Si l’on n’a égard qu’à l’action de
la valeur de
du no 46 donne
![{\displaystyle {\begin{aligned}&y{\frac {\partial {\rm {R}}}{\partial x}}-x{\frac {\partial {\rm {R}}}{\partial y}}=m'(x'y-xy')\times \\&\qquad \left({\frac {1}{\left(x'^{2}+y'^{2}+z'^{2}\right)^{\frac {3}{2}}}}-{\frac {1}{\left[(x'-x)^{2}+(y'-y)^{2}+(z'-z)^{2}\right]^{\frac {3}{2}}}}\right),\\\\&z{\frac {\partial {\rm {R}}}{\partial x}}-x{\frac {\partial {\rm {R}}}{\partial z}}=m'(x'z-xz')\times \\&\qquad \left({\frac {1}{\left(x'^{2}+y'^{2}+z'^{2}\right)^{\frac {3}{2}}}}-{\frac {1}{\left[(x'-x)^{2}+(y'-y)^{2}+(z'-z)^{2}\right]^{\frac {3}{2}}}}\right),\\\\&z{\frac {\partial {\rm {R}}}{\partial y}}-y{\frac {\partial {\rm {R}}}{\partial z}}=m'(y'z-yz')\times \\&\qquad \left({\frac {1}{\left(x'^{2}+y'^{2}+z'^{2}\right)^{\frac {3}{2}}}}-{\frac {1}{\left[(x'-x)^{2}+(y'-y)^{2}+(z'-z)^{2}\right]^{\frac {3}{2}}}}\right).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fdca600fedef9f0742346c37f6fb02d3df0dd98)
Soit maintenant
![{\displaystyle {\frac {c''}{c}}=p,\qquad {\frac {c'}{c}}=q\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb23c4e03c4b7b16c1fcdcf05983ab8168cb7a35)
les deux variables
et
détermineront, par le no 64, la tangente de l’inclinaison
de l’orbite de
et la longitude
de son nœud, au moyen des équations
![{\displaystyle \operatorname {tang} \varphi ={\sqrt {p^{2}+q^{2}}},\qquad \operatorname {tang} \theta ={\frac {p}{q}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8e737ab09dc4646cd4a9f206a6b03213cd9b564)
Nommons
ce que deviennent
et
relativement aux corps
; on aura, par le no 64
![{\displaystyle z=qy-px,\qquad z'=q'y'-p'x',\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62fb60b4a356290fbe0161475c80045e6df36904)
La valeur précédente de
différentiée, donne
![{\displaystyle {\frac {dp}{dt}}={\frac {1}{c}}{\frac {dc''-pdc}{dt}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5b23878cfe596420a5d0a0c81f6b6c1bd5b8259)
en substituant au lieu de
et de
leurs valeurs, on aura
![{\displaystyle {\frac {dp}{dt}}={\frac {m'}{c}}\left[(q-q')yy'+(p'-p)x'y\right]\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/695bdfa4ce86dd5e9f8ca8cf93808f84a89247b8)
![{\displaystyle \left({\frac {1}{\left(x'^{2}+y'^{2}+z'^{2}\right)^{\frac {3}{2}}}}-{\frac {1}{\left[(x'-x)^{2}+(y'-y)^{2}+(z'-z)^{2}\right]^{\frac {3}{2}}}}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2247a318697cbfbabe7569ea7da73dcca03d9f9a)
on trouvera pareillement
![{\displaystyle {\frac {dq}{dt}}={\frac {m'}{c}}\left[(p-p')xx'+(q'-q)xy'\right]\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e6d172d48df3db0bbd1043d4b75cda1f6ef4dee)
![{\displaystyle \left({\frac {1}{\left(x'^{2}+y'^{2}+z'^{2}\right)^{\frac {3}{2}}}}-{\frac {1}{\left[(x'-x)^{2}+(y'-y)^{2}+(z'-z)^{2}\right]^{\frac {3}{2}}}}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2247a318697cbfbabe7569ea7da73dcca03d9f9a)