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(meteorobs) photo meteor orbits?
The following message appeared the other day on the ASTRO mailing list. Leigh's
post brought up some questions for me as to exactly how two-station photography
is used to determine meteor orbits. It seems clear why the apparent angular
velocity would be needed: I assume that the track of a meteor on each piece of
film is treated as a "straight line" in 3-D terms (segment of a great circle in
skydome terms) for simplicity. I'm guessing that the straight line is assumed as
a coplanar tangent segment to the meteoroid's original orbit, whose eccentricity
can then be calculated based on a finally derived heliocentric velocity?
But is this straight-line path also an artifact of a meteoroid's entry into the
atmosphere? And are TWO angular velocities actually required to calculate the
geocentric velocity? It seems like once a straight path in the atmosphere is
found, EITHER station's apparent angular velocity (or anyone else on earth's for
that matter) should be enough to derive the true geocentric velocity. Oder? ;>
[PS: I know all of this can probably be found in the IMO Photographic Handbook,
but please help me save book money by answering the question here instead! :)]
Thanks in advance for hard answers to goofy questions!
Lew
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To: ASTRO
From: palmer@SFUdot ca (Leigh Palmer)
Date: Tue, 27 Aug 1996 18:34:38 -0700
Subject: Re: Meteor on film
>This method is how to get 3-D positions. Of cource, one must observes
>same meteor at another distant site. If you get positions and a velosity
>of the meteor, then one calculates this meteor's orbit in solar-system.
>One more I say that one guesses at the falling point of meteorite.
The orbit can only be determined if one has some indication of time on the
film. This is most commonly achieved by chopping the light with a shutter
so that the meteor trail appears as a series of dashes. In such a case the
position of the track is not determined from the positions of stars at the
time of the meteor (the exposure is not usually terminated immediately).
Lacking this information the only clue would come from curvature of the
track, and I can't imagine that would be easy.
You are certainly correct in saying that one can only guess at the impact
point from such data. A network of such cameras here in Canada did succeed
in finding one such meteorite (Innisfree?). It also helped that when the
object landed in Saskatchewan the ground was covered in snow - the pieces
of the meteorite were easy to spot lying on top of the white surface! The
network is no longer operating. Since the discovery that the Antarctic ice
sheet concentrates meteorites that has provided the most cost effective
method of finding them.
Leigh
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