Stellar quadrupole disturbing function (${\mathcal{R}}_{\star }=G{M}_{\star }{r}^2{P}_2(\cos \psi )/a_p^3$)

For a particle far inside a planet's stellar orbit, the planet-centred disturbing function is $G{M}_{\star }/|{\mathbf{r}}_{\star }-\mathbf{r}|-G{M}_{\star }\mathbf{r}\cdot {\mathbf{r}}_{\star }/r_{\star }^3$. Expanding the first term cancels its dipole against the indirect disturbing function. Dropping the spatially constant monopole leaves $G{M}_{\star }{r}^2(3{\cos }^2\psi -1)/(2r_{\star }^3)$, where $\psi$ is the angle between the two offset vectors. The omitted instantaneous terms are $O(G{M}_{\star }{r}^3/r_{\star }^4)$. The gradient of the retained expression is the leading differential stellar tidal acceleration.

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 $\mathbf{k}$ has $⟨{\hat{r}}_i{\hat{r}}_j⟩=({\delta }_{ij}-k_i{k}_j)/2$. Contracting the two ring tensors gives $⟨{\cos }^2\psi ⟩=[1+(\mathbf{k}\cdot {\mathbf{k}}_p{)}^2]/4$, proving the displayed exact angular factor. For small orbital inclinations relative to a common plane, $I_m^2=I^2+I_p^2-2I{I}_p\cos ({\Omega }_p-\Omega )+O(I^4)$. Thus the secular angular factor is $1/4-3(I^2+I_p^2)/8+3I{I}_p\cos ({\Omega }_p-\Omega )/4$. This potential generates the slow stellar contribution to satellite nodal precession.