8.2 Kepler’s laws

The three laws of Kepler can be stated as follows:

• Law of Orbits: Kepler’s law of orbits states that each planet revolves around the sun in an elliptical orbit, with the sun as one of the ellipse’s foci.

In the above figure, \(A P\) represents the major axis of the elliptical orbit of the planet, and \(B C\) represents the minor axis of the elliptical orbit of the planet. \(S_1\) is one of the foci of the ellipse where the sun is located.

1. Perihelion: It is the nearest point of the planet’s orbit from the sun. In the above figure, perihelion is represented by point \(P\)
2. Aphelion: It is the farthest point of the planet’s orbit from the sun. In the above figure, aphelion is represented by point \(A\).

• Law of areas: Kepler’s second law, or the law of equal areas, states that the imaginary line that joins any planet to the sun sweeps equal areas in equal intervals of time. The imaginary line joining the planet and the sun is called the radius vector.

The change in the linear speed of a planet while moving around the sun keeps the areal velocity of the planet constant. Consider a planet moves from \(A\) to \(B\), then from \(C\) to \(D\) and then from \(E\) to \(F\) in the same time interval \(\Delta t\), then
Area \(A S B=\) Area \(C S D=\) Area \(E S F\)

Let us assume that the sun is located at the origin, \(\vec{r}\) be the position vector and \(\vec{p}\) be the momentum of the planet. The area swept out by the planet having mass \(m\) in time interval \(\Delta t\) given by

\(
{\Delta A}=\frac{1}{2} \times {(\vec{r} \times \vec{v})} \times {\Delta t}
\)
\(
\Rightarrow \frac{\Delta A}{\Delta t}=\frac{1}{2} \times \frac{(\vec{r} \times \vec{p})}{m}
\)
(We know that momentum is the product of mass and velocity. So, \(\vec{v}=\frac{\vec{p}}{m}\) )
\(
\frac{\Delta A}{\Delta t} = \frac{\vec{L}}{2 m}
\)

Where \(\vec{L}\) is the angular momentum, and it is given by
\(
\vec{L}=\vec{r} \times \vec{p}
\)

For a central force, which is directed along \(\vec{r}, \vec{L}\) is a constant as the planet goes around. Hence, \(\Delta {A} / \Delta t\) is a constant according to the last equation. This is the law of areas. Gravitation is a central force and hence the law of areas follows.

Kepler’s second law is based on the law of conservation of angular momentum. Let us assume that the velocity of the planet is \(v_1\) when it moves from \(A\) to \(B\) in time interval \(\Delta t\), and the radius vector \(r_1\) has swept and area \(A S B\). Similarly, the velocity of the planet is \(v_2\) when it moves from \(C\) to \(D\) in the same time interval \(\Delta t\), the radius vector \(v_2\) has swept the area CSD. Then according to the law of conservation of angular momentum, we get \(m v_1 r_1=m v_2 v_2\)
Where \(m=\) mass of the planet.
\(
\begin{aligned}
&v_1 r_1=v_2 r_2 \\
&\Rightarrow \frac{v_1}{v_2}=\frac{r_2}{r_1}
\end{aligned}
\)
From here, we can conclude that the planet’s velocity is inversely proportional to the distance of the planet from the sun. Thus, planets appear to move faster when they are closer to the sun, whereas they appear to move slower when they are farther from the sun.

• Law of periods: Kepler’s third law states that the square of the time period \((T)\) of the revolution of a planet around the sun is directly proportional to the cube of its semi-major axis \((a\) )
  \(
  \begin{aligned}
  &T^2 \propto a^3 \\
  &\Rightarrow \frac{T^2}{a^3}=\text { constant }
  \end{aligned}
  \)
  Let us assume a planet of mass \(\mathrm{m}\) is moving around the sun of mass \(M, T\) is the period of the revolution of the planet and \(R\) is the average distance of the planet from the Sun which revolves in an almost circular orbit with constant angular velocity \(w\), then by considering the law of gravitation, we can write
  \(
  \begin{aligned}
  &G=\frac{M m}{R^2}=m R w^2 \\
  &\Rightarrow G=\frac{M m}{R^2}=m R \frac{4 \pi^2}{T^2} \\
  &\Rightarrow T^2=\left(\frac{4 \pi^2}{G M}\right) R^3
  \end{aligned}
  \)
  Where \(\frac{4 \pi^2}{G M}\) is a constant. Therefore, we can say that \(T^2 \propto R^3\)

According to Kepler’s third law, if the distance of the planet from the sun increases, then the planet’s time to orbit the sun increases. For example, Mercury is the nearest planet to the sun, so it takes 88 days to revolve around the sun, whereas Neptune is the farthest planet from the sun, and it takes 165 Earth years to revolve once around the sun.

Example 1: A Saturn year is \(29.5\) times the Earth year. How far is Saturn from the sun if the Earth is \(1.50 \times 10^8 \mathrm{~km}\) away from the sun?

Solution:

Let \(T_s\) be the time period of the revolution of Saturn, and \(T_E\) be the time period of the revolution of Earth.
Now, the orbital radius of the Earth, \(R_E=1.50 \times 10^8 \mathrm{~km}=1.50 \times 10^{11} \mathrm{~m}\)
Let the orbital radius of Saturn be \(R_s\)
According to Kepler’s law of periods,
\(
\left[\frac{T_s}{T_E}\right]^2=\left[\frac{R_s}{R_E}\right]^3 \ldots \ldots(1)
\)
It is given that a Saturn year is \(29.5\) times the Earth year which means
\(
\frac{T_s}{T_E}=29.5
\)
By substituting this value in equation \((1)\), we get
\(
\begin{aligned}
&(29.5)^2=\left(\frac{R_s}{1.50 \times 10^{11}}\right)^3 \\
&R_s=1.5 \times 10^{11} \times(29.5)^{\frac{2}{3}}=1.43 \times 10^{12} \mathrm{~m}
\end{aligned}
\)
Therefore, Saturn is \(1.43 \times 10^{12} \mathrm{~m}\) far away from the sun.

Example 2: The time period of a satellite of Earth is 8 hours. If the separation between the Earth and the satellite is increased to two times the previous value, find the new time period of the satellite.

Solution:

Given the time period of the Earth, \(T_1=8 \mathrm{~h}\)
Let the radius of the orbit of the satellite \(=r_1\)
The time period of the satellite \(=T_2\)
The new radius of the orbit of the satellite, \(r_2=2 r_1\)
According to Kepler’s law of periods,
\(
\begin{aligned}
&{\left[\frac{T_2}{T_1}\right]^2=\left[\frac{r_2}{r_1}\right]^3 \quad \ldots(1)} \\
&\Rightarrow T_2=T_1\left(\frac{r_2}{r_1}\right)^{\frac{3}{2}}=T_1\left(\frac{2 r_1}{r_1}\right)^{\frac{3}{2}} \\
&\Rightarrow T_2=8(2)^{\frac{3}{2}}=16 \sqrt{2} \mathrm{~h}
\end{aligned}
\)
Therefore, the time period of the satellite after increasing the distance between it and the Earth is \(16 \sqrt{2}\) hours.