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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 a spherical grain of density $\rho$, the radiation-pressure coefficient isThe $r^{-2}$ dependence of radiation pressure cancels that of stellar gravity. Here radiation-pressure efficiency includes absorption and the momentum-transfer part of scattering, averaged over the stellar spectrum. In the geometrical optics regime, $D$ exceeds the important stellar wavelengths, $Q_{\mathrm{pr}}$ is of order unity, and $\beta \propto D^{-1}$: cross-sectional area grows as $D^2$, whereas mass grows as $D^3$. Around the stellar wavelength, Mie scattering can produce a broad maximum and material-dependent structure. In the Rayleigh scattering regime, absorption gives $Q_{\mathrm{abs}}\propto D$, while scattering gives $Q_{\mathrm{sca}}\propto D^4$. Thus small absorbing grains approach an approximately constant $\beta$, whereas nearly transparent grains have $\beta \propto D^3$. A universal fall to zero at small $D$ is therefore inappropriate without specifying the optical properties. The sketch shows both possible small-grain limits; its vertical normalization is illustrative.

Assume release with negligible velocity relative to the planet, negligible planetary gravity after release, and a constant radiation-pressure coefficient. Put $\mu =G{M}_{\star }$ and ${\mu }_d=(1-\beta )\mu$. The inherited speed and specific angular momentum are $v_p=\sqrt{\mu /a_p}$ and $h=a_p{v}_p$. The new specific orbital energy isFor $0\le \beta <1/2$, the dust orbit released from a circular parent ring is an ellipse. Using ${\epsilon }_d=-{\mu }_d/(2a_d)$ and $h^2={\mu }_d{a}_d(1-e_d^2)$ givesFor $\beta >0$, release is at periapsis, since $a_d(1-e_d)=a_p$. At the radiation-pressure blowout threshold, $\beta =1/2$, the orbit is parabolic. For $1/2<\beta <1$ it is hyperbolic: the signed semimajor axis is negative and $e_d>1$. For $\beta \ge 1$ the central force ceases to be attractive, so the elliptical interpretation and the positive-eccentricity formula cannot be extended unchanged.

With ${\theta }_d$ measured from release, it is the true anomaly. The polar equation of a Kepler orbit and conservation of specific angular momentum giveDefine the signed dust-tail angular lag by $\Delta \theta ={\theta }_d-{\theta }_p$, with ${\theta }_p=n_{p}t$ and both angles unwrapped. For fixed ${\theta }_d$, expansion to first order in $\beta$ givesso $\Delta \theta =2\beta (\sin {\theta }_d-{\theta }_d)+O({\beta }^2{\theta }_d)$. The dust trails the planet, hence the signed lag is negative. Its slope is $-4\beta {\sin }^2({\theta }_d/2)$; it decreases monotonically, has horizontal tangents at successive release-direction passages, and has a sinusoidal ripple about $-2\beta {\theta }_d$. At the end of orbit $j$, ${\theta }_d=2\pi j$ and $\Delta \theta \simeq -4\pi \beta j$. At five orbits this is $-0.2\pi$ for $\beta =0.01$ and $-0.8\pi$ for $\beta =0.04$. The upper-$\beta$ sketch remains a first-order approximation; its accumulated error is $O({\beta }^2{\theta }_d)$.

For the subsequent positive-distance statements, set $s=-\Delta \theta ={\theta }_p-{\theta }_d\ge 0$. The mean motion of the grain isA full relative wrap, $s=2\pi$, therefore takes $t_{\mathrm{wrap}}\simeq \pi /(\beta n_p)$ after averaging over the orbital ripple. During this time Poynting–Robertson drag changes the semimajor axis byThe signed change is negative. This estimate treats $a_d\simeq a_p$ as fixed in the rate and computes the drag accumulated over the radiation-pressure wrap time. Since $v_p/c\ll 1$, drag causes little migration over that time. For an arbitrarily tiny $\beta$, however, this small migration can itself alter the relative phase appreciably: neglecting that feedback on the wrap time additionally requires $v_p/c\ll \beta$.

Rapid dust sublimation is a plausible sink near a hot planet: a grain can evaporate long before it completes a relative wrap. Destructive collisions can also remove visible grains. Poynting–Robertson drag alone, with the estimate above, does not explain immediate removal from a short tail. If the destruction time is $t_{\mathrm{loss}}$, a short tail requires roughly $2\beta n_p{t}_{\mathrm{loss}}\ll 2\pi$.

