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Hyperbolic motion in special relativity is rectilinear motion with constant proper acceleration. A representative positive-$x$ world line is $x=c{t}_0\cosh (\tau /t_0)$, $t=t_0\sinh (\tau /t_0)$, with proper acceleration $c/t_0$. This describes a hyperbola in spacetime, not a hyperbolic Kepler orbit.

Specific orbital energy, orbital eccentricity and the distribution. Let $\mu =G{M}_{\star }$, $v_K=\sqrt{\mu /a}$ 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 areUsing ${\epsilon }^{\prime }=-\mu /(2a^{\prime })$ and $e^{\prime 2}=1+2{\epsilon }^{\prime }{h}^{\prime 2}/{\mu }^2$ therefore givesThe denominator must be positive for an elliptic Kepler orbit. A vanishing denominator describes a parabolic Kepler orbit; negative $a^{\prime }$ is the signed semi-major axis of a hyperbolic Kepler orbit.

Differentiating the squared orbital eccentricity givesIn the small-kick regime $0<\gamma <1$, $1+\gamma c>0$. Thus the minimum is at $c=-\gamma /3$, withThe 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 $e^{\prime }=0$; the displayed stationary value is then not the global minimum.

For $0<\gamma <\sqrt{2}-1$, every kick direction gives a bound prograde Kepler orbit. The two endpoint maxima, one at each end of the accessible curve, areThe positive endpoint is the global maximum. There is one interior minimum, at the location just found. Uniform $\theta$ gives the kick-orbit distributionBoth signs of the radial kick occupy the same $e^{\prime }$–$a^{\prime }/a$ 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 $e^{\prime }=1$, $a^{\prime }/a\to \pm \infty$. In particular, a sketch of a wholly elliptic population presupposes the bound-kick restriction above.

The eccentricity vector at release has radial component $v_t^2/v_K^2-1$ and tangential component $-v_r{v}_t/v_K^2$. With the release radius chosen as zero longitude,This also directly verifies the squared orbital eccentricity above and fixes the sign convention for the longitude of periapsis.

The 1:1 encounter window. Exact equality of orbital periods requires $a^{\prime }=a$, 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 $2\pi (a^{\prime }/a{)}^{3/2}$. Its longitudinal miss distance is, to first order,Thus $d<\Delta$ gives the two-sided cosine windowIntegrating $f_c$ over that window gives, to leading order for a narrow window away from $\gamma =2$,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 $0\le \theta <2\pi$ 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.

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 epicycleThese solve $\ddot{x}-2n\dot{y}-3n^{2}x=0$, $\ddot{y}+2n\dot{x}=0$ 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,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 $\gamma s\ll 1$ for the drifting cases.

Let $\mu =M_p/M_{\star }$, $q=\sqrt{1-\beta }$, and restrict first to $0\le \beta <1$. The radiation-pressure coefficient reduces the grain's central gravitational parameter to $G{M}_{\star }{q}^2$, whereas the planet's is approximately $G{M}_{\star }$. Their mean motions obey $n/n_p=q(1+ϵ{)}^{-3/2}$. Setting this ratio to $j/(j+1)$ gives the radiation-pressure shift of a mean-motion resonance:It lies interior to the planet exactly when $q(j+1)<j$, or $j>q/(1-q)$. Equality puts the frequency commensurability at the planet's radius. For $\beta =0$ every such commensurability is exterior.

There is a frame qualification in the printed encounter formula. The circular inertial speed of the grain at $a_p(1+{ϵ}_1)$ is $v_1=q{v}_p(1+{ϵ}_1{)}^{-1/2}$. Subtracting the planet's inertial velocity at conjunction gives the local planet-relative inertial speedThis is the velocity used in the question's two-body gravitational assist calculation. In a frame actually rotating about the star at $n_p$, the grain's tangential velocity is instead $v_1-n_p{a}_p(1+{ϵ}_1)$, with magnitude $v_p[1+{ϵ}_1-q(1-{ϵ}_1/2)]$. In particular its nonradiating shear is $3v_p{ϵ}_1/2$, rather than $v_p{ϵ}_1/2$. Thus the requested expression uses a local translating, inertial encounter convention despite the source's rotating-frame wording. The following derivation consistently uses that convention, and its coefficients are an encounter estimate rather than an exact rotating-frame result.

