Work done by forces (part 2)

0:00- [Instructor] A planet of mass M is attracting a very tiny object of negligible mass, small m. Let's say at time T equal to zero, that small m is at rest. And right now it's at a distance of Ri, R initial from the center of the planet. The center of masses are at a distance of Ri.

0:18Now, as time passes by what's going to happen? Well, it's attracted and there are no other forces acting, imagine that this is in vacuum, then it will accelerate towards the mass, right? It'll keep accelerating. Its velocity will keep increasing. And eventually after some time it will come somewhere over here.

0:34Let's call that distance to be Rf, the separation distance at the final point of our interest. The question that we will try to answer in this video is when it's over here, what is its speed going to be? Our goal is to find an expression for the speed in terms of Ri, Rf, the masses, and the appropriate constants.

0:56So how do we do this? Well, one way is to use Newton's second law. If we know what the net force is, and here there's only one force acting on it, the force of gravity, we can use that to calculate acceleration as a function of time. And from acceleration, I can calculate the velocity as a function of time and then I can use that function to figure out the speed at this particular point.

1:17But that's gonna take slightly longer. So we're gonna take a slightly different approach over here. We're gonna use the work energy theorem. Okay? So remember work energy theorem? It says that the change in the kinetic energy of an object will always equal the total work done on that object.

1:38We're gonna use this to try and calculate the velocity over here. Okay? All right, so let's do this. What is the change in the kinetic energy? Well, it's the final kinetic energy minus the initial kinetic energy. So that'll be the total work done. What is the final kinetic energy?

1:55Well, the final point is over here. This is what we are interested in. So the final kinetic energy is half M into V square. We don't know what v is, right? So we can just call that as half MV square, minus what's the initial kinetic energy? Initial kinetic energy is this point.

2:09This is the initial point. Well, it's at rest. So initial speed is zero. So initial kinetic energy is also zero. So you get minus zero and that equals the total work done on this tiny mass. And so now the big question is how much is the total work done? How do we figure this out?

2:30Well, just to recap, remember, work done by any force is given as a dot product of the force and the displacement. So to find the total work done, we have to find the work done by each force acting on this mass from here to here, and then just add them up. But there's only one force acting on our object, right?

2:49There is only the force of gravity acting on an object. So the work done by gravity will be the total work done. And so the main question now for us is what is the work done by gravity when the mass comes from here to here? So here's my attempt, okay? So the work done by gravity would be the force of gravity times the displacement times the cos of the angle between them.

3:18So what is the force of gravity? Let's just call it as Fg. And what is the displacement? Well, the displacement is from here to here. So that's Ri minus Rf. So it's magnitude is Ri minus Rf times cos of the angle between the two. What's the cos of the angle between the two?

3:36What is the angle between them? Well, the force is towards the left and the displacement is also towards the left. So the angle between them is zero. So you get cos zero. cos zero is just one. And there we have it. That is the work done. That's my attempt. The question for you is, do you think this is accurate?

3:54Want you to pause the video and think about this, Well, let's see. This expression for calculating the work done only applies when the force is a constant over the entire displacement. So the question for us is, is the force of gravity a constant over the entire displacement?

4:11And the answer is no. We know that farther away the force of gravity is smaller, it's weaker, but as we go closer and closer, the force of gravity increases. So clearly we cannot directly apply this expression. This is wrong. Okay, so what do we do? Here's what we can do.

4:34The force of gravity keeps increasing as I go from here to here, right? How about we divide this entire displacement into say, five equal parts? And let's assume that the force in each section is a constant. Now sure, even within the section, the force varies, but it won't vary as much as it does over this entire displacement.

4:56So we can approximate, we'll take some approximate average value and for each section, let's assume the force to be a constant. Then we can calculate the work done in each section separately. Here's what I mean. So let's divide this displacement into five equal parts.

5:13Let's call each one as delta r. Now what we can do is we can calculate the work done in this section as this force multiplied by this displacement multiplied by the cos of the angle between them, that is the angle is zero because they're in the same direction.

5:29So it's just going to be F times delta r, and it'll gimme some value of work done for this particular section. And then I'm gonna repeat over here, I'm gonna calculate the work done in this section separately, it's gonna be slightly larger than here because the force of gravity over here is slightly stronger than over here.

5:46And I'll just repeat and then I will add all of them up and that would give me the total work done. So when the force is varying, in general, our strategy is going to be like this. We divide the displacement into some number of equal parts. And then to calculate the work done, what we do is for each section we calculate the work done individually by multiplying the force in that section and the displacement.

6:13So we take the dot product of the force and the displacement in that section, and then we repeat this for each section and then we add them all up. In other words, we take a summation. That is how you calculate the work done when the force is varying. But this is not accurate, isn't it?

