partant,
![{\displaystyle r=h\sin p\sin q\pm {\sqrt {\begin{aligned}&a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q\\+&2\alpha a^{2}\varphi \left[h\cos \theta +r\sin p\sin(\theta -q)\right]\end{aligned}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77ae1a148f58bac523429ec3ff1be817249898dd)
ou
![{\displaystyle {\begin{aligned}r=h\sin p\sin q&\pm {\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}\\&\pm {\frac {\alpha a^{2}\varphi \left[h\cos \theta +r\sin p\sin(\theta -q)\right]}{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cced2fddac8cbdd2f4457f04c05ed64db0b4d1b5)
Il est aisé de voir que l’on aura
en prenant le radical en
et en substituant sous le signe
au lieu de
la valeur que l’on aurait en supposant
et prenant le radical en
ce qui donne
![{\displaystyle r'=h\sin p\sin q+{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad94de65456cf59ff31de2917d709784cf4d8e0b)
![{\displaystyle +\alpha a^{2}{\frac {\varphi \left[h\cos \theta +\left(h\sin ^{2}p\sin q+\sin p{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}\right)\sin(\theta -q)\right]}{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8222d9b52bd6fa4769dc9adbc9c84031321e1eac)
on trouvera pareillement
![{\displaystyle r=h\sin p\sin q-{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c8d3910c115a5a6049327879442702a9fbf2de8)
![{\displaystyle -\alpha a^{2}{\frac {\varphi \left[h\cos \theta +\left(h\sin ^{2}p\sin q-\sin p{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}\right)\sin(\theta -q)\right]}{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52eb89ca93ec9a0f78610553df2fa9cc52136479)
On aura ainsi, pour l’attraction du sphéroïde suivant ![{\displaystyle \mathrm {TC} ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5750b54b160960e3228b4b0631988f088b31a242)
![{\displaystyle \iint dpdq(r'+r)\sin ^{2}p\sin q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7733d6929f5819b8cf408c6fc7f90f0a543860f4)
![{\displaystyle =\iint 2hdpdq\sin ^{3}p\sin ^{2}q+\iint {\frac {\alpha a^{2}dpdq\sin ^{2}p\sin q}{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/626d8e83987e790eacfd08213e55a47370e49091)
![{\displaystyle \times \left\{{\begin{aligned}&\varphi \left[h\cos \theta +\left(h\sin ^{2}p\sin q+\sin p{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}\right)\sin(\theta -q)\right]\\-&\varphi \left[h\cos \theta +\left(h\sin ^{2}p\sin q-\sin p{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}\right)\sin(\theta -q)\right]\end{aligned}}\right\}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c78175fb6af0cf3f97c847319a0fe3c5507aa38)
et pour l’attraction suivant ![{\displaystyle \mathrm {TI} ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/419349815b7ec0977db00b91bf1bb3b27130f979)
![{\displaystyle \iint dpdq(r'+r)\sin ^{2}p\cos q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5dc1c2cb9e1ad99423e6fc4fc9c6803b5a5b20c2)
![{\displaystyle =\iint 2hdpdq\sin ^{3}p\sin q\cos q+\iint {\frac {\alpha a^{2}dpdq\sin ^{2}p\cos q}{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f2c59bd94f87b0bfde149189cc7922f69288ccf)
![{\displaystyle \times \left\{{\begin{aligned}&\varphi \left[h\cos \theta +\left(h\sin ^{2}p\sin q+\sin p{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}\right)\sin(\theta -q)\right]\\-&\varphi \left[h\cos \theta +\left(h\sin ^{2}p\sin q-\sin p{\sqrt {a^{2}-h^{2}+h^{2}\sin ^{2}p\sin ^{2}q}}\right)\sin(\theta -q)\right]\end{aligned}}\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/881ce73fba883153cdfcf98351e934c97fee21fa)