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MIT 8 01 - Lecture Notes

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MIT OpenCourseWare http://ocw.mit.edu 8.01 Physics I: Classical Mechanics, Fall 1999 Please use the following citation format: Walter Lewin, 8.01 Physics I: Classical Mechanics, Fall 1999. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/termsMIT OpenCourseWare http://ocw.mit.edu 8.01 Physics I: Classical Mechanics, Fall 1999 Transcript – Lecture 14 If you are standing somewhere on Earth... this is the Earth, the mass of the Earth, radius of the Earth, and you're here. And let's assume for simplicity that there's no atmosphere that could interfere with us, and I want to give you one huge kick, an enormous speed, so that you never, ever come back to Earth, that you escape the gravitational attraction of the Earth. What should that speed be? Well, when you're standing here and you have that speed, your mechanical energy-- which we often simply call E, the total energy-- is the sum of your kinetic energy-- this is your mass; this is your escape velocity squared-- plus the potential energy, and the potential energy equals minus m Mg divided by the radius of the Earth. So this is your kinetic energy and this is your potential energy-- always negative, as we discussed before. Mechanical energy is conserved, because gravity is a conservative force. So no matter where you are on your way to infinity, if you are at some distance r, that mechanical energy is the same. And so this should also be one-half m v at a particular location r squared minus m M earth G divided by that little r. And so at infinity, when you get there-- little r is infinity, this is zero, potential energy at infinity is zero-- and if I get U at infinity with zero kinetic energy, then this term is also zero. And that's the minimum amount of energy that I would require to get you to infinity and to have you escape the gravitational pull of the Earth. If I give you a higher speed, well, then, you end up at infinity with a little bit net kinetic energy, so the most efficient way that I can do that is to make this also zero, so you reach infinity at zero speed. So this is for r goes to infinity. And so this E equals zero then.And so this term is the same as this term for your escape velocity. And so we find that one-half m v escape squared equals m M earth G divided by the radius of the Earth. I lose my little m, and I find that the escape velocity that I have to give you is the square root of two M earth G divided by the radius of the Earth. And this is enough, is sufficient to get you all the way to infinity with zero kinetic energy. If you substitute in here the mass of the Earth and the radius of the Earth, then you will find that this is about 11.2 kilometers per second. That is the escape velocity that you need. It's about 25,000 miles per hour. Again, we assume that there is no air that could interfere with you. If the total energy when you leave the Earth with that velocity-- if the total energy is larger than zero, you do better than that. You reach infinity with kinetic energy which is a little larger than zero. We call this unbound orbit-- larger or equal. If E is smaller than zero, that the total energy that you have is negative, then you will never escape the gravitational pull of the Earth, and you will be one way or another in what we call a bound orbit. Let's pursue the idea of circular orbits. Later in the course we will cover elliptical orbits, but now let's exclusively talk about circular orbits. Now, this is the mass of the Earth, and in a circular orbit is an object with mass m, a satellite, and m is way, way, way smaller than the mass of the Earth. And the radius of the orbit is R, and this object has a certain velocity v, tangential speed. The speed doesn't change, but the direction changes, and there has to be a gravitational force to hold it in orbit, and the gravitational force is exactly the same as the centripetal force-- we've discussed that many times before. And so the gravitational force which is necessary to make it go around-- I could also say the centripetal force is necessary to make it go around in a circle-- that gravitational force equals m M earth G divided by R squared.This is now the distance from the Earth to the satellite, and that must be equal to m v squared divided by R, and that is that tangential speed that you see here, which a little later in time, of course, would be here. I lose my m, and so you see now that the orbital speed--not to be mistaken for escape velocity--the orbital speed is exactly the same what we have there except the square root of two divided by R. This is now R. And there it was R earth. If you know R, then you can calculate the speed in orbit. If you know the speed in orbit, you can calculate R. And so the period of going around in the orbit, T equals two pi R divided by the orbital speed, and when you do that, you get two pi. You get an R to the power three-halfs, and you have downstairs the square root of G M earth. Let me move this in a little. Two pi R to the power three-halfs. So again, if you know the radius, if you know how far you are away from the Earth, the period follows uniquely. If you know the period, then the distance to the satellite follows uniquely. If we take the shuttle as an example of a near-Earth orbit, so we have the shuttle. The shuttle may be 400 kilometers above the Earth's surface. So we have to add to the radius of the Earth 400 kilometers, so you end up with about 6,800 kilometers for the radius of the orbit of the shuttle, and you substitute that in here, the mass of the Earth and the gravitational constant, you'll find that T is about 90 minutes. It's about 1Ω hours. The shuttle takes about 1Ω hours to go around, and the speed, that tangential speed, is very close to eight kilometers per second. And that holds for all near-Earth-orbit satellites. Whether they are 400 or 500 or 600 kilometers, that doesn't change very much. If you take the moon--the moon is much further away than the shuttle, and you take the distance to the moon--which is some 385,000 kilometers--you substitute that in this equation, you will find that the period for the moon to go around the Earth is about 27Ω days. And its speed is only one kilometer per


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