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About this Lesson
 Type: Video Tutorial
 Length: 10:54
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 117 MB
 Posted: 07/01/2009
This lesson is part of the following series:
Physics (147 lessons, $198.00)
Physics: Momentum (8 lessons, $16.83)
Physics: Momentum and Its Conservation (5 lessons, $9.90)
This lesson was selected from a broader, comprehensive course, Physics I. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/physics. The full course covers kinematics, dynamics, energy, momentum, the physics of extended objects, gravity, fluids, relativity, oscillatory motion, waves, and more. The course features two renowned professors: Steven Pollock, an associate professor of Physics at he University of Colorado at Boulder and Ephraim Fischbach, a professor of physics at Purdue University.
Steven Pollock earned a Bachelor of Science in physics from the Massachusetts Institute of Technology and a Ph.D. from Stanford University. Prof. Pollock wears two research hats: he studies theoretical nuclear physics, and does physics education research. Currently, his research activities focus on questions of replication and sustainability of reformed teaching techniques in (very) large introductory courses. He received an Alfred P. Sloan Research Fellowship in 1994 and a Boulder Faculty Assembly (CU campuswide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for NonPhysicists: a Tour of the Microcosmos” and “The Great Ideas of Classical Physics”. Prof. Pollock regularly gives public presentations in which he brings physics alive at conferences, seminars, colloquia, and for community audiences.
Ephraim Fischbach earned a B.A. in physics from Columbia University and a Ph.D. from the University of Pennsylvania. In Thinkwell Physics I, he delivers the "Physics in Action" video lectures and demonstrates numerous laboratory techniques and realworld applications. As part of his mission to encourage an interest in physics wherever he goes, Prof. Fischbach coordinates Physics on the Road, an Outreach/Funfest program. He is the author or coauthor of more than 180 publications including a recent book, “The Search for NonNewtonian Gravity”, and was made a Fellow of the American Physical Society in 2001. He also serves as a referee for a number of journals including “Physical Review” and “Physical Review Letters”.
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Rocket ships are a really dramatic demonstration of lots of physics. It's so impressive that we can design and build rocket ships that really, they work, they get to the moon, and we've sent probes out beyond the edges of our solar system. And we know where they're going to go, how fast they're going to be going. People use the expression `rocket science' along with `brain surgery' as a kind of a representation of the epitome of human intellect and achievement. And it's kind of cool that, just knowing introductory physics, one can understand an awful lot of rocket science. So I'd like to think a little bit about how rockets work and whether or not we can just use principles of physics, Newton's Law and conservation of momentum, and figure out, for instance, what's going to be the velocity of a rock as a function of time.
Now, what's the principle of a rocket? The principle is very simple: suppose you're out in deep space and you're holding a bunch of little pebbles in your hand. And you take one pebble and you toss it out the back. So what's happening? You are applying a force to the pebble and, by Newton's Third Law, the pebble applies an equal and opposite force back on you. So the pebble accelerates one way and you accelerate the other way. You accelerate, and now you toss another pebble, and another and another. Each time you toss one, you accelerate a little bit more. That's what a rocket's doing. Instead of pebbles, it's just throwing little molecules of fuel out the back, and each one that it throws out, it's accelerating a little bit forward in the opposite direction. So that's the principle, it's very simple.
How about calculating? How do you figure out velocity as a function of time for a rocket? The principle of physics there, the quantitative principle, is just conservation of momentum. You're out in deep space, there's no external forces on you, and so, if you just figure out at any moment in time what's your momentum, and then a moment later, you can consider your momentum plus the momentum of the little chunk of fuel that you just sent out. And they will be equal, momentum before and momentum after will be the same. So that's the principle we're going to use to solve the equations.
Let me draw some pictures, because, I've got to admit, it is rocket science. It's not the easiest story in the world, and you've got to be as careful as can be to get the equations right. Here's a rocket out in deep space and I'm going to call it's mass, at this instant in time, capital M plus little m. So this represents the mass of the rocket and this represents the mass of that little chunk of fuel that we're just about to send out the back. At this moment in time, as observed by somebody in a fixed reference frame, say, on earth, the rocket is moving with velocity v. Now the rocket tosses out a little chunk of fuel, so here's the picture a moment later. The rocket has accelerated a little bit. It's got velocity v plus v. We'd really like an equation for v. The rocket's mass is now capital M. It's thrown this little m out the back.
