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

The stated one-sided temperature contrast is $\beta =0.00105$, so the Doppler dipole givesThe 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.