Physics - Gravitation Question with Solution | TestHub

w.r.t. COM K.E. $=\frac{1}{2}($ red mass $){\mathrm{v}}_{\text{rel }}{}^2$

$$\text{ w.r.t. COM Angular momentum }=\frac{\mathrm{mr}}{2}{\mathrm{v}}_{\mathrm{rel}}$$

∴ Equating energy

$$\frac{1}{2}\frac{\mathrm{m}}{2}{\mathrm{v}}_0^2-\frac{{\mathrm{Gm}}^2}{{\mathrm{r}}^0}-\frac{1}{2}\frac{\mathrm{m}}{2}{\mathrm{v}}_{\mathrm{rel}}^2-\frac{{\mathrm{Gm}}^2}{\mathrm{r}}$$

(Here ${\mathrm{v}}_{\text{rel }}$ is relative velocity ⟂ to line as ${\mathrm{v}}_{\text{rel }}$ along the line joining is zero when separation is either min. or max.) Angular momentum conservation

$$\begin{array}{rlr}& \frac{{\mathrm{mr}}_0}{2}{\mathrm{v}}_0=\frac{\mathrm{mr}}{2}{\mathrm{v}}_{\mathrm{rel}}& ={\mathrm{v}}_{\mathrm{rel}}=\frac{{\mathrm{r}}_0{\mathrm{v}}_0}{{\mathrm{r}}_2}\end{array}$$

Solving $3{\mathrm{r}}^2-4{\mathrm{rr}}_0+{\mathrm{r}}_0{}^2=0$

$$\begin{array}{rlr}& \therefore r_{\max }=r_0{r}_{\min }=\frac{r_0}{3}& \text{ Ratio }=3.00\end{array}$$