Question 6.16: Determine the Coriolis deflection of a projectile fired east...

Since the projectile is fired due east (the j direction in Fig. 6.28), the angle of elevation is  β.  Then  \alpha=\pi / 2, \gamma=\frac{\pi}{2}  –  \beta,  and (6.115b) becomes

First consider the case when the Earth’s rotation is neglected. With Ω = 0, (6.116a) reduces to the classical elementary solution for projectile motion:

\mathbf{x}(P, t)=V t \cos \beta \mathbf{j}  +  V t\left(\sin \beta  –  \frac{g t}{2 V}\right) \mathbf{k}.               (6.116b)

The time of flight  t^*=(2 V \sin \beta) / g  for which  z\left(t^*\right)=0  is then used to find the projectile’s range  r \equiv y\left(t^*\right),  namely,

r=\frac{V^2}{g} \sin 2 \beta \text {. }             (6.116c)

Now consider the Earth’s rotational effect. Equation (6.116a) indicates a lateral (i-directed) Coriolis deflection of the projectile normal to its east directed range line, toward the south in the northern hemisphere and toward the north in the southern hemisphere. To find the deflection, we need the projectile’s time of flight t* given by   z\left(t^*\right)=0  in (6.116a). To the first  order in Ω, we find for  V \Omega / g \ll 1,

t^*=\frac{2 V \sin \beta}{g}\left(1  +  \frac{2 V \Omega \cos \beta \cos \lambda}{g}\right).                     (6.1 16d)

The lateral deflection  x^* \equiv x\left(t^*\right)  and the range  r^* \equiv y\left(t^*\right)  are now determined by the remaining components in (6.1 16a). The projectile’s Coriolis deflection to first order in Ω, with  V \Omega / g \ll 1,   is thus given by

x^*=\frac{4 \Omega V^3 \sin ^2 \beta \cos \beta \sin \lambda}{g^2}                 (6.116e)

The reader will explore the range effect in the exercise.