Question

In a planetary motion, the areal velocity of the position vector of a planet depends on angular velocity ($\omega$) and distance (r) of the planet from the sun. The correct relation for areal velocity $\left( {\dfrac{{dA}}{{dt}}} \right)$ is
${\text{A}}{\text{. }}\dfrac{{dA}}{{dt}} \propto \omega r$
${\text{B}}{\text{. }}\dfrac{{dA}}{{dt}} \propto {\omega ^2}r$
${\text{C}}{\text{. }}\dfrac{{dA}}{{dt}} \propto \omega {r^2}$
${\text{D}}{\text{. }}\dfrac{{dA}}{{dt}} \propto \sqrt {\omega r}$

Answer 1

Kepler’s second law states that the area vector of planets sweeps equal amount of areas for equal duration of time. The rate of change of area is proportional to the ratio of angular momentum of the planet and mass of the planet.

Formula used:
The angular momentum of a particle is given as
$L = m{\text{v}}r$
where L is the angular momentum of the particle whose mass is m and is moving with velocity v and r signifies the distance of the particle from origin about which the particle is rotating.
The linear velocity is given in terms of angular velocity as follows:
${\text{v}} = r\omega$
$\therefore L = m{r^2}\omega {\text{ }}...{\text{(i)}}$

Complete step by step answer:
Kepler’s laws of planetary motion describe the various properties of the planets orbiting the sun in our solar system. There are three laws of planetary motion given by Kepler.
1. Law of orbits: All planets revolve around the sun in closed elliptical orbits while the sun is situated at one of the foci of the ellipse made by planetary motion.
2. Law of areas: The area vector of the planets sweeps equal amounts of area in equal intervals of time. Mathematically, it is given as
$\dfrac{{dA}}{{dt}} = \dfrac{L}{{2m}}$
where L signifies the angular momentum of the planet and m is the mass of the planet.
3. Law of periods: The square of the time period of revolution is directly proportional to the cube of the length of the semi-major axis of all planets.
${T^2} \propto {a^3}$
$\Rightarrow \dfrac{{{T^2}}}{{{a^3}}} = {\text{constant}}$
where T is the time period of the planet and a signifies the length of the semi-major axis of the planet.
The mathematical expression for the second law of area is given as:
$\dfrac{{dA}}{{dt}} = \dfrac{L}{{2m}}{\text{ }}...{\text{(ii)}}$
Using the equation (i) in equation (ii), we get
$\dfrac{{dA}}{{dt}} = \dfrac{{m{r^2}\omega }}{{2m}} = \dfrac{{{r^2}\omega }}{2}$
$\Rightarrow \dfrac{{dA}}{{dt}} \propto {r^2}\omega$
This is the required solution and the correct answer is option C.

Note:
The second Kepler’s law signifies that the planets move faster when they are closer to the sun and move slower when they are away from the sun. There is more kinetic energy when the planet is at perihelion and less kinetic energy when the planet is at aphelion.