 Hi, I'm Zor. Welcome to a new Zor education. I would like to solve a few problems related to gravitational field. Now we know what gravity is about and we can solve a few problems. I would like to actually emphasize some difference between the gravitational field and its energy and the previously researched and started nuclear energy and chemical energy. In those two cases, we were not really too precise, so to speak, in our laws because it's kind of complicated. It's on microscopic level of atoms and molecules. Now the the gravitation field is much more, well, it's palpable. It's close to us. It's on the macro level. So we can actually solve a few problems related to certain laws, which we definitely know exist, like the law of universal gravity and stuff like this. So these laws we will use, definitely, in solving these problems. I have four problems. Now this lecture, obviously, as everything else, has very detailed notes. So if you will go to unizor.com physics 14, that's the course and you will go to energy and gravitational field energy. This is where you can find all these notes. Actually, I do suggest you to use the website rather than lecture itself, which you might have found on YouTube or somewhere else, because, again, it's a course. So there are prerequisites. There are detailed notes. There are exams and I do recommend you to do basically the whole course rather than just one particular lecture. In any case, so let's consider a few problems. Now the problem number one is related to the fact that there is a concept of gravitational potential, which we were studied before, and we use the formula that on the radius r from the center of the spherical gravitational field produced by a point mass, there is this concept called gravitational potential, which is equal to minus g capital M divided by r, where r is the radius of distance from the point mass, which is originator of the gravitational field, g is universal constant, and m is the mass of the point mass, which is at the center of the gravity. Now what's strange might be, I mean, it's not strange, but might seem strange from the first look, that this formula depends only on radius. Now the physical sense of the gravitational potential is, so if you have this source of the gravitational field, and if you will take some kind of a probe, probe object, and bring it from infinity to a certain place on on the radius r from the center, so the work, which is done in case of probe object, is of unit mass. This work from infinity to radius r is actually the gravitational potential. But now the question is what if I will change the trajectory? I mean, I can go this way, or I can go, let's say, this way to the same point. Is the work exactly the same? I mean, the work depends on the only on the final radius, which is just on the distance from the, what if I will go to this point, also on the distance r, is again the work exactly the same, which gravitational field has to actually do to bring the object from infinity to radius r from the from the source. Well, apparently, yes. And the first problem is, okay, can I basically support this statement that it really depends, this work depends only on the well, source and gravitational constant and the radius, the distance from the source. Okay, so that's exactly what I'm going to prove right now. So basically the statement is prove that this work is independent on the trajectory as long as the final radius is r. Okay, so how can I do it? Here it is. Now, what is work? Work is a scalar product of the force and the distance this force is acting. Now, in case the force is actually variable, we can't really say what's the distance this force actually is acting, but we have to do, we have to go to differentials and these are vectors. And this is a scalar product. So this is the definition of the differential of work as the object is moved by this force along some kind of a trajectory where the force is actually dependent on trajectory. And the force does depend in this particular case because the force according to the universal law of gravity, f is equal to gmm divided by r square, right? So it depends on r. Okay, fine. Now, let's consider this particular scalar product. So the force is variable and this is the differential, differential, let's say we are going this way. Well, this is my differential. Increment of the distance covered by this object as it is attracted by the gravity field. Now, I can always represent this ds as sum of two displacements. ds as a vector which is basically according to this for instance trajectory, it's always directed along the trajectory, basically along the tangential line, but it's an infinitesimal increment along the trajectory. I can always represent it as a sum of radial and tangential displacement. So this is my radial, this is dsr and this is my tangential dst. So this is a simple representation of the vector as a sum of two perpendicular to each other because tangential perpendicular is, tangential is perpendicular to to the radial. So what's interesting now is that after this representation, my differential of the work is equal to vector of force, which is a scalar, it's a scalar product of two different vectors. Now, scalar product is distributive. So I can definitely say this, this is radial. Now, this tangential is perpendicular to radius. Now, the force is always radial. We know that, right? Because this is the gravity, gravity always directed towards the source. So it goes always radial, which means that f and ds tangential are perpendicular to each other, and that's why as a perpendicular to each other vectors, their scalar product is zero, and we have only this. So