Angular momentum decay under uniform linear drag ($L(t)=L(0)e^{-kt/m}$)

Consider particles of the same constant mass $m$, interacting through equal-and-opposite central internal forces, with external forces $F_i=-k{\dot{r}}_i$ for constant $k$. About a fixed point $a$, internal pair torques cancel and the external torque is $G_a=-k{\sum }_i(r_i-a)\times {\dot{r}}_i=-(k/m)L_a$. Hence ${\dot{L}}_a=-(k/m)L_a$ and the displayed exponential decay follows. The same result holds for angular momentum about the centre of mass, because the moving-origin term $-\dot{R}\times P=-\dot{R}\times M\dot{R}$ vanishes. Unequal masses do not generally permit factoring out a single $k/m$.