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Next: Bibliography Up: Proper elements for Earth Previous: 7. Conclusions and future

8. Appendix


  
Figure: An illustration of the discontinuity extraction procedure. Top left: the graphic of the derivative of the integral average of $1/D$ with respect to $\omega $, as a function of $\omega ,e$; there are two lines of discontinuity, corresponding to crossings at the ascending node with the Earth and at the descending one with Venus. Top right: the derivative of the integral average of $1/d$ with respect to $\omega $; note the discontinuity at the ascending node with the Earth. Bottom left: the difference between the two previous functions shows that the discontinuity due to the Earth has been removed; note that the difference function is continuous, but not differentiable along the ascending node crossing line. The discontinuity at the descending node with Venus is still present. Bottom right: the level curves of the integral average of $1/D$.
\begin{figure*}
\centerline{\psfig{figure=figure/remjump.eps,height=9.5cm}}
\end{figure*}

This appendix contains the analytical formulas representing the discontinuity of the derivatives of the averaged perturbing function $\overline {R}$.

If we choose, for instance, the direction from the $\{d^+_{nod}>0\}$to the $\{d^+_{nod}<0\}$ area, the difference in the twofold definition of the derivatives of the averaged perturbing function are

 \begin{eqnarraystar}\mbox{Diff}\biggl({\partial\overline{R}\over \partial G}\big...
...iff}\biggl({\partial\overline{R}\over \partial Z}\biggr)&&= 0
\end{eqnarraystar}



where


\begin{displaymath}det(\underline{\underline {\bf A}}) = {a^2 a'^2\over
(1-e^2)}\cdot(\sin^2I (1 +2 e\cos\omega + e^2) +
e^2\cos^2I \sin^2\omega)
\end{displaymath}

\begin{eqnarraystar}\mbox{Diff}\biggl({\partial\over \partial e}\vert d^+_{nod}\...
...mega}\biggr] =
-{2a(1-e^2)e\sin\omega\over(1+e\cos\omega)^2}
\end{eqnarraystar}



(see [Gronchi and Milani 1998] for the notation).

In case of crossing at the descending node, we have to modify the previous formulas as follows:

\begin{displaymath}det(\underline{\underline {\bf A}}) = {a^2 a'^2\over
(1-e^2)}\cdot(\sin^2I (1 -2 e\cos\omega + e^2) + e^2\cos^2I
\sin^2\omega)
\end{displaymath}

\begin{eqnarraystar}\mbox{Diff}\biggl({\partial\over \partial e}\vert d^-_{nod}\...
...artial\omega} =
{2a(1-e^2)e\sin\omega\over(1-e\cos\omega)^2}
\end{eqnarraystar}




next up previous
Next: Bibliography Up: Proper elements for Earth Previous: 7. Conclusions and future
G.-F. Gronchi
2000-05-15