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A binary black hole is a binary system whose two components are black holes. Gravitational waves carry away orbital energy and angular momentum, gradually reducing the separation; sufficiently close systems progress through inspiral, merger and the settling of the final black hole. The increasing orbital angular velocity produces a signal whose gravitational-wave frequency rises, called a chirp. In GW150914, two LIGO interferometers recorded a signal sweeping from about 35 to 250 hertz in September 2015. Comparing that measured gravitational wave strain with general relativity waveforms inferred the component masses and the merger's properties. This permits the study of systems that need no luminous surrounding gas to be detected.^[ref]

Launching at the correct circular orbital speed but at in-plane angle $\alpha$ from the tangent leaves specific orbital energy $-GM/(2R)$, hence semi-major axis $R$. The specific angular momentum becomes $\sqrt{GMR}\cos \alpha$, so $j^2=GMa(1-e^2)$ gives $e=|\sin \alpha |$. The actual periapsis and apoapsis are $R(1\mp e)$, equally far from the intended circular orbit. The maximum geometric radial separation is therefore $Re$. This is not an equal-time separation between two moving satellites.

For a Hohmann transfer orbit from radius $r_0$ to $\eta r_0$, with $\eta >1$ and initial circular speed $v_0=\sqrt{GM/r_0}$, conservation of orbital energy and angular momentum gives transfer endpoint speeds $v_p/v_0=\sqrt{2\eta /(1+\eta )}$ and $v_a/v_0=\sqrt{2/[\eta (1+\eta )]}$. The required prograde impulses areIf the spacecraft mass is constant and fuel use is proportional to impulse magnitude, these also give the fuel fractions relative to a budget with total speed change $v_0$. Their sum is a sum of magnitudes, even though the burns occur at different positions and directions.

For an initially circular binary losing mass instantaneously, let $f$ be the retained total mass fraction and $q$ the kick speed in units of the initial relative orbital speed. A kick exactly parallel or opposite to the exploding star's velocity leaves negative relative orbital energy precisely for the title inequality. Equality is marginal escape. For an opposite kick $q=1$, the relative velocity and angular momentum vanish: negative energy then produces a radial collision, rather than a persistent finite-pericentre binary.

Orbital energy is the mechanical energy associated with relative motion and mutual attraction. In the isolated Newtonian two-body problem, let $\mu =m_1{m}_2/(m_1+m_2)$ be the reduced mass, $r$ the separation and $v$ the relative speed. ThenDividing by $\mu$ gives specific orbital energy. With zero potential energy at infinity, negative $E$ gives a bound Kepler orbit, zero gives the parabolic limit and positive $E$ gives an unbound encounter.

For an elliptic orbit, $E=-G{m}_1{m}_2/(2a)$ with semi-major axis $a$. Energy loss can therefore shrink a bound orbit, whereas a sufficiently large gain can unbind it. External forces or interactions with additional bodies invalidate conservation of the isolated pair's orbital energy.

Let $M_0=M_1+M_2$, $M^{\prime }=M_{\mathrm{NS}}+M_2=M_0-\Delta M$, and let the initial circular separation be $A$. The initial relative speed is $v_0=v_1+v_2$, with $v_0^2=G{M}_0/A$. Instantaneous symmetric mass loss changes neither the separation nor either surviving star's velocity at that instant. In the new centre-of-mass frame, use the reduced mass ${\mu }^{\prime }=M_{\mathrm{NS}}{M}_2/M^{\prime }$; the orbital energy isConsequentlyLess than half the total initial binary mass lost gives a bound orbit, not necessarily a circular one. Exactly half gives zero orbital energy and the marginal parabolic escape case. The threshold concerns the total mass of both stars, not just the exploding star. Using the old centre-of-mass kinetic energy without subtracting the new systemic motion would give the wrong binding criterion; the relative-velocity form performs that subtraction automatically.

Write $f=M^{\prime }/M_0$ for the retained fraction and $q=v^{-}/v_0$ for an opposite kick. With more than half the mass lost, $0<f<1/2$. A same-direction kick cannot bind the pair, because $1+v^{+}/v_0\ge 1>\sqrt{2f}$. An opposite kick can reduce the relative speed enough to leave negative energy:Both endpoints are marginal zero-energy cases. Kicks somewhat larger than $v_0$ can still satisfy this interval, but then the direction of the relative orbital motion is reversed.

At $v^{-}=v_0=v_1+v_2$, the two surviving stars have identical instantaneous velocities, so their relative velocity and orbital angular momentum vanish. Their energy is nevertheless negative,They fall radially toward each other from rest in their new centre-of-mass frame. In the point-mass idealization the collision orbit has semimajor axis $A/2$ and zero pericentre; real finite-size stars collide or interact before reaching zero separation. It is not a stable circular binary. Zero relative angular momentum, rather than the negative energy alone, determines this plunge outcome.

