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/49

 $$r=\frac{p}{2 e \cos \frac{1}{2}(v-\psi) \cos \frac{1}{2}(v+\psi)}$$

For $v=0,$ the factors $\cos \frac{1}{2}(v-\psi),$  and $\cos \frac{1}{2}(v+\psi),$  in the denominator of this fraction become equal, the second vanishes for the greatest positive value of $v,$  and the first for the greatest negative value. Putting, therefore,

$$\frac{\cos \frac{1}{2}(v-\psi)}{\cos \frac{1}{2}(v+\psi)}=u$$

we shall have $u=1$ in perihelion; it will increase to infinity as $v$  approaches its limit $180^{\circ}-\psi;$  on the other hand it will decrease indefinitely as $v$  is supposed to return to its other limit $-\left(180^{\circ}-\psi\right);$  so that reciprocal values of $u,$  or, what amounts to the same thing, values whose logarithms are complementary, correspond to opposite values of $v.$

This quotient $u$ is very conveniently nsed in the hyperbola as an auxiliary quantity; the angle, the tangent of which is

$$\tan \frac{1}{2} v \sqrt{\frac{e-1}{e+1}}$$

can be made to render the same service with almost equal elegance; and in order to preserve the analogy with the ellipse, we will denote this angle by $\frac{1}{3} F.$ In this way the following relations between the quantities $v, r, u, F$  are easily brought together, in which we put $a=-b,$  so that $b$  becomes a positive quantity. I. $b=p \operatorname{cotan}^{2} \psi$

II. $r=\frac{p}{1+e \cos v}=\frac{p \cos \psi}{2 \cos \frac{1}{2}(v-\psi) \cos \frac{1}{2}(v+\psi)}$

III. $\tan \frac{1}{2} F=\tan \frac{1}{8} v \sqrt{\frac{e-1}{e+1}}=\tan \frac{1}{2} v \tan \frac{1}{2} \psi=\frac{u-1}{u+1}$

IV. $u=\frac{\cos \frac{1}{2}(v-\psi)}{\cos \frac{1}{2}(v+\psi)}=\frac{1+\tan \frac{1}{2} F}{1-\tan \frac{1}{2} F^{\prime}}=\tan \left(45^{\circ}+\frac{1}{2} F^{\prime}\right)$ V. $\frac{1}{\cos F}=\frac{1}{2}\left(u+\frac{1}{u}\right)=\frac{1+\cos \psi \cos v}{2 \cos \frac{1}{2}(v-\psi) \cos \frac{1}{2}(v+\psi)}=\frac{e+\cos v}{1+e \cos v}.$

By subtracting 1 from both sides of equation $V.$ we get,

VI. $\sin \frac{1}{2} v \sqrt{ } \frac{\sin \frac{1}{2} F}{} \sqrt{\frac{p}{(e-1) \cos F}}=\sin \frac{1}{2} F \sqrt{\frac{(e+1) b}{\cos F}}$

$$=\frac{1}{2}(u-1) \sqrt{\frac{p}{(e-1) u}}=\frac{1}{2}(u-1) \sqrt{\frac{(e+1) b}{u} .}$$