Coriolis force in free fall

Does the coriolis force has any measurable effect in free fall from large heights?

Take for example the sky diving experiment by F. Baumgartner who started from a height of about 40 km above New Mexico. Does the coriolis force has any measurable effect in this experiment?

I think there should be a deviation to the east and another one to the south (which should be smaller in size). But I don't have an idea how large the deviation would be.

How can one calculate the size of the deviations?

• forgive my ignorance but isn't the coriolis force only a matter of changing the frame of reference? The coriolis force shouldn't er... have a physical effect on Baumgartner right? – dearN Oct 21 '12 at 23:30
• @drN, the coriolis force won't have an effect on Baumgartner in an inertial frame. But it will cause him to hit the ground at a different point than directly below where he began his fall. – Colin McFaul Oct 22 '12 at 1:10
• @ColinMcFaul True that! – dearN Oct 22 '12 at 12:46

Yes it does. If you fix a (non inertial) frame of reference whose origin is on Earth's surface, at latitude $\lambda$, then a freely falling particle of mass $m$ has the equation of motion, $$m\frac{d^2\vec r}{dt^2}=m\vec g_{ef}-2m\vec\omega\times\vec r,$$ where $$\vec g_{ef}=\vec g-\vec\omega\times\left[\omega\times(\vec R+\vec r)\right]\approx \vec g-\vec\omega\times(\omega\times\vec r),$$ is the effective gravity which takes into account the centrifugal force and $\vec R$ is the vector from the center of the Earth to the origin of the non inertial frame. The vector $\vec omega$ is the angular velocity of the Earth. Solving for $\vec r$ one gets $$\vec r(t)=\vec v(0)t+\frac{\vec g_{ef}t^2}{2}-2\omega\times\int_0^t \vec r(t)dt.$$ This equation can be soved order by order, the first order being $$\vec r(t)=\vec v(0)t+\frac{\vec g_{ef}t^2}{2}-2\omega\times\left(\frac{\vec v(0)t^2}{2}+\frac{\vec g_{ef}t^3}{6}\right).$$ If the particle has $\vec v(0)=\vec 0$, the deflection relative to the effective vertical (defined by the direction of $\vec g_{ef}$) is given by $$\Delta x=\frac{\vec\omega\times\vec g_{ef}t^3}{3}=\frac{g_{ef}\omega t^3\cos\lambda}{3}\hat i,$$ where $\hat i$ is directed east (tangentially to the circle of latitude). Plugging in the approximate time to fall from the height $h$, $t=\sqrt{2h/g_{ef}}$, $$\Delta x=\frac{1}{3}\left(\frac{8h}{g_{ef}}\right)^{1/2}\omega\cos\lambda\approx \frac{1}{3}\left(\frac{8h^3}{g}\right)^{1/2}\omega\cos\lambda,$$ since $g_{ef}$ already is $g_{ef}=g+\mathcal O(\omega^2)$. The angular velocity of the Earth is $\omega\approx 7.3\cdot 10^{-5}\, \mathrm{s}^{-1}$, so a free fall of order $10^4\, \mathrm m$ would give a deviation of order $10^1\, \mathrm m$. Notice however that this not take the drag force into account.