# 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?

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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? – drN 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! – drN Oct 22 '12 at 12:46

The Coriolis "force" isn't a proper force. It's probably better called the Coriolis effect. It's named after a French mathematician and engineer.

In a rotating frame of reference, like the earth's, when we differentiate a dynamical variable (like the position of an object -- here, Baumgartner) with respect to time we get two terms. The first is just the (infinitesimal) change in position with respect to the (infinitesimal) change in time in the rotating reference system. The second term, not present in an inertial frame (a rotating frame is not inertial (why?)) is proportional to the (instantaneous) angular velocity of the rotating frame with respect to an (instantaneous) inertial frame of the object. The second term is the Coriolis effect and it arises due to the fact that the even when the object isn't moving with respect to the non-inertial frame, it's position in the inertial frame changes due to the fact that the position vector (say, $\mathbf{r}$) changes due to the fact that we chose a moving coordinate system. The relevant equation is \begin{align} \frac{d\mathbf{r}}{dt}\Big|_I &= \frac{d\mathbf{r}}{dt}\Big|_N + \boldsymbol{\omega}\times\mathbf{r}, \end{align} in a hopefully obvious notation. Of course, if you use a reference frame at rest with respect to, say, the center of the earth, you don't have any Coriolis terms in your equations. But you do have to take into account that the target, here the earth, is moving.

If you open a decent undergraduate or graduate classical mechanics book like Davis, Symon, or (egads!) Goldstein you can find a full development. But we can do an order of magnitude calculation, quick like. Baumgartner fell for 4m, 19s. Let's call this 250s, for ease in figuring. The magnitude of the angular velocity of the earth, which we'll take (incorrectly, but this gives an upper bound) as perpendicular to a position vector from the center of the earth to Baumgartner, is 2\pi / 24 hours. Since 24h = 86,400s the magnitude of the Coriolis effect on Baumgarnter results in an additional velocity of about 10^3/10^5 ($\omega$ in $s^{-1}$) * 10 (height in miles), giving about 1/10 of a mile/second. So Baumgartner would have had to correct for this effect. And this can be done with high precision.

I would guess that it would be a little tougher to correct for wind!

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MarkWayne, would you agree that: Assuming no wind and that Baumgartner was stationary above a specific Point A when he jumped, the Coriolis effect would result in him landing slightly more to the East of Point A? Would you also agree that the effect would have been the same in both North and South Hemispheres? – Pieter Müller Oct 31 '12 at 12:59
@PieterMüller: with your assumptions, no. The Coriolis effect implies that the landing point is to the west of the point the object would land at if $\omega = 0$. This is because the earth rotates about it's axis with a tangential velocity to the east. The angular velocity of the earth is a vector whose direction points from south to north. The radius vector (in either frame) points from the center of the earth to the object. With this information you can calculate using the relevant equation above that the second term for $d\mathbf{r}/dt\Big|_N$, the cross term, points to the west. – MarkWayne Nov 3 '12 at 5:25
@MarkWayne: When a "stationary" object above point A at a height is dropped, it should indeed land to the east of point A, not west, since the object starts with a faster eastward velocity then point A. No? See for example [here][1]: grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/… – Ben Schwehn 2 hours ago