 Alright friends, so in the last session towards the end, so we were discussing about variation of accession due to gravity with height and with depth, isn't it? So a small variation to it as in the accession due to gravity can vary depending on where you are on the earth surface as well. For example, if let's say this is the equator, okay, this is equatorial plane and this is the axis of rotation of the earth suppose, fine, so what I am saying is, this is, I hope you have done your geometry properly in class 9, 10, 10th, you understand what is equatorial plane, right? So this is the axis of rotation of the earth, let's say this axis of rotation, earth is rotating with angular velocity omega, okay. Can you tell me quickly what is the angular velocity of the earth's rotation? You don't need to find a value, just tell me the idea how, I mean how you can get the angular velocity. The time period of revolution of the earth is how much? Time period of revolution of the earth is 24 hours, okay. So omega is 2 pi divided by time period, okay, T where T is, in terms of second if you want to write, it will be 24 hours, 60 minutes, 60 seconds. So this many seconds will be the time period of the earth, fine, so this is how you get the angular velocity of the earth itself. So let us see whether accession due to gravity can be changed because of the angular velocity, okay. Let's say an object is placed on the earth surface over here, okay. So it is on the earth surface, all right and let us say that this angle, okay, the angle this line makes with the equatorial plane it is lambda, all right and the radius of earth is let us say r, okay. Now can you tell me what will be the net force towards the center on this mass or you do one thing, you just represent all the forces on this mass, forces and acceleration on this mass. Can you do that? Let's say mass is m, okay, as I tell me one thing is this mass rotating in a circle, yes or no? Is this mass moving in a circle, right and what is the radius of that circle? What is the radius of that circular motion? This mass moves in a circle like this, isn't it? What is the radius of the circle? Radius circle is this, right? So let's say this is the radius of circle r on which this mass is rotating. The small r has to be equal to r cos of lambda, how you get that? Just drop a perpendicular on the equatorial plane, okay, then this distance from here to here this is the value of r also, okay, this is the value of r which is r cos lambda, all of you clear about it, clear, okay. Now that mass is moving in a circle of radius r cos lambda, so it will experience a centrifugal force, let's say in this direction, okay, since it is moving in a circle, how much that will be equal to m omega square into radius of revolution of, right, so that is r cos of lambda, fine, okay. And then there will be a force towards the center of the earth, how much will be this force? This force will be equal to m into g only, right, whatever is adjacent to gravity on the surface towards the center it will experience the force, right, it will be mg, okay, clear, right, all of you clear about it, any doubts? Somebody asked me the topic name is accession due to variation in accession due to gravity because of the rotation of the earth, okay, the variation in g due to rotation, rotation of the earth, clear, right, no doubts, no doubts anyone, okay, now let's say what is, let's see what is the net force along the, along this direction which is towards the center of the earth, okay, so this angle is also lambda, okay, this angle is also lambda because that is lambda, right, so there will be a component of this, component of that force, okay, which is equal to what? Component of that force is equal to m omega square r cos lambda into cos lambda, so that will be equal to cos square lambda, okay, and there will be another component perpendicular to this, like this, okay, that will be m omega square cos lambda into sin lambda, okay, but my interest is just to find out what is the net force towards the center of the earth, which is equal to mg minus m omega square r into cos square lambda, fine, so let's call it as m into g effective, right, whatever is the force towards the center that should be equal to m into the value of g effective, right, that is how we have been deriving g effective or variation of g till now, okay, so g effective becomes equal to g minus omega square radius of earth into cos square lambda, where lambda is latitude angle, okay, so this is how you get the variation of g with respect to the, you know, latitude angle and rotation of the earth, okay, so usually we ignore the variation of g because of the rotation of the earth because the variation is very little, but then of course there will be variation, tell me where the value of g will be maximum, the g will be maximum when? g will be maximum when cos lambda is 0, okay, or lambda is 90 degree, that happens at the pole, okay, so g will be maximum over here, no, lambda is equal to 0 is not the point where g is maximum, okay, lambda should be 90 degree, then only cos of lambda will be 0, okay, so that is north pole and south pole, so at north pole and south pole the value of g will