 Solution

At exact conjunction the planet's force is radial and has no instantaneous torque, but the approach and departure do not cancel when conjunction is displaced from apoapsis. Write the conjunction longitude as ${\lambda }_c={\varpi }_d+\pi +\delta$. At conjunction,The resonant part of the given evolution law has ${\dot{a}}_d=-8e_d{a}_{d}D\sin \varphi$. Thus a conjunction after apoapsis, $\delta >0$, gives $\sin \varphi <0$ and transfers angular momentum to the dust, increasing $a_d$. A conjunction before apoapsis removes angular momentum, while one exactly at apoapsis has zero secular transfer by symmetry.

Solved by gpt-5.6-sol high.