6:30Because remember, we assumed that the force is a constant in each section. It's not. the force is varying even within each section, right? So what we have gotten is an approximation. And just to make this point clear, we can also look at this geometrically. If you plot a graph of the magnitude of the force of gravity versus the distance r, where r represents the distance from the center of mass of this planet, then the magnitude of the force of gravity as you go closer and closer to the planet increases.

7:00This is how the magnitude of the force of gravity increases. And if you now consider the work done between some initial and some final points over here, then that work done geometrically represents the area under this curve. And the question is how do you find the area under this curve, because it's not a standard shape?

7:21But what we can do is we can divide the shape into say, five equal parts. That's exactly what we did over here. And then approximate each part to be a rectangle. And then we can find the area of each rectangle individually, which is basically height into width.

7:40Over here the width represents delta r and the height represents the force at that particular location. And so you can see clearly over here in each section we are approximating the force to be a constant, but of course the force is changing. And so when we sum it up, we get an approximate value of the area.

7:57It's not exact. As you can see, there are some gaps over here and there and there are places where we have overshot. So this is the approximate answer. But now comes the question, how can we make this approximation better? Want you to pause the video and think about it.

8:12Well, instead of five parts, divide this entire section into 10 equal parts. Now notice you'll get a much better approximation. So you see where we're going with this. The more number of parts you divided with, the smaller the delta r you choose, the better your approximation becomes.

8:34And then eventually under the limiting case, when delta r is approaching zero, our approximation approaches the exact value. So technically the work done would be equal to the summation under the limiting case, when delta r approaches zero. And we call this the integral.

8:58And of course the integral will have some bounds from some initial position to final position. Such integral in which a function is evaluated over a specific path, we give a name to it, we call it a line integral. So work done is most fundamentally and generally defined as a line integral.

9:18But of course when this dot product between the four centers of displacement is a constant over the entire path, well it simplifies to give us this expression. So coming back over here, what's the work done by gravity? Well, it's going to be the line integral of Fg dotted with dr along the object's path.

9:34And what are the bounds? The lower bound is the initial position, Ri and the upper bound is the final position Rf. So that simplifies to the left hand side becomes half mv square. And now we need to evaluate this integral. So let's clean things up a little bit.

9:50Okay, so we have two things that we need to figure out. What is the force of gravity vectorially? Well, for that we can use Newton's universal law of gravity. An object at any distance r from the center of the mass of a planet, will experience a force of gravity, which is given by in magnitude, it's going to be the constant G times the product of the masses divided by the square of the distance between the center of masses.

10:16Okay? So that's the magnitude. But what about its direction? Well notice the direction is, well, it's attractive. So the direction over here is to the left. But how do we write that vectorially? Well, we can say that the vector, the direction is in the opposite direction of my position vector r, right?

10:36The position vector is from the center of mass of the planet to this point. So the direction of the force is in the opposite direction. But how do I write that vectorially? We write it as negative r hat. r hat is a unit vector, meaning its magnitude is one. And that vector will be in the direction of the position vector over here.

10:58So since our direction is to the left, we'll use minus r hat. And since its magnitude is one, it doesn't add anything to the magnitude. So it's only useful to represent the direction. So this is the vector expression for the force of gravity. So we have this. Okay, next we need the vector expression for the displacement dr.

11:18How do we represent this? Remember dr is an infinitesimal displacement, right? How do we represent that? Well again, vectorially, dr would be equal to the magnitude of the displacement, which is just going to be, we'll just write it as dr, times the direction, and the direction we can use again r hat to represent the direction.

11:39But what will be the direction? Will it be in the positive direction or the negative direction? That's the important part. Well, I should think, hey, you know this object is going from, it's moving towards the left. So it's direction should also be negative r hat, but that would be wrong because the direction of the displacement is taken care by the bounds.

12:00Since I've already input the bounds as Ri to Rf, that takes care of the direction and therefore I don't have to think about the direction over here. I'm just gonna multiply it by r hat. Does that make sense? What I mean to say is we don't have to worry about the direction of the displacement over here because it's being taken care by the bounds.

12:18For example, in our case to represent that the object is going closer to the planet. Look, the lower bound is a bigger number compared to the upper bound. If you wanted to represent the object going farther away from the planet, the lower bound would be a smaller number compared to the upper bound.

12:32Does that make sense? So it's the bounds that take care of the direction of the displacement. So we don't have to worry about the displacement direction over here. Okay, so we have both of them. We can plug this in. So it's going to be the integral of the force, which is this thing over here, dot that displacement vector, which is over here.

12:51Okay, let's simplify this. So what do we get? Well dot product only applies to vectors, so we can separate out the magnitude. So minus G Mm by r squared times dr. We'll separate that out. Times the dot product of r hat and r hat. What's this dot product equal to?

13:09Well remember dot product equals the magnitude of r hat, which is just one, times the magnitude of the other r hat, which is also equal to one, times the cos of the angle between the two r hats. Well the two r hats are in the same direction, so the angle is zero, so cos zero is one.

