 All right friends welcome good afternoon can you type in your names quickly a quick fact as in if you have a Chromecast or if you have a this facility of projecting YouTube on your television you can project it on your television as well okay welcome very few people have joined in is it because tomorrow is your physics UT Ashutosh you know that Ashutosh knows lunch timing of others morning Bharat has morning you just woke up is it Bharat hello okay so I think examined it recently how was it how was I came on test okay fine so let us start anyways this session is getting recorded as well so we can start fine so last first of all let me put a disclaimer here that we are having this session for your UT preparation right because your UT is tomorrow right and gravitation is coming so my focus will be to complete this topic gravitation which is useful for your school exam preparation okay of course there is there are lot more than what we will be discussing just now okay so whatever is other things than the school level that will be taking up next class okay fine so our focus is that you get good marks in school uts right now okay fine so let us continue talking about this particular chapter in the first session we have I think towards the end we derive the variation of accession due to gravity good good afternoon Preeti good afternoon Omkar okay so in the last session towards the end good afternoon Sri Ramya 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 be you know it can vary depending on where you are on the earth surface as well okay 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 9th and 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 rotation you don't need to find a value just tell me the idea how I mean how you can get 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 efficient 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 alright and let us say that the this angle okay the angle this line makes with the equatorial plane it is lambda alright 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 accession 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 quickly ask 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 who all are present is it clear to all of you ask doubt okay ask doubts you can WhatsApp me also your doubts this probably will not be asked in your school exams okay but it is like it may be asked so it totally depends on who is setting the paper in the school exam so this could be asked but usually it is not asked okay let's 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 accession due to gravity okay it's like we were talking in terms of you know it's 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 we 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's simply half mv square okay it's 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 one and that is point number two okay so if I move to point number two like this or if I go like that or whatever path you take like this okay it doesn't matter what path you take if your work done is independent of the path taken then your force for which you're 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 doesn't 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 doesn't 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's say if g is constant if g doesn't 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 forces how much oh that is wrong 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 okay so if you assume the accession due to gravity as constant then it is simply equal to change input why you have retracted your messages 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 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 you know 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 it if it is a prop routine a 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 okay so let's say this is earth okay the earth has radius of earth and mass of earth me okay now let's say this is the path along which I am moving an object okay so suppose I am going from this point point number one to point number two okay this distance is let us say r1 okay and that distance from the center I am finding the distance okay this distance is r2 right now can you use this definition of potential energy which is delta u is equal to minus of negative of the work done by the gravity force to find the change in potential energy change in potential between point one and two so this will be delta u delta u is what u2 minus u1 okay and try to use this formula work done is integral of f dot dr okay this f is the force of gravitation all right now try to do it on your own quickly distances are very large so you can't say gravitation force is mg because g itself is changing let me know once you are done how much is the gravitational force the value of gravitational force at a distance of r is how much let's say mass is m this is small m how much is the gravitational force g m into mass of earth okay divided by r square okay this is the gravitational force all of you clear about it okay say change is always final minus initial doesn't matter the delta u delta x delta y delta t whatever it is it will be always final minus initial find this is the gravitational force isn't it so the work done by the gravitational force is integral of this into dr r1 to r2 is this correct writing like this is this correct is this the work done no there is a mistake there's a mistake over here can you tell me what is a mistake this is an error the direction of force is like this and you are moving in this direction okay there's a dot product between force and displacement okay you are moving in opposite direction of the direction of force so that is a negative sign will come in okay so what you have done is you have taken a random point at a distance of r found out the gravitational force and you have assumed a small displacement along this path let us say this distance is dr okay so you can assume that for that small distance dr you can assume force is constant because dr is very very small okay although force is not constant it keeps on changing the distance r but for small dr you can assume that force is a constant so for dr amount of movement you can find the work done as simply force into dr displacement because dr you can assume force is constant for that dr okay and then you need to integrate that total work done okay is there any doubt till now any doubts till now quickly tell me this is a usual question they ask you in schools derive the expression for potential energy for the gravitation in case of this this kind of scenario okay switching the limits should you have if you switch the limits