= Solution
The conventional complete-flyby <impulse approximation> uses a straight incoming path $x=x_0$, $y=-Sx_0t$, with $S>0$. Integrate the satellite's transverse acceleration over the whole encounter. For $x_0>0$,
$$
\Delta v_x=-\frac{GM_s}{x_0^2}\int_{-\infty}^{\infty}(1+S^2t^2)^{-3/2}dt
=-\frac{2GM_s}{Sx_0^2}.
$$
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 means
$$
\boxed{\Delta v_y\simeq\frac{2(GM_s)^2}{S^3x_0^5}.}
$$
The 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 $GM_s/(S^2|x_0|^3)\ll1$. 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 $GM_s/(Sx_0^2)$ 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 normalization>
$$
\Delta v_y=\chi\frac{(GM_s)^2}{S^3x_0^5}:
\qquad\chi=\tfrac12\ \text{for the supplied model},\quad\chi=2\ \text{for the complete-flyby model}.
$$
The 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 https://academic.oup.com/mnras/article/468/4/4610/3098191[the primary coplanar impulse calculation summarized by Chametla and collaborators].
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