be maximum, okay, and at the equator the value of g will be minimum where lambda is 0, is it clear to all of you, okay, let us go to the next concept, so till now we have been talking about forces, right, we have been talking about forces and the variation of adjacent due to gravity, okay, it is like we were talking in terms of, you know, it is like laws of motion of gravitation, okay, something like that, now we are talking, now we are going to discuss about energy consideration or you can say work by energy inside the gravitation, okay, so in order to understand the energy consideration we need to define the potential energy, okay, so potential energy in gravitation we are trying to define now, once you define the potential energy it becomes straightforward, all you have to do is to use work energy theorem then, okay, work done is u2 plus k2 minus u1 plus k1, so you will be using this only again and again, once you are clear how to write potential energy in case of gravitation because work done is very clear, right, work done is force multiple of f into dr integral of that or if it is a constant force then it is f dot displacement and kinetic energy is also very clear, right, if it is a rigid body you can write its kinetic energy, if it is a point object it is simply half mv square, okay, it is only about potential energy where we need to pay attention how to write the potential energy, okay, now do you guys remember the definition of potential energy, how we define, how we define potential energy, first of all potential energy is defined only for conservative force, conservative forces, okay, so we have discussed about conservative forces in work by energy chapter, right, what are conservative forces, conservative forces are the forces in which the work done by the force do not depend on the path taken, it only depends on what is the initial point and what is the final point, suppose this is the point number 1 and that is point number 2, okay, so if I move to point number 2 like this or if I go like that or whatever path you take like this, okay, it does not matter what path you take, if your work done is independent of the path taken then your force for which you are finding the work done that force is conservative force, okay, so you can say that the work done by that force is a state function, okay, just like internal energy, you guys remember internal energy, yes or no, in case of internal energy the change in internal energy only depends on the change in temperature, it does not matter what is the process you are going through, it will be always equal to n Cv delta T for gases, okay, similarly these are the special kind of forces for which the work done does not depend on the path taken, okay, since it is a state function you can define a state function with respect to the work done and that state function is you know we call it as the potential energy, okay, now the definition of potential energy is this, change in potential energy should be equal to the negative of the work done by the conservative force, okay, now let us say if G is constant, if G does not vary, okay, then assuming that you lift a block of mass m from here and you have lifted up to a height of H, okay, so this is your mass, so you have lifted the block up to a height of H, okay, can you tell me what is the amount of work done by the gravitational force, assuming that G is constant and gravitational force is mg, always downward mg, right, so can you tell me work done by the gravitational force is how much, mg is acting downward and displacement is upwards, okay, so work done by gravity is minus of mg H, okay, so change in potential energy which is negative of the work done by conservative force which is mg will be equal to mg H, so if you assume accession due to gravity to be constant, then change in potential energy will be equal to mg H and that is what we have been using throughout, okay, that is what we have been using in work by energy chapter and wherever you ignore the variation of accession due to gravity with height that is how you find the potential energy, okay, but in this chapter we are talking about scenarios where you take a mass let's say 100 kilometer above the earth surface or thousands of kilometer from the center above, okay, so when you talk about those kind of distances you cannot ignore variation in accession due to gravity, okay, and since you cannot ignore variation in accession due to gravity, hence you need to use this particular definition of potential energy to define the potential energy again, this is an approximation, okay, when you say mg H, mg H is an approximation, it's not the exact thing, okay, so let's try to define the potential energy in case of gravitation in a more let's say accurate manner, but whenever you see a mechanics question and if you think that whatever is happening is, I mean if it is a routine mechanics question, don't take variation in accession due to gravity, it is assumed that everything is near the earth surface and G is constant, okay, so looking at that question you can easily make out, you know, that whether you should consider gravitation potential energy as mg H or what we are going to derive now,