 Those who are online please type in your names, morning, so your school UTs are still going on is it, school UTs are over or it is still going on and if it is going on what is the next test and when, how was your physics UT, how was your physics this time, where the concepts were same just the problems will be different, very easy. So today's session, today's session what we will do is, we will first start with problem practice, problems on whatever we have done last, okay, problems on whatever we have done till now in gravitation we will take up those problems, okay and then we will discuss few of the concepts which are beyond your school curriculum but they are, you know, they are there in J syllabus, okay. So we will start with the problem solving, all of you ready, should we start or you have any specific doubt in this chapter which you want to discuss first, okay. So I think let us start, fine I will start taking up your questions from your school book, okay first, once we are done with your school level questions then only we will go to the next level, okay. So you can see this question right in front of your screen, question number 8.5 you see this, please solve this, good to know that Devdas, all those who attended Sunday must but very few people attended the Sunday's class, anyways please attempt this question number 8.5, good morning, good morning everyone, so in this chapter you will see that concepts are straightforward when it comes to the school level questions but calculation is bigger, so you should be good at calculation and let me tell you a fact, in 2019 the common feedback is that, common feedback in January J mains is that the paper was calculation intensive, okay. So many students could attempt only 20 out of 30 questions, fine, the attempt was less because they could not calculate very fast, getting it, so you must, you must not use calculator in these type of calculations, you must do the hand calculation otherwise, you know the speed will take a beating, your speed will be very less and because of that even though you know how to solve it, you will not be able to attempt the full question paper, okay. So don't make that error which, you know I have again told the same thing to your seniors also but few listened, few chose not to listen, so no surprises who suffered at the end, okay. Anyone is able to get the answer? Bharat may not be able to join in today's class, so many of you might be missing Bharat today. R is not the radius of Phobos, R could be the orbital radius of the Phobos. What concept will you use here, which concept will you use? You will use T square is proportional to R cube, isn't it? So T1 by T2 whole square will be equal to R1 by R2 whole cube, fine, all right. So Phobos has a period of 7 hours, 39 minutes and orbital radius of Phobos is given. So let's say T1 is for Phobos, okay. We need to find the, okay, this is, I think, this is not used right now. The first one, the first part is something else. Phobos has time period given and the orbital radius is given. We need to find mass of the Phobos, okay. So we know that the, when you equate mass of Phobos into orbital velocity square divided by R, this should be equal to G times, G times mass of mass into mass of Phobos divided by R square, okay, yes. So from here, you will see that mass of Phobos get cancelled out, all right. So velocity will come out to be a root over G mass of mass divided by R, okay. This is the velocity, okay. Time period is 2 pi R divided by velocity, okay. So you will get 2 pi root GM, G mass of mass, okay, into R raised to power 3 by 2, okay. So we need to find the mass of the mass, right. So just rearrange the term. First you should get what is mass of, what is the mass of the mass in terms of other variables, okay. So root over mass of mass is equal to 2 pi divided by TG, R raised to power 3 by 2, right. Then square both sides. So mass of mass will become equal to 2 pi divided by GT square into R cube, all right. Once you get this, then you substitute the values, okay. Then you substitute the values to get the answer, okay. T must be in seconds. So convert 7 hours, 39 minutes into seconds. How will you convert that 7 hours, 60 minutes, 60 seconds, okay. Plus 39 minutes, so plus 39 into 60 seconds. This is the value of T which you should substitute over here. R cube, you convert this in meters, okay, and then cube it. And the value of G is known, okay. Yeah, 2 pi is squared, so this squared, fine. Oh, Shravan, Shravan calculated, is it? No, Shravan, we are calculating mass. We are not calculating the time period right now. Mass will be 5.97 into transfer 24 kg. All of you clear about this one? Now the second part of this question is assume that Earth and mass move in a circular orbit around the Sun. Martian orbit is 1.52 times the orbit radius of the Earth. What is the length of Martian year in days? One year should be what? One year should be the time taken by the planet to complete one revolution around the Sun. That is how one year for every planet is defined, okay. So here you need to use this Kepler's law. Time period of Earth around the Sun, which is one year, is given as 365 days, okay. So let's say T1 is for the Earth, okay, R1 is radius of revolution for the Earth and R2 is for the mass, okay. So I can say that T2 is equal to T1 times R2 by R1 times 3 by 2, okay. R2 by R1 is 1.52, okay. 1.52 raised to power 3 by 2, you have to do it here. Now you can't find out 1.52 raised to power 3 by 2 using hand calculation. So you must be very much familiar with the lock tables, okay. So try using lock tables. If you don't have a lock table, please get a copy of that. So using lock tables, you'll get 1.52 raised to power 3 by 2 and then multiply that with T1. T1 is what, 365 days. Clear all of you, is this thing clear? Any doubt, please ask quickly, okay, fine, great. This is 8.5, 8.6. Mass is this, 6.4 into transport 23. This one, 8.6, answer I've already shown you. Okay, can you get the expression, like how will you find? Forget about the final answer, okay. Can you tell me using what all two, using what all two different methods, can you get the mass of earth, okay? Can you tell me the two different methods in which you can get the mass of earth? Don't worry about the final answer. Tell me the concept, actually, what all two concepts will you be using to get the mass of earth? Yes, this is one method. This is one method. The value of accession due to gravity, G is given, right? So, since G is given, I can use the value of G to find the mass of earth because G must be equal to G into mass of earth divided by radius of earth square. The radius of earth is also given, okay? What is the other method? Other method is time period, okay? But time period revolution of moons orbit is given. It's not the earth's time period that is given. So, how will you do that? This is suppose moon that is orbiting the earth, okay? This is the earth, fine? So, if moon has a velocity of, let's say, U, okay? And this is the orbit of the moon. Distance of the moon is given as this. So, this is the orbit of moon also, right? So, that is R and let's say this is moon, mass of earth is M E and mass of moon is M M. What did I say last class? Any situation in which something is moving in a circle, the first formula you should write is this. M V square by R should be equal to G times M M divided by R square, okay? So, from here, you'll get this velocity, velocity of the orbit, okay? And once you get the velocity of orbit, okay? Time period will be 2 pi R divided by V, okay? And you can see that the velocity of orbit depends on the mass of the earth, all right? So, when you substitute V over here, mass of earth will come in the expression, okay? So, the time period is also given, 27.3 days that you convert in seconds, okay? How do you convert in second? 