370(4): General Dynamics of the Orbiting Asymmetric Top

In general the lagrangian is Eq. (2), where the potential energy is a function of r, theta, phi and chi. The problem must be solved with a model for the dependence of U on the Euler angles, so the general solution is found by solving Eqs. (15), (17), 20), (23) and (28) to (30) simultaneously. It is known that this kind of problem can be solved by Maxima on a desktop, so it looks promising. In general the nutations and precessions of the earth, regarded as an asymmetric top, are superimposed on its orbital motion. This kind of approach is completely general, and ought to give patterns reminiscent of the Milankovitch cycles. The earth modelled as a dumbbell has already been solved by Horst Eckardt, and those results might also show some patterns reminiscent of the climate cycles. The extra ingredient in this note is that the potential is assumed to depend on the Euler angles. This assumption means that the nutations and precessions of the earth affect its position in an orbit. If it is assumed as usual that U depends only on r, the problem simplifies. The lagrangian (2) simplifies to that of a symmetric top with one point fixed when r dot = 0 and U = mghcos theta, and when I1 = I2. Here g = – MG / R squared is the acceleration due to gravity, M being the mass of the earth and R its radius. The minus sign denotes attraction. The problem of the gyroscope subjected to an extra potential can also be approached by modelling the potential of the extra force. The extra force will change the nutations and precessions. When the extra force balances the force due to gravity the net potential is zero. So it all comes down to modelling the potential energy.


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