Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 4 / 10B (Dynamics and Relativity)

Apply the rotating-frame derivative formula first to position and then to its inertial velocity:The angular velocity is constant, so there is no term involving $\dot{\mathbf{\omega }}$. By Newton's second law in the inertial frame, the inertial acceleration is $\mathbf{F}/m$. Rearrangement gives the equation of motion in a rotating frame:The two additional accelerations are the Coriolis acceleration and centrifugal acceleration.

At $P$, use the local right-handed basis ${\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U$ pointing east, north and vertically outward. The Earth's rotation vector has componentsInitially $\dot{\mathbf{r}}=0$ in the rotating reference frame, so the Coriolis acceleration is zero. Evaluating the centrifugal acceleration givesThe gravitational acceleration is $-g{\mathbf{e}}_U$. HenceThe small angle $\delta$ from the downward vertical therefore satisfiesFor the northern hemisphere, $0\le \theta \le \pi /2$, this is the printed expression without absolute values. More generally, the printed signed meridional expression specifies the direction of the tilt; an unsigned angle uses its absolute value. The horizontal component is toward the equator and vanishes at both poles and at the equator.

Neglect the centrifugal acceleration, which is of order ${\omega }^2$, and use the local flat-surface approximation. To zeroth order in $\omega$, the height above $P$ and flight time areInserting this vertical velocity into the Coriolis acceleration gives the eastward componentWith $x(0)=\dot{x}(0)=0$, two integrations yieldto first order. Horizontal velocity is already of order $\omega$, so its effect on vertical acceleration is of order ${\omega }^2$. Thus the flight time remains $T_0$ at this order. The Coriolis deflection of a vertical projectile at landing isThe displacement is westward, and its magnitude is $4\omega v^3\sin \theta /(3g^2)$. The horizontal speed remains westward during the flight: $\dot{x}=-2\omega \sin \theta z_0(t)$. This explains why the eastward acceleration on descent does not cancel the earlier displacement.