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## 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, [itex]T=\sqrt{2h/g}[/itex]. Perform a calculation in second approximation (retain terms in [itex]w^2[/itex]) 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:

[tex]C_i\frac{h^2}{g}\omega^2sin(\lambda)cos(\lambda)[/tex]

with C1=2/3, C2=5/6, C3=5/2.

## Homework Equations

[tex]\vec{F}=\vec{S}+m(\vec{g_o}-\vec{\omega}\times(\vec{\omega}\times\vec{r}))-2m\vec{\omega}\times\vec{v_r}[/tex]

[tex]\vec{g_o}=-G\frac{M_e}{(R_e+h)^2}[/tex]

## 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.

[itex]\lambda[/itex] is the latitude.

[itex]\vec{S}=0[/itex], Assuming no external forces like air resistance.

[tex]\vec{\omega}=-\omega cos\lambda\vec{e_x}+\omega sin\lambda\vec{e_z}[/tex]

[itex]R_e[/itex]+h is in the positive z direction

[tex]\vec{\omega}\times(\vec{R_e}+\vec{h}})=(R_e+h)\omega cos\lambda \vec{e_y}[/tex]

[tex]\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}[/tex]

For [itex]v_r[/itex] 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 [itex]-g_ot-t\omega^2(R_e+h)cos^2\lambda[/itex]? Or is it just -gt?