On a ensuite
![{\displaystyle d\zeta =ndt,\qquad d\zeta '=n'dt,\qquad d\zeta ''=n''dt\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a5219b1371753510fb9c6e0e7ee7474022af001)
partant,
![{\displaystyle {\frac {d^{2}\zeta }{dt}}=-{\frac {3}{2}}n^{\frac {1}{3}}{\frac {da}{a^{2}}},\qquad {\frac {d^{2}\zeta '}{dt}}=-{\frac {3}{2}}n'^{\frac {1}{3}}{\frac {da'}{a'^{2}}},\qquad {\frac {d^{2}\zeta ''}{dt}}=-{\frac {3}{2}}n''^{\frac {1}{3}}{\frac {da''}{a''^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0a0eb6d58080541dccbbbff233e6c45def1bafd)
On a vu dans le no 61 que, si l’on n’a égard qu’aux inégalités qui ont de très-longues périodes, on a
const.
![{\displaystyle ={\frac {m}{a}}+{\frac {m'}{a'}}+{\frac {m''}{a''}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fea7db3f3bff04a29cbda01fe5a46baf5d2c1ba)
ce qui donne
![{\displaystyle 0=m{\frac {da}{a^{2}}}+m'{\frac {da'}{a'^{2}}}+m''{\frac {da''}{a''^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9845f1b78c63752c5e380b382ade3fa46461b9b9)
On a vu dans le même numéro que, si l’on néglige les carrés des excentricités et des inclinaisons des orbites, on a
const.
![{\displaystyle =m{\sqrt {a}}+m'{\sqrt {a'}}+m''{\sqrt {a''}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48505910013baf54cdd3e77642f0a0e4ea13de85)
ce qui donne
![{\displaystyle 0=m{\frac {da}{\sqrt {a}}}+m'{\frac {da'}{\sqrt {a'}}}+m''{\frac {da''}{\sqrt {a''}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22f04c107c94efffb456633a614a83b032092cec)
De ces diverses équations il est aisé de conclure
![{\displaystyle {\begin{aligned}{\frac {d^{2}\zeta }{dt}}&=-{\frac {3}{2}}n^{\frac {1}{3}}{\frac {da}{a^{2}}},\\\\{\frac {d^{2}\zeta '}{dt}}&={\frac {3}{2}}{\frac {mn'^{\frac {1}{3}}}{m'n}}{\frac {n-n''}{n'-n''}}{\frac {da}{a^{2}}},\\\\{\frac {d^{2}\zeta ''}{dt}}&=-{\frac {3}{2}}{\frac {mn''^{\frac {1}{3}}}{m''n}}{\frac {n-n'}{n'-n''}}{\frac {da}{a^{2}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/670b4a54958bb865707fe3063607c8e629ef303d)
Enfin l’équation
du no 64 donne
![{\displaystyle -{\frac {da}{a^{2}}}=2d{\rm {R}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f40bd47c58976a67fb9e822ec825d122a4451f6)
Il ne s’agit donc plus que de déterminer d{\rm R}.