Page:Newton's Principia (1846).djvu/119



From C, the centre of the semi-circle, let the semi-diameter CA be drawn, cutting the parallels at right angles in M and N, and join CP. Because of the similar triangles CPM, PZT, and RZQ, we shall have CP² to PM² as PR² to QT²; and, from the nature of the circle, PR² is equal to the rectangle $$\scriptstyle QR\times\overline{RN+QN}$$, or, the points P, Q, coinciding, to the rectangle $$\scriptstyle QR\times2PM$$. Therefore CP² is to PM² as $$\scriptstyle QR\times2PM$$ to QT²; and $$\scriptstyle \frac{QT^{2}}{QR}=\frac{2PM^{3}}{CP^{2}}$$, and $$\scriptstyle \frac{QT^{2}\times SP^{2}}{QR}=\frac{2PM^{3}\times SP^{2}}{CP^{2}}$$. And therefore (by Corol. 1 and 5; Prop. VI.), the centripetal force is reciprocally as $$\scriptstyle \frac{2PM^{3}\times SP^{2}}{CP^{2}}$$; that is (neglecting the given ratio $$\scriptstyle \frac{2SP^{2}}{CP^{2}}$$), reciprocally as PM³. Q.E.I.

And the same thing is likewise easily inferred from the preceding Proposition.

And by a like reasoning, a body will be moved in an ellipsis, or even in an hyperbola, or parabola, by a centripetal force which is reciprocally as the cube of the ordinate directed to an infinitely remote centre of force.

Suppose the indefinitely small angle PSQ to be given; because, then, all the angles are given, the figure SPRQT will be given in specie. Therefore the ratio $$\scriptstyle \frac{QT}{QR}$$ is also given, and $$\scriptstyle \frac{QT^{2}}{QR}$$ is as QT, that is (because the figure is given in specie), as SP. But if the angle PSQ is any way changed, the right line QR, subtending the angle of contact QPR