= Solution
A <star> of <declination> $\delta=\theta$ crosses the <zenith>. Its direction rotates around the <celestial pole> at $\Omega_{\rm sid}$, but its actual <angular velocity> on the <celestial sphere> is $\Omega_{\rm sid}\cos\theta$. Consequently the small-angle crossing-gap estimate becomes
$$
\boxed{\phi_{\rm full}\simeq\frac{\pi\cos\theta}{\omega T_{\rm sid}},\qquad
\phi_{\rm radius}\simeq\frac{\pi\cos\theta}{2\omega T_{\rm sid}}.}
$$
For clarity, the factor $\cos\theta$ is the radius of the <star>'s daily circle on a unit <celestial sphere>; it does not modify the sidereal <hour angle> rate.
The estimate assumes the required slew is nearly $180^\circ$. A more exact ideal symmetric reacquisition calculation is possible. Let the endpoints have <hour angles> $-H,+H$ and suppose $0<\theta<\pi/2$, $H$ small enough that both endpoints are above the <astronomical horizon>. The horizontal direction components are $n=\sin\theta\cos\theta(1-\cos H)$ and $w=\pm\cos\theta\sin H$. Their shortest <azimuth> separation is
$$
\Delta\eta(H)=2\arctan\!\left(\frac{\cot(H/2)}{\sin\theta}\right).
$$
The minimum ideal gap satisfies $2H\Omega_{\rm az}/\Omega_{\rm sid}=\Delta\eta(H)$; its angular separation is $\phi_{\rm full}=2\arcsin(\cos\theta\sin H)$. Expanding for a fast drive gives the boxed expression. At an equatorial site $\Delta\eta=\pi$ exactly. At a geographic pole the direction with $\delta=\theta$ is stationary, so the crossing argument is inapplicable and the limit is zero. A full near-<zenith> rate-limited footprint still requires specifying the trajectory and drive model.
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