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/76

 We present the equation between the time $t$ and the auxiliary quantity $u$  in. the following form :-

$$(e-1)\left(\frac{1}{20}\left(u-\frac{1}{u}\right)+\frac{9}{10} \log u\right)+\left(\frac{1}{10}+\frac{9}{10} e\right)\left(\frac{1}{2}\left(u-\frac{1}{u}\right)-\log u\right)=k t\left(\frac{e-1}{q}\right)^{\frac{3}{2}}$$

in which the logarithms are hyperbolic, and

$$\frac{1}{20}\left(u-\frac{1}{u}\right)+\frac{9}{10} \log u$$

is a quantity of the first order,

$$\frac{1}{3}\left(u-\frac{1}{u}\right)-\log u$$

a quantity of the third order, when $\log u$ may be considered as a small quantity of the first order. Putting, therefore,

$$\frac{6\left(\frac{1}{2}\left(u-\frac{1}{u}\right)-\log u\right)}{\frac{1}{20}\left(u-\frac{1}{u}\right)+\frac{9}{10} \log u}=4 A, \frac{\frac{1}{20}\left(u-\frac{1}{u}\right)+\frac{9}{10} \log u}{2 \sqrt{ } A}=B,$$

$A$ will be a quantity of the second order, but $B$  will differ from unity by a difference of the fourth order. Our equation will then assume the following form:-

$$B\left(2(e-1) A^{\frac{1}{2}}+\frac{2}{15}(1+9 e) A^{\frac{3}{2}}\right)=k t\left(\frac{e-1}{q}\right)^{\frac{3}{2}}$$. . . . . [2]

which is entirely analogous to equation [1] of article 37. Putting moreover,

$$\left(\frac{u-1}{u+1}\right)^{2}=T$$

$T$ will be a quantity of the second order, and by the method of infinite series will be found

$\frac{A}{T} = 1 + \frac{4}{5} A + \ frac{8}{175} A^2 - \frac{8}{525} A^3 + \frac{1896}{336875} A^4 - \frac{28744}{13138125} A^5 +$ etc.

Wherefore, putting

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

$C$ will be a quantity of the fourth order, and

$$A=\frac{(1+C) T}{1-\frac{1}{8} T}$$

Finally, for the radius vector, there readily follows from equation VII., article 21,

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