13:23So it's one times one times one, which is one. So this dr product is just one. We're done with vectors. We are only left with the magnitudes now. Okay, so what did that give me? Well, I can pull the constants out, minus G Mm, I'll pull it out and I'll be left with the integral of one over r squared, dr.

13:43What is the integral of one over r squared dr? Well, that's just negative one over r. And of course we have the bounds over here. Well, from here we just have to simplify. So it will be a great idea to pause and see if you can derive the expression for the speed yourself.

14:00Okay, so the minus sign cancels out and so now I can plug in the bounds. We first plug in the upper bound minus the lower bound, so I'll get one over Rf minus one over Ri. The masses cancel out. And from here V squared equals rearranging two times this thing. And if you take squared on both sides, I will find my V to be the square of two GM into one over Rf minus one over Ri.

14:27We have found the expression for the speed at this point. So the final thing we need to do is make a sense check over here. First of all, does it make sense that M is not a part of our equation? Yeah, because we have seen before that the acceleration of an object due to gravity is independent of its mass.

14:44And so the speed it would've gained is also independent of its mass. We also notice if the mass of the planet increases, then again the speed increases. That makes sense. If this was a much more massive planet, the force of gravity would be much stronger, it'll be much more acceleration, and so the speed would be higher.

15:00Another thing you can notice is that if Ri was the same as Rf, if these two values were exactly the same, then this will give me zero and the speed would be zero. Does that make sense? Well, yeah, if Ri equals Rf, then notice its position has not changed at all, which means it's at time to equal to zero itself and the object is at rest.

15:20So its speed is obviously zero. Another check we can do is what if Rf becomes larger? Well then this number becomes smaller and so the speed also becomes smaller. Does that make sense? Well, yeah, if Rf becomes larger, then we are letting the mass only accelerate for a smaller distance, and so it wouldn't have gained as much speed as before.

15:40So you would expect the speed to be lower. That makes sense. And look, what if Rf was larger than Ri? Then this number becomes smaller than this number, giving you a negative number under a square root. What does that mean? Well, if Rf is larger than Ri, we are asking what is the speed when the mass goes farther away from the planet?

16:03And that's impossible. When an object starts from rest, it'll always be attracted towards the planet, and so it'll only move closer to the planet. It can't move farther away. So it's basically saying that situation does not apply over here. And finally, you can also check for the units and make sure the units on the left is the same as units on the right.

16:20That is something that you can try yourself. Finally, before we end the video, I wanna talk about a cool application of our hard work. Something that blew my mind when I first saw it. Ready? Okay, we can use this to estimate the speed at which an asteroid can hit a planet.

16:36Here's how we can do it. Let's make a copy of this. Okay, so if an asteroid were to come and hit our planet, what would be the positions Ri and Rf? Oh, Rf would be over here. So it will be just the radius of the planet. So for Rf, it will plug capital R, which is the radius of the planet.

16:53What would be Ri? What would be the initial position of the asteroid? Well, an asteroid very, very far away would make Ri to be very large. If Ri is very large compared to the radius of our planet, then this number is gonna be very small compared to this number.

17:07So we can just neglect that. And there we have it. This is the speed at which an asteroid would hit a planet. And notice that only depends on the mass of the planet and the radius of the planet. Now to be clear, we are assuming that the asteroid starts at rest from very far away, but that need not be true.

17:26The asteroid could be moving towards our planet already from far away, which means the speed could be higher than this. In other words, what we have found is the minimum speed at which an asteroid could hit a planet. I find this incredible because if I know the mass of a planet and the radius of the planet, I can actually find a value.

17:46Let's do that for Earth. If you plug in the mass of the Earth and the radius of the Earth, this number comes to be approximately 11 kilometers per second. So as mind boggling as it sounds, just by sitting over here at our desk using a pen and paper, we can calculate the speed at which, a minimum speed at which an asteroid could hit the Earth.

18:05That's incredible. Right? Now, of course there are some nuances. If the asteroid is very tiny, then when it enters our atmosphere, it gets slowed down or it can even just burn up. But of course, if the asteroid is large enough where the slowdown effect is much negligible, then this is a very good estimate of the minimum speed at which asteroids could hit our planet.

18:26And here's the final thing that blows my mind. Are you ready? If this represents the minimum speed with which an asteroid very far away would come and hit a planet, the reverse must also be true. This is also the minimum speed at which you should launch an asteroid or any object for that matter, from the surface of the planet so that it can go very far away.

18:47In other words, it can escape the planet's gravitational pull and therefore this is also the escape speed of the planet. And so the escape speed from the surface of the Earth is roughly about 11 kilometers per second. Again, ignoring the effects of atmosphere.

19:04This is an important number if you are working with rockets and space graphs.