 Hello everyone, welcome to our video lecture on orbital equations. Myself, K. R. Biradar, assistant professor, department of electronics and telecommunication engineering, Walsh and the Institute of Technology, Sallapur. Let us start with the learning outcomes first. At the end of this session, students will be able to illustrate on what principle satellite remains in stable orbit. Introduction. In this video, it will be covered about the equations which are related to orbital motion. According to Newton's law of motion, force is equal to mass into acceleration, where f is a force acting on the object, mass m is the mass of the object and a is acceleration. This equation helps to understand the motion of the satellite in stable orbit, centrifugal force and the centripetal force. In a stable orbit, there are two main forces acting on a satellite. Those are centripetal force and centripetal force. If these two forces are equal, the satellite remains in a stable orbit. Forces acting on a satellite. We can see this diagram. At the centre, you can find the earth. So, satellite is revolving in this circular orbit. The forces acting on that satellites are 2. One is towards the earth station. One is going out of the earth station. The force which is acting towards the earth station is called the centripetal force. One force which is trying to escape out of this orbit that is called the centrifugal force. Let us see those definitions. Centripetal force f 1 is due to gravitational attraction of the planet about which the satellite is orbiting. That is what? The earth is a planet here. The satellite is revolving in this direction. So, there is a gravitational force towards the planet for this satellite which attempts to pull the satellite towards the planet. So, the always this earth gravitational force present in the earth will try to pull the this earth towards it due to gravitational force. One is called centrifugal force. We shall see now centrifugal force f 2 is due to kinetic energy of the satellite which attempts to pling the satellite into a higher orbit. That means, the satellite is revolving in this direction. So, it will try to pling it out of its orbit. This is when these two forces are equal, then it satellite remains in its stable orbit. We can see the analogy for this one. So, take a ball and tie a rope to it, try to hold it in your fingers and try to rotate. One force is towards the finger and other force is going out of the finger. The force towards the finger is called centripetal force that is due to this rope. One which is going away from this position that is called the due to the kinetic energy. So, it will try to pull out of this that is path that is also called centrifugal force. Now, we shall see mathematically these two forces. Mathematically the centrifugal force f 1 acting on satellite due earth can be written as f 1 is equal to capital G into capital M into small m divided by r square where G is a universal gravitational constant which is equal to 6.673 into 10 to the power of minus 11 Newton meter square divided by kg square. Capital M is the mass of the earth which is equal to 5.918 into 10 to the power of 24 kg. Small m is the mass of the satellite and r is the capital R is the distance from satellite to the earth centre of. Similarly, centrifugal force can be written as f 2 is equal to small m v square divided by r where v is the orbital velocity of the satellite. Small m again you have seen that is m is a mass of the satellite and r is the distance from satellite to earth centre or centre of the earth. Arbital velocity when both centrifugal and centrifugal forces are balanced to each other that means when I say f 1 and f 2 are equal in centrifugal and centrifugal forces. So, one is due to kinetic energy one is other is due to gravitational force those two are equal. Now, equate these two f 1 and f 2 now already we have seen f 1 equal to G m into small m divided by r square whereas, f 2 we have seen m v square divided by r. When I equate these two so, f 1 equal to f 2 that is equal to capital G into capital m into small m divided by r square which is equal to small m into v square divided by r. One r is on both side that is at the denominator get cancels. One small m is on both side numerator get cancels. So, remaining is capital G into capital M divided by r which is equal to v square that means G is a gravitational constant capital M is the mass of the earth and r is a distance from the satellite to the earth centre v is a velocity. Now, you simplify and separate it out v is now equal to square root of capital G into capital M divided by r. So, that is also equal to square root of mu divided by r where mu is is equal to G into m. This is velocity which always depends on distance from the satellite to the earth centre not on G and m because G is a gravitational constant this is a constant m is a mass of the earth that is also constant. So, only distance can change from satellite to the earth centre that is r. So, the velocity depends on the r time period of a satellite t. What is time period? How much time it takes to complete one path that is called the time period of a satellite. For example, earth also rotates or completes in one path it takes around 365 days. Here also satellite takes its own time to complete its path or one complete path. If the orbit is circular the distance travelled by a satellite in one orbit around a planet is given by 2 pi r that is a 2 pi r is a circumference of a circle where r is a radius of the circle. Now, in this example where r is the distance between satellite in the centre of the earth. Now, time period is equal to distance divided by velocity. So, distance is a total distance travelled by the satellite which is equal to 2 pi r and v is the velocity 2 pi r divided by v that is also equal to 2 pi r divided by v you can replace by square root of mu by r where mu is a equal to G into m r is a distance from satellite to the centre of the earth. Further, if you simply simplify it becomes t equal to 2 pi r is to 3 by 2 divided by mu is to 1 by 2 by taking r at the numerator. Let us see one problem. A satellite is orbiting earth in a uniform circular orbit at a height of 630 kilometer from the surface of the earth assuming the radius of the earth and its mass to be 63070 kilometer and 5.98 into 10 to the power of 24 kg respectively determine the velocity of the satellites. He has given radius and mass of the earth radius is around 63070 kilometer and height of the satellite is 630 kilometer and also he has given mass of the earth equal to 5.98 into 10 to the power of 24 kg. He asked you to find the velocity of the satellite. Assume gravitational constant mu is equal to 6.67 into 10 to the power of minus 11 Newton meter square. Pass the video and try to solve this problem. I think you might have solved this problem. Let us see the solution. Given mass of the earth capital M which is equal to 5.98 into 10 to the power of 24 kg earths gravitational constant which is equal to 6.67 into 10 to the power of minus 11 Newton meter square radius of the earth r is equal to 63070 kilometer height of the satellite h is equal to 630 kilometer height of the satellite means from the surface of earth to the satellite. Kepler's coefficient mu is equal to capital G mean capital G into M. Substitute G and M it comes out to be 3.9886 into 10 to the power of 5 kilometer cube per second square. Radius of the orbit r is equal to radius of the earth plus height of the satellite. So, radius of the earth he has given 63070 kilometer height he has given 630 kilometer it comes out to be 7000 kilometer. The velocity of the satellite is given by V is equal to square root of G into M divided by r this is also equal to square root of mu by r. So, if I substitute all the values it comes out to be 7.54 kilometer per second. Note for a given radius the velocity of the satellite is to be 7.54 kilometer per second since G and M are constants. These are the references used to prepare the above ppt.