A particle of mass m is subjected to an attractive central force of magnitude $\mathrm{k}/{\mathrm{r}}^2$, k being a constant. If at the instant when the particle is at an extreme position in its closed orbit, at a distance a form the centre of force, its speed is ($\frac{\mathrm{k}}{2\mathrm{ma}}$), if the distance of other extreme position is b. Find a/b.

$\mathrm{F}=-\mathrm{k}/{\mathrm{r}}^2$ (negative sign is for attractive force)

Potential energy $\mathrm{U}=-\int F\mathrm{Fr}=\int \frac{\mathrm{k}}{{\mathrm{r}}^2}\mathrm{dr}=\frac{-\mathrm{k}}{\mathrm{r}}$

Conservation of energy gives (let at other extreme position r = b)

${\mathrm{k}}_1+{\mathrm{U}}_1={\mathrm{k}}_2+{\mathrm{U}}_2$

$\frac{1}{2}{\mathrm{mv}}_1^2-\frac{\mathrm{k}}{\mathrm{a}}=\frac{1}{2}{\mathrm{mv}}_2^2-\frac{\mathrm{k}}{\mathrm{b}}..........\left(i\right)$

$\text{ where }{\mathrm{v}}_1=\sqrt{\frac{\mathrm{k}}{2\mathrm{ma}}}$

Conservation of angular momentum gives $m{v}_{1}a=m{v}_{2}b$

${\mathrm{v}}_2=\frac{\mathrm{a}}{\mathrm{b}}{\mathrm{v}}_1=\frac{\mathrm{a}}{\mathrm{b}}\sqrt{\frac{\mathrm{k}}{2\mathrm{ma}}}$

Therefore, from eq.(i)

$\Rightarrow \frac{1}{2}\mathrm{m}\frac{\mathrm{k}}{2\mathrm{ma}}-\frac{\mathrm{k}}{\mathrm{a}}=\frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^2\frac{\mathrm{k}}{2\mathrm{ma}}-\frac{\mathrm{k}}{\mathrm{b}}$

$\frac{-3k}{4a}=\frac{ak}{4b^2}-\frac{k}{b}\Rightarrow b^2-\frac{4a}{3}b+\frac{a^2}{3}=0$

$\text{ Hence, }\mathrm{b}=\frac{\mathrm{a}}{3}\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=3$