Gravitation - JEE Advanced CC

Good day, Hello everyone. Can you hear me? All of you? >> Yes sir. >> Yes sir. Okay. Yes sir. >> Should we wait for others? Let's wait for 1 minute. Others will join. All right. Hello. Fine. So, let us start today's session on problem solving of gravitation. Okay. So we there will won't be any theory discussion uh directly while discussing the questions there can be some relevant theory discussion. Okay. So that is the idea and uh I have taken questions uh some of the questions are which you might have seen before but majority of them are little tricky in nature. So if you are not able to solve many of them doesn't matter okay because I have ensured that many question you should not be able to solve so that we learn from this session okay so solve this question. You can DM your answers. Okay. Shall we discuss now or should I wait? You can speak up. You don't need to be on mute. >> Yes, I think we can discuss this. Okay. See uh there is a satellite. Now the satellite is not kept on earth. So satellite's angular velocity can be different from the earth's angular velocity. So earth is also rotating. So we need to account for that. All right. So it is revolving in equatorial plane. So earth also rotates like that wherein this is the equator plane and earth rotates like this and earth rotates again from uh west to east direction. All right. So uh the omega of the earth is 2 pi by u we can write in hours omega 2 pi by 24 hours 2 pi by time period right this is of the earth and omega of the satellite is uh 2 pi by 8 all right now we need to basically see uh with respect to earth Right? Because it is observed vertically overhead a longitude. So basically there's a person standing on the earth who is observing the satellite. Now that person is rotating with the earth. So the person is looking at the relative angular velocity of the uh satellite. So the relative angular velocity of the satellite because you know if let us say both the angular velocities are equal then the person won't see the satellite moving at all. It'll become like a geostationary satellite. Okay. So that is why relative anglo have to be seen. So 2 by 8 minus 2 by 24 that you equate to 2 pi by time period. Time period as in how much time it takes to revolve. Right? So from here you'll get the answer. Are we getting 12 everybody? >> Yes sir. Okay. So you know one very important thing that we sometime ignore is that the earth is rotating. All right. So once in a while that fact is used and time period of rotation uh is 24 hours. So omega you can easily find for the earth. Do this. Yeah, these initial questions everybody should be getting it as a warm-up exercise. We can I think many of you have already answered. Shall I wait for others? Anybody? Okay. Now one very important relation which is used so many times is t² is proportional to r cube. Right? So uh this is for the planet for the satellite. Yeah, somebody is saying something. Okay. Right. So you know whenever appropriate we should be using this law. So t_sub_1 by ts2 because it's like the question is framed in such a way that it's a comparison of the time period of revolution is equal to uh r / 4r cq right so t_1 by t2 is uh 1 by8 right 4 is 2 square. So square square get cancel 2 cube. So 1x 8. So time period of this satellite B if it is 8 hours then for A it is 1 hour. That way you can think. Now in 1/4 year 1 quarter year is 8 / 4 it becomes 2 hours. How many revolution A will make? 1 hour it makes one full revolution. So in 2 hour it'll make two full revolutions. So answer is two. Okay. So you may have your own way of reasoning it but this is how you do it. Okay. Any doubts anybody? Nothing. No doubts. >> Okay. Good. One more. There's also a straightforward question. Do it. If you do gravitation properly then you know electrostatics, laws of motion, work by energy many chapters automatically are getting you know revised. So don't think that it is only gravitation you are studying. Soon you'll realize that many concepts from different chapters are utilized. Only one person has answered till now. What else? Okay, three, four, four people have answered and all four of you got it correct. That's good to see. I'll do it. All right. See, uh it is somewhat similar to the previous question. But in previous question, if you see uh both the planets are revolving around the same center point which is sun. Okay. But here uh the sun orbits sun is orbiting and uh the earth is orbiting sun. So basically there are two center points of revolution. So you cannot use t² is proportional to r cube like that here because there two different scenarios. But uh you know there is an expression between because you can see the options also and the question is framed in such a way that it is about time period and the uh radius of the revolution. So we need to basically get the exact expression wherein yeah wherein there is an in uh there is an equal to sign because we can't use proportional to because proportionality constant for both the scenarios will be different as there one is revolving around Milky Way and one is revolving around the sun. So I mean do do you guys remember that exact expression or not or you tend to derive it >> the time period one for orbit? >> Uh t² is equal to what into r >> I derived it. >> You derived it. Okay. Yeah even I would have derived it. I don't even remember people's name. Forget about the formulas. So this is the m instead of v square I'll use omega² m omega square r right so m and m get cancelled so um um what you can say omega² is equal to in fact >> 4 pi square by t² right >> uh thank you so this is 4 pi² square t² r. So then it becomes r cq is equal to g m by 4² t². Right? So this is where you know this m is coming. So in the first sentence where it is revolving on the galaxy uh this is the mass of the galaxy and this is capital t and this is capital r. And in the second situation instead of capital R it is small R and this is mass of the sun now and this is small T². So you know since there is an equal to sign you you don't need to assume that proportionality constant should be equal. So now you can divide it and get a ratio of mass of galaxy with mass of sun. Okay. In both the scenarios we are using t² and r cube relation only but here you can use just proportionality and divide so that constant get cancelled but here they're two different revolutions so proportionality constants are not same so you need to get an equal to sign so this is how you solve it and probably a is the answer because majority of you have said that not 100% sure but this is how you do it okay shall I go forward Yes sir. >> Okay. Now get ready. We'll just uh go one level up. Okay. And do it uh maybe uh you should be very very I can say honest and if you're not getting you're not getting at least you'll learn something but if you get the answer from somewhere somehow then you'll not learn also. Okay. So just focus on the learning. Do this. Um just draw the free body diagram show the forces and the horizontal force should be less than or equal to mu time normal reaction. That's all you have to use. Grav intensity of gravitational field is nothing but acceleration due to gravity. Force divided by mass is accession to gravity only that is intensity. I'll do it. One second. Yes, I can do. Okay, some of you got some answer. Okay, let's do it then. So assume that this angle is theta. Okay, let me take it horizontal only like that. Yeah, this angle is theta. So there is a this B is here. It is evolving