 Hi, I'm Zor. Welcome to the news or education. Now, I promise to consider another relatively simple example of the work performed by variable force. In this case my example is the force of gravity. Well, first of all, why is it variable? Well, we all know that there is a universal law of gravity which basically says that the force of gravity depends on two masses which interact and it's inversely proportional to square of distance. So basically the universal law of gravity tells us that the force is equal to some universal constant, gravitational constant, times two masses and divided by radius square, which is the difference between the centers of these masses. Now, I will put as f of r because it's a function. So it's a function of the distance between these. So if these are point masses, then it's basically different the distance between these point masses. If it's two planets, for instance, like the Earth and the Moon, then it's the distance between their centers of masses. All right? So this is the universal formula and obviously it depends on the distance. Now, if we are talking about the planet Earth and the very small object which we are lifting on a very small height, we probably can assume that the force is relatively constant because it's a very small height relative to the whole radius of the Earth, which is like 6,000 kilometers, whatever. And in this case, we know that since the force is constant and if you would like to lift it by height h, the force is equal to m times g, where g is the acceleration of free fall. So this is the force. This is the height, which is distance, distance which we are lifting. So the work must be equal to mgh force times distance, right? So that's kind of a formula which is obvious for these small distances. However, in this particular case, I would like to consider a little bit more general case when the height, which we are lifting our object of mass m, is really significant relative to the radius of the Earth. And we cannot really say that the force is constant. Force depends on the radius and the radius in the beginning is very large. I mean, it's 6,000 kilometers when we are on the surface. But then if we are lifting by, I know, a thousand kilometers, it's substantial. All right. Now, the highest mountain is like 10 kilometers and relatively to the 6,000 kilometers of the radius, it's really not a lot. So even on the Mount Everest, the force will be probably almost the same as on the sea level. But again, if we go to a lunar orbit, for instance, that's substantial difference. Okay, so how can we calculate the work performed by the force which lifts the object of mass m to the height h in a more general case? Well, it's basically very, very similar to whatever we were talking about the spring. It's just the formula is different. So the pendency of the force on a distance, of a distance from from the center of the planet. So let's assume this object of mass m is a point object. So it doesn't have any radius. It's just a point, but the earth is relatively large. So the beginning position of our object is on the level, on the surface level of earth, which has a radius r. So let's assume our planet has a radius r. And then we are trying to lift it to the difference. So the new distance from the center would be the r, the radius of the planet plus the height by which we have lifted our object. So that's what we are doing. r, lowercase r, is changing from capital R to capital R plus h. So again, differential of the work, which is performed on the distance r from the center, is the force times the infinitesimal increment of the distance from the center, right? So which is g m m r square g r. Now what should we do now? Well, we should integrate it. We should summarize all these little differentials of the work as we are going up from 0 to h. Well, more precisely from r to r plus h, right? So it's integral from r to r plus h of this thing. Okay, so what is the function derivative of which is equal to 1 over r square? Well, the function 1 over r gives derivative minus 1 over r square. So we need to add minus to this. So it's minus g m m divided by r and we have to limit from r to r plus h. So first we have to put the upper limit and that will be with a minus sign and lower with a plus sign. So which is equal to g m m. So with a plus sign would be minus 1 over r minus would be 1 over r plus h. Well, this is basically the final formula. Now what I will do, I will simplify it a little bit. Well, I'm not sure it's simplification, but just transform into something a little bit more relatable for the i. So I'll have a common denominator. So it would be g m m r plus h minus r. It will be h times, I'm sorry, r r plus h. So this is a total formula for the work performed by some force against the gravity to lift the object off the surface of a planet of a radius r by the height h. Okay, let's go back to earth. And I would like to know if this formula does correspond to this one with a small h. So if the height is not very big, then it should basically correspond, right? Now what we know about this force? Well, this force, if it's on the surface of the earth, is equal to mg, right? Mass times the acceleration of the free fall. This is the force of r. So this is g m m divided by capital R square. Now let's just think about it. What I will do, I will replace this. Okay, I will put w with an index zero, which means it's basically very close to the surface of the earth. So I will replace this with mg r square and then h and divide by r r plus h. Now what is it equal to? It's equal to mg h r divided by r plus h, right? Now r and r. So it's h and another r. Now let's talk about this one. With a very small h, this is very close to one, right? If r is really very, very big, then this is very close to one because the numerator and denominator are very close to each other. It's slightly less than one but anywhere very close. So for a very small h, actually on the surface of the earth it's equal to one but as we are going a little bit higher, it's slightly less than one but still very, very close to one and this corresponds to this. So that's what exactly kind of explains that this is a correct formula with certain approximation based on very small height. So when the height is very small, this formula is fine. When the formula needs to be a little bit more precise, it's this but in this particular case g doesn't make as much sense as this. This is a universal constant which is like the same everywhere, g is something which we kind of used to have on earth. So this just proves that our formula, our universal formula on the surface of the earth and close to this surface for a very small height is really the same as we kind of used to know from assuming that the force of gravity is constant. But this is the main formula which basically tells the work which is needed to perform to lift an object from the surface of the planet of radius r by the height h. Again, it does not really depend on how we do it. It depends only on the final result and the final result is our object on the height h above the surface. So there is no dependency on anything else, speed with which we are lifting or direction or maybe we're doing something like this instead of straight line. I mean, no matter what it is, this is the right formula for the work. Okay, that's it for gravity. Again, the previous lecture about elasticity and this one about gravity, they're just illustration of how to deal with variable force, variable force when you still have to calculate the work which is performed by this force. It's all based on dw is equal to f times gs instead of w equals f times gs when f is constant. Here we just divide the whole distance we are moving an object into small differentials and on each differential we have this formula and then we integrate it to summarize these little increments of of the work. Okay, that's it. Thank you very much. I recommend you to go to theunisor.com to this particular lecture's textual description. Read it. It's always a good exercise. It's like reading the textbook. But after you listen to the lecture it's much easier actually. It will be just a very plain exercise. That's it. Thanks and good luck.