 Hi, I'm Zor. Welcome to a new Zor education. We are talking about gravitation and we have actually considered a couple of simple cases. We are considering the point mass field of gravitation and also we have considered a case when you have two point masses and we were talking about gravitational potential of the field created by these two masses and we have proven that the gravitational potential is additive function, which means that the gravitational potential of two masses at any point in the space around them is exactly the same as the sum of two potentials with which each of them, each of these two point masses, creates in its own field. So the fields are superpositioning each other. They're adding the potential, the gravitational potential is additive function. So to calculate what's the gravitational potential at any point, if you have more than one source of gravity, you can calculate what's the gravitational potential of each and it's a scalar. It's not a vector, potential is a scalar. So each of objects which are the source of gravity and then add these potentials together. Now today I would like to consider, instead of point masses, I would consider a little bit more complicated object, a thin rod which has mass distributed along its length. And I would like to basically calculate what is the potential, gravitational potential at any point around this rod. Okay, so that's the plan for today. Now this lecture is part of the physics 14's course which is presented on unidisor.com. I do recommend you to watch this lecture from the website because it contains detailed notes. Plus it's a course which means every lecture has its predecessors which you should really know before you approach anything else. So if you just found this lecture somewhere on YouTube by itself, just keep in mind that this is part of the course. Plus there is a prerequisite course on the same website unidisor.com. It's called mass for teens. And mass is definitely needed. Now today, in whatever I'm going to discuss today, the calculus is a must because we're talking about integration obviously. So if you have certain hesitations about using integrals, then please refresh this from the mass and then go to this particular lecture. Okay, so you would like to know what is the gravitational potential of the rod. So we have a rod which has mass linearly distributed, uniformly distributed along its length. And it's a very thin, well, mathematically speaking, this is the infinitely thin rod. Right? And as much as mathematically speaking, point mass is actually a point that has zero dimension. Now in this particular case, it's a one-dimensional object with a mass distributed along its length. And I would like to know what is the gravitational potential at any point in space around it. Okay, now, first of all, it's obvious that if I will have a plane which contains this line and this point, and line and point outside of this line, do determine a plane. And I will consider only this plane. I will have only two-dimensional task instead of three-dimensional. So when I was talking about space, the whole thing is just to put the proper coordinates. So we'll consider this particular plane as the xy system of coordinates with a z perpendicular to this particular plane. And obviously, nothing happens in the z, right? So all the potential at any point in space can be actually determined using these calculations made in this plane, basically, because we can always, for any other plane, for any other point, we can do this plane and basically calculate what exactly the potential is. So it would be probably even easier if I will put this along the x-axis from point A to point B. So let's say our rod has legs L from A to B. And our point where we are interested to find out the potential, it has certain coordinates PQ on this plane. So I have to find out what is the gravitational potential at this point if I have a mass distributed, uniformly distributed along this rod. Well, what obvious solution to this is, let's just have a very small piece of this rod and calculate what's the gravitational potential of this particular piece. That will take another piece and another piece and another piece and add them up together. Now, speaking more mathematically, we will divide our segment AB into infinitesimal number of infinitesimally small, sorry, infinite number of infinitesimally small segments. So this is, let's say, x and this is dx. And we will integrate. I mean, after we will find what is my potential d differential of my potential based on x, then I will just integrate it from A to B. And that would be my gravitational potential of the entire rod. So all I have to do is to find this function and then integrate. Simple. All right, now what is this function? Well, we know that if you have an object of mass m, then on a distance r from it, if it's a point object, this is the gravitational potential. Right? Now, we have addressed this in the previous lectures. But g is universal constant, m is mass of the point object, and r is the distance from this source of gravitation to a point where we are measuring our gravitational potential. Now, what is in our case this thing is? Well, mass is very easily calculated for this thing because if we know that the mass of the entire rod is m, let's say, and the length is l, then the density of the mass along the linear dimension would be m divided by l. That's the mass of the unit of lengths. So if I will multiply it by dx, I will actually have the mass of this thing. We'll call it dm equals to m divided by l dx. So this is the mass of this particular thing. And in our formula, we have to use this one. So g we have, mass we have, this is dm, which is m dx divided by l. And now we have to divide by r. Now, r is the distance between here and here. Well, if this is x and this is p, then this catatose of this triangle is p minus x. And another catatose is q. Right? So the distance is p minus x square or x minus p, that minus p doesn't really matter, since it's a square, x minus p looks better, plus q square. That's it. This is a potential, this is a potential of this infinitesimal segment of our rod. It has a gravitational constant. It has the mass. And it has, in the denominator, it has the distance from this, as we consider this to be an equivalent of the point mass, obviously, since it's a very small, it's infinitesimally