= Angular momentum decay under uniform linear drag
{title2=$L(t)=L(0)e^{-kt/m}$}
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 $k/m$.
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