 Hi, I'm Zor. Welcome to Unisor Education. I would like to talk about potential energy and kinetic energy as well in one particular case when some object is just falling on the earth from space. Now this lecture is part of the course called Physics for Teens presented on Unisor.com. This is the website. On this site you will find not only this course but also course of mathematics called Math for Teens. And also there are some civics related courses, not by myself, basically by another person. Now the prerequisites for Physics for Teens is Math for Teens, especially you have to be familiar with calculus and vector algebra. I do suggest you to refresh these parts of the Math for Teens course if you don't really feel comfortable. Okay, so now also if you found this lecture somewhere on YouTube or anywhere else, this will be only the lecture. In case of Unisor.com you will have the complete course and every lecture actually supplemented with very detailed notes, including this one of course. And also there are exams for those who want to challenge themselves. Now the site is completely free, there are no advertising, even logging in is not necessary, although if you would like to save your exams and make sure that somebody else like your parents or supervisors see the exams, the results, etc. and supervise your process of learning, then you probably have to just establish a sign-in and use it. All right, so we are talking about potential energy and in particular I'm talking about, in this case, if this is the Earth and somewhere in space I have an object which is falling down to Earth. Now I consider this object to be far enough, so the gravity actually is changing. I mean, if the object is somewhere near the surface of the Earth we usually consider that the gravity is constant. In this case I don't. In this case I would like actually to apply the full force of universal gravity law to this particular problem. So I will try to find out what exactly is the potential energy and kinetic energy of the object which is falling. Now so let's consider the radius of the Earth is r. The initial position is capital H, so it's H above the ground. All right, now I assume that this is the beginning of the motion, so the initial speed is zero and we just let it go and the object falls down onto the Earth. So the mass of the Earth is m, the mass of the object is lowercase m and now we can actually start. So I assume that at a certain moment there is a height H of this object as it falls down, where H obviously is changing from capital H to zero. And I would like to establish what is my potential and later on kinetic energy at this particular point H for whatever, for any value of H between H and zero, between capital H and zero. Okay, now in this particular case we have obviously the only force which is acting on this object is the gravity. And the universal law of gravity tells us that the force is equal to the gravitational constant. It's directly proportional to each mass, mass and mass, and inversely proportional to the distance between the centers of the mass. Now the centers are at this distance. Inversely proportional I meant to say to the square of the distance, right? This is the universal law of gravity. r plus H is exactly the distance between the center of Earth and the center of this object. Well, object obviously is assumed to be a physically very small material point. So we know the force. Okay, now what is the potential energy of the object which is at point lowercase H above the ground? Well, that's basically the work which the force of gravity performs as it pulls the object down to the surface of the Earth. Now we know the general formula for the work which is force times the distance, right? But that's in case we know the force and its constant during the whole distance. In this case the force is not constant. The force is changing and we have to do it slightly differently. And what's the way how people, what's the way how we approach this? Obviously through integration. So we are dividing this whole segment into infinitesimal segments, right? Each one of them has lengths of DH differential of H, right? Now on during this distance DH my force can be assumed as constant and equal to this one. And now what we have to really do, we have to integrate it from H is equal to capital H to zero. Now what is slightly more convenient for me is I will just put X is equal to R plus H, DX is equal to DH, obviously. So I will put X here. Now I have a little bit simpler X square. Now if H is equal to capital H, my X is equal to R plus H. And if H is equal to zero, my capital, my X is equal to capital R. Okay, so that would be a better choice. Now if I will integrate this, which means I will just summarize all these infinitesimal things, well don't forget that 1 over X is equal to minus 1 over X square. So the derivative of the 1 over X is equal to minus 1 over X square. And we have exactly this type of thing. This is 1 over X square with a sign obviously. So I can use my indefinite integral from 1 over X square as 1 X, 1 over X. So my total integral would be g m m. And then using the formula of Newton-Ladenitz I will have 1 over R minus 1 over R plus H. Right? That would be my