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For a fixed mass on a Keplerian circular orbit, $L_s=M_s\sqrt{G{M}_{\ast }{r}_0}$ has derivative $M_s{r}_0\Omega /2$. The torque on the satellite is minus the torque $\Gamma$ on the disk. This gives the displayed drift law. In the density-slope satellite torque model, ${\dot{r}}_0=-(8\chi /9)\beta q({\Sigma }_0{r}_0^2/M_{\ast })r_0\Omega (H/r_0{)}^{-2}$.

The odd mass density difference is $2{\Sigma }_0\beta x/r_0$. Integrating $x^{-3}$ from $H$ to infinity givesFor Keplerian shear, $S=3\Omega /2$ and $G{M}_s=q{\Omega }^2{r}_0^3$. Thus the density-slope satellite torque isUsing the complete-flyby normalization multiplies this supplied-model result by four. The linear mass density law is a local Taylor approximation, not a nonnegative mass density on the entire infinite real line. With an outer local cutoff $L\gg H$, the same calculation replaces $H^{-2}$ by $H^{-2}-L^{-2}$; the extension to infinity is the leading local result. Small $|\beta |H/r_0$ keeps the mass density perturbation small in the dominant encounter region.

Outer particles gain angular momentum and inner particles lose it. The satellite transmits angular momentum from inner to outer material. If $\beta >0$, the denser outer side receives the larger torque: the disc gains net angular momentum and the satellite loses it. If $\beta <0$, the net transfer reverses. At zero slope these exchanges cancel in the satellite's total torque.