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A strip $dx$ at $x>0$ passes the satellite at relative speed $Sx$, so its encounter mass flux is $Sx\Sigma (x)dx$. Each unit mass gains specific angular momentum $r_0\Delta v_y$. Therefore the one-sided impulse torque on a disk isPutting $\chi =1/2$ gives exactly the requested expression. Inner strips have the opposite signed impulse but the same positive encounter flux. The net disc torque is consequentlyAn even surface density makes the integrand vanish pointwise, so the total torque is zero although each one-sided torque can be nonzero. The cutoff must keep the encounters in the weak-deflection regime; an appropriate physical value is at least of order the thickness or the relevant strong-scattering radius. The infinite-sheet integral also assumes convergence or a specified outer truncation.

For a straight shearing-sheet encounter, $V=S|x|$, and the outer and inner longitudinal changes have opposite signs. The complete-flyby gravitational impulse gives $\chi =2$. Some simplified prescribed kick models use a different coefficient; carrying it explicitly makes the corresponding one-sided impulse torque on a disk and migration rates unambiguous. This coefficient does not assert an exact solution of the full rotating Hill encounter.