54, 55] 



LAW OF INVERSE SQUARE. 



57 



Lemma. 



Given a point P, a tangent PT 7 , a focus S, and the focal chord of curvature 

 PQ, one conic, and only one, can be 

 described, and this conic is an ellipse, 

 parabola, or hyperbola according as 



Let U be the middle point of PQ. 

 Draw PG at right angles to PT, and 

 UG parallel to PT ; draw UO and OK 

 at right angles to SP meeting PG and 

 jSP in and K respectively. 



Then by similar triangles OPU, 

 UPG, GPK we have 



OP : PU=PU : PG = PG : PK. 



Whence OP=^l. >o 



rK Fig. 36. 



Now describe a conic with focus S 



and axis SG to touch PT 7 at P, G is the foot of the normal, and PK is the 

 semi-latus rectum. Hence is the centre of curvature. 



Since SG : SP = eccentricity, the conic is determinate and unique. 



Since a semicircle on PU as diameter passes through G, we have when 

 SP>%PU, SG<SP ; wheuSP<$PU,SG>SP; when SP=%PU, SG=SP. 



Thus the conic is an ellipse, parabola, or hyperbola according as 

 SP>, =--, or<\PU. 



Now let a point move from P with velocity V in direction PT 

 and have an acceleration //./(distance) 2 towards S. 



Find Q in PS produced so that 

 T7*_o M 



Then by Example 7 on p. 54, PQ is the chord of curvature of 

 the path in direction PS. 



With S as focus describe a conic touching PT at P, and 

 having PQ for focal chord of curvature at P. 



Let a point describe this conic as a central orbit about S 

 starting with velocity V at P, the two moving points have at 

 starting the same position, velocity, and acceleration, and their 

 accelerations are always the same when their distances from S are 

 the same, they therefore describe the same orbit. 



