122.] CENTRAL FORCES. 63- 



120, To find the equation of the orbit, it is only necessary to eliminate 

 t in each case. 



In the case of attraction, this elimination can be performed by solv- 

 ing for cos Kt, sin K/, squaring and adding. The result is 



OiJF - ^i*) 2 + (a*y - b&Y = (a A - A) 2 , 

 and this represents an ellipse, since 



(af + a/) W + tt) ~ ("A + a AY = (a A ~ <*AY 

 is always positive. The centre of the ellipse is at the origin, and the 

 lines ay = d^x, a 2 y = bgx are a pair of conjugate diameters. 



121, In the case of repulsion, solve for e Kt and e~ Kt , and multiply. 

 The resulting equation, 



(&\y - b\x) (fa <*zy) = (a A <*A) 2 , 

 represents a hyperbola whose asymptotes are the lines a l y = b l x, 



122. It is worthy of notice that the more general problem of the 

 motion of a particle attracted by any number of fixed centres, with forces 

 directly proportional to the distances from these centres, can be reduced 

 to the problem of Art. 119. 



Let x, y, z be the co-ordinates of the particle, r i its distance from the 

 centre O i ; x it y it z t the co-ordinates of O t ; and K ( V ( the acceleration 

 produced by O t . Then the .^-component of the resultant acceleration is 



and similar expressions obtain for the y and z components. Hence, the 

 equations of motion are 



As these expressions are linear in x, y, z, there is one, and only one, 

 point at which the resultant acceleration is zero. Denoting its co- 

 ordinates by x, y, z, we have 



' = 



The form of these equations shows that this point of zero acceleration, 

 which is sometimes called the mean centre, is the centroid of the centres 



