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/50

 In the same manner, by adding 1 to both sides, it becomes

VII. $\cos \frac{1}{2} v \sqrt{ } r=\cos \frac{1}{2} F \sqrt{\frac{p}{(e+1) \cos F}}=\cos \frac{1}{3} F \sqrt{\frac{(e-1) b}{\cos F}}$

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

By dividing VI. by VII. we should reproduce III.: the multiplication produces VIII. $r \sin v=p \operatorname{cotan} \psi \tan F=b \tan \psi \tan F$

$$=\frac{1}{2} p \operatorname{cotan} \psi\left(u-\frac{1}{u}\right)=\frac{1}{2} b \tan \psi\left(u-\frac{1}{u}\right) \text {. }$$

From the combination of the equations II. V. are easily derived

IX. $r \cos v=b\left(e-\frac{1}{\cos F}\right)=\frac{1}{2} b\left(2 e-u-\frac{1}{u}\right),$ X. $r=b\left(\frac{e}{\cos F}-1\right)=\frac{1}{2} b\left(e\left(u+\frac{1}{u}\right)-2\right).$

22.
By the differentiation of the formula IV. (regarding $\psi$ as a constant quantity) we get

$$\frac{\mathrm{d} u}{u}=\frac{1}{2}\left(\tan \frac{1}{2}(v+\psi)-\tan \frac{1}{2}(v-\psi)\right) \mathrm{d} v=\frac{r \tan \psi}{p} \mathrm{~d} v$$

hence,

$$r r \mathrm{~d} v=\frac{p r}{u \tan \psi} \mathrm{d} u$$

or by substituting for $r$ the value taken from $\mathrm{X}.$

$$r r \mathrm{~d} v=b b \tan \psi\left(\frac{1}{2} e\left(1+\frac{1}{u u}\right)-\frac{1}{u}\right) \mathrm{d} u$$

Afterwards by integrating in such a manner that the integral may vanish at the perihelion, it becomes

$$

\int r r \mathrm{d}v = b b a \tan \psi \left ( \frac{1}{2} e \left ( u - \frac{1}{u} \right ) -\log u \right ) = kt \sqrt{p} \sqrt { \left ( 1 + \mu \right ) } = kt \tan \psi \sqrt{b}  \sqrt{ 1 + \mu } $$

The logarithm here is the hyperbolic; if we wish to use the logarithm from Brigg’s system, or in general from the system of which the modulus $=\lambda,$ and