 So, let us take a case of spherical shell, we are going to discuss how you deal with case of spherical shell, because this is very common. See how nicely I draw the sphere, can you draw such sphere. So, let us say that the radius of this spherical shell is r and the mass m is uniformly distributed on this spherical shell, we need to suppose find out the force on a mass, let us say a point mass is kept over here inside somewhere, let us say point mass is m, we need to estimate the force between of course, you cannot derive exactly, but can you estimate the force between small m and capital M and its direction, small m is inside capital M, 0, do you know it as a fact or you are if you are applying some logic to arrive that, I want you to apply logic, I do not care about the final answer, so try to arrive at answer 0 by applying some logic. Logically it is 0, draw a line like this that passes through small m and cuts the entire spherical shell. Now, take one mass of this entire spherical shell this side and one mass on that side. So, you will see that this mass will get attracted towards that and it will get attracted towards this, focus here, these are the two forces because of this point and that point, so the vertical component of forces will cancel out themselves, so only force that remains is horizontal, so resultant between this and that will be horizontal in this direction, that is correct. Now, similarly you will see that if you take two elements on the left side of this sphere, the resultant of these two will be coming off in this direction horizontally. Now, the left hand side the mass is more, but then the distance of that mass from this small m is more, so the force will be weaker because it is further away, the mass is further away, but they are more in number and this side the masses are closer to this small m, but they are less in number. So, net net what will happen is that this force will cancel out that force, so the net force becomes 0. So, remember this if a mass is inside this spherical shell, the force between the mass and the spherical shell will be 0, no matter where it is placed inside this spherical shell the force will be 0 always. So, this is case 1 spherical shell m is inside. Now, case 2 spherical shell m is outside, for outside will not get into the visualization will quickly, so m is here, small m is here it is given that it is at a distance of r from the center, the radius of the shell is capital R, the mass of the shell is m. Now, in this case when the mass is outside this spherical shell, in this case the entire capital M behaves as if it is a point mass located at the center. Now tell me what will be the force between this and that, quickly tell me this should be equal to what. You can directly use a universal law of gravitation, gm1 m2 by r square, since this entire mass behaves as if it is located at the center only. So, it will be the scenario is like this only, capital M is over here and small m is over there and distance between them is r. So, it will be behaving like this small m and capital M, ok, distance is r. The force will be g times mm by r square how can I, is it clear? Clear, right? All of you understood this thing, you understood these two cases, all of you those who have not typed in the answer please tell me, ok. So, we are going to use these two cases to analyze this solid sphere now. Guys, let me tell you one thing, if you do the gravitation chapter properly, you know, comprehensively if you do each and every part of it, your class 12th will become a cakewalk. Anyways, one of the lengthy chapters, ray optics is now in class 11th only, your class 12th will become very very simple and because the gravitation resembles very closely with electrostatics, ok, which is considered to be the difficult topic in class 12th. So, if you do gravitation properly even electrostatics becomes straightforward to you, then you know class 12th will, very very simple class 12th will be for you, ok. So, next year we are planning to finish the entire curriculum by probably mid of September or at max starting of the October itself, ok. But then we will be running, you know, problem practice sessions every day there will be classes after mid of October. So, if you do gravitation properly, the pace of class 12th will help you to align yourself because we need to finish the syllabus in October since J mains is now happening in January, ok. So, gravitation even though is a small topic in class 11th and I can finish the gravitation school level very quickly, just couple of hours. But then I do not want to do that, I will be talking about gravitation in much greater detail, I will be introducing gravitation potential, gravitational field and all that I will be introducing and will be doing derivations, all that will be very useful when you go to the class 12th, ok. Alright, so now let us talk about this solid sphere. This is your solid sphere, ok. The mass is there everywhere, ok, starting from center to the periphery, the mass is distributed in the volume, ok. Total mass of this sphere is capital M, solid sphere, alright. The radius of this sphere is capital R, alright. You need to find out what is the force of attraction between a mass that is kept at a distance of small r, let us say this small m is kept at a distance of small r from the center. So, between small m and capital M, what is the gravitational force of attraction, ok. The hint is, the hint is you can consider entire sphere to be made up of shells, entire sphere to be made of spherical shells of different radius, ok. So, like this the hollow spheres are stacked, like that you can consider the entire scenario, ok. So, that is the hint. Now, try to get the answer. The gravitational force between small m and the capital M is what? Small m is kept at a distance of small r and small r is less than capital R. That is absolutely dimensionally incorrect. All of you take your time, ok. You have about a minute or a half, minute and a half, ok. Another hint is that you consider an imaginary boundary like this, ok. This has a radius of small r, ok. And if you are considering the entire solid sphere to be made up of the spherical shells, then all the shells which are between