B is here. Okay. So this is the gravitational force between this mass and B which is mass capital M. Then it has mg over here. It has normal reaction. Okay. And uh you can assume a plane which passes through all these forces and this is your friction force. Do you all understand the I mean any doubts in free body diagram? >> No sir. Nothing. >> Nothing. Okay. Now uh you have uh vertically you can say n + fg sin theta minus mg is equal to zero. Right? So n is equal to mg minus gravitational force sin theta. And uh friction force should be um less than or equal to mu * normal reaction right and uh horizontally this is the normal force horizontally there is uh I can write like this f minus fg G cos theta is horizontal right this should be uh equal to zero. So friction force should be equal to fg cos theta and friction force is this fg cos theta should be uh less than or equal to mu time normal reaction. So m g minus f_sg cos theta. So sin theta sir >> sin theta sorry sin theta is this fine till now the range of >> yes sir I reached this equation too I didn't know what to do after this >> between so that you remains motionless okay so what is the value of FG. FG is g mm by R². G mm by R² cos theta less than equal to mu mg plus this. So we need to basically see theta is a variable we need to get rid of it. Now so value of theta is there in the diagram >> what >> the by 2 minus theta we can write it as sine of by 2 - theta is >> no no no anything is missing here friction is equal to fg cos theta Uh >> no sorry I got the same equation >> I tried making it as an equation in terms of mu and then finding its like minimum value or something so that mu can be greater than that >> but that didn't work. >> It should work. >> The derivative is very weird. >> You don't need to take derivative everywhere. Keep quiet one second. This now we need to get rid of theta. So see theta is a variable. So can anybody tell me what is the maximum value of this? What is the maximum value of this >> under root 1 + mu square? >> Under root 1 + mu square. So the maximum value of this should be less than mu mg. All of you agree or not? Huh? >> Yeah. So g mm by r² maximum value is 1 + mu² should be less than or equal to mu mg and when I'm gone so this will get you the value of mu is it clear now >> yes sir >> has anybody done like this anyone Okay. All right. Fine. So, uh, as far as I know, the answer is C. If you simplify, you should get C. Let me check. Final answer C. Final answer is C. Correct. Okay, shall I move ahead? See, this is where you know I have seen many Olympiad questions in which they get something like this you know cos theta plus some constant time sin theta and you have to take the maximum value of that. So that that that is how it it should click to you as well. Okay, it's not that you can substitute the value of theta to be something. Uh basically you can use calculus or whatever you want to because this is the only thing that is changing. So you need to maximize this. You can maximize using trigonometry as well. Fine. Shall I go to next? Anybody has any doubts? Any doubts? Anybody? Okay. Go to next. This one. You need to think freely. Then only you will be able to these able to do these kind of stuffs. You can type in your final answer. I I'm giving you maybe four or five minutes. I should not disturb you. So four five minutes I'm giving Collinear means that u they are all rotating together like a stick. Consider a straight line and uh entire the entire straight line is rotating. Okay. And uh the both the stars and ships are on that straight line. Uh checking whether orbit is stable or not is beyond our scope. So ignore the second statement. Actually it is not stable. Shall we? I hope uh some of you are in middle of thing. Any physics question, any question starts with a diagram. A proper well-labelled diagram is half the question done. Should I write the final answer? Final answer I have m1 by m_sub_2 R2. This is the final answer. Anybody fighting still to get the answer? About 2 minutes. So you got the answer got some answer. >> No sir. >> Van >> no sir. >> What is stopping you? You don't know from where to start. Is that the issue? >> No, I kind of did find a place to start. The the entire system if it's coina would have the same angular velocity about some point, right? >> What is that point? >> I tried to find that point. That's what I'm not getting. >> Are you sure? I mean can anybody help? What is what could be that point about which uh entire thing is rotating? What do you think? Pune. I'm in a class. So probably the center of mass the binary star system. >> Exactly center of mass. So everything you look at with respect to center of mass that is a good starting point. And then haven't I told you uh in 11th and 12th that if anything is rotating in a circle what equation you should write first? What is that first equation? Enter your first one. >> Anything rotating in a circle, you have to write force toward the center is m omega² into radius. That is the first equation you should write. And that's all you have to do in this question. And what is that center? Center for all three is what? center of mass. Should I do it now? We just write the equations. A can can you do one thing? Draw the diagram. All of you represent all these values and then uh I will just write down the equations. Do it quick. >> You have a doubt in the diagram. diagram. Let me let also draw the diagram and then we'll discuss right now. What do you understand? Draw it. Okay. So, I hope all of you have drawn it. So, this is that line. This is one star. This is the other star. One is m_sub_1 and m_sub_2. Now there is a uh spaceship also. Now spaceship mass is so little compared to star that we can assume center of mass doesn't shift anywhere. So center of mass is let's say here. This is our center of mass. Okay. Distance spaceship from the star is R1 and R2. So this is your R1. This is your R2. But the revolution is happening uh with from know uh the center is center of mass. So R1 and R2 they are not the distance from center of mass. So you need to assume something. So you can say this is a this is b you can say c also but you don't need distance of center mass from uh the uh spaceship because that you can get now b minus r1 that is okay so yeah tell me what is the issue with the diagram >> sir how are you so sure that the spaceship has to be between the stars and not some place How? What? >> Like why does it have to be between the stars? >> That is the condition. They are colinear. >> But they can be collinear if the spaceship is not in between the stars also. >> How it can be colinear? If it is here, how will you make a straight line connecting? >> On the same line, but on one of the sides on the same side. >> Okay. Okay. Okay. Yeah. >> But then it wouldn't be much of a spaceship, would it? If it can't see one of the stars. >> No, no. Let it be that side also. Let it be. It doesn't matter. Uh if you if you keep it here, okay, and then rotate it that is also fine. But then you know since in the question it is not specified whether the spaceship is in between or outside. So clearly it doesn't matter. But then you have to solve and if you have to solve then you have to assume one of the two scenarios and whatever scenario you assume the answer should come out to be same right so you can't get confused for that because it doesn't depend on it since it is not given is it clear Arita >> uh yes sir >> so I I hope you understand how it is revolving everybody this