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About this Lesson
- Type: Video Tutorial
- Length: 12:50
- Media: Video/mp4
- Use: Watch Online & Download
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- Posted: 07/01/2009
This lesson is part of the following series:
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 campus-wide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for Non-Physicists: 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 real-world 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 Non-Newtonian 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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Newton's Universal Law of Gravity: F = . This is a constant of nature, this is the mass of the two objects that you're considering. r is the distance between them and the formula tells you the gravitational attraction, it's the force on either one of them. It's an attractive force, heading from one towards the other. It's a very simple looking formula, it is, in many respects a very simple formula, but it's worthwhile thinking about a few details of it. For instance, when I say the distance between the objects, let's look at a specific example.
Here's the earth and there's some satellites sitting over here. If I want to plug into that formula in order to deduce the force on the satellite, what's the r that I'm supposed to plug in? What do I mean by that? You might imagine that it's the distance to the surface of the earth, that's how high up the satellite is. You might that I'm supposed to plug in the distance all the way to the middle, is it the distance to the backside, what is it?
Newton was worried about this question, he thought about it long and hard. Of course, the real answer is just calculated. I mean, you break the earth up into a bunch of little chunks and you use the principle of super position. The net force of gravity on the satellite is the sum of the forces to each of each and every little chunk in this sphere. And, so you do get a weaker force from the little chunks on the backside, and a stronger force from the chunks on the front side. And they're pointing in different angles.
It's a calculus problem. A fairly nasty calculus problem and you have to remember, poor Isaac Newton, he wasn't handed calculus, he had to invent it in order to solve problems like this. So, he spent a long time and finally proved, that yes indeed, the force of gravity in his formula, what you should do is to plug in the distance between the centers of the objects. And that's true even if the center of the earth is very dense, and the outside is less dense, which is in fact the way the earth is. The only assumption you have to make is that it's spherically symmetric. It can't be different density on one side, and on the other side.
So, that's how you use the formula. And it's what I said before - if you've got two objects and they're spherically symmetrical, the force of gravity is computed simply by using that formula with the radius between their centers in the denominator, squared.
So, let's go back to this problem of the earth, and just for the fun of it, let's compute what is the force of gravity for an arbitrary object of mass, little m. Right here at the surface, actually can be anywhere on the surface. I'm just going to plug numbers into the equation. So, I've got to look up some numbers. The constant capital G, 6.67 times 10^-11, it's a little tough calculator exercise because these number are both tiny and huge. M is the mass of the earth, so that's something like six times 10^24 kilograms. Dividing by r^2, that's the radius of the earth, it's the distance from the center of the earth to the object at the surface. And that's about 6,000 kilometers, or six million meters. And that's what I've got to plug in here. And when you do that, I'm leaving M as unknown. I get the force of gravity is... so plug away.
You've got all the crazy numbers in your calculator and when the dust settles and you look at the units, you see that you get a nice ordinary looking number: 9.8 is meters per second squared, times the mass (I can call it little m or M, whatever I want) 9.8 meters per second squared times mass and I flop myself on the head because, of course, the force of gravity at the surface of the earth is MG, right? This is just our little g, 9.98 meters per second squared, and, it better be, Newton's new universal law of gravity, has to work. It has to agree with what we know, that the force of gravity of any mass, M, right here, near the surface of the earth is supposed to be M times the number little g.
So, that's nice and it says as we already knew, that the force of gravity is proportional to the mass. So, if you have only gravity acting, than the F = Ma tells me that the acceleration is going to be 9.8 meters per second squared no matter what M is. So you drop a penny and a feather, any two objects, and if the only force acting on them is gravity, then their acceleration is going be the same, as long as you're near the surface of the earth.
So, now you say, "hang on a second, I'm getting nervous because this formula says force of gravity is constant. That's what we've been using all along. And the new formula, the correct formula says it's got this crazy r dependency. It's not constant at all. It varies as you move away from the center of the earth. What if I move a meter away from the center of the earth? Aren't I going to get a different formula?"
Well, technically yes. If you plug in the new radius, it's not six million. You have to add a meter. So, what happens when you add a meter? It's not seven times 10^6, it's 6.000001 times 10 to the sixth meters. And you square that on your calculator and you hardly even notice that this number is not 9.8, in fact this is already approximate, it's rounded to two thick figs. So, you simply can't care, at this level of accuracy, what exactly the radius is: one meter, 10 meters, even 8,000 meters. Go up to the top of Mount Everest, plug in the new radius so it's six million plus 8,000, roughly, and square it, and even then to two thick figs, you going to get the same number as you got down here. So, yes, if you had a very sensitive measurement of gravitation force, or equivalently the gravitational acceleration of an object in free fall, yes, you could tell the difference. It's possible, with sensitive equipment. But for practical purposes, it's fine to use what we've been using near the surface of the earth.
