Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/374

286 L, A, S3, B, (Fig. 3.) erect the perpendicualrs Ll, Aa, Ss, Bb, of which uppoe Ss equal to SI; and through the point s, to the aymptotes Ll, LB, decribe the hyperbola asb meeting the perpendiculars Aa, Bb, in a and b; and the rectangle 2ASI ubducted from the hyperbola AasbB, will leave ANB the area ought.

If the centripetal force tending to the everal particles of the pheres decreae in a quadruplicate ratio of the ditance from the particles; write $$\scriptstyle \frac {PE^4}{2AS^3}$$ for V, then $$\scriptstyle \overline \sqrt {2PS + LD}$$ for PE, and DN will become as $$\scriptstyle \frac {SI^2 \times SL}{\sqrt {2SI}} \times \frac 1{\sqrt {LD^2}} - \frac {SI^2}{2 \sqrt {2SI}} \times \frac 1{\sqrt {LD}} - \frac {SI^2 \times ALB}{2 \sqrt {2SI}} \times \frac 1{\sqrt {LD^2}}$$ Thee three parts drawn into the length AB, produce o many areas, viz. $$\scriptstyle {2SI^2 \times SL}{\sqrt {2SI}}$$ into $$\scriptstyle \overline {\frac 1{\sqrt {LA}} - \frac 1{\sqrt {LB}}}$$; $$\scriptstyle \frac {SI^2}{\sqrt {2SI}}$$ into $$\scriptstyle \overline {\sqrt {LB - LA}}$$; and $$\scriptstyle \frac {SI^2 \times ALB}{3 \sqrt {2SI}}$$ into $$\scriptstyle \overline {\frac 1{\sqrt {LA^2}} - \frac 1{\sqrt {LB^2}}}$$. And thee after due reduction come forth $$\scriptstyle \frac {2SI^2 \times SL}{LI}$$, $$\scriptstyle SI^2$$ and $$\scriptstyle SI^2 + \frac {2SI^2}{3LI}$$. And thee by ubducting the lat from the firt become $$\scriptstyle \frac {4SI^2}{3LI}$$. Therefore the entire force with which the corpucle P is attracted towards the centre of the phere is as $$\scriptstyle \frac {SI^2}{PI}$$, that us reciprocally as $$\scriptstyle PS^2 \times PI$$. Q. E. I.