Question

A planet of mass $m$ moves in an elliptical path 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 center of the sun. Mass of the sun is $M$.

• A. $m\sqrt{\frac{GM{r}_1{r}_2}{(r_1+r_2)}}$
• B. $m\sqrt{\frac{2GM{r}_1{r}_2}{(r_1+r_2)}}$
• C. $m\sqrt{\frac{2GM{r}_1^2{r}_2^2}{(r_1+r_2{)}^2}}$
• D. $m\sqrt{\frac{GM{r}_1{r}_2}{2(r_1+r_2)}}$

Answer 1

The correct option is B $m\sqrt{\frac{2GM{r}_1{r}_2}{(r_1+r_2)}}$


Let,
$m$ be the mass of the planet,
$v_1$ be the speed of planet at the farthest point,
$v_2$ be the speed of planet at the closest point.

Conserving angular momentum of $m$ about the Sun.

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

$\Rightarrow v_2=\left(\frac{r_1}{r_2}\right)v_1...\left(1\right)$

Conserving mechanical energy between the closest and the farthest point.

$P.E_1+K.E_1=P.E_2+K.E_2$

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

$\Rightarrow \frac{1}{2}[v_1^2-v_2^2]=GM[\frac{1}{r_1}-\frac{1}{r_2}]$

As $v_2=\left(\frac{r_1}{r_2}\right)v_1,$

$\Rightarrow [v_1^2-v_1^2\times \frac{r_1^2}{r_2^2}]=GM[\frac{1}{r_1}-\frac{1}{r_2}]$

$\Rightarrow v_1^2\frac{[r_2^2-r_1^2]}{r_2^2}=2GM\left[\frac{r_2-r_1}{r_1{r}_2}\right]$

$\Rightarrow v_1^2\left[\frac{(r_2+r_1)(r_2-r_1)}{r_2^2}\right]=2GM\frac{(r_2-r_1)}{r_1{r}_2}$

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

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

Angular momentum

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

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

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

Hence, option (b) is the correct answer.

Why this question? To make students understand the application of conservation of angular momentum and the conservation of mechanical energy in the same problem. This method is used across various types of problems and hence important to understand well.