Physics - Gravitation Question with Solution | TestHub

Using conservation of angular momentum about O.

$\text{m}{\text{V}}_{\text{P}}{\text{r}}_{\text{P}}=\text{m}{\text{V}}_{\text{A}}{\text{r}}_{\text{A}}=\text{m}{\text{V}}_0{\text{r}}_0\text{cos}\theta$

${\text{V}}_{\text{A}}{\text{r}}_{\text{A}}={\text{V}}_{\text{P}}{\text{r}}_{\text{P}}=\frac{3{\text{v}}_0{\text{r}}_0}{5}$......(1)

using conservation of energy

$\frac{1}{2}\text{m}{\text{V}}_{\text{A}}^2+\frac{-\text{GMm}}{{\text{r}}_{\text{A}}}=\frac{-\text{GMm}}{{\text{r}}_0}+\frac{1}{2}\text{m}{\text{V}}_0^2$

$\Rightarrow \frac{9{\text{V}}_0^2{\text{r}}_0^2}{50{\text{r}}_{\text{A}}^2}-\frac{{\text{V}}_0^2{\text{r}}_0}{{\text{r}}_{\text{A}}}+\frac{{\text{V}}_0^2}{2}=0$$\left[\text{Let}\frac{{\text{r}}_0}{{\text{r}}_{\text{A}}}=\text{x}\right]$This is a quadratic equation in$\left(\frac{r_0}{r_a}\right)$

$\Rightarrow 9x^2-50x+25=0\Rightarrow x=5 or x=\frac{5}{9}$



$\Rightarrow r_A=\frac{9}{5}{r}_o \dots \dots .\left(1\right)$

$r_p=\frac{r_o}{5} \dots \dots .\left(2\right)$

using (1)&(2)

$\Rightarrow \frac{{\text{V}}_{\text{P}}}{{\text{V}}_{\text{A}}}=\frac{{\text{r}}_{\text{A}}}{{\text{r}}_{\text{P}}}=9$