230 ANSWERS. 



(6) U= 



(9) Compare J. C. MAXWELL, Electricity and magnetism, Vol. I., 

 Arts. 118-123, and the plates I.-III. 



(10) Taking the axis of z vertically upwards and denoting by r^ r% 

 the distances from C l} C 2 , we have by the principle of kinetic energy 

 d(mv 2 ) = mgdz Kmd*\/r? + K'mr. 2 dr 2 ; hence the equipotential sur- 

 faces are (c + gz ^ K 'r 2 2 ) 2 r l 2 =K 2 . If *' = o, the equation becomes for 

 Ci as origin (x*+y 2 + z 2 ) (c + gz) z = * 2 ; if K = o, we find for C 2 as 



origin the concentric spheres x 2 +f + f z 



Pages 52, 53. 



+ Ooj'o J'o^o) 2=0. 



(2) F i = Kmmfi has the ^-component X t = Kmm^ (x x t )/r t 

 = -Kmm { (xxj) ; hence 2X= Km^.mj(xx^, '%= Km^n^yy^, 

 '%Z= Km^m^z Zf) . Equating these to zero, the position of equili- 

 brium is found as the centroid 



= = _ 



^m { ' ^m t ' m t 



of the masses m t . Taking this point as origin, the equation of the plane 

 of motion is the same as in Ex. (i) ; and the resultant force has the 

 components Km^m { - x, Km^m^y, Km^m^z. 



Pages 65, 66. 



(2) The equation of the orbit given in Ex. (i) is satisfied not only 

 by (x , y ) , but also by (X (] /K, }'O/K) ; i.e. the orbit passes not only 

 through the initial point P Q , but also through the point Q, which is 

 the extremity of the radius vector OQ V Q /K parallel to V Q ; OP Q and 

 OQ are the conjugate semi-diameters whose equations are x y=yox, 

 x Q y = yox. 



(4) The problem reduces to that of constructing the axes of a conic 

 from a pair of conjugate diameters. 



(8) The equation of a central conic can be written in the form 



--- + -T r> 

 /~ a* P aW 



where / is the perpendicular from the centre to the tangent ; the upper 

 sign gives the ellipse, the lower the hyperbola. Apply (n), Art. IT 6. 



