Lagrangian Mechanics Made Simple
Part 4
Orbital Motion and Kepler's Equation
Ron Steinke <rsteinke@w-link.net>
Last month we introduced the problem of a system two particles whose potential energy
depended only on the distance
between them. The dynamics of this system reduced to
(2) |
Kepler's Equation
For the force of gravity, ^{1}the potential is
(4) |
This solution is unique up to a constant shift in the angle , .
Eq. (5) describes a conic section,
(6) |
(7) |
The Lagrange Stability Points
The three body problem is not solvable in the general case. However,
some approximate solutions are quite interesting. Consider
the case of a small body of mass
orbiting two larger bodies of masses
and
. For
simplicity we will assume that the two large bodies orbit each other circularly with angular velocity
, and
are unaffected by their attraction to the smaller body. If
and
are
the distances of the two bodies from their common center of rotation, then
where is the distance between the two bodies,
Here, is Newton's gravitational constant.
The two larger bodies lie at positions
and
in a coordinate system with angular velocity
. The Lagrangian for the lighter particle at position
is
(10) |
(11) |
This substitution gives the Lagrangian the simpler form
(12) |
To find metastable points, we look for solutions where
is zero. We do this by examining the effective potential,
There exist points for which the centripetal acceleration is balanced by the
attraction of the larger bodies. At these points, the force,
where we have dropped the common factor of .
It is possible to solve the second equation in (15) by setting to zero. The value of is then found by solving the first equation. There are three of these solutions, one with , between the two larger bodies, and one each with and , off to each side. These three points are saddle points. If you perturb the small mass off one of these points by a small amount along the direction of its orbit, it will tend to fall back into place. If you perturb it into a higher or lower orbit, however, it will tend to move away from the stability point, either towards one of the two heavier bodies, or pulled away by the centripetal component of the potential (13).
There are also two solutions where
is not equal to zero. These are
The force in the direction is proportional to , indicating that it is pulling against the displacement. Any small perturbation of a particle in the direction will result in a force which tends to pull it back to . The same is not true of perturbations in the and directions, however. The effect of and on the force are determined by a portion of the matrix in (17). The eigenvalues of this submatrix are
(18) |
About this document ...
Lagrangian Mechanics Made SimplePart 4
Orbital Motion and Kepler's Equation
Footnotes
- ... gravity, ^{1}
- This section leans heavily on J. Marion and S. Thornton, Classical Dynamics of Particles and Systems, 3rd ed., pp. 256-260
Ron Steinke 2002-08-11
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