yeah that is correct what are you saying you learn about it later on but right now let's keep it simple only Krishna Bajaj welcome no doubts till now right till now no doubts all right integral of dr by r square what it is integral of dr by r square is minus of 1 by r okay so we are going from r1 to r2 fine so limits are from r1 to r2 so this will give you the minus minus gets neutralized it's a multiplication so this will be gm me by r2 minus of gm me by r1 okay so this is what work done by the conservative force fine and how the potential energy was defined change in potential energy which is u2 minus u1 is negative of the work done by the conservative force right so minus of gm me by r2 okay minus of minus gm me by r1 okay this is u2 minus u1 okay now can you tell me what is u2 and what is u1 all of you tell me what is u1 and what is u2 can I say that u2 is this this can I say that u2 is minus of gm me by r2 is this correct others quickly type in my simple question is is this correct saying like this is it correct this is not correct okay this is not correct all right fine all right this is not correct but when I say potential energy is mgh when I say potential energy is mgh in work by energy chapter when we said potential energy is mgh we always assume 0 is something and compared to that 0 the potential energy is mgh okay so we never found out the absolute potential energy we first assumed what is 0 and then compared to that potential energy we say it is mgh and whatever is the case okay and hence in this case also we need to define what is 0 potential energy okay but the problem is that now it is not just about earth surface are you getting it we are talking about stars we are talking about planets we are talking about interaction between sun and other planets so I cannot say that let us assume 0 potential energy as a surface of the earth fine it has to be something which is unique to everything all right so that is the reason why we need to define the 0 potential energy in a very careful manner so that that is universal okay so let us try to do that let's say r2 is tending towards infinity if r2 tends towards infinity this term will go towards 0 okay so I can say that u infinity minus u1 is equal to minus of or plus of gme by r1 okay now till now we haven't assumed anything all right now if you assume okay if you assume u infinity to be 0 then you can say that potential energy at 0.1 is minus of gme by r1 okay so when you say that this is a potential energy between the two masses m and capital Me you already assumed that the potential energy between these two masses will be 0 when they are separated by an infinite distance okay so write it down in your notebook that the 0 potential energy the 0 potential energy is a situation in which the masses are separated by infinite distances fine so in work by energy chapter you try to I mean depending on what kind of problem it is you always try to find out where appropriately you can define 0 potential energy right every time you define 0 potential energy in work by energy chapter but in gravitation chapter there is this unique situation in which gravitation potential energy is 0 you don't need to define every time every new problem what is 0 gravitation potential energy here we are assuming that gravitation potential energy is a situation in which the two masses are infinitely separated from each other is this thing clear to all of you any doubts till now anything okay all right so what we have learned just now is the potential energy between the two masses is minus of gmm by r okay so this is the potential energy between two point masses or two uniform spheres separated by a distance r okay so once you define the potential energy you can you can use the work energy theorem okay anywhere you may want to okay now understand one thing that this is negative there is a negative sign over here so if r decreases what will happen to potential energy it will increase or decrease potential increase or decrease if r decreases correct potential energy will go down okay so earlier if it is let us say minus 100 joule if you decrease the r it may become minus 200 so minus 200 is less than minus 100 fine there is a negative sign over there all right and also you need to remember few things like potential energy you got this formula this formula is valid for only two masses fine but then there can be multiple masses okay for example this situation all of you draw this suppose you have a square okay this is a square let's say you have four masses masses m and the distance between the adjacent masses is a it's a square all right can you find out the potential energy of this system okay the formula you have is only for two masses but right now you have four masses okay now can you try to find out the potential energy of the given arrangement quickly do that the hint is you can take two masses at a time and add all the potential energy okay potential energy is a scalar quantity okay you don't need to worry about directions and all you have four masses yes you need to take care of diagonals as well potential energy will be potential energy between one and two potential energy between two and three between three and four four and one one and three and two and four okay these kind of combinations can be there in fact you can easily apply the combinator x four c two is six now six combinations will be there okay now u12 is equal to u23 u34 u41 so just find out one and multiply with four so this will be equal to minus of four times gm square by a this is equal to sum of these four okay then you need to take care of the diagonal ones as well okay you need to take care of the diagonal ones these two and those two you may think that I have already counted mass two how can it have an energy with four I have already counted two with one two with three how come it has with four also the point is it doesn't matter how many masses are around the forces between the fourth and the second one is unaffected by what are the masses around between these two masses the force will be g m1 m2 by r square and because of that force only the potential is coming in right so it doesn't matter how many masses are around you can always count one mass multiple times when you are talking about the potential energy fine now potential between one and three is what between one and three the distance is a root two okay the diagonal length is a root two this is a this is a the distance between one and three is a root two same as the case with two and four as well so these two are equal so you can say that minus of two times gm square divided by a root two fine and then you can simplify it