 Hi, I'm Zor. Welcome to Inuzor Education. I would like to consider another problem related to gravitational field and its potential. I would like to find out the gravitational potential of the solid sphere. So, before, in the previous lecture, we were talking about the spherical shell, empty inside. So, all the mass is on the surface of this shell. Now I would like to mass to be distributed evenly. It's a uniform spherical shell, so evenly within the whole volume of the sphere. Now, this lecture is part of the course called Physics for Teens. It's presented on Unizor.com and that's where I suggest you to watch this lecture because, first of all, it has prerequisites because this is a course. Plus, on the same website, you can find another course called Mass for Teens, which is also a prerequisite, especially things like calculus and vector algebra. It's all necessary components, necessary prerequisites before you start learning physics on this level, at least. So, okay, so let's get back to this. Let's get back to our problem. Our problem is, again, to calculate the gravitational potential of a solid sphere and, right now, the first problem is to do it outside of the sphere. So, let's say you have some kind of a point on the distance h of the sphere of radius r and mass m. All right, so I have the sphere, radius r, this distance is h, h is greater than r and we have to determine the gravitational potential at this particular point. Now, I will use the results of the previous lecture. Now, the previous lecture basically determined the final result is the following, that the potential on the distance h from the empty spherical shell of mass m is equal to minus gm divided by h. So, that was my formula. Now, that was a spherical shell empty inside and the mass is distributed only on the surface. Here we have a different story. We have a solid body of object, right? So, that's different and yet I would like actually to apply the results of the previous lecture and we'll do it in the following manner. We will split this sphere into an infinite number of concentric empty shells. So, we will have spheres of different radius inserted one inside another, all concentric and all centered at point all and in our case I will consider the radius of a sphere which I have inside as variable x, let's say. Now, what if I will consider two different spheres, concentric spheres, very, very close to each other from radius x to radius x plus dx. Now, dx is a differential, obviously, you understand that. The x is the radius of the inner sphere. So, there are two spheres very close to each other. x is, well, it's a variable, we will integrate it in the future, obviously. So, let's consider a sphere of radius x and the sphere concentric of radius x plus dx where dx is infinitesimally small increment of the radius x. Well, this object looks very much like the one which we were considering in the previous lecture which is an empty shell and an empty spherical shell where the mass is concentrated only on the surface. So, as long as dx is infinitesimally small, the thickness of this object between two spheres is extremely small, infinitesimally small. That's why we can apply the formulas which we have derived in the previous lecture to this particular object between these two infinitesimally close to each other spheres. So, all we have to do is... Now, h, by the way, is exactly the same for all these spheres. So, h is the constant. So, the only thing which is variable is mass. The mass of this layer, spherical layer, I should say, probably, or a shell and obviously it depends on the radius x, etc. We have to really calculate it. But we can definitely calculate this v of h and of the radius x of the spheres. This is a potential, gravitational potential of this infinitesimally thin spherical shell. So, I will calculate this and then I will integrate it by x. And that would give me the entire gravitational potential at this particular point. Okay, so, let's do it. So, all we need is to find out m. Well, in our case, obviously we should not talk about m. This is mass, right? Now, in our case, since mass is actually distributed evenly within the whole big spherical object of radius r, what we have to do is we have to find out the volume of this thin spherical shell of the radius x and multiply by the density of the matter inside this spherical object. What is this density? Well, obviously the density is equal to total mass of the object divided by its volume, right? The volume of the sphere of the radius r is 4 third pi r cube, right? So, it's 4 third goes here, pi r cube. So, that's my density. Now, if this is my density, all I have to do is to find out the volume of this thin spherical shell, multiply by density and I will have the mass. Knowing the mass, I will use this formula instead of this m and I will get exactly what's my potential at the point p, which is only derived by one particular shell of radius x. So, what is the value? Well, considering these are two concentric shells which are spherical surfaces which are very, very close to each other, well, infinitesimally close to each other, I can basically consider this to be the volume to be equal to the area of the sphere multiplied by its thickness, which is infinitesimally small. Now, the