binary star revolves around center of mass like this. Okay. Whereas this m1 can revolve about center of mass in a bigger circle like this. So both of them are revolving around the center of mass as a center. Now I have to erase this. >> Excuse me sir. Um I thought why can't the satellite be at the center of mass itself that becomes a specific scenario that is not a uh you can keep it answer won't change still >> okay >> but I try to solve like this okay fine uh if the spaceship is at the center of mass then then it is always colinear isn't it? it'll be always colinear then uh whatever is a value of m1 and m2 it will be colinear only because androath doesn't move and if what goes like this and m1 is here m2 is there yeah it'll be always colinear anyways let's solve for this scenario so first of all um for this center of mass for this to be center of mass All of you agree that moment of masses should be equal this equation. All of you know this or not? >> Yes. >> Yes sir. >> Okay. Then the constraint relation A + B should be equal to R1 + R2. This is second. So using these two you can get either A or B in terms of R1 and R2. Right? So that is one. Then u let us say angular velocity is omega. This is omega. This is omega. So the force on m_sub_2. What is the force on m_sub_2? G m_sub_1 m_sub_2 divided by r1 + r2². This is the force toward the center of mass. Right? I hope all of you agree. I'm ignoring the gravitational pull by the spaceship on the star. Is it negligible? This should be equal to m_sub_2 omega² into what? What should I write anyone? >> Sir, you're doing this for m2, right? So, it's distance from the center of mass. >> So, what is that? That is what I >> a now somebody will somebody else will answer the next one. So for the m1 we can write g m_sub_1 m_sub_2 by see for m1 we should not write because it becomes a redundant equation. You can see here if you write for m1 it'll become like this m1 omega² b right? So if you combine three and four it becomes one. So it gives you redundant equation. Equate right hand side of three and four it becomes redundant. So we don't write redundant equation unnecessarily. But now we will write the equation for the spaceship. Can you all write that equation? All of you please write. Let me know once you're done. Done. Others are you guys done for the spaceship? Spaceship is getting pulled this way and this way. So towards the center of mass force you have to take that is g m_sub_2 m is a mass of spaceship divided by what is the distance r2² minus g m_sub_1 m by r1² this should be equal to m omega² distance from the center of mass is What? For the spaceship, what should I write here? >> B - R1. B minus R1. I told you let somebody else tell. Anyways, so these are the four equations. You have to play with them and get the answer. M1id M2. Okay. So mathematically it appears that it is not an easy equation to solve but you know I think you should be able to the these three and four you uh there is b and a right so you can write a as r1 + r2 minus b here and then substitute b as uh you can get b in terms of r1 and r2 by writing uh this a as m_sub_1 by m_sub_2 * b is b is equal to r1 + r2. So you can get B in terms of R1 R2. Substitute here. A you write R1 + R2 minus B. Substitute here also. And then just play with it. You get the answer. Okay. I'll move on to the next question. Right. Okay. Are you finding it tough? >> Interesting is a better word, sir. What I >> mean it's more interesting than just tough. >> Okay. Should I give you mo much more interesting ones? Everybody >> I don't think so. Sir, I think this is enough for today. >> I have these kinds do this. I'm giving you another five minutes to solve. What is that first thought that comes in mind when you read this question in order how to solve it? First step >> may >> uh B >> escape velocity will be different for every particle depending on how far away from the planet it is. Correct. >> So you're just taking the escape velocity clue from here and trying to fit in how do I solve with respect to escape velocity using that as a starting point. Basically I'm trying to find the that distance from the planet below which every particle gets captured into orbit. >> That velocity below which every particle get captured. >> That distance from the planet because all of them have the same velocity. >> Uhhuh. >> There will be some distance from the planet. I said velocity will be sufficient to escape orbit completely and leave the planet not get captured. Anything below that will get captured. So if I can find that value of r then that gives me the thickness. I already have the length and density. >> Great. So is saying what Van is saying that this is the cloud. Uh right this is the cloud. So he's trying to find out how far the particle can be in order for it to get captured by the planet. Right? That's what you're saying. So how will you basically get that how far it is? What is that limiting condition for which it should get captured? You understand my question? >> Yes sir. >> How do you get that this is the farthest away the particle will be for it to get captured with the planet? So should the velocity in the direction of the planet be greater than the escape velocity? Velocity in the direction of the planet as a component this component. >> Yes. >> It is coming towards the planet. Right? So whether it is greater or not it will fall on the planet. It's not going away. If it is going away then escape velocity comes in picture. Think more. >> The point at which escape velocity becomes zero of a particle. >> Escape velocity becomes zero is infinite. So that is like entire cloud will get captured. in don't get fixated on escape velocity that is what I would say because particles are moving towards the planet it's not that it is going away I mean I know even I when I was solving before the session I was uh reading too much on this SK velocity thing but it's not about SK velocity it is about something else how the >> about conserving momentum >> how the particles will come towards the planet how do you think they will travel they will get attracted by the gravitational pull right oh so these two straight away get into like this suppose this particle how this particle will go to the planet it travels like this and falls like that isn't it it falls like that now can I say can I say that all the particles falling on the planet has the same velocity is it true or not >> yes sir >> you can conserve energy and you can see that initially they are moving with what velocity vot and kindic plus potential is constant. So you know that uh just before reaching the planet every particle's velocity will be equal. So it's not about uh that some particle will have a different velocity compared to other but still some particles will miss the planet. Why? Why they will miss the planet? Because they will go like this instead of falling on the planet. they will get attracted by the planet. So that is why they get deviated but it doesn't fall on the planet. Just imagine a particle is just falling on the planet. Then what will happen to the particle? Particle go like this and tangentially touches the planet like this. Do you all agree? Not getting it. If let us say this is the planet. Okay. This is the planet. So this particle the limiting case the particle will travel like this and will just go tangentially like this and falls on the planet. Do you all agree? The particle just >> just above the particle will miss the planet. It'll go like this. >> Agree? All of you? Can