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## Kepler

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**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**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**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**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**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**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