Past exam of the physics B course of the University of Cambridge / 2010 / Paper 1 / Section A / 2 / Solution

Use local east, north and up axes. At latitude $\lambda$, $\mathbf{\Omega }=(0,\Omega \cos \lambda ,\Omega \sin \lambda )$ and eastward velocity is $(v,0,0)$. The Coriolis acceleration isIts horizontal part is southward, so the lift must acquire a northward component: bank left, with the north wing lower. For the requested local Coriolis correction, let the vertical lift balance $mg$. The Coriolis bank angle for eastward flight isUse $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\Omega =7.2921\times 10^{-5}{\mathrm{s}}^{-1}$. Including the upward Coriolis term replaces $g$ by $g-2\Omega v\cos \lambda$ and changes the answer by only about $0.1\%$. This is the horizontal Coriolis compensation, not a full calculation of a curved constant-latitude flight path. The rotation rate follows Earth's sidereal period.^[ref]