27.3 into 24 into 60 into 60, okay? Now, it is a good idea to remember how many seconds one day is, how many second one day is. It is a good idea to remember this and also one year has how many seconds. So, if you remember this, it will really help you. So, I hope all of you are clear. Any doubts? Any doubt? Quickly type in, okay? Okay, let's leave this one. This question, 8.8. The hint is you need to use conservation of energy, conservation of energy, okay? If final energy E2 is more than E1, so how much energy you need to give? You need to give E2 minus E1, right? Energy cannot be created. You have to give the energy. Can anyone take a pic of your UT paper and send me a cross over WhatsApp when you're free? Anyone got the answer? Just do the first part, okay? Just till here. Should I do it? Yeah, I've got the UT paper. Others don't send it, now I have it. Thank you. All right, so let us see. The potential energy is what? Potential energy is minus of g M1, M2 by R. This is a potential energy, right? And we have already derived that kinetic energy is minus of potential energy by two, right? So total energy is what? Total energy is potential energy plus kinetic energy, which is U by two, right? So always remember, if radius of revolution is R, total energy will be always g M1, M2 divided by two R. This is the total energy. You don't need to find out the kinetic energy and potential energy separately. So it's a good idea to remember that if radius of revolution of an object is R, okay? Then the total energy will be minus g M1, M2 by two R, okay? Now don't calculate right now. So too much text is there. Where should I write? So initial energy, initial energy, when this satellite is at that orbit will be equal to what? Minus of g mass of Earth into mass of satellite divided by two times of two R e, that is four R e. This is your initial energy, initial total energy. And final total energy will be minus of g mass of Earth into mass of satellite divided by two times of the new orbit radius, which is four R e. So that is eight times R e, okay? Now tell me is T2 greater than or less than T1? Is T2 more or less than T2, T1? T2 is more than T1 or not, right? T2 is more, right? It is closer to, I mean the absolute value of T2 is less but when you put a negative sign with it, okay? Then T2 becomes more than T1, all right? So if T2 is more than T1, then T2 minus T1 extra energy is required to have T2, okay? Because one thing is very, you should note a point that if a satellite is revolving around a radius of R, then it's potential energy, it's kinetic energy and total energy, everything can be written in terms of R only. So if R is fixed, its energy is fixed, okay? If its energy increases, automatically its radius will change, fine? So suppose you have a, let's say a satellite which is revolving around the radius, okay? And if you want to change its orbital radius, then what do you do? You burn some fuel, you can put it like a rocket, you can increase its velocity. So if you increase its velocity, its kinetic energy automatically increases, okay? So if energy increases, automatically it will spiral into the larger radius, having more energy, getting it? So if you want to decrease its radius, you decrease its velocity, velocity of satellite, if you, as soon as you decrease the velocity of satellite, its orbital radius will, orbital radius will change because its total energy is changing, fine? So total energy is a direct function of a radius of revolution. All of you understood this one? Once you know what is the change in the total energy, you can find the change in kinetic and change in potential energy as well. Clear, all of you? Okay, I don't know when will be our next, the classroom session will be because every time Tuesday comes, there's a holiday, holiday or a bund. So I think it's been a month now. Nirenjan thinks that I am like 100% only one batch is there. There are other batches Nirenjan. Anyways, I have taken a lot of extra sessions, even though it is online. Okay, so let's try to solve few of your school questions. Can you solve 8.12? Okay, you're doing it or should I, should I attempt it? Yes, yes, all of you get the value, okay? Learn from your seniors mistake. Don't avoid calculation. In fact, you should welcome it. You should feel good that, okay, fine. It should be like a challenge, right? That let me quickly find the correct answer. Then only you will improve on calculation. Check from your back of the book. Is it correct? Whatever answers you're getting, you can check the back of a book 8.12 question. Is that the correct answer? All of you tried, right? Now I can solve. Should I solve? Now Ashutosh got 0.1 into 10 is for eight meters deviation, which is like 10 is for seven meters. 10 is for seven meters is like 10 is for four kilometers. That's like 10,000 kilometers. More he got. Okay, others are also getting the similar thing. All right, let me solve this. A rocket is fired from the earth towards the sun. At what distance from the earth center is the gravity of force on the rocket, zero. Mass of sun is given, mass of earth is given. And yes, so this is earth. The sun is vast. It'll be very, very big. But compared to the distance between earth and sun, it will size of the sun and earth will be like a point mass, okay? So rocket is fired like this into fine distance where the gravity of force on the rocket is zero. So I think this is a straightforward question. So let's say this is the distance at which the gravity of force due to earth equals the gravity of force due to the sun. So they will be equal and opposite, right? So all you have to do is equate them. So G times mass of earth into mass of rocket divided by mass of sun is given, mass of earth. Orbital radius is this. Okay, so let's say this distance is D. Distance from the earth is r's, right? So let's say this is D and that distance centered to this point distance will become r minus D, okay? So this should be equal to G times mass of sun into mass of rocket divided by r minus D whole square, right? So all you have to do equate this and mass of rocket will get canceled out, fine? So from here, you get the value of D. Even G is gone, fine? So D will be equal to r minus, okay? R minus D under root of M E by M