you type in or speak? >> Yes sir. >> Right. So the limiting condition is that what is that condition? That the particle when it reaches the planet its velocity is tangential. That is the limiting condition. Now can anybody guess what could be that uh physical concept we should be using to capture how far the particle should be from here? What law can we use for that situation? Because this distance like what Van said if you get this distance then I can just imagine a circular cross-section here then all the particles have the same scenario and I need to just get volume of the cylinder multiply with the the density I'll get the answer right. So how to connect this velocity, this distance and this velocity and some distance. Which concept? >> Projectile motion >> which law >> we can conserve angular momentum. >> Angular momentum. You have to conserve angle moment about what? About this point. About this point the particle is experiencing no torque. Isn't it? Only this planet is applying gravitational pull this way. This way all the pull is passing through the center. So if you can conserve the angular momentum about this point, you can connect the initial point and final point. That is how you have to connect, right? You you know this you know some of the initial condition you know its velocity you know some of the things about the final condition. So if you connect them you get an equation and to connect them you have to use conservation of angular momentum. Is it clear making sense to everybody? Speak up. Is it clear? >> Yes, sir. >> Yes. Who else has to say? Okay. >> Yes. >> Yes. Okay. Fine. So, should I do it or you want to do it? You do it. Get the equations. Complete. Don't you don't need to solve it. Get the equation which will be used to solve and let me know once you're done. type in once you're done writing the equations. You have to write three equations conservation of angular momentum I'll write mass of the particle let's say mp mp v into d angular momentum about this point. How will you get get the perpendicular distance perpendicular to velocity the distance is d. So this should be equal to mass of the particle v into radius of the planet is r. What happened? V into r v into r. So v into r should be equal to v into d. That is the first equation. Then you have uh conservation of energy half mass of the particle v² minus g um is it very far in from dust particular planet from great distance. So potential energy initially is zero. This should be equal to half mass of the particle into v² minus of g mp by r. Now mass of the planet is not given but escape velocity is given. That is where they want you to use that stuff. So you know that escape velocity is roo of 2 g m by r isn't it? So gm by r becomes uh sk velocity squar divid by 2. So this is sk velocity squar divid by two. This is the second this is the third. Then you have to basically find mass. How much mass of dust particle has been collected? Collected mass once you get the value of d is pi d² into l. This is the answer. Okay. Any doubts? Anybody? You can speak up. Final answer I don't have for this but I I hope you got the questions right T. Okay. All right. Shall we move ahead? >> Yes. >> Ready. 5 minutes for this. Yeah, rod has mass, bead has mass. Look at the distances carefully. Rot and L and X. R is extremely large. Did I do it? Okay. So let us discuss this one. So uh prana how will you do this? How will you start? What is the thought process? Maybe because the force there's there's a very very tiny difference in distance across the rod. So maybe that would affect the acceleration of the rod compared to the bead. And we try to figure that out. >> That will affect what the relative acceleration of the bead and the rod. So you're basically trying to find relative acceleration of bead and the rod. H >> that's what you're trying to good that is what you have to do find out the accation of the rod find out the accation of the bead and then get the relative acceleration okay because they're asking how much time it takes for the bead to come out right uh so bead is moving on the rod so broad is also moving so to get the relative acceleration for that I need to have I need to get the force on the rod. So to get force on the rod you have to consider at a distance of x dx thickness of the rod mass per needle length lambda. So the df force is g mass of the earth into lambda dx mass of the small mass of that into divided by x² right this is a df integrate this you'll get the value of force on the rod right and force on the rod should be equal mass which is lambda into L time accation of the rod. So from here you'll get acceleration of the rod. That is clear right? This will give you accation of the rod. Now acceleration there is an approximation probably involved here. So let me do it completely. So, g m lambda um it goes from r to r + l. So 1x r - 1x r + l. So I'm assuming the force on the rod to be constant. Is it true that it is constant? Because r can keep changing. Yes or no? >> Yeah. Yeah. So it'll vary as it gets it'll increase as it gets closer, right? >> Huh? So but why I'm assuming it to be constant? >> Because r not is much much larger. >> Rot is extremely large and we are talking about the time in which the bead will come out. That time would be in seconds or minutes or in hours let us say. But this 4 into 10^ 8 m nothing will happen to it. So that is why we are ignoring the variation of the force between that time interval when the beat comes out. So this is G M R L / R + L. This is expression of the rod. Now acceleration of the beat. Acceleration of the bead is u a force we can write directly g m is a mass of the bead let us say divided by r + x² this should be equal to m into accation of the bead. So accation of the beat is g m / r + x². Now x note is very small compared to r. So you ignore x from here. So it will be g m / r². So this accation of the beat, right? I'm adding x not to r. It doesn't matter. So the relative acceleration is acceleration of the bead minus acceleration of the rod. This is your relative acceleration. And once you get relative acceleration, you know x not is equal to half a relative t². So from here you get the value of t. Clear to all of you? So in this question you have to think in a very simple manner and use approximation to get the answer. And you need to appreciate the distances involved. If you don't acknowledge the fact that distances are very small compared to R not, you'll make it extremely complicated. And some of you might have done that. Right? All of you clear about this? Shall we? Okay. 5 minutes for that this one. Okay. So we will discuss. Okay. One answer to two people got the answer. Who will uh explain the thought process? Any what do you think how will you how will you proceed? So first I try to draw the diagram diagram. It is found or it has atmosphere right diagram 10 then Hello. >> Am I audible? Any >> Oh yes sir. I was uh on mute. >> After that I tried to uh like >> I tried to like understand the situation and like write some equations. I wrote the equation with uh I wrote the equation of the force uh I took like a small volume in the atmosphere. >> Okay. So you you you use a small volume to do what? How will that help you? It gave me like a proper mass and object point object so that I could write my equations easily. >> What equations? That's what I'm asking. What are you trying to write? What are you trying to achieve using? >> I wrote the equation of force using >> force. Correct. So you are writing