Question

The orbital angular momentum of a satellite of mass $m$ revolving at a distance $r$ from the centre is $L$. If the distance is increased to $16r$, then the new angular momentum will be (Take $M$ as mass of the Earth)

• A. $16L$
• B. $64L$
• C. $\frac{L}{4}$
• D. $4L$

Answer 1

The correct option is D $4L$

The magnitude of angular momentum is given by $$L=mvr$$ Where, the orbital speed, $v=\sqrt{\frac{GM}{r}}$

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

$\Rightarrow L=m\sqrt{GMr}$

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

Let, the new angular momentum be $L^{\prime }$

$\therefore \frac{L^{\prime }}{L}=\sqrt{\frac{16r}{r}}$

$\Rightarrow L^{\prime }=4L$

Hence, option (d) is the correct answer.

Key Concept: The orbital speed of a satellite is given by $$v=\sqrt{\frac{GM}{r}}$$