Question

A planet of mass $m$ moves along an ellipse around the Sun so that its maximum and minimum distances from the Sun are equal to $r_1$ and $r_2$ respectively. Find the angular momentum $M$ of this planet relative to the centre of the Sun.

Answer 1

As the planet is under central force (gravitational interaction), its angular momentum is conserved about the Sun (which is situated at one of the focii of the ellipse)
So, $m{v}_1{r}_1=m{v}_2{r}_2$ or, $v_1^2=\frac{v_2^2{r}_2^2}{r_1^2}$ (1)
From the conservation of mechanical energy of the system (Sun+planet),
$-\frac{\gamma m_{s}m}{r_1}+\frac{1}{2}m{v}_1^2=-\frac{\gamma m_{s}m}{r_2}+\frac{1}{2}m{v}_2^2$
or, $-\frac{\gamma m_s}{r_1}+\frac{1}{2}{v}_2^2\frac{r_2^2}{r_1^2}=-\left(\frac{\gamma m_s}{r_2}\right)+\frac{1}{2}{v}_2^2$ [Using (1)]
Thus, $v_2=\sqrt{\frac{2\gamma m_s{r}_1}{r_2(r_1+r_2)}}$ (2)
Hence $M=m{v}_2{r}_2=m\sqrt{\frac{2\gamma m_s{r}_1{r}_2}{(r_1+r_2)}}$