further you can say that root two cancelled it will be like this so minus of four plus root two gm square by a any doubts till now anything anything you want to ask no doubts no doubts okay you need to now find out find out the work done to create the system this is the system I am talking about you are creating this this system by bringing these masses from infinity so initially they are separated infinitely apart okay so you are bringing these masses from infinity and making this arrangement okay now in order to make this arrangement how much work is done by you is what you need to find out okay properly solve it okay properly solve it use work energy theorem so whenever someone is asking you to find the work done immediately this should come in your mind work energy theorem okay we are talking about the work done to create the system okay so the first point is at when the masses are infinitely separated from each other right and our gravitational potential will be zero if masses are infinitely separated from each other gravitational potential is zero u1 is zero initially when they are at infinity even k1 is infinity and then you are bringing them slowly okay so finally when they are static like this then also k2 is zero okay so whatever work you need to do is equal to the final potential energy itself okay so you need to actually do the negative work okay you need to do this much work whatever is a potential energy to build the system to create the system so work done is equal to potential energy itself is this thing clear to all of you any doubts no doubts now find out work done to destroy the system destroying the system means you are taking these masses from this system and placing it in infinity so totally dismantling the system from wherever you have brought it work done will be negative yes because the forces are attractive right so they are trying to come close to each other you have to move slowly so you need to apply force in opposite direction of the attraction so that is why your direction of force is in opposite direction of the displacement so that is why work done will be negative now work done to destroy is how much now your initial point is this and final point is infinity okay so the work done which is equal to u2 plus k2 minus u1 plus k1 okay so first point is this and second point is infinity so u2 is zero k2 is zero k1 is zero so the work done will be negative of the potential energy which is this only okay right so these kind of questions are very common in school exams all right so they they ask you I mean you can understand that it is like to create a system to destroy the system and what is a potential energy these things are in a way they are trying to ask the same thing as in what is the potential energy of the system so depending on that you will need to do some amount of work to destroy it or to create it all right so whenever like there is a question in school where they're trying to ask you or they are asking you how much work is done to create a system first step is to find the potential energy of the system okay and the next step is to use the work energy theorem okay so these are straightforward questions which are routinely asked in school no doubts till now right nothing 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'll 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 till 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 all right radius of the earth is r e 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 potentials will be there when the mass is at the surface of earth and you're launching it it'll 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 all right 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 all right 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'll 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 all right so from here v is equal to under root of two 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 two g r e same thing because g is this you can write this as like that all right 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 we 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 nct book itself okay so here goes the numerical this question came in 1997 j e advanced okay so i think after that only nct 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 4m the center to center distance is 6r 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 4m okay so these are the two spheres kept at a distance of 6r 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 tried i said 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 zero is that what will be the condition for minimum velocity is the minimum velocity guys i'm 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 zero 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 got how how you got that answer that is wrong then you can't 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 all right is 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 zero okay why that is not the condition why this is not the condition away your don't try to prove that point again and again my question is something different okay my question is something different okay can this ever happen can this ever happen that when it reaches the other sphere its velocity is just becoming zero it can never happen okay it's like you're falling from the sky and you're when you're about to reach the earth your velocity becomes zero okay it's absurd it'll 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 if let's say this is f1 and this is f2 okay if if f1 is greater than f2 all right 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'll 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 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 zero this mass will automatically pull that mass inside okay so basically the velocity becomes velocity should become zero 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 all right now f1 is what f1 is g mm divided by let's say r square r small r is a distance where the forces are balanced so let's 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 force are balanced all right and what is that point you'll solve this expression you'll 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 advance level question I'm 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 zero okay so I can use this work energy theorem between the initial point and the final point okay the work done is zero what else is zero k2 is zero when it reaches that point r is equal to 2r all right now k1 is half m into v square all right u1 is what u1 is potential g between small m and capital M plus potential g between small m and 