Take impact parameter $b\simeq a_p{ϵ}_1$ and a small gravitational scattering angle $\theta$. Expanding the supplied hyperbolic scattering relation gives $\theta \simeq 2G{M}_p/(b{v}_{\infty }^2)$. Initially the relative tangential velocity is $-v_{\infty }$; after the deflection it has tangential component $-v_{\infty }\cos \theta$. The relative speed is unchanged in the local hyperbolic Kepler orbit, but the stellar-frame kinetic energy changes. Taking the dot product with the planet's velocity givesThe specific orbital energy of the grain is $-G{M}_{\star }(1-\beta )/(2a)$. Comparing it immediately before and after the encounter at the same position givesConsequentlyThe encounter increases the grain's semi-major axis within this geometry. These steps assume ${ϵ}_1\ll 1$, $\theta \ll 1$, and an orbital kick small enough for the subsequent linearization.

Choose the initial conjunction longitude as zero. The synodic period is $2\pi /(n_p-n)$, and the planet's longitude accumulated by the next conjunction isThese are unwrapped longitudes, counting complete revolutions. The second uses the new mean orbital period while neglecting the encounter's immediate phase offset and the difference between true and mean conjunction. A comparison reduced modulo $2\pi$ could not implement the stated one-turn sensitivity test.

For $\beta =0$, $v_{\infty }\simeq v_p{ϵ}_1/2$, so $\delta ϵ\simeq 32{\mu }^2{ϵ}_1^{-5}$ and ${\lambda }_c\simeq 4\pi /(3ϵ)$. Conjunction-longitude sensitivity to an encounter givesRequiring this to exceed $2\pi$ gives ${ϵ}_1^7<64{\mu }^2/3$. At a high-index first-order mean-motion resonance, ${ϵ}_j=(1+1/j{)}^{2/3}-1\simeq 2/(3j)$, henceThe interpretation of the resonance-overlap encounter estimate is loss of coherent encounter phases: a small kick changes the next synodic encounter time by more than one planetary revolution. Resonant protection by repeatedly meeting at controlled orbital phases can then fail, allowing irregular kicks and diffusion. This sensitivity criterion motivates an unstable region; it is not a proof that every orbit or resonant phase in it must escape.

For a quantitative sketch with radiation, put $h=1-q(1-ϵ/2)$ and $d=1-q(1+ϵ{)}^{-3/2}$. Differentiating the conjunction formula and substituting the kick givesEvaluate this at the positive ${ϵ}_j$ and take $\mathcal{S}=1$ as the critical-index curve. At small $\beta$, increased encounter speed weakens planetary scattering, moving the threshold to larger $j$. As $\beta$ rises further, the radiation-pressure shift of a mean-motion resonance brings each commensurability closer to the planet, strengthening the close encounters. The curve turns over and approaches the exterior-limit boundary $j=q/(1-q)$. Larger planetary mass produces stronger kicks and a lower critical index. These competing effects are shown below; the curves retain the factors in $\mathcal{S}$, so their $\beta =0$ intercepts differ slightly from the leading high-$j$ formula. The boundary for disappearing exterior commensurabilities is geometric; divergence of the encounter approximation near $ϵ=0$ should not be read as a precise physical scattering rate.

In the center of mass frame, the velocities are ${\mathbf{v}}_1=(M_2/M)\mathbf{v}$ and ${\mathbf{v}}_2=-(M_1/M)\mathbf{v}$. Hence the kinetic energy is $T=\frac{1}{2}\mu v^2$, while the Newtonian gravitational potential energy is $U=-G{M}_1{M}_2/r=-GM\mu /r$. The two-body orbital energy is thereforeThe relative equation of Newtonian gravity is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. It is a central force, so conservation of angular momentum confines the motion to a plane and preserves the specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|=r^2\dot{\theta }$.

For completeness, derive the polar equation of a Kepler orbit. Let $w=1/r$ and use primes for $\theta$ derivatives. Then $\dot{r}=-h{w}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{w}^2(w^{\prime \prime }+w)$. The radial equation gives the Binet equationChoose the angular origin at closest approach. Its solution is $w=(GM/h^2)(1+e\cos \theta )$, soFor the ellipse, the orbital eccentricity satisfies $0\le e<1$. Its extreme separations are $r_p=l/(1+e)$ and $r_a=l/(1-e)$. The semi-major axis is half their sum, giving $a=l/(1-e^2)$ and thus $l=a(1-e^2)$. The true anomaly $\theta$ runs through a full $2\pi$; the endpoints are the same position, and for a circular Kepler orbit the angular origin is arbitrary.

The radial and transverse relative speeds are $\dot{r}=(GM/h)e\sin \theta$ and $r\dot{\theta }=h/r$. Substituting these into the specific orbital energy givesTherefore the conserved orbital energy isThe negative specific orbital energy and fixed angular momentum characterize the bound Kepler orbit; neither should be assumed unchanged through a mass-ejecting explosion.

