 Hi, I'm Zor. Welcome to InDesert Education. We'll talk about gravitation. In particular, we have already researched how gravitational potential looks like in some cases like the thin rod, the ring, and the disc. Today's lecture is about gravitational potential of a spherical thin shell. So it's like a ball, but nothing is inside. So everything, the mass, is spread only around the surface. Now, this lecture is part of the course called Physics for Teens, presented on Unisor.com. On the same side, there is a prerequisite, Mass for Teens, which is definitely a very needed course of mathematics for this particular course of physics, because I'm using calculus and vector algebra all the time in physics. Now, if you found this lecture on YouTube or on any other source, but Unisor.com, I do recommend you actually to go to Unisor.com rather, because it's a course, which means all lectures are logically connected. So there is a prerequisite to this particular lecture. I will use some results from the previous one. So it's a course. So you have to really take it in logical steps one after another. Now, the site is free, by the way. There are no advertisements, no strings attached at all. So let's start. We are talking about gravitational field of empty spherical shell. All right. So obviously, as in all other cases, we will use integration. The only thing is, we have to really do it smartly. So our integration will be relatively simple. And obviously, I will use the previous lecture's results. Okay, so let's consider this is a spherical shell. Let me just put it this way. Now you see it's definitely a sphere. Now this is its center. Now obviously, we will take the system of coordinates, where the center of the sphere is origin of coordinates. This is the point of interest P. Right now, consider it's outside of the shell, outside of this sphere. We will consider inside case separately after this one. All right, so let's put some dimension. So consider the radius is r. And the distance of P is equal to h. Okay, now I am just thinking about how to break this, how to slice this particular spherical shell into pieces, so I can reduce my problem to something which I already know. Primarily, I will use the ring gravitational field in this case. So here is what I'm proposing to do. Let's cut the sphere with infinite number of planes perpendicular to OP. Now, each one will cut a circle. Now, let's consider you have this circle. And the center of this circle is, well, let's call it X, for instance. So the difference from O to X is X, if you wish. Now, what's important right now for me, and that's the way how I'm proposing to approach this problem, I would like to take as a variable this angle. Let's call it phi. So whenever my angle is changing from zero to pi, my cutting plane will be either on this side of the sphere or on that side of the sphere. So basically, whenever I'm changing my angle from zero to pi, my corresponding section plane will just move from right to left, completely from one pole to another pole. Now, let's increment my angle phi. So this is phi plus d phi. Now, this is an infinitesimal increment of the angle. And I have another cutting section. All right. So this angle is differential of phi. Now, what's in between these two planes? Well, if angle d phi differential is infinitesimal, the thickness of this figure will be infinitesimal, small, which means it will be basically a ring, right? Because this is my empty spherical shell. So if I cut it with a plane, I will have a circle and another circle. So what's in between these two circles? Well, a ring, a ring which has infinitesimal thickness. And obviously, we can consider the diameters of both rings since my angle d phi is infinitesimal. So the diameter will be infinitesimally close to each other in value. Now, let's recall, well, if this is a ring, or I mean it's infinitesimally close to the ring, right? So I can use all the formulas which I have derived when I was talking about the rings. Now, in case of a ring, I have a formula which I will put here. The gravitational potential, if you have a ring and you have a distance to the point along the perpendicular to the plane of the ring through its center, if you have a distance h and you have the radius of the ring r, then it would be the formula which I have derived in the previous lecture. This is a potential of a ring, gravitational potential of the ring. So this is the ring, this is h, and this is r, and the mass is m. So this is the formula which I have derived for this particular case. Now, we have a different case obviously here. Well, let's start from the beginning. What is r? What is the radius of the ring? Well, the radius of the ring is the distance from here to here, which is r times sine. So instead of r, I should put r sine phi. Now, h is the distance from here, from the center of the ring, to the point of interest. Now, from all to p is h, from all to x is r times cosine. So the difference between h and r cosine phi, this supposed to be instead of this h. So this is h minus r cosine phi. Now, what's m? m is the mass of the ring. Well, the mass of the ring is basically a differential because it's infinitesimally thin. And I can find this mass by basically checking what's the surface of the ring relative to the surface of the whole spherical shell. So if I know the mass of the shell and I know the surface of this ring and divided by surface of the