Page:Theory of the motion of the heavenly bodies moving about the sun in conic sections- a translation of Gauss's "Theoria motus." With an appendix (IA theoryofmotionof00gaus).pdf/73

 $$A=\frac{T-\frac{6}{8} T^{2}+\frac{9}{1} T^{8}-\frac{12}{8} T^{4}+1 \frac{5}{1} T^{6}-\text { etc. }}{1-\frac{8}{18} T+\frac{7}{25} T^{2}-\frac{8}{36} T^{18}+\frac{9}{45} T^{4}-\text { etc. }},$$

in which the law of progression is obvious. Hence is deduced, by the inversion of the series,

$$\frac{A}{T}=1-\frac{4}{5} A+\frac{8}{175} A^{2}+\frac{8}{526} A^{3}+\frac{1}{33 \frac{8}{8} \frac{9}{8} \frac{6}{75}} A^{4}+\frac{28744}{1313 \frac{4}{81} \frac{4}{26}} A^{5}+\text { etc. }$$

Putting, therefore,

$$\frac{A}{T}=1-\frac{4}{5} A+C,$$

$C$ will be a quantity of the fourth order, which being included in our table, we can pass directly to $v$  from $A$  by means of the formula,

$$\tan \frac{1}{3} v=\sqrt{\frac{1+e}{1-e}} \sqrt{\frac{A}{1-\frac{c}{6} A+C}}=\frac{\gamma \tan \frac{1}{2} w}{\sqrt{\left(1-\frac{c}{6} A+C\right)}},$$

denoting by $\gamma$ the constant

$$\sqrt{\frac{5+5 e}{1+9 e}} .$$

In this way we gain at the same time a very convenient computation for the radius vector. It becomes, in fact, (article 8, VI.),

$$r=\frac{q \cos ^{2} \frac{1}{2} E}{\cos ^{2} \frac{1}{2} v}=\frac{q}{(1+T) \cos ^{2} \frac{1}{2} v}=\frac{\left(1-\frac{4}{5} A+C\right) q}{\left(1+\frac{1}{6} A+C\right) \cos ^{2} \frac{1}{2} v}$$

41.

Nothing now remains but to reduce the inverse problem also, that is, the determination of the time from the true anomaly, to a more expeditious form of computation: for this purpose we have added to our table a new column for $T.$ $T,$  therefore, will be computed first from $v$  by means of the formula

$$T=\frac{1-e}{1+e} \tan ^{2} \frac{1}{2} v$$

then $A$ and $\log B$  are taken from our table with the argument $T,$  or, (which is more accurate, and even more convenient also), $C$  and $\log B,$  and hence $A$  by the formula

$$A=\frac{(1+O) T}{1+\frac{4}{3} T}$$

finally $t$ is derived from $A$  and $B$  by formula [1], article 37. If it is desired to call into use the Barkerian table here also, which however in this inverse problem