Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 316 / 1 / v / Solution

At periapsis, the particle's mean longitude equals its longitude of periapsis: $\lambda =\varpi$. Hence the pericentre geometry of an exterior resonance givesThe resonant argument therefore fixes the angular separation from the planet whenever the particle reaches periapsis. With $r_{\mathrm{peri}}=a(1-e)$ and planetary radius $a_p$, the actual separation isThus a restricted libration phase can prevent close encounters even when the radial ranges of the two orbits overlap: this is resonance protection.