= Gravitational quadrupole potential of a point-mass binary
{title2=$\Phi_2=-G\mu[3(\mathbf x\cdot\mathbf r)^2-x^2r^2]/(2x^5)$}
For a <binary star> viewed far from its <centre of mass>, let $\mathbf r$ be the separation and $\mu$ the <reduced mass>, $M=M_1+M_2$ the total <mass>, $r=|\mathbf r|$ and $x=|\mathbf x|$. The <Taylor expansion> of the point-mass <Newtonian gravitational potential> has zero dipole because $\sum_A M_A\mathbf x_A=0$. Its second moment is $\sum_A M_Ax_{Ai}x_{Aj}=\mu r_i r_j$, so
$$
\Phi=-\frac{GM}{x}-\frac{G\mu}{2x^5}\left[3(\mathbf x\cdot\mathbf r)^2-x^2r^2\right]+O(GMr^3/x^4).
$$
This holds for an instantaneous separation in any orbit, not only a <circular orbit>. The <mass quadrupole moment> convention $Q_{ij}=\mu(r_ir_j-r^2\delta_{ij}/3)$ gives $\Phi_2=-3GQ_{ij}x_ix_j/(2x^5)$.
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