 Hi, I'm Zor. Welcome to Unizor education. Today we will talk about new type of energy, the energy of gravitational field. So right now we are talking about gravitational field as a concept. Now this lecture is part of the course called Physics for Teens offered on Unizor.com. If you found this lecture somewhere else, like on YouTube, just stand alone. You just have to understand that this is a lecture which is part of the course. There are prerequisites. There are, you know, prior lectures. By the way, one of the prerequisites of this course is another course, which is called Math for Teens, which I consider to be necessary to understand the physics because, especially calculus and vector algebra, because I'm always using these concepts in physics. Well, actually I can say that the whole calculus as a part of the mathematics was invented because of the certain things in physics, which people like Newton had to understand and describe, etc. So math is very important. Okay, so we are talking about gravitational fields right now. Before we were talking about different forces, which are acting on a very short distance, like for instance the horse pulls the wagon. So there is a connection between them, they're touching each other. So these are forces, which we are talking about acting on a very short distance and there is a definite material connection between the source of the force and the object it acts upon. So it's like pushing or pulling or whatever. Now, gravitational forces are felt or are acting upon the distance. So for instance, the comet is flying somewhere near the Sun and it changes its orbit because the Sun pulls it. There is no physical connection, no material connection between them. So the gravitational force is interesting exactly because of this absence of any material connection between the source and the object participating in the action. So that's very important quality. And physicists for a very long time did not really understand how to approach this until they just came up with this concept, which is called field. Which is, well, the field is the part of the space, the area of the space where certain forces are felt by certain objects. Something like gravitational force is felt by any object. So any object which is moving in the gravitational field will change its direction because of gravitational force. Now there are some other fields like electromagnetic fields and they're acting only on certain objects which have these electromagnetic properties, doesn't act on a stone, for instance. So let's talk about the definition. Again, we are talking about gravitational force as something which is basically an area of space where the gravitational field is an area of space where gravitational force acts on a distance without any kind of physical contact between the source and the object. Sometimes I will use the word probe instead of the object. So there is a source of gravitational field, like for instance our planet Earth or Sun. And there is a probe, some kind of an object which is positioned somewhere near this particular source of gravitational field and it feels this force, the gravitational field, and its direction of movement depends on this force or its changes because of this force. So now let's talk about numerical representation of this force. Now we were talking about gravitational force when we were talking about dynamics in the mechanics. So we actually explained what happens and the force was expressed as proportional to masses of the source and the probe object. It's inversely proportional to square distance between them and this is the universal gravitational constant. Now let me explain a little bit this formula. We did talk about this before but I just want to repeat it. These masses are gravitational masses. So it's a measure of how gravitational force acts upon this particular object. Now before in mechanics we were talking about inertial masses. Remember this second law of Newton? Now this is inertial mass. Now experimentally it was discovered that they are actually proportional to each other and that's why we are using one concept mass which is used in both inertial and gravitational meaning. Now in this particular case that was historically the first. So the unit of mass, the unit of acceleration and the unit of force were established in such a way that you can have this equation without any kind of coefficients. Now since this is done our mass is already established. It's grams, kilograms, whatever else. So we're using the same units to measure the gravitational mass and that's why the formula actually doesn't look as pretty as this one. We need this coefficient to align the units of measurement. So this is the same Newton's for instance as in this case and this is the same kilograms as this one and this is the same meters and time is in seconds as here as meters and times in seconds. So that's why we need this coefficient to basically equalize. But what's important is that the mass participating in gravitational activity is proportional to the mass participating in activity related to plane dynamics. So that's about masses. Now let's talk about the square of the distance between these masses. Well first of all I assume that masses are point masses. So it's easier than to talk about the distance between them because it's the distance between two points. Now why is it square? Well I think I already offered this explanation. It's not a proof. It's an explanation basically that if you have a source of gravity then the force actually is spread around the whole sphere around it. And it's the same force which is spread around bigger circle, bigger sphere. So the same mass positioned here and there. Well since the same force is spread around a bigger surface then to a particular