A particle of mass m is subjected to an attractive central force of magnitude `k//r^(2)`, 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. Then a/b is

Answer

3

Explanation

`F=-K//r^(2)` (negative sign is for attractive force)
Potential energy `U=-intFdr=intK/(r_(2))dr=-K/r`
Conservation of energy gives (let at other extreme position `r=b`)
`K_(1)+U_(1)=K_(2)+U_(2)`
`1/2mv_(1)^(2)-K/a=1/2mv_(2)^(2)-K/b`………i
where `v_(1)=sqrt(k/(2ma)`
Conservation of angular momentum gives
`mv_(1)a=mv_(2)b`
`v_(2)=a/bv_(1)=a/bsqrt(K/(2ma))`
Therfore, from eqn i
`implies1/2m k/(2ma)-K/a=1/2m(a/b)^(2)k/(2ma)-K/b`
`=(3K)/(4a)=(aK)/(4b^(2))-K/bimpliesb^(2)-(4a)/3b+(a^(3))/3=0`
Hence `b=a//3, impliesa//b=3`