 Hello everyone, welcome to a video lecture on Kepler's Laws for Planetary Motion. Myself, K. R. Biradhar, Assistant Professor, Department of Electronics and Telecommunication Engineering, Walton Institute of Technology, Sallapur. Let us start with the learning outcomes first. At the end of this session, students will be able to describe Kepler's Laws for Planetary Motion. Introduction, Earth revolve around the sun in a definite orbit. This is a fact. We know, Earth always revolve around the sun in a particular orbit. Likewise, the same way, artificial satellites which are launched by us, which revolve around the Earth in a definite orbit. So, the principles which are applied for Earth revolve around the sun and artificial satellites or man-made satellites revolve around the Earth are same. Kepler gave three most important law in satellite communication theory. Those are called Kepler's Laws for Planetary Motion. These are popularly known as Kepler's Laws. Kepler's first law and its statement. It states that the path followed by planets around the sun will be an ellipse. Kepler's first law, he gave, he stated the path followed by planets. Maybe Earth is a planet which revolve around the sun and it follows a definite path that is ellipse. It follows elliptical path. We can see this diagram, figure 1A, it consists of a two focuses, F1 and F2. And there is an elliptical path. In the elliptical path, the planets will be revolving. The sun will be present at one of the focus. See this animation. The sun is at one focus that is at F1. The Earth is revolving around the sun, but it is taking a elliptical path. That states the Kepler's first law. Kepler's second law, that statement is Kepler's second law states that for equal time intervals, the area covered by satellite will be same. You can see this animation. There are two parts, here you can see shaded line. It is one day to travel. So, it is taking the time to travel from A to B. In the same area will be covered in the other place, but time it takes is same. This can be explained with one figure. You shall see that in the next slide. Let us see illustration of Kepler's second law. See there are two focus in this ellipse. So sun is at one focus. So, planet is revolving in a definite orbit and it follows the elliptical path as per Kepler's first law. There are two areas will be covered here. One is area one and area two. The time it takes to travel from P1 to P2, it is T and here to cover area two, the time it takes to travel from P3 to P4 is also T. When these two areas are equal, that is area one and area two are equal for a given time T. Third law, the Kepler's third law, the square of the time period of orbit is proportional to the cube of the mean distance between the two bodies. That means, time period we know, time period is a time taken to complete one orbit by the planet. Square of the time period or time taken is proportional to the cube of the mean distance between two bodies that is also called the semi major axis. Mathematically, it can be written as T square is proportional to A cube. Here T is a time taken to complete one path or that is time period and A is a semi major axis. T square is proportional to A cube. See this diagram, it consists of a sun at one focus, the planet is revolving and taking the elliptical path as per Kepler's first law and time to complete one particular path is T and the distance between this end and this end is called the major axis, half of it that is one of the end and its center that is called the semi major axis that is A. Here T square is proportional to A cube, T is the time taken and A is the semi major axis as shown in this diagram. But when you consider that is a semi major axis, this side also semi major axis it becomes major axis. By removing now proportionality constant because the T square is proportional to A cube, if I remove the proportionality constant this becomes T square is equal to 4 pi square divided by mu into A cube where T time period to complete one path, A is a semi major axis and the mu is the Kepler's coefficient. Same thing it defined here, but mu is Kepler's constant its value is 3.98004 into 10 to the power of 5 kilometer cube per second square. If the orbit is circular then the radius of this circle is equal to semi major axis A, r is equal to A therefore, T square now becomes equal to 4 pi square by mu into R cube where the A is there that will be replaced by R. This is a total time period calculation formula. Let us see one example. Calculate the orbital period of this satellite moving in an elliptical orbit having Kepler's coefficient is 3.98006 into 10 to the power of 5 kilometer cube per second square and major axis of 50,000 kilometer. What he has given here? Kepler's coefficient that is mu 3.98006 into 10 to the power of 5 and major axis that is 2A he has given that is equal to 50,000 kilometer. Pass the video think and try to solve this problem. I think you might have solved this problem. Let us see the solution. What he has given? Mu he has given mu is equal to 3.98006 into 10 to the power of 5 kilometer cube per second square. Major axis he has given 2A is equal to 50,000 kilometer. Now if I calculate minor axis, major axis is 2A, minor axis is A. So if I take two other side it becomes A equal to 25,000 kilometer. Now we have a formula T square is equal to 4 pi square A cube by mu. If I substitute all the values, so T becomes equal to 39,344 seconds. I need to convert this one in terms of our minute and seconds. First you need to divide it by 60 that comes in minutes and then again you need to divide it by 60 that comes in hours. So it comes around 10 hour 55 minutes and 44 seconds. These are the references I used to prepare the above PPT. Thank you.