force equation so that you get accession due to gravity. >> Yeah. As to gravity is total gravitational force divided by mass. >> Yeah. >> So you're trying to find force on that imaginary point mass kept at a distance of x away. All right. Now the force can you divide into two parts? One from the planet and one from the uh this thing atmosphere. Right? Now what you can say see for the planet it is straightforward right? The force from the planet you can directly write g mass of the planet into mass of the particle divid by x². But how will you consider the uh force due to the atmosphere? How will you take care of that? Should I ask somebody else? Okay, I have troubled you enough. >> So, is it only because of the inner part of the atmosphere and not the outer? >> So, you can draw one imaginary spherical shell and you know the shell theorem only the mass that is within this green sphere can apply the force and it'll apply the force as if it is located at the center. So you just need to find how much is this mass and whatever mass comes here you can direct write g into m uh gmm by x² that's all. So mass of this atmosphere which will apply the force mass of the atmosphere again you you all know you have done it enough number of times 4 pi r² d r row into dv right integral so 4 pi row integral of r² d r integral will go from where to There >> so row is not constant right >> sorry sorry sorry 4 pi integral row is sigma by r so 1 by r r² d r so limit will be from where to where >> to x >> from so r to x - r >> r X right so it'll become 4 pi pro becomes X² - R² by 2. So this will become mass of the atmosphere. So the force total force is g mm by x² plus g mass of the atmosphere which is 2i sigma x² - r² m divided by x² now gravity g is f by m. So m and m gone. So you can write this as g m by x² - 2 g sigma by x² into r² this I'll keep it separate plus 2 pi sigma g. Now this G should be independent of X because it is said that it should be constant throughout. It'll be independent of X only when this bracket term is zero. So equate it to zero. Then it becomes G becomes constant which is this clear to all of you. Speak up guys. Is it clear? Yes sir. >> Okay. >> Yes sir. Clear. Final answer comes out to be final answer. Someone got the answer. Wait m / 2 pi r. Oh that is your answer only. Same answer also got. >> Okay. All right. Only the part one 5 minutes for There. Okay. So shall we discuss? What does Huns Say >> okay from Vipure is it? >> Yes. >> Okay. Okay. All right. So who will T? >> Yes sir. >> Tell how to start. Um, atmospheric pressure is the total force applied on the earth by the atmosphere divided by the area of the earth. >> Mhm. >> And we can find the force applied by the atmosphere on the new planet by the gravitational force. >> But what is you're trying to achieve ratio of atmospheric pressure on the surface of the planet to the earth. So given whatever are the scenario whatever whatever parameters are given here uh mean density mass mean density why why do you think mean density is given >> uh like average density so you don't have to you just multiply it with the volume >> exactly right diameter of the planet is given its mean density mass of the atmosphere dor not and I'm not okay. Mean density of the atmosphere is same. So atmosphere's density is same. Now suppose atmosphere density is given and the height of the atmosphere is given. What could be the atmospheric pressure? Then >> uh you would need to integrate for different points cuz the gravitational force >> you're assuming density to be constant mean density of atmosphere. So >> and then you have to multiply with the volume of the atmosphere as well to find the total mass of the atmosphere. >> Are you sure? Mean density is given. So I can just use ro gh yeah we can do that as well. >> I can directly I don't need to integrate that is why you know the mean values are given here. So if let us say I have to solve the first question it should be like uh uh mean density of the atmosphere is same right so let's say that is sigma kn sigma kn g sigma kn g1 h1 divided by sigma g2 h2. So the answer should be G1 H1 divided by G2 H2. I do not know the value of G1 H1 and H2. So this is now I will try to find out G1 H1 and G2 H2. All of you agree everyone this is Earth. This is planet. Okay. Now how will you get H1 and H2? What do you think everyone >> of the atmosphere? >> What? >> They give us the mass of the atmosphere. >> Huh? So how will you do that? It is the mass of the atmosphere of the planet is 10 times the mass of atmosphere of the earth. So how will you use that >> volume? because density is constant. >> H. So what should be the equation? What should I write? >> So it's given that height of atmosphere is very small compared to radius. >> So instead of using the volume of a cube thingy volume of a sorry sphere, you can just take surface area into height. >> What should I write here? So surface area will be 4 into r² or p into d² where d is the diameter of the planet time h is height of the atmosphere will be equal to the volume of the atmosphere >> and mass by volume is constant for both planets fine so you can do that and you can equate the uh so let's try to do that so m not is basically what you are saying is 4 pi radius of the earth. You have to write 4 pi r² into this is h2 that into uh density of the atmosphere sigma knot. Now the 10 m should be equal to 4 pi radius of the planet² time h1 into sigma kn. Are you able to understand everybody this equation what I'm writing here this is m not right? So I'm writing 10 m as 10 into that. So this equation all of you get this equation you type in I mean you don't need to do this approximation what you could could have done is uh sigma kn into 43 4 by 3 into r earth + H 2² H2 cq minus R earth cube and then use binomial approximation you get the same thing because h is very small compared to so you can use binary approximation take outside okay can you type in is this equation clear or not everybody Ready? Is this equation clear? Okay fine. So radius of planet is how much time of radius of earth? 10 times. So we'll get something here. So we'll get 100 into h1 should be equal to 10 into h2. So h2 is 10 * h1. that is you get h1 by h2. Now g1 by g2 that is I guess easier uh diameter mean density is this. So basically you need to get the value of g in terms of density this is g m divided by r². So m is 4x3 r cq into density divided by r². The small g is 4 g r row by 3. So g1 is this g1 is for the planet. uh density is 4 g by3 radius is 10 * radius of the earth and the density is row node by 4 so 10 x4 is 5x2 so 5x2 * of g2 g1 right so we have g1 G1 by G2 5x2 H1 by H2 1x 10. So multiply uh multiply so you'll get 1x4 that should be the answer clear to everybody. This is not Olympiad. Okay. This is I think one of those J advanced questions only. The last two three questions were not olympiad. Is it clear to everybody type in anybody has any doubts you can speak up. No doubts nothing. I'll move ahead. Okay. Five minutes for this. You'll do both parts take six to seven minutes. both parts. Anyone? About to get anyone about to get the answer? Okay. How to proceed? Sumat is there? You're on mute. Samar is not there. Okay. Shorty, how will you do this? >> So for this question, we are talking about a satellite which is pretty close to earth and now we are saying that this collides Indian accent. Then exit. >> So sir, we are saying that there is a satellite that is hovering over the earth and there's a cloud of dust particles that