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For the $5:3$ exterior mean-motion resonance, $p=3$, $q=2$. The truncated dissipative resonant eccentricity equilibrium givesFor weak drag and $C>0$, take the resonant-argument libration centre ${\varphi }_0\simeq \pi$. The pericentre geometry of an exterior resonance then places the three pericentres at angles $\pi /3$, $\pi$, and $5\pi /3$ relative to the planet. The trajectory closes after three particle orbits and five planet orbits. It moves retrograde in the rotating frame on average but develops a prograde loop near each periapsis, because the periapsis angular speed exceeds the planet's angular speed.

For a finite drag offset ${\varphi }_0=\pi +\arcsin S$, rotate the whole particle pattern through $\arcsin S/3$. The sketch uses the eccentricity requested by the second-order model; $e\simeq 0.555$ is not small, so the curve is an illustration of that approximation rather than a quantitatively accurate integration of the full dissipative resonance.

The specific angular momentum is $h=\sqrt{G{M}_{\star }a(1-e^2)}$, so the periapsis angular speed isAt the exterior mean-motion resonance, neglecting radiation pressure, $n/n_p=p/(p+q)$. Thus at the limiting orbital eccentricity $e=\sqrt{2q/(5p-q)}$,The condition for overtaking the planet in angle isFor the formal second-order equilibrium, it holds for every physical bound value $0<q<5p/3$. To see this, rewrite $q/p=5e^2/(2+e^2)$, so $p/(p+q)=(2+e^2)/(2+6e^2)$. For $0<e<1$, $\sqrt{1+e}/(1-e{)}^{3/2}>1+2e$, whileThe instantaneous angular motion is therefore prograde relative to the planet near periapsis, although its mean relative motion is retrograde. The range $q\ll p$ is the regime where the eccentricity expansion is most reliable.