Question

A planet of mass $m$ moves in an elliptical orbit around an unknown star of mass $M$ such that its maximum and minimum distances from the star are equal to $r_1$ and $r_2$ respectively. The angular momentum of the planet relative to the centre of the star is

• A. $m\sqrt{\frac{2GM{r}_1{r}_2}{r_1+r_2}}$
• B. 0
• C. $m\sqrt{\frac{2GM\left(r_1+r_2\right)}{4_1{r}_2}}$
• D. $m\sqrt{\frac{2GMm{r}_1}{(r_1+r_2)r_2}}$

Answer 1

Answer: (A)

According to the law of conservation of angular momentum,
$m{V}_1{I}_1=m{V}_2{I}_2$
$\Rightarrow V_2=\frac{V_1{r}_1}{r_2}$ ......(i)
From the law of conservation of total mechanical energy.
$\frac{-GMm}{r_1}+\frac{1}{2}m{v}_1^2=-\frac{GMm}{r_2}+\frac{1}{2}m{v}_2^2$ ......(ii)
From Eqs. (i) and (ii), we get
$V_1=\sqrt{\frac{2GM{r}_2}{\left(r_1+r_2\right)r_1}}$
Angular momentum,
$L=m{v}_1{r}_1=m\left(\sqrt{\frac{2GM{r}_2}{\left(r_1+r_2\right)r_1}}\right)\times r_1$
$L=m\sqrt{\frac{2GM{r}_1{I}_2}{r_1+r_2}}$