205 
the motion is that of a particle first going away from the centre 
and then returning when r= Aa/(A—1). The value 2Aa/(2A—1), 
of 7, for which the acceleration becomes zero, is smaller than 
Aa/(A—1). The particle ascends (during which there is first repulsion); 
at a given moment the acceleration becomes zero for 7 —= 2 A«/(2A—1); 
then we get attraction, which for r== Aa@/(A —1) has exhausted 
the motion and makes it return; the acceleration of the reversed 
motion is first positive, then becomes negative for 7 = 2Aa/(2A—1) 
and the motion stops (infinitely slowly) for =a. In case that i 
lies between O and 3, so that 7 can have all values, there is no 
point where the acceleration becomes zero. According to (20) there 
is then always repulsion; the- velocity is maximum at an infinite 
distance viz., according to (18), 1 — A which lies between 4V 2 and 1. 
8. We now return to the general case, where neither 7 nory are 
continually. zero. We must then take equations (18) and (14) as a 
starting point; by eliminating dt we find 
l B: WE RN PD \ a 
EREN dy HPS i Thies bod. 
pase le 
7” 7 
Expressing dy in r and dr we obtain 
Bdr 
ie (0-5) 
ne T: 
a 
Putting now —=—2, we get 
7 
¥ (LA) 
Pet é 
|: 1 ot RB? 
So gp becomes an elliptic integral in the VANG and 7 therefore 
an elliptic function of y. Of 
dp. = 
dp = 
Ad? (le Aja? 3 
Hv v B: © ali B: -——— 
let z,, 2,, 2, be the roots, so that 
Adc’ (A—1)a? db 
e,te,t+a;=1 , vw, J 1 ba ad od ber » 2, aos B wa) 
then we can introduce as constants of integration the quantities 
Ls Ts, ©, (connected by the relation «7, -+ 2, +2, = 1) instead of 
A and B. 