4m okay so this is a potential g you need to add it up at point one okay at point one potential g 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 sent from center you need to measure the distances okay so this is 5r all right now u2 is what u2 is again potential g between small m and capital M plus potential g between small m and 4m when the particle reaches at point 2 point 2 is at a distance of 2 times capital R from the center of the first sphere so potential g between small m and capital M is minus of gmm divided by 2r okay and then potential g between small m and capital M will be what gm into 4m divide by what quickly tell me divide by what will come over here distance 4r you getting distances wrong all right so just substitute these values you 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 right so next let us talk about another very important concept in this chapter that is satellites so today we'll be talking about this as the last topic for today satellites what are satellites anyone what are the satellites satellites are the objects or a body that revolves around what yeah that revolves around a satellite sorry that revolves around a planet okay so it's very simple an object revolving around planet you call that as a satellite right and what is planet a planet is an object that revolves around the sun okay like that you can understand it is just a definition of it okay so now let us try to analyze the behavior of satellite okay let's say this is the orbit of the satellite satellite will be moving in a circular orbit okay we are assuming that that it will be moving in a circular orbit if it comes under the influence of neighboring planets that there will be slight distortion in the orbit will be there but more or less it will be circular only so all the motion in this particular chapter we are assuming it to be circular only whether it planet revolving around the sun or satellite revolving around the planet so we are assuming it to be circular only okay so let's say this is earth okay we are talking about earth satellite let us say okay the radius of the revolution for the satellite let's say this is my satellite okay this satellite is moving okay this is moving like this with velocity v okay mass of the planet let's say earth is that planet let's say me and mass of this satellite is small m okay by the way these satellites can be natural can be artificial also for example moon is a natural satellite okay and whatever satellites are launched let's say if isro launches a satellite or gps satellites all those are artificial satellites okay satellites have its own importance so there are different purposes satellites are used because there once you are above the earth surface your field of vision is enlarged you can literally scan through entire earth so it can be used for military purposes or you know there are multiple things like for example you can use it for predicting the weather or movement of the monsoon there are so many things you can do with the satellites okay but we'll not get into the all that right now let us first try to analyze the motion of a satellite okay let's say its mass is small m the velocity is v okay so can you write down the force equation first as in the gravitational force is equal to centrifugal force can you write that equation all of you quickly do that okay so toward the center if you see there will be a gravitational force like this okay and then there will be a centrifugal force like that centrifugal force will be equal to mv square by r right and this gravitational force fg will be equal to g mass of earth into mass of planet divided by r square fine so this force should be equal to centrifugal force mv square by r right so this is just force balance this must be happening if a satellite is moving around in a circular orbit okay all right now from here a very interesting thing will come up so you'll see that if r get cancelled off you'll get m into v square is equal to g m m e divided by r all right if you multiply half both sides this is nothing but kinetic energy okay all right and potential energy is what potential energy between the satellite and the earth this will be minus of g mass of satellite into mass of earth divide by the distance between them right so this is the potential energy and that is kinetic energy which is equal to this okay so you can see that kinetic energy is nothing but minus two times of the potential energy yes or no all of you able to appreciate okay wait it is minus one by two times minus half times the potential energy yes Bharat Sushant and Abhay said yes earlier itself okay is this thing clear to all of you any doubts please ask quickly okay I realize feeling the pressure of ut tomorrow you have a lot of pressure so you can say potential energy is minus of two times of kinetic energy okay good to feel the little bit pressure little bit pressure is good okay so but then I think you'll do good only because school level exams they will not be too much involved as in but then of course there will be some let's say theoretical questions will be there then some derivations will be there okay so you will also realize that in school exams you lose most amount of marks you lose most amount of marks by by you know by for suppose there were one or two marks question two marks question three marks question one mark question so you get those one or two marker wrong that is where most of your marks are gone okay so you'll hardly get any five marker wrong okay so most of the time it will be one or two markers so I would suggest that when you prepare for the school exams okay do not read the theory from the book again and again that will never help okay so you you might have some reference book for your school exams for example uh Pradeep is there I guess then Aurora all these kind of books are there you just scan through those one or two marker questions okay they are standard questions which are repeated every now and then okay so once you scan through those questions you will realize that if you sit in school exams the questions from those only will be asked right so it's not like when you prepare for J you need to be very open minded and you should have a very very vast amount of problem practice that's not what is required in school they test you on the theory derivations and there's some theoretical aspects okay that that you will not be able to master if you read the chapter only okay reading chapter is counterproductive you're wasting your time when you're preparing for your exams okay reading of chapter or learning a concept should be done in someone teaches you and then after that you should just do