}{dt^{2}}}-{\frac {\mathrm {K} (dx'dx+dy'dy)}{dt^{2}}}{\frac {dx'}{ds'}},\\{\frac {\partial \mathrm {R} }{\partial y}}=&{\frac {(\mathrm {K'-K} )ds'dy'}{dt^{2}}}-{\frac {\mathrm {K} ds'dy}{dt^{2}}}-{\frac {\mathrm {K} (dx'dx+dy'dy)}{dt^{2}}}{\frac {dy'}{ds'}},\\{\frac {\partial \mathrm {R} }{\partial z}}=&-{\frac {Kds'dz}{dt^{2}}}\,;\end{aligned}}$

or on a par l’article I

${\begin{aligned}-{\frac {\partial \mathrm {R} }{\partial u}}-{\frac {s}{u}}{\frac {\partial \mathrm {R} }{\partial s}}=&{\frac {1}{u}}\left(x{\frac {\partial \mathrm {R} }{\partial x}}+y{\frac {\partial \mathrm {R} }{\partial y}}\right),\\{\frac {\partial \mathrm {R} }{\partial v}}=&x{\frac {\partial \mathrm {R} }{\partial y}}-y{\frac {\partial \mathrm {R} }{\partial x}},\\{\frac {\partial \mathrm {R} }{\partial s}}=&{\frac {1}{u}}{\frac {\partial \mathrm {R} }{\partial z}},\end{aligned}}$

d’où l’on tire, en négligeant l’excentricité de l’orbe terrestre,

${\begin{aligned}-{\frac {\partial \mathrm {R} }{\partial u}}-{\frac {s}{u}}{\frac {\partial \mathrm {R} }{\partial s}}=&{\frac {(\mathrm {K'-K} )dv'^{2}}{u^{2}u'^{2}dt^{2}}}\sin(v-v')+{\frac {\mathrm {K} dv'du}{u^{4}u'dt^{2}}}\\&-{\frac {\mathrm {K} dv'dv}{2v^{3}u'dt^{2}}}\sin(2v-2v')+{\frac {\mathrm {K} dv'du}{u^{4}u'dt^{2}}}\sin ^{2}(v-v'),\\\\{\frac {\partial \mathrm {R} }{\partial v}}=&{\frac {(\mathrm {K'-K} )dv'^{2}}{uu'^{2}dt^{2}}}\cos(v-v')-{\frac {\mathrm {K} dvdv'}{u^{2}u'dt^{2}}}\\&-{\frac {\mathrm {K} dv'dv}{u^{2}u'dt^{2}}}\cos ^{2}(v-v')+{\frac {\mathrm {K} dv'dv}{2u^{3}u'dt^{2}}}\sin(2v-2v'),\\\\{\frac {\partial \mathrm {R} }{\partial s}}=&{\frac {\mathrm {K} sdv'du}{u^{3}u'dt^{2}}}-{\frac {\mathrm {K} dv'ds}{u^{2}u'dt^{2}}}.\end{aligned}}$

L’équation $(f)$ de l’article I donnera donc, en négligeant les carrés de l’inclinaison et de l’excentricité de l’orbe lunaire, et en substituant