Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/268

there will ariſe RGC - RFF + TFF to $\scriptstyle bT^{m}+cT^{n}$ as -FF to $\scriptstyle -mbT^{m-1}-ncT^{n-2}+{\frac {mm-m}{2}}bXT^{m-2}+{\frac {nn-n}{2}}cXT^{n-2}$ &c. And taking the laſt ratio's that ariſe when the orbits come to a circular form, there will come forth GG to $\scriptstyle bT^{m-1}+cT{n-1}$ as FF to $\scriptstyle bT^{m-2}+ncT^{n-1}$ and again GG to FF as $\scriptstyle bT^{m-1}+cT{n-1}$ to $\scriptstyle bT^{m-2}+ncT^{n-1}$. This proportion. by expreſſing the greateſt altitude CV or T arithmetically by unity, becomes, GG to FF as b + c to mb + nc, and therefore as 1 to $\textstyle {\frac {mb+nc}{b+c}}$. Whence G becomes to F, that is the angle VCp to the angle VCP as 1 to $\textstyle {\sqrt {\frac {mb+nc}{b+c}}}$. And therefore ſince the angle VCP between the upper and the lower apſis, in an immovable ellipſis, is of 180 deg. the angle VCp between the ſame apſides in an orbit which a body deſcribes with a centripetal force. that is as $\textstyle {\frac {bA^{m}+cA^{n}}{A^{2}}}$ will be equal to an angle of $\textstyle 180{\sqrt {\frac {b+c}{mb+nc}}}$ deg. And by the ſame reaſoning if the centripetal force be as $\textstyle {\frac {bA^{m}-cA^{n}}{A^{3}}}$ the angle between the apſides will be found equal to $\textstyle 180{\sqrt {\frac {b+c}{mb+nc}}}$ deg. After the ſame manner the problem is ſolved in more difficult caſes. The quantity to which the centripetal force is proportional. muſt always be reſolved into a converging