La 3ème loi de Kepler... toujours elle, permet d'écrire :

{T^{2}\over a^{3}} = {4\pi^{2}\over { {\mathcal{G}}}M} \iff T = 2\pi\left({a^{3}\over {\cal G}M}\right)^{1/2}

\Rightarrow m\sin i = V _{\mathrm{\parallel}}\left[{M^{2}\over 2\pi{\cal G}}2\pi\left({a^{3}\over {\cal G}M}\right)^{1/2}\right]^{1/3} = V _{\mathrm{\parallel}}\left({Ma\over {\cal G}}\right)^{1/2}

\Rightarrow a = {1\over V _{\mathrm{\parallel}}^{2}}{{\cal G}\over M}(m\sin i)^{2}