Potential energy and conservative forces (part 2) | AP Physics | Khan Academy

[00:00:00]If you take a tiny object like a stone, for example, from the surface of a planet and raise it to some height, then the change in the gravitational potential energy of this system is given by this expression, where m is the mass of this tiny object, g is the acceleration due to gravity everywhere, and delta y represents the change in position. This concept is super useful in solving certain kinds of problems very quickly. But, this expression is only valid when we're dealing with a very tiny object close to a very massive object like a planet or a star or a moon for that matter, because only then we can assume the force of gravity to be uniform everywhere, and therefore the value of g would be the same everywhere.

[00:00:44]But, look what happens when we go very, very far away from that massive object.

[00:00:48]Now, the force of gravity is no longer the same, and therefore g varies, and so this expression is no longer valid. So, the question we want to try and answer in this video is what is the expression for gravitational potential energy in general?

[00:01:03]So, how do we do this? Well, let's ask ourselves how do you calculate potential energy in general for any conservative force? I don't like to rememberize any expressions, so here's how I'm thinking.

[00:01:15]It all starts with work-energy theorem, which says that the change in kinetic energy of any object, in our case it would be this particular object, um equals the total work done on that object. Work done by all the forces or work done by the net force on that object. Now, we know how to calculate the work done by a net force. It's going to be the line integral of that net force dot dr. But, what if there's only a single force acting on our mass, and that force happens to be conservative, like in our case, gravity is a conservative force?

[00:01:47]Then, instead of the net force, we have the conservative force, because that's the only force acting. That's the net force. But, when a conservative force acts on an object, the total mechanical energy is conserved. In other words, if the kinetic energy of this object increases, the system's potential energy will reduce by the exact same amount.

[00:02:08]And if the kinetic energy decreases, the system's potential energy will increase by the exact same amount. In other words, the change in kinetic energy is the negative of the change in potential energy, and that's how we can bring in the potential energy into the picture.

[00:02:23]The negative sign is basically saying when kinetic energy increases, the system's potential energy decreases and vice versa. And rearranging, this is the expression to calculate the change in potential energy for any conservative force acting on an object. So now, what we need to do is find the expression for our conservative force, which is basically the force of gravity, and dr vector, and then plug it in.

[00:02:47]Okay, so let's start with our conservative force, which is the force of gravity. How do we write that vectorially? Well, we know that the force of gravity, um, acting on this particular mass is, you know, it's always attractive, so it's towards the left. And we know from Newton's universal law of gravity that the magnitude of that force is GMM / r squared, the inverse square law. But how do we write this vectorially? Well, vectorially, we can add a unit vector. A unit vector is a vector that has magnitude one. It's only useful in representing direction. Now, look, we have defined the origin of our coordinate system to be here. So this is the object's position vector. And since our force vector is in the opposite direction of the position vector, we're going to use negative r hat to represent the direction. R hat represents the unit vector in the direction of the position vector, so in this diagram, it's to the right. And since this is to the left, we use negative r hat. So this is how we write our force of gravity vectorially.

[00:03:48]Okay, how do we write dr vector?

[00:03:51]Well, in a in a in a similar way, dr is an infinitesimal displacement of this tiny mass. So, dr vector would be the magnitude of dr, which is basically dr we can write, times a unit vector. But, what's the direction of that unit vector? Well, over here we're assuming just one-dimensional motion. We are assuming the mass is only moving along the radial direction. So, even here the unit vector is going to be r hat. But, is that going to be positive r hat or negative r hat?

[00:04:21]Well, that completely depends upon your bounds. For example, if A was over here and B was over here, then we are moving this way. So, displacement is in this direction. But, if A was over here and B was over here, then it would be to the right. So, the displacement would be towards the right. So, it's the bounds that take care of the direction of the displacement, so we don't have to worry about it. So, we can just multiply it with the unit vector r hat. All right, so we can now plug this in and calculate the change in potential energy. So, let's do that. So, the change in potential energy is going to be negative from our position A to position B. Our conservative force is over here dot dr, which is basically this one.

