# Kepler

Télécharger la présentation

## Kepler

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

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