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Homework Statement:

A ball is dropped from the top of the tower of Pisa (which of course is not vertical). Neglect air resistance and take into account Earth's rotation for a) and b).
a) Show that its path deviates from that of a plumbline and that the ball is drifted to the east a distance (##d_p## stands for traveled distance from the tower of Pisa):
$$d_p = \sqrt{\frac{8 h^3}{9g}}\Big(\Omega \sin \beta \Big)$$
b) A ball is now dropped from a vertical tower. Show that the traveled distance (##d_v## stands for traveled distance from the vertical tower) is
$$d_v = \sqrt{\frac{2 h^3}{g}}\Big(\Omega \sin \beta\Big)$$
Note: ##\Omega## is the angular velocity with respect to the inertial frame, the angle ##\beta## is the colatitude of the origin ##O## and ##h## is the height from which the ball is released. The origin ##O## is taken at the point where the ball is released.
Reference: Gregory's Classical Mechanics; Rotating reference frames (chapter 17)
Relevant Equations:

Projectile equation on a rotating Earth:
$$m\Big( \frac{d \mathbf v}{dt} + 2 \mathbf \Omega \times \mathbf v \Big) = mg \mathbf k$$
Where ##\mathbf k## is the unit vector in the apparent vertical direction.
Earth's angular velocity (please see attached image if you want to know why):
$$\mathbf \Omega = (\Omega \sin \beta \mathbf) \mathbf i + (\Omega \cos \beta) \mathbf k$$
a) Let's first analyse the plumbline.
When it is in equilibrium, it follows from the projectile equation on a rotating Earth that:
$$\mathbf 0 = mg \mathbf k + \mathbf T$$
Thus, the tension in the plumbline is ##mg## and the string is parallel to the apparent vertical.
Let's now analyse the ball. Suppose that the apparent vertical through ##O## meets the ground at the point ##(0, 0, h)##. Let us have the following initial conditions: ##\mathbf v = \mathbf 0## and ##\mathbf r = \mathbf 0## when ##t = 0##. As ##\mathbf \Omega## is constant, we can integrate projectile equation on a rotating Earth with respect to time to get:
$$\frac{d \mathbf r}{dt} +2 \mathbf \Omega \times \mathbf r = gt \mathbf k + \mathbf C$$
Applying initial conditions we get:
$$\frac{d \mathbf r}{dt} +2 \mathbf \Omega \times \mathbf r = gt \mathbf k$$
If we integrate again with respect to time we end up with:
$$\mathbf r (t) =  \frac 1 2 g t^2 \mathbf k  2 \mathbf \Omega \times \int_{0}^{t} \mathbf r(t') dt' \ \ \ \ (1)$$
Let's get a solution for the above equation by using the mathematical method called iteration.
The zeroth order approximation corresponds to the nonrotating Earth case. Thus:
$$\mathbf r_0 (t) =  \frac 1 2 g t^2 \mathbf k$$
In order to get the first order approximation we just have to plug the zeroth one into ##(1)## and integrate to get:
$$\mathbf r_1 (t) =  \frac 1 2 g t^2 \mathbf k  2 \mathbf \Omega \times \int_{0}^{t} \mathbf r_0 (t') dt'$$
$$\mathbf r_1 (t) =  \frac 1 2 g t^2 \mathbf k + \frac 1 3 g t^3 (\mathbf \Omega \times \mathbf k)$$
If we compute the cross product we get
$$\mathbf \Omega \times \mathbf k = (\Omega \sin \beta) \mathbf j$$
Thus the first approximation is:
$$\mathbf r_1 (t) =  \frac 1 2 g t^2 \mathbf k + \frac 1 3 g t^3 (\Omega \sin \beta) \mathbf j$$
Here comes the meat:
The interesting thing is that the first order approximation predicts that the ball will drift to the +ive ##\mathbf j## direction. Is this direction east? (based on the provided diagram). My book states so but I do not see it...
Now we get the ##\mathbf j## component of the distance. What we could do is solving the time for the zeroth approximation, ##t= \sqrt{2h/g}##, and then plug it into the first approximation. This should be OK, as both approximations are small. We get:
$$\frac 1 3 g \Big(\frac{2h}{g}\Big)^{3/2} (\Omega \sin \beta) = \sqrt{\frac{8 h^3}{9g}} (\Omega \sin \beta)$$
That's nice! The only thing I am missing is understanding why my book states it drifts to the east.
b) For this section I know I need to modify somehow the first order approximation to take into account that now the ball is released from a vertical tower, but I do not see the trick. Could you give me a hint in this one?
Thanks.