 Solution

The printed line-density statement is incorrect for a phase-mixed orbit. A steady line density on a Kepler orbit is inversely proportional to speed: for a cross-sectional-area current $J_{\sigma }$, the area per unit arc length is $d\sigma /ds=J_{\sigma }/v$. Equivalently, phase mixing gives $d\sigma =({\sigma }_{\mathrm{tot}}/P)dt$, uniform in mean anomaly, where $P=2\pi \sqrt{a^3/\mu }$ is the orbital period.

For an optically thin population of blackbodies in radiative equilibrium, a fragment absorbs $\sigma L_{\star }/(4\pi r^2)$ and reradiates the same luminosity. Thus the fractional luminosity of a phase-mixed eccentric wire isThe specific angular momentum gives $dt/r^2=d\theta /h$, so the integral is $2\pi /h$. Since $Ph=2\pi a^2\sqrt{1-e^2}$,This derives the intended result after explicitly correcting the density to $1/v$.

The error is consequential. If one instead imposes the literal density $d\sigma /ds\propto v$, its time weighting is $v^{2}dt$. Using $⟨v^2⟩=\mu /a$, $⟨r^{-2}⟩=[a^2\sqrt{1-e^2}{]}^{-1}$ and $⟨r^{-3}⟩=[a^3(1-e^2{)}^{3/2}{]}^{-1}$ givesAt $e=1/2$ this is $5/3$ times the printed result. The inverse-radius averages can also be obtained directly with eccentric anomaly $E$, $r=a(1-e\cos E)$ and $dt=(1-e\cos E)dE/n$.