Question

A planet of mass $m$ moves in an 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

• A. $m \sqrt{\frac{2GM r_{1}r_{2}}{r_{1} + r_{2}} } $
• B. 0
• C. $m \sqrt{\frac{2 G M\left(r_{1}+r_{2}\right)}{4_{1} r_{2}}}$
• D. $m \sqrt{\frac{2GM m r_{1}}{(r_{1} + r_{2}) r_2 } } $

Answer 1

The Correct Option is A

According to the law of conservation of angular momentum, $m V_{1} I_{1} =m V_{2} I_{2} $ $\Rightarrow V_{2} =\frac{V_{1} r_{1}}{r_{2}}$ ......(i) From the law of conservation of total mechanical energy. $\frac{-G M m}{r_{1}}+\frac{1}{2} m v_{1}^{2}=-\frac{G M m}{r_{2}}+\frac{1}{2} m v_{2}^{2}$ ......(ii) 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, $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} $ $L=m \sqrt{\frac{2 G M r_{1} I_{2}}{r_{1}+r_{2}}}$