comes into contact with this particular satellite. >> So because of this uh it appears that it experiences it experiences some kind of a resistive force which decreases its velocity. And we know that since this is going in a circular orbit, omega cannot change. So if velocity decreases, then something else must make up for it. Meaning the radius also decreases and we proceed from there. >> How will you find force? >> So I did dm by dt into velocity. >> Right? So you might have seen some scenario where you must have encountered some question with bullets are hitting the wall and you have to find the force or uh or you might have seen scenario where photons are hitting the surface and you need to find the force. So there rate of change of momentum is what you need to find. So um and in case of fluids also you might have seen force we have to here mass is changing right as it keeps sticking on it mass is changing so force is v dm by dt force is m dv by dt plus v dm by dt also okay now velocity is changing very very slowly so we are ignoring that so it is just simply v dm by dt I hope this is clear to everybody. >> Yes sir. >> Uh don't count. >> So I have a doubt here. >> H >> uh we said that velocity is changing very slowly but it's still changing right. So do we have to write it as a function of mass over here or we just uh assume that it's the same velocity throughout? We're assuming it is same. I mean it's it's that uh variable mass scenario. So we are ignoring that change in velocity extremely less >> sir. But can't we find that velocity if we uh conserve momentum? >> You can do you can do why not? You can do all that. But you're not supposed to do. We need to all understand that what you supposed to do what you're not supposed to do. So sometimes we are running behind highly accurate scenario mathematically unnecessarily. For example, here the mean radius is given. So simply we have to write row GH. But the first thought that comes in our mind to integrate then uh what else there was this scenario also here we ignored x not compared to r not so physics is not about being 100% accurate because u you know if you complicate it and try to include everything probably you will not be able to solve the equation itself. Okay, but then we'll talk about change in velocity in the second part. Anyways, it changes but it changes extremely small rate. So, we are ignoring that. Fourth is V DM by DT and DM is what? It is area of cross-section S dx is dv dt into row. So it is row s v² s is a cross-section area which so basically as it is moving the dust is sticking on it and rate at which dust is sticking that is what is dm by dt okay so this is the force now how will you find the velocity with what velocity it sticks how you get it's The orbital velocity don't you think so? So it'll be g m by radius of the earth² that should be equal to m v² by r. So v² is equal to g m by radius of the earth that will go there and you'll get the value of force. This is your first part. Okay. And it is no longer moving in a circle. Its radius shifts continuously. And if dust is there, it will slowly falling into the falling in here. Right? So basically you can't even use this equation. You can argue that how can you use this equation? This equation is valid only for the circular motion. But it is not perfectly circular. So how will you get the second? How will you solve second part? Any ideas? Anybody? Anybody wants to tell? Second part. How will you do it? So since torque is torque is zero about all the center. Do we conserve the angular momentum and try to find the new radius? Find the change of velocity and the radius of the server path. But to find the new radius you need to know the new velocity. Isn't it? And do you think angular momentum is conserved? Are you sure? Isn't just creating that external torque with respect to center? >> Can we conserve linear momentum in the collision? >> There is a force. The force is only you found drag force is there. Right? So if there is an external force, think more. Think more how you find change in velocity. So can you find acceleration like the tangential acceleration using the >> using this you find the tangential acceleration >> this is the force this force equal to mass into tangential then what >> so and they've said that it's one revolution so you know the time period for one revolution anyways Yes, you can that you can do that ways. Any other way can think of anyone see there are only two ways to handle either you use the force or torque or you use the energy. Okay. So in conjunction to the momentum conservation, momentum conservation cannot completely solve your question until unless something is already given to you. So either in mechanics you'll be using force or energy. So there are always two ways of solving question. You can use energy. But in order to use work energy theorem you have to use work done is equal to change in the uh total energy. Right? Now why I'm putting negative total energy work done over here is it positive work done or negative work done in one revolution when it is revolving >> negative it >> is negative so work done is minus of row s v² now here also approximation I'm assuming this force to be unchanged in one revolution but the fact is slowly the force is you know decreasing uh as it is uh going in sorry it is increasing because velocity has to probably increase as its radius is decreasing. So but then we need to keep that uh approximation and this is this one and uh total energy of a satellite is what? Do you remember minus of G mm by 2R? Do you remember this kind energy plus potential energy is this? >> Yes sir. >> Right. So this is equal to uh change in the energy. Now this is E. So delta E is G mm by 2 R² into delta R because delta R is very small I can use derivative concept here. So G mm by 2R² delta R this will give you delta R. Delta R is negative because it is coming in. So delta R is negative. >> So won't delta E be caused by change in velocity also. >> Yes, it does. Velocity is a function of R. Velocity is a function of R. So this is kindinetic energy plus potential energy sum of both. >> Okay. So >> see uh let me complete. See this is the delta E. Once you know delta E let's say you got the value of delta R using this. What you can do next is this. This same delta E same E you can also write in terms of the uh kindinetic energy which is minus of/ M into V². Do you remember this? It looks like some of you have forgotten that. So I hope you know that G mm by R² is equal to M V² by R. So R R get cancelled. So half MV² is equal to G mm by 2R. Now you remember this. So honey. Yes sir. >> Right. So total energy is minus GMM by 2R total energy of the satellite system. So delta E once you get that is also equal to uh negative of E is equal to negative of/ MV². So delta E in terms of velocity also you can write which is M 2V delta V. So in terms of velocity it is minus of MV delta V. So what if you get delta R you can relate it to delta V. Equate these two. Is it clear? Clear to all of you. Everyone type in. See it is the change is so little that we are assuming it is going in a server path and change is so little that in one revolution delta R is like dr extremely less. So that is why all of these approximations are used here to get the solution all of you is it clear? See I have chosen questions which are not regular types so that you can see what all things can happen in gravitation chapter. Okay. So if you're not getting most of it or anything that is good you're learning