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Unformatted text preview: v 1 , we f nd v 1 = p 2 GM/R e . (3) We can eliminate G and M by realizing that at the earths surface, the force of gravity is simply expressed as GMm R 2 e = mg, (4) where g is the familiar gravitational acceleration (9.8 m/s 2 ). Thus, Eq.(3) simpli f es to v 1 = p 2 gR e . (5) The earths radius is approximately 6400 km, so Eq.(5) yields the numerical result v 1 = 11 . 2 km/s, (6) or (since 1 km = 0.6214 mi) v 1 = 6 . 96 mi/s. (7) 1...
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This note was uploaded on 12/19/2010 for the course PHYSICS ph 409 taught by Professor Wilemski during the Fall '10 term at Missouri S&T.
 Fall '10
 Wilemski
 mechanics, Energy, Kinetic Energy, Mass

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