= Solution
At latitude $\theta_0$, resolve the planetary angular velocity as
$$
\boldsymbol\Omega=(0,\Omega\cos\theta_0,
\Omega\sin\theta_0).
$$
For velocity $(u,v,w)$, the <Coriolis acceleration> is
$$
2\boldsymbol\Omega\times\mathbf u
=2\Omega
\left(
w\cos\theta_0-v\sin\theta_0,
u\sin\theta_0,
-u\cos\theta_0
\right).
$$
The <traditional approximation> drops the terms involving the horizontal rotation component $\Omega\cos\theta_0$. The retained horizontal force is therefore
$$
f\widehat{\mathbf z}\times\mathbf u_h,
\qquad
f=2\Omega\sin\theta_0.
$$
Its direct velocity-scale requirement is
$$
W\cos\theta_0\ll U\sin\theta_0.
$$
For an <incompressible flow>, $W/U=O(H/L)$, so a sufficient condition is
$$
\boxed{\frac HL|\cot\theta_0|\ll1},
$$
together with the small aspect ratio that makes <hydrostatic pressure> the leading vertical momentum balance. The approximation consequently becomes delicate near the equator.
The <beta plane> is the local <Taylor expansion>
$$
f(y)=2\Omega\sin(\theta_0+y/a)
=f_0+\beta y+O\!\left(\Omega L^2/a^2\right),
$$
where
$$
f_0=2\Omega\sin\theta_0,
\qquad
\beta=\frac{2\Omega\cos\theta_0}{a},
$$
and $a$ is the planetary radius. It requires a local Cartesian region,
$$
\boxed{L/a\ll1},
$$
and $H/a\ll1$. Away from the equator, treating the variation as a perturbation of an <f-plane> also requires $\beta L/|f_0|\ll1$; near the equator one instead retains the linear term as the leading Coriolis parameter.
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