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Kepler
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Inverse Square Force Force can be derived from a potential. < 0 for attractive force Choose constant of integration so V ( ) = 0. m2m2 r1r1 F 2 int r2r2 R m1m1 F 1 int r = r 1 – r 2
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Kepler Orbits Right side of the orbit equation is constant. Equation is integrable.Equation is integrable. Integration constants: e, 0Integration constants: e, 0 Equation describes a conic section. init orientation (often 0) init orientation (often 0) e sets the shape: e 1 hyperbola. e sets the shape: e 1 hyperbola. s is the directrix. s is the directrix. focus r s
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Kepler Lagrangian The lagrangian can be expressed in polar coordinates. L is independent of time. The total energy is a constant of the motion.The total energy is a constant of the motion. Orbit is symmetrical about an apse.Orbit is symmetrical about an apse. constant
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Apsidal Position Elliptical orbits have stable apses. Kepler’s first lawKepler’s first law Minimum and maximum values of r.Minimum and maximum values of r. Other orbits only have a minimum.Other orbits only have a minimum. The energy is related to e: Set r = r 2, no velocitySet r = r 2, no velocity r s r1r1 r2r2
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Angular Momentum Change in area between orbit and focus is dA/dt It is constantIt is constant Kepler’s 2 nd law Area for the whole ellipse relates to the period. semimajor axis: a=(r 1 +r 2 )/2. Kepler’s 3 rd law r dr
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Effective Potential Treat problem as a one dimension only. Just radial r term.Just radial r term. Minimum in potential implies bounded orbits. For > 0, no minimumFor > 0, no minimum For E > 0, unboundedFor E > 0, unbounded V eff 0 r 0 r unbounded possibly bounded
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