small. And the distance from here to the point of interest, where we are measuring our potential. So from this point on, it's just pure technicality. It's just exercise and calculus. And here it is. We don't need any more of this. All right. Well, first of all, to tell you the truth, if you are attempting to take this integral just by yourself, it might be a little bit difficult. Now, we do have something which is called table of integrals. And there are a lot of simple integrals which look more or less simply already been found, determined, proven, etc., etc. So I personally just went to the website where this table of integrals is presented. And I just found that integral which I'm interested in, which is this one, is equal to natural logarithm of t plus square root of t square plus c square plus c. Okay. So again, I didn't do it myself. I just took it from the website. But, however, I don't really trust too much anybody in this particular case. So I took a derivative of this. And derivative of this is equal exactly this. And that's very, very easy to prove. What's the derivative of logarithm? It's one over this thing, right? So it's t divided by square root of t square plus c square times the derivative of the inner function. Inner function is this one. So its derivative is one. That's the derivative of t is one. Derivative of square root is one over two square roots times derivative of internal function. Internal function is t square. So I have to multiply it by 2t, right? Derivative of t square is 2t. And what happens now? Well, first of all, these two are canceling. If you will take this into common denominator, you will have square root of t square plus c square plus t over square root of t square plus c square. And here is t plus t square plus c square. So they will cancel each other and you will have one over this, which is this. So I checked it. Alright, fine. So everything is fine. And I can use this particular expression as an indefinite integral, as an indefinite integral which I need for this particular case. Okay, so this is equal to, well obviously g times m divided by l goes out. Now here I have x minus p square. So I will substitute t is equal to x minus p to make it simpler. Then my limits of integration would be a minus p b minus p dt. dt and dx are the same because it's a constant divided by square root of t square plus q square, which is exactly this one where q is c and t is t. So nothing strange with this. So let me wipe out this one. So I will do this. So my answer is logarithm of t plus square root of t square plus q square in limits from a minus p to b minus p. So that's my result. I don't mean the constant obviously because now it's a definite integral, which is equal to logarithm if we substitute b minus p instead of t minus logarithm when we substitute a minus p. So difference between logarithms is logarithm of their division, result of their division. So I will put b minus p plus square root of b minus p square plus q square divided by a minus p plus square root of a minus p square plus q square. Well, not such a simple expression mind you, right? Actually, I don't need that. Absolutely. So it's not such a simple expression and obviously it would be nice if you can somehow check if this is exactly the right thing. And there are many different checks. For instance, you can check a particular case when our point is exactly on the perpendicular bisector to this particular rod. And check again against maybe something which you can find on the internet, etc. Another case which is which is researched on the internet in many, many, many different websites I checked is when the point is here on the same line with this one. So there are many different particular cases where you can check this formula. So my point here is that from the qualitative standpoint problems like this one when you have some kind of an object which is not a material point like rod in this case or maybe maybe a ring or something like this, these are straightforward. So qualitatively, they are not really very, very difficult problems. However, when you start really doing this thing, you might find that it really tends to formulas really complicated like this one, for instance. It should not really scare you. And I wanted actually to mention that these calculational difficulties are just, again, technicalities. They are just calculations. Computers can probably do something like this very easily. What is important is the ability to break the big problem like in this particular case, we have to break the problem of determination of the gravitational potential of the entire rod into smaller pieces which we already know how to do like individual material point which is a point mass that we know how to do it. And then considering that gravitational potential is an additive function, you can integrate them together. And again, granted, integral can be complicated. I agree with that. However, I mean, that should not really discourage you because, again, these are technical problems. What's most important is to find out the approach. Approach is divide and conquer. Divide your rod or your solid object into smaller pieces and integrate around the whole object. So that was the point of this particular lecture. I was trying to present this particular problem. I don't know why, but for some reason, I was trying to find similar problem on the Internet. And I found the only particular case when the point is on the same axis as the rod. And in one particular case, when it's on the perpendicular bisector, but only one actually. So that's kind of strange because it's really straightforward thing and obviously, but probably people got a little bit scared of these integrals because the formulas are kind of complex and people don't like the complex formulas. In any case, I hope it was educational. And I do suggest you to read the notes for this lecture. They are very detailed. And whatever I was just writing, it's all presented in the notes. So read it again. I mean, it's always a good exercise. And I will probably provide a little bit more examples of integration because you see, this is gravity integration one. I do intend to have at least one more where I will try to present some problem related to solid objects. So we have a gravitational field of solid objects. Okay, so that's it for today. Thank you very much and good luck.