Critical belt mass for Yarkovsky removal
= Critical belt mass for Yarkovsky removal
{title2=$M_y=\frac{3^{3/4}}4 AB^{-3/4}C^{-1/4}$}
For $t_c=(A/M)D^{1/2}$, the ratio $t_y/t_c$ has its minimum at $D_*=\sqrt{3C/B}$. Removal beats collisions somewhere exactly when $M<M_y=(3^{3/4}/4)AB^{-3/4}C^{-1/4}$, provided the relevant diameters are present. This criterion uses the collision-only cascade as a reference.