S. So from here, just simplify it further, you get the value of D, okay? All of you clear? I think this should be something just straight forward. Let's see the next question. This one, 8.17. Yes, conservation of energy only. You're correct, 8.17. Anyone got it? Okay, so here all you have to do is to apply work energy theorem, okay? So a rocket goes from here, okay? It will return to Earth if its velocity becomes zero at some point. Let's say at this level, its velocity becomes zero. So rocket reaches here, its velocity becomes zero. So then it will come down, getting it, all right? So we need to find how far from the Earth. When somebody asks you like this, how far from the Earth, what do you actually find? You find the distance from the surface or from the center? When I ask you how far from the Earth, you should find the distance from the center or from the surface. You have to get the height, okay? You need to find this value, how far from the Earth, okay? From the surface, correct? Let's say velocity is given, let's say velocity is V, okay? Velocity V is given as five kilometer per second. So that you need to convert in meter per second. Find the velocity is 5,000 meter per second, all right? So everything else is given. All we have to do is substitute in this expression. W is equal to U2 plus K2 minus U1 plus K1, okay? Now, only gravitation is there for which you're considering potential energy. W is zero, even K2 is zero, all right? K1 is simply half M into V square, fine? What is U1? Potential energy on the Earth's surface, which is minus of G mass of Earth into mass of this rocket divided by radius of Earth. It is at a distance of RE from the center, fine? U2 will be equal to minus of G mass of Earth into mass of rocket divided by radius of Earth plus H. When you write the potential energy, you need to find the distance from the center, not from the surface, okay? The good thing is that the mass of the rocket get canceled away. It is independent of mass of rocket, all right? Using this expression, you'll get the value of H. All of you clear? Okay, let's move to the next question. This one. You get the expression, okay? Let's say, escape velocity is VE, okay? In terms of VE, can you get the answer? Let's say, escape velocity is V only. So V is the escape velocity. In terms of V, what is the answer? Again, you have to use work as a theorem, only same expression, same way, exactly the same approach will be there, okay? We're going to use this only, right? W is zero. Final potential energy is zero. It has gone to infinity, right? Very far away, okay? Now, K1 is what? Half M into V square, all right? U1 is what? Let's say K2 is half M into V1 square, okay? U1, in terms of expression, it is just GM into mass of Earth divided by, on the surface of Earth, RE, okay? So you can cancel out M, like this, all right? And you will get, are you saying that its velocity will increase at infinity? You're saying that it will have more velocity than what you have thrown with at the infinity. That should not be correct, right? The Earth will attract itself, sorry, Earth will attract the mass and it will decrease the velocity, the escape velocity. If it is V, it should be less than V, the final velocity. V1 square by two, that's fine. We also know that escape velocity, it's a good idea to remember the expression for escape velocity. Otherwise, you'll be, you know, deriving it again and again. This is two GM by RE, fine? So if we square it, V square will be equal to two GM by RE, where this V is three times the escape velocity, three times. So this is, sorry, I thought of something else, so. Let me write that, so this is three V square, fine? So V square by two is GM by RE, fine? So this two is gone, the two will come over here. So V square is equal to two GM by RE. So V1 square will be equal to nine times V square minus V square. So V1 is root over eight times V. So that is also equal to two root two times V. Yeah, yeah, yeah. Correct, I miss that. All of you clear about it? Should we go to the next one? This one. Brinda and Sri Ramya, you got the material, right? All three books you got, right? Okay. All right, anybody else got the answer? Again, you have to use the concept of energy only, nothing else. See, initial energy will be whatever is energy at that orbit. What will be the final energy, all of you? Final potentiality will become what? When it leaves the earth's influence, final potentiality should become what? Zero, right? And it just reaches the infinity, so final kinetic energy is also zero. So total energy initially is minus of G mm divided by two R, right? This is how we derive the total energy. We have done this earlier as well. Finally, the total energy should be equal to zero. Okay? So the change in the total energy is what is required. That is zero minus of G mm by two R. So this is G m into mass of earth divided by two R. This much energy is required. This is mass of satellite. Capital M is mass of earth. Okay, G, you know what it is. R is the distance from the center. So that is 400 kilometer from the earth's surface, right? So four into 10 raised to the power of five plus the radius of earth. So that is 6.4 into 10 raised to the power of six. This is the value of R. D e is not that, Sriram, it is a plus that. It is not negative, okay? Since initial energy is negative, final energy is zero. So difference is zero minus of this. Minus, minus becomes plus. So change in energy has to be positive. Then only you will give the energy to achieve the, achieve zero. You have, suppose minus 10, you want to become zero. So you have to add 10 over it, right? Fine, let's go to the next one. Other, let me create a hypothetical question similar to this only. Suppose you have two stars. This is mass M. This is mass 4M. Radius of that is R. Radius of that is 4R, okay? In the universe only these two things are there. The distance between them is 10R, all right? So now they are left. And once they are left, they will start moving towards each other. Fine, okay? You need to find with what velocity they will collide each other. What will be the velocity of each of these two stars when they collide? Solve it, whatever concept you want to apply, but nothing will be beyond what we have learned already. Anyone? Are you in between? Should I solve this? Shravan, are you telling me both of them will have same velocity? Suppose they collide over here, okay? So suppose they collide over here. So at this moment, we need to find their velocities, okay? Let's say that their velocities are, let's say this one is V1 and the velocity of that star is V2, okay? So I can conserve energy, right? I can apply the work energy theorem between the points when they were at a distance of 10R and now. So the work done can be written as U2 plus K2 minus U1 plus K1, okay? So this is zero. Now you are applying this work energy theorem for both stars together. Okay, so when you write counting energy, you need to write counting energy for both, okay? Potential energy will be anyway be there for both. So