[meteorite-list] spiraling meteors
From: Matson, Robert <ROBERT.D.MATSON_at_meteoritecentral.com>
Date: Thu Jul 15 19:55:27 2004 Message-ID: <AF564D2B9D91D411B9FE00508BF1C86904EE5E3B_at_US-Torrance.mail.saic.com> Hi All, I echo Marco's position -- meteors absolutely do not corkscrew. You are talking about objects travelling at a minimum of 25,000 mph. The force necessary to divert an object with reasonable mass from a straight line trajectory is enormous. Take a typical line-of-sight range of 80 km, and assume that you can detect a zig-zag motion if the peak-to-peak deviation is a tenth of a Moon diameter (0.05 degrees). This translates into a minimum cross- track motion of 70 meters. To create the type of zigzag that would be visually noticeable, the meteor must move back and forth quite quickly, otherwise you'd end up with a very stretched out sine wave that would hardly be noticeable. I'll generously assume that the along track motion is 40 times longer than the side-to-side motion during one cycle. So the meteor travels 2.8 km while zigging 70 meters and zagging back 70 meters. So what kind of acceleration are we talking about? The lateral motion is 35 meters in one quarter of the cycle time. Assuming uniform acceleration: 35 meters = 0.5 * A * T^2 A = 70/T^2 (meters/sec^2) where T is in seconds. How long does it take our sample meteor to travel a quarter cycle? Assuming a very leisurely entry velocity of 12 km/sec, the full cycle distance of 2.8 km is covered in .2333 seconds. So the quarter cycle distance is covered in only .0583 seconds. The lateral acceleration works out to: A = 70 meters/(.0583 sec)^2 = 20600 meters/sec^2 This is 2100 g's! For even a tiny meteoroid, there is no mechanism available to impart the necessary force to cause this kind of acceleration. For comparison, if the force of pressure in the meteor's direction were sufficient to cause a 2100-g deceleration, the meteor would come to a dead stop in 0.58 seconds. --Rob Received on Thu 15 Jul 2004 07:49:50 PM PDT |
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