class7a - U =-GMm r (77) and in our case of the particle U...

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example: A particle of mass m can move under the gravitational attraction of two other particles of equal mass M , Fxed at the points ( ± a, 0 , 0). Show that the origin O is a position of equilibrium, but that it is not stable. solution: When the particle is at O , the total force on it is zero, so the origin is indeed an equilibrium position. Whether it is stable or unstable is determined by whether the potential energy function U ( x,y,z ) has a minimum at O . Consider the masses M to be on the x -axis: then F = GMm r 2 (76) implies generally speaking
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Unformatted text preview: U =-GMm r (77) and in our case of the particle U ( x, , 0) =-GMm a-x-GMm a + x =-2 amMG a 2-x 2 (78) Dierentiating gives U x =-2 amMG [-( a 2-x 2 )-2 (-2 x )] =-4 amMGx ( a 2-x 2 ) 2 (79) indeed this is zero when x = 0. Dierentiating again: 2 U x 2 =-4 amMG ( a 2-x 2 ) 2-16 amMGx 2 ( a 2-x 2 ) 3 (80) and evaluate at x = 0: 2 U x 2 =-4 mMG a 3 (81) which indicates unstable equilibrium. 11...
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This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.

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