point or a small square which is positioned here the amount of these forces are inversely proportional to the area. And area of the sphere is as we know 4 pi r square. So that's why the greater the radius the greater the surface in square proportional to r to radius square. So the same force is basically distributed on a bigger surface. And that's why it's inversely proportional to this one. And again everything which is related to dimensions to bring them into numerically equivalent values are hidden in this constant. Okay so this is the force which acts upon a point object on a distance r of mass m from the source of the gravity m. Well to tell you exactly obviously this object acts upon the source as well. I mean it's mutually attracted. But in our case when we are imagining the source of gravity being let's say sun or earth and this probe object being probably something small we can probably say that okay whatever the force is acted upon this particular object is the force which has a source as an earth or a sun. And we basically ignore how this object attracts the sun. Okay now how from this particular expression of the force I can go into some characteristic of the gravitational field which basically characterizes the field. Now there is a very convenient method using which we can characterize the strengths of the field. And here it is. Let's just think about the field from the energy standpoint. Well if you have an object which is the source of gravitational field and a probe object let's say this is fixed at zero zero coordinates. Now this one has certain coordinate x. Well then the object will be attracted and the force directed towards the origin of coordinates right. Okay fine now what I would like to do is I would like to measure how much work the field will do to basically move this particular object from one location to another. Well let's consider the location is r1 this location is r2. So we are moving from r1 to r2 so all the time the field acts upon this particular object right and the strength of the field is this one right. So the function which describes the force as the function of distance from the origin of coordinates is this right. So direction of the force is against the x coordinates. Now what should I do to find out amount of work which field does by moving from here to here. Well the force is variable right it depends. So I can't really simply multiply force by the distance because the force is changing. However I can break it into infinitely small segments each of them has the the size dx differential of x. Now on this distance my force is this one. So I have to basically integrate force times dx from r1 to r2 to summarize all these little infinitely small segments because this represents the amount of work the field does when we are moving from x to x plus dx right. Okay fine so let's just do it let's calculate this integral it will be this. First of all I'll put minus here because the force acts against the positive direction of the x because this is actually supposed to be a scalar product of two vectors but vector dx is over this way vector f at x always this way so that's why I can just simply put the minus sign and put dx here put integral from r1 to r2 and this is a very easy integral by the way because as we know the derivative of 1 over x is minus 1 over x squared right. So knowing that you can very easily integrate this sink and it will be gm m over x from r1 to r2 which is equal to gm m 1 over r2 minus 1 over r1. So that's the value of this integral and this is amount of work which field does to move from position r1 to position r2. Now here I will do something which I rarely do I'll state something without the proof. You see we are talking about a spherical field right so if you have let's just consider we are in space right this is the source of the gravity and we can move anyway in the gravitational field. All right now the statement which I would like to make is the following amount of work which is needed to move from this point to this point depends only on difference basically it depends only on on distances the beginning in the distance and the ending distance and here is why. Now I talked about this as amount of energy field spans well if we are talking about outside forces it's amount of and I would like to know how much outside forces have to spend energy to move from one point to another. Now in this case field actually helps us right from r1 we are going closer to r2 field pulls this particular object into this direction it's not outside forces which do the job it's the field helps us to do the job which means that if we are talking from the position of outside of this system then the work which is needed should actually be well this value but with a minus sign so whenever we are moving our object closer we are making we are we are working basically gaining energy not spending this whenever we are pulling from the source of gravity further then we are spending energy so that's why whenever we are moving to a very complicated trajectory so whenever we are moving a little bit closer we are gaining energy because the field actually helps us whenever we are trying to move away we spend this energy and eventually all I'm saying is that pluses and minuses are negating each other I mean it's obvious in the one-dimensional case right if you're moving if you're moving this way there then it's negative energy if you're moving this way it's it's positive energy which we have to spend right but eventually if we go back to the same place it will be zero right because nothing's changed and so in one-dimensional case it's it's obvious in three-dimensional case it's not obvious but mathematically it's very easy to show basically all I'm saying and I don't want to overload