something new. You are utilizing your time properly because every second every minute you are learning something new. That is what counts. Okay. Solving number of questions which you can do in first attempt is a waste of time. Now do this. These are not Olympiads, okay? They are like proper J advance questions. But there is a twist in the tail. So Ry didn't understand what we have to prove. >> You have to basically get a relation between small n and rest of the thing. That's all. >> How the variables are connected. >> No, like that last point. uh >> remain about the same point on the equator. What same point? >> See, same point meaning that it is uh revolving with the same angular velocity with the earth. Okay. Relative to earth, it is I mean it is revolving with the earth. Is it clear to everybody? >> Got it. >> It's like a geostationary satellite type of thing. See again it is a circular motion. So just find the force towards the center. And >> is NR from the surface or from the center of the earth? >> Is it not clear? Uh just above the surface of the earth out to a radius NR. NR is the radius center. Whenever something is moving in a circle just write net force toward the center is m omega² r. That's all is there in this question most probably got it. The elevator is not touching the earth. Okay. It is written that elevator is just above the surface of the earth. So there is no question of normal reaction. Anybody is getting Okay, Prano got the answer. Uh, yes. Anybody else got the answer? >> No, sir. >> Okay. Fine. So, this is the Earth. This is the rod. It's rotating with omega kn. So even the rod has to rotate with omega kn and uh the rod is not touching the earth. Average density is row posted mega space out the radius nr. This is nr. This is r h. So the uh this is the center center of the rod right. So first we need to find the force towards the center. Uh so you know this is like a rigid body rotating in a circular path. So we just need to bother about the what happens to the center of mass of the rod. Center of mass is moving in a circle of this radius. Right? So this radius is how much? This radius is this is NR. This distance is NR - R / 2. Total length is NR - R. That is NR - R / 2 and add R to it. that is a distance of the center of mass distance of the center of mass from the center of the revolution. So that is n r + r by 2. This is the radius. Then we have to find the net force. So this is like let's say at a distance of x this is dx something similar we did so f g m lambda x dx / x² and then x goes from r to n R - R. So this is G M lambda. Those shouldn't go to they're measuring X from the center of the fourth ray. >> Oh, correct. Correct. Correct. R to NR right G M by lambda 1x R - 1 by NR F is equal to G M the lambda can be written as lambda is total mass divided by length which is nr - r. This should be equal to m M is m omega² into the radius of revolution. This equation you just cancel out the unnecessary terms simplify get the answer. We have not done anything great here. Just found the force found the radius of revolution and wrote the centime force is equal to m omega square r. That's all clear to all of you any doubts anybody? Nothing I'm going to next question. to this 5 to 6 minutes. Um both the parts 6 to 7 That's Do you guys get breaks during the crash course? No, sir. We'll discuss in couple of minutes. Uh let's get one hint. How will you find relate uh maximum distance? What concept to be used here? Energy consation. >> Energy conservation. How will you identify the point which is having the maximum distance from the center? What is that identification? >> Are we not saying that the entirety of the kinetic energy gets converted to potential energy at that particular point? >> No, need not be. It will still have some velocity. The point at which relative velocity with the center becomes zero. >> Uh no but somehat like that >> the direction of the center rather so it's not going further away now it's coming closer. >> Correct. It is the velocity component >> uh that that's what I meant component sorry >> uh component away from away along the line joining is zero. It'll keep on going away away and then the velocity become perpendicular to the line joining entire velocity. Right? Now think over it. Which concept will you use to relate that point with whatever point you know initially? >> So SHM >> not SHM it comes out of the tunnel right it goes out. It is moving outside >> at that specific point. It's in circular motion at that instant because velocity is perpendicular to the center. >> So is it conservation of angular momentum? >> Angular momentum. So the these are I mean you you should be now uh tuning your thoughts uh that whenever you get such scenario you have to use conservation of angular momentum because that is where I mean we used to think that angular momentum is like not so useful and not so fundamental equation. It's just like mathematical correlation with velocity and perpendicular distance. But we forget to use it like how we use conservation of linear momentum. Start using it whenever you see that about a point about any point let us say torque is zero. You can use the conservation of angular momentum. For example here once it comes out first of all once the particle is inside there's a normal reaction also due let's say if it is here there's a normal reaction which will have torque about the center. So that is why angular momentum is not considered. Oh angular momentum is written also right. So you need to get the hint from here itself. Anyways once it comes out like this then normal direction becomes zero. Now angular momentum is conserved because only other force is gravitational force which is passing through the center. So about the center torque is zero. So it goes like this and like that right? No, it will come back. Sorry. Goes like this and tries to again come back and fall maybe. Okay. So maximum distance appears to be somewhere here. It is here where the velocity is making 90° with the line joining. All right. So between this point number two and point number one you can conserve the angular momentum. Okay. Right. But you need to first get this velocity. This is v_sub_1. This is v_sub_2. How will you get v_sub_1 and v_sub_2? Right now it is at a distance of rx2 away. So you need to get uh only this much part of the earth is creating the force. Oh, it will be that formula. Do you know the potential gravitational potential inside the solid sphere? Anybody remembers that expression inside the solid? G mm by 2 RQ into 3 R² - small R² >> by 2 R cube into 3 small R² >> no capital R² minus small R square >> this is correct >> and minus sign like overall >> this is a minus okay great so basically first step is to get V_sub1 so between this point this is point O point 0 and one we'll use conservation of mechanical energy why it will not perform SHM so suppose it is uh restricted inside the funnel then it'll perform SHM but suppose at point number one suppose you use conservation of mechanical energy okay and you're getting velocity at point number one to be zero then the maximum distance is this distance only Sorry that distance distance from the center they're asking that is r only that is the answer but suppose you you're getting greater than zero velocity it means it will come out so you need to check whether it will be confined within or not so let's not assume anything let's find v_sub1 so you substitute small r equal to rx2 here uh potential energy is mass into potential