Kepler’s Laws of Planetary Motion: Nicolaus Copernicus suggested that all planets, including the Earth, travel in a circular orbit around the sun. Galileo later backed this view as well. Johannes Kepler 1571 – 1640 developed three laws of Planetary Motion based on observations and analyses of the data provided by these astronomers. Let’s find out more about Kepler’s rules of planetary motion.

Kepler’s Laws of Planetary Motion – Introduction

The three laws of Kepler describe the motion of a planet around the sun. Kepler’s first law describes the orbit of a planet around the sun. Kepler’s second law is known as the law of equal areas. The third law explains the relation between the orbital period of a planet and the planet’s distance from the sun.

What are Kepler’s Three Laws of Planetary Motion?

a. Kepler’s First Law

Kepler’s first law is also known as the 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, \(AP\) represents the major axis of the elliptical orbit of the planet, and \(BC\) 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 \(A.\)

b. Kepler’s Second Law

Kepler’s second law, or the law of equal areas, states that the planet’s areal velocity around the sun is constant. In other words, the imaginary line joining 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
\({\rm{Area}}\,ASB\,{\rm{ = }}\,{\rm{Area}}\,CSD\,\,{\rm{ = }}\,\,{\rm{Area}}\,ESF\)

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

\(\Delta A = \frac{1}{2}\left({\overrightarrow r \times \overrightarrow v \Delta t} \right)\)
\( \Rightarrow \frac{{\Delta A}}{{\Delta t}} = \frac{1}{2} \times \frac{{\left({\overrightarrow r \times \overrightarrow p } \right)}}{m}\)

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

Practice Exam Questions

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

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 \(ASB.\) 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.
\({v_1}{r_1} = {v_2}{r_2}\)
\( \Rightarrow \frac{{{v_1}}}{{{v_2}}} = \frac{{{r_2}}}{{{r_1}}}\)

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.

C. Kepler’s Third Law

Kepler’s third law is also known as the law of periods. Kepler’s third law states that the square of the time period \(\left( T \right)\) of the revolution of a planet around the sun is directly proportional to the cube of its semi-major axis \(\left( a. \right)\)

\({T^2} \propto {a^3}\)
\( \Rightarrow \frac{{{T^2}}}{{{a^3}}} = {\text{constant}}\)

Let us assume a planet of mass \({\text{m}}\) 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

\(G = \frac{{Mm}}{{{R^2}}} = mR{w^2}\)
\( \Rightarrow G = \frac{{Mm}}{{{R^2}}} = mR\frac{{4{\pi ^2}}}{{{T^2}}}\)
\( \Rightarrow {T^2} = \left({\frac{{4{\pi ^2}}}{{GM}}} \right){R^3}\)

Where \(\frac{{4{\pi ^2}}}{{GM}}\) 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.

Solved Examples on Kepler’s Laws of Planetary Motion

Q.1. A Saturn year is \(29.5\) times the Earth year. How far is Saturn from the sun if the Earth is \(1.50 \times {10^8}\,{\text{km}}\) away from the sun?
Ans: 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, \({R_E} = 1.50 \times {10^8}\,{\text{km=1}}{\text{.50}} \times {\text{1}}{{\text{0}}^{11}}\,{\text{m}}\)
Let the orbital radius of Saturn be \({R_s}\)
According to Kepler’s law of periods,
\({\left[{\frac{{{T_s}}}{{{T_E}}}} \right]^2} = {\left[ {\frac{{{R_s}}}{{{R_E}}}} \right]^3}……\left( 1 \right)\)
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 \(\left( 1 \right),\) we get
\({(29.5)^2} = {\left({\frac{{{R_s}}}{{1.50 \times {{10}^{11}}}}} \right)^3}\)
\({R_s} = 1.5\, \times \,{10^{11}} \times {(29.5)^{\frac{2}{3}}} = \,1.43\, \times {10^{12}}\,{\rm{m}}\)
Therefore, Saturn is \(1.43\, \times {10^{12}}\,{\rm{m}}\) far away from the sun.

Q.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.
Ans: Given the time period of the Earth, \({T_1} = 8\,{\text{h}}\)
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,
\({\left[{\frac{{{T_2}}}{{{T_1}}}} \right]^2} = {\left[ {\frac{{{r_2}}}{{{r_1}}}} \right]^3}\quad \ldots (1)\)
\(\Rightarrow {T_2} = {T_1}{\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^{\frac{3}{2}}} = {T_1}{\left({\frac{{2{r_1}}}{{{r_1}}}} \right)^{\frac{3}{2}}}\)
\(\Rightarrow {T_2} = 8{(2)^{\frac{3}{2}}} = 16\sqrt 2 ~{\text{h}}\)
Therefore, the time period of the satellite after increasing the distance between it and the Earth is \(16\sqrt 2 \) hours.

Summary

Kepler’s laws of planetary motion are empirical because they are based on experiments and observations. These laws have no theoretical foundation. From Kepler’s laws of planetary motion, we can conclude that all planets move around the sun in elliptical orbits, with the sun situated at one of the foci of the ellipse. The areal velocity of a planet around the sun is constant, and the square of the time period of revolution of a planet is directly proportional to the cube of the semi-major axis of the elliptical orbit.

FAQs on Kepler’s Laws of Planetary Motion

Q.1. How many laws did Kepler use to define planetary motion?
Ans: Kepler uses three empirical laws to define planetary motion. These three laws of Kepler’s are as follows:
1. Law of orbits
2. Law of areas
3. Laws of periods

Q.2. What is Kepler’s third law?
Ans: Kepler’s third law states that the square of the time period \(\left(T \right)\) of the revolution of a planet around the sun is directly proportional to the cube of its semi-major axis \({\rm{(a)}}.\)
\({T^2} \propto {a^3}\)

Q.3. What is the first law of planetary motion?
Ans: Kepler’s first law of planetary motion states that each planet revolves around the sun in an elliptical orbit, with the sun as one of the ellipse’s foci.

Q.4. What is Kepler’s second law?
Ans: Kepler’s second law of planetary motion states that the imaginary line joining any planet to the sun sweeps equal areas in equal intervals of time, or the areal velocity of a planet around the sun is constant.

Q.5. What are Kepler’s third laws called?
Ans: Kepler’s third law is also known as the law of periods.

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