thickness is dx, right? And the area of the sphere of radius x is 4 pi x square. So, this is my volume. This is the volume of a very thin shell of radius x and the thickness of dx. Using this volume and the density, I can find out the mass. So, it's differential, I will have gm of x. It's equal to this times that, right? So, it's 4 pi, by the way, will go away. So, I will have 3m x square dx divided by r cubed. All right, 3m and yes, 3m divided by r cubed. I think that's what it is. Great. So, now we can use this instead of the m in this formula and we will have an increment of the gravitational potential that this particular thin shell contributes. So, that's why I would probably call it dv of x. Well, h is an index. It's a constant, basically. It's a distance, all right? So, it's minus g divided by h times n instead of m, we should substitute this. So, it would be 3m x square dx divided by r cubed. 3m x square dx, yeah, seems to be a fine. Now, what should I do now? Well, I should integrate it by x from 0 to r, right? So, all these concentric shells must be summarized together because if you remember, gravitational potential is an additive function, right? So, all I have to do now, I have to integrate it from 0 to r by x and that gives me, so what is the constant? 3m, 3gm, 3gm divided by h r cubed is out. These are all constants. Integral from 0 to r x square dx, right? Well, integral of x square is x cubed divided by 3. That's an indefinite integral which I have to substitute from 0 to r. So, with r it will be r cubed. With 0 it will be 0. So, the whole thing actually is equal to r cubed, which makes my integral which is basically the entire gravitational potential v of h equals to r cubed, oh, I'm sorry, r cubed divided by 3, sorry. One third, x square integral would be x cubed divided by 3. So, this is by 3 and we substitute r instead of r cubed and we'll get this. So, what happens? r cubed cancels, 3 cancels and I have, obviously I forgot the minus sign, so I would have minus gm divided by h. Surprise, surprise. We have exactly the same formula as the one for empty shell. So, this is kind of a universal formula and it's exactly the same as if the whole mass is concentrated in the center, if it's a point mass. So, basically there is no difference between point mass or a shell with the center in this particular point of the same mass or even a solid sphere of uniform solid sphere with the same mass with the center in the same point. So, these objects, a point mass of mass m, a shell of mass m with the same center or the solid sphere of the mass m, they all act exactly the same at this particular point. So, the gravitational field at this point will be exactly the same, whether it's a point, a shell or a solid sphere. That's quite remarkable to tell you the truth. Okay, so that's basically the end of this problem and now we have to go to the second part of this problem. Well, what is this point P, not outside by inside the solid sphere? Remember, when we were talking about spherical shelves, the previous lecture, we also had these two cases. One case was outside and obviously the outside was this formula, right? So, as age grows, the distance, the whole potential goes down. But if the point is inside the spherical shell, if you remember, it was minus g m divided by r, where r is the radius of the shell and it does not depend on where exactly on which, on what distance from the center point is located. So, this is again quite a remarkable thing. So, inside a spherical shell, the potential is exactly the same at any point, regardless of its distance to the center. As long as it's inside the shell, all these gravitational forces are nullifying each other basically and the potential remains exactly the same. Now, let's talk about solid sphere. Now, we are talking about this point to be inside. Well, you can imagine, let's say our sphere is some kind of a planet, which is, let's consider this uniformly, mass inside this planet is uniformly arranged and we make some kind of a, we drill down to the center some kind of a tunnel. So, we are very small tunnel, so it doesn't really change the whole mass. And now we are traveling inside this tunnel to the center of the planet. How the gravitational field will change in this particular case based on the distance of the center, of course. Now, again, if you remember, in case of the gravitational potential of the spherical shell, the gravitational potential is exactly the same. Now, will it be the same in case of a sphere? Well, let's find out. That's my second problem. Okay, let's just wipe it out. So, this is our sphere and now we have our point here. Let's say this is radius R. The radius of the whole sphere is R, capital R. Now, being in this particular position, we experience the gravitational field from two different objects. One object is a sphere of the radius lower case R and we are on the surface of this sphere, right on the surface. Another object is whatever remains from the whole big sphere if we will remove this nucleus, let's call it nucleus. So, this part is nucleus and this part is whatever is left from the nucleus. Now, in this particular case, this is a little bit more complex than before because before we were talking about infinitesimally thin