Southerly Deflection by Motion of the Earth

[bob1182006]

Marion and Thornton Classical Dynamics of Particles and Systems 5th ed. Chapter 11 #13.

Homework Statement

(Refers to an example where they calculated the horizontal 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=\sqrt{2h/g}. Perform a calculation in second approximation (retain terms in w^2) and calculate the southerly deflection.
There are three components to consider (a) Coriolis force to second order (C1), (b)variation of centrifulgal force with height (C2), (c) variation of gravitational force with height (C3).

Each component gives a result:
C_i\frac{h^2}{g}\omega^2sin(\lambda)cos(\lambda)
with C1=2/3, C2=5/6, C3=5/2.

Homework Equations

\vec{F}=\vec{S}+m(\vec{g_o}-\vec{\omega}\times(\vec{\omega}\times\vec{r}))-2m\vec{\omega}\times\vec{v_r}
\vec{g_o}=-G\frac{M_e}{(R_e+h)^2}

The Attempt at a Solution

The coordinate system has z radially outward from the Earth. x in the southerly direction, y in the easterly direction.
\lambda is the latitude.
\vec{S}=0, Assuming no external forces like air resistance.

\vec{\omega}=-\omega cos\lambda\vec{e_x}+\omega sin\lambda\vec{e_z}
R_e+h is in the positive z direction
\vec{\omega}\times(\vec{R_e}+\vec{h}})=(R_e+h)\omega cos\lambda \vec{e_y}
\vec{\omega}\times(\vec{\omega}\times(\vec{R_e}+\vec{h}))=-\omega^2(R_e+h)sin\lambda cos\lambda \vec{e_x}-\omega^2(R_e+h)cos^2\lambda \vec{e_z}

For v_r I think it should be just -gt in the radial direction. But my book says that the Coriolis Force produces a small velocity component in the y and z directions that they neglected.
Would the component of the velocity of the z direction be -g_ot-t\omega^2(R_e+h)cos^2\lambda? Or is it just -gt?

 

[nickjer]

I believe your problem asks for only terms up to \omega^2. So you can treat the velocity as -gt in the z-direction. But you might want to check this.