📂Classical Mechanics

Kepler's Second Law: The Law of Equal Areas

Kepler’s Second Law: Law of Equal Areas¹

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

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The Law of Equal Areas is the second of Kepler’s laws of planetary motion. However, this is not a special law that occurs only between the Sun and the planets, but a general law that is valid for any object (particle) moving under a central force.

Proof

Angular Momentum and Area Speed of a Particle Moving in a Central Force Field

Since it is a rotational motion, it is convenient to describe the motion of the particle in polar coordinates. The velocity in polar coordinates is as follows.

$$ \mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta}\hat{\boldsymbol{\theta}} $$

Then, the magnitude of the particle’s angular momentum is as follows.

$$ | \mathbf{L}| = |\mathbf{r} \times m\mathbf{v}|=|r \hat{\mathbf{r}} \times m(\dot{r} \hat{\mathbf{r}} + r \dot{\theta}\hat{\boldsymbol{\theta}})| $$

At this time, given $|\hat{\mathbf{r}} \times \hat{\mathbf{r}}|=0$ and $|\hat{\mathbf{r}} \times \hat{\boldsymbol{\theta}}|=1$ and the angular momentum of a particle moving under a central force is conserved, the following equation is valid.

$$ \begin{equation} L=| \mathbf{L}|=mr^{2}\dot{\theta}=\text{constant} \label{eq1} \end{equation} $$

Because of $\left| \hat{\mathbf{r}} \times \hat{\mathbf{r}} \right|=0$, any motion in the radial direction does not affect the angular momentum. This result generally applies to any particles moving under a central force, including the planets orbiting the Sun due to its gravity. Now, to calculate the area speed, let’s look at the figure below.

The triangle made by the position vector $\mathbf{r}$ of a particle moving $d \mathbf{r}$ in $dt$ time under a central force is as shown in the figure above. Therefore, if we calculate the area of the triangle, it is as follows.

$$ \begin{align} dA &= {\textstyle \frac{1}{2}} \left| \mathbf{r} \times d \mathbf{r} \right| \nonumber \\ &= {\textstyle \frac{1}{2}} \left| r \hat{\mathbf{r}} \times(dr \hat{\mathbf{r}} + rd\theta \hat{\boldsymbol{\theta}}) \right| \nonumber \\ &= {\textstyle \frac{1}{2}}r^{2}d\theta \label{eq2} \end{align} $$ Since $\left| \hat{\mathbf{r}} \times \hat{\mathbf{r}} \right|=0$, any motion in the radial direction does not affect the area $dA$. Now, from $\eqref{eq1}$ and $\eqref{eq2}$, if we calculate the ‘area speed’, it is as follows.

$$ \begin{align*} \frac{ d A}{ d t } &= {\textstyle \frac{1}{2}} r^{2 \frac{ d \theta}{ d t }} \\ &= {\textstyle \frac{1}{2}}r^{2}\dot{\theta} \\ &= \frac{L}{2m} \\ &= \text{constant} \end{align*} $$

Therefore, the area speed is proportional to the angular momentum, and the object (particle) moves the same area during the same period of time.

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