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For an elliptic Kepler orbit, Kepler's equation relates mean anomaly $M$ and eccentric anomaly $E$ by

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.