Question

The angular momentum of a planet of mass $M$ moving around the sun in an elliptical orbit is $\overrightarrow{L}$. The magnitude of the areal velocity of the planet is

• A. $\frac{L}{M}$
• B. $\frac{2L}{M}$
• C. $\frac{L}{2M}$
• D. $\frac{4L}{M}$

Answer 1

The correct option is C

$\frac{L}{2M}$

Step 1: Given data

Mass of planet=$M$

The angular momentum of a planet=$L$

Step 2: Formula used:

We know that the formula of angular momentum is

$L=Mvr$ Where M is the mass of the planet, V is the velocity and r is the radius of the orbit.

$v=r\omega$ where, $\omega$ is angular velocity

Step 3: Determine the magnitude of the areal velocity of the planet

Consider the following figure:

From the formula,

$L=Mvr----\left(1\right)$

The area covered in the figure is -

Suppose $d\theta$ is the angle covered for making area $dA$

From

$$\begin{gathered} angle=\frac{arc}{radius} \\ d\theta =\frac{dl}{r} \\ dl=rd\theta \end{gathered}$$

The area will be calculated as-

$$\begin{gathered} dA=\frac{1}{2}r\left(rd\theta \right) \\ dA=\frac{1}{2}{r}^{2}d\theta \end{gathered}$$

On differentiating with respect to $t$-

$\begin{array}{rcl}\frac{dA}{dt}& =& \frac{1}{2}{r}^2\frac{d\theta }{dt}\\ & =& \frac{1}{2}{r}^2\omega .................\left(2\right)as\omega =\frac{d\theta }{dt}\end{array}$

The areal velocity of the planet is $\frac{dA}{dt}$.

From equation (1)-

$$\begin{gathered} L=Mvr \\ L=M\left(r\omega \right)r \\ L=M{r}^2\omega \\ {r}^2\omega =\frac{L}{M} \end{gathered}$$

Now putting the values from equation (1)-

$\frac{dA}{dt}=\frac{L}{2M}$

So, the magnitude of the areal velocity of the planet will be $\frac{L}{2M}$.

Hence, option C is the correct answer.