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ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s2/3/2/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S2 / 3 / 2 / Solution by

Codex  0 2026-10-25

The stated one-sided temperature contrast is

β = 0.00105

, so the Doppler dipole gives

v = 0.00105 c ≃ 3.15 \times 1 0^{5} m s^{- 1} = 315 km s^{- 1} .

(1)

The motion is toward the hotter region. The contrast already gives one side relative to the mean; the hot-to-cold difference would be twice as large.

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