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:
(sun is stationary.)

• A. $M\sqrt{\frac{2Gmab}{(a+b)}}$
• B. $mb\sqrt{\frac{2GM}{a+b}}$
• C. $m\sqrt{\frac{2GMab}{(a+b)}}$
• D. $M\sqrt{\frac{2Gma}{b}}$

Answer 1

The correct option is C $m\sqrt{\frac{2GMab}{(a+b)}}$



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. So, we can apply angular momentum conservation on the sun- planet system when planet at maximum and minimum distance from the sun.

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

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

$\Rightarrow v_1=\frac{v_{2}b}{a}..........\left(1\right)$

Since, gravitation field is a conservative field, so 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

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

$\Rightarrow v_1^2(b^2-a^2)=2b^{2}GM\frac{(b-a)}{ab}$

$\Rightarrow v_1^2=\frac{2GMb}{a(a+b)}$

$\Rightarrow v_1=\sqrt{\frac{2GMb}{a(a+b)}}$

Therefore, angular momentum of planet will be

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

$\Rightarrow L=ma\sqrt{\frac{2GMb}{a(a+b)}}$

Hence, option $\left(c\right)$ is correct.

Why this question ? Tip: 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