# [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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