 Critical belt mass for Yarkovsky removal

For $t_c=(A/M)D^{1/2}$, the ratio $t_y/t_c$ has its minimum at $D_{\ast }=\sqrt{3C/B}$. Removal beats collisions somewhere exactly when $M<M_y=(3^{3/4}/4)A{B}^{-3/4}{C}^{-1/4}$, provided the relevant diameters are present. This criterion uses the collision-only cascade as a reference.