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Write $u=e^2$ and use the torque balance from part (ii) to eliminate the resonant argument. The resonant contribution isAdding the Poynting–Robertson drag contribution gives the dissipative resonant eccentricity equilibrium equationwhereFor $5p>q$, integrating with $u(0)=e_0^2$ givesHere $a$ is the fixed resonant semi-major axis and the expression describes averaged evolution, neglecting the small resonant-argument libration. The physical bound-orbit interpretation also requires $B<1$; a quantitatively controlled second-order calculation needs $B\ll 1$.

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.