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Unformatted text preview: 1.2 Planetary motion REFERENCE: Arya, Section 7.7 to 7.9 1.3 Rutherford Scattering The other very important problem involving inverse-square forces is the scattering of charged particles in a Coulomb or electrostatic field. The potential for this case is V ( r ) = k r (23) where k is k = q 1 q 2 2 π (24) which is positive or negative depending on the sign of the charges of the two particles. Now, what is the orbit of the alpha particle (a positively charged particle) incident to a positively charged nucleus? If we look at the eccenticity = 1 + 2 EL 2 μk 2 . (25) suggests that, since E > 0, > 0 and the motion is hyperbolic. Our equation for the angle that the path of the particle traverses from its minimum radius (between the two Foci) to r → ∞ is, from our analysis of central force motion Δ α = r max r min ( L/μr 2 ) dr 2 μ E- V ( r )- L 2 2 μr 2 (26) where, since the particle is scattering from long distances where L = bm 1 u 1 (27) and the energy E = K = 1 2 mu 2 1 (28) Substituting into our equation for the angle...
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- Spring '10
- Electric charge, Rutherford Scattering, sin θ sin, positively charged particle, scattering cross section, trig identity sin