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For an attractive central force of magnitude $c/r^5$, a circle of radius $R$ through the force center has polar equation $r=2R\cos \theta$. With specific angular momentum $h$, substitution into the Binet equation gives $c=8m{h}^2{R}^2$. Consequently $R$ is inversely proportional to the magnitude of the angular momentum, and the angular-time integral gives $T=2\pi R^2/|h|\propto R^3$. This is the finite travel time along the punctured circle from one approach to the singular center to the next; the force law alone does not specify continuation through the collision.

An attractive inverse-square force together with radial drag $-k\dot{r}/r^2$ preserves angular momentum $h=r^2\dot{\theta }$ but modifies the Binet equation for $u=1/r$ by a first derivative. For $k/|h|<2$ and $h\ne 0$, the reciprocal radius is a decaying oscillation about $p/h^2$. Forward angular distance $\mathrm{sgn}(h)(\theta -{\theta }_0)$ makes its damping sign explicit for either orbital orientation.

For radial acceleration $-GM(1+\gamma /r)/r^2$, conserved specific angular momentum $h$ gives an angular radial frequency $\alpha =\sqrt{1-GM\gamma /h^2}$ through the Binet equation. Successive pericentres differ by $2\pi /\alpha$, so a small perturbation produces apsidal precession $\Delta \varpi \simeq \pi GM\gamma /h^2$. Writing the unperturbed semilatus rectum as $p=h^2/(GM)$ gives the displayed inverse relation. For a nearly circular orbit of period $P$, use $p\simeq [GM(P/(2\pi ){)}^2{]}^{1/3}$. Positive $\gamma$ means prograde precession; $\gamma$ is a length. A measured rate per century must first be converted to an angle per orbit.

Constant areal velocity gives constant specific angular momentum $l=r^2\dot{\theta }$, hence zero transverse acceleration and a central force. For a focus-based ellipse $u=1/r=(1+\epsilon \cos \theta )/A$, one has $u_{\theta \theta }+u=1/A$. The Binet equation then gives the displayed radial acceleration, proving an attractive inverse-square law. An orbit's shape alone does not specify its acceleration without a time law.

For nonzero angular momentum $L$, put $u=1/r$ and $\lambda =1+2mB/L^2$. The Binet equation is $u_{\theta \theta }+\lambda u=0$. Positive $\lambda$ gives cosine scattering branches, zero gives linear reciprocal radius, and negative gives exponential combinations, with energy distinguishing sinh, exponential and cosh branches. Only intervals with $u>0$ are physical.

For radial acceleration $-\mu /r^2+C/r^3$, the Binet equation gives $u^{\prime \prime }+(1+C/h^2)u=\mu /h^2$, where $h$ is specific angular momentum. A regular bound orbit has $h^2+C>0$, $a=-\mu /(2\mathcal{E})$ and $e^2=1+2\mathcal{E}(h^2+C)/{\mu }^2$. Its radius is $a(1-e^2)/[1+e\cos (\alpha \theta )]$, with $\alpha =\sqrt{1+C/h^2}$. The apsidal precession per radial period is $2\pi ({\alpha }^{-1}-1)$, which reduces to the displayed expression for a small perturbation. Negative $C$ produces prograde precession; positive $C$ produces retrograde precession. The dimensionless coefficient is $C/(\mu a)$, since $C$ has units of length to the fourth power divided by time squared.

For radial acceleration $-A/r^2-B/r^3$, conserved specific angular momentum $l$ and $u=1/r$ reduce the Binet equation to the displayed linear equation. If $l^2>B$, its angular oscillation frequency is $\sqrt{1-B/l^2}$; attraction proportional to $r^{-3}$ changes the apsidal angle. A zero of $u$ ends the physical positive-radius branch and corresponds to escape, not passage to a negative radius. A spatial self-intersection need not be a periodic return because velocity may differ.