so that is your u0. So u0 +/ m v² v not is given that should be equal to u1 u1 you can write very easily g mm minus of gmm minus of gmm by r +/ m into v1² Now most probably V1 will come out to be greater than zero. It is a J advanced question. It can't be just this much. So V_sub_1 is greater than zero. Now once you get v_sub_1 then you need to do one more circus here because this about that point perpendicular distance is not r right of this velocity perpendicular distance from this point line of velocity is rx2 okay theta and all will not come so angular momentum of that v_sub_1 at point number one is m v_sub_1 into rx2 that should be equal to m v2 into this distance d. So m and m get cancel. This is your first equation. Now you have to do conservation of mechanical energy between 1 and 2. So minus of g mm by r +/ m v1² should be equal to minus of g mm by d +/ m v2². So solve the 1 and 2 equation to get the value of d. This is how you do it. Sir in the first part of the question they asked if angular momentum is conserved in the tunnel right so since at both the middle point and at the end point the perpendicular distance remains same from the center >> so wouldn't the velocity be same if angular momentum is conserved >> angular momentum is not conserved as it is inside the tunnel because there's a normal reaction also applied that normal reaction torque is not zero about Yes sir. >> Once it is outside of the tunnel then only the force is passing through the center of the uh earth. So you can say that outside of the tunnel angular momentum is conserved. Inside the tunnel normalization is there angular momentum is not conserved. But then yes if it would have been conserved then velocity will not change. You're correct to point out that. >> So, >> but the particle inside the tunnel doesn't necessarily have to touch the tunnel, right? For the normal reaction to act. >> Normal is a contact force. You need a contact. >> Okay. >> And it will touch. It has to touch. it can't uh I mean when you give the velocity projectile is fired along a tunnel from the center that velocity how will you give suppose you are like okay no I'll give the velocity if it is inside the tunnel in the air I'll get so it becomes a projectile and you'll say that it will come out uh before it falls on the ground but tunnel distance is rx2 given so you need to assume that tunnel is very very uh thin So you even if you move little bit in vertical direction you will touch. So you can't ignore the contact. >> Okay sir. >> Uh this is this is like one of those regular straightforward questions. this one. How much is the time? Okay, you have 6 to 7 minutes. This uh you know whatever question I'm giving now, it is pretty much doable. Please attempt it with full energy. Don't put your energy down. done. >> Yes sir. >> Hey, is this a regular thing where 20 people join and at the end 13 left? So this is a good thing I'd say because normally there are about six people left by the time the class gets over. >> Oh, so this only happened like the last physics class and this one >> physics actually J main physics is easiest of the three subjects but J advanced physics is toughest of all three. So this changes colors very fast. So but then the basic theme remains the same. If you think in a simple manner, mathematics can be tough. But if your thought process is simple and straightforward, you'll be able to crack. Don't complicate your thought process that some random theorem you know or some expression you remember. J advance is not about knowing 100 things. It's about just knowing few things and able to use them. Don't run behind new theorems, new equations and all that. It is all useless. Okay. So shall we do it? Anybody about to get the answer? Do we have more questions to simple? I think that's it. This is after that we have just straightforward questions. This is also not tough but I don't know why you are not doing it anyways. Shall I do it? Somebody did. Is it? No. Okay. Vital got something. Okay. Yeah. Yeah. Seems to be correct. Vital got it. Got it. Let's see anything moving in a circle. What did I tell you? Force toward the center is mv² by r or m omega square r. So let's see that particle of mass m moves under the action of central force whose potential is this. If this is the potential uh the field is what gravitational field do you remember this? >> Yes sir. gravitational field is minus of 3 k r². Now just like charge into electrical field is electrical force mass into gravitational field is gravitational force. So uh the force gravitational force is 3 m kr². I'm ignoring the negative sign just as a direction. This should be equal to u m v ² by r or what energy? Okay. So m v² is equal to 3 m kr cq. So kinetic energy is 3x2 mkr cq. This is kindinetic energy. What about potential energy? How do you find potential energy? Anyone? >> Mass into potential. >> Mass into potential. Just like mass into electrical potential is electrical potential energy. Mass into gravitational potential is gravitational potential energy. So minus of m K r cq. So total energy is sum of these two. So which is 1 by 2 m k r cq. Now put r equal to a. You'll get the answer for the total energy. And the angular momentum is m vr right m into v into r. Now how will you get velocity? So from here from here get the velocity in terms of r substitute here you'll get angular momentum and then period also you'll get 2 pi r by v. again V you can get from here clear to everyone this is a proper J advanc question this is how it will be can't be easier than this fine let me see what I what else I have this question though you might have seen so many times should not even bother to equal to half the did I leave anything oh above no sure or take just for completeness we'll do these questions also solve to get the correct answer and quickly. Hey, So I think uh you all are able to do this after doing some tricky ones. Artificial satellite is moving in a circle orbit around the earth with a speed equal to half the magnitude of the escape velocity. Okay. Is escape velocity half of it is this. This is the velocity. Determine the height. So, MV² by R height you have to determine not the radius. So, R is GM by V². So gm by v² is 1 by 4 2 gm by r so it is 2 r 2 r you know uh so height is r 2 r minus r okay if it is an mcq there will be an option of 2 r also so if you are in a hurry you'll mark 2 r and move ahead There go there go your th00and to 2,000 ranks. Okay. So be very careful in J1s. It's not about speed. J1s don't hurry up things. If a satellite is stopped suddenly in its orbit and allowed to fall freely speed at which it hits the surface of the earth, it stops. So velocity becomes zero. It has a potential energy. So, I hope you can do all of this. I don't need to do it. That's it, I guess. Right. So, we'll end the class here. And uh anybody has any doubts in gravitation from anywhere, anything you can ask now. Nothing. Okay. >> No sir, no doubts. >> Take care of your chemistry guys. I have seen chemistry percentiles of yours are not great. And remember the rankers their chemistry is the best. So you know learn chemistry. Forget about I mean it's boring and all that. It's don't make such excuses. It's not about boring or something. It's about your future. It's about your ranks. So attempt chemistry properly and learn. There is a time that you can learn you you can basically uh see what all tricky things you don't understand of chemistry and learn them. Okay. Cho. Bye. >> Thank you sir. >> Thank you sir. Bye.