spherical shell. Now, this is a thick spherical shell which has certain substantial widths, capital R minus lower case R, the radius, right? So, that's kind of a different thing. So, if we are talking about the inner sphere on which surface our point is, that's easy. This is something which we have just done before. The potential at this point which is a result of the field of the inner sphere is, as we know, minus g m1 divided by R. So, this is my first component, my first component of the gravitational potential depends on the mass of the inner sphere, nucleus. Now, this R is the distance from the center. So, I'm just using whatever I just derived in the previous part of this lecture, right? So, this is the gravitational potential of this point from the inner nucleus of this sphere, from the inner sphere. Now, let's now calculate the second component. But first of all, speaking about the m1. Well, m1 is easy. Again, our density is 3m divided by 4 pi r cubed, right? Now, if we will multiply this density times the volume of the nucleus, inner sphere, we will have m1, mass of this sphere equals to rho, which is 3m divided by 4 pi r cubed, times the volume of the inner sphere, right? So, we have the density, we will have the volume. Volume is 4 third pi r cubed, lower case r cubed, right? That's the volume of this inner thing. It remains, well, 3, 3, 4, 4, 5, 5. So, it's actually m r cubed divided by r cubed. So, that's my m1. That's m times r cubed divided by r cubed. Okay, and so, let's just change. If we will put this, r cubed will cancel with this r, it will be r squared. In the numerator and r cubed in the denominator. So, we will have this r squared r cubed. Got it. So, that's my potential at this point from the gravitational field of the inner nucleus. Now, let's talk about this thick, now we're talking about thick empty shell, thick shell. Well, here I don't really have any formula, since it's a thick one. I can't really use anything before. Well, I have to resort again to integration. So, what I will do exactly the same as before, I will have two concentric spheres. And I will calculate the gravitational potential at this point only from this sphere. From this spherical shell of infinitesimal widths. And then I will integrate. Right? So, consider this is x, the distance to our concentric shells. Now, again, let's calculate, now we know this, that inside the shell, the gravitational potential is the same. And it's equal to g times mass of the shell divided by radius of the shell. Well, radius of the shell, I have to replace with x, right? That's radius of my infinitesimally small shell. Now mass, mass of this particular infinitesimally thin shell is equal to, we just calculated it a second ago. So, it's rho times the volume which is 4 pi x square dx. So, this is the area of the sphere and this is the width of the, this is the thickness of this shell. So, if I will multiply, that's what I will get, which is equal to 3m divided by 4 pi r cube times 4 pi x square dx. What happens here is 4 pi goes out, right? So, I have 3m x square divided by r cube. So, mass, mass should be 3m x square dx divided by r cube. So, my gravitational potential of this shell, inside of this shell, because this point is inside, inside any shell, whatever I can cut from, from the sick shell. So, the gravitational potential, let's call it dv of x, it's differential, it's infinitesimally small, small one because the shell has infinitesimally small mass. So, it's equal to minus g, instead of m, we do this, 3m x square dx divided by r cube. And instead of r, we have to put x, because that's the radius of the shell, right? So, then all I have to do is integrate it from x is equal to lower case r to x is equal to uppercase r. Because we have to integrate from all these concentric shells, have the minimum radius of lower case r, maximum radius of capital R. So, that's what we have to do. Now, this is x square, this is x. So, I basically have minus 3gm divided by r cube, okay? Integral of x dx, which is integral of, indefinite integral of x is x square divided by 2. So, I will have here x square divided by 2 in limits from lower case r to capital R, which is equal to minus 3gm r square minus r square divided by 2r cube. Am I right? Yeah. So, this is the gravitational potential v2 of r, lower case r. That's where our point is located, okay? So, let me wipe out everything else. Oops, I did something wrong. So, it was gm r square minus r square. So, this is gravitational potential at this point caused by inner sphere, inner solid sphere. This is gravitational potential of this thick shell at this particular point. But we have to do now, we have to add them together. This is the total gravitational potential. Now, considering this is 2r, I can do this 2 as well. So, I have 2r cube in both cases and gm in both cases. So, I can put gm divided by 2r cube outside the parentheses. And what do I have here? I have 3r square minus 3r square and this is 2r square, which is 3r square minus 1r square. So, this is the answer. This is a gravitational potential at this particular point inside a solid sphere. Let's just think about this a little bit. What happens if my point is right on the surface of the whole sphere? So, what if lower case r is equal to