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Kepler

Kepler. Force can be derived from a potential. k < 0 for attractive force Choose constant of integration so V (  ) = 0. Inverse Square Force. F 2 int. m 2. r = r 1 – r 2. m 1. R. r 2. F 1 int. r 1. The Lagrangian can be expressed in polar coordinates. L is independent of time.

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Kepler

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  1. Kepler

  2. Force can be derived from a potential. k < 0 for attractive force Choose constant of integration so V() = 0. Inverse Square Force F2int m2 r = r1 – r2 m1 R r2 F1int r1

  3. The Lagrangian can be expressed in polar coordinates. L is independent of time. The total energy is a constant of the motion. Orbit is symmetrical about an apse. Kepler Lagrangian

  4. The right side of the orbit equation is constant. Equation is integrable Integration constants: e, q0 e related to initial energy Phase angle corresponds to orientation. The substitution can be reversed to get polar or Cartesian coordinates. Kepler Orbits

  5. The orbit equation describes a conic section. q0 init orientation (set to 0) s is the directrix. The constant e is the eccentricity. sets the shape e < 1 ellipse e =1 parabola e >1 hyperbola Conic Sections r q s focus

  6. Elliptical orbits have stable apses. Kepler’s first law Minimum and maximum values of r Other orbits only have a minimum The energy is related to e: Set r = r2, no velocity Apsidal Position r r1 q r2 s

  7. The change in area between orbit and focus is dA/dt Related to angular velocity The change is constant due to constant angular momentum. This is Kepler’s 2nd law Angular Momentum dr r

  8. The area for the whole ellipse relates to the period. semimajor axis: a=(r1+r2)/2. This is Kepler’s 3rd law. Relation holds for all orbits Constant depends on m, k Period and Ellipse r r1 q r2 s

  9. Effective Potential • Treat problem as a one dimension only. • Just radial r term. • Minimum in potential implies bounded orbits. • For k > 0, no minimum • For E > 0, unbounded Veff Veff r r 0 0 possibly bounded unbounded

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