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The impact gives fragments a spread of orbital energy and specific angular momentum. Their resulting spread of mean motion lets Keplerian shear stretch a compact dust clump into an arc and then a ring, while the spread of orbital frequencies causes phase mixing. Size-dependent radiation-pressure coefficients immediately place small grains on different eccentric or even radiation-pressure blowout orbits; Poynting–Robertson drag and stellar-wind drag then alter their orbits on longer timescales. Further collisions grind or disperse the clump, and planetary perturbations can accelerate mixing.

These processes depend strongly on $D_f$. Small grains have larger radiation-force-to-gravity ratios and generally shorter collisional or drag lifetimes, while larger fragments remain closer to the parent orbit but can preserve a velocity-dispersion-driven clump for longer. The lifetime also depends on collision location, ejection velocities, optical depth, and orbital radius. A universal fixed $\Delta t$ is therefore a useful population-model approximation, not a literal property of every collision; a size- and event-dependent lifetime distribution is more realistic.

Since the binary separation and mean motion are both one, the Kepler third law givesThe center of mass condition puts the bodies atin the rotating reference frame. Thus the sketch has $M_1$, $O$, and $M_2$ in that order along the $x$ axis, with $P=(x,y)$, andIn the inertial sketch this entire configuration rotates uniformly about $O$.

If an exterior planetesimal reaches conjunction just before apoapsis, its radius is increasing. The closer pre-conjunction pull from the trailing inner planet removes more specific angular momentum than the more distant post-conjunction pull restores. Its semi-major axis falls, its mean motion rises, and the next conjunction moves later in its orbit toward apoapsis. A conjunction just after apoapsis produces the reverse imbalance: the stronger post-conjunction pull adds angular momentum, raises the semi-major axis, and shifts the next conjunction earlier. The conjunction phase is therefore restored toward apoapsis.

The symmetric configuration has one of the three conjunction branches at apoapsis and the other two symmetrically placed. It hasso the resonant argument librates about $180^{\circ }$.

The radiation-pressure coefficient reduces the dust grain's effective stellar gravitational parameter to $(1-\beta )G{M}_{\ast }$. At the exterior 5:4 mean-motion resonance, $n_d/n_p=4/5$, so mean motion givesThis is a first-order mean-motion resonance. Its leading disturbing-function term is therefore linear in the small orbital eccentricity:Thus its dimensionless strength is of order $(M_p/M_{\ast })e_d$, up to the Laplace-coefficient combination $f_{5:4}$, and its resonant argument varies slowly near commensurability.

The specific angular momentum of an elliptic Kepler orbit is $h=\sqrt{\mu a(1-e^2)}$. At periapsis, $r=a(1-e)$, soThe mean motion understates instantaneous angular speed near periapsis. A particle in an exterior mean-motion resonance can briefly overtake the planet in angle despite completing full orbits more slowly.

A grain with radiation-pressure coefficient $\beta <1$ feels effective central gravitational parameter $G{M}_{\star }(1-\beta )$. Combining its mean motion with the planet's mean motion at an exterior-type $(j+1):j$ frequency commensurability gives the displayed semi-major axis. The commensurability lies geometrically exterior only for $j<\sqrt{1-\beta }/(1-\sqrt{1-\beta })$. Radiation changes the grain's attraction, so the same frequency ratio need not imply the usual radial ordering.

A secular tidal torque is the torque remaining after averaging the changing tidal interaction over fast orbital and rotational motion. The misaligned tidal deformation lets the companion's gravity change the spin angular momentum; the orbit receives the opposite torque. Tidal dissipation converts mechanical energy into heat, while total angular momentum is conserved in an isolated system. Thus the spin-orbit coupling exchanges angular momentum rather than destroying it. ^[ref].

For intuition near synchronous rotation on a circular orbit, hold mean motion $n$ and moment of inertia $I$ fixed and adopt the illustrative linear dissipative lawHere $\omega$ is spin angular velocity, $\theta$ is a marked surface direction in an inertial frame, and $K$ is a response coefficient, not a universal material constant. A faster spin is slowed and a slower prograde spin is accelerated. The solution and the associated mechanical power loss areThe orbital contribution to this power is $-n⟨\tau ⟩$, and the spin contribution is $\omega ⟨\tau ⟩$; their sum becomes heat. Holding $n$ fixed is a local approximation to the coupled evolution. Actual secular tidal torque depends on the deformation response and orbital eccentricity, so this simple law does not establish which spin-orbit resonance an eccentric body captures into.

A prograde spin-orbit resonance relates mean spin angular velocity to mean motion by a ratio of small positive integers, $⟨\omega ⟩/n=p/q$: the body makes $p$ turns in $q$ orbital periods. Synchronous rotation is $1:1$; Mercury's $3:2$ spin-orbit resonance is nonsynchronous. The relevant combination of spin and orbital angles must stay bounded around a preferred orientation, not merely have a near-integer frequency ratio. For an elongated body on an eccentric orbit, the companion's gravitational torque on the permanent shape can sustain the resonance while tidal dissipation permits capture. ^[ref].

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.