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The conventional complete-flyby impulse approximation uses a straight incoming path $x=x_0$, $y=-S{x}_{0}t$, with $S>0$. Integrate the satellite's transverse acceleration over the whole encounter. For $x_0>0$,The hint's integral is one over a half encounter, and two over the full encounter. The leading longitudinal impulse vanishes by oddness. In weak elastic scattering, conservation of the relative speed $V=S|x_0|$ gives the next-order change along the original velocity as $-\Delta v_x^2/(2V)$. With the original outer flow along negative $y$, this meansThe same signed expression holds for inner particles and has the opposite sign there. This is the complete-flyby gravitational impulse. Its weak-deflection requirement is $G{M}_s/(S^2|x_0{|}^3)\ll 1$. Rotation and tidal dynamics retained throughout the encounter give a more detailed response; this impulse model does not claim to solve the full Hill scattering problem exactly.

The printed coefficient $1/2$ is four times smaller than this complete-flyby value. It is obtained if the single-sided transverse impulse $G{M}_s/(S{x}_0^2)$ is inserted into the same quadratic longitudinal estimate, leaving out the other half. Thus the scaling and sign agree, but that numerical coefficient is not derived by the usual full-encounter prescription. To keep the subsequent requested formulas unambiguous, write the satellite impulse normalizationThe following parts are derived for general $\chi$ and specialize to the supplied value. The complete-flyby one-sided torque coefficient $8/27$ is also the normalization used in the primary coplanar impulse calculation summarized by Chametla and collaborators.

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.