For an attractive central force, the transverse acceleration vanishes. In polar coordinates,Thus the specific angular momentum $h=r^2\dot{\theta }$ is constant. Assume $h\ne 0$, so angle can parameterize the orbit, and put $u=1/r$. Primes denote angle derivatives. The chain rule givesThe outward radial component of the attractive force is $-f(u)$. Its Newton's second law equation is therefore $m(\ddot{r}-r{\dot{\theta }}^2)=-f(u)$, giving the Binet equationThe sign reflects $f$ being the positive inward force magnitude. Zero angular momentum produces radial motion and is outside this angular parametrization.

A central force has zero torque, so $h=r^2\dot{\theta }$ is constant and the angular momentum is $mh$. The areal velocity is $h/2$. Put $u=1/r$ and let primes denote angular derivatives. Then $\dot{r}=-h{u}^{\prime }$, $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$, and $r{\dot{\theta }}^2=h^2{u}^3$. The radial equation with inward force magnitude $f(u)$ is $m(\ddot{r}-r{\dot{\theta }}^2)=-f(u)$, giving the Binet equationChoose the angular origin along a diameter of a radius-$R$ circle through $O$. Its polar equation is $r=2R\cos \theta$, so $u=(2R{)}^{-1}\sec \theta$ on $-\pi /2<\theta <\pi /2$. Since $(\sec \theta {)}^{\prime \prime }+\sec \theta =2{\sec }^3\theta$, substitution gives $u^{\prime \prime }+u=8R^2{u}^3$ and henceThe force is attractive. For fixed force coefficient $c$ and mass, $R|mh|=\sqrt{mc/8}$, so $R$ is inversely proportional to the magnitude of the angular momentum. The time along the circle isThis circular orbit through an inverse-fifth-power singularity reaches the force center at the two endpoints of this angular interval. The integral is finite; continuation through the singular collision is not specified by the force law alone.

Write $k=GM$ and $r=|\mathbf{r}|$. The gravitational central force has zero torque, so the specific angular momentumis constant. For a nonradial orbit $h=|\mathbf{h}|>0$, motion lies in the plane normal to $\mathbf{h}$, and plane polar coordinates give $r^2\dot{\theta }=h$. With $u=1/r$ and primes denoting $\theta$ derivatives,The radial equation $\ddot{r}-r{\dot{\theta }}^2=-k/r^2$ is therefore the Binet equationIts general solution, after choosing the polar origin of angle, isThe origin is a focus of this conic section. To identify which conic section occurs, use the conserved specific energyA bound, nonradial orbit has $\epsilon <0$, hence $0\le e<1$ and is an ellipse. For example, putting $x=r\cos \theta$, $y=r\sin \theta$ in $r+ex=p$ givesan explicit ellipse equation. This is the intended low, bound Kepler orbit. The inverse-square equation alone does not imply an ellipse: an initially tangential speed larger than $\sqrt{2GM/r}$ gives $\epsilon >0$, $e>1$ and a hyperbolic Kepler orbit. Zero angular momentum instead gives radial motion. Those qualifications are necessary for the elliptic conclusion.

In a circular orbit, centripetal acceleration balances the gravitational acceleration:Here $\Omega$ is the positive angular speed; either sense of circulation has this period.

With the stated drag force, use the full angular momentum $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. Gravitational torque is still zero, while the drag torque isTherefore $\boxed{\mathbf{L}(t)=\mathbf{L}(0)e^{-At/m}}$ for constant $A>0$. Its direction remains fixed, and its magnitude has exponential decay with time constant $m/A$; the orbit remains planar although its shape is no longer a fixed ellipse.

Set $\mu =GM>0$ and interpret $L$ in the displayed equation as specific angular momentum, $L=r^2\dot{\psi }$, rather than total angular momentum. The gravitational potential is per unit mass and $a$ has dimensions of length. Spherical symmetry fixes the orbital plane and conserves the vector angular momentum, while the autonomous specific orbital energy is also conserved.

