= Solution
Put $m=m_1+m_2$, $\mu=Gm$, $n=\sqrt{\mu/a^3}$, $v=na$, and $\theta=nt$. Work instantaneously in the inertial frame in which the <centre of mass> is at rest. The unperturbed circular relative orbit has
$$
\mathbf r=a(\cos\theta,\sin\theta),\qquad
\mathbf v=v(-\sin\theta,\cos\theta).
$$
The individual velocities are $\mathbf v_1=-(m_2/m)\mathbf v$ and $\mathbf v_2=(m_1/m)\mathbf v$. Linear <gas drag> gives $\mathbf F_i/m_i=-(\mathbf v_i-v_g\hat{\mathbf y})/t_i$, where $t_i=t_{si}$ are the <aerodynamic stopping times>. Define
$$
\kappa=\frac{m_1/t_2+m_2/t_1}{m},\qquad
\Delta=\frac1{t_2}-\frac1{t_1}.
$$
Then the <gas drag on a planetesimal binary> is
$$
\boxed{f_x=\kappa v\sin\theta,\qquad
f_y=-\kappa v\cos\theta+\Delta v_g.}
$$
Equivalently, $\mathbf f=-\kappa\mathbf v+\Delta v_g\hat{\mathbf y}$. If the <centre of mass> moves, $v_g\hat{\mathbf y}$ here is its relative gas velocity, held approximately fixed during one short binary orbit.
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