the radius 0 to small r are inside this small m, ok. And all the shells that are between capital R and small r that will be outside the small m, ok. Now, tell me one thing. Shells between small r and 0, ok. I am talking about this, this region, ok. That will behave like what? That will behave like, like a what? Behave like a point mass, right, because that small m is outside all the shells, ok. Now, it will behave like a point mass, but what is the mass of this zone? That you need to find out, ok. It is not capital M, right. So, the mass of 0 to r, this you can find out by multiplying density with the volume, density into 4 by 3 pi r cube will be the mass, let us say m 1 which will be inside small m, ok. And density is what? Density is mass divided by total volume 4 by 3 pi capital R cube, this into 4 by 3 pi r cube, this is m 1, right. So, m 1 will come out to be capital M times r divided by capital R whole cube, ok. Understood all of you? So, this mass will behave as if it is located at the center for small m, alright. Is it clear to all of you? This is very important, so that is why I am doing it slowly. Tell me any doubts? Now, the mass that is between, mass between small r and capital R, how it will act for small m? Force because of that will be what? Because of this mass, what will be the force on small m? How much that will be? It will be 0, right? It will be 0, ok. Because the small m is lying inside all those shells between radius small r and capital R, ok. So, I do not need to even bother about that. I just need to find out what is the gravitational force between m 1 and m that is the total force between capital M and m because the other mass is not exerting any force. So, total force, now this mass will be as if located at the center, right. So, this force will be g times m 1 into m divided by r square, fine. This force can be written as g times m 1 is m times r by r whole cube into small m divided by r square. So, this will be equal to g times m by r cube times small r. Why I have put that in the inside the bracket is because this bracket term is a constant, ok. So, the force inside the solid sphere is directly proportional to the distance from the center. So, as you move away from the center inside the solid sphere, your force will increase linearly. Are you getting point here? If you plot a graph between the force and distance from the center, you will get a straight line passing through the origin. At the center force will be 0. This is the point where small r becomes equal to capital R, ok. I will take a pause here. Tell me if you have any doubts till now. Anyone of you have any doubts? Ok, great. Now, can you find out the same force in terms of density? Suppose rho is given. Suppose density rho is given, mass is not given. Then what will be the same thing? I want to find out the force, I want to find out the force between the solid sphere and small m, ok, where the small m is located inside the solid sphere of density rho. Mass is not given, density is given. Density is rho and radius is r. You need to find the force between small m and this sphere of density rho and r. Let me know once you are done. This is not part of any theory. I am doing it because you will encounter such scenarios many a times when you solve the numericals. Same approach, of course, same. A y approach should be changed. It is a same numerical, right, same thing. So, now m1 is rho into 4 by 3 pi r cube, alright. So, the force is g times m1 m divided by r square, alright. So, when you substitute the value of m1, you will get 4 rho 4 by 3 pi r cube into m that into g divided by r square, fine. So, r square you can cancel out from the numerators. So, 4 will be there now. This 4g rho, where is pi Shoshant, so pi is there, pi m into r, ok. So, it is a good idea to remember these direct expressions. So, it will save lot of your time when you solve numericals. Clear all of you? Now, let us move to the next topic. Next topic now will be very straightforward to you is variation of acceleration due to gravity. Can you quickly find out what is the value of, what is the value of acceleration due to gravity at this surface of the earth? Mass of earth is m e and the radius of earth is r e. Mass of earth, let us keep it as m only. Can you quickly get the value of g on the surface of earth? How much it is? Alright, divide by 3. Yeah, divide by 3. Yeah, yeah, 4 by 3. Sorry for that. It will be 4 by 3. So, the gravitational force between the mass on the surface of the earth and the earth will be given by this, gm into small m divided by r e square. Why? Because mass is kept on the surface of the earth, earth being a solid sphere, you can safely claim that the mass is outside the sphere. So, this entire mass of earth will be behaving as if it is located at the center of the earth. So, this is the force between a mass kept on the surface and the earth. And if acceleration due to gravity is g, this force should be equal to m into g. Again, it has to be m into g. Now, using this, you can find out that acceleration due to gravity on the surface is g times m divided by r e square. So, you must remember this expression also. It is used many a times in problem solving, those higher-level problem solving. So, this is on the surface. So, acceleration due to gravity is given in terms of gm and r e square. In fact, you can use this expression many a times you will get, for example, when you solve a numerical, you will get terms like gm divided by r e cube. Suppose this term you get while solving numerical, any random numerical. So, you can say that this term is equal to gm by r e square multiplied by 1 by r e. And you can just substitute this as acceleration due to gravity g which is 9.8. This entire thing is nothing but 9.8 divided by radius of earth. So, if you remember this, it will help you in calculations as well. So, it is a good idea to remember and try to use it every now and then. Now, let us see how g will vary if you go inside the earth and if you go, if you move above the surface of the earth. All right. So, let us take one by one. So, this is all we will be doing. We will be just talking about variation of gravity, acceleration due to gravity today and after that we end the