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/35

 8.

Let us now collect together those relations between the anomalies and the radius vector which deserve particular attention, the derivation of which will present no difficulties to any one moderately skilled in trigonometrical analysis. Greater elegance is attained in most of these formulas by introducing in the place of $e$ the angle the sine of which $=e.$  This angle being denoted by $\varphi,$  we have

$$\begin{aligned} & \sqrt{1-ee}=\cos \varphi, \quad \sqrt{1+e}=\cos(45^{\circ}-\tfrac{1}{2} \varphi) \sqrt{2}, \\ & \sqrt{1-e}=\cos (45^{\circ}+\tfrac{1}{2} \varphi) \sqrt{2}, \quad \sqrt{\frac{1-e}{1+e}}=\tan (45^{\circ}-\tfrac{1}{2} \varphi), \\ & \sqrt{1+e}+\sqrt{1-e}=2 \cos \tfrac{1}{2} \varphi, \quad \sqrt{1+e}-\sqrt{1-e}=2 \sin \tfrac{1}{2} \varphi. \end{aligned}$$

The following are the principal relations between $a,$ $p,$  $r,$  $e,$  $\varphi,$  $v,$  $E,$  $M.$

1. $p=a \cos ^{2} \varphi$

2. $r=\frac{p}{1+e \cos v}$

3. $r=a(1-e \cos E)$

4. $\cos E=\frac{\cos v+e}{1+e \cos v},$ or $\cos v=\frac{\cos E-e}{1-e \cos E}$

5. $\sin \frac{1}{2} E =\sqrt{\frac{1}{2}(1-\cos E)}=\sin \frac{1}{2} v \sqrt{\frac{1-e}{1+e \cos v}}$ $\phantom{\sin \frac{1}{2} E}=\sin \frac{1}{2} v \sqrt{\frac{r(1-e)}{p}}=\sin \frac{1}{2} v \sqrt{\frac{r}{a(1+e)}}$ | $\cos \frac{1}{2} E=\sqrt{\frac{1}{2}(1+\cos E)}=\cos \frac{1}{2} v \sqrt{\frac{1+e}{1+e \cos v}}$ $\phantom{\cos \frac{1}{2} E}=\cos \frac{1}{2} v \sqrt{\frac{r(1+e)}{p}}=\cos \frac{1}{2} v \sqrt{\frac{r}{a(1-e)}}$ | $\tan \frac{1}{2} E=\tan \frac{1}{2} v \tan \left(45^{\circ}-\frac{1}{2} \varphi\right)$ | $\sin E=\frac{r \sin v \cos \varphi}{p}=\frac{r \sin v}{a \cos \varphi}$ | $r \cos v=a(\cos E-e)=2 a \cos \left(\frac{1}{2} E+\frac{1}{2} \varphi+45^{\circ}\right) \cos \left(\frac{1}{2} E-\frac{1}{2} \varphi-45^{\circ}\right)$ | $\sin \frac{1}{2}(v-E)=\sin \frac{1}{2} \varphi \sin v \sqrt{\frac{r}{p}}=\sin \frac{1}{2} \varphi \sin E \sqrt{\frac{a}{r}}$ | $\sin \frac{1}{2}(v+E)=\cos \frac{1}{2} \varphi \sin v \sqrt{\frac{r}{p}}=\cos \frac{1}{2} \varphi \sin E \sqrt{\frac{a}{r}}$ | $M=E-e \sin E.$ |undefined