capital R? Well, I will have 3r square minus r square, which is 2r square. This is 2r cube. So, in this case, if r is equal to r, my formula would be minus gm divided by r, which is correct formula for gravitational potential of the solid object outside it. Well, on the surface means outside it. So, that would be the same formula as this one. Now, what if my r is equal to 0? This is the center of the planet. So, what happens with gravitational potential at the center of the Earth, let's say, if you reach it in some way or another? Well, using this formula, you will have minus 3 second gm divided by r. So, as you see, it's about 50% greater than the gravitational potential on the surface. So, inside a center of the sphere, gravitational potential is about 50% greater than the gravitational potential on the surface. Now, let's just consider that we are more interested, not exactly in gravitational potential, but in the gravitational force. Now, in case of empty shell, we know that the gravitational potential doesn't depend on the position inside this particular shell. It's always the same, and it's always the same. It means that the force, which is basically a derivative of gravitational potential, is equal to 0. So, inside the empty shell, you have exactly zero force, which is acting upon the probe object. Alright, so let's do it in this particular case. Now, well, here we have two cases and two different formulas. One formula is this one when the object is inside, when the lower case r is less than the capital, this formula. This is only for the outer shell. This formula is. If lower case r is less than capital R, this is the formula for gravitational potential inside. And we can always find its derivative by r to find the force. Okay, so now this is the constant. So, all I have to do is gm divided by 2r cubed and the derivative of r squared. It will be the plus and two. It's 2r. The derivative of r squared is 2r. 2 and 2 will, so it will be gm divided by r cubed r. Right? That would be my force, which depends on r. And I have to multiply it, obviously, by the mass of the probe object. I can consider the probe object as a unit mass. Anyway, so this is the force which is acting on the object inside. And you see it's proportional to r, which means that in the very center the force is equal to zero, which is kind of understandable because all masses are gravitating in all the different directions in the same fashion, right? You are inside the center of the solid sphere. Now, but as you are going further from the center, your gravity, your weight, if you wish, which you are experiencing, will grow linearly up to the point when r is equal to, lower case r is equal to capital R when we are moving to a surface. So when r is here, force is zero. As soon as the r starts increasing, my force is increasing to the point that f of the force at the radius capital R is equal to gm m divided by r squared. Well, which is a known formula for gravity based on Newton's universal gravity law, right? We know that. Proportional to masses and reverse proportional to square of this. Okay, that's fine. Now, what if we are outside of the object, outside of the solid sphere? Now, whenever we are outside of this sphere, we know that this is something which I have considered in the previous problem. And we know that in this particular case, we have to basically equate the potential with the one from the point mass of the same mass, right? So if r is greater than r, my potential would be equal to minus gm m divided by r. And derivative, not potential, sorry, potential, no mass. But derivative of this would be equal to, well, now it would be gm divided by r squared. And if I multiply it by mass, I will get the force. So here, as we see, our gravitational force is diminishing as r is growing. You see, here the gravity is increasing linearly. Now, after we have reached the surface, it will start decreasing because we are getting further and further from the mass. So my graph, if you wish, would be first, it would be like this, and then decrease, say, it increases linearly and then decreasing inversely proportional to the square of the radius. So this is the graph of the force, or weight, if you wish, which you experience. If you are traveling from zero, from the point inside the very center of Earth to the surface, your weight will grow linearly up to this weight, which is your normal weight. But then, from this on, as soon as you start moving further from the planet Earth, then your gravity will be decreasing again, the same way as the Newton's law of gravity is. So this is a very interesting graph. This is how your experience of weight will be if you will travel from the center of the Earth down to its radius and then further into the space. Well, that's it. I do suggest you to read the notes for this lecture, check my calculations. I think I'm correct, actually. I think it's a very interesting observation. Basically, it's how exactly you feel if you're traveling from the inside the planet to its outside surface. So gradually, your weight will increase from zero to your normal weight and then it will start decreasing as soon as you're going further and further from the planet. All right, so that's it for today. Thank you very much and good luck.