K1 is what? All of you quickly tell me K1 is what? When they're just started moving towards each other, K1 is zero, right? There was no counting energy. What will be U1? U1 is minus of G M into 4M divided by 10R. This is U1, okay? And K2 can be split into two parts, half M into V2 square plus half into 4M, mass is 4M now, that into V1 square. All of you understanding whatever you have done till now? Anything you want to discuss? U2 is what? Quickly tell me U2 is what? U2 is zero. U2 will be zero when both the stars are infinitely away from each other. U2 is when they are about to collide, they're center to center distance is what? The center to center distance is 4R plus R, 5R. Earlier it was 10R, okay? So U2 will be equal to minus of G M into 4M divided by 5R, okay? Now this equation has how many variables? It has V1 and V2, but you have only one equation. How will you get the value of V1 and V2 separately? Okay, that is one. If you take the system M and 4M, is there any external force on M and 4M? No external force, so you can conserve the momentum, right? Initial momentum is zero. Final momentum is M into V2 minus of 4M into V1, fine? So this is your second equation. The first equation is conservation of energy and the second equation is conservation of momentum. So here work energy theorem, sorry, work by energy chapter is used. So this chapter can utilize concept from multiple chapter. Even rigid body motions will be used at times. So we got V2 is equal to 4V1 from conservation of momentum. So you have to substitute the value of V2 in terms of V1 and get the value of V1. All of you understood this one? The only concept from the gravitation used to solve this particular question is the expression for potential energy. It's GMM by R, that's it. Everything else is from the other chapter. All of you understood? What is the doubt, Gaurav and Shushant? No doubts, right? Okay, now I will talk about few concepts which will be used to solve questions which are beyond your school level, okay? The first one is field, gravitation field. So I'll start with why we need this particular term or this particular concept, okay? So we are done with your school requirement, okay? Now I'm going beyond your schools. So suppose two point masses are there. Okay, two point masses are there. M1 and M2 are the masses and distance between them is R, okay? What will be the force on M1 due to M2? How much that will be? It is given by GM1 M2 by R square, simple, right? Now I tell you this now. Suppose you have a rod, rod of mass, capital M and length L, okay? And you have a mass over here. Let's say mass is M, okay? This distance is R. Can you directly get the expression for force between these two? Between the rod and mass, small m, can you directly get the value of the force? Just like you got the above expression, can you get it directly? Not possible, right? Okay, Shravan wrote something. Now Shravan, have you applied any fundamental while, okay? Do those who are answering, have you used any fundamental or you have just guessed it? It's a wild guess. It's just a guess, okay? It is not necessarily correct, okay? So if you consider the rod made of many point masses, let's say one point mass is over here, okay? The other one is there, all right? So we have multiple point masses at different locations. Suppose this is a point mass, which is part of the rod only. I am dividing the rod into different different masses. If this distance is X, okay? And this point masses mass is dm. The force will be equal to g into m into dm divided by X square. Now I can use this expression. This expression is valid only for point masses, okay? But in order to get the total force, this is small force df. In order to get total force, I need to integrate and find out, okay? Now there are two variables, dm and X. So I need to write one in terms of the other to get the value of total force, all right? Now the problem is that at times there could be rod, there could be some other irregular shape. There could be so many other different kinds of scenario, okay? So we need to define something which can be helpful for us to determine the effect of the distribution of mass. For example, what is the effect due to the rod at this point, okay? So basically what I'm trying to say here that the force depends on both the masses, right? Let's take an example of this only. Suppose you have two masses, m1 and m2, okay? The force between them is a function of both m1 and m2, okay? But I want to find a function which is only of m1. So basically what I'm trying to say here is that suppose m2 is there, so m2 will experience a force, okay? Why it experiences a force? Because of the presence of m1, okay? Because if you remove m1, f will become zero, fine? Now similarly if you remove m2, force will be zero because m2 is not there. But even if m2 is not here, there must be something happening over here because of m1, because of which if I place a mass over here, immediately it starts feeling of force in this direction, okay? So there must be something here which depends only on m1, okay? So basically I'm trying to define the effect of m1, okay? The force is also effect of m1 only, but the problem is that the effect of m1 cannot be a function of both m1 and m2, okay? So effect of m1 should be a function of only m1, okay? So I am trying to define a function which only depends on m1, okay? Now I will define a function, let's say, okay? Let me scroll. So the force is gm1 m2 by r2, right? So force divided by m2 is gm1 by r2, okay? So force per unit mass is a function of only m1, okay? So this particular thing, this particular expression, gm1 by r2, which only depends on m1, can be referred as field due to m1 at a distance of r, okay? m1 is point mass, okay? So this is gravitational field, all right? So if I define a gravitational field, if I know gravitational field is x at a particular location and if I place mass m, then all you have to do is multiply x with m, I'll get the value of force because field is force per unit mass, okay? So we have just now defined gravitational field because of a point mass. Is this clear to all of you? Any doubts? Please feel free to ask, okay? Don't hesitate. Is this thing clear? See, I want to define an expression which I can treat as an effect of mass m1, okay? The effect of mass m1 is also the gravitational force, okay? Which is gm1 m2 by r2, but that effect depends on m2 as well, okay? I want to define the effect of m1 as only the function of m1, okay? All right? So I am using the force expression only. I am using the expression of force only to define a function which depends only on m1. What I'm doing is I'm dividing the force by m2, okay? So force is gm1 m2 by r2. So if I define the force by m2, I get gm1 by r2. Now, this expression depends only on m1, okay? Fine? So this expression, we call it as field due to m1 at a distance of r. So force per unit mass is the field because of m1 at a distance of r. Yes, Shushant, you can say that. See, this thing will be more clear as you proceed