But what if you get far away. Clearly this formula takes over. This is the right one, and you need to use it. Newton needed to test this equation. I mean, he came up with it from various arguments, I'm not even sure exactly how he arrived at this equation, but you certainly have to test it. And g is such a tiny number, it's very difficult to come up with laboratory experiments to directly measure the force of gravity. It was not really a hundred years after Isaac Newton before somebody was able to do that. I'll tell you about that in just a minute.
So, what could Newton do to check this formula? Here's a clever trick, which he thought of. Let me figure out the force of gravity on an apple. The force on an apple is equal to g, M of the earth, M of the apple divided by, what's the radius-well it's an apple right here in my orchard-it's the radius of the earth, squared. And let me also figure out the force on the moon. The force on the moon is given by g, the mass of the earth, the mass of the moon divided by what radius? It's the distance between the center of the earth and the center of moon, so I might call that r[E-M]^2: the earth, moon distance.
Now Newton, remember, doesn't know what capital G is, Newton doesn't know what the mass of the earth is, Newton doesn't know what the mass of the moon is. So, you're thinking how does he check these equations? Smart guy! So, he says rather than comparing the forces, let me think about the acceleration. After all, the apple, let me let it fall out of the tree, it's in free fall. F = Ma where M is the mass of the apple. So, let me figure up the acceleration of the apple. I simply divide this formula by Ma. And so I get the acceleration of the apple, is just this, and similarly for the moon. I'm going to assume that the moon is feeling no force except for gravity. There's nothing else touching it. So, the acceleration of the moon must be F divided by M (from Newton's second law), and I get a very similar looking formula. The acceleration of the apple is equal to g, M, earth over r, earth squared. And the acceleration of the moon is equal to g, M of the earth divided by r-earth/moon squared. And let me divide these two equations. Why do I do that? Because then capital G cancels so I don't need to know it. And so does the mass of the earth. And what I get is the ratio of the earth/moon distance to the radius of the earth, both squared.
In Isaac Newton's time, the radius of the earth was known, in fact that had been known for many, many years. And the earth/moon distance had been deduced by astronomers, and so Newton knew this ratio. In fact, the earth/moon distance is about sixty times the radius of the earth. And so, this is about 60 squared.
Let me look at the accelerations over here. What's the acceleration of the apple? I know that. It's a free falling apple, 9.8 meters per second squared-Newton knows that too. How about the acceleration of the moon? Do we know that? Newton does. Smart fellow! I think he's saying to himself: look, the moon is an object, it's moving around in orbit. I know the radius of that orbit. r-earth/moon. I can figure out the acceleration because this is simple circular motion. Acceleration is just v^2, where r is r-earth/moon. And what's the velocity of the moon. Velocity is distance divided time. It's traveling 2r-earth/moon divided by-how much time does it take the moon to go around the earth once-twenty-seven days; that's been known for an awfully long time.
So, Newton knows all these numbers. That's the speed of the moon, plugs it in here, figures out the acceleration and goes back and he checks. And sure enough, 9.8 is to the acceleration of the moon as sixty is to one, quantity squared, 3,600. It worked! His formula really is working. It's a beautiful check. And it teaches him immediately, the moon is just an object. It's obeying Newton's Laws. It's obeying the universal law of gravity, it's obeying the kinematical rules, it's obeying F = Ma. Everything is working just fine. There's nothing heavenly about the moon, it's just another object.
I said I was going to tell you how capital G is measured. Newton hasn't measured it; He's canceled it out. Capital G was measured in 1798, roughly a hundred years later, by a fellow named Cavendish, in a laboratory experiment. He took known masses, m and known masses, big M, these were just chunks of material, and he tied them together with the rod and he basically measured them between them. You can sort of see from the picture they're gravitationally attracted to one another which is going to cause this thing to want to twist. And you can measure the force of twist on this wire. It's a very sensitive experiment, and a very difficult one. And so what Cavendish is doing is he is measuring the force, he's knowing what the M and M are, they were known masses. And he measured the distance r between their centers and from that he can deduce capital G.
And he actually got a very good value for it. The brilliance of Isaac Newton was not in his ability to understand the complexity of the world, in my opinion. The brilliance of Newton was in seeing the simplicity of the world. Newton's Three Laws of Dynamics, the ones that we've been using and talking about, plus Newton's Universal Law of Gravity explains everything. It explains Lunar Orbits, and Earth Orbits and everything else that we've been talking about. Spinning gyroscopes and horses on carts and all the fantastic complexity of the world. It all boils down to just these very simple laws. Gravity is not the only fundamental force of nature. But there's not very many. Nowadays, we know about a few others. Electric, or magnetic forces, we call it electro-magnetic forces. The strong nuclear force and another force called the weak force. And that's it. And those forces plus Newton's Laws really explain in a fantastic way, all this complexity that we see in the world around us.
Force of Gravity
Gravity on Earth Page [3 of 3]
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