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For a circular Kepler orbit, neglecting stellar spin, the orbital angular momentum of a binary star is $J=\mu a^2\Omega =\mu \sqrt{GMa}$, where $M$ is total mass, $\mu$ is the reduced mass, and $a$ is separation. Kepler's third law also gives $J=G^{2/3}{P}^{1/3}{M}_1{M}_2/[(2\pi {)}^{1/3}{M}^{1/3}]$. Distances from the center of mass are $a{M}_2/M$ and $a{M}_1/M$; adding their individual orbital angular momenta proves the formula.

On a steady phase-mixed orbit, line density on a Kepler orbit is proportional to $1/v$, so cross-sectional area is uniform in time. For an optically thin wire of blackbodies in radiative equilibrium,The proof uses specific angular momentum: $dt/r^2=df/h$, and $Ph=2\pi a^2\sqrt{1-e^2}$. If one instead imposes a line density proportional to $v$, area is weighted by $v^{2}dt$, yielding $f={\sigma }_{\mathrm{tot}}(1+e^2)/[4\pi a^2(1-e^2{)}^{3/2}]$. The two distributions agree only for a circular Kepler orbit.

For an impulsive supernova kick in a binary star initially in a circular Kepler orbit, let $\beta$ be the remaining total mass fraction and $\alpha$ the kick speed divided by the original relative speed. The displayed post-kick specific orbital energy is negative for a bound pair and positive for escape. Over all directions, the velocity factor ranges from $(1-\alpha {)}^2$ to $(1+\alpha {)}^2$. Some direction leaves a bound pair exactly when $2\beta >(1-\alpha {)}^2$; some direction permits positive-energy escape exactly when $2\beta <(1+\alpha {)}^2$. Equalities are marginal parabolic Kepler orbits. The circular initial orbit and impulsive approximation are essential.

Let $M=M_1+M_2$, and let $\mu =M_1{M}_2/M$ be the reduced mass. The stellar distances from the center of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. For a circular Kepler orbit, ${\Omega }^2=GM/a^3$, so summing the two orbital contributions gives the circular-binary orbital angular momentumThe separation $a$ is inside the square root. This also has the required dimensions of angular momentum.

For the homologous rotating collapse, label a fluid element by $s=r/R$ and its polar angle $\theta$. Its enclosed mass $m(s)$ is constant during stellar homology, as are the dimensionless density profile and the inertia coefficient $\alpha$. Conservation of angular momentum and solid-body rotation giveThe magnitude of the centrifugal acceleration is ${\Omega }^{2}r\sin \theta$, while the Newtonian gravitational field has magnitude $Gm(s)/r^2$. ThereforeUsing only the radial component of centrifugal force replaces $\sin \theta$ by ${\sin }^2\theta$ and leaves the scaling unchanged. The centrifugal force vanishes on the rotation axis; the central point is understood through the limiting field rather than a ratio of two zero forces.

At the equator of the cloud, critical rotation of a spherical cloud means ${\Omega }_c^2{R}_c=GM/R_c^2$, where $R_c=R_{\mathrm{crit}}$. ConsequentlyImmediately after the first fission, each daughter has mass $M/2$ and radius $a/2$ because the spheres touch. Their orbital moment of inertia is $M{a}^2/4$, while their combined spin moment of inertia is $\alpha M{a}^2/4$. With a common spin and orbital frequency ${\Omega }_f=\sqrt{GM/a^3}$, the synchronous fission model for binary formation givesThe common frequency can change during fission; corotation requires equal instantaneous frequencies, not an unchanged frequency from before the rearrangement. Equating the two values of $J$ yieldsFor a nonzero positive inertia coefficient, the necessary and sufficient condition in this algebraic model for $a<R_c$ isFor example, a moment of inertia of a uniform solid sphere has $\alpha =2/5$ and gives $a/R_c=64/49>1$; it fails the required compact-fission condition. A sufficiently centrally concentrated cloud can have smaller $\alpha$.

For the next collapse, neglect exchange of spin with the unchanged outer orbit, so each daughter's spin angular momentum is separately conserved. Immediately after the first fission that spin isFor a daughter of mass $m=M/2$ to reach critical rotation of a spherical cloud at radius $R_{c,d}$, the same relation gives $S_d=\alpha m\sqrt{Gm{R}_{c,d}}$. HenceEquivalently, its centrifugal-to-gravity ratio starts at $1/4$ when its radius is $a/2$ and reaches one after contraction by a factor four. Applying the same fission rule to this daughter produces an inner pair with separationCorotation at this second fission is local to each newly formed inner pair; it is not a single common frequency for the entire four-star hierarchy.

