Question

A planet of mass $m$ moves along an ellipse around the sun of mass $M$ so that its maximum and minimum distances from sun are $a$ and $b$ respectively. The angular momentum $L$ of this planet about the centre of sun will be:
(Assume, sun is stationary and $a=2b)$

• A. $m\sqrt{\frac{GMb}{3}}$
• B. $2m\sqrt{\frac{GMb}{3}}$
• C. $m\sqrt{3GMb}$
• D. $4m\sqrt{\frac{GMb}{3}}$

Answer 1

The correct option is B $2m\sqrt{\frac{GMb}{3}}$



The planet is revolving around the sun due to gravitational pull.

Since, we know that ${\tau }_{ext}$ about the centre of sun is zero on the system of sun and planet. We can apply angular momentum conservation on the sun- planet system , when planet is at maximum and minimum distances from the sun.

$\therefore L_i\text{at max. distance}=L_f\text{at min. distance}$

$\Rightarrow m{v}_{1}a=m{v}_{2}b$

As, $a=2b$

$\Rightarrow v_1=\frac{v_2}{2}.....\left(1\right)$

Since, gravitation field is a conservative field, we can apply energy conservation.

$\therefore U_1+K_1=U_2+K_2$

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

Substituting from eq. $\left(1\right)$, we get

$\frac{1}{2}m[v_1^2-(2v_1{)}^2]=GMm[\frac{1}{a}-\frac{1}{b}]$

$\Rightarrow \frac{1}{2}m[-3v_1^2]=-GMm[\frac{1}{2b}]$

$\Rightarrow v_1^2=\frac{GM}{3b}$

$\Rightarrow v_1=\sqrt{\frac{GM}{3b}}$

Therefore, angular momentum of planet will be

$L=m{v}_{1}a=m{v}_{2}b$

$\Rightarrow L=ma\sqrt{\frac{GM}{3b}}$

$\Rightarrow L=m\times 2b\times \sqrt{\frac{GM}{3b}}$

$\Rightarrow L=2m\sqrt{\frac{GMb}{3}}$

Hence, option (b) is the correct answer.

Why this question ? The gravitational force exerted by Sun will be directed along line joining centre of planet and sun. Thus, ${\tau }_{ext}=0$ on system about centre of sun, and we can apply angular momentum conservation.