further, okay? It will not be clear. It will not be that everything will be clear right now itself, but I want you to be, at least be, appreciate the fact that we need something which depends only on m1, okay? Otherwise, what will happen is that every time, suppose there is only, let's say, or you can say that suppose black hole is there, okay? Suppose there's a black hole, okay? I want to find out what will be the effect of this black hole at the neighboring places, fine? So in order to find the effect, you don't put a mass over here and then try to find the force with which black hole will pull it, right? So if you know the mass and radius of this black hole, you can find the gravitational field of the black hole, okay? Which only depends on what the black hole is, what is the size and dimension, what is the mass of that black hole. So there has to be a function which only depends on the mass which is surrounding. Okay, now let me proceed further. It will be more clear, okay? Otherwise, we'll just keep on talking about the same thing over and over again. So you'll appreciate quickly why we need something like gravitational field, okay? Let's say you have a rod, okay? You have a rod which has mass M and length L. Mass is distributed uniformly over the length, okay? Now I want to find out the field at this point because of the rod. That point is at a distance of, this is at a distance of, let's say D, okay? At this point, what is the field because of this rod, okay? So once you get the field, let's call it as E, the field multiplied by the mass will be the force on the mass because the field is force per unit mass, right? But the field which you have defined, G M by R square, this is the field only for the point mass, but the rod is not a point mass, okay? So now try to attempt this one. See, it will be wrong to say that field is same as acceleration because there can be other forces also. Net force may not be equal to just gravitational force, okay? Others, okay, let me do this now. So we have the expression for the field due to a point mass, okay? But then in front of us, there is a rod, all right? So what we'll do is that we can divide this entire rod into point masses, different different point masses. So let's say there is this small length DX which can be treated like a point mass and that DX, let us say that is at a distance of X from the point where you're trying to find the field, okay? So this DX can be treated like a point mass, right? Total mass is M. Can you tell me what is the mass of that DX? DM is what? If total mass is M, the mass of this DX element is what? DX length is what? So mass per unit length, which is M by L into DX, this is DM, fine? So the field over here because of DX, because of DX, the field over here will be what? I can directly use this formula because DX is a point mass. So the field, let's call it as DE because that is only because of the DX, right? So this is G, DM, which is this divided by X square, okay? So this can be written as GM by L into DX by X square, all right? Now tell me one thing, is the field scalar or vector? Field is a scalar or a vector? See, listen how it is. The mass into field should be force, right? Force is a vector, mass is a scalar. So field has to be a vector. Then only scalar times vector is a vector, okay? So the field has to be a vector only, okay? Now tell me what is the direction of the field? Which direction the field will be? Direction of field is same as the force would be, okay? Because mass into field is force. So it is along this direction. Now the thing is whichever DX element you take, wherever you go in this rod, all of them will have the same direction, the electric field, sorry, the gravitational field. So the total field will be just integral of this. And the limits will be from where to where? Limits will be from D, it starts from D, the rod starts from D and it goes from D to D plus L, okay? So the total field E will be equal to minus of GM by L, integral of one by X square is minus one by X, minus I have taken outside. So this can be written as G times M by L, one by D minus one by D plus L, okay? This is the field, okay? All of you clear till now? Direction of force, direction of field will be direction of force which a point mass kept over here will experience. It will experience a force in this direction. That is the direction of field. Is this unclear how the field due to the rod we have derived? Field due to rod is this. So once you know the field, the force at that point, suppose if field is E, if I keep a mass over M over here, the force will be what? Force will be M into E only, simply, okay? So it's a good idea to derive the expression for field for different kind of shapes and sizes so that if you know the expression for field at a point for a particular geometry, if you place a mass M over here, the force will be simply mass M into field, okay? The unit for the field will be Newton per kg. Any doubts, guys, on this? Whatever we have done just now in front of your screen, is there any doubt? Ashutosh, there is no logic as such why it should be like this, okay? We'll not assume that. Yes, we can derive similarly for other objects and that is what I am right now will be asking you to do. I'll be asking you to do that only right now. Okay, so let's try to find out. Did you all understand the utility of finding the field? Once you know the field, the force at a particular location, the magnitude of the force at a particular location will be straight forward. You just have to multiply mass with that field to get the force, okay? Fine, let's try to do the further calculations now. Suppose you have, okay, it will be mathematically a little bit involved. So, okay, let's do the simpler one first. You have a ring, okay? You have a ring. Let's first take this type of ring, which is in the plane of your screen, this ring, all right? Can you tell me what is the field at the center of this ring? Total mass is m and radius is r. What is the field at the center? See here, you can draw lines like this. Now you can see that the field because of this point mass, which is located over here, will be in which direction? This direction, right? Due to this point mass, the field will be in this direction. These two fields will cancel out each other. Similarly, anywhere you'll see, if you try to find the field, there will be a point, diametrically opposite point will be there for which the field will be equal and opposite. So all these fields will cancel out and total field will come out to be zero. Getting it? So due to symmetry, clear? So if you place a mass at the center of the ring, it will not experience any gravitational force because field is zero. So mass into field, which is force, that is also zero. Clear all of you? Okay, now try to find out the field because of the sector of the circle. Okay, let's say you have, this is a center, let's say radius is r and this angle is theta. Okay, mass is m, total mass is m. It's a ring, sector of a circle, sector of a circle. What does it mean? Oh