Re: [APML] The Kolb Fireball?

[Joe Pedit <pedit@email.unc.edu>]

Glenn Shaw wrote:
> 
> The "glow" from incoming meteors is about 80 kilometers, or 50 miles in height.
> Thus the triangulation "error" between the two observers is 50/1500 or 30 in
> difference. This corresponds to 3 degrees of difference, or about 6 times the
> diameter of the full moon or the Andromeda Galaxy.
> 

Is this calculation correct?  Assume the earth is flat and one observer
sees the meteor directly overhead at an altitude of 50 miles.  How high
above the horizon would an observer 1500 miles away see the same
meteor?  Inverse tangent of 50/1500 is equal to 1.9 degrees above the
horizon.  The second observer would see the same meteor (90 - 1.9) =
88.1 degrees away from the zenith.  Throw in the effects of the
curvature of the earth and ignore atmospheric refraction and you will
find that the meteor would have to be at an altitude of ~300 miles for
the second observer to see it on the horizon.

Joe

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