Question

A cosmic body $A$ moves to the Sun with velocity $v_0$ (when far from the Sun) and aiming parameter $l$ the arm of the vector $\overrightarrow{v_0}$ relative to the centre of the Sun (figure shown above). Find the minimum distance by which this body will get to the Sun.

Answer 1

At the minimum separation with the Sun, the cosmic body's velocity is perpendicular to its position vector relative to the Sun. If $r_{min}$ be the sought minimum distance, from conservation of angular momentum about the Sun (C).
$m{v}_{0}l=mv{r}_{min}$ or, $v=\frac{v_{0}l}{r_{min}}$ (1)
From conservation of mechanical energy of the system (sun $+$ cosmic body),
$\frac{1}{2}m{v}_0^2=-\frac{\gamma m_{s}m}{r_{min}}+\frac{1}{2}m{v}^2$
So, $\frac{v_0^2}{2}=-\frac{\gamma m_s}{r_{min}}+\frac{v_0^2}{2r_{min}^2}$ (using 1)
or, $v_0^2{r}_{min}^2+2\gamma m_s{r}_{min}-v_0^2{l}^2=0$
$r_{min}=\frac{-2\gamma m_s\pm \sqrt{4{\gamma }^2{m}_s^2+4v_0^2{v}_0^2{l}^2}}{2v_0^2}=\frac{-\gamma m_s\pm \sqrt{{\gamma }^2{m}_s^2+v_0^4{l}^2}}{v_0^2}$
Hence, taking positive root
$r_{min}=\left(\frac{\gamma m_s}{v_0^2}\right)[\sqrt{1+{\left(\frac{l{v}_0^2}{\gamma m_s}\right)}^2}-1]$