no, sector of a circle means part of disk. So it's an arc, okay? It's a circular arc which makes angle theta at the center. Find out. Okay, Niranjan has got some answer, others. Niranjan, that is dimensionally incorrect. Okay, let's see how we can do this. First, make full use of symmetry, okay? Divide it into two equal parts and then take this small mass over here and take another small mass symmetrically located over there. Okay, you will see that the field because of these two will be in this direction, okay? So these two point masses will have field along these two point masses direction. Okay, let's say this is dm. This is also dm, all right? I am taking this as phi, okay? So this is also phi. I am taking symmetrically located dm, right? Let's see whether your answer is correct. These are the two fields. So you can see that horizontal component of these two fields get canceled out, okay? Only vertical component remains. Integration is a scalar addition. You can't have a vector addition when you integrate, okay? So that is why you can integrate only components along vertical and along horizontal direction in this case. Okay? So the horizontal component of the field get canceled out and every point mass you consider over here, you'll have a symmetrically located point mass at the other part of the ring, other part of the arc. So your field will get canceled out horizontally every time, okay? So you'll have only net field along y direction, okay? So that will be e times cos of phi, all right? So let's say I'm considering small field due to the small mass dm, okay? So if I take two masses at a time, the total field dEy will be two times dE cos phi. So dE cos phi because of this and dE cos phi because of that. dE is what? G times mass is dm divided by radius square, that into cos of phi. Now the problem is that now you have two variables, m and phi. So I need to write down mass in terms of phi. Okay? So if you zoom it a bit, you'll see that this dm will occupy the small length of the arc because mass is distributed on the arc, right? So if you consider a small mass, it will occupy small length also. So let's say this length is dl, okay? So dm will be equal to mass per unit length which is m divided by the arc length is what? R into theta, right? So this into dl is dm, okay? And from the definition of angle, dl should be equal to R into d theta. All of you are understanding, right? Is there any doubt? So dm is why you are retracting the messages one by one? It's okay to be wrong. So m by theta into d theta. Actually it is d phi, d phi, m by theta into d phi, okay? Because I'm taking phi as a variable, okay? So this will be equal to 2g dm which will be equal to m divided by theta that into R square to cos of phi d phi, okay? So total field will be integral of this. Are you able to see the screen? Are you able to see now? Now tell me the limits of integral from where to where should I integrate? What should be the limits of integral? Zero to theta by two you have to integrate, okay? Because when you say two times you're, every time you're considering one mass here, one mass there, okay? So if you go from here to here, you're automatically considering corresponding masses from here to here. You're taking two masses at a time, okay? So you'll integrate from zero to theta by two. But if you don't do two times, then you can integrate from minus theta by two to plus theta by two, okay? Because when you take two, multiply with two, you're saying that I'm taking two masses at a time. So the gravitation field will be integral of cos phi will be sin phi, okay? So gravitation field will become equal to two times gm divided by r square theta into sin of theta by two, right? So if I keep a mass m over here, if I keep a mass small m over here, the force experienced by the small m will be m into e, okay? Once you know the field. Any doubts, anything you want to discuss? Quickly tell me, this field is along j direction, right? So it will be j cap, it's a vector. See, even if you're understanding 60%, that is good enough right now to start with. So nobody will understand everything right from scratch, if you're listening this for the first time. So all you have to do is after the class, sit with whatever we have done today and you'll be able to understand everything. You did not understand how we took the limit of integration. There you go. This dm you have to take here, then here, then here, everywhere you have to take dms, then only you have to cover entire arc, right? So that is why you need to go from this point to that point along this line to cover the entire mass, okay? Now what you're doing is, when you're considering mass over here, you're considering another mass over there as well. So by the time you travel from here to there, you have already covered automatically all these masses, okay? So the limits will be whatever angle this point mass over here makes to whatever angle this point mass makes from the vertical. Total angle is theta, this angle is theta by two, this angle, angle with the vertical because this vertical line divides it into equal parts. So this will be theta by two. So dm starts from zero, it goes to theta by two. So theta, the angle phi starts from zero and goes to theta by two, okay? Clear now? You have a ring, okay? You have a ring, this ring, half of it is inside your screen and half of it is coming out of your screen, okay? So basically the plane of the ring is perpendicular to your screen, okay? This distance is z, from this point, the center of the ring, the distance is z, okay? The radius of the ring is r, mass is m, okay? You need to find field at this point because of the ring. Point is along the axis of the ring. The point is along the axis of the ring, okay? The ring is perpendicular to the plane of your screen, okay? All of you try it, should I do it? I'll do it now. All right, so let us do this now. So the field because of this, let's say dm will be along this line, right? This field is, let us say, dE, okay? This angle is theta, let's say, okay? Now, can I say that angle theta will be same for whatever mass you take along the ring? Along the ring, whichever point you go, angle theta will be same. Is this correct statement, okay? And is it also correct that the component, let's say dE is component along this axis. Along the y axis will be dE sine theta. This one will be dE cos theta. Now, if you join this one, okay? That once will also have a same magnitude of electric field, sorry, the gravitational field. So it will have a vertical component downward. So the only component that remains will be along the axis. Is that also correct? The only component dE should be equal to dE cos theta because sine theta component gets canceled anyways. When you sum it up, only component which is along the axis will remain. Is this a statement correct? Okay? So dE is what? dE is G times dM divided by this distance is square from here till this point, okay? That is Z square plus R square, okay? This into cos of theta. Now, cos of theta can be written