Is the Jacobi Theorem Valid in the Singly Averaged Restricted Circular Three-Body Problem?

Abstract

C. Jacobi found that in the general N-body problem (including N = 3), the negativity of the total energy of the system was necessary for the Lagrangian stability of any solution. For the restricted three-body problem, this statement is trivial, since a zero-mass body makes zero contribution to the system energy. If we consider only the equations describing the motion of a zero-mass point, then the energy integral disappears. However, if we average the equations over the longitudes of the main bodies, the energy integral appears again. Is the Jacobi theorem true in this case? It turns out that it is not. For arbitrarily large values of total energy, there are bounded periodic orbits. At the same time, the negativity of the energy turned out to be sufficient for the boundedness of an orbit in the configuration space.

APPENDIX

Let

$$W = \frac{{{\mathbf{K}}(k)}}{{\sqrt \varphi }},\quad {{k}^{2}} = \frac{{4c\varrho }}{\varphi },\quad 1 - {{k}^{2}} = \frac{\psi }{\varphi },$$

(A.1)

where

$$\varphi = {{\varrho }^{2}} + {{z}^{2}} + {{c}^{2}} + 2c\varrho ,\quad \psi = {{\varrho }^{2}} + {{z}^{2}} + {{c}^{2}} - 2c\varrho .$$

Clearly, 0 $\leqslant$ k $\leqslant$ 1. The equality on the left is achieved only on the z axis (at $\varrho$ = 0), and the equality on the right is achieved only on a special circle (at $\varrho$ = c, z = 0).

The derivative of the elliptic integral is known [4]

$$\frac{{\partial {\mathbf{K}}}}{{\partial ({{k}^{2}})}} = \frac{1}{{2{{k}^{2}}}}\left( {\frac{1}{{1 - {{k}^{2}}}}{\mathbf{E}} - {\mathbf{K}}} \right) = \frac{\varphi }{{8c\varrho }}\left( {\frac{\varphi }{\psi }{\mathbf{E}} - {\mathbf{K}}} \right).$$

The equalities

$$\frac{{\partial ({{k}^{2}})}}{{\partial \varrho }} = \frac{{4c({{z}^{2}} + {{c}^{2}} - {{\varrho }^{2}})}}{{{{\varphi }^{2}}}},\quad \frac{{\partial ({{k}^{2}})}}{{\partial z}} = - \frac{{8c\varrho z}}{{{{\varphi }^{2}}}}$$

allow us to calculate the desired derivatives using the cylindrical coordinates $\varrho$ and z:

$$\frac{{\partial {\mathbf{K}}}}{{\partial \varrho }} = \frac{{{{z}^{2}} + {{c}^{2}} - {{\varrho }^{2}}}}{{2\varrho \varphi }}\left( {\frac{\varphi }{\psi }{\mathbf{E}} - {\mathbf{K}}} \right),\quad \frac{{\partial {\mathbf{K}}}}{{\partial z}} = - \frac{z}{\varphi }\left( {\frac{\varphi }{\psi }{\mathbf{E}} - {\mathbf{K}}} \right).$$

Now it is easy to obtain the components of the gradient of the function W in the cylindrical coordinates

$$\frac{{\partial W}}{{\partial \varrho }} = \frac{1}{{2\varrho \sqrt \varphi }}\left( {\frac{{{{z}^{2}} + {{c}^{2}} - {{\varrho }^{2}}}}{\psi }{\mathbf{E}} - {\mathbf{K}}} \right),\quad \frac{{\partial W}}{{\partial z}} = - \frac{z}{{\psi \sqrt \varphi }}{\mathbf{E}}$$

and the virial

$$\varrho \frac{{\partial W}}{{\partial \varrho }} + z\frac{{\partial W}}{{\partial z}} = \frac{1}{{2\sqrt \varphi }}\left( {\frac{{{{c}^{2}} - {{\varrho }^{2}} - {{z}^{2}}}}{\psi }{\mathbf{E}} - {\mathbf{K}}} \right).$$

(A.2)