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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. Right side of the orbit equation is constant. Equation is integrable.

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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. Right side of the orbit equation is constant. Equation is integrable. Integration constants: e, q0 Equation describes a conic section. q0 init orientation (often 0) e sets the shape: e < 1 ellipse, e =1 parabola, e >1 hyperbola. s is the directrix. Kepler Orbits r q s focus

  4. 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 constant

  5. 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

  6. Change in area between orbit and focus is dA/dt It is constant Kepler’s 2nd law Area for the whole ellipse relates to the period. semimajor axis: a=(r1+r2)/2. Kepler’s 3rd law Angular Momentum dr r

  7. 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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