 Now, let's talk about something which is very unique, escape speed, see what happens is when you throw a stone in upward direction it will come down, so when you throw it with higher velocity it will go further up, but again it will come down, okay. So escape speed is the minimum value of the speed which is required, okay to leave the earth gravitational pull, so what will happen is that the mass will just go to infinity it will not come back, okay and it does not matter at what angle you throw it, okay, so even if you throw it horizontally from the earth surface from here, it will slowly go to infinity, if you throw it with that speed, fine, so we need to find out what is that minimum speed with which if I throw an object it will go to the infinity or it will reach the infinity, okay, let's say that minimum velocity is V, alright, radius of the earth is Re, okay, mass of the earth is capital M and mass of this object is also supposed given small m, fine, now can you find out what is the minimum velocity which is required by this mass to reach the infinity, the hint is try using work energy theorem, potential energy when the object reaches at infinity is zero, okay, but potential will be there when the mass is at the surface of earth and you are launching it, it will have some potential energy, the work done will be zero because there is no other external force other than gravity, okay and for gravity force you are anyway considering the potential energy, so you will not double count it when you talk about the work done, so work done is zero, okay, K2 is what when the point number two is infinity, point number one is surface of the earth, okay, so at infinity kinetic energy is zero, it has just reached the infinity, alright, now U2 is what all of you quickly tell me, U2 is what when the object reaches infinity what will be the potential energy between small m and the capital M, it will be zero, okay, K1 is what kinetic energy initially which is half m into V square, alright and U1, U1 is minus of g M M by radius of earth, okay, now just substitute the values, when you substitute the values you will see that everything else is zero, so these two terms when you equate to zero small m get cancelled, so you will have half times V square to be equal to g M by r e, alright, so from here V is equal to under root of 2 g M by r e, fine, now we know that g M by r e square this is efficient due to gravity, fine, so you can also write this term as root 2 g r e, same thing because g is this you can write this as like that, alright, so when you substitute all the values you will get escape velocity to be equal to 11.2 kilometers per second, fine, so if you throw an object with that velocity, okay, you throw an object with that velocity the object will not be able to come back, the object won't be able to come back, fine, all of you clear, right, fine, so let us take a numerical on whatever we have learned just now, in fact this is from your NCIT book itself, okay, so here goes the numerical, this question came in 1997 J E advanced, okay, so I think after that only NCIT has adopted this question in the textbook, it's a nice question, all of you draw the diagram with me, radius is r, okay, the mass over here is m and here the mass is 4 m, the center to center distance is 6 r, okay, the question goes like this, you have a mass small m, mass small m is there kept on the surface of the first sphere, okay, it is kept on the surface of the first sphere, it is projected from the surface with velocity v, it is projected with velocity v from the surface, you need to find the minimum velocity v, you need to find minimum velocity v for which mass m can reach the other sphere, 4 m, okay, so these are the two spheres kept at a distance of 6 r, small m mass is projected from the first sphere, you need to find minimum velocity v for which the small m can reach the other sphere, try to do this your own, all of you try it, as I tell me what will be the condition for minimum velocity, okay, what will be the condition, will the minimum velocity will be such that when it reaches the other sphere, the velocity becomes 0, is that what will be the condition for minimum velocity, is the minimum velocity, guys I am asking you something, the condition for minimum velocity is what, simple question, the condition for minimum velocity is what, is is the minimum velocity that velocity for which it reaches the other sphere with 0 velocity, just reaches the other sphere, is that the condition for minimum velocity, first answer me that, is that the condition for minimum velocity, so Niranjan how you got that answer, that is wrong then, you cannot get that answer using that, why that is not the condition, why that is not the condition, the question is that for minimum velocity, what is the condition, is this the condition that minimum velocity is that velocity for which it just reaches the second sphere, when it reaches the second sphere velocity becomes 0, okay, why that is not the condition, why this is not the condition, okay, can this ever happen, can this ever happen that when it reaches the other sphere, its velocity is just becoming 0, it can never happen, okay, it is like you are falling from the sky and you are when you are about to reach the earth, your velocity becomes 0, okay, it is absurd, it will never happen, okay, so you need to understand one thing that this mass is under the influence of two gravity, okay, the one gravity, one sphere is trying to pull it this way, the other one is trying to pull it that way, okay, so if let us say this is f1 and this is f2, okay, if f1 is greater than f2, if f1 is greater than f2, then it will automatically attract this mass m, even if this mass somehow reaches this sphere, but since f1 is greater than m2, as soon as it hits this sphere it will come back because f1 is greater than f2 all the while, okay, so if this is valid for entire path from this point to that point, then this situation will never happen, because it will ultimately get pulled back on the first sphere, okay, but if f1 becomes equal to f2 in between and that point onwards f2 becomes greater than f1 because you know forces of gravitation depends on the distance, so closer the mass reaches near the second sphere, the larger the force of the second sphere will be, so this force will slowly increase and this force will slowly decrease, so there will be a point when these two forces become equal and after that point onwards f2, f2 will become greater than f1, so if you cross that point even if your velocity is 0, this mass will automatically pull that mass inside, okay, so basically the velocity should become 0 where f1 is equal to f2 and after that point onwards the second sphere will anyway pull it towards itself, understood? All of you understood? Now the first step is to find the point where f1 become equal to f2, alright, now f1 is what? f1 is gmm divided by let us say r square, small r is a distance where the forces are balanced, so let us say this is small r, okay, this when become equal to g, now capital M is 4m into small m divided by 6r minus r the whole square, right? So when you equate the forces you get a point where the four gravitation forces are balanced, alright and what is that point? You will solve this expression, you will get 6r minus r is equal to plus minus 2r, okay, so you will get r to be equal to 2r, understood? Is there any doubt till now? Any doubt? This is actually an advanced level question, I am not sure what it is doing in NCRT but good to see such question in NCRT, okay? So I have got a point which should be at a distance of 2r from the center of the first sphere where the velocity should become 0, okay, so I can use this work energy theorem between the initial point and the final point, okay? So fine, the work done is 0, what else is 0? k2 is 0 when it reaches that point r is equal to 2r, alright? Now k1 is half m into v square, alright? u1 is what? u1 is potential energy between small m and capital M plus potential energy between small m and 4m, okay? So this is potential energy you need to add it up at 0.1, okay? At 0.1, potential energy between small m and capital M is gmm by the radius r, okay? Minus of gm into 4m, divide by what? Quickly tell me, this divide by what? What is the distance between small m and the second sphere? How much it is? Look at the diagram, 5r, right? So from center from here till this point, the distance is 5r, distance is 5r, from center you need to measure the distances, okay? So this is 5r, alright? Now u2 is what? u2 is again potential energy between small m and capital M plus potential energy between small m and 4m when the particle reaches at 0.2. 0.2 is at a distance of 2 times capital r from the center of the first sphere. So potential energy between small m and capital M is minus of gmm divided by 2r, okay? And then potential energy between small m and capital M will be what? gm into 4m divided by what? Quickly tell me, divide by what will come over here? Distance, 4r, you are getting distances wrong, alright? So just substitute these values, we will get the answer, okay? So these kind of questions, you know, these are little bit involved question, you are not expected to solve these questions right from where like you have learned the concept and you will not be able to immediately start solving these kind of questions. Although this is a solved problem, so you know, if you have book open in front of you, you can easily refer to that, okay? But I don't expect you to get it immediately after, you know, learning a concept. So you may not get such kind of questions in school exams, but then since it is, it's a part of your solved exercise in your NCRT, I thought of taking it up.