HW1_prob4_GPS12_G02

HW1_prob4_GPS12_G02 - v 1 , we f nd v 1 = p 2 GM/R e . (3)...

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Problem 1.5 (3 rd ed. 1.12) The energy conservation theorem reads T 1 + V 1 = T 2 + V 2 , (1) where the labels 1 and 2 refer to di f erent points in space. The kinetic energy is T = mv 2 / 2 where m is the particle mass and v is its speed. The gravitational potential energy is V = GMm/r ,wh e r e G is the Newtonian constant of gravitation, M is the mass of the earth, and r is the distance from the earth’s center to the particle. Let point 1 lie on the earth’s surface and point 2 be an in f nite distance from the earth. In this case, we have T 2 = V 2 =0 ,and T 1 + V 1 = 1 2 mv 2 1 GMm R e =0 , (2) where
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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.

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