For the orbit equation for combined inverse-square and inverse-cube attraction, defineA regular nonradial bounded oscillation requires $L\ne 0$ and $L^2-2\mu a>0$. Solving the linear Binet equation then givesFor a noncircular bound orbit $0<e<1$; its specific orbital energy is $E={\mu }^2(e^2-1)/[2(L^2-2\mu a)]$. The effective potential $[L^2-2\mu a]/(2r^2)-\mu /r$ has a centrifugal barrier and a minimum in this case. If $L^2-2\mu a\le 0$, the oscillator frequency is zero or imaginary and there is no regular finite-radius bound oscillation: the attractive centre lacks a centrifugal barrier. Unbound branches with $e\ge 1$ are not closed ellipselike trajectories.

Successive pericentres differ in angle by $2\pi /\nu$. The full phase space state returns after $p$ radial oscillations and $q$ revolutions precisely whenwhere $p,q$ are coprime positive integers; the second formula applies when $a\ne 0$. Thus $a>0$ requires $0<p/q<1$ and $L^2>2\mu a$, whereas $a<0$ requires $p/q>1$. An irrational $\nu$ advances the apsidal direction through a dense set of angles and gives a nonclosed rosette, with only the four regular spherical invariants needed to label its two-dimensional closure.

To see the additional resonant label explicitly, let $u_c=\mu /(L^2-2\mu a)$ and $w=u-u_c-i{u}^{\prime }/\nu$. Then $w=C{e}^{i\nu (\psi -{\psi }_0)}$, so on a family with fixed resonant $L$,Its argument fixes the remaining apsidal phase modulo the symmetry of the closed curve; its magnitude is already determined by specific orbital energy and $L$. Choosing reference axes in the fixed orbital plane makes this single-valued under $\psi \mapsto \psi +2\pi$. This is the fifth independent orbit label on the resonant family described by closed orbits with an inverse-cube perturbation.

For $a\ne 0$, $\nu$ varies with $L$. The orbital resonance condition holds only on selected angular momentum surfaces, so this construction is not a fifth global smooth integral on an open set of arbitrary nearby $L$. If the question asks for a maximally superintegrable Hamiltonian system on an open region of noncircular bound orbits, the required case is $a=0$: then $\nu =1$ for every $L$, and the conserved Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{v}\times \mathbf{L}-\mu \hat{\mathbf{r}}$ fixes the apsidal direction globally. The relations $\mathbf{A}\cdot \mathbf{L}=0$ and $A^2={\mu }^2+2E{L}^2$ leave five independent constants. For the usual orbit-wise interpretation, noncircular bound orbits with rational $\nu$ have five isolating labels; irrational $\nu$ has four. Circular orbits are closed for any allowed $\nu$ but have no independent apsidal phase, so the generic noncircular count cannot simply be applied to them.

For completeness, regular unbound branches are different. When $e\ge 1$ and $\nu >0$, positive radius restricts the radial phase to $|\nu (\psi -{\psi }_0)|<\arccos (-1/e)\le \pi$. The apsidal invariant on an unbound central-force branch isThe principal argument is continuous on this scattering branch because it never reaches an apocentre across the negative-real-axis cut. Thus the periapsis direction can be a fifth single-valued scattering label for any positive $\nu$, including irrational $\nu$; the rationality condition applies to the noncircular bound case. For plunging or collision branches with $L^2-2\mu a\le 0$, local trajectory labels can still be defined away from the singular centre, but the preceding periodic-orbit criterion is inapplicable and no smooth continuation through $r=0$ is asserted.

