Question

A planet of mass $m$ having angular momentum $L$ is revolving around the sun. The aerial velocity of the planet will be

• A. $L/m$
• B. $L/2m$
• C. $2L/m$
• D. $L/4m$

Answer 1

The correct option is B

$L/2m$

Step 1. Given data

A planet of mass $m$ having angular momentum $L$ is revolving around the sun.

Step 2. Finding the aerial velocity

If $d\theta$ be the small angle formed by the small arc shown in the figure, $dA$ be the area covered by the planet for the small angle $d\theta$, $r$ be the radius, $L$ be the angular moment and $\omega$ be the angular frequency then,

For small $d\theta$ area covered by the planet, $dA=\frac{1}{2}\left(r^{2}d\theta \right)$

Aerial velocity is the rate at which area is being swept out by the particle relative to the center.

The aerial velocity of the planet, $\frac{dA}{dt}=\frac{1}{2}\left(r^2\frac{d\theta }{dt}\right)$

$$\begin{gathered} \frac{dA}{dt}=\frac{1}{2}{r}^2\omega \\ \frac{dA}{dt}=\frac{L}{2m} \end{gathered}$$

[ $L=m{r}^2\omega$, Angular velocity is the rate of change of the position angle of an object with respect to the time, $\omega =\frac{d\theta }{dt}$]

Hence, option ($B$) is correct.