# Derive Kepler's equal area law.

## Question:

Derive Kepler's equal area law.

## Kepler's Laws:

Mechanical motion of our solar system obeys gravitational law, and planets are orbiting around the sun whose mass is heaviest. The There are three empirical laws by Kepler to explain the motion of planets in our solar system. The first law is the law of ellipse, the second law is the law of area, and the third law is the law of harmony.

The angular momentum L of the orbiting celestial body around the sun is written as

{eq}\vec L=\vec r \times \vec p {/eq}

Here, {eq}\vec r {/eq} is the position vector and {eq}\vec p {/eq} is the linear momentum of the body. That is,

{eq}\vec p= m\vec v {/eq}

Here, m is the mass of the body and {eq}\vec v {/eq} is the velocity vector.

Please note, if {eq}\vec r {/eq} and {eq}\vec p {/eq} are parallel, then {eq}\vec r\times\vec p=0 {/eq}

After differentiating the angular momentum equation, we have

{eq}\begin{align} \dfrac{d\vec L}{dt}&=\vec v\times\vec p +\vec r \times\dfrac{d\vec p}{dt}\\ &=\vec r\times\vec F \end{align} {/eq}

Please note, F is a gravitational force and it is colinear with the position vector r.

So, L is constant in time. The angular momentum is conserved.

Let, the component of v perpendicular to r be V, then

{eq}\dfrac{L}{m}=rV=\rm constant {/eq}

Also, the total area A of the ellipse and the orbital period P is given by

{eq}\dfrac{A}{P}=\dfrac{rV}{2}=\rm constant {/eq}

Thus, the radius vector to a planet sweeps out equal areas in equal intervals of time. When a planet is closer to the sun, it revolves faster, and, on the other hand, when a planet is further away from the sun, it revolves slower.