[00:05:04]It'll be a great idea to pause the video and see if you can simplify this yourself.

[00:05:10]All right, first of all the minus signs cancel each other out. And now what I can do is I can separate the magnitudes G M M by r squared dr and separate out the unit vectors. So, you get r hat dot r hat over here.

[00:05:24]What's r hat dot r hat? Well, that equals the magnitude of r hat, which is one because it's a unit vector, times magnitude of this r hat, which is again one, times cos of the angle between them. The angle between them is zero because they're in the same direction.

[00:05:38]So, cos of zero is one. So, you get one times one times one. So, this is just one. So, we are only left with this part.

[00:05:45]Now, how do we simplify that? Well, I can take the constants out. G, M M are constants. They don't change as we move from one point to another. So, let's pull them out. And what we're left inside is the integral of dr / r².

[00:06:01]What's the integral of 1 / r² with respect to r?

[00:06:04]Well, it's -1/r.

[00:06:08]So, you get -1/r.

[00:06:10]And you have the bounds. Let's pull out the minus sign over here, and then we can plug in. And now we can plug in the upper bound first. You get 1/b minus the lower bound. And there we have it. This is the expression for the change in potential energy. But you might say, "Well, I want the potential energy expression for a specific point at some distance r from the center of this planet." This is the change in potential energy. How do I get potential energy at r from this expression?

[00:06:40]Well, let's see.

[00:06:42]First of all, change in potential energy is the final potential energy minus the initial potential energy. The final potential energy is the potential energy at the upper bound, which is at position b, wherever that is. And the initial potential energy is the potential energy at the lower bound, which is at a. So, I can write this as U of b, potential energy at b, minus potential energy at a. That equals this term. Now, from here, how do I find expression for potential energy at a specific point?

[00:07:12]Well, for that, we need to define a reference where the potential energy is zero. And that is completely our choice.

[00:07:20]For example, one non-standard way of doing this, just to make a point that, you know, it's completely our choice, is that we could say, "Hey, when this mass is on the surface of this planet, let's choose that configuration as zero potential energy." Because we are completely free to choose which configuration we want as zero potential energy. So, we can say, "Let's choose, this is our choice, U of R, meaning when the two masses have a center of and their centers of masses are separated by distance R, let's choose that as our zero.

[00:07:52]So, now what I can do with this reference is I can say, okay, now that we have this reference, let's put B as small r, what we want, and let's choose A as capital R, which is our reference.

[00:08:03]And if you plug that in, you'll get U of small r minus U of capital R, and that equals you will get This will become small r and this will become capital R.

[00:08:12]And since this is zero, we get U of R.

[00:08:17]This is how you find expression for potential energy at a specific location.

[00:08:25]However, I said that this is a non-standard way of doing it. Why did I say that? Well, that's because choosing the potential energy to be zero when, you know, this mass is on the surface of the planet is quite arbitrary if you think about it. I mean, imagine there are multiple planets in general, and you're calculating potential energy of the entire system. How does it make sense to choose that the potential energy to be zero for a specific planet's surface? Right? That sounds very arbitrary, isn't it? And so, it's not the most convenient, you know, choice of our reference.

[00:08:57]Although it's perfectly accurate, it's not wrong, it's not really convenient.

[00:09:01]So, here's the standard way of doing it.

[00:09:03]Here's a more convenient reference choice reference choice that we usually have.

[00:09:07]We're going to choose when the two masses are infinitely far away, meaning when their gravitational force between them is negligible. That's where That's where we'll choose the potential energy to be zero. In other words, let's choose a reference where U of infinity is zero, meaning when they're infinitely far apart, let's choose that to be zero.

