Re: beta angle?

From: Sebastian Stabroth (
Date: Thu May 23 2002 - 16:48:09 EDT

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    > But sin(Launch Azimuth) = cos(i) / cos(lat)  (90 degrees when i=lat)
    > so here Launch Azimuth is 45 degrees.
    Right, this is the inertial launch azimuth. Because the launch site has a
    velocity component (the Earth rotates with 0.409 km/s to the east at KSC) the
    actual launch azimuth would be around 2 deg more to the north, for a coplanar
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