For a fixed central gravitational source, the radial and transverse components of Newton's second law in polar coordinates areThe second equation gives $r^2\dot{\theta }=h$, the constant specific angular momentum. Take $h>0$ by choosing the orientation of $\theta$, and put $u=1/r$. Writing $u^{\prime }$ and $u^{\prime \prime }$ for derivatives with respect to $\theta$,Substituting into the radial equation and dividing by $-h^2{u}^2$ proves the Binet equationAt perihelion the velocity is tangential. The escape velocity is $v_0=\sqrt{2GM/r_0}$, so $h=r_0{v}_0$ and $h^2=2GM{r}_0$. Consequently $k=1/(2r_0)$. The initial conditions are $u(0)=1/r_0$ and $u^{\prime }(0)=0$. Solving the linear ordinary differential equation gives the parabolic Kepler orbitUsing angular momentum conservation again,On the outgoing branch $r=2r_0$ is first reached at $\theta =\pi /2$. Integrate from release at perihelion, using $w=\tan (\theta /2)$:More generally the same integral gives $t=(2r_0^2/h)(w+w^3/3)$, the Barker equation for this parabolic orbit.

Multiplying the tangential equation of motion by $r$ givesTherefore $h=r^2\dot{\theta }$ is constant. It is the signed specific angular momentum; reverse the angular orientation if necessary to take $h>0$ for a nonradial orbit. Set $u=1/r$ and denote differentiation with respect to $\theta$ by a prime. ThenSubstitution into the radial equation of motion and division by $-h^2{u}^2$ give the Binet equationIts general solution is $u=GM/h^2+A\cos \theta +B\sin \theta$. Write $A=(GM/h^2)e\cos {\theta }_0$ and $B=(GM/h^2)e\sin {\theta }_0$, where $e\ge 0$. This produces the Kepler orbitHere $e$ is the orbital eccentricity, and ${\theta }_0$ points toward periapsis when $e>0$. Only portions with $u>0$ describe positive radial distances. The expression assumes $h\ne 0$; a zero-angular momentum trajectory is radial and must be treated separately.

Write $\mu =G{M}_{\odot }$, and interpret the specified radial force field as acceleration, or force per unit test-particle mass. Since it is a central force, the specific angular momentum $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ is conserved: $\dot{\mathbf{h}}=\mathbf{r}\times \mathbf{F}=0$. For $h=|\mathbf{h}|\ne 0$, motion is confined to the fixed plane perpendicular to $\mathbf{h}$. Choose the orientation of the polar angle so that $h=r^2\dot{\theta }>0$.

To derive the Binet equation here, let $u=1/r$ and differentiate with respect to $\theta$. Then $\dot{r}=-h{u}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{u}^2(u^{\prime \prime }+u)$. Equating this to the radial acceleration givesPut $K=h^2+C$ and $\alpha =\sqrt{K/h^2}$. For $K>0$, the solution isThe radial potential energy per unit mass is $V(r)=-\mu /r+C/(2r^2)$, so the conserved specific orbital energy isSubstitute the displayed $u$ and $\dot{r}=-h{u}^{\prime }$ to obtain $\mathcal{E}={\mu }^2(e^2-1)/(2K)$. Therefore defineFor regular bound motion, $K>0$ and $-{\mu }^2/(2K)\le \mathcal{E}<0$, giving $a>0$ and $0\le e<1$. Since $K/\mu =a(1-e^2)$, choose the angular origin at a pericentre, so ${\theta }_0=0$, and obtainThe turning radii are $r_{\mathrm{p}}=a(1-e)$ and $r_{\mathrm{a}}=a(1+e)$. Thus $a$ is the arithmetic mean of the turning radii; when the orbit precesses it should not be interpreted as the semimajor axis of a fixed ellipse. If $K\le 0$, the centrifugal barrier needed for a regular two-turning-point bound orbit is absent, and this bound-orbit parametrization does not apply.