[00:09:31]Again, this is our choice. Completely We are completely free to choose this. Why is this more convenient? Well, let's see. Again, let's substitute B is equal to small r, what we want, and A is equal to infinity, and let's see what we get.

[00:09:44]So, we will get U of r minus U of infinity equals, you know, minus GMM by into 1 over r minus 1 over infinity. Now, remember, infinity represent Think of infinity as a very large number compared to small r.

[00:09:57]Then, this number will be very tiny compared to this one, and so look, our expression simplifies, and we get minus GMM divided by r.

[00:10:04]And since this is zero, this becomes our expression for potential energy. Isn't this a much simpler expression compared to this one? That's why we like to choose this as our reference. Again, remember, there's nothing wrong with this reference. This expression is perfectly valid for this given reference, but, you know, this is the most convenient one, and therefore we choose this one.

[00:10:27]All right. So, now we have established what we wanted, but let's do some sense check and see if we can get some intuition behind, you know, what this expression is saying. First of all, we know intuitively when you throw a stone up, for example, the kinetic energy of the stone reduces, so the potential energy of the system increases. Same thing should apply applies over here. As this stone goes farther and farther away, the potential energy of the system should increase. So, as r increases, potential energy of the system increases. Let's check whether this expression also agrees with that. As r becomes bigger, this number becomes smaller, but there's a negative sign, so this entire value ends up becoming larger. So, yes, this expression also agrees and says that as as r becomes larger, potential energy value becomes larger. That makes sense.

[00:11:14]Secondly, if I were to plug M or small m as zero, potential energy goes to zero.

[00:11:19]Does that make sense? Well, yeah. I mean, if either of the masses were to vanish, then you no longer have a force of gravity. There is no conservative forces. There is no potential energy.

[00:11:28]So, that makes sense as well. But the last thing that could be annoying over here is the negative sign. This expression says that regardless of what value of R we choose, the potential energy, gravitational potential energy, will always be negative.

[00:11:44]What does that mean?

[00:11:46]Well, think about it this way.

[00:11:48]We know that as we go farther and farther and farther away, the potential energy of the system increases.

[00:11:54]So, when is the potential energy maximum? Well, when they are extremely far away, when they are infinitely far away.

[00:12:02]So, U of infinity is actually the maximum potential energy. And look, we assigned that maximum value as zero.

[00:12:11]Now, if the maximum potential energy itself is zero, then for any other configuration, the potential energy is going to be less than the maximum. So, it's going to be less than zero, so it's going to be negative. That's why there is a negative sign. That's why gravitational potential energy under this reference will always be negative.

[00:12:32]This is analogous to saying that the top floor of a building is zero. Then, all the other floors below it would be negative. But, nothing physically changes. It's just the way we chose the reference. Of course, that's not the most common choice when we are very close to the planet's surface, but here it makes a lot of sense.

[00:12:53]But, finally, this expression only works when you're dealing with two masses.

[00:12:57]What if you have three or four or n number of masses? Now, what happens?

[00:13:02]Well, now we can look at this in much more general. First of all, let's call this m1, m2, m3, and so on. And let's call the distance between these two masses, their centers of masses, r12.

[00:13:15]Similarly, between these two centers of masses, r23, and so on and so forth.

[00:13:19]Now, how do we write this?

[00:13:20]Well, first we write down the potential energy only due to the contribution of these two masses. So, it'll be -G m1 m2 / r12.

[00:13:29]Then, we add the contribution of the potential energy due to just these two masses.

[00:13:34]It'll be minus G m2 m3 / r23, where r23 is the distance between centers of masses of these two masses. Then finally, we can add the contribution for these two masses. It'll be minus G m1 m3 or m3 m1, where r31 is the distance between their centers of masses.

[00:13:52]And so on and so forth. So, if there are more than two masses, you calculate the contribution of each pair and then add them up. Which means we have found an expression for potential energy for n objects gravitationally interacting with each other. That's incredible, if you ask me.