session. So, first is variation of g with, yes Gaurav, you can do that. Let us say that you have moved up a distance of edge from the center. So, total distance, sorry, edge from the surface. So, if radius of earth is r e, then the force between this m and the mass of the earth, between the earth and this small m is what, can you quickly tell me? What is the force? This mass is outside this solid sphere. The solid sphere will behave like a point mass located at the center, isn't it? So, the force will be g times m m divided by r e plus h whole square. Fine, distance from the center is r e plus h. So, this is what the universal law of gravitation is. So, I am using that. And if the accession due to gravity at that point is g, if it is g, then this force should be equal to m into g only. So, from here you will get the value of g as g m divided by r e plus h whole square. Right? You can further simplify it by taking r e outside. So, r e square 1 plus h by r e whole square. All of you comfortable till whatever I have done? Any doubts till now? Now, you will see that this g m by r e square, this g m by r e square, this you can write as accession due to gravity on the earth surface. So, that let us say is g naught. So, g is equal to g naught 1 plus h by r e whole square. Getting it or you can write it in a more concise way like this. g is equal to g naught 1 plus h divided by r e raise to power minus 2. Understood? Now, till now we have not approximated anything. This is the exact formula. This is the exact value. Now, if h is very less compared to radius of earth, only then you can use binomial approximation. What is binomial approximation? I think we have already encountered that many a times. This kind of expression can be written as approximately 1 plus x into y if mod of x is very less than 1. It is like 0.001. So, this h by r e will be very less than 1 close to 0 only when h is very less compared to r e. So, if this is valid then only you can write g is equal to approximately g naught 1 minus 2 h by r e minus 2 will come inside the bracket. But always remember this is an approximation. This formula is an approximation. Suppose I asked you to find out accession due to gravity. If I asked you to find out accession due to gravity at height equals to r by 3, you cannot use this because height is 30 percent of the radius. Then which formula you should use? You should use this one. If I tell you height is let us say r by 100 then you can approximately use this binomial approximation. But if I tell you h is r by 3 or r by 2 or 2 times the radius like that if I tell you then you cannot use this approximated formula. This is an approximation only when height is very less compared to the radius of the earth. And one more thing you should remember always and be cautious is that h is the height from the surface. The distance will be h plus r e. So, many times I have seen students making that error. Is it clear to all of you variation of g with height? Is it clear? How g varies with the height? Now let us talk about variation of g with depth. You are going inside the earth. So, suppose this is the earth and you have gone inside the earth. Now you are measuring the depth. You are not measuring the distance from the center. You are measuring the depth. So, suppose this is the depth. Depth is let us say d from the surface this is the depth. And the radius of the earth is r e. Can you first find out the force between the earth which has mass m and radius r e and this small m? What is the force? I think we have just derived such scenario. We have derived it to be g m capital M divided by radius of earth cube into small r. We have derived it already. All of you remember? Now what is r over here in terms of d? r is radius of earth minus d. So, I can also write this force as g m m. Let me first equate this to m into g. This is the force. If accession due to gravity is g at that location, this can be written as m into g. By the way, when they will ask you in school to derive the variation of accession due to gravity with the depth, you cannot start from here. So, whatever we have done to arrive at this expression, you need to do that and then start from there. Since I have done already, so I am starting from this point. When you equate it, small m will get cancelled away. So, g will be equal to g to capital M divided by r e cube. So, r e cube can be written as r e square into r e. So, this divided by r e. Why I have written like this? Because this bracket is nothing but accession due to gravity at the surface. So, g naught into r e minus d divided by r e. So, that is 1 minus d by r e. So, this is how accession due to gravity will change as you go inside the earth by a distance of d from the surface. This is the variation of accession due to gravity from the surface if you go inside at a distance d. And remember, there is no approximation when we talk about the depth. It is not an approximation. No approximation. Is it? It is exact. So, you do not need to worry about when you talk about accession due to gravity variation with the depth. But when you talk about accession due to gravity with height, that final formula is an approximation. So, if you go too far away from the center, always try to use the exact one not the approximate one. And it is a good idea to derive every time you encounter such scenario in case you have any doubts. All right. So, that is it for today, guys. So, we need one more three to four hour session to complete the gravitation. I will try to take it on Sunday and definitely we will be finishing it up before your UT. So, we will be doing the gravitation next session that is on Sunday. So, what you can do is, since your physics UT is on Monday, so you should prepare everything till now whatever we have done. We have done waves and gravitation till this variation of accession due to gravity we have done. So, when you come to the next class on Sunday, probably, then you should be done with whatever we have done till now for your UT, so that whatever new we will be doing on Sunday, you have sufficient time to master that as well. Okay. Yeah, Sunday you have came on, I know that, but came on is only for three hours. All right, guys. So, see you soon. Bye-bye.