Initially the circular Kepler orbit has $r=a$ and $v^2=GM/a$. Treat the supernova kick in a binary star as impulsive: the relative position does not change, the companion's velocity is unchanged during the impulse, and the new neutron star receives $\mathbf{u}$ with $|\mathbf{u}|=\alpha v$. Thus ${\mathbf{v}}^{\prime }=\mathbf{v}+\mathbf{u}$, and with $\psi$ the angle between these vectors before the kick,Write $M^{\prime }=M_1^{\prime }+M_2$, $\beta =M^{\prime }/M$ and ${\mu }^{\prime }=M_1^{\prime }{M}_2/M^{\prime }$. The post-explosion specific orbital energy isUsing ${\epsilon }^{\prime }=-G{M}^{\prime }/(2a^{\prime })$ givesHere $a^{\prime }>0$ for a bound Kepler orbit. For a hyperbolic Kepler orbit, this formula uses the signed energy parameter $a^{\prime }<0$, rather than the positive geometric magnitude of the hyperbola's semi-major axis. At the parabolic Kepler orbit boundary, $1/a^{\prime }=0$.

The kick-direction binary survival criterion is bound if $S<2\beta$ and unbound with positive asymptotic speed if $S>2\beta$. To prove the printed sufficient disruption condition, choose a perpendicular kick, $\cos \psi =0$: then $S=1+{\alpha }^2$, so $M^{\prime }<\frac{1}{2}(1+{\alpha }^2)M$ produces positive specific orbital energy. This is sufficient, not the sharp existence threshold. Sincethe complete kick-direction binary survival criterion isThe second inequality follows by taking a kick directly opposite to $\mathbf{v}$. All directions remain bound if $2\beta >(1+\alpha {)}^2$, while every direction has positive escape energy if $2\beta <(1-\alpha {)}^2$. For $\alpha >0$, the marginal direction obeys $\cos \psi =(2\beta -1-{\alpha }^2)/(2\alpha )$ when the right-hand side lies in $[-1,1]$. Equalities give marginal parabolic Kepler orbits for the relevant extreme direction. With zero kick, losing more than half the original total mass unbinds the circular Kepler orbit.

For an escaping pair, the Newtonian gravitational potential energy approaches zero at infinite separation. Conservation of the post-explosion two-body orbital energy then gives $\frac{1}{2}{\mu }^{\prime }{V}^2={\mu }^{\prime }{\epsilon }^{\prime }$, or the asymptotic relative speed of a disrupted binaryThis is the relative recession speed, not the velocity of the new center of mass or either star's individual velocity in the original inertial frame. A marginal parabolic Kepler orbit separates with $V=0$ at infinity.

For unit mass, the inverse-square force has potential energy $-k/r$, with zero potential at infinity. Conservation of angular momentum gives $r^2\dot{\theta }=h$. Differentiate the polar equation of the Kepler orbit:Consequently the total mechanical energy iswhere $k=h^2/\ell$. This also explains the sign distinction between an elliptic orbit, a parabolic Kepler orbit, and a hyperbolic Kepler orbit.

At the original periapsis, the parabolic Kepler orbit has $e=1$, $r=d=\ell /2$, and purely tangential speed $V$. ThusThe outward radial impulse has zero moment about the Sun, so it preserves angular momentum and hence $\ell$. Its additional velocity is perpendicular to the original velocity, so the new mechanical energy is $E^{\prime }={\alpha }^2{V}^2/2$. Applying the energy formula givesFor $\alpha >0$, the new orbit is a hyperbolic Kepler orbit; $\alpha =0$ leaves the parabola unchanged.

To fix the orientation of the sketch, put the Sun at the origin, the impulse point at $(d,0)$, and take the original motion there upwards. The new velocity is $(\alpha V,V)$. The eccentricity vector isIt points towards the new periapsis, which is rotated clockwise through $\beta =\arctan (2\alpha )$ from the original periapsis direction. Thus, measured using the original polar angle $\theta$, the new orbit isIn particular, the impulse point is already on the outgoing part of the new orbit and is not its new periapsis. The old parabola satisfies $x=d-y^2/(4d)$. Both orbits pass through $(d,0)$ and have the same focus, but the hyperbola's new periapsis axis is tilted.

