Kepler Problem - Solution of The Kepler Problem

Solution of The Kepler Problem

The equation of motion for the radius of a particle of mass moving in a central potential is given by Lagrange's equations

m\frac{d^{2}r}{dt^{2}} - mr \omega^{2} = m\frac{d^{2}r}{dt^{2}} - \frac{L^{2}}{mr^{3}} = -\frac{dV}{dr}


and the angular momentum is conserved. For illustration, the first term on the left-hand side is zero for circular orbits, and the applied inwards force equals the centripetal force requirement, as expected.

If L is not zero the definition of angular momentum allows a change of independent variable from to

\frac{d}{dt} = \frac{L}{mr^{2}} \frac{d}{d\theta}

giving the new equation of motion that is independent of time

\frac{L}{r^{2}} \frac{d}{d\theta} \left( \frac{L}{mr^{2}} \frac{dr}{d\theta} \right)- \frac{L^{2}}{mr^{3}} = -\frac{dV}{dr}

The expansion of the first term is

\frac{L}{r^{2}} \frac{d}{d\theta} \left( \frac{L}{mr^{2}} \frac{dr}{d\theta} \right) = \frac{L^{2}}{mr^{5}} \left( \frac{dr}{d\theta} \right)^2 - \frac{L^{2}}{mr^{4}} \frac{d^{2}r}{d\theta^{2}}


This equation becomes quasilinear on making the change of variables and multiplying both sides by

\frac{du}{d\theta} = \frac{-1}{r^{2}} \frac{dr}{d\theta}
\frac{d^{2}u}{d^{2}\theta^{2}} = \frac{2}{r^{3}} \left( \frac{dr}{d\theta} \right)^{2} - \frac{1}{r^{2}} \frac{d^{2}r}{d\theta^{2}}

After substitution and rearrangement:

\frac{d^{2}u}{d\theta^{2}} + u = -\frac{m}{L^{2}} \frac{d}{du} V(1/u)

For an inverse-square force law such as the gravitational or electrostatic potential, the potential can be written

V(\mathbf{r}) = \frac{k}{r} = ku

The orbit can be derived from the general equation

\frac{d^{2}u}{d\theta^{2}} + u = -\frac{m}{L^{2}} \frac{d}{du} V(1/u) = -\frac{km}{L^{2}}

whose solution is the constant plus a simple sinusoid

u \equiv \frac{1}{r} = -\frac{km}{L^{2}} \left

where (the eccentricity) and (the phase offset) are constants of integration.

This is the general formula for a conic section that has one focus at the origin; corresponds to a circle, corresponds to an ellipse, corresponds to a parabola, and corresponds to a hyperbola. The eccentricity is related to the total energy (cf. the Laplace–Runge–Lenz vector)

e = \sqrt{1 + \frac{2EL^{2}}{k^{2}m}}

Comparing these formulae shows that corresponds to an ellipse (all solutions which are closed orbits are ellipses), corresponds to a parabola, and corresponds to a hyperbola. In particular, for perfectly circular orbits (the central force exactly equals the centripetal force requirement, which determines the required angular velocity for a given circular radius).

For a repulsive force (k > 0) only e > 1 applies.

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