Page:Newton's Principia (1846).djvu/185

 between the upper and the lower apsis will be equal to $$\scriptstyle \frac{180}{2} $$deg., or 90 deg. Therefore the body having performed a fourth part of one revolution, will arrive at the lower apsis, and having performed another fourth part, will arrive at the upper apsis, and so on by turns in infinitum. This appears also from Prop. X. For a body acted on by this centripetal force will revolve in an immovable ellipsis, whose centre is the centre of force. If the centripetal force is reciprocally as the distance, that is, directly as $$\scriptstyle \frac{1}{A}$$ or $$\scriptstyle \frac{A^{2}}{A^{3}}$$, n will be equal to 2; and therefore the angle between the upper and lower apsis will be $$\scriptstyle \frac{180}{\sqrt{2}}$$ deg., or 127 deg., 16 min., 45 sec.; and therefore a body revolving with such a force, will by a perpetual repetition of this angle, move alternately from the upper to the lower and from the lower to the upper apsis for ever. So, also, if the centripetal force be reciprocally as the biquadrate root of the eleventh power of the altitude, that is, reciprocally as $$\scriptstyle A\frac{11}{4}$$, and, therefore, directly as $$\scriptstyle \frac{1}{A\frac{11}{4}}$$ or as $$\scriptstyle \frac{A\frac{1}{4}}{A^{3}}$$, n will be equal to ¼, and $$\scriptstyle \frac{180}{\sqrt{n}}$$ deg. will be equal to 360 deg.; and therefore the body parting from the upper apsis, and from thence perpetually descending, will arrive at the lower apsis when it has completed one entire revolution; and thence ascending perpetually, when it has completed another entire revolution, it will arrive again at the upper apsis; and so alternately for ever.

. 3. Taking m and n for any indices of the powers of the altitude, and b and c for any given numbers, suppose the centripetal force to be as $$\scriptstyle \frac{bA^{m}+ca^{n}}{A^{3}}$$, that is, as $$\scriptstyle \frac{b\ into\ \overline{T-X}|^{m}+c\ into\ \overline{T-X}|^{n}}{A^{3}}$$ or (by the method of converging series above-mentioned) as

$$\scriptstyle \frac{\frac{bT^{m}-cT^{n}-mbXT^{m-1}ncXT^{n-1}}{}+\frac{mm-m}{2}bXXT^{m-2}+\frac{nn-n}{2}cXXT^{n-2},}{A^{3}}$$ &c.

and comparing the terms of the numerators, there will arise RGG - RFF + TFF to bTm + cTn as - FF to - mbTm-1 - ncTn-1 + $$\scriptstyle \frac{mm-m}{2}$$ bXTm-2 + $$\scriptstyle \frac{nn-n}{2}$$ cXTn-2, &c. And taking the last ratios that arise when the orbits come to a circular form, there will come forth GG to bTm-1 + cTn-1 as FF to mbTm-1 + ncTn-1; and again, GG to FF as bTm-1 + cTn-1 to mbTn-1 + ncTn-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