8 Proceedings of the Royal Irish Academy. 



the equations of motion become 

 and, by symmetry, 



£-«-* 



Thus the problem is reduced to that of motion under the central force \hp, 

 varying as the distance. The equation of energy becomes 



*\$h®)\ ->+><***■ 



3. For elliptic motion, we have h = - fi/2a = - |m 2 « 2 . A solution 

 consistent with the last equation is therefore 



% = \/(l - e)a cos \na,T, 



ij = y/{l + e)a sin \naT, 

 to which correspond 



x = a os liar - ac, 



y = a ^/l - e 2 sin war, 

 and r = £ 2 + jj 2 = a - ae cos war. 



Hence m< = n J ?y?t = war - e sin war, 



which is the simple solution in terms of nar = E, the eccentric anomaly. 



For parabolic motion, h = ; and the motion in the (£, v) plane is 

 rectilinear. A solution consistent with the equation of energy is 



or x = \{$ - fiT 2 ), y = v/T^J . r. 



He- . - ft(, ♦ rf* - (flf^, ♦ j^}. 



which becomes the ordinary solution when we notice that 



tan |0. 



GT- - 



For hyperbohc motion h = + fi/2a, and the appropriate solution is 



4 = y^e - l)a cosh = ( - j r, rj = ^/(e + l)a sinh - ( - I r, 

 corresponding to 

 a; = ae - acosh(-) r, y = u^/e 2 - lsinhl- ]r, r = aecoshf-j r - a. 



which is the analogue of Kepler's equation for hyperbolic motion. 