as Z divided by R square plus Z square raised to power half. Till now, you're comfortable? Till now, whatever we have done, you're okay? Okay. Now you'll see that wherever you go, wherever this dM goes along the ring, Z is constant, R is constant, G is anyway constant. So all this expression comes out of integral. This thing is equal to GZ to Z square plus R square raised to power three by two, okay? Integral of dM. And integral of dM is total mass only, okay? So zero to capital M, total mass, fine? So this will be equal to G times MZ divided by R square plus Z square raised to power three by two, okay? And it will be along the axis towards the center. What happened to cos theta? They go, drop a perpendicular from here, okay? This will become right angle triangle, this one, okay? So cos of theta in this right angle triangle will be what? Z divided by habitanous, habitanous is this, which is using Pythagoras theorem. You'll get a Z square plus R square raised to power half. All of you understood how we get the gravitation field because of the ring. Now if you place a point mass at that point, let's say M you place over there, the force will be M into field because field is force per unit mass. Shall we take up the next one? Next situation is this. You will learn a lot of integration before starting integration in mathematics itself. Total mass is M for the rod, okay? Mass is M of the rod and here is a point. This point is at a perpendicular distance D from this straight, this thing, mass, straight wire mass, okay? What else is given is the angle theta one and this angle is theta two, okay? So basically we have been given the angles, the two ends of this rod makes with this point and the perpendicular distance D, okay? You need to find gravitation field E. So none of this will be as in school, okay? Don't worry about it. Just focus on whether you learned something new today. Find the gravitation field at point P due to the mass M, which is a rod. Should I solve it? Atm stick at the integral. What is Bharat saying? Atm stick, okay? I am stuck. Okay, so let me solve this now. All of you focus. It's not very easy, okay? So don't worry if you are not able to get it. It's all right. Focus on how we are doing it, okay? Again, the usual process, I will find out the small mass along the rod, okay? Are able to see the screen? Let's say this is, see the thing is the length of the rod is not given, right? Just the values of theta one, theta two is given. So in terms of theta, let me try to find out the integral over here, okay? So let's say this is phi, okay? This, the small angle, let's say, this can be written as d phi, okay? This, this if I say that this is equal to let's say x, okay? This will be dx. dx is the length of a small mass, which I'm considering, okay? This is a perpendicular sense, d, okay? Now, field because of just this dm mass is what g into dm divided by this length. Let's say this length is r, r square. This is de. All of you clear? This is de. Now, there is no direct symmetry over here. So we'll not try to look for it unnecessarily. So we'll just integrate dex. This will give us ex, okay? And e y will be integral of dey. This will give us field along y direction, okay? Once you get these two field, total field will be root over ex square plus e y square, okay? And tan of angle it makes with the x axis will be tan theta equal to e y by ex, all right? So let's try to first get the integral dex. Dex will be g into dm by r square into cos of phi, okay? Now there are three variables, dm, r and phi. So everything let's try to write in terms of phi only, okay? So r is equal to d divided by cos of phi, isn't it? Is this incorrect? R is d by cos phi, okay? Or you can say that this is d sec of phi, all right? And dm, dm is equal to, let's assume the total length is l. So m by l into dex, okay? Find mass per unit length into dex, right? Now x is equal to what? x is equal to d tan of phi, d into tan of phi. So dex by d phi is d into sec square phi, right? So from here, I'll get dex is equal to sec square phi into d into d phi. This d is making, d is not a differential, okay? So that is why let's take it as capital D. This is capital D, then it'll be easier to visualize. This is capital D. Anywhere else, D is there. This one. This is capital D, okay? So this is dex and this is the expression for dex. So dex will become equal to g times, dm is what, m by l into dex, right? So m by total length into dex, which will be sec square phi d into d phi, this is dm, all right? Divided by r square, r is d tan theta. So it will become dex square sec square phi, all right? And this into cos of phi. We're able to see the screen. So dex, so dex will be integral of this, integral of that, okay? It has become a little involved, okay? But sec square theta goes away. So this will become equal to, one d is also gone. This will be gm by dl integral of cos of phi d phi, fine? And the phi value goes from minus theta one to plus theta two, okay? So ex integral of cos phi sine phi, right? So sine of theta two minus of sine of minus theta one. So you'll get gm by dl or ld sine of theta two plus sine of theta one. All of you understood how ex has come? All of you understood how ex has come? L is the length of the rod, let's say length is given. If it is not given, you can always find also in terms of d and theta, but let's say l is given. All of you clear? Okay, can you get the value of ey now? Get the value of ey, how much is ey? Since ey, rather than this cos phi, sine phi will come, that's the only difference, yes or no? That is the only difference. So integral of sine phi d phi will come, limits will be exactly the same, okay? So ey will come out to be equal to yes or one. gm by dl cos of theta one minus cos of theta two, okay? This is what you get ey. Looks like I'm teaching mathematics today. I can feel that. Okay, so tell me if the rod has infinite length, if rod has infinite length, what will the value of theta one and theta two? Theta one is pi by two, theta two is also pi by two, right? The angle theta one and theta two are increasing. Yes, as you increase the length, theta will become more and more and slowly and slowly this line will become vertical. Okay, so this line will become perfectly vertical if the line, this rod is infinite length, okay? So for infinite length wire or infinite length rod, ey is zero and that has to be because then it becomes symmetrical about x-axis, okay? So theta one equal to theta two equal to 90 degree. So this ey becomes zero and ex is theta two plus theta one become two, pi by two, pi by two if you substitute. Two gm by dl, okay? So like that you can get the value of field because of the small length of the small mass length, okay? Understood all of you? Okay, now do one thing. Okay, not this. You have a disk, okay? Mass is m and radius is r, okay? Shushant, not now but slowly and slowly you'll automatically remember. Don't just wake up every day in the morning and start mugging up, okay? You should not be doing that but you will remember it