= Solution
\b[<Specific orbital energy>, <orbital eccentricity> and the distribution.] Let $\mu=GM_\star$, $v_K=\sqrt{\mu/a}$ and $c=\cos\theta$. Choose positive $\sin\theta$ to mean an outward radial kick. At release the tangential and radial <velocities> are $v_t=v_K(1+\gamma c)$ and $v_r=\gamma v_K\sin\theta$. The <specific orbital energy> and <specific angular momentum> are
$$
\varepsilon'=\frac{\mu}{2a}(-1+2\gamma c+\gamma^2),\qquad h'=av_K(1+\gamma c).
$$
Using $\varepsilon'=-\mu/(2a')$ and $e'^2=1+2\varepsilon'h'^2/\mu^2$ therefore gives
$$
\boxed{\frac{a'}a=\frac1{1-2\gamma c-\gamma^2},\qquad
e'^2=1-(1-2\gamma c-\gamma^2)(1+\gamma c)^2.}
$$
The denominator must be positive for an <elliptic Kepler orbit>. A vanishing denominator describes a <parabolic Kepler orbit>; negative $a'$ is the signed <semi-major axis> of a <hyperbolic Kepler orbit>.
Differentiating the squared <orbital eccentricity> gives
$$
\frac{d(e'^2)}{dc}=2\gamma^2(3c+\gamma)(1+\gamma c).
$$
In the small-kick regime $0<\gamma<1$, $1+\gamma c>0$. Thus the minimum is at $c=-\gamma/3$, with
$$
\boxed{e'_{\min}=\gamma\sqrt{1-\frac{\gamma^2}{3}+\frac{\gamma^4}{27}},\qquad
\left.\frac{a'}a\right|_{e'_{\min}}=\frac1{1-\gamma^2/3}.}
$$
The entire expression, including the $\gamma^4/27$ term, lies under the square root in the PDF. This minimum is not a universal statement for arbitrary kick size. For example $\gamma=2$, $c=-1$ reverses the circular velocity and gives a retrograde <circular orbit> with $e'=0$; the displayed stationary value is then not the global minimum.
For $0<\gamma<\sqrt2-1$, every kick direction gives a bound prograde <Kepler orbit>. The two endpoint maxima, one at each end of the accessible curve, are
$$
\begin{array}{c|c|c}
c&a'/a&e'\\ \hline
-1&(1+2\gamma-\gamma^2)^{-1}&2\gamma-\gamma^2\\
+1&(1-2\gamma-\gamma^2)^{-1}&2\gamma+\gamma^2 .
\end{array}
$$
The positive endpoint is the global maximum. There is one interior minimum, at the location just found. Uniform $\theta$ gives the <kick-orbit distribution>
$$
f_c(c)=\frac1{\pi\sqrt{1-c^2}},\qquad
f_\alpha(\alpha)=\frac1{2\pi\gamma\alpha^2\sqrt{1-c(\alpha)^2}},\qquad
c(\alpha)=\frac{1-\gamma^2-\alpha^{-1}}{2\gamma},\quad \alpha=\frac{a'}a.
$$
Both signs of the radial kick occupy the same $e'$–$a'/a$ curve; they have opposite apsidal orientations. The distribution is concentrated near its endpoints, with integrable square-root singularities. For larger kicks retain the parametric curve and its bound portion $c<(1-\gamma^2)/(2\gamma)$; an unbound branch begins at $e'=1$, $a'/a\to\pm\infty$. In particular, a sketch of a wholly elliptic population presupposes the bound-kick restriction above.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-59-kick-distribution.png]
{title=Eccentricity and semimajor-axis distribution for isotropic planar velocity kicks of magnitude 0.2 times the circular speed}
{height=440}
The <eccentricity vector> at release has radial component $v_t^2/v_K^2-1$ and tangential component $-v_rv_t/v_K^2$. With the release radius chosen as zero longitude,
$$
\boxed{e'\cos\varpi'=\gamma c(2+\gamma c),\qquad
e'\sin\varpi'=-\gamma\sin\theta(1+\gamma c).}
$$
This also directly verifies the squared <orbital eccentricity> above and fixes the sign convention for the <longitude of periapsis>.
\b[The 1:1 encounter window.] Exact equality of <orbital periods> requires $a'=a$, hence $c_0=-\gamma/2$. There are two kick directions with this cosine when $0<\gamma<2$. To quantify a finite window one must specify a return time: near exact commensurability, consider the particle's first complete return to its release point. During that time the source advances through $2\pi(a'/a)^{3/2}$. Its longitudinal miss distance is, to first order,
$$
d\simeq2\pi a\left|\left(\frac{a'}a\right)^{3/2}-1\right|
\simeq3\pi a\left|\frac{a'}a-1\right|.
$$
Thus $d<\Delta$ gives the two-sided cosine window
$$
|c-c_0|<\frac{\Delta}{6\pi a\gamma}.
$$
Integrating $f_c$ over that window gives, to leading order for a narrow window away from $\gamma=2$,
$$
\boxed{f_{\mathrm{return}}\simeq\frac{\Delta}{3\pi^2a\gamma\sqrt{1-\gamma^2/4}}.}
$$
The printed fraction is half this value. It is obtained by retaining only one side of the <semi-major axis> window, or only one of the two kick-angle branches. Neither restriction is in the question. Thus the printed coefficient cannot be shown for the stated uniform $0\leq\theta<2\pi$ population and a symmetric first-return distance criterion. This already exhibits the discrepancy under the usual longitudinal approximation; minimizing the distance over the encounter interval introduces a further velocity-direction correction, not a missing branch. With arbitrarily long observation times, noncommensurate bound trajectories can also return arbitrarily close to the common release point, so an eventual-encounter fraction is a different, time-dependent question.
\b[Rotating-frame sketches.] In the <rotating reference frame> of the source put $x$ radially outward, $y$ along the source's motion and $s=nt$, where $n=v_K/a$. The linear <Hill equations> for a stellar <Kepler orbit>, with initial displacement zero, give the <velocity-kick epicycle>
$$
\boxed{\begin{aligned}
\frac{x}{\gamma a}&=2\cos\theta(1-\cos s)+\sin\theta\sin s,\\
\frac{y}{\gamma a}&=\cos\theta(4\sin s-3s)+2\sin\theta(\cos s-1).
\end{aligned}}
$$
These solve $\ddot x-2n\dot y-3n^2x=0$, $\ddot y+2n\dot x=0$ and reproduce the initial kick. For $\theta=0$, the particle starts forward, moves outward and drifts backward; the curve has a loop each orbital cycle, with net $\Delta y=-6\pi\gamma a$ per cycle. For $\theta=\pi$, both coordinates reverse: it moves inward and drifts forward by $6\pi\gamma a$. For $\theta=\pi/2$, the leading trajectory is a closed <epicyclic ellipse>,
$$
\left(\frac{x}{\gamma a}\right)^2+
\left(\frac{y+2\gamma a}{2\gamma a}\right)^2=1.
$$
It starts at the top of the ellipse moving radially outward and returns to the source after one period to this order. Its exact <semi-major axis> differs from $a$ at order $\gamma^2$, so exact closure is not implied. The local sketches require $|x|,|y|\ll a$, in particular $\gamma s\ll1$ for the drifting cases.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-59-rotating-kicks.png]
{title=Particle trajectories after tangential prograde, tangential retrograde and outward radial kicks in the source rotating frame}
{height=420}
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