 Hi, I'm Zor. Welcome to Unisor Education. I would like to continue solving relatively simple problems related to gravitational potential. So today we will talk about the ring and the disc as the sources of gravitation. Now, this lecture is part of the course called Physics for Teen, presented on Unisor.com. I suggest you to watch this lecture and all others which are presented on this website, from the website itself. Because it's a course, which means there are prerequisites, there are certain logical order between the lectures, obviously I'm using in any subsequent lecture, whatever I was using before. Also, on the same website, there is a course called Math for Teen, which you really have to know whatever is in that course to understand the physics for teen, especially the calculus, because the whole physics is based on the calculus. Alright, so, going back to gravity integration 2, and again, two problems. They will be by themselves in this lecture, but obviously I'm doing it for the purpose. And my purpose is the next lecture, where I will use the solid objects as the source of gravitation. So, right now we will talk about, first is about the ring. So, let's consider we have a ring, which has certain mass m, it's infinitely thin, it's radius r, and on the perpendicular line through the center perpendicular to the plane where the ring is actually located. I will have certain point of interest, which is on the height h above this ring. Now, obviously I will introduce the system of coordinates, which is such that the disk is in the x, y plane, and obviously the z axis is the one which contains the point p on the height h above the x, y, z plane. Now, I'm interested in the gravitational potential at point p. So, point p coordinates are 0, 0, h. Now, as we know, the gravitational potential is an additive function, which depends obviously on the gravitational field. What it means is, if you have two sources of gravitational fields, and basically they are combined, and in any particular point we can measure the gravitational potential, and it will be equal to the sum of gravitational potential of this field alone plus this field alone. I have proven this in one of the previous lectures. So, we will use this additive function of the gravitational potential to derive the total gravitational potential of the ring. Now, the obvious way how we do it in this and in any other case similar, we will divide actually the ring into small pieces. So, this is an infinitesimal piece, let's say it's length dL. I will measure the potential from this piece at this point, and then I will integrate along this circumference of the ring. Okay, now, what's simplification of, what is a very simple property, simple characteristic of this problem is that all these little dLs are on the same distance from the point of interest. Now, we know the general formula that for a point mass, the gravitation depends on gravitational constant, mass of this point mass, and divided by distance to point of interest from this point of mass. Now, it actually seems natural that in this particular case, since every little infinitesimal piece is on the same distance, and then when we will summarize this, we will probably have to summarize and get the total mass, basically. So, most likely I will have formula similar to this, where r is the distance from the point on the ring to a point of interest on the line, on the z line, which is actually square root of r squared plus h squared, right? So, intuitively thinking, I think I have to have this formula, g times mass of the ring divided by this square root of r squared plus h squared, that's obviously the hypotenuse, and this is right triangle. However, let's just not rely on our intuition, let's try to derive it strictly mathematically. Okay, so I'm going to basically have the same formula, let me put it here, in this form, and we will see how it's derived in a relatively rigorous way. So, what I will do is I will have the central angle from the horizontal axis to a radius on the ring. Again, the ring is in the x-y plane. So, let's have it as phi, it's a variable. And I will increment phi to phi plus d phi. So, this is my segment, so this is phi, and the dotted line is phi plus d phi. Now, that actually cuts this infinitesimal, this is infinitesimal increment of the angle, this will be infinitesimal increment of the length of the arc, right, which is equal to, dL is equal to radius times angle, we know that, right, from geometry. So, the line, the lengths of the arc depends on the angle using this formula. Well, again, that's actually why I was saying that the mass for teens is a prerequisite, you have to know all these little things. Alright, so we have the dL, now, we need the mass. Now, we have the total mass of the ring is m, so the density of the mass, rho, is equal to m divided by the length, so this is 2 pi r. So, 2 pi r is the length of the ring, and if I divide mass by the total length, I will have the density per unit of length. Now, if I have the density, and now I have the length, dL, my dm, which is infinitesimal mass of this little piece, is equal to rho density times dL, which is m r d phi divided by 2 pi r, and r is out, so we have m d phi divided by 2 pi. Okay, so this is the mass. Now, if I know the mass of this, and I know the distance, distance is equal to square root of h squared plus r squared, right? I can determine the gravitational potential of this point, at this point, from this infinitesimal piece of my ring. So, it's dv, d is gravitational potential, and dv is an increment of the gravitational potential from this infinitesimal piece of the ring. It's equal to g mass of this divided by distance r, which is, which is g, dm is this, and divided by r, and again r is square root of r squared plus h squared. Now, what do I have to do to determine the total gravitational potential of an entire ring? Well, obviously I have to integrate my dv, well, v is actually a function of phi, right? We all know that. So, that's why dv is actually this. From 0 to 2 pi, right? My angle should be rotating entire circle, which is equal to integral from 0 to 2 pi. Instead of dv, we will put this, gm d phi divided by 2 pi square root of r squared plus h squared. What's interesting is that these are all constants. So, we can go outside of the integral, and what will be the integral itself? Integral from 0 to 2 pi of d phi, which is actually equal to 1, right? It's equal to phi from gm 2 pi square root of r squared plus h squared. The indefinite integral is phi, right? Because the derivative from the phi will be, the differential from the phi will be d phi from 0 to 2 