1. Coulomb Scattering

2. The trajectory from an inverse square force forms a conic section. e < 1 ellipse e =1 parabola e >1 hyperbola. The force center is at a focus. Hyperbolic Orbits q attractive focus a r a ae repulsive focus

3. Orient the incident axis to horizontal. Scattering mass forms a hyperbolic trajectory. Reoriented View q attractive focus a r a ae repulsive focus h v0 q m b x

4. The potential is defined at infinity. The impact parameter would be closest approach for no force. Compare to angular momentum Impact Parameter h v0 q m x b

5. Scattering cross section is based on the potential. Impact parameter Differential impact parameter The result is the Rutherford scattering cross section Rutherford Cross Section q b y

6. Problem A spaceship of mass m moving with velocity v0 approaches the Moon. The impact parameter is b. The velocity v0 is perpendicular to the orbital velocity V of the moon. Show that if the spaceship passes behind the Moon it gains kinetic energy as it leaves the Moon. Lunar Miss q M b x m

7. The moon is much more massive. Lunar frame is CM Incident and final energies same in CM Lunar Frame b m1 v0 q x b M

8. In the Moon’s frame the velocity components come from the scattering angle. In the observer’s frame the velocity is boosted by the moon. Energy Boost next