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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-316/1/vii/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 316 / 1 / vii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

Use an edge-on orbit and an equatorial, effectively central chord. Near the foreground crossing the radial velocity is along the line of sight; the projected transverse speed is

v_{⊥} = h / r

. In the small stellar-angular-radius limit,

r ≫ R_{⋆}

, the crossing time is

2 R_{⋆} / v_{⊥} = 2 R_{⋆} r / h

The cross-sectional-area current therefore places an area

σ_{front} = (2 f h) \frac{2 R _{⋆} r}{h} = 4 f R_{⋆} r

(1)

in front of the stellar disk. For an optically thin wire and a uniformly bright stellar disk, transit dimming by an optically thin orbital wire gives

\frac{Δ F}{F _{⋆}} = \frac{σ _{front}}{π R _{⋆}^{2}} = \frac{4}{π} \frac{r}{R _{⋆}} f .

(2)

This is a linear occultation estimate, requiring negligible overlap and a fractional dimming much smaller than one. A finite impact parameter shortens the chord; limb darkening, finite wire thickness and variation of

r

across a large stellar angular extent modify the coefficient. The estimate cannot be extrapolated to dimming greater than unity.

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