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The central force has no transverse component. Its angular equation is thereforeThus the specific angular momentum is the constantThe radial equation isSubstitute $\dot{\theta }=l/r^2$:Consequently the radial motion for $r>0$ is exactly a unit-mass one-dimensional motion in the effective potentialIts conserved energy is $\frac{1}{2}{\dot{r}}^2+V_{\mathrm{eff}}(r)$; the second term in the effective potential accounts for the angular kinetic energy.

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.

A central force has no component in the angular direction. The angular component of the acceleration in polar coordinates therefore gives $2\dot{r}\dot{\theta }+r\ddot{\theta }=0$, orFor unit mass, $h=r^2\dot{\theta }$ is the signed angular momentum perpendicular to the orbital plane, so its magnitude $|h|$ is conserved. Equivalently, the torque $\mathbf{r}\times \mathbf{F}$ vanishes, which also fixes that plane.

The given potential energy is $k/r$, whose radial force is $k/r^2$: positive $k$ is repulsive and negative $k$ attractive. The kinetic energy in polar coordinates is $({\dot{r}}^2+r^2{\dot{\theta }}^2)/2$, so conservation of energy givesFor $h\ne 0$ and $k>0$, the effective potential is positive, strictly decreasing from $+\infty$ to $0$, and has no stationary point. For $k=-\kappa <0$, it still tends to $+\infty$ at zero, crosses zero at $r=h^2/(2\kappa )$, and has its unique minimum atIt then approaches zero from below. The minimum corresponds to a stable circular orbit. If $h=0$, the centrifugal barrier disappears; the attractive effective potential is simply $-\kappa /r$ and has no minimum.

A central force has the form $\mathbf{F}=f(r)\hat{\mathbf{r}}$, so it always points along the line from a fixed centre to the particle. Its torque about the centre vanishes:Thus angular momentum is conserved. Since $\mathbf{r}\cdot \mathbf{L}=0$, the trajectory lies in the fixed plane through the origin perpendicular to $\mathbf{L}$.

The area swept in time $dt$ is $dA=\frac{1}{2}|\mathbf{r}\times d\mathbf{r}|$, sowhich is constant. This proves Kepler's second law.

The central force conserves angular momentum. The initial impact parameter gives $L=ub$, while at closest approach the velocity is transverse and $L=wp$. Thus $u=wp/b$. Conservation of energy givesUnder $u^2\gg 2k/R$, neglect the initial potential term and substitute $u=wp/b$:Consequently

Motion under a central force is planar. In plane polar coordinates, the radial component of the acceleration is $\ddot{r}-r{\dot{\theta }}^2$, soThe conserved specific angular momentum is $h=r^2\dot{\theta }$. Substitution of $\dot{\theta }=h/r^2$ yieldsThe last term is the centrifugal part of the radial effective potential.

A central force conserves angular momentumPut $u=1/r$. Since $\dot{\theta }=\ell u^2/m$,where primes now denote differentiation with respect to $\theta$. Substitution into the radial equation gives the Binet equation

The force is central force, so its torque about the star vanishes. The specific angular momentum is therefore constant:Taking the scalar product of $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$ with $\dot{\mathbf{r}}$ givesThus conservation of energy gives the second constant