371
THÉORIE DES SATELLITES DE JUPITER.
puissance de ces forces doivent disparaître d’eux-mêmes, on aura
![{\displaystyle {\begin{aligned}0=&{\frac {d^{2}\left(r\delta 'r+{\frac {1}{2}}\delta r^{2}\right)}{dt^{2}}}+{\frac {(1+m)r\delta 'r}{r^{3}}}-{\frac {(1+m)\delta r^{2}}{r^{3}}}\\&+2\int \operatorname {d} \mathrm {R} +x{\frac {\partial \mathrm {R} }{\partial x}}+y{\frac {\partial \mathrm {R} }{\partial y}}+z{\frac {\partial \mathrm {R} }{\partial z}},\\\\0=&r{\frac {d^{2}\delta 'r}{dt^{2}}}+\delta 'r{\frac {d^{2}r}{dt^{2}}}+\delta r{\frac {d^{2}\delta r}{dt^{2}}}-2r^{2}{\frac {dv}{dt}}{\frac {d\delta 'v}{dt}}-2r\delta 'r{\frac {d^{2}v}{dt^{2}}}\\&-r^{2}\left({\frac {d\delta v}{dt}}\right)^{2}-4r\delta r{\frac {dv}{dt}}{\frac {d\delta v}{dt}}-\delta r^{2}\left({\frac {dv}{dt}}\right)^{2}\\&-{\frac {(1+m)r\delta 'r}{r^{3}}}+{\frac {(1+m)\delta r^{2}}{r^{3}}}+x{\frac {\partial \mathrm {R} }{\partial x}}+y{\frac {\partial \mathrm {R} }{\partial y}}+z{\frac {\partial \mathrm {R} }{\partial z}},\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8e2c2e2828f88fdf142a0b8f7fe644ac8c75822)
en observant de ne conserver dans les fonctions
![{\displaystyle \int \operatorname {d} \mathrm {R} \quad {\text{et}}\quad x{\frac {\partial \mathrm {R} }{\partial x}}+y{\frac {\partial \mathrm {R} }{\partial y}}+z{\frac {\partial \mathrm {R} }{\partial z}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ae11a35187a56af8dc575d1ffb9ab6d59289426)
que les termes dépendants des carrés et des produits des forces perturbatrices.
Les propriétés du mouvement elliptique donnent
![{\displaystyle {\begin{aligned}{\frac {r^{2}dv}{dt}}=&{\sqrt {(1+m)a\left(1-e^{2}\right)}}\\{\frac {rd^{2}v}{dt^{2}}}=&{\frac {d^{2}r}{dt^{2}}}+{\frac {1+m}{r^{2}}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8a2a238077a789555b10b9bab8ba6b77607e52c)
en substituant ces valeurs dans la seconde des deux équations différentielles précédentes, et en y mettant, au lieu de
sa valeur tirée de la première, on aura
![{\displaystyle {\begin{aligned}{\frac {2d\delta 'v}{dt}}&{\sqrt {(1+m)a\left(1-e^{2}\right)}}\\=&{\frac {d(rd\delta 'r-\delta 'rdr)}{dt^{2}}}+3{\frac {d^{2}\left(r\delta 'r+{\frac {1}{2}}\delta r^{2}\right)}{dt^{2}}}\\&-{\frac {2(1+m)\delta r^{2}}{r^{3}}}+6\int \operatorname {d} \mathrm {R} +4\left(x{\frac {\partial \mathrm {R} }{\partial x}}+y{\frac {\partial \mathrm {R} }{\partial y}}+z{\frac {\partial \mathrm {R} }{\partial z}}\right)\\&+{\frac {\delta rd^{2}\delta r-r^{2}(d\delta v)^{2}-4rdv\delta rd\delta v-dv^{2}\delta r^{2}}{dt^{2}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a7d2aba07e9b4d55197fe2410d0ca5fb103e787)