and would you say that in step B-4 on that link that it is using a 5th order expansion?
Let's do this systematically. Consider the 5th order expansion for the 'equation of centre':
[math]
\nu - M = \alpha_1\,\sin(M)+\alpha_2\,\sin(2\,M)+\alpha_3\,\sin(3\,M)+\alpha_4\,\sin(4\,M)+\alpha_5\,\sin(5\,M)
[/math]
where
[math]
\alpha_1 = 2\,e-\frac{1}{4}\,e^3+\frac{5}{96}\,e^5
[/math]
[math]
\alpha_2 = \frac{5}{4}\,e^2-\frac{11}{24}\,e^4
[/math]
[math]
\alpha_3 = \frac{13}{12}\,e^3-\frac{43}{64}\,e^5
[/math]
[math]
\alpha_4 = \frac{103}{96}\,e^4
[/math]
[math]
\alpha_5 = \frac{1097}{960}\,e^5
[/math]
To calculate the coefficients of the expansion, we need to know [MATH]e[/MATH], Mars' orbital eccentricity. From
http://ntrs.nasa.gov/archive/nasa/ca...0000097895.pdf (the second of the two papers I listed earlier), we see that the orbital eccentricity is given as:
[MATH]e = 0.09340 + 2.477 \times 10^{-9}\,d^{-1}\,\Delta t_{J2000}[/MATH]
This basically is a numerical least-squares fit of a linear approximation of the orbital eccentricity to JPL's high precision ephemeris. The first term is the value of the orbital eccentricity at epoch J2000. The second term is a 'drift' term indicating that the Mars orbital eccentricity is not constant with time.
For a moment, let's forget about the drift component and just work with the 0.09340. If we plug this value into the above, we get:
[math]
\alpha_1 = 0.186597
[/math]
[math]
\alpha_2 = 0.0108696
[/math]
[math]
\alpha_3 = 0.000877903
[/math]
[math]
\alpha_4 = 0.0000816495
[/math]
[math]
\alpha_5 = \text{8.122126994772126$\grave{ }$*${}^{\wedge}$-6}
[/math]
But these values are expressed in radians whereas B4 expresses the coefficients in terms of degrees. So, to compare apples with apples we need to multiply all of the above coefficient values by [MATH]180/\pi[/MATH]. If we do this (and round after the 4th Decimal place, we get:
[math]
\alpha_1 = 10.6912\,{}^{\circ}
[/math]
[math]
\alpha_2 = 0.6228\,{}^{\circ}
[/math]
[math]
\alpha_3 = 0.0503\,{}^{\circ}
[/math]
[math]
\alpha_4 = 0.0047\,{}^{\circ}
[/math]
[math]
\alpha_5 = 0.0005\,{}^{\circ}
[/math]
So that the 5th order expansion for the 'equation of centre' becomes:
[math]
\nu - M = 10.6912\,{}^{\circ}\,\sin(M)+0.6228\,{}^{\circ}\,\sin(2\,M)+ 0.0503\,{}^{\circ}\,\sin(3\,M)+0.0047\,{}^{\circ}\,\sin(4\,M)+0.0005\,{}^{\circ}\,\sin(5\,M)
[/math]
Clearly, this is essentially the same as the values of B4 - although the consistency of some of the rounding used in that expression is a little dubious in my view.
But what about the 'drift term'? If, as per B4, we include the first order (in [MATH]\Delta t[/MATH]) term in the coefficient of [MATH]\sin(M)[/MATH], we find the that the better expression for [MATH]\alpha_1[/MATH] is:
[math]
\alpha_1 = (10.6912 + 2.82918\times 10^{-7}\,\Delta t_{J2000})\,{}^{\circ}
[/math]
And that, I think, explains the drift term in the coefficient of [MATH]\sin(M)[/MATH] appearing in B4.
So, yes, B4 is using a 5th order expansion.
Note, though that there is an additional 'PBS' term in B4. You might ask: what is this about? Well, if you go back to the source papers, it easy to see that the authors are using a simple mean orbital parameter model for the motion of Mars around the Sun. In practice, Mars' motion is more complicated than this - largely due to the combined influence of Jupiter and Saturn - and the 'PBS' term is an attempt to take into account the major perturbations of the 'equation of centre' induced by those planets.
If you have a comment on how what your showing me compares to what the link is saying in steps B-4 and B-2, B-5 I would be interested.
B2 calculates the 'fictitious mean sun' longitude.
B3 calculates the effect of major planetary perturbations on Mars' 'equation of centre'
B4 calculates the 'equation of centre' using (in the main) the 5th order expansion for the equation of centre and then adding the perturbation terms
B5 adds the fictitious mean longitude of the Sun to the equation of centre.
