= Solution
On the corrected steady <phase-mixed orbit>, each fragment crosses a fixed orbital longitude once per <orbital period>. Hence the <cross-sectional-area current> through that point is
$$
J_\sigma=\frac{\sigma_{\rm tot}}{P}.
$$
From the <fractional luminosity of a phase-mixed eccentric wire>, $\sigma_{\rm tot}=4\pi a^2\sqrt{1-e^2}\,f=2Phf$. Therefore the area rate at the foreground crossing of the line of sight is
$$
\boxed{\dot\sigma=J_\sigma=2fh}.
$$
Only the foreground segment blocks the <star>; the far-side intersection does not add another occultation current. The orbital area current is constant because the line density varies as $1/v$, even though the local orbital speed varies.
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