(meteorobs) Meteor Parallax

Swift, Wesley R. (MSFC-NNM05AB50C)[RAYTHEON] Wesley.Swift at nasa.gov
Tue Jan 16 17:23:48 EST 2007


David,

	Ok.  Try this:

*  Assume observer 1 sees a meteor at his zenith at an altitude of h
(say 100km)
*  Assume observer 2 sees the meteor just at his horizon
*  Observer 1, observer 2 and the center of the earth, C, form a right
triangle 
        with the right angle at observer 1(the horizon is normal to a
line to C) 
*  Observer 1 to C is Re=6378km, meteor to C is Re + h = 6478 km and
angle c = arc cos(Re/(Re+h)  
*  The distance from Obs 1 to Obs 2 is the arc, S, where S = Re * c
where c is in radians.
*  For the case where both see the meteor at the horizon, distance is
2*S

>From this one can draw out and solve for several special cases of
interest.

Wes


-----Original Message-----
From: meteorobs-bounces at meteorobs.org
[mailto:meteorobs-bounces at meteorobs.org] On Behalf Of David Oesper
Sent: Saturday, January 13, 2007 9:25 PM
To: meteorobs at meteorobs.org
Subject: (meteorobs) Meteor Parallax

If two observers seeing the same meteor are separated by a distance D in
miles, then they will see the location of the meteor shifted against the
background stars by approximately S degrees, as given by the following
equation:

S = 2 arctan (D/100)

This equation assumes that the meteor is seen when it is at an altitude
of 50 miles, and the curvature of the Earth is negligible.

Being curious about how far apart two observers might be and still see
the same meteor, it occurs to me that one cannot ignore the curvature of
the Earth.  However, I am at a loss to figure out the exact mathematical
relation.  Can anyone point me in the right direction on this?

Thanks much,

David Oesper
Dodgeville, Wisconsin

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