NEWTON S PRINCIPIA. 99 



and next the force, with the velocity, when the curve is 

 given. 



For this purpose, let P be the perpendicular S C to the 

 plane from S the given centre, this being the shortest 

 line from the point to the plane ; D the distance S P from 

 the centre to any point P of the curve; the distance C P = d 

 of that point P to the centre in the plane, that is, to the 

 point C where p falls on the plane; and let F, the central 

 force, be represented by R S. It is evident that the force 

 R S, acting in the line P S (without the plane), is com 

 pounded of two, R K and S K, of which R K only can 

 have any effect on the motion in the plane, the other S K 

 which tends to draw the body out of the plane being by the 

 supposition nothing, because the body moves wholly in the 



Q Ti /^ T&amp;gt; 



plane C P B E. But by similar triangles, R K = p-g 



= -^ ; therefore if the proportion of the centripetal force 



to the distance be known, that is, if F = D M , R K = d. 

 D&quot;- 1 . But D 2 = d 2 + p 2 , and D = Vd 2 +p* ; therefore 

 R K, the force acting at P towards the centre C, is d x 

 (d 2 + p 2 ) ^, which gives it in terms of the distance C P, 

 and the given line S C. Thus if the central force is as 

 the distance S P, the force acting towards the centre 

 becomes equal to d, or as the distance on the plane. So if 

 the central force is inversely as the distance, then n 1, 



and the force to the centre on the plane is ^ r 

 -= 2&quot;, and if it is inversely as the square of the dis- 



tance, the force on the plane is ; -. But the 



&amp;lt;&amp;gt; + p*)? 

 central force being given in the plane, the investigation 



H 2 



