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MÉCANIQUE ANALYTIQUE.
donne, par la différentiation,
![{\displaystyle {\frac {d\theta }{dt}}={\frac {\sqrt {\cfrac {\mathrm {g} }{a^{3}}}}{1-e\cos \theta }}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6acd6afde2123502a87b9ca8b7352923cd7b510e)
donc on aura
![{\displaystyle \mathrm {X} '=-{\sqrt {\cfrac {\mathrm {g} }{a}}}\ {\frac {\sin \theta }{1-e\cos \theta }},\qquad \mathrm {Y} '=-{\sqrt {\cfrac {\mathrm {g} \left(1-e^{2}\right)}{a}}}\ {\frac {\cos \theta }{1-e\cos \theta }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/543b4f053f960b64d317bad46fee3a0c0f76eeaf)
Il faut maintenant différentier ces formules en faisant varier les trois constantes
nous dénoterons par la caractéristique
les variations relatives à ces constantes ; ainsi l’on aura d’abord
![{\displaystyle \delta \theta ={\frac {(t-c){\sqrt {\cfrac {\mathrm {g} }{a^{3}}}}+\sin \theta de-{\sqrt {\cfrac {\mathrm {g} }{a^{3}}}}dc}{1-e\cos \theta }}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/130886bbf7c7e0e0d8b0f6c6f2f90a25b42ff2d6)
ensuite
![{\displaystyle {\begin{aligned}\delta \mathrm {X} =&-a\sin \theta \delta \theta +\cos \theta da-d(ae),\\\delta \mathrm {Y} =&a{\sqrt {1-e^{2}}}\cos \theta \delta \theta +\sin \theta d\left(a{\sqrt {1-e^{2}}}\right),\\\delta \mathrm {X} '=&-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\cos \theta -e}{(1-e\cos \theta )^{2}}}\delta \theta \\&-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sin \theta \cos \theta }{(1-e\cos \theta )^{2}}}de-{\frac {\sin \theta }{1-e\cos \theta }}d{\sqrt {\frac {\mathrm {g} }{a}}},\\\delta \mathrm {Y} '=&-{\sqrt {\frac {\mathrm {g} }{a}}}{\sqrt {1-e^{2}}}{\frac {\sin \theta }{(1-e\cos \theta )^{2}}}\delta \theta \\&+{\sqrt {\frac {\mathrm {g} }{a}}}{\sqrt {1-e^{2}}}{\frac {\cos ^{2}\theta }{(1-e\cos \theta )^{2}}}dc+{\frac {\cos \theta }{1-e\cos \theta }}d\left({\sqrt {\frac {\mathrm {g} }{a}}}{\sqrt {1-e^{2}}}\right).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f3d27c178e2cb2aba3579ff1130e652ea67a77e)
On peut faire ici
mais il est plus simple de faire
ce qui donne aussi
ainsi l’on aura, en changeant
en
![{\displaystyle {\begin{aligned}\delta \theta =&-{\sqrt {\frac {\mathrm {g} }{a^{3}}}}{\frac {dc}{1-e}},\\\delta X=&(1-e)da-adc,\\\delta Y=&a{\sqrt {1-e^{2}}}\delta \theta =-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sqrt {1-e^{2}}}{1-e}}dc,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32e680761c41e043567915b960fbba814d370588)