Page:Gauss - Théorie du mouvement des corps célestes, traduction Dubois, 1864.djvu/360

341

NOTE XI.

${\frac {1}{r}}+{\frac {1}{r'}}={\frac {2}{p}}+{\frac {2(p\cos g-\cos f{\sqrt {rr'}})\cos f}{p{\sqrt {rr'}}}}$

ou

$p{\frac {{\sqrt {rr'}}(r'+r)}{rr'}}=2{\sqrt {rr'}}\sin ^{2}f+2p\cos g\cos f$

on en déduit

(m)

${\frac {2rr'}{p}}={\frac {(r+r')-2\cos g\cos f{\sqrt {rr'}}}{\sin ^{2}f}}$

Mais, des relations

$\left\{{\begin{aligned}{\frac {1}{r}}-{\frac {1}{r'}}&={\frac {2e}{p}}\sin f\sin \mathrm {F} \\{\frac {1}{r}}+{\frac {1}{r'}}&={\frac {2}{p}}+{\frac {2e}{p}}\cos f\cos \mathrm {F} ,\end{aligned}}\right.$

on déduit aussi,

$\operatorname {tang} f\operatorname {tang} \mathrm {F} ={\frac {r'-r}{(r+r')-{\dfrac {2rr'}{p}}}}$

ou, à cause de la relation (m),

${\begin{aligned}\operatorname {tang} f\operatorname {tang} \mathrm {F} &={\frac {r'-r}{(r+r')-{\dfrac {(r+r')-2\cos g\cos f{\sqrt {rr'}}}{\sin ^{2}f}}}}\\&={\frac {(r'-r)\sin ^{2}f}{2\cos g\cos f{\sqrt {rr'}}-(r'+r)\cos ^{2}f}},\end{aligned}}$

d’où enfin,

$\operatorname {tang} \mathrm {F} ={\frac {(r'-r)\sin f}{2\cos g{\sqrt {rr'}}-(r'+r)\cos f}}$

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Note XI (art. 99).

Pour démontrer la relation

$\mathrm {Z} ={\frac {(1+2z){\sqrt {z+z^{2}}}-\log({\sqrt {1+z}}+{\sqrt {z}})}{2\left(z+z^{2}\right)^{\frac {3}{2}}}},$