This spin assumption matters. If tidal locking instead kept each shrinking daughter synchronized to the fixed outer orbital period throughout the intervening collapse, its angular velocity would remain fixed. Its centrifugal-to-gravity ratio would then scale as $R_d^3$ and decrease, so it would never reach the proposed second fission. This is a counterexample to extending the corotation assumption through the whole contraction. The printed ratio describes the separately spin-conserving interpretation of the repeated fission hierarchy.

The model can generate a compact hierarchy algebraically, but is not a general physical account of multiple-star formation. In the allowed range $\alpha <1/3$, one has $a^{\prime }/a<1/8$, suggesting well-separated inner and outer scales. However, rapidly rotating gas deforms, touching daughters are tidally distorted, and pressure, gas flows, dissipation and spin-orbit torques cannot generally be ignored. The assumed identical profiles, equal mass splits, instantaneous corotation and later torque-free contractions are restrictive. Star formation can involve gravitational fragmentation and redistribution of angular momentum; the toy budget neither proves that fission occurs nor predicts the population of real multiple systems.

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.

The transverse velocity is $v_{\theta }=h/r$. Subtracting its square from the vis-viva equation, and putting $u=a/r$, givesConsequentlyThe positive branch is outward motion and the negative branch is inward motion. The PDF's unsigned expression gives the radial speed, or the outward branch, rather than the signed radial velocity everywhere. For example, on the inward part of any noncircular elliptic orbit, $\dot{r}<0$.

The same result follows from $\dot{r}=(\mu e/h)\sin f$, with $f$ the true anomaly. The radicand is nonnegative precisely on $1/(1+e)\le u\le 1/(1-e)$; it vanishes at periapsis and apoapsis. For a circular Kepler orbit, $e=0$, $u=1$ and $\dot{r}=0$.

Put $A=\beta \mu /c$. The Poynting–Robertson drag acceleration has $\bar{R}=-2A\dot{r}/r^2$ and $\bar{T}=-A{v}_{\theta }/r^2$. Its work is strictly dissipative:For an unperturbed elliptic orbit, use $dt=r^{2}df/h$, $\dot{r}=(\mu /h)e\sin f$ and $v_{\theta }=(\mu /h)(1+e\cos f)$. Averaging over its orbital period $P$ givesSince $\dot{a}=2a^2\dot{\epsilon }/\mu$, $P=2\pi a^{3/2}/\sqrt{\mu }$ and $h^3={\mu }^{3/2}{a}^{3/2}(1-e^2{)}^{3/2}$, the Poynting–Robertson decay of a circumstellar orbit isFor a circular Kepler orbit this becomes $-2A/a$. This is a secular first-order average, not an exact instantaneous decay law. If static radiation pressure is appreciable, use a bound conservative orbit with central parameter ${\mu }_{\mathrm{eff}}=(1-\beta )\mu$, requiring $0\le \beta <1$, and define its orbital elements accordingly; repeating the work calculation cancels ${\mu }_{\mathrm{eff}}$ and leaves the same coefficient $A=\beta G{M}_{\star }/c$.

At the encounter radius $r=a_p$, the planet's circular Kepler orbit has speedThe comet's speed follows from the vis-viva equation:Its component parallel to the planet's prograde tangential velocity is fixed by its specific angular momentum,The relative velocity therefore satisfiesand hence

Put the star at the origin, take $\hat{\mathbf{z}}$ along the line of sight toward the observer, and let $\hat{\mathbf{x}}$ point North in the plane of the sky. The $\hat{\mathbf{y}}$ direction is then $90^{\circ }$ anticlockwise from North. Draw the ascending node on the sky plane at position angle $\Omega$, tilt the circular Kepler orbit through the orbital inclination $I$ about that nodal line, and place the planet an angle $f$ along the orbit from the ascending node. The direction of increasing $f$ must cross the sky plane toward $+\hat{\mathbf{z}}$ at $f=0$.

