Question

If areal velocity of a planet of mass $M$ going around sun in elliptical orbit is $\frac{dA}{dt}$ , its angular momentum about sun is

• A. $\frac{1}{2}\cdot M\frac{dA}{dt}$
• B. $M\frac{dA}{dt}$
• C. $\pi M\frac{dA}{dt}$
• D. $2M\frac{dA}{dt}$

Answer 1

The correct option is D $2M\frac{dA}{dt}$

Areal Velocity, $A=$ Area swept by radius vector/Time taken

Time period, $T=\frac{2\pi }{\omega }$

$dA=\frac{r^2}{2}d\theta$

$\frac{dA}{dt}=\frac{r^2}{2}\cdot \frac{d\theta }{dt}=\frac{r^2}{2}\cdot \omega$

We know that angular momentum $\left(L\right)$ is given by, $L=I\omega$

$L=M{r}^2\omega =2M\frac{dA}{dt}$