A planet of mass $m$ moves in a elliptical orbit around an unknown star of mass $M$ such that its maximum and minimum distances from the star are equal to $r_1$ and $r_2$ respectively. The angular momentum of the planet relative to the centre of the star is

• Correct answer
  A

  $m \sqrt{\frac{2 G M r_1 r_2}{r_1+r_2}}$
• B

  0
• C

  $m \sqrt{\frac{2 G M\left(r_1+r_2\right)}{r_1 r_2}}$
• D

  $\sqrt{\frac{2 G M m r_1}{\left(r_1+r_2\right) r_2}}$

Solution

According to the law of conservation of angular momentum, $$ \begin{aligned} m v_1 r_1 & =m v_2 r_2 \\ \Rightarrow \quad v_2 & =\frac{v_1 r_1}{r_2} \end{aligned} $$ From the law of conservation of total mechanical energy. $$ \frac{-G M m}{r_1}+\frac{1}{2} m v_1^2=-\frac{G M m}{r_2}+\frac{1}{2} m v_2^2 $$ From Eqs. (i) and (ii), we get $$ v_1=\sqrt{\frac{2 G M r_2}{\left(r_1+r_2\right) r_1}} $$ Angular momentum, $$ \begin{aligned} & L=m v_1 r_1=m\left(\sqrt{\frac{2 G M r_2}{\left(r_1+r_2\right) r_1}}\right) \times r_1 \\ & L=m \sqrt{\frac{2 G M r_1 r_2}{r_1+r_2}} \end{aligned} $$

Hence, the correct answer is option: 'A'.