basically, what I'm saying is we can completely ignore all movements which are perpendicular to the radius, and we can only consider the movement towards the radius, and then, if you would like to move to this point, for instance, well, we move to this point and then this point, and this perpendicular to the radius movement doesn't really produce any work by the gravitational field. Because whenever you are moving without changing the the radius, your work is equal to zero because you are moving perpendicularly to the force. So the force doesn't really make, doesn't really do any work. So that's why regardless of whatever your trajectory is, any kind of a perpendicular to the radius can be completely ignored, and only the movement along the radius is important. And that's what actually results in this formula which is independent on the trajectory or the place where I'm ending, as long as it's on the radius R from this particular place. For instance, how can I get to this place? Well, I can go this way and then along the circle always perpendicularly to the radius, move to this one, and this doesn't generate any extra work. So I can go to any point by doing all the radial movements and tangential to the radius. And the second one doesn't really produce any work. So that's why this is as a representation of the work which is done by the field by bringing it from the infinity to the radius R, if it's a unit object, unit mass object. So that's why it's completely independent of trajectory and the place where exactly we are ending our movement as long as it's on the radius R from the source. Okay, so that was my first problem which basically justifies the formula as it is. So this is amount of work which field does by bringing, by attracting the object of the unit mass from infinity to a concrete distance R from the center. Okay, so let me go to the second problem. I'll just wipe it out. Okay. Second problem would be a little bit more involved mathematically. Why do we actually pay so much attention to the gravitational potential? Well, obviously I was explaining that it really defines the field. At every point, if I know the gravitational potential, I can find out exactly where the gravitational force and what's the direction of the gravitational force and what's its magnitude, etc. However, there is one much more important quality of this concept of gravitational potential. It's additive, which means if you have two different sources of gravitation, let's say you have the moon and the earth. And now somewhere around this pair of bodies, the gravitational force obviously exists, but separately earth attracts something and moon attracts something. How can I do whatever the calculations are required to find out what's my combined gravitational force? And this is always a very important point to basically to spend some time on because obviously all our space traveling depends on many different space objects like sun and then moon and earth and some planets, etc. Each one of them contributes their own gravitational field and what is exactly the result? What's the resulting gravitational field produced by many objects? So my point is that if you know separately the gravitational potential of each individual component of each individual object in space, which is a source of gravitation, then the gravitational potential of all of them together acting at any point is a sum of each individual gravitational potential. So it's extremely important for all the calculations. Well, now I have to somehow prove it, right? Now the proof in a very general case is kind of cumbersome. Whatever I'm going to do right now is basically to prove this in two for two equal in mass objects and not in every point around them, but on a perpendicular bisector between them. So it's an easier problem and it's easy enough to present it to you. And I would like to say that basically the other more complicated problems when the objects are not of equal mass, the sources and you are not actually going in the middle between them, but somewhere else. These are all calculated in a very similar fashion, just more cumbersome calculation. So I will concentrate on this relatively simple case. So I will have two masses of equal size and I would like to have a perpendicular bisector between them. And what I'm saying is that if I will take and calculate the gravitational potential at any point on this bisector then its gravitational potential equals to some of the gravitational potentials of components. In a way, it's similar to the way how we add forces. If you have two forces, two vectors, we can just add them together. And basically the whole proof is based on the addition of vectors. So this particular, let me just reverse it a little bit. So it will be exactly like in my notes. So my masses are here. So this is, let's say, R and this is R and I'm going to a point on the distance H from the midpoint between them. So this is mass and this is mass. Now, again, I would like to calculate the gravitational potential in this particular point. Now, let's go back to the definition of gravitational potential. It's the work done by the gravitational field, now the combined gravitational field to bring this object from infinity to this particular point. So my object here let's have its mass, M, actually at the very end I'll put M is equal to 1 because gravitational potential is related to the work done by the gravitational field to bring the object of the unit mass to this position. So let's say it's M right now. Now, there are two forces which are acting along this line, along this line. They are the same. So if my position is X somewhere here, here, here, we're