A convenient orthonormal basis in the orbital plane isHere ${\mathbf{e}}_1$ points along the nodal line and ${\mathbf{e}}_2$ fixes the required sense of motion. The planet lies at $\mathbf{r}=a(\cos f{\mathbf{e}}_1+\sin f{\mathbf{e}}_2)$; this relation also specifies all the labels required in the sketch.

The parent body's circular Kepler orbit has speed $v_K=\sqrt{\mu /a_p}$. Release without a velocity impulse therefore givesThe release point is the dust orbit's point of closest approach, where $\theta -\varpi =\pi$. Substituting $r=a_p$ in the orbit equation givesThe longitude of periapsis $\varpi$ specifies the symmetry axis of the conic. With this signed repulsive-force convention, it points opposite the release direction, so the closest point is at $\theta =\varpi +\pi$.

Let $v_K=\sqrt{\mu /a}$ be the speed of the circular Kepler orbit at radius $a$. The release velocity relative to the planetesimal isThe particle starts at the same position as its parent, so the change in specific orbital energy isUsing the velocity components from part (c),Since $C=-\mu /(2a)$ and $C^{\prime }=-\mu /(2a^{\prime })$, rearrangement gives

For $T_p=5/2$ and $I=0$, the fixed-Tisserand parameter curve begins atIt falls smoothly toand then rises asymptotically back toward $e=1$ as $a/a_p\to \infty$. Since $T_p<3$, this coplanar locus never reaches a circular Kepler orbit.

Uniform ${\rho }_0,p_0,$ and ${\mathbf{B}}_0=B_0\hat{\mathbf{z}}$ obey the unperturbed ideal magnetohydrodynamics equations whenIndeed $({\mathbf{v}}_0\cdot \nabla ){\mathbf{v}}_0=0$, while the Coriolis and tidal terms cancel:

For perturbations proportional to $e^{i(\omega t-kz)}$, the horizontal velocity and magnetic perturbations decouple from the compressive variables. Their linear equations areEliminating $B_x,B_y,v_x,v_y$ and writing the Alfvén speed as $v_A^2=B_0^2/(4\pi {\rho }_0)$ givesSince $d{\Omega }_0^2/d\log r=-2q{\Omega }_0^2$, a root has ${\omega }^2<0$ precisely when the constant term is negative:This is the magnetorotational instability criterion.

For a circular Kepler orbit, $q=3/2$. With $K=k^2{v}_A^2$, the unstable branch isMinimizing it givesand henceThe e-folding time is of order the dynamical time, so the growth is rapid: several e-foldings occur in one orbit.

Instability requires $k^2{v}_A^2<3{\Omega }_0^2$, making the critical wavelength of the magnetorotational instabilityFor a thin isothermal disk, $H=c_{s,0}/{\Omega }_0$. Setting ${\lambda }_{\mathrm{crit}}=2H$ gives $v_A^2=3c_{s,0}^2/{\pi }^2$, and thereforeAbove this field strength the shortest unstable vertical MRI wavelength exceeds the full disk thickness, so no such vertical mode fits inside the disk. Magnetic tension then stabilizes this local mode; the result corresponds to a magnetic-to-gas pressure ratio $B^2/(8\pi p)=3/(2{\pi }^2)$.

With $A=\beta G{M}_{\star }/c$ constant, average the drag work $\dot{\epsilon }=-A(2{\dot{r}}^2+v_{\theta }^2)/r^2$ over a bound elliptic orbit. Using $dt=r^{2}df/h$ gives the integral $\pi (2+3e^2){\mu }^2/h^2$, where $\mu$ is the conservative central parameter. Combining with $\dot{a}=2a^2\dot{\epsilon }/\mu$ givesThis is a weak-drag secular average. It also holds when the conservative parameter is $(1-\beta )G{M}_{\star }$, provided a bound orbit exists; the conservative parameter cancels but the drag amplitude remains $A$. For a circular Kepler orbit, $a^2(t)=a_0^2-4At$.

A satellite remaining above a fixed point of a uniformly rotating spherical planet must have a prograde equatorial circular Kepler orbit with angular speed $2\pi /T$. Balancing gravitational acceleration and centripetal acceleration gives the displayed radius and speed $v=2\pi r/T$. The satellite's own mass cancels. A physically realizable external orbit additionally requires this radius to exceed the planet's radius.