Work in the planet's rest frame, neglecting its recoil and tidal forces during the close encounter. A hyperbolic Kepler orbit has positive specific orbital energy $\epsilon =v_{\infty }^2/2$ and specific angular momentum $h=b{v}_{\infty }$, where $b$ is the impact parameter. Its orbital eccentricity isAt the pericentre distance, $r_{\min }=h^2/[\mu (1+e_h)]$. Substituting $h^2={\mu }^2(e_h^2-1)/v_{\infty }^2$ yieldsThe two asymptotes have true anomalies $\pm f_{\infty }$, where $\cos f_{\infty }=-1/e_h$. Their directions of motion give the gravitational scattering angle $\theta =2f_{\infty }-\pi$, so $\sin (\theta /2)=-\cos f_{\infty }=1/e_h$. ThereforeAs a useful independent form, $b^2=r_{\min }^2+2\mu r_{\min }/v_{\infty }^2$, displaying gravitational focusing. Here $\mu =G{M}_p$ to the stated small-comet accuracy; the exact relative-coordinate formula replaces it by $G(M_p+M)$.

The repulsive force is $\mathbf{F}=k\mathbf{r}/r^3=-\nabla (k/r)$, so the Hamiltonian isThe Hamilton's equations give $\dot{\mathbf{r}}=\mathbf{p}/m$ and $\dot{\mathbf{p}}=k\mathbf{r}/r^3$. Consequentlyso the angular momentum is conserved. Moreover,It follows that the Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{p}\times \mathbf{L}+mk\hat{\mathbf{r}}$ is conserved. By the Hamiltonian conservation law, these time derivatives are exactly $\{\mathbf{L},H\}$ and $\{\mathbf{A},H\}$, so both Poisson brackets vanish.

The integrals of motion are the energy $H$, the components of $\mathbf{L}$, and the components of $\mathbf{A}$, subject toTaking the dot product of $\mathbf{A}$ with $\hat{\mathbf{r}}$ givesTherefore the polar equation of a repulsive inverse-square orbit iswhere $\lambda =L^2/(mk)\ge 0$ and $e=A/(mk)\ge 0$. For a nonradial physical scattering orbit $H>0$, the identity above gives $e>1$, as expected for a hyperbolic Kepler orbit.

Let $h=r^2\dot{f}$ be the specific angular momentum. The radial Kepler orbit equation has semi-latus rectum $p=h^2/\mu$, so comparison withgives $h^2=\mu a(1-e^2)$ for an elliptic orbit and $h^2=\mu a(e^2-1)$ for a hyperbolic Kepler orbit. At either apsis, $\dot{r}=0$ and $v=h/r$. Substituting the apsidal radius and angular momentum into part (i), or equivalently using the vis-viva equation, givesThus the signs are $C=\mp \mu /(2a)$ in the convention of the question.

In the planetocentric hyperbolic Kepler orbit, the speed at infinity is $\Delta v_{pc}$ and the impact parameter is $b$. Conservation of specific angular momentum givesConservation of specific orbital energy at the moon's radius giveswhere $v_m^2=G{M}_p/a_m$. ThusThe moon moves tangentially at $v_m$, soTherefore the correctly dimensioned reading of the displayed result is

For some orbital phase of the moon to permit an impact, the comet's planetocentric hyperbolic Kepler orbit must at least cross the moon's circular orbit. At the limiting case $a_m$ is the comet's periapsis, so the radial velocity in part (iv) vanishes. ConsequentlyThis is gravitational focusing by the planet; an actual collision additionally requires the moon to occupy the crossing point at the right time.

At periapsis, $r=a(1-e)=a\delta$. The grain's energy from part (ii) is nonnegative whensoThe strict inequality gives a hyperbolic Kepler orbit, while equality gives a parabolic Kepler orbit. The printed eccentricity formula yields $\beta >\delta /(2+\delta )$, which has the same requested lowest-order limit $\beta >\delta /2$.

In polar coordinates in the orbital plane, $h=r^2\dot{\theta }$. Setting $u=1/r$, the Binet equation for the effective inverse-square force isIts general solution isand henceThis is a branch of a hyperbolic Kepler orbit; because $1-\beta <0$, its physical branch has a negative denominator.

For an instantaneous in-plane change of velocity of magnitude $\gamma v_K$ on a circular orbit, with angle $\theta$ measured from the forward tangential direction, the new specific orbital energy gives $a^{\prime }/a=(1-2\gamma \cos \theta -{\gamma }^2{)}^{-1}$. The specific angular momentum gives $e^{\prime 2}=1-(1-2\gamma \cos \theta -{\gamma }^2)(1+\gamma \cos \theta {)}^2$. The signed semi-major axis becomes negative for an unbound hyperbolic Kepler orbit.

For a Kepler orbit with semi-major axis $a$, the vis-viva equation isFor the convention that a hyperbolic Kepler orbit has positive semi-major-axis magnitude $a$, its final term instead has a plus sign.