automatically if you solve a lot of questions of J advance level. You're not able to see the screen. This distance is z, okay? Along the x-axis, distance is z, okay? Now, again the screen is gone. So find out the field at this point. At this point what is the gravitational field because of the disk of mass m and radius r, okay? The hint is that you can consider the entire disk as if it is made up of concentric circles of different radii. The point is above the plane, yes, along the x-axis. I can't draw 3D on the screen so that is why it may not be very clear but it is along the x-axis. Yes, yes, you should consider the entire disk as if it is made up of multiple concentric rings. So for a ring, you already know the expression, right? Use that expression over here. For a ring, we have derived the expression as G z divided by, this is for the ring. Yes, let me know should I solve? Okay, now let me solve this. Small field, small amount of field because of a ring of radius r having width dr is at this point will be equal to G times z divided by, now tell me what will be capital R? What I should write instead of this? Small r, right? Because that is a ring. So small r square raise to the power 3 by 2 m is dm, okay? dm is the mass of the ring of width dr, okay? So total mass is capital M that is distributed in an area of a circle of radius capital R. So dm will be what? Mass per unit area that is m divided by pi r square into the area of the ring which is 2 pi r into dr. All of you clear, right? In case of any doubts, immediately ping me, okay? Don't wait. 2 m by r square into r dr, this is dm, okay? So small amount of field will be G times z 2 m by r square all right into r dr, yes. Not able to see the screen. I think we didn't have break today, right? So continuously three hours, maybe YouTube doesn't support, okay? I think today was very heavy, a mathematically very heavy class. A lot of concept we could discuss. Anyways, so only 10 minutes. Oh, it's already 120, okay? We'll end at 130, all right? So dE will be equal to 2gmz by r square r dr divided by, so I've taken all the constant terms outside of the integral. Let me take two inside of this, okay? So the value of r, the value of r, small r will go from where to where? Zero to capital R, right? You need to cover the entire disk. So the rings will have radius starting from zero. It'll go all the way up to r, capital R, right? So let's say r square is t, okay? So 2r dr by dt is dt, no, it is one, good, this is one. So 2r dr will be equal to dt, okay? Let me put z square plus r square itself as t. z is a constant, so when you differentiate, it will come out to be like this. So overall integral E will become equal to gmz divided by r square, 2r dr is dt, okay? And z square plus r square is t, so it is t raised to power three by two, okay? But the limits will be changed. Now when r is zero, t is z square. So this will be z square, and when r is capital R, t is capital R square plus z square. Understood till now? So this will come out integral of t raised to power minus three by two dt. How much is this? This integral is t raised to power minus three by two plus one divided by minus three by two plus one, okay? So this will come out to be two by, this will come out to be one divided by root t and the denominator will have, this is minus two by root t, right? Minus two by root t. So this is gmz, okay? Don't gmz by r square integral is minus two by root t. Integral is minus two by root t. Is there any sillier that we have done? Just quickly check. I think not. Two r dr is dt, no? Direct, this is gmz by r square, one by, instead of t put z square, so it becomes z minus two will come out, let me take two outside. z square plus r square, okay? So if I take this z inside the bracket, I'll get two gm by r square, one minus z divided by root over z square plus r square. Fine, so this is the field. This is the field at a distance z along the axis of the disk, okay? So if you place a point mass at that location where field is e, the force will be m into e, okay? Isn't that cos theta, z divided by z square minus r square. Theta, where? Theta, it depends how you define theta. If you put it like this and that, z divided by this distance. If you say that this, if this angle is given, that is very nice. If this is theta, okay, if theta is given, z is not given, then sorry, z should be given. What should not be given, z is? Anyways, if theta is given, you can write in terms of theta also. Yeah, z need not be given. If theta is given, z need not be given, then e will be equal to just two gm by r square into one minus cos theta, okay? And one minus cos theta is two cos square theta by two. So this will be four gm by r square cos square theta by two. But theta should be given. Okay, usually this formula is used. All of you got it. All right, so similarly, we need to also define, say this is field, okay? So another one-hour class is remaining where we define something called gravitational potential. Gravitational potential. We'll do it next class. So just like field is force per unit mass, gravitational potential is potential energy per unit mass. So we'll be defining potential, similarly for ring, for disk, for wire, for sphere, for sector, all that we'll be doing it in the next class. Niranjan, how much integration practice you need? I think we have done enough, right? We have, like in bits and pieces, we have done a lot of integration practice already. We have already done one, I think two-hour class on integration sometime back. And for, you were there in the bridge course where we did integration. One minus cos theta is two sine square theta by two. Yeah, yeah, yeah, yeah, yeah, yeah. Small correction, guys. This is sine. See, we can do one thing. See, the thing is class 11th is going to get over soon, and we are thinking of starting class 12th. Class 12th will start, I think, second week of March, roughly. So then we can discuss exclusively, we can spend one full week, one or two weeks, we'll spend only on application of integration in physics. Okay? Right now, anyway, class 11th is going to get over. Yeah, yeah, March. Second week of March. So after your class 11th exams get over, after three, four days, we'll start class 12th. Isn't that exciting? Have you seen the planner, all of you? Go through the planner, see how it is. We'll tweak a little bit here and there. All right, guys, so that's it for today. Thanks for coming. We did a lot of advanced level stuff today. I hope it did not scare you, but at the same time, don't take it lightly. Make sure you spend one or two hours reading what our concept we have discussed today because nobody gets it 100% while sitting in just in a class, right? You need to struggle at home, solve a few questions, then only you'll master the concept. Okay? Fine, so hopefully next Tuesday, we'll be having, we'll be having classroom discussion, not an online, unless somebody calls for a band. Okay, bye-bye, see you. And like the YouTube link, okay? In the channel, hit the like button and subscribe.