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The sudden potential change leaves the star's position and velocity unchanged. Its new energy is thereforeThus $E_1>0$ precisely when $r<a$. Parametrize the orbit by eccentric anomaly $\eta$:A phase-mixed orbit is uniform in the mean anomaly ${\theta }_r$. For $e>0$, the condition $r<a$ is $-\pi /2<\eta <\pi /2$ modulo one period. The corresponding mean-anomaly interval has lengthHenceThe energy increase $G(M_0-M_1)/r$ is largest near periapsis, so stars are more readily unbound there than near apoapsis. A perfectly circular orbit is the measure-zero marginal case $E_1=0$ at every phase.

At conjunction, the planetesimal's mean anomaly isIf ${\varphi }_1=\pi +\delta$ with $|\delta |\le \Delta {\varphi }_1\le \pi$, the conjunction branch that can come closest to periapsis hasLet $E_{min}$ solve Kepler's equationAt that phase the orbital radius is $r_{min,c}=a(1-e\cos E_{min})$. A geometrical close encounter is possible only ifwhere $R_{\mathrm{enc}}$ may be chosen as the planet's Hill radius or another encounter distance. With a point planet, set $R_{\mathrm{enc}}=0$. This implicit inequality is the requested eccentricity constraint as a function of $\Delta {\varphi }_1$.

The weaker necessary condition that the orbits cross isIt becomes sufficient for phase access only as $\Delta {\varphi }_1\to \pi$, when a conjunction can approach periapsis. Smaller libration amplitude keeps conjunctions farther from periapsis and requires a larger eccentricity than this orbit-crossing bound.

For a Kepler orbit with pericentre distance $q$, apocentre distance $Q$, and semi-major axis $a=(q+Q)/2$, a uniform mean anomaly givesThe two radial passages per orbital period are included, and ${\int }_q^Q{p}_r(r)dr=1$.

For an eccentric outer body and circular inner perturber, the exterior resonant argument is $\varphi =(p+q){\lambda }_{\mathrm{out}}-p{\lambda }_{\mathrm{in}}-q\varpi$. At equal mean longitudes $\Lambda$, conjunction phases areThe synodic period and coprime $p,q$ determine the visitation order, with advance $2\pi p/q$. For actual true alignment, write $x=f$ and $y=M(f)$ for the eccentric body's true anomaly and mean anomaly. Then $\varphi =(p+q)y-px$ instead of $qx$. The evenly spaced directions are a leading-small-eccentricity approximation.

True longitude is the conventional orbital angle formed from the longitude of ascending node, argument of periapsis and true anomaly. For a coplanar Kepler orbit it is the actual polar longitude of the position vector. For an inclined orbital plane the sum of angles is generally not the projected azimuth. The mean longitude uses mean anomaly $M$ in place of $f$, so ${\ell }_{\mathrm{true}}-\lambda =f-M$. For small orbital eccentricity, Kepler's equation gives $E=M+e\sin M+O(e^2)$, while the eccentric-to-true anomaly relation gives $f=E+e\sin E+O(e^2)$; hence $f-M=2e\sin M+O(e^2)$. This periodic difference matters when separating a slow resonant argument from a rapidly varying geometric angle.