 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 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. So 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. Finds velocity is 5000 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 are 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 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 cancelled 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 energy theorem only. Same expression. Same way. Exactly the same approach will be there. Okay. We're going to use this only. Right. W is 0. Final potential energy is 0. 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 2, 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 2 GM by RE. Fine. So if we square it, V square will be equal to 2 GM by RE, where this V is 3 times the escape velocity, 3 times. So this is, sorry, I thought of something else. So 3V square. Fine. So V square by 2 is GM by RE. Fine. So test 2 is gone. 2 will come over here. So V square is equal to 2 GM by RE. So V1 square will be equal to 9 times V square minus V square. So V1 is root over 8 times V. So that is also equal to 2 root 2 times V. Yeah, yeah, yeah. Correct. I missed that. All of you clear about it? Should we go to the next 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 GMM divided by 2R, 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 GMM by 2R. So this is GM into mass of earth divided by 2R. 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 4 into 10 to the power 5 plus the radius of earth. So that is 6.4 into 10 to the power 6. This is the value of R. DE 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 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. 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 that universe, only these two things are there. The distance between them is 10R. Okay. 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 magic 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 magic 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 have just started moving towards each other, K1 is zero. Right. There was no kind of 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 we 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, their center to center distance is what? Their 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 per 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 is 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.