ALONG THE CIRCUMFERENCE OF A CIRCLE. 65 



whence also 



NC = (NZ ,ti^ F)2 ; 4NF . NC = (NZ + NF) 2 . 

 4NF ' 



14. Having thus determined the position of the point C, we can determine 

 also the intensity of the gravitation for, putting for NI the value just found, 



^• NZ 2 Xz-NF) ^ NE(NZ + Nr) 



whence G y = NZ . NE ; so that the intensity of the gravitation for 7 must be a 

 mean proportional between those for a and ft. 



15. A descent from C under the influence of gravitation of the intensity 

 NZ . ZE would cause a heavy body to have at Z and at N the very velocities 

 which the moveable point 7 has at those places ; and we have now to inquire 

 whether the same influence would give to that body, when at any intermediate 

 point, the corresponding velocity. Before treating of this matter generally, it 

 may be instructive to inquire into the velocity of the point 7 when it is at M the 

 middle of FZ ; the moveable point 7 is at M when a is at Z and ft at F, now at 

 that instant the velocity of ft is zero, while the velocity of a, proportional to the 

 square root of NZ . ZA, is represented by NZ . ZE, so that the velocity of 7 must 

 thenbeiNZ.ZE. 



But the velocity of the heavy body when at M is given by the general formula, 



J(G y . NZ . QO) = V{G r . NZ (NC - NQ)f . 



Substituting for Gy the value above found we have 



vj = NZ . NE (NP - NM 2 ) 



but since M is the middle of the arc FZ 



2NZ : NZ + NF or 2NE : NE + NZ : : NZ 2 : NM 2 



wherefore 



= NZ 2 NE + NZ ; but NI 2 = NZ 2 < NE + NZ > 2 

 2NE 4NE.NZ 



wherefore 



NI 2 - NM 2 - NZ 2 pE 2 +2NE.NZ + NZ 2 _ 2NE.NZ + 2NZ 2 } 

 I 4NE.NZ 4NE.NZ J 



NE 2 - NZ 2 ZE 2 



= NZ 2 ,~L ™„ = NZ 2 - 



4NE . NZ 4NE . NZ 



and consequently 



vj = \ . NZ 2 . ZE 



or 



v =iNZ . ZE 



