Space Shuttle Math

From: Jonathan T Wojack (tlj18@juno.com)
Date: Thu Oct 12 2000 - 09:23:44 PDT

  • Next message: Jonathan T Wojack: "ISS Observation, no Shuttle"

    O.K.:
    
    Calculating that the Shuttle was 550 "earth-km" from my eyes, and 538
    "sight-km", I then input these figures into the classic
    hypotenuse/pythagoras formula.  X=114.263, then, for shuttle altitude. 
    Through e-mail, I've learned that the shuttle was actually 102 km above
    the earth at that time.  That's a disparity of about 12 kilometers: so,
    would it be correct to assume that this disparity is due to the curvature
    of the earth?
    
    If so, then does that mean that if I shot a laser beam at ~ 0 degrees
    altitude directly under the shuttle, then when the laser beam arrived
    directly under the shuttle (where the laser beam would be only a couple
    of meters above the ground, assuming the earth is perfectly flat for
    illustration purposes only), the laser photons would actually be 12 km
    above the surface of the earth (actually, the Atlantic Ocean)? 
    
    --------------------------------
    Jonathan T. Wojack                                         
    tlj18@juno.com
    
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