55-57] LAW OF INVERSE SQUARE. 59 



towards 0, i.e. we shall have 



X = fjbX . 



Since the radius vector OP describes areas uniformly we can 

 utilise the figure to express the position in terms of the time. 



Let angle AOP = 0, and let t be the time of going from A to 

 N. Then 



~ Ar OP 2 (2a cos 0) 2 

 x = ON = -z = v s = 2a cos 2 0, 

 2a 2a 



and fa = twice the curvilinear area AOP 



twice the sector ACP + twice the trianle OCP 



thus * 4- (20 + sin 20). 



V/* 



Thus the coordinate x and the time t are both expressed in 

 terms of a parameter 6. 



57. Examples. 



1. The same results may, of course, be arrived at by integrating the 

 equation x = 2 with the conditions that when =0, x = 2a, x=0. 



Multiplying both sides by xdt and integrating, we find 



# 2 = + C, where C is an arbitrary constant ; 



00 



putting 2 = 0, we have C - ^- . 



Thus 

 Hence 



By putting # = 2acos 2 in this deduce the result in the text. 



2. Find the time of falling to 0. 



3. Prove that as N approaches the velocity increases without limit. 

 [We shall see hereafter that when a natural system is devised in such a 



way that a point of a body moves as here described, either the body cannot 

 pass through the point corresponding to 0, or before it reaches the formula 

 for the acceleration changes and the velocity at is finite.] 



