 Solution

Using $q=a(1-e)$ and $Q=a(1+e)$, differentiate each apsidal distance. The apsidal evolution under Poynting–Robertson drag isBoth the pericentre distance and apocentre distance decrease. Since $a=(Q+q)/2$ and $e=(Q-q)/(Q+q)$, the same rates areTheir ratio isIn particular $Q\gg q$ gives $dQ/dq\simeq (5/3)(Q/q{)}^2$, whereas the circular limit gives slope $1$.