= Solution
The odd <mass density> difference is $2\Sigma_0\beta x/r_0$. Integrating $x^{-3}$ from $H$ to infinity gives
$$
\Gamma=\chi\frac{\beta\Sigma_0(GM_s)^2}{S^2H^2}.
$$
For <Keplerian shear>, $S=3\Omega/2$ and $GM_s=q\Omega^2r_0^3$. Thus the <density-slope satellite torque> is
$$
\Gamma=\frac{4\chi}{9}\,\beta\frac{q^2}{(H/r_0)^2}\Sigma_0r_0^4\Omega^2,
\qquad
\boxed{\Gamma=\frac29\,\beta\frac{q^2}{(H/r_0)^2}\Sigma_0r_0^4\Omega^2\quad(\chi=\tfrac12).}
$$
Using 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>.
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