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.

• A. $m\sqrt{\left[\frac{GM{r}_1{r}_2}{2(r_1+r_2)}\right]}$
• B. $2m\sqrt{\left[\frac{2GM{r}_1{r}_2}{(r_1+r_2)}\right]}$
• C. $m\sqrt{\left[\frac{3GM{r}_1{r}_2}{(r_1+r_2)}\right]}$
• D. $m\sqrt{\left[\frac{2GM{r}_1{r}_2}{(r_1+r_2)}\right]}$

Answer 1

The correct option is D $m\sqrt{\left[\frac{2GM{r}_1{r}_2}{(r_1+r_2)}\right]}$

Angular momentum of any particle at distance $r$ from a point is given by $mvr$ where $v$ is the velocity at that point.

Angular momentum at point 1 $L_1=m{v}_1{r}_1$

Angular momentum at point 2 $L_2=m{v}_2{r}_2$

Applying Angular momentum conservation i.e $L_1=L_2$

$m{v}_1{r}_1=m{v}_2{r}_2$

$⟹v_1{r}_1=v_2{r}_2$

Energy will also remain conserved between point 1 and 2.

energy at point 1 $E_1=\frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}$

energy at point 2 $E_2=\frac{1}{2}m{v}_2^2-\frac{GMm}{r_2}$

$E_1=E_2$

$\frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}$$=\frac{1}{2}m{v}_2^2-\frac{GMm}{r_2}$

$v_1=\sqrt{2GM(\frac{1}{r_1}-\frac{1}{r_2})+v_2^2}$

using $v_1{r}_1=v_2{r}_2$ from above result,

$⟹v_1=\sqrt{2GM(\frac{1}{r_1}-\frac{1}{r_2})+(\frac{v_1{r}_1}{r_2}{)}^2}$

$v_1^2=\frac{2GM{r}_2}{r_1(r_1+r_2)}$

$v_1=\sqrt{\frac{2GM{r}_2}{r_1(r_1+r_2)}}$

angular momentum of planet $m{v}_1{r}_1$

$L=m\sqrt{\frac{2GM{r}_2{r}_1}{(r_1+r_2)}}$