Now, what velocity should I associate with that little m? A rocket shoots fuel out the back with some velocity via exhaust. And it's sort of a characteristic number of the engine and you might think of it as a constant. The problem is I'm a little hesitate to just put it in the picture, because vex is the velocity of the fuel with respect to the rocket. In this picture, I'm drawing all my arrows as velocities with respect to the observer on the ground. If a rocket is cruising by fast and it ejects a little piece of fuel out the back with relative velocity vex, what do you see from the ground? You see a velocity v of the rocket plus vex, so what you see here is v plus vex. Now that's plus an arrow. The vex arrow is to the left, so, of course, this arrow will be shorter than the initial velocity. In other words, if you're throwing stuff out the back and it's got a huge exhaust velocity, this would come out to be a negative arrow, and so the fuel really would be flying backwards. On the other hand, if vex was some tiny number, if the rocket was just sort of sputtering fuel out the back, it's cruising along at a big v, the rocket fuel is sort of behind it, it's going a little bit more slowly, but still moving in the same rocket as the rocket, just with a relative velocity of vex. So that's one little subtle thing that you just have to think about and convince yourself that this is correct.
Now, we're all set. Here's the initial situation, here's the final situation, no external forces. This is my system, it's still my system here, and so I can set initial momentum equals final momentum. So M times v plus m times v, that's the initial momentum. v times m plus vex times m plus m times v plus m times v, that's the grand total momentum afterwards. Set them equal. Write it down, you'll see a bunch of terms, and then you'll realize that many are common on both sides and you can cancel them. And you will end up with what's known as a rocket equation. In fact, if you take that rocket equation and you divide by t, you'll get the following very interesting equation, so let's write it down: M dv dt is equal to minus vex dM dt. Now, where this minus sign comes from, if you're looking at your equations, you probably have a plus sign there, but you also have a d little m, so there's another little thinker. If you are ejecting fuel out the back, your rate of change of mass is negative if you are ejecting positive m. In other words, dM is negative of dm. So that's just a little subtle point. It's just our definition of the sign of the mass of the fuel being thrown out the back.
So this is the equation. It's the rocket equation and look at it, just think about it for a second. At the instant in time that we're considering, we've got mass times acceleration, dv dt of the rocket. So that should be the force on the rocket: F equals ma and there it is. That's the formula. We call it the thrust on the rocket. vex is typically a number and dM dt, that's how rapidly you are throwing fuel out the back, how many kilogram per second is being thrown out. So these are typically constants, so this looks like a constant force equation and you think, "Hey, that's easy! Constant force means constant acceleration, doesn't it? I could just use vf equals vi plus a times time." No, it is, at this moment in time, this righthand side is a constant, but M is changing with time. The rocket is constantly losing mass, it's throwing out fuel. And so this is really not a good simple constant acceleration story and you can't use the old constant acceleration kinematics equations. If want to know vf in terms of vi, you've just to solve this little differential equation. It's a math problem, so I'm not going to go through the details too carefully. Physicists tend to do nifty things, like cancel dt from both sides, and let's see, I've got a dM and I'll divided by M, so I'll get dv is equal to vex dM divided by M. That's my equation rewritten, after having canceled out dt, and now what do you do with an equation like this? Integrate both sides. This is a constant, so constants come out of integrals. I get the integral from vi till vf and, over here, this integral is over mass, it's from Mi till Mf and I ran out of room there. This is the integral from Mi to Mf. What's the integral of 1 over M dM? That's a natural logarithm. This is just an easy integral, it's vf minus vi. Here's the answer, we've really got it: vf is equal to vi plus vex times the logarithm of Mi over Mf. I did a little trick here. What I really had was logarithm of Mf minus logarithm of Mi and I used an identity, log of a minus log of b is log of a over b, and then I used another identity of logarithms to turn it upside down and get rid of my minus sign. Just some math tricks.
Look at this formula; it makes some sense. The logarithm of Mi over Mf, you started off with a big mass and, after you've ejected most of the fuel, you've got a rather small final mass, so you're going to be taking the logarithm of a pretty big number, so that's going to be some positive number. You multiply it by vex. So if you want to go fast, send out your fuel with a large exhaust velocity. Remember the thrust was vex dm dt. Which do you want to make big? If you make dM dt big, you're going to lose mass really fast, you're going to get rid of all your fuel right away. So you'd rather make vex as big as possible, and then this equation says that's what you'll want to do to get a big final velocity. This is what the rocket engineers are working on when they're designing better rockets, one of the things is to make the exhaust speed very large. Mi over Mf can be a big number. Unfortunately, the natural logarithm is a function, which, when you take the log of a big number, it gives you a much smaller number, typically, when you've got big numbers here, like the log of 1000 is just a little bit under 7. So that's too bad, it means that you need a huge amount of fuel in a rocket and a very small amount of payload if you want to get up to any kind of reasonable speeds.
This is rocket science. It's not the easiest physics in the world, but we're certainly completely equipped to understand the basic ideas and do the calculations. There's a few little technical details you've got to think about, but it's all stuff we've already learned, like relative velocities and conservation of momentum. It's kind of impressive, mere mortals, like me and you, can really understand rocket science.
Momentum
Momentum and Its Conservation
Rocket Propulsion Page [2 of 2]
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