For a steady dust-tail continuity equation, assume a constant injection rate $\dot{N}$, grains with the same fixed $\beta$, negligible release-speed dispersion, and negligible destruction within the segment being calculated. A finite destruction lifetime can terminate that segment, or multiply its density by a survival probability. To first order,To obtain the explicit trigonometric profile, make the additional secular phase approximation for a dust tail, $s\simeq 2\beta {\theta }_d$, discarding the bounded $-2\beta \sin {\theta }_d$ term in the angle-to-age map while retaining the periodic instantaneous drift speed. ThenSince the square is even, the same written profile applies to the signed $\Delta \theta$ if number intervals are interpreted with positive orientation. The result is an approximate phase-substitution profile, not a uniformly valid consequence of $\beta \ll 1$. The consistent first-order description is instead parametric:In particular, immediately after release, $s\sim \beta {\theta }_d^3/3$ and $n\propto s^{-2/3}$, whereas the phase-substitution expression would give $s^{-2}$. Nor does $s\ll 2\pi$ by itself justify neglecting the orbital ripple. The formal enhancements where $\dot{s}=0$ are dust-tail caustics; a spread of radiation-pressure coefficients, finite release velocities, and destruction smooth them. With constant lifetime $\tau$, the parametric profile acquires $\exp [-t({\theta }_d)/\tau ]$. These qualifications state precisely the extra assumptions behind the explicit profile.

Orbital energy. Treat the star as stationary to leading order in $m/M$, with Newtonian gravitational parameter $\mu =GM$. For velocity components in plane polar coordinates, $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$. The kinetic energy is $m{v}^2/2$. Integrating Newton's law of universal gravitation with zero gravitational potential energy at infinite separation gives $U(r)=-GMm/r$. Thuswhere $J=m{r}^2\dot{\varphi }$ is the satellite's angular momentum. The first term is radial kinetic energy; the second is transverse kinetic energy, also the centrifugal contribution to the radial effective potential; the last is the attractive gravitational potential energy. These are not three independent sources of energy: the centrifugal term is already part of the ordinary kinetic energy.

For an elliptic Kepler orbit of semi-major axis $a$ and orbital eccentricity $e$, the polar equation of a Kepler orbit $r=p/(1+e\cos \varphi )$ has semilatus rectum $p=J^2/(GM{m}^2)=a(1-e^2)$. Substituting $\dot{r}=(Je/(mp))\sin \varphi$ into the preceding expression reduces it toFor a circular Kepler orbit, $a=R$, $v_0^2=GM/R$, $K=GMm/(2R)$, and $U=-GMm/R$. Negative total energy describes a bound orbit.

Rapid, symmetric mass loss. Assume the mass is removed on a timescale short compared with the satellite's orbital period, without an impulsive kick or appreciable wind drag on the satellite. The post-loss gravitational parameter is ${\mu }^{\prime }=G(1-f)M$ with $f=0.01$. At the change, $r=R$, $\dot{r}=0$, and $v=v_0=\sqrt{GM/R}$ are unchanged. Spherical symmetry keeps the force central, so the specific angular momentum $j=J/m=\sqrt{GMR}$ is conserved. Orbital energy is not conserved across the changing gravitational field:The new semi-major axis is consequentlyThe original speed now exceeds the new circular speed at $R$, and the outward radial acceleration is $v_0^2/R-{\mu }^{\prime }/R^2=fGM/R^2>0$. Thus the release point is the new pericentre, not the apocentre. The impulsive orbital expansion under isotropic mass loss givesFor the stated one-percent loss,An independent angular momentum check is $j^2={\mu }^{\prime }{a}^{\prime }(1-e^{\prime 2})=GMR$. The impulsive orbit stays bound for $f<1/2$; loss of half the central mass is the zero-orbital energy limit.

What the wind timescale changes. The source specifies a spherically symmetric wind but does not give its duration. The preceding nearest and furthest distances require the rapid-loss assumption. In the opposite, slow limit, adiabatic orbital expansion under isotropic mass loss preserves an initially circular orbit to leading adiabatic order. Conserved specific angular momentum then gives $j^2=GM(t)r(t)$, hence $M(t)r(t)=MR$. After the same loss,For a finite intermediate timescale, the orbit depends on the mass-loss history; the percentage alone does not select a unique pair of turning distances. Neglecting wind recoil and wind drag is common to both idealizations.