Kepler’s laws of planetary motion | Definition , Diagrams , Equations ,Facts and Solved Examples .

Johannes Kepler used empirical observations, and sophisticated calculations to arrive at the famous Kepler’s laws of planetary motion , published in 1609 and 1619. What is particularly significant about Kepler’s laws is that they challenged the prevailing world’s view, with the Earth in the center of the universe (a geocentric theory ) and the Sun and all the planets and stars orbiting around it, just as the Moon does .

Fig. 1 Solar system

Kepler and other pioneers, particularly Nicolaus Copernicus and Galileo Galilei, changed this geocentric view into a heliocentric ( Sun- centred ) cosmology . The planetary system of the star Sun called solar system consists of nine planets. Eg. Mercury, Earth , Venus , Mars, Jupiter, Uranus, Saturn , Neptune and Pluto . Kepler’s complete analysis of planetary motion is summarized in three statements known as Kepler’s laws .


Statement of Kepler’s first law is ” Each planet revolve in an elliptical orbit around the sun and the sun fixed at one focus of the ellipse . This law is also popular as the law of elliptical orbits and definitely it gives the idea about the shape of the orbits of the planets round the sun .

Fig . 2

Figure 2 shows the geometry of an ellipse , which is similar to the elliptical orbit of a planet . Mathematically an ellipse can be defined by choosing two fixed points F1 and F2 , each of which is a called a focus , and then drawing a curve through these points in such a way that sum of distance of any points from focus should be same . Means the sum of the distances r1 and r2 from F1 and F2 respectively , should be constant. The longest distance through the center between two points on the ellipse ( passing through both the focus) is called the major axis . In the above figure 2 , the major axis is drawn along the x direction .

Such a way , the shortest distance through the center between two points on the curve of ellipse is called the minor axis . In in above figure 2 , manor axis is along Y axis .

Note – In the elliptical orbit of a planet around the Sun , the Sun is at one focus of the ellipse and planet revolve around the sun in the elliptical curve . There is nothing at the other focus of the ellipse .

Now imagine a planet in an elliptical orbit as shown in figure, with the Sun at focus F. When the planet is at the far right in the diagram, the distance between the planet and the Sun at this point, called the aphelion, the planet is at its maximum distance from the Sun. ( For an object in orbit around the Earth, this point is called the apogee.) Conversely, when the planet is at the left end of the ellipse, the distance between the planet and the Sun at this point, called the perihelion (for an Earth orbit, the perigee ), the planet is at its minimum distance from the Sun .


According to Kepler’s second law The radius vector, drawn from the sun to a planet, sweeps out equal areas in equal time, i.e. its areal velocity (or the area swept out by it per unit time) is constant . This is referred to as the law of areas and gives the relationship between the orbital speed of the planet and its distance from the sun . According to this law, planet will move slowly when it is farthest from sun and move rapidly when it is nearest to sun. It is similar to law of conservation of angular momentum .

Fig. 3

Prove of  Kepler’s second law 

Consider that a planet of mass m is revolving around the sun of mass M in a circular orbit of radius r. Let v be its orbital velocity .

Fig. 4

If dA is small area (shown shaded) swept by the line joining the planet to the sun in time dt , then
dA = area of ∆ABS = base x perpendicular/2 

or dA = AB×AS/2       ..( i )

If dθ is the angular displacement of the planet in time dt i.e. ,when it moves from point A to B , then AB=r dθ . Substituting for AB (=rdθ ) and AS (=r) in Eq. ( i ) , we have

dA = rdθ×r /2 = r2dθ/2    …( ii )

Dividing both sides of the Eq. (ii) by dt ,

we haved A/dt = r2dθ/2dt

or dA/dt = r2ω/2 …( iii )

where  ω = dθ/dt  is the angular speed of the planet in its orbit and dA/dt is the areal velocity of the planet. Multiplying and dividing RHS of Eq. (iii) by m (the mass of the planet), we have

dA/dt = mr2ω/2m
Since mr2ω = L , the angular momentum of the planet about the axis through the sun , we have
dA/dt = mr2ω/2m   … ( iv ) 

As there are no any external torque acting on the planet during its orbital motion so , its angular momentum (L) must remain constant during motion . Since here , both m and L are constant quantities , the Eq. (iii) becomes

dA/dt = L/2m = constant

Hence, we can easily say that when a planet moves around the sun , its areal velocity remains constant . It proves Kepler’s second law of planetary motion .


Kepler’s third law say that ” The square of time period of revolution (T) of any planet around the sun will be directly proportional to the cube of the length of semi-major axis of the orbit . This equation is valid for both type of motion of planet , either we consider circular motion or elliptical motion . In case of Circle there is no major or minor axis because in circle we consider radius .

T2 ∝ a3 or T2 = Ka3 …. ( i )

where a = semi-major axis or if we consider circular motion then ‘ a ‘ will be radius of circle .

Here in the above formula , K is the proportionality constant and is independent of the mass of the planet , that’s why Equation (i) is valid for any planet . If will will change the star about which the planets are revolving then value of constant K will be change .

Note – If we were to consider the orbit of a satellite such as the Moon revolves about the Earth , the constant would have a different value .

Newton was able to show that the same relationship holds for an elliptical orbit , with the orbit radius r replaced by semi-major axis a ,

 thus         T = 2πa3/2 / ( GMs )1/2 
Here , Ms is the mass of the sun .

Solved Examples

  • Problem 1. Find the no of years required Mars to make one revolution about the Sun if the mean distance of Mars from the Sun is 1.524 times that of the earth from the Sun .

Solution – According to Kepler’s law ,

T2 = kR3

 Let , R1 is the distance of mars from Sun and R2 is the distance of earth from Sun . 
And let T1 is the time required for mars ( in year )to make one revolution about the Sun  and T2 is same for earth .
T12/T22 = R13/R2

T1/T2 = ( R1/R2 ) 3/2 

T1 = T2 ( R1/ R2 )3/2 

T1 = 1 ( 1.524 ) 3/2

T1 = 1.88 years

  • Problem 2. If the earth is at one – eight of it’s present distance from the Sun , the duration of the year as compare to present year will be .

Solution – We know that according to Kepler’s law , T2 = K R3 

T12/T22 = R13 / R23

T2/T22= R3/( R/8 )3 = ( 8 )3 

T22= T2/512 
T2 = T/16√2   ( now year will 1/16√2 of the present year ) .

  • Problem 3.  During a small time intervals T1 and T2 , A planet sweeps areas 2A and 3A respectively , Find the value of T1/T2 .

Solution – According to Kepler’s 2nd law we can say that , when a planet revolution around that the sun , the area swept out by it per unit time is constant .


T1 /T2 = 2A / 3A 

T1/T2 =  2/3

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