bringing from infinity to this point H. So regardless of where exactly this, if this is X, let's calculate this force and this force. Then we will add them together. We will add them together as vectors. So addition would be here, obviously on the same bisector because these forces are obviously equal in magnitude. Now, if this angle is phi, so what can I say about some of these forces? Well, first of all the magnitude of each of those forces is G, M, M divided by the square of a distance, right? This is the universal law of gravity. Square of a distance, if this is on the height X and this is R, that would be X square plus R square. Now, this is force. Now, this is one force. Now, this is another force. When we have to add them together, we basically have to add them again as a components. Each force can be represented as sum of this and this. So this is this force and this one is sum of this plus this. Now, these two are opposite in direction and equal in magnitude, so they nullify each other. These forces, the components along this axis are adding together. So each one of them is, if this is F, now this is phi, this is phi. So this is F times sine phi. And then we have another F times sine phi. So we have two F times sine phi, right? So this is the force which is acting as a result of addition of two vectors. Vector of force this, vector of force that. When we add them together, this is the result, a magnitude. And it's always directed down because, again, these forces are equal in size, angles are equal, everything is equal, so that's why it goes perpendicularly down. So my object is moved from infinity from over there down, down, down, down to this point at the height H. How can I calculate the work? Well, if I know the force and I know basically the trajectory, so I have to just now, force is function of X, obviously, where X is the distance from the midpoint. So what I have to do, I have to multiply it by differential of X and integrate from infinity to H. So my total work would be integral from infinity to H to F, which is this, g m m divided by X square, okay? That's my work. That's my force, I mean. And if I will multiply this force, sorry, plus R square. Now I have to multiply it by sine of F. Now, what is the sine of F? Sine of F. Well, this is X divided by square root of X square plus H, plus R square, right? So sine of F would be X divided by square root of X square plus R square. So this is sine of this F, right? As the point moves, X is its location, its distance from, if points move from infinity down to H, at any point the sine is equal to X divided by hypotenuse, which is square root of X square plus R square. By the way, I use lowercase R, uppercase. All right. And I have to, now this is the force, okay? Now I have to multiply it by distance d x. That's a displacement. So this is the force acting on this distance, and now I integrate it to get a complete work. So this is my work. Okay, from now on it's just pure mathematical manipulations. I can rewrite it as 2 g m m integral infinity H X dx divided by X square plus R square to the power three seconds, right? This is one, and this is half. So it's one and a half, it's three seconds. All right, how can I calculate this particular integral? Well, it happens that this particular integral is really very easy. Sometimes you can get integrals where when it's very difficult to do. Here it's just a very simple substitution. So if you will substitute Y equals to X square plus R square. Now dy is equal to 2x dx, right? So you see 2x dx, and this is Y. So what we have is g m m integral dy dy, which is this 2x dx divided by Y to the power three seconds. That's a very simple thing, right? The only thing I have to think about is my limits. Well, if X is equal to infinity, Y is equal to infinity, that's easy. Now if X is equal to R, Y is equal to 2R square, right? Okay. So now we don't need this picture. Now we can integrate this thing. Well, the indefinite integral of Y to the power of minus three seconds. Now minus, you see this is denominator, so I put it on the top, so I put minus here. dy. Now what is indefinite integral of this? Well, this is well, again, integral of Y to the k dy is equal to Y to the power of k plus one divided by k plus one, right? Plus C. So in this particular case we will have k plus one, it's minus one second. So it's Y one one one second divided by minus one second. Plus C. Well, C we don't really need. C because we are talking about definite integral in these limits. So we have g m m minus two, right, divided by one half, it's multiple. And this is one over square root of Y, right? Something like this. From infinity to to r square. Why r square? I'm sorry. To h. H, not r, I have to substitute. So it's h square plus r square. Not to r square. Sorry about that. Okay, so this is my result. So let's just convert it into a more compact formula. Which means Well, first of all, obviously the infinity would give me zero here. So we can disregard it. So we only have to substitute the first one and the result would be two. One over square root of r square plus h square. Well, this is a great formula because Let's return back to our drawing. What is the potential at this point of only this particular mass? Well, that's g m divided by the distance. And the distance is square root of r square plus h square. Same thing here. Now, but with a minus sign, obviously. Now, if m is our unit mass, because we're talking about the work which performed by on the unit mass, we will have here exactly double here. One and another. So potential at this point of two point masses is exactly the same as sum of potentials of each one of those. And that's