shell, this is the ratio which I have to decrease. So what's my dm is equal to original m, mass the entire spherical shell. Now, I have to multiply by area of this ring and divide by area of the entire spherical shell. So what's the area of this ring? Well, it's a circumference times the length of this arch, which is circumference is, if I know the radius, which is r sine phi. So the circumference is 2 pi r sine phi. And what is this particular arc? Well, that's basically considering I know the the angle differential of the angle d phi, this would be r times d phi. So r would be square and then d phi. Now, this cancels. So I have m sine phi d phi divided by 2. So this is the mass of my ring. Now, I'm ready to substitute into this formula the mass. So this goes to here, instead of this mass. So this is the formula for the ring of this type. But instead of r, I should put this. Instead of h, I should put this. And instead of m, I should put this. Okay, so let me write it down. Now, what will be the result if I will substitute into this formula? This would be the gravitational potential of this infinitesimally thin ring, which is actually a differential of gravitational potential, because then I will have to integrate it for the whole range of angle phi from 0 to pi, right? So this is equal to minus g. Instead of m, I put this one m sine phi d phi square root instead of r, r square sine square phi. And instead of h, it would be h minus r cosine phi square. So this is my infinitesimal gravitational potential of this ring. Well, let's simplify it a little bit. So it's minus g m sine phi d phi divided by, if I will open this square, it would be h square minus 2 r h r cosine plus r square cosine square, right? So I have r square sine square plus r square cosine square. r square would be outside of the parentheses. Inside would be sine square plus cosine square, which is 1, which means I will have square root of r square minus 2 h r cosine phi plus h square, h square plus h square minus. So r square plus r square sine and cosine square would be 1, h square here and minus h r cosine. So this is my formula for differential of gravitational potential, which to be exact is a function of phi, obviously, right? So first we fix the angle phi, then we cut one plane at angle phi, another phi plus d phi, and this would be my gravitational potential of this narrow ring. Now I have to integrate this thing, this formula, by phi from 0 to pi, right? Okay, this is a game. Now we're talking about pure technicality. So just try to follow me, it's not really that difficult, it's really easy actually. So if I will do this integral from 0 to pi d v of phi. Now this is kind of complicated, right? However, if you notice, if I will take this whole square root of something, and I will substitute, by the way, I think I missed 2 here. I missed 2, yeah, that was 2. It was, it was 2 divided by 4. Right, sorry about that. Now, what I would like to do is to integrate that thing, I would like to make a simple substitution. Now, why do they do this? Because if I will differentiate it, what will I do by phi? Well, the derivative of a square root is 1 over 2 square roots of this same kind times derivative of the inner function. Inner function, this is constant, this is constant. So this is the only thing which is here, so it's minus 2 h r. And derivative of a cosine is minus sine of phi. So I do this one. Okay, now I can get rid of this, cancel them out. So what do I see now? Well, this is basically exactly what I need. I need sine and I need this square root, etc. So I can definitely say that dy divided by h r is equal to this part. Am I right? dy, which is derivative of y times dy, it's, so derivative of y times dy. So dy is equal to times dy. All right? If my y is this function, then differential is equal to derivative times differential of the argument. So I can say that I can substitute this piece, you see, this piece exactly the same. So dy divided by h r, let's put it this way. Is equal to this. Correct? So now if I would like to integrate this, I can put integral, minus gm obviously is a constant. Minus gm is a constant. Now 2 is also a constant. So minus 1 over 2. Okay? Then instead of this, I will have to put h r again minus gm 2 h r. That would be my factor. Minus gm over 2 and h r. So whatever is left will be equal to dy. Am I right? Which I have to integrate from. When angle phi is equal to 0 to angle phi to equal to pi. When it's 0, cosine is 1. So I have r square plus h square minus 2 h r which is h minus r square. And then the square root would be h minus r. Now remember h minus r because we're talking about the point outside of the sphere. Remember this is our sphere, this is h and this is r. Now when cosine is equal to pi, cosine of pi is minus 1. So it would be plus here. So it's r plus h square and square root. So it would be h plus r. Okay? Now as you see, after this substitution, my integral is actually trivial. And that was the purpose. As in many other cases in integration, a good substitution solved the problem very easily. So integral of dy is what's indefinite integral of dy is y plus constant c. So in this particular case, I don't need this. So in this particular case, it's equal to minus gm divided by 2 h r. So it's y plus constant c from h minus r to h plus r. Obviously, we can forget about c because it's a constant and it will cancel each other plus c minus c. So upper minus gm