Page:Annales de mathématiques pures et appliquées, 1813-1814, Tome 4.djvu/187

177

D’ASTRONOMIE

$+\left({\frac {\operatorname {d} \phi }{\operatorname {d} \lambda }}\right)\left(pq\operatorname {Cos} .^{2}\lambda \operatorname {Cos} .^{3}\mu \left(2\operatorname {Cos} .\phi -\operatorname {Sin} .\lambda \operatorname {Cos} .^{2}\phi +\operatorname {Sin} .\lambda \operatorname {Sin} .^{2}\phi \right)\right),$

$\left({\frac {\operatorname {d} M}{\operatorname {d} \mu }}\right)=-3pq\operatorname {Cos} .^{2}\lambda \operatorname {Cos} .^{2}\mu \operatorname {Sin} .\mu \operatorname {Sin} .\phi \left(2-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi \right)$

$+\left({\frac {\operatorname {d} \psi }{\operatorname {d} \mu }}\right)\left(pq\operatorname {Cos} .^{2}\lambda \operatorname {Cos} .^{3}\mu \left(2\operatorname {Cos} .\phi -\operatorname {Sin} .\lambda \operatorname {Cos} .^{2}\phi +\operatorname {Sin} .\lambda \operatorname {Sin} .^{2}\phi \right)\right),$

$\left({\frac {\operatorname {d} N}{\operatorname {d} \lambda }}\right)=-3pq\operatorname {Cos} .^{2}\lambda \operatorname {Sin} .\lambda \operatorname {Cos} .^{2}\mu \operatorname {Sin} .\psi \left(2-\operatorname {Sin} .\mu \operatorname {Cos} .\psi \right)$

$+\left({\frac {\operatorname {d} \phi }{\operatorname {d} \lambda }}\right)\left(pq\operatorname {Cos} .^{3}\lambda \operatorname {Cos} .^{2}\mu \left(2\operatorname {Cos} .\psi -\operatorname {Sin} .\mu \operatorname {Cos} .^{2}\psi +\operatorname {Sin} .\mu \operatorname {Sin} .^{2}\psi \right)\right),$

$\left({\frac {\operatorname {d} N}{\operatorname {d} \mu }}\right)=-pq\operatorname {Cos} .^{3}\lambda \operatorname {Cos} .\mu \operatorname {Sin} .\psi \left(4\operatorname {Sin} .\mu +\operatorname {Cos} .^{2}\mu \operatorname {Cos} .\psi -2\operatorname {Sin} .^{2}\mu \operatorname {Cos} .\psi \right)$

$+\left({\frac {\operatorname {d} \psi }{\operatorname {d} \mu }}\right)pq\operatorname {Cos} .^{3}\lambda \operatorname {Cos} .^{2}\mu \left(2\operatorname {Cos} .\psi +\operatorname {Sin} .\mu \operatorname {Sin} .^{2}\psi -\operatorname {Sin} .\mu \operatorname {Cos} .^{2}\psi \right).$

Reste donc à trouver les expressions littérales de $\left({\frac {\operatorname {d} \phi }{\operatorname {d} \lambda }}\right),\left({\frac {\operatorname {d} \psi }{\operatorname {d} \mu }}\right)$ $\operatorname {d} \lambda$ et $\operatorname {d} \mu$, et à effectuer ensuite les développemens. Or, ayant déjà exprimé $\operatorname {d} \phi$ en $\operatorname {d} t$ et $\operatorname {d} \lambda$, de même que $\operatorname {d} \psi$ en $\operatorname {d} t$ et $\operatorname {d} \mu$, on n’aura qu’à substituer, dans l’une de ses expressions, la valeur de $\operatorname {d} t$ en $\operatorname {d} \lambda$ et $\operatorname {d} \mu$ : on aura ainsi la différentielle complète de $\operatorname {d} \phi$ ou $\operatorname {d} \psi$, d’où on conclura

$\left({\frac {\operatorname {d} \phi }{\operatorname {d} \lambda }}\right)={\tfrac {p\operatorname {Sin} .\phi \operatorname {Cos} .^{2}\lambda (1+P)Q^{2}}{qP^{2}\operatorname {Cos} .^{3}\mu -pQ^{2}\operatorname {Cos} .^{3}\lambda }},\quad \left({\frac {\operatorname {d} \psi }{\operatorname {d} \mu }}\right)=-{\tfrac {q\operatorname {Sin} .\psi \operatorname {Cos} .^{2}\mu (1+Q)P^{2}}{qP^{2}\operatorname {Cos} .^{3}\mu -pQ^{2}\operatorname {Cos} .^{3}\lambda }}.$

34. Après avoir effectué ces développemens, on pourra procéder ; sans difficulté, à la détermination des rapports différentiels ${\frac {\operatorname {d} ^{2}t}{\operatorname {d} \lambda ^{2}}},{\frac {\operatorname {d} ^{2}t}{\operatorname {d} \lambda \operatorname {d} \mu }},{\frac {\operatorname {d} ^{2}t}{\operatorname {d} \mu ^{2}}}$. Ayant $2\varpi \left({\frac {\operatorname {d} t}{\operatorname {d} \lambda }}\right)={\frac {M}{F}},\ 2\varpi \left({\frac {\operatorname {d} t}{\operatorname {d} \mu }}\right)=-{\frac {N}{F}},$ il en résultera

$2\varpi F^{2}\left({\frac {\operatorname {d} ^{2}t}{2\operatorname {d} \lambda ^{2}}}\right)=F\left({\frac {\operatorname {d} M}{\operatorname {d} \lambda }}\right)-M\left({\frac {\operatorname {d} F}{\operatorname {d} \lambda }}\right),$