HW7_prob5_GPS3.31_G3.26

HW7_prob5_GPS3.31_G3.26 - Problem GPS 3-31 (2nd ed. 3.26)...

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Problem GPS 3-31 (2 nd ed. 3.26) Start with Eq.(3.97) (in both 2 nd and 3 rd editions), relating the scattering angle Θ ( s ) to the interaction potential V ( r ) and the impact parameter s , Θ ( s )= π 2 s Z u m 0 ¡ 1 V (1 /u ) /E s 2 u 2 ¢ 1 / 2 du. (1) For the repulsive central force f = kr 3 ,we f nd easily that V (1 /u )=( k/ 2) u 2 . (2) With Eq.(2), Eq.(1) becomes Θ ( s )= π 2 s Z u m 0 ¡ 1 ( k/ 2 E + s 2 ) u 2 ¢ 1 / 2 du, (3) which we can integrate as soon as we determine the value of u m =( r m ) 1 . The distance of closest approach ( periapsis ) r m is the value of r at the turning point of the orbit, i.e., where · r =0 . This value is easily found from the energy equation evaluated with · r =0 and r = r m : E = k 2 r 2 m + l 2 2 μr 2 m . (4) Upon replacing the angular momentum l by the impact parameter, l = s 2 μE , and solving for r m ,wefnd r m =( u m ) 1 = p s 2 + k/ 2 E, (5) which allows us to further simplify Eq.(3) to read
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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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HW7_prob5_GPS3.31_G3.26 - Problem GPS 3-31 (2nd ed. 3.26)...

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