to h r times upper h plus r minus lower h minus r. Now this is h cancel. So it's 2 r. So instead of 2 r and 2 r here, so it cancels this as well. So I will have minus gm divided by h as what's remarkable actually. We have such a simple formula as a result. Not only that, if you have a point mass of mass m and on a distance h, you have a probe object where we would like to measure the gravitational potential. Remember, it's equal to minus gm divided by h. So for a point mass, we have this formula. And we have exactly the same formula for a shell where distance h is measured from the center of a shell. So you see, it's quite remarkable as I was saying. It means that we can consider the gravitational field of the shell, spherical shell in this particular case. The same way as if the whole mass is concentrated in the center. So whenever we are outside of the spherical shell, we can consider this spherical shell actually to be as good as a point mass with all the mass concentrated in its center. The gravitational field, the gravitational potential at each point outside of the shell will be exactly the same whether it's a shell or it's a point mass with a mass concentrated in its center. That's why in many cases related to gravitation and something like space exploration, etc., we also, in many, many cases, consider all these round planets as point masses. Granted, this is proven for a shell, not for a solid sphere. That would be the subject of the next lecture and you will see that the result will be exactly the same. However, for this particular lecture, I have another continuation. Since we are talking about an empty spherical shell and we know what will be the gravitational potential outside of the shell, question is what will be inside. Well, inside it's quite interesting actually. To tell you the truth, I was surprised myself when I knew the result. Okay, if you remember, our substitution was this. Yes, h square minus 2 hr cosine phi. That was our substitution. Now, when we were integrating from phi is equal to 0 to phi is equal to pi. Now, the problem is this is an arithmetic square root which means it's a positive value and without any kind of a hesitation I put as this limit h minus r. Why? Because h is greater than r and if cosine of phi is equal to 1, for phi is equal to 0, cosine is equal to 1, I have r square plus h square minus h to hr and obviously it's r minus h square. But if you will extract the square root, it would be absolute value of h minus r, which if h is greater than r is this. But what if h is smaller than r? When my point is inside this sphere, the distance from the center is less than the radius of the sphere. In this particular case, what would be then the value of y when the cosine is equal to 0? Well, in this case my absolute value of h minus r is actually equal to r minus h, not h minus r. Right? Because r is greater than h, so it's positive. Rithmetic square root is always positive. And my integral would be, instead of this, would be r minus h. So it would be r minus h. So what happens here? r would be cancelled each other because it's plus and minus and this h would be with a plus, it would be 2h and it's 2h would be cancelling out and r remains here. And what's remarkable about that is it's independent of h, which means wherever you are inside this spherical shell, no matter where you are, no matter what your h distance to a center is, the gravitational potential will be exactly the same. Now, why it's remarkable? Well, it's remarkable because if gravitational potential is the same, it means that the force which acts on any probe object inside the spherical shell is equal to 0 because what is the gravitational force? If you remember, it's related to, obviously, the mass and derivative of the gravitational potential. We spoke about this in one of the previous lectures. And if the gravitational potential is constant throughout inside of the sphere, then the derivative in any direction by x, by y, by z, whatever, on any direction, the change would be equal to 0. So, you know, what is derivative? It's delta function divided by delta argument. If argument is the distance from the center and the function is constant, obviously the difference is equal to 0, derivative is equal to 0 and the force is equal to 0. So, there is no gravitational force inside a spherical shell. That's what's very important. You will be basically in weightless position if you are inside that thing. Well, so, we were considering two cases. One is outside when you can basically replace the spherical shell with a point mass of the same mass, concentrated in the center. And the second case when you are inside the spherical shell, in which case the potential will be exactly the same everywhere in all the points from which follows that it's a weightless state. There is no gravitational, I mean, there is a gravitational force which is equal to 0, if you wish. So, the gravitational force is equal to 0 inside this spherical shell. And obviously, I will use this for the next lecture about solid sphere as an object which we are considering the gravitational field of. But that will be the next one. Now, I do suggest you to read the notes for this lecture. Again, the notes are always on the same website where you can find the video of this lecture. So, you can watch the video. You can read the notes. Notes are like a textbook. That's it for today. Thank you very much and good luck.