All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Science
From a Balloon (Posted on 2005-07-07) Difficulty: 3 of 5
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?

No Solution Yet Submitted by Erik O.    
Rating: 3.0000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution The answer is East, | Comment 30 of 33 |

For the problem to make any sense, one must assume that 1) the balloon is stationary above a point on the earth, and 2) that the word “still” refers to ‘zero” wind.

<o:p> </o:p>

1)      This can be true in a real sense since as the earth rotates with some angular velocity, w, at the very least the lower portions of the earth’s atmosphere, the oceans etc (all fluids) move with it in a steady state motion.  Otherwise, how could we ever have a “calm” day?

2)      Using this definition of “still” implies that there is no wind induced force acting on the stone to deflect its path as it falls, or to move the balloon after the stone is released.

<o:p> </o:p>

Given the above, the earth, balloon and stone all have the same initial angular velocity, w.  In the instant prior to the release of the stone, the following is true:

<o:p> </o:p>

The magnitude of the velocity of a point on the earth directly below the balloon is Vearth= w*Re, where Re=radius of the earth at that point.  Likewise, the magnitude of the velocity of the balloon (and stone prior to release) is Vstone=Vballoon=w*(Re+Ab) where Ab=altitude of the balloon above the earth).  These are called tangential velocities.

<o:p> </o:p>

The key is that these tangential velocities are perpendicular to the radius of the earth (actually, they are the cross product of the angular velocity and the radius).  Being perpendicular to the radius of the earth, they are also perpendicular to the force of gravity, and, therefore, gravity cannot affect these velocities, at least in magnitude.

<o:p> </o:p>

Conclusion:  In the absence of any other forces (such as wind), (more later on this), the stone will maintain its larger tangential velocity and therefore will traverse east as the stone approaches the surface of the earth.

<o:p> </o:p>

Interestingly enough, as the stone drops with its constant tangential velocity, it drops through layers of the atmosphere that are moving at lower tangential velocities as the stone approaches the earth.  This would tend to reduce the amount of eastward migration of the stone, essentially “blowing” the stone a bit to the west.  However,I would believe that this is a very small force given the relatively small differences in tangential velocities.

<o:p> </o:p>

Example:  Balloon at 1 mile high, radius of the earth approx. 4000 miles, then the difference in tangential velocities is only 1 part in 4000.  This would likely create inconsequential aerodynamic forces, since aero forces a dependant on relative velocity to the second power.


  Posted by Kenny M on 2005-11-20 20:03:15
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information