= Double-averaged circular tidal potential
{title2=$\overline{\mathcal R}_\star=\frac{GM_\star a^2}{8a_p^3}(3\cos^2I_m-1)$}
For two circular <Kepler orbits> with <mutual inclination> $I_m$, independently average the <stellar quadrupole disturbing function> over both <mean anomalies>. A circular ring with unit normal $\boldsymbol k$ has $\langle\widehat r_i\widehat r_j\rangle=(\delta_{ij}-k_i k_j)/2$. Contracting the two ring tensors gives $\langle\cos^2\psi\rangle=[1+(\boldsymbol k\cdot\boldsymbol k_p)^2]/4$, proving the displayed exact angular factor. For small <orbital inclinations> relative to a common plane, $I_m^2=I^2+I_p^2-2II_p\cos(\Omega_p-\Omega)+O(I^4)$. Thus the secular angular factor is $1/4-3(I^2+I_p^2)/8+3II_p\cos(\Omega_p-\Omega)/4$. This potential generates the slow stellar contribution to satellite <nodal precession>.
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