Page:Annales de mathématiques pures et appliquées, 1813-1814, Tome 4.djvu/181

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D’ASTRONOMIE.

 Les corrections sont expliquées en page de discussion

porter uniquement sur les deux excentricités $\lambda$ et $\mu ,$ et que le temps, doit être regardé comme exempt de différentiation. On aura ainsi

${\begin{aligned}{\frac {\operatorname {d} \phi }{\operatorname {d} \lambda }}=-{\frac {\operatorname {Sin} .\phi (2-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi )}{\operatorname {Cos} .\lambda }},&\quad {\frac {\operatorname {d} r}{\operatorname {d} \lambda }}=a\operatorname {Cos} .\lambda \operatorname {Cos} .\phi ,\\{\frac {\operatorname {d} \psi }{\operatorname {d} \mu }}=-{\frac {\operatorname {Sin} .\psi (2-\operatorname {Sin} .\mu \operatorname {Cos} .\psi )}{\operatorname {Cos} .\mu }}\,;&\quad {\frac {\operatorname {d} s}{\operatorname {d} \mu }}=b\operatorname {Cos} .\mu \operatorname {Cos} .\psi .\\\end{aligned}}$

22. Enfin, de l’expression de $\operatorname {Tang} .\nu \!\nu ,$ donnée ci-dessus, on tire l’expression générale de $\operatorname {d} \nu \!\nu ,$ ainsi qu’il suit

$\operatorname {d} \nu \!\nu ={\frac {\begin{aligned}-r^{2}\operatorname {d} \phi +rs\operatorname {d} \phi &\operatorname {Cos} .(\alpha -\beta -\phi +\psi )-s\operatorname {d} r\operatorname {Sin} .(\alpha -\beta -\phi +\psi )\\-s^{2}\operatorname {d} \psi +rs\operatorname {d} \psi &\operatorname {Cos} .(\alpha -\beta -\phi +\psi )+r\operatorname {d} r\operatorname {Sin} .(\alpha -\beta -\phi +\psi )\\\end{aligned}}{r^{2}-2rs\operatorname {Cos} .(\alpha -\beta -\phi +\psi )+s^{2}}}\,;$

ce qui donnera, pour les deux coefficiens partiels ${\frac {\operatorname {d} \nu \!\nu }{\operatorname {d} \lambda }},\ {\frac {\operatorname {d} \nu \!\nu }{\operatorname {d} \mu }}$

${\frac {\operatorname {d} \nu \!\nu }{\operatorname {d} \lambda }}={\frac {\operatorname {Cos} .\lambda }{r^{2}-2rs\operatorname {Cos} .(\alpha -\beta -\phi +\psi )+s^{2}}}\left\{{\tfrac {a^{2}\operatorname {Cos} .^{2}\lambda \operatorname {Sin} .\phi (2-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi )}{(1-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi )^{2}}}\right.$

$-{\frac {ab\operatorname {Cos} .^{2}\mu \operatorname {Sin} .\phi (2-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi )}{(1-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi )(1-\operatorname {Sin} .\mu \operatorname {Cos} .\psi )}}\operatorname {Cos} .(\alpha -\beta -\phi +\psi )$

$\left.-{\frac {ab\operatorname {Cos} .^{2}\mu \operatorname {Cos} .\phi }{(1-\operatorname {Sin} .\mu \operatorname {Cos} .\psi )}}\operatorname {Sin} .(\alpha -\beta -\phi +\psi )\right\}\,;$

${\frac {\operatorname {d} \nu \!\nu }{\operatorname {d} \mu }}={\frac {\operatorname {Cos} .\mu }{r^{2}-2rs\operatorname {Cos} .(\alpha -\beta -\phi +\psi )+s^{2}}}\left\{{\tfrac {a^{2}\operatorname {Cos} .^{2}\mu \operatorname {Sin} .\psi (2-\operatorname {Sin} .\mu \operatorname {Cos} .\psi )}{(1-\operatorname {Sin} .\mu \operatorname {Cos} .\psi )^{2}}}\right.$

$-{\frac {ab\operatorname {Cos} .^{2}\mu \operatorname {Sin} .\psi (2-\operatorname {Sin} .\mu \operatorname {Cos} .\psi )}{(1-\operatorname {Sin} .\mu \operatorname {Cos} .\psi )(1-\operatorname {Sin} .\mu \operatorname {Cos} .\psi )}}\operatorname {Cos} .(\alpha -\beta -\phi +\psi )$

$\left.+{\frac {ab\operatorname {Cos} .^{2}\mu \operatorname {Cos} .\psi }{(1-\operatorname {Sin} .\mu \operatorname {Cos} .\psi )}}\operatorname {Sin} .(\alpha -\beta -\phi +\psi )\right\}.$

24. Pour en tirer les deux coefficiens $B,B',$ il faudra faire, dans les deux expressions, $\lambda =0,\ \mu =0,\ r=a,\ s=b,\ \phi ={\frac {2\varpi t}{p}},\psi ={\frac {2\varpi t}{q}}\,;$ on aura ainsi