Write $\mu =GM$ and let $v_c=\sqrt{\mu /r_0}$ be the original circular orbital speed. An impulse leaves the orbital angular momentum unchanged if it is radial: its angular impulse $\mathbf{r}\times \Delta \mathbf{p}$ is zero. The tangential velocity remains $v_c$, and an added radial component $w$ givesThus kinetic energy increases even though angular momentum does not. For a Radial kick from a circular Kepler orbit, the new specific orbital energy and squared specific angular momentum areThe orbital eccentricity is $e=\sqrt{1+2\epsilon j^2/{\mu }^2}=\sqrt{\beta -1}$, while the semi-latus rectum is $p=j^2/\mu =r_0$. HenceFor $\beta =2$ the trajectory is a parabolic Kepler orbit and has no finite furthest radius; its periapsis is $r_0/2$. For $\beta >2$ it is a hyperbolic orbit, with the same formula for $r_{\min }$ but unbounded radius:One must not use the negative denominator in the bound-orbit expression as a negative physical apocentre. On a departing unbound branch the conic's periapsis may already lie in the past; the future minimum from the kick event is then $r_0$. These are ideal conic orbits; a periapsis inside a real planet would cause impact before the ideal continuation.

For a fixed-speed orbital launch-direction error, the initial radius and speed are correct, so the specific orbital energy is still $\epsilon =-GM/(2R)$ and the new semi-major axis is still $a=R$. Resolve the launch velocity into radial and tangential components, taking $\alpha$ as the in-plane deviation from the intended tangent:For a nondegenerate orbit $\cos \alpha \ne 0$, the orbital eccentricity follows from $j^2=GMa(1-e^2)$:The intended circular orbit has radius $R$ everywhere, so the largest radial distance between the orbit curves occurs at either apsis:This is the geometric separation of the two orbit curves, not the distance between two satellites compared at equal elapsed times. The latter also depends on orbital phase. A purely radial launch is a degenerate zero-angular-momentum trajectory through the central body and requires separate treatment of the collision.

To escape the Solar system in the single-Sun approximation, the heliocentric specific orbital energy must reach zero. At Earth's orbital radius, the required heliocentric escape speed is $\sqrt{2}{v}_E$. The smallest change from Earth's initial velocity is prograde, parallel to its motion:Combining the terrestrial and solar energy requirements givesThe asymptotic heliocentric trajectory is a parabolic Kepler orbit; larger launch speeds can give positive solar orbital energy.

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.

The source's tangential arrows $a,c$ respectively point along and against the original motion; $b,d$ point radially outward and inward. Write $r_0$ for the initial orbital radius, $v_c=\sqrt{GM/r_0}$ for the circular orbital speed, and $\delta >0$ for the small added speed. The resulting trajectories follow the velocity kick on a circular Kepler orbit formulas. Here orbital energy includes kinetic energy and gravitational potential energy; angular momentum refers to the central object.

Adding tangential velocity along the existing motion increases both orbital energy and angular momentum:For a sufficiently small kick the Kepler orbit remains an ellipse. At the kick point its radial velocity vanishes and its tangential speed exceeds the circular orbital speed, so that point becomes the pericentre. The ellipse extends farther outward than the original circle.

A radial outward kick has no angular impulse about the centre. ThereforeThe initial position is on the outward-moving portion of a bound elliptic Kepler orbit, between pericentre and apocentre, rather than at either turning point. The eccentricity is $e=\delta /v_c$ and the semilatus rectum remains $r_0$. The orbital energy change is second order: it vanishes to first order in the impulse but is positive for any nonzero kick.

The inward radial kick also leaves angular momentum unchanged and increases orbital energy by $m{\delta }^2/2$. Its eccentricity and semi-major axis equal those in case $b$, but the orientation of its eccentricity vector is reflected about the initial radius. The satellite initially moves inward on its ellipse, toward pericentre. Neither radial kick decreases the orbital energy.

For a differentiable central potential, the force is $\mathbf{F}=-V^{\prime }(r)\hat{\mathbf{r}}$. The angular momentum is $\mathbf{J}=m\mathbf{r}\times \dot{\mathbf{r}}$, soThis proves conservation of angular momentum. For $J\ne 0$, the trajectory lies in the fixed plane perpendicular to $\mathbf{J}$. In plane polar coordinates, $J=m{r}^2\dot{\varphi }$. Substituting this into the kinetic energy gives the radial energy equationThe second term is the centrifugal contribution to the effective potential. Its radial derivative reproduces $m\ddot{r}=-V^{\prime }(r)+J^2/(m{r}^3)$. Purely radial motion is the $J=0$ case.

For the satellite, write $\mu =GM$, with $M$ Earth's mass and $m$ the satellite's mass. At apogee and perigee, the radial velocity is zero, so angular momentum conservation and conservation of energy giveEliminating $v_{\mathrm{P}}$ yieldsFor an elliptic orbit, $r_{\mathrm{A}}+r_{\mathrm{P}}=2a$. Therefore the vis-viva equation at apogee isSubstitution into the orbital energy gives $E=-GMm/(2a)$. The circular case follows by continuity, or directly from the radial force balance.