For $\alpha =1$, one has $C=0$ and the orbit is a closed Kepler orbit: an ellipse of eccentricity $e$ with the Sun at a focus. In addition to $\mathcal{E}$ and the three components of $\mathbf{h}$, the Laplace-Runge-Lenz vectoris conserved. Indeed,so $\dot{\mathbf{A}}=0$. It points toward the fixed pericentre. The constraints $\mathbf{A}\cdot \mathbf{h}=0$ and $A^2={\mu }^2+2\mathcal{E}{h}^2$ leave five generically independent isolating integrals. One may takechoosing a component $A_j$ whose differential is independent of the other four on the local region under consideration. The extra integral fixes the apsidal direction in the already fixed orbital plane.

For $\alpha \ne 1$, successive pericentres are separated by $2\pi /\alpha$ in polar angle. The generic bound orbit is a rosette in that same plane. Its perihelion advance per radial cycle isFor $\alpha <1$ this is prograde; for $\alpha >1$ it is retrograde. If $\alpha$ is rational, the orbit closes after a finite number of radial cycles; if it is irrational and $0<e<1$, its successive radial phases densely fill the accessible annulus. A generic independent set of single-valued isolating integrals isThe apsidal direction is no longer fixed. For irrational frequency ratio, any continuous integral of motion must be constant on the whole two-dimensional orbital torus, because one orbit is dense there; there is no further independent continuous integral specifying a fixed apsidal direction. Exceptional closed resonant orbits do not provide a fifth global integral on an open region of phase space. Circular and radial orbits are degenerate cases of these generic counts.

The plotted precession is enlarged for visibility. Both examples have $e=0.5$ and the same turning radii; the second has $\alpha =\sqrt{0.9}$.

For the precession rate, it is useful to derive the radial period rather than assume an unperturbed angular period. Set $r=a(1-e\cos \chi )$ in the radial energy equation. On a complete radial cycle,Hence $n=2\pi /P_r=\sqrt{\mu /a^3}$. Also $K=\mu a(1-e^2)$ and $h^2=K-C$, so ${\alpha }^{-1}=\sqrt{1-C/K}$. For $|C|\ll K$, inverse-cube perturbation of Kepler apsidal motion givesThere is a dimensional issue with the printed parameter: $C$ has units $L^4/T^2$ and $\mu$ has units $L^3/T^2$, so $\eta =aC/\mu$ has units $L^2$. Retaining that definition literally, the requested expression isA dimensionless perturbation coefficient would instead be $\tilde{\eta }=C/(\mu a)=\eta /a^2$. In that convention the rate is $-n\tilde{\eta }/[2(1-e^2)]$. These two conventions must not be conflated.

Use the stated residual precession and period. The advance in one radial cycle isWith $e=0.206$, the small-perturbation estimate becomesThe negative sign is required: an additional attractive inverse-cube acceleration advances perihelion, whereas the positive-$C$ term would make it regress. If a length-unit estimate of the printed $\eta$ is desired, $P_r=0.24\mathrm{yr}$ and the solar value $\mu \simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ give $a\simeq 5.78\times 10^{10}\mathrm{m}\simeq 0.386\mathrm{au}$ through the radial-period law. ThusThe dimensionless value is the estimate obtainable directly from the supplied period, eccentricity and advance; the dimensional value additionally uses the solar gravitational parameter to set the length scale.

A central force has no angular component, so the supplied polar acceleration formula and Newton's second law give $d(r^2\dot{\theta })/dt=0$. Thus $l=r^2\dot{\theta }$ is the conserved specific angular momentum. For a nonradial orbit, $l\ne 0$ allows $\theta$ to be used as a parameter. With $u=1/r$ and a prime denoting $d/d\theta$,Also $r{\dot{\theta }}^2=l^2{u}^3$. The radial equation $\ddot{r}-r{\dot{\theta }}^2=-f(r)$ therefore becomes the Binet equationFor a purely radial trajectory $l=0$, the angular parameter is unavailable, and one instead uses the direct radial equation. The orbit in part (b) is nonradial.

