Question

A cosmic body $\text{A}$ moves to the sun with velocity $v_0$ (when far from the sun) and aimimg parmeter $l$ the arm of the vector ${\overrightarrow{v}}_0$ relative to the center of the sun. Find the minimum distance by which this body will get to the Sun. Mass of the Sun is $M$.

• A. $\frac{2GM}{v_0^2}[\sqrt{1+{\left(\frac{l{v}_0^2}{GM}\right)}^2}-1]$
• B. $\frac{GM}{v_0^2}$
• C. $\frac{GM}{v_0^2}[\sqrt{1+\left(\frac{l{v}_0^2}{GM}\right)}-1]$
• D. $\frac{GM}{2v_0^2}[\sqrt{1+{\left(\frac{l{v}_0^2}{GM}\right)}^2}-1]$

Answer 1

The correct option is C $\frac{GM}{v_0^2}[\sqrt{1+\left(\frac{l{v}_0^2}{GM}\right)}-1]$


Let,
$v$ be the speed of the body when it is about to reach the Sun.
$m$ is the mass of the body and $r$ is the distance of closest approach.
At the closest distance, the velocity of the body will be perpendicular to the line joining the sun and the body.
Conserving angular momentum about the Sun.

$m{v}_{0}l=mvr$

$\Rightarrow v=v_0\frac{l}{r}$

Conserving mechanical energy,

$P.E_i+K.E_i=P.E_f+K.E_f$

$\Rightarrow 0+\frac{1}{2}m{v}_0^2=-\frac{GMm}{r}+\frac{1}{2}m{v}^2$

The body is for away initially $\therefore P.E_i=0$

$\Rightarrow \frac{1}{2}m(v^2-v_0^2)=\frac{GMm}{r}$

As, $v=v_0\frac{l}{r}$

$\Rightarrow m(v_0^2\frac{l^2}{r^2}-v_0^2)=\frac{GMm}{r}$

$\Rightarrow \frac{1}{2}{v}_0^2(\frac{l^2}{r^2}-1)=\frac{GM}{r}$

$\Rightarrow v_0^2{r}^2+2GMr-v_0^2{l}^2=0$

$\Rightarrow r=\frac{-2GM\pm \sqrt{4G^2{M}^2+4v_0^4{l}^2}}{2v_0^2}$

$\Rightarrow r=\frac{-GM\pm \sqrt{G^2{M}^2+v_0^4{l}^2}}{v_0^2}$

Neglecting the negative value,

$\Rightarrow r=\frac{GM}{v_0^2}[\sqrt{1+{\left(\frac{v_0^{2}l}{GM}\right)}^2}-1]$

Hence, option (c) is correct answer.

Why this question? Note : The velocity of the particle at the closest or the farthest point is perpendicular to the line joining the particle and the other mass. This is because at these points the velocity of approach or separation will become zero.