integral. Well, except, I'm sorry, I'm not integrating to capital H. I'm integrating to lowercase H. So that would be lowercase H. And that would be lowercase H. So at the point where H, lowercase H is equal to 0, which means on the surface I will have 1 over R minus 1 over R which is 0. So potential energy at this point is equal to 0 because the object is already lying on the ground. Now if the object is on the capital H distance in the very, very beginning of my motion, of my falling down to Earth, it will be 1 over R minus 1 over R plus capital H. So this is basically it for potential energy. Now if, by the way, my position, my position H is very, very far, so this is like infinity, 1 divided by infinity, then my potential energy would be GMM divided by capital R. So this is like a maximum. Obviously, if the object starts at motion further away, the potential energy would be greater, but anyway, it's not really infinitely growing potential energy. It has the limit. Limit is GMM divided by R. So it's asymptotically moving to this particular value. So this is all for potential energy. Now just one very small comment. If, so let's consider what's the potential energy at the very beginning. At the very beginning it would be GMM 1 over R minus 1 over R plus capital H. Now what if this H is really small relative to the radius? Now you know that if I will use the common denominator, it would be this. R plus H. It would be R plus H minus R, which is H, which is H, right? And the problem is that approximately R times R plus H, if H is really very, very small relatively to R, this is almost like R square, almost. Approximately, obviously. And this thing is something which we know as the G, free fall acceleration on Earth. This is mass of the Earth. This is radius of the Earth. And this is universal constant. And in which case our potential energy is equal to what? M times G times H, which is the typical kind of expression for potential energy of the object at the height H, because Mg is weight. So that's basically the force which we consider to be constant when the height is really small relative to the radius of the Earth. So that's just a side consideration which basically proves that this formula is the result of this more complex formula with the height very, very small relatively to the radius of the Earth. Now, let's just remember this formula. Let me put it somewhere else. It's Gm M1 over R minus 1 over R plus H. Now, this is potential energy at the height lowercase H above the ground. Now, what I would like to do next is I would like to talk about kinetic energy. Now, what is kinetic energy of the falling object? So again, I assume that we start motion at the very beginning at the height H from the ground. And we just let this particular object go with zero initial velocity towards the Earth, right? Okay, fine. Now, our kinetic energy, so this is my potential, EP. Now, my kinetic energy, this is basically MV square divided by 2, right? Well, the problem is obviously V because V is variable and it's a variable speed and it's not the variable speed with constant acceleration. Acceleration is also a variable because the force is variable, it depends on the H, right? Since the force is variable, acceleration is variable, so it's a little bit more complicated here. Okay, so here is the way how I suggest to do this type of thing. What do we know? We know the force. If we know the force, we basically know the acceleration. So, let's put acceleration, it's equal to force divided by acceleration is equal to force divided by mass, right? Because mass times acceleration gives me force. At any time, it doesn't matter that A is variable and F is variable, M is constant actually. But at any given time, whatever the variables are, this equation is the second law of Newton, right? So, there is no problem with this. So, in this particular case, I can say that A is equal to F divided by M, which is GM divided by R plus H square, right? Okay, the acceleration is going this way. So, I might as well put a minus sign here, right? Just to show that because we are talking about this is zero and this is a capital H, so when our variable lowercase h is moving, this is the positive direction. But since acceleration goes this way and force is going this way, so this is negative. In theory, I have to really put this minus here as well. So, this is basically an absolute variable of F. All right, anyway. Next. So, I know this. Now, what is acceleration? Acceleration is the second derivative of the distance by time. But we need speed. Speed is the first derivative. How can I get from the second derivative to the first derivative? Well, here I am applying a little trick because in theory, in theory, I need not just the second derivative, I need second derivative square, right? Speed is, I mean, the first derivative, sorry, the first derivative square. I have a second derivative, but I need the first derivative square. So, here's how I approach this. So, let's have, if H is a function of time, that's my distance basically, right? So, I can say that this is h second derivative of time is equal to minus gm divided by r plus h square. Well, h is function of time. Let me put it here as well, h of time. Now, I