54, 55] 



LAW OF INVERSE SQUARE. 



57 



Lemma. 



Given a point P, a tangent PT, 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 

 /&amp;gt;$&amp;lt;, =, or &amp;gt;4SP. 



Let U be the middle point of PQ. 

 Draw PQ at right angles to PT, and 

 UG parallel to PT ; draw UO and GK 

 at right angles to SP meeting PG and 

 SP in and A&quot; respectively. 



Then by similar triangles OPU, 

 UPG, GPK we have 



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



Whence 



Fig. 36. 



Now describe a conic with focus $ 



and axis SG to touch PT 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&amp;gt;%PU, SG&amp;lt;SP-, when SP&amp;lt;%PU, SG&amp;gt;SP ; when SP=%PU, SG=SP. 



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

 P&amp;gt;, =,or 



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

 and have an acceleration ^/(distance) 2 towards 8. 



Find Q in PS produced so that 



F*-2 ^ PQ 

 Z SF&amp;gt; 4 



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

 the path in direction PS. 



With 8 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 8 are 

 the same, they therefore describe the same orbit. 



