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/46

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18.

In the PARABOLA, the eccentric anomaly, the mean anomaly, and the mean motion, become ${\textstyle =0;}$ here therefore these ideas cannot aid in the comparison of the motion with the time. In the parabola, however, there is no necessity for an auxiliary angle in integrating ${\textstyle r\mathrm {~d} v;}$ for we have

and thus,

$$rr\mathrm {~d} v={\frac {pp\mathrm {~d} v}{4\cos ^{4}{\frac {1}{2}}v}}={\frac {pp\mathrm {~d} \tan {\frac {1}{2}}v}{2\cos ^{2}{\frac {1}{2}}v}}={\frac {1}{2}}pp\left(1+\tan ^{2}{\frac {1}{2}}v\right)\mathrm {d} \tan {\frac {1}{2}}v$$

$$\int rrdv={\frac {1}{2}}pp\left(\tan {\frac {1}{2}}v+{\frac {1}{3}}\tan ^{3}{\frac {1}{2}}v\right)+{\text{ Constant. }}$$

If the time is supposed to commence with the perihelion passage, the Constant ${\textstyle =0;}$ therefore we have

$$\tan {\frac {1}{2}}v+{\frac {1}{3}}\tan ^{3}{\frac {1}{2}}v={\frac {2tkv(1+\mu )}{p^{\frac {3}{2}}}}$$

by means of which formula, ${\textstyle t}$ may be derived from ${\textstyle v,}$ and ${\textstyle v}$ from ${\textstyle t,}$ when ${\textstyle p}$ and ${\textstyle \mu }$ are known. In the parabolic elements it is usual, instead of ${\textstyle p,}$ to make use of the radius vector at the perihelion, which is ${\textstyle {\frac {1}{2}}p,}$ and to neglect entirely the mass ${\textstyle \mu .}$ It will scarcely ever be possible to determine the mass of a body, the orbit of which is computed as a parabola; and indeed all comets appear, according to the best and most recent observations, to have so little density and mass, that the latter can be considered insensible and be safely neglected.

19.

The solution of the problem, from the true anomaly to find the time, and, in a still greater degree, the solution of the inverse problem, can be greatly abbreviated by means of an auxiliary table, such as is found in many astronomical works.