pi. So, if we will substitute formula, Newton-Leibnitz formula, we will substitute from the phi we will have 2 pi minus 0, which is 2 pi, and we will cancel 2 pi, and we will have gm and square root, which is exactly what we have to prove. So, our intuition is right, basically. Now, this is extremely simple case of, well, relatively complicated object, the ring. But why is it simple? All the distances are exactly the same. And that's my first problem, which I wanted to present today. My second problem would be slightly more complicated. It will be a disc instead of a ring. So, let me wipe out this. Now, my drawing will be probably more or less the same, except now, again, now this is x, mass is m, and this is y. But now, this is a solid, although infinitely thin, disc. So, mass is now distributed along a disc, not along the ring, right? Which means my density is equal to mass divided by pr squared. Right? Because this is the error. So, my mass is distributed not along the circumference of the ring, but along the entire surface of the disc. So, this is, now, how do I approach this problem? In exactly the same fashion. When I had a ring, I divided by little segments of the ring, little arches, right? Now, if I have a disc, what I will divide it into, I will divide it into concentric rings. Now, each one will be very, very thin. So, my variable, which I'm going to use, will be r, which will be from 0 to capital R, capital R is the radius of the entire disc. Now, r is the radius of one of the infinite number of the concentric rings. Right? Now, my widths of the ring will be infinitesimal. It will be from r to r plus dr. So, it will be a very, very thin ring. And I would like, actually, to use the formula which I have derived before. Now, remember for the ring, it was mass G, mass divided by square root of h square plus r square. Now, in this case, my ring is r square, because there are many rings. I'm talking about any ring on a distance, on a radius r. And instead of m, I will have, basically, the dm, because it's an infinitesimal widths of the ring. Right? Now, how can I find out dm? r, basically, I understand what r is. Now, h is also not exactly h. h would be my... Now, h would be the same h. Sorry. h would be the same h. It's r which is different. r is changing. So, what's my mass of the ring? Well, if I know the density, I have to just know the area of the ring. So, what is the area of the ring? If this is r, and this is r plus dr. Now, my area of the ring is pi r plus dr, square minus pi r square, right? The total area of the circumference of the incremented ring minus this one, which is equal to pi times r square plus 2 r dr plus dr square minus pi r square, which is equal to 2 pi r dr plus something which we can actually ignore because it's an infinitesimal of the higher order. See this square? So, whenever we integrate it, it will just disappear. Basically, my area is this, and since I know the density, I can calculate the mass. So, dm, which is mass of the ring, which is equal to m divided by pi r square times 2 pi r dr. So, I have the mass. Well, pi obviously goes out. So, we have m r dr 2 divided by r square. Now, that's my m. Now, my dv, which is differential of gravitational potential of the ring only. I will use this formula. It would be g. Now, dm, so it's 2m r dr divided by r square and divided by square root of h square plus r square. All I have to do now is to integrate it by r. This is all function of r. r is my radius of variable radius of the small ring. So, r is changing from 0 to capital R of dv. That's my potential. All right, let's do this. So, now it's just technicality. And as you see, physics is a lot of mass. Now, here is very simplifying, basically, consideration. You see, this is r square and this is 2r dr. Now, we know that r square derivative is equal to 2r. So, basically, we can consider this, this and this as differential of r square. Differential of r square would be derivative, which is 2r times differential of r dr. However, I will do it even better since I can always add the constant to a function without changing derivative. I can say that this is equal to 2r dr, right? Because, you know, the derivative of sum is sum of derivative, so I still have 2r and this is 0 because it's a constant. What my integral now becomes is from 0 to r of gm divided by r square should be outside of the integral because they are constants. Now, inside, I will have d of r square plus h square divided by square root of r square. That would be easier. I changed the order. It doesn't really matter. Equals. Now, what makes sense now is to substitute, let's say, y is equal to r square plus h square. So, I have gm r square. gm is not a general motor. It's the multiplication of a universal gravitational constant by the mass of the disk. Now, integral. Now, if I will replace this, then if r is equal to 0, then y is equal to h square. But if r is equal to r square, that would be r square plus h square. d y divided by square root of y. Now, we are in a familiar territory of a very simple integral. Now, again, square root of y derivative is equal to, this is very familiar. It's y to the power of 1 half, right? It's the derivative is this. So, we can guess that basically our integral is equal to 2 square root of y. Because if I will do the derivative from y, it will be 1 over 2 square root of y. 2 and 2 will cancel. And I will have just 1 over square root of y. So, that's my derivative. So, that's my independent, an infinite integral, which has to be from h square to r square plus h square. And obviously, this multiplier still exists. And the answer is, my potential of the disk is equal to times m divided by r square. So, m is a mass of the disk. r is its radius. Then I have 2. And then in parentheses, I will have square root of r square plus h square minus square root of h square, which is h. And that's the answer. This is a gravitational potential of the disk. Okay, now, before ending this letter on this note, I would like to tell why I'm doing this exercise. Well, first of all, it's obviously useful by itself. It's something which really develops your fluency in using your mathematical apparatus to physical problems. But I also have another goal in mind. On the next lecture, I will consider a solid ball as the source of gravity. And I will divide it into... I will slice it actually into individual disks of infinitesimal widths. And I will use this formula, obviously, for gravitational potential of the disk. And then I will integrate it among all the different disks. So, that would be the next one. And for now, that's it. Thank you very much. And good luck, actually.