An instantaneous tangential retrograde burn at apogee leaves the position unchanged. For the speed change $v\mapsto (1-ϵ)v$, the orbital energy change isDifferentiating $E=-\mu m/(2a)$ gives $\Delta E=\mu m\Delta a/(2a^2)$ to first order. Since $r_{\mathrm{A}}=a(1+e)$, the vis-viva equation implies $v^2=\mu (1-e)/[a(1+e)]$. HenceThe factor of two distinguishes the major axis from the semimajor axis. As a finite-change check, the new semimajor axis is exactlywithin the instantaneous-burn, unchanged-mass model. Expanding this expression gives the stated result.

Retain the unchanged-angular momentum assumption of the preceding part. The initial orbit has radius $r_o=6400+500=6900\mathrm{km}$, and the desired pericentre distance is $r_p=6400+100=6500\mathrm{km}$. Using $r_p=r_o/(1+e)$ givesThe required fractional increase in initial kinetic energy is thereforeHere $\Delta E=\Delta K>0$ because the instantaneous kick does not change potential energy. Since the initial total orbital energy is $E_0=-K_0$, an alternative convention for the requested fractional energy change givesThe total orbital energy becomes less negative. Explicitly specifying its denominator avoids interpreting a positive energy input as a decrease of energy.

Classification of attractive inverse-square orbits. Write the potential energy as $U=-\kappa /r$, with $\kappa =GMm>0$ and zero at infinity. For nonzero angular momentum $J$, the Kepler orbit is a conic,Negative total orbital energy gives an elliptic orbit, $0\le e<1$. Its special case $e=0$ is the circular Kepler orbit, at the minimum energy permitted for that $J$, $E=-m{\kappa }^2/(2J^2)$. Zero energy gives a parabolic Kepler orbit, $e=1$, and positive energy a hyperbolic Kepler orbit, $e>1$. For $J=0$, the motion is radial: these are degenerate limiting trajectories and may strike the centre or the finite planetary surface.

The scattering geometry. The impact parameter $b$ is the perpendicular offset of the undeflected incoming asymptote from the centre, not the actual periapsis radius $r_{\min }$. Gravity bends the incoming path toward the planet.

The grazing threshold. At infinity, conservation of energy and angular momentum give $E=m{v}_{\infty }^2/2$ and $J=mb{v}_{\infty }$. At closest approach the radial speed is zero, so the velocity is tangential andEliminating $v_p$ givesThis increases monotonically with $r_{\min }>0$. Setting $r_{\min }=R$ gives the grazing impact parameter in gravitational focusing,Here $v_{\mathrm{esc}}=\sqrt{2GM/R}$ is the surface escape speed. The expression has units of length. Equality is the tangency limit; a strictly noncolliding trajectory requires $b>b_{\min }$. At high speed, $b_{\min }\to R$, while at low speed $b_{\min }\sim \sqrt{2GMR}/v_{\infty }$. Thus gravitational focusing enlarges the collision cross-section to $\pi b_{\min }^2$.

Deflection angle. For this hyperbola,The asymptotes occur at $\cos {\varphi }_{\infty }=-1/e$. The change of velocity direction is $\chi =2{\varphi }_{\infty }-\pi$, giving $\sin (\chi /2)=1/e$. Hence the gravitational scattering angle satisfiesFor the grazing trajectory this isIt is the attractive counterpart of the Rutherford impact-parameter relation; the sign of the transverse bending changes, but the positive deflection magnitude has this same form.

Range of angles. Along the grazing family, varying $v_{\infty }>0$ givesThe limits are $\chi \to 0$ as $v_{\infty }\to \infty$ and $\chi \to \pi$ as $v_{\infty }\to 0^{+}$. Neither endpoint is reached at finite positive speed. The low-speed limiting orbit still has a finite periapsis, but an increasingly distant incoming asymptote.

Choose zero gravitational potential energy at infinite separation. The spacecraft's conserved total orbital energy isAt periapsis and apoapsis, the radial velocity vanishes, so the velocity is perpendicular to the radius. The conserved angular momentum magnitude is thereforeThe angular momentum vector is normal to the fixed orbital plane. The respective units are joules and $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. These expressions use Newtonian gravity and the approximation $M\gg m$.

At closest approach the velocity is tangential. The conserved specific orbital energy and specific angular momentum areThe negative energy gives an elliptic orbit. Its semi-major axis isSince $r_p+r_a=2a$,This treats the star as fixed, appropriate to $m\ll M$. Equivalently, equate the energy at the two turning points using $v_a=\ell /r_a$. The planet mass cancels; a circular-orbit relation would be inappropriate here.