Set $\theta =0$ initially and choose its positive sense to agree with the initial tangential motion. The inward radial component is $\dot{r}(0)=-4\cos (\pi /3)=-2$, while the tangential speed is $4\sin (\pi /3)=2\sqrt{3}$. HenceSubstitution in the Binet equation gives $12(u^{\prime \prime }+u)=3+9u$, or $u^{\prime \prime }+u/4=1/4$. This is an orbit equation for combined inverse-square and inverse-cube attraction. Solving the constant-coefficient equation and applying both initial conditions givesOn $0\le \theta \le 2\pi$, the sine is nonnegative, so $r$ stays finite and positive. At $\theta =2\pi$, $u=1$, hence $r=1$ and the particle is back at its original spatial position after one revolution. But $u^{\prime }(2\pi )=-1/\sqrt{3}$, so its radial velocity is now $+2$, rather than the original $-2$. The return is not a periodic return in position and velocity.

After that revolution, the first zero of $u$ is at $\theta =8\pi /3$, where $\sin (\theta /2)=-\sqrt{3}/2$. The physical branch therefore has $r\to \infty$ as $\theta \uparrow 8\pi /3$; it must not be continued into negative $u$. Since $dt/d\theta =1/(l{u}^2)$ and $u$ has a simple zero, reaching infinity takes infinite time. As a check, the specific potential energy is $\Phi (r)=-3/r-9/(2r^2)$ and the conserved energy per unit mass is $4^2/2+\Phi (1)=1/2$, giving escape speed $1$ at infinity, also obtained from $-l{u}^{\prime }(8\pi /3)=1$.

For inferring an inverse-square force from Kepler laws, first use the area swept by an infinitesimal radius sector:Kepler's second law makes the areal velocity constant, so $l=r^2\dot{\theta }$ is constant. Then $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$, meaning that the acceleration has no transverse component. Newton's second law therefore gives a central force directed along the Sun–planet line.

For the radial dependence, set $u=1/r$. Since $\dot{\theta }=l{u}^2$,The radial acceleration is the Binet equation expressionThe focus-based ellipse has $u=(1+\epsilon \cos \theta )/A$, hence $u_{\theta \theta }+u=1/A$. ThusFor a planet of mass $m$, the force is $F_r=-m{l}^2/(A{r}^2)$: it is attractive and obeys the inverse-square law. This deduction uses both the shape of the orbit and its constant areal velocity; the orbit shape alone would not determine its time-dependent acceleration.

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.

In centre-of-mass and relative coordinates, the kinetic energy separates into centre-of-mass motion plus $\mu v^2/2$, while the mutual potential is $-G{M}_1{M}_2/r=-GM\mu /r$. In the centre-of-mass frame the orbital energy is therefore

The force is central, so the specific relative angular momentum $h=r^2\dot{\theta }$ is conserved. With $u=1/r$, the radial equation becomes the Binet equationwhose solution isHere $\theta$ is the true anomaly measured from periapsis and $e$ is the orbital eccentricity. Substitution into the energy, or evaluation at an apsis, givesand hence

Immediately before the supernova the circular relative speed obeys $v^2=GM/a$. The impulsive supernova kick in a binary star changes it toApplying the energy formula just after the explosion at the unchanged separation $r=a$ givesUsing the original circular-speed relation,The post-explosion system is bound exactly when $1+2\alpha \cos \psi +{\alpha }^2<2M^{\prime }/M$. The left side ranges from $(1-\alpha {)}^2$ to $(1+\alpha {)}^2$, so every kick direction remains bound ifwhereas every direction unbinds the stars if

For an unbound orbit, conservation of relative specific energy at infinity gives $V^2/2=v^{\prime 2}/2-G{M}^{\prime }/a$. Therefore

The transverse equation of motion is $r^{-1}d(r^2\dot{\theta })/dt=0$, sois constant. With $u=1/r$,Differentiating once more and substituting into the radial equation $\ddot{r}-r{\dot{\theta }}^2=-f(r)$ gives the Binet equation