problem practice okay so try that in this ut uh just scan through a lot of questions yourself from the book rather than reading the the theory from the book again and again okay anyways coming back to this potential energy is minus of two times of the kinetic energy so total mechanical energy is what potential energy plus kinetic energy okay so when you add it up you will have potential is minus of two times of kinetic energy right so plus kinetic energy so total energy is negative of kinetic energy this is total energy of the satellite all right so if you know that total energy is t kinetic energy will be negative of t and potential energy will be negative of the two times of the kinetic energy all right our potential energy is two times of the total energy. So like that you can relate kinetic energy, potential energy and you can play around with that. No doubts, right? No doubts. Now let us try to derive the Kepler's law of orbits. Things are clear now. Any doubts? Crystal clear okay. Now for the satellite, this is treated like a numerical okay. For a satellite of mass m revolving around radius of r about a planet of mass capital M, alright. These things are given for this you need to find the time period. Find out, draw the diagram okay and then try to find the time period quickly. By the way in your NCRT they always have considered distance from the center. They write it as RE plus height from the surface okay. So do not get confused okay. We are taking distance from the center. So that is why we are writing small r when you read from NCRT or I do not know how they have taught in school. They will write radius of earth plus h okay. So I have directly taken it as r. But a g is not constant, g is the variable. Yeah so g depends on r, you need to substitute that as well. I want only as a function of r, small r. What is the time period? See always guys remember whenever something is moving in a circle okay. If any object moves in a circle, the first equation you write is the central force. The force towards the center should be equal to mv square by r okay. That equation you should write just close your eyes and write that equation before even thinking of anything okay. So central force is gravitational force, gm1 m2 by r square should be equal to mv square by r. So that is where you start solving the question whenever something is moving in a circle okay. Alright let me solve this now. So this is small m, the satellite has mass small m, it moves with velocity v okay. So I can write here gm into capital M divided by r square this is equal to mv square by r alright. So from here I will get velocity or the magnitude of the velocity to be equal to under root of gm by r okay. This is the velocity or the speed with which the satellite is revolving around the planet okay. Time period of this satellite is 2 pi r which is distance divided by the speed v okay. So you can see that it will be equal to 2 pi r multiplied by this r divided by gm right. So t will be equal to 2 pi root of gm r raise to power 3 by 2 right. Now square both sides you will get t square is equal to 4 pi square divided by gm into r cube right. This which you can say that t square is proportional to r cube alright. So although you have derived t square proportional to r cube for the case of satellites but same kind of equations you can write for a planet also a planet revolving around the sun exactly same thing will come out okay. So you can see that Kepler has arrived at this particular expression without knowing any formula any equations just purely by observation he came up with this particular relation that t square is proportional to r cube fine. So I think I have covered almost I mean I have covered majority of the chapter with respect to your school curriculum okay. There will be small side topics that are there. So we could not cover it but I think we have covered enough so that you can do well in your schools yeah Ashutosh that is another one small topic weightlessness is there. So you want should we cover now itself you have time you have time or you want to sell study for your school exams now in the range and one study till four okay I can teach you till six also. So I will quickly talk about weightlessness okay just give me one minute I will just quickly talk about it okay. So suppose you are orbiting in a spaceship okay or you are inside the satellite this is your satellite okay very small satellite it is okay. So here is an object inside okay. Now we are trying to see what is the force the net force this object inside the spaceship will feel okay. Now since this object is revolving around the radius of r okay it must be feeling a centrifugal force outward which is mv square by r right and the same object must feel gravitation pull also okay which is equal to g mm by r square okay. So the net force net force on the object is what gm and by r square minus mv square by r this is the effective force okay which should be equal to 0 which should be equal to 0. So I mean is this thing clear is this thing clear all of you quickly type so net force is 0. So the object inside will not experience any force I mean the net although individually there are forces this force and that force but they add up to 0 and hence you do not feel any force so even when you place your feet on the floor of this spaceship the spaceship will exert zero normal reaction because net force is anyway balanced so normal reaction is not required fine. So if normal reaction is 0 it will be like a free fall situation or you can say that weightlessness you will not feel any pull towards the earth clear. So this is what the weightlessness is about all right. So all of you do well in school UTs okay that is something which should be taken care automatically okay it should never happen that you you are able to do good in let's say J or J advanced preparation but you're not getting marks in school UTs then there is something wrong in which something wrong the way you are studying okay if your concepts are clear you should be able to get marks at all levels okay. So when you prepare for the exams like school which have lesser scope compared to the competition all you can do is for theoretical question just browse through or read through those one or two marker question where you lose most amount of the marks okay because five marker you I don't think you'll lose mark in five marker questions because they are not many they're just couple of them you can try to understand how the derivations and all those things are done and you can easily score marks in five markers all right okay guys thanks for coming on Sunday afternoon we will meet again soon bye bye