with these three-dimensional variations of this particular problem I'm explaining on one-dimensional case and just asking you to believe me that in a three-dimensional case it's exactly the same so amount of work which outside forces must spend depends only on the distance the beginning distance and the ending distance and this amount of work is equal to this with a minus sign so whenever it's a plus sign we have just calculated it's the field which is doing if we are looking at this from the outside outside should have it with a minus sign okay that's fine now we will introduce a very useful function which does not depend on the probe object it's a function which describes the field only you see what actually here is dependent on the object the mass right so let's introduce the function which is equal to g m divided by r with a minus sign now if r1 is infinity then we will have basically this right without m so if we will take a unit an object of a unit mass and move it from infinity towards location on a distance r from the source of gravity that's amount of work which we have to do and again it's negative because it's actually the field which helps us it's not me however if we want to move from r to infinity it will be exactly the same but with a plus sign obviously right so this is called gravitational potential it's a characteristic of the field this particular in this particular case we are talking about a spherical gravitational field which is produced by a point object of mass capital m so on any distance r from the position from the center of gravity from the position of this fixed source of gravitational field this is a potential and whenever we want to know amount of energy needed to move an object from one place to another well this amount of energy is equal to multiplied by mass so it's a difference between potentials and this is the potential function times mass so potential gravitational potential is a characteristic of the field at any point within the gravitational field around this particular point object of mass m this is a characteristic of the field the strengths of the field use whatever you want to call it so it's called gravitational potential and again from this we can always find amount of energy needed to move from one position within the gravitational field to another by the probe object with a mass m all right what else did i not talk about let me check okay right so basically i would like to emphasize one more gravitational potential is a very important characteristic of the gravitational field now in this particular field we are talking about the spherical field now there might be more complicated field consider the solar system we have the planets and the planets also have their own gravitational field and gravitational field of the sun is one thing and gravitational field of the earth or or Jupiter or something else is another they're all interpositioned somewhere somehow so we can't really use this centrally symmetrical spherical asymmetrical model but what we can do is at any given moment of time the gravitational field somewhere in the solar system has its potential because it's a sum of all the different gravitational fields and the potential describes the field and all we need to know to find the amount of work to move from one position in the solar system to another is to know potential here and potential there and then multiplying their difference by m we know amount of energy which is needed regardless of the trajectory well this is obviously an ideal situation obviously again there are some losses of energy etc etc but in any case in ideal case in a purely theoretical sense the potential the gravitational potential function regardless of how it's expressed right now for instance it's a complicated in case we have more than one mass right but whatever whatever we have discovered as a function we can use it to calculate the amount of energy now what else is important here is the following now if this is the potential let's let's do this let's take the first derivative of this this is one over r with a minus sign the derivative one over r square right minus and minus so we will have exactly what we have here multiplied by m so what I want to say is that knowing potential knowing gravitational potential not only we can find out very easily amount of work needed to move an object from one point to another we also know the force which is acting at this particular point r and here I would like actually to mention one thing force is a vector potential is a constant it's a scalar now in a one-dimensional case it's kind of easy because what is the derivative derivative is value of one point minus value of another point divided by the distance between these points right in a three-dimensional it's a three-dimensional vector because again what we do is we have one point in three-dimensional and another point the distance is a vector so whenever we are calculating this it's called gradient actually of the gravitational field so it's a vector which actually shows which direction the gravitational potential grows so we will talk about this maybe a little bit more details in another lecture but I just want you to know that we can derive from this gravitational potential we can derive energy needed to move from one place to another and we can determine the strength of the force and direction the strength and direction of the force acting upon the the object all right which means that this is actually like acceleration and now we are going into a second law of Newton right okay so that's it for today I do recommend you to read the textual material for this particular lecture it just helps you to digest this concept of a field gravitational potential etc that's it thanks very much and goodbye