Coriolis force effect on skydivers?

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dwightlathan
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Galileo observed that objects dropped from height hit ground to the east of a plumb bob. I guess this is due to the Coriolis force. How would this distance be calculated for something like a baseball off the Empire State Building or a skydiver?
 
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dwightlathan said:
Galileo observed that objects dropped from height hit ground to the east of a plumb bob. I guess this is due to the Coriolis force. How would this distance be calculated for something like a baseball off the Empire State Building or a skydiver?
I doubt that Galileo measured this. If you compute the shift, it is of less than a millimeter for a height of 10 meters. Of course, it is well known that Galileo played on the leaning tower of Pisa and its 55 meters. For this height the shift is of 8 mm and could be measurable if you could stop the wind. I think that it is part of the Galileo myth. For a spherical form sky diver in no wind conditions the shift would be of 6 meters for a drop of 4000 meters. Of course this is for a jump from a stationary helicopter.

Please, leave Coriolis alone!
 
Curiosity is a powerful thing :)

You may be right about Galileo's observations being myth, Lpfr. Whether he did or didn't Galileo certainly isn't credited for discovering the Coriolis effect.

Can anyone post the equations used to determine the horizontal movement? To many people it would be a concrete example of an abstract phenomenon.
 
Fcor = 2mr' x Omega = 2mv x Omega

Omega depends upon the location of the object since the Earth isn't a perfect sphere. But it's generally seen as about 7.3 x 10^-5 s^-1.

When Coriolis is added to free fall, the equation becomes r''= g + 2r' x Omega.

r'= (x',y',x') and Omega (0,Omega sin(theta), Omega cos(theta))

The equation of motion for the x-dimension becomes 2Omega(y'cos(theta) - z'sin(theta)). Y-dimension= -2Omega x' cos(theta) and z-dimension= -g + 2Omega x' sin (theta).

Working through approximations, working those approximations back into the original equations and so forth, you get a rough approximation that x= 1/3 Omega(gt^3)sin(theta).

Let's say you put theta at 90 degrees and let an object fall for about 20 seconds with g at 10 m/s/s. The x displacement is only about 2.2 cm, while it fell 100 meters.