Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/103

95

SECONDE PARTIE. — SECTION VII.

ensuite

${\frac {\partial \chi }{\partial \chi }}=1,\qquad {\frac {\partial \pi }{\partial h}}=-\cos k\sin i,\qquad {\frac {\partial \pi }{\partial i}}=\sin k,$

${\frac {\partial \sigma }{\partial h}}=\sin k\sin i,\qquad {\frac {\partial \sigma }{\partial i}}=\cos k,$

et toutes les autres différences partielles seraient nulles, ce qui donne

${\frac {\partial \pi }{\partial h}}{\frac {\partial \sigma }{\partial i}}-{\frac {\partial \pi }{\partial i}}{\frac {\partial \sigma }{\partial h}}=-\sin i.$

De là on aura

${\begin{alignedat}{3}\,[a,h]=&0,&[a,i\ ]=&0,&[a,\chi ]=&0,\\\,[b,h]=&0,&[b,i\ ]=&0,&[b,\chi ]=&-{\frac {1}{2}}{\sqrt {\frac {\mathrm {g} }{b}}},\\\,[c,h]=&0,&[c,i\ ]=&0,&[c,\chi ]=&0,\\\,[h,i]=&-{\sqrt {\mathrm {g} b}}\sin i,\qquad &[h,\chi ]=&0,\qquad &[i,\chi ]=&0.\end{alignedat}}$

74. Ces valeurs, jointes à celles que nous avons déjà trouvées (art. 69), donneront enfin

${\begin{alignedat}{2}{\frac {\partial \Omega }{\partial a}}dt=&{\frac {\mathrm {g} }{2a^{2}}}dc,&{\frac {\partial \Omega }{\partial b}}dt=&-{\frac {1}{2}}{\sqrt {\frac {\mathrm {g} }{b}}}d\chi ,\\{\frac {\partial \Omega }{\partial c}}dt=&-{\frac {\mathrm {g} }{2a^{2}}}da,&{\frac {\partial \Omega }{\partial h}}dt=&-{\sqrt {\mathrm {g} b}}\sin i\,di,\\{\frac {\partial \Omega }{\partial i}}dt=&{\sqrt {\mathrm {g} b}}\sin i\,dh,\qquad &{\frac {\partial \Omega }{\partial \chi }}dt=&{\frac {1}{2}}{\sqrt {\frac {\mathrm {g} }{b}}}db\,;\end{alignedat}}$

d’où résultent ces expressions très simples des variations des éléments elliptiques

${\begin{alignedat}{2}da=&-{\frac {2a^{2}}{\mathrm {g} }}{\frac {\partial \Omega }{\partial c}}dt,&dc=&{\frac {2a^{2}}{\mathrm {g} }}{\frac {\partial \Omega }{\partial a}}dt,\\db=&{\frac {2{\sqrt {b}}}{\sqrt {\mathrm {g} }}}{\frac {\partial \Omega }{\partial \chi }}dt,&d\chi =&-{\frac {2{\sqrt {b}}}{\sqrt {\mathrm {g} }}}{\frac {\partial \Omega }{\partial b}}dt,\\dh=&{\frac {1}{{\sqrt {\mathrm {g} b}}\sin i}}{\frac {\partial \Omega }{\partial i}}dt,\qquad &di=&-{\frac {1}{{\sqrt {\mathrm {g} b}}\sin i}}{\frac {\partial \Omega }{\partial h}}dt.\end{alignedat}}$

75. On aurait des formules un peu moins simples, si, à la place du paramètre $b\,;$ on voulait conserver l’excentricité $e.$ Alors, à cause de