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 `sqrt(k//2ma)`, if the distance of other extreme position is b. Find `a//b`.

Options

A. 2
B. 3
C. 4
D. 5

Answer

B

Explanation

(b) `F=-k//r^(2)` (negative sign is for attractive force )
Potential energy `U=-intF dr=int(k)/(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)/(2)mv_(1)^(2)-(k)/(a)=(1)/(2)mv_(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)/(b)v_(1)=(a)/(b)sqrt((k)/(2ma))`
Therefore, from eq. (i)
`implies (1)/(2)m(k)/(2ma)-(k)/(a)=(1)/(2)m((a)/(b))^(2)(k)/(2ma)-(k)/(b)`
`-(3k)/(4a)=(ak)/(4b^(2))-(k)/(b)implies b^(2)-(4a)/(3)b+(a^(2))/(3)=0`
Hence, `b=a//3 implies a//b=3`