exactly what I wanted to prove. This is that the gravitational potential is additive. If you add more sources of gravity, each one contributes certain gravitational potential. And the gravitational potential of the combined system of all these bodies is exactly the sum of each individual gravitational potential. So that's what's very, very important. Gravitational potential is additive. Now, knowing that, we can probably calculate more or less what is exactly my gravitational potential at any point in space. Because, for instance, not far from the moon, we have gravitational potential of moon and earth. And if you want to to land on the moon, you have to really take into account everything, right? And everything means you have to really add the potentials of each body of earth separately from the moon, add them together, and that would be the potential at any point. And knowing gravitational potential, you basically know where exactly the force of gravity is directed and what's its magnitude, because everything is the function from the gravitational potential. Now, I proved it only in this particular case. I told you in the very beginning, this is the kind of a simple case. Now, obviously, you can make it more difficult by having different masses in this particular case. And this body not being somewhere in the middle line, perpendicular bisector, but somewhere else. So it will be m1, m2, and the point will be somewhere here, which can be defined by something, whatever the definition is, by distances, for instance, or angles, or whatever. All I'm saying is we can go through detailed calculations and do the same thing exactly, and you will see exactly the same result. So, gravitational potential is additive. That's very, very important property, which is used everywhere, wherever you are calculating anything related to gravity of more than one source. All right. And now, for a much simpler case, I have two, basically one algebraic manipulation needed problems. Now, my problem number three is I would like to express the mass of the source of gravity as a function of the free fall acceleration on its surface and the radius of that surface. Now, I'm assuming that we're talking about the spherical sources, like planets, basically. And I do consider that we're talking about ideal case. So the planet is ideal sphere with a point mass concentrated in the very center of this planet. And now, we are on the surface, which has a radius r, and we can measure the free fall acceleration. How can we find out the mass of the planet if we know that on the radius r, the acceleration is such and such? Well, very simply. We can measure the weight of any object knowing the free fall acceleration. This is the force, actually, which pushes down my probe object, which is on the surface of the planet. At the same time, we know that the force of gravitation depends on gravitational constant, mass of the planet, mass of the probe object divided by square of the distance between them. And distance between them is the radius right now, right? So this is the planet. We are here. But we are thinking that all mass is concentrated in the center. And this is the radius. Well, from here, we obviously don't need this. And our mass is equal to g r square divided by g. So this is the free fall acceleration. This is the universal constant. And this is the radius. So this is basically how we express mass in terms of, and that's how we can actually weight the Earth, how much, what exactly the mass of the Earth is. Knowing geometrical characteristics such as radius, which probably is not so difficult to find out, and just free fall acceleration, which we can always measure on the surface. All right. Next, as easy. So next, I would like to express in the same terms the gravitational potential of the gravitational field based on, again, the same thing, based on something which we can measure. So why am I exercising? Because g and r can be measured with our own technology. Mass of the Earth is not so easy to measure. So that's why I wanted to have this problem. And gravitational potential also can be expressed in these terms. Now, again, we know that the gravitational potential is equal to g m divided by r. So that was basically the result of the definition, which is a work which is performed by gravitational field to bring the probe object of the unit mass from infinity to the radius r, to the distance r from the center. Now, we already know what m is from the previous problem, right, which is g r square divided by capital G. So that means g m, which is g r square divided by g, oops, capital G, and r. So this, this, this, this, g r. So multiplication of acceleration of the free fall on the surface of Earth times the radius of the Earth gives us the magnitude of the gravitational potential on the surface of Earth. Okay, these are four problems. I do suggest you to, to try, especially the problem number two, where I have two bodies. Just try to do yourself the same problem. If you are, if you want to challenge, then you can try even the more complicated problem to prove that gravitational potential is additive in general case, like not necessarily to equal masses and not necessarily along the bisector between them. So that would be great if you can do it. In any case, read the notes for this lecture. It's always useful and it's presented on Unisor.com in Physics 14's course. Now this website is completely free, there are no advertisements, and it also contains some other courses, the mass, mass 14's, which is prerequisite for this course, and there are some civic courses. So I do suggest you to visit the website. That's it. Thank you very much and good luck.