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 ________________________________________________________________ YOU'RE PAYING TOO MUCH FOR THE INTERNET! Juno now offers FREE Internet Access! Try it today - there's no risk! For your FREE software, visit: http://dl.www.juno.com/get/tagj. ----------------------------------------------------------------- Unsubscribe from SeeSat-L by sending a message with 'unsubscribe' in the SUBJECT to SeeSat-L-request@lists.satellite.eu.org http://www2.satellite.eu.org/seesat/seesatindex.html
This archive was generated by hypermail 2b29 : Thu Oct 12 2000 - 19:02:09 PDT