A_{\rm Bondi}/B \sim 1/h > 1$$ and consumption dominates repulsion in setting the gap depth, independent of planet mass in the sub-thermal regime. On the other hand, for $A=A_{\rm Hill}$, $$\label{eqn:ab_Hill}
A_{\rm Hill}/B \sim m^{-4/3} h^3 $$ which says that for super-thermal planets that are massive enough, repulsion dominates consumption ($A_{\rm Hill}/B < 1$).

We may also solve for the relative accretion rates: $$\begin{aligned}
\frac{\dot{M}_{\rm p}}{\dot{M}_+} &\sim \frac{A\Sigma_{\rm p}}{\Sigma_+\nu} \sim  \frac{A/\nu}{1+(A+B)/\nu} \label{eq:m_dot_ratio} \\
 \frac{\dot{M}_-}{\dot{M}_+} &\sim \frac{\Sigma_-}{\Sigma_+} \sim \frac{1+B/\nu}{1+(A+B)/\nu} \,. \label{eqn:mdot-0}\end{aligned}$$ A couple example limiting cases of (\[eq:m\_dot\_ratio\]) and (\[eqn:mdot-0\]) are as follows. If we take $A/B = A_{\rm Bondi}/B \sim 1/h > 1$ and further assume that $B/\nu > 1$ so that the inner gap contrast is significant (equation \[eqn:oom1\]), we find $$\begin{aligned}
    \frac{\dot{M}_{\rm p}}{\dot{M}_+} &\sim 1 - B/A_{\rm Bondi} \sim 1 - h \label{eqn:mdotp}\\
    \frac{\dot{M}_-}{\dot{M}_+} &\sim B/A_{\rm Bondi} \sim h \label{eqn:mdot-}\end{aligned}$$ which says that the planet consumes nearly all of the mass supplied to it by the outer disc, leaving behind a fraction $h$ to feed the inner disc. If instead we take $A/B = A_{\rm Hill}/B$ and further assume $B > A_{\rm Hill} > \nu$ (repulsion-limited and deep gap), then $$\begin{aligned}
    \frac{\dot{M}_{\rm p}}{\dot{M}_+} &\sim A_{\rm Hill}/B \sim m^{-4/3} 