Question

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

Answer 1

The angular momentum
$L=mvr.....\left(1\right)$
Conservation of angular momenta at point $1$ and $2$;
$m{v}_1{r}_1=m{v}_2{r}_2$
$\Rightarrow v_1{r}_1=v_2{r}_2.....\left(2\right)$
Conservation of energy at $1$ and $2$;
$\frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}=\frac{1}{2}m{v}_2^2-\frac{GMm}{r_2}....\left(3\right)$
Using equation $\left(2\right)$ and $\left(3\right)$ we obtain
$v_1=\sqrt{2GM(\frac{1}{r_1}-\frac{1}{r_2})+v_2^2}$
$\Rightarrow v_1=\sqrt{2GM(\frac{1}{r_1}-\frac{1}{r_2})+{\left(\frac{v_1{r}_1}{r_2}\right)}^2}$
$\Rightarrow v_1^2[1-{\left(\frac{r_1}{r_2}\right)}^2]=2GM\left(\frac{r_2-r_1}{r_1{r}_2}\right)$
$\Rightarrow v_1=\sqrt{\frac{2GM{r}_2}{r_1(r_1+r_2)}}$
$\therefore L=m{v}_1{r}_1=m\sqrt{\frac{2GM{r}_1{r}_2}{r_1+r_2}}$.