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Anderson Video - Potential Energy and Gravity

Professor Anderson
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>> Hello class. Professor Anderson here. Let's talk a little bit about potential energy and gravity. This is something that you're familiar with already. When we talked about the earth and we said we're going to lift a box from here to here, a distance h, we say oh that box has potential energy in it. And you told me a while back what that was. It's just mgh. Alright? What's the potential energy of the box? It's mgh. But how did we get to that? The way we got to that was you have to do work when you lift the box. How much work do you have to do? Well, you have to integrate the force over the distance. How much do you lift it with? Well, you lift it with mg. How far do you go? Eh, you go from zero to h. But mg is just a constant. So that comes out in front. We have the integral, zero to h of dy. But that just becomes y, evaluated from zero to h. And so we just get mgh. But there was a crucial step that we took there. The crucial step was right here. Right? We went from this integral to this step because mg was constant. And if mg is constant, it comes out of the integral. You can put it in front of the integral. But what if mg is not constant? If mg is not constant, you can't do that step. You cannot take it out of the integral. And what we know is, when you go very, very high up in altitude, all of a sudden that gravitational force is no longer constant. And so we have to be a little bit more careful about that. OK? So let's take a look at that. Let's say we do the following. We have some object that is near the earth, and we're going to lift it up to a distance that is much farther away. And let's say that these two things are initial position, final position relative to the center of the earth. How do I calculate how much work it took to do that? Well, we know that. Work is the integral of the force over the distance. OK? And specifically, there's a dot product there. But we know that we're going to have to apply a force in that direction. It's going to move in the same direction, so the dot product is in fact going to go away. How much force do we have to apply? We have to apply at least as much as gravity, which we know is g mass of the earth, mass of the object, divided by r squared. What does our variable become? It becomes dr. OK? How far are you from the center of the earth is r. So we're integrating along dr. And then we're going to integrate from an initial position to a final position. Alright. We can probably do that. Let's see. We've got a g, m e, m. Those are all constants. We can put those out in front of the integral. We have a one over r squared dr. And we know how to do that, right? Integral of one over r squared is just negative one over r. So let's put the negative sign out in front. This becomes a one over r. And we are evaluating that one over r from ri to rf. And so this becomes the following. We're going to go up here. It becomes negative GMem one over rf minus one over ri. So this is in general how much work it takes to move an object from one r sub i to a different r sub f. But we like to make an approximation where we do the following. Let's set our reference point at infinity. OK? So ri is going to be infinity. Remember, potential energy we said you can set your zero point wherever you like. Let's set our zero point out at infinity. And therefore, the work that you do or the gravitational potential energy is just this. We're only worried about that final position r sub f. And if we write it in general, this is the gravitational potential energy of our system. OK? It's negative because we set zero at infinity. And so anything closer to the earth or closer to some other massive object will have negative energy.