Question

An angular moment of a satellite revolving round the earth in a circular orbit at a height $R$ above the surface is $L$. Here $R$ is radius of the earth.The magnitude of angular momentum of another satellite of the same mass revolving very close to the surface of the earth is

• A. $\frac{L}{2}$
• B. $\frac{L}{\sqrt{2}}$
• C. $\sqrt{2}L$
• D. $2L$

Answer 1

The correct option is B $\frac{L}{\sqrt{2}}$

The orbital velocity of a satellite in a circular orbit of radius $r$ is given by

$v=\sqrt{\frac{GM}{r}}$

The angular momentum of the satellite will be

$L=mvr=m\left(\sqrt{\frac{GM}{r}}\right)r$

Since mass $m$ of both satellite are same and, $G$ and $M$ are constant.

$\therefore L\propto \sqrt{r}$

So, $\frac{L_2}{L_1}=\sqrt{\frac{r_2}{r_1}}=\sqrt{\frac{R}{2R}}=\frac{1}{\sqrt{2}}$

Here, $r_1=2R;r_2=R;L_1=L$

$\Rightarrow \frac{L_2}{L}=\sqrt{\frac{R}{2R}}=\frac{1}{\sqrt{2}}$

$\therefore L_2=\frac{L}{\sqrt{2}}$

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