= Solution
Use local east, north and up axes. At latitude $\lambda$, $\boldsymbol\Omega=(0,\Omega\cos\lambda,\Omega\sin\lambda)$ and eastward <velocity> is $(v,0,0)$. The <Coriolis acceleration> is
$$
-2\boldsymbol\Omega\times\mathbf v=(0,-2\Omega v\sin\lambda,2\Omega v\cos\lambda).
$$
Its horizontal part is southward, so the lift must acquire a northward component: \b[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> is
$$
\tan\theta=\frac{2\Omega v\sin\lambda}{g},\qquad
v=\frac{800}{3.6}=222.22\,\mathrm{m\,s^{-1}},\qquad
\boxed{\theta\simeq0.00310\,\mathrm{rad}=0.178^\circ.}
$$
Use $g=9.81\,\mathrm{m\,s^{-2}}$ and $\Omega=7.2921\times10^{-5}\,\mathrm{s^{-1}}$. 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.<Earth Fact Sheet by National Aeronautics and Space Administration>[Orbital parameters: sidereal rotation period 23.9345 hours]{ref}
Back to article page
