If a stone is dropped from a balloon on a still day, does the stone fall directly below the balloon, or to the west or east of it?

(In reply to My thoughts by Jer)

I'm sure Jer has the intended solution here. The baloon, being at some height above the earth's surface and turning at the same angular rate as the earth, is moving faster than the surface, eastward.  Therefore the dropped stone retains this linear motion while picking up vertical motion, and falls slightly east of the bottom of the baloon.

This is called a coriolis effect. But I'm sure the amount is so small as to be outweighed by the fact that the stone has to be dropped in one direction or another over the side of the balloon's basket. If dropped from the west side of the balloon, I'm sure it would move only slightly toward the point directly under the center of the balloon. If dropped from the east side, then only slightly farther east out from the point directly under the center of the balloon.

Let's say the balloon is over the equator, just to make things simple.  Let's also say that the balloon is 1 mile up.  The earth's surface is moving about 1000 mi/hr at the equator and the balloon would be moving at 4001/4000 of that speed, or an additional 1/4 of 1 mph.  It would take about 18 seconds for the stone to reach the ground. At 1/4 mph, that would take it about 6 or 7 feet east of the point directly under the point of drop.  This disregards both the vertical and horizontal motion being resisted by the windless (relative to the ground) air.

At other latitudes the effect would be proportional to the cosine of the latitude.  The horizontal distance covered would also vary with the square root of the height of the balloon as that affects the time: 16 t^2 = h, in feet and seconds.

Posted by Charlie on 2005-07-07 20:25:59