Question

A planet of mass $m$ revolves in elliptical orbit around the sun so that its maximum and minimum distances from the sun are equal to $r_a$ and $r_p$ respectively. Find the angular momentum of this planet relative to the sun.

• A. $L=m\sqrt{\frac{GM{r}_p{r}_a}{(r_p+r_a)}}$
• B. $L=m\sqrt{\frac{2GM{r}_p{r}_a}{(r_p+r_a)}}$
• C. $L=M\sqrt{\frac{2GM{r}_p{r}_a}{(r_p+r_a)}}$
• D. $L=M\sqrt{\frac{(r_p+r_a)}{Gm{r}_p{r}_a}}$

Answer 1

Answer: (B)

For a planet moving around the sun in an orbit, angular momentum $\left(L\right)$ is constant.

$L=m{v}_a{r}_a=m{v}_b{r}_b\Rightarrow v_a=v_b\frac{r_b}{r_a}$

By conserving energy between the two points, farthest and nearest.

$\frac{-GMm}{r_b}+\frac{1}{2}m{v}_b^2=\frac{-GMm}{r_a}+\frac{1}{2}m{v}_a^2$

$\Rightarrow GM(\frac{1}{r_a}-\frac{1}{r_b})=\frac{1}{2}(v_a^2-v_b^2)$

Substituting $v_a$ from momentum equation.

$\Rightarrow GM(\frac{1}{r_a}-\frac{1}{r_b})=\frac{1}{2}[v_b^2\left(\frac{r_b^2}{r_a^2}\right)-v_b^2]$