will do something very, very artificial. It's a trick, if you wish. I will multiply it by two h first derivative of time and here. Now, why did I do it? Well, for a very simple reason. What is a derivative of this? This is the speed, right? The first derivative. What is derivative of the speed, sorry, derivative of the speed square? We need the square. It's equal to 2 h times derivative of the inner function, which is the second derivative. You see this? The first derivative of the square of the speed is 2 times the speed times acceleration just according to the normal rules of differentiation, right? Same thing as derivative of 2, sorry, derivative of x square of t derivative is 2x of t times x derivative of t, right? I mean, that's normal differentiation. And this is exactly what I have here. So, on the left, I have my first derivative square and derivative. You know what? Instead of this, I will put d by dt, okay? So, d by dt, which is differential by time of the first derivative square is equal to 2 times whatever under the square and whatever the inner derivative of the inner function. And now on the right, what I have. You see, it's a trick, but it gives me a lot. Now, what is the derivative of 1r plus h of t? What is derivative of 1 over this expression? It's equal to derivative of 1 over something, it's minus 1 over something square times derivative of the inner function, which is h, derivative of h by t, which is here. You see, everything is here. Such a nice trick. So, instead of this, I can write d by dt of gm and then 1 over r plus h of t. This is very fortunate that I was able, by this artificial trick, by multiplying, by the way, there is a 2 here as well. I forgot the 2. I multiply by 2h prime and 2h prime here. So, I need 2 here. All right. Now, if my two derivatives are the same, it means my functions are different only by constant, right? So, let me write out this thing. And I can write that my functions without derivative differ by constant. Some constant. Now, where can I get this constant? Well, very simple reason. First of all, when t is equal to 0, I know that h of t is equal to h, capital H, right? In the beginning, I'm in this position and my first derivative is equal to 0. So, if I will substitute 0 here, I will have h prime of 0, which is 0, equals to 2gm r plus h of 0 is h plus c. So, what my c is equal to? Well, my c is equal to, right? So, I can write it here as m minus. You know what I do? I will put 2 here and I open the parenthesis, put 1 here, minus 1 over r plus capital H. Okay, almost done. Now, m times square of the speed divided by 2. So, I will multiply by m divided by 2. So, I will have m times h, h of t square divided by 2 equals 2. Divide by 2, it will be gm m 1 over r plus h minus 1 r plus capital H. Well, this is h of t, of course. Now, that's great. So, this is an expression of my kinetic energy. And now, well, the climax point, what is the climax point? E, full mechanical energy is equal to kinetic energy plus potential energy, right? Which is gm m here and here. And what will be in the parenthesis? r plus lower case h and minus r plus lower case h. So, what left is 1 over r minus 1 over r plus h equals constant. It's not dependent on lower case h. So, on any position during this fall, my total energy, my full mechanical energy is constant. My potential energy is going down. My kinetic energy is going up. Now, potential energy goes down because the height goes down, right? The h goes down, which means this is getting bigger. So, we subtract bigger and bigger, right? So, the whole thing becomes smaller and smaller. Kinetic energy, h becomes smaller, then this component becomes greater. So, my kinetic energy is growing. So, potential energy is going down. Kinetic energy goes up, but their sum, full mechanical energy, is remaining the same. Now, if you remember in the previous lecture about the spring, I also came up with the same conclusion. So, whenever something is moving, potential and kinetic energy are changing, obviously, but the total sum of them, the full mechanical energy, remains constant. Obviously, this is a particular case of the general law of conservation of energy, which we will talk about. But in this case, we have two very distinct cases of full mechanical energy to be constant, whatever the movement is. So, the system, if it's a closed system, there are no outside influences, has certain amount of mechanical energy and it does not change, no matter how we have all these components of the system are moving. I'm actually making right now much more general statement than I have actually physically proven. But anyway, this is the principle, the conservation of energy is a principle which cannot be proven in all the possible cases, but whatever the theory allows us and whatever the practical measuring allow us, that principle actually is held. So, this is the quintessential part of this particular lecture. We came up with the potential, kinetic energy and the full energy, which is constant. All right. So, I do recommend you to read the notes for this lecture on Unisor.com. It will probably help you just to digest this material a little bit better. Well, that's it for today. Thank you very much and good luck.