# Kepler's Laws

Kepler's laws are based on planetary motion. Kepler's laws identify the fact about the orbits of planets and their motion in those orbits. The Kepler's three laws are:

1. The planets move in elliptical orbits around the Sun at one of the foci of the ellipse.
2. The line joining the Sun and the planet sweeps out equal areas in equal time intervals.
3. The square of the time period of a planet is proportional to the cube of the semi-major axis length of the elliptical orbit.

## Kepler's First Law

The Kepler's first law identifies the type of orbit in which a planet moves. It says that all planets move in elliptical orbits with the sun at one of the foci of the ellipse. A circle is also a special case of an ellipse. So the orbit of a planet is an ellipse. In Figure 1 a planet $p$ moves in an elliptical orbit with the sun at one of the foci $S$ of the ellipse.

The axis $AB$ joining the two foci passing through the centre of the ellipse $O$ is the major axis of the ellipse. So in Figure 1 the length of the major axis is $AB = 2a$ and $a$ is the length of semi-major axis of the ellipse. And $CD$ is the minor axis of the ellipse which passes through the centre being perpendicular to the major axis. The minor axis has length $2b$ and $b$ is the length of semi-minor axis. The distance from one of the foci to the centre of the ellipse is $c$. The ratio $c/a$ is called eccentricity $e$ of the ellipse. For a circle $c = 0$ and the eccentricity is also zero. Therefore, a circle is an especial case of an ellipse.

## Kepler's Second Law

According to Kepler's second law a planet moving in an elliptical orbit sweeps out equal areas in equal time intervals. To prove this consider a planet moving in an elliptical orbit around the Sun at one of the foci of the ellipse as in Figure 2. Figure 2(a) A planet moves in an elliptical orbit around the sun at one of the foci of the ellipse.

Let the initial position vector of the planet from the Sun be $\vec r$ and in a very small time interval $dt$, the final position vector joining the Sun and the planet is $\vec r + \vec dr$. The perpendicular component of velocity $\vec v$ of the planet with the position vector is $v_\bot = v \sin \theta$. But from Figure 2 in a small time interval $dt$ you can find $rd\phi = v\sin \theta dt$. The small area $dA$ swept out by the position vector joining the Sun and the planet in time $dt$ is

$dA = \frac{1}{2}r(r{\kern 1pt} d\phi ) = \frac{1}{2}r(v\sin \theta {\kern 1pt} dt ) \tag{1} \label{1}$

Therefore, the rate of change of area is

$\frac{{dA}}{{dt}} = \frac{1}{2}r{\kern 1pt} v\sin \theta = \frac{1}{2}|\vec r \times \vec v| \tag{2} \label{2}$

In the above equation the quantity $dA/dt$ is the rate of change of area of the ellipse called areal velocity or sometimes called sector velocity. Note that the only force between the planet and the Sun is due to the gravitational interaction. The gravitational force $\vec F$ the Sun exerts on the planet is directed towards the centre of the Sun while the position vector $\vec r$ of the planet from the Sun is in opposite direction to $\vec F$; both vectors lie in the same line but in opposite directions. Therefore, the torque produced by the gravitational force of the Sun on the planet must be zero, that is,

$|{\vec \tau }| = |{\vec r \times \vec F }| = 0$

If $\vec L$ is the angular momentum of the planet around the Sun, the torque is the rate of change of angular momentum that is, $d\vec L/dt$. Since the torque is zero, the the rate of change of angular momentum is zero and hence the angular momentum is conserved or constant. You can rewrite Eq. \eqref{2} as

$\frac{dA}{dt}=\frac{|\vec{r}\times {{M}_{p}}\vec{v}|}{2{{M}_{p}}}=\frac{|\vec{L}|}{2{{M}_{p}}} \tag{3} \label{3}$

where $M_p$ is the mass of the planet. The above equation shows that the areal velocity of the planet around the Sun is constant which proves Kepler's second law. The conservation of angular momentum of planets is also the fact that why the planets move in plane around the Sun.

## Kepler's Third Law

Now the same planet of mass $M_p$ as in the above discussion of second law revolves in an elliptical orbit around the Sun at one of the foci of the ellipse. The Kepler's third law tells us that the square of its time period of revolution around the Sun is equal to the cube of the semi-major axis length of its elliptical orbit. The area of the ellipse is $\pi ab$ and the areal velocity is given by Eq. \eqref{3}. We represent the magnitude of angular momentum as $L$. Now the time period of revolution of the planet around the Sun is the ratio of the area of the elliptical orbit to the areal velocity, and hence, the time period $T$ is:

$T=\frac{\pi ab}{L/2{{M}_{p}}}=\frac{2{{M}_{p}}\pi ab}{L} \tag{4} \label{4}$

Squaring both sides of the above equation gives:

${{T}^{2}}=\frac{4M_{p}^{2}{{\pi }^{2}}{{a}^{2}}{{b}^{2}}}{{{L}^{2}}} \tag{5} \label{5}$

If $l$ is the semi-latus rectum of the elliptical orbit or ellipse, then ${{l}^{2}}={{b}^{2}}/a$ which implies ${{b}^{2}}=l{{a}^{2}}$. Therefore, putting the value of ${{b}^{2}}=l{{a}^{2}}$ in Eq. \eqref{5}, we get,

${{T}^{2}}=\frac{4M_{p}^{2}{{\pi }^{2}}l{{a}^{3}}}{{{L}^{2}}} \tag{6} \label{6}$

The above equation shows that the square of the time period of revolution of the planet around the Sun in its elliptical orbit is proportional to the cube of the semi-major axis length of the elliptical orbit.