A particle of mass m is subjected to an attractive central force of magnitude k/r2, k being a constant. If at the instant When the particle is at an extreme position in its closed orbit, at a distance a from the centre of force, its speed is (k/2ma), if the distance of other extreme position is b. Find a/b.

F=−K/r2 (negative sign is for attractive force)Potential energy U=-∫Fdr=∫Kr2dr=−KrConservation of energy gives (let at other extreme position r = b)K1+U1=K2+U212mv12−Ka=12mv22−Kb  ………………..(i)where v1=K2maConservation of angular momentum givesmv1a=mv2bv2=abv1=abK2maTherefore, from Eq. (i)⇒12mK2ma−Ka=12mab2K2ma−Kb−3K4a=aK4b2−Kb⇒b2−4a3b+a23=0Hence, as b < a, b=a/3⇒a/b=3