Question

If the angular momentum of a planet of mass m, moving around the sun in a circular orbit is L, about the centre of the Sun, its areal velocity is:
A. $\dfrac{L}{m}$
B. $\dfrac{{4L}}{m}$
C. $\dfrac{L}{{2m}}$
D. None of these

Answer 1

The areal velocity is the change in the area swept by the planet in unit time t. Recall the formula for the period of revolution of the planet and angular momentum of the planet. Rearranging these equations, you can find the areal velocity in terms of angular momentum of the planet.

Formula used:
Angular momentum, $L = mvr$
where, m is the mass of the planet, v is the orbital velocity and r is the radius of orbit.
Period of revolution, $t = \dfrac{{2\pi r}}{v}$
where, v is the orbital velocity.

Complete step by step solution:
As we know the area velocity is the total area swept by the planet to the total time taken. Therefore, we can express the area velocity as,
${v_A} = \dfrac{A}{t}$ ………………(1)
Here, A is the area and t is the total time taken.
We have the formula for the angular momentum of the planet,
$L = mvr$ …… (2)
Here, m is the mass of the planet, v is the orbital velocity and r is the radius of orbit.
The total area swept by the planet while revolving in the circular orbit of radius r is,
$A = \pi {r^2}$ …… (3)
Also, we know that the total time period for the planet to complete one revolution around the sun is expressed as,
$t = \dfrac{{2\pi r}}{v}$ …… (4)
Here, v is the orbital velocity.
Substituting equation (3) and (4) in equation (1), we get,
${v_A} = \dfrac{{\pi {r^2}}}{{\dfrac{{2\pi r}}{v}}}$
$\Rightarrow {v_A} = \dfrac{{vr}}{2}$
But from equation (2), we can write,
$\therefore{v_A} = \dfrac{L}{{2m}}$

So, the correct answer is option C.

Note: As we can see, the areal velocity is proportional to the angular momentum of the planet. But, we also know that the angular momentum of the planet in elliptical or circular orbit remains constant. Therefore, the areal velocity also remains constant. This is the discovery by famous astronomer Johannes Kepler.