= Solution
{c}
Use the positive rotation-rate convention implicit in the northward <planetary vorticity> gradient. One <Pluto> day is $T_P=6.4\times86400=552960\,\mathrm s$. At the equator,
$$
r=T_P^{-1}=1.8084\times10^{-6}\,\mathrm{s^{-1}},\qquad
\beta=\frac{2\Omega}{a_p}=\frac{4\pi}{T_Pa_p}=1.9129\times10^{-11}\,\mathrm{m^{-1}\,s^{-1}}.
$$
The width of the <western boundary current> is therefore
$$
\boxed{\delta_W=\frac r\beta=\frac{a_p}{4\pi}\simeq9.45\times10^4\,\mathrm m\simeq95\,\mathrm{km}.}
$$
Here $b=a_p\pi/3\simeq1244\,\mathrm{km}$ is the full meridional scale, and $L=a_p(50\pi/180)\simeq1037\,\mathrm{km}$. For these data the exact exponents give $\lambda^{-1}\simeq89.7\,\mathrm{km}$, $\lambda L\simeq11.56$ and $\gamma L\simeq0.593$. The western layer is narrow, but $\gamma L\ll1$ is only marginal. The leading estimates below should therefore be distinguished from evaluation of the exact linear model.
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