Refer to Example 10.3 concerning the deflection from the plumb line of a particle falling in Earth's gravitational field. Take g to be defined at ground level and use the zeroth order result for the time-of-fall, T=2h/g. Perform a calculation in second approximation (i.e., retain terms in ω2) and calculate the southerly deflection. There are three components to consider: (a) Coriolis force to second order (C1), (b) variation of centrifugal force with height (C2), and (c) variation of gravitational force with height (C3). Show that each of these components gives a result equal to
Cigh2ω2sinλcosλ
with C1=2/3, C2=5/6, and C3=5/2. The total southerly deflection is therefore (4h2ω2sinλcosλ)/g.