hi complete lesson in gravitational field uh see the description of this video already have done uh several videos related to different units uh related to unit five already I have done uh thermodynamics uh cosmology complete lesson in cosmology complete lesson in uh thermodynamics and three different lessons which covers the whole portion of the nuclear Decay so that uh you URL or the link for these videos are given in the description of this video uh also by using the playlist of this channel you can navigate to any video uh you would like to uh watch right okay so uh in this lesson I'm going to discuss uh the unit five topic gravitational field already cosmology is discussed already I have done it already Okay so uh first you should know what is gravitational field same thing like electric field electric field we Define the region where a charge experiences electrostatic force here we Define region where a mass experiences gravitational force is defined as gravitational field region where Mass experiences gravitational force is defined as gravitational field okay so that you know that second next you should know the how to define the strength of the gravitation ation field which is called gravitational field strength and it is denoted by G simple G comparable analogy you can compare with electric field strength how do we Define the electric field strength uh the electrostatic force per positive po of charge electrostatic force per colum positive colum charge is defined as the electric field strength at a point but here uh we don't have positive mass or negative Mass so here we Define the gravitational field strength as gravitational force per unit Mass gravitational force per unit mass is defined as gravitational field strength at a point but normally uh here the point masses we don't consider I'll come to that point later but here the gravitational field strength is defined as gravitational force gravitational force per unit Mass means we are divide by mass so the notation symbol G gravitational force F mass m so FAL mg the first equation the unit of gravitational field strength the unit of [Music] G will be unit of force which is Newton Mass per kilogram if you want you can give it in uh s unit you give it sorry base units Newton means I call Newton Second Law FAL M A okay so Newton means f = ma so Newton means kog M second - 2 per kog you'll get m per second squ okay this gives an important uh conclusion or it gives a new IDE idea that means gravitational field strength the unit is same as m/s squ which is the unit of acceleration okay so now we are going to compare the gravitational field strength and acceleration of free normally you know the questions related to gravitational field we don't consider a point mass or we don't say normally the posi you can say object a exerts gravitational force on object B we don't consider because gravitational force is weaker much weaker because there we are going to get an equation you already would have done it f = g M1 M2 r² where G is a very small numerical constant so normally when we consider gravitational force there should be between two objects one must be one of the objects must be a planet or a moon or a Galaxy or a star the other one could be anything even a pen doesn't matter but one of the objects must be massive something like a planet or moon or star or a Galaxy like that because to exert a significant gravitational force one of the object should be massive the other one could be uh lighter no problem uh so there we normally consider the gravitational field or gravitational field strength created by a planet or moon or star or a Galaxy so there we can compare the acceleration of free fall of an object towards the planet or acceleration of freeall of an object towards the star or Accel freefall of an object towards a moon or Accel freeall towards a Galaxy but not much applications uh we are going to compare the ACC freeall and the gravitational field strength are they same or different we'll get both are the same I'll show that now okay so already we got an equation that is from the definition of gravitational field strength uh which is FAL mg already we got the equation FAL mg is the first equation now G we defined as gravitational field strength which is uh equal to gravitational force per unit Mass so and when we found the base unit of G we got m/s squ which is the unit of acceleration so now we are going to compare the gravitational field strength and Accel freeall of an object towards a planet are they same or different okay I told them the questions related to gravitational field strength in this topic there will be a planet or moon or star or Galaxy the other one could be an object so normally we consider the gravitational field strength of a planet at a distance or gravitational field strength of a moon at a particular distance like that we don't consider gravitational field created by a pin at a distance gravitational fields created by a cricket ball at a distance no we don't consider those things because they are much much weaker right we'll come to that point okay now we'll consider whether they are same so for that we'll consider a planet or moon or uh Galaxy whatever it is its mass Let It Be capital M so in this topic remember that always the distances are measured from the center of Mass to center of mass in uh Electric field point charges we don't consider the dimension of a point charge so there's no problem when we use the equation FAL k q1 q r² or eal KQ r² R is always measured from the point charge Point charge means it which has no Dimension but when we consider a uniformly Charged uh sphere there I told we assume the whole charge is concentrated at the center this all about unit four electric field it's comparable analogy we can compare the most of the equations in electric field and the gravitational field but here all the distances Whenever there is a question Rel distance any equation but we are going to get those all equations whenever there's a distance in the equation those distances are always measured from the center of mass okay so here there's a planet uh it's Mass planet or moon or Star whatever it is its mass is capital M okay so the gravitational field strength of this particular the planet at Point P Let It Be G that means if I keep an object now I can say object one of the uh you know uh we are going to learn Newton's uh law of gravitation where we'll get an equation for the force between two uh masses one Mass should be related to a massive object like planet or moon or something other one could be a small object no problem okay so here is a massive object which is a planet there's a small object which has mass m and the gravitational field strength of this planet at Point p is G and at that point the mass m is kept means the gravitational force is always attractive so the gravitational force acting on this Mass according to this definition F equal m so according to the definition of gravitational field strength the gravitational force exert from the point a small Mass kept at Point P at Point P the gravitational field strength of this planet is G so the gravitation force acting on the mass m at point B according to this definition is FAL mg let it be first equation right now we'll think about Newton's Second Law you know the object which has mass m has only unbalanced force we'll neglect all the drag Force nothing is there no drag Force so the only force acting on the object is the gravitational force F so the object has an unbalanced force towards the planet so if I use the Newton Second Law now f ma if I use FAL ma if I use Newton Second Law the only force acting on the object is f and its mass is M the acceleration Free Fall is a second equation where a is the acceleration of Free Fall here G is the gravitational field so I got the first equation by using the definition of the gravitational field strength second equation I'm getting by using Newton Second Law completely different laws this is the definition of g this is the Newton Second Law now compare the forces the object has only one force which is f that is the gravitational force so the force acting on the object is the same Force so the left hand side is the same there for right hand side should be the same so that means ma a equal to mg where a is the acceleration of Free Fall G is the gravitational field strength so M and M get cancel a is equal to G that means the acceleration of free fall of an object towards a planet is equal to the gravitational field strength of the planet at that point again I'm saying the acceleration free fall of a object towards a planet at a point is same as or equal to the gravitational field strength of the planet at that point so this idea we don't have in electric field but here it's a New Concept so a is equal to gation field for and the gravitational field strength of a planet or moon at a point is the same they are the same you must know this uh is not a derivation but you should be able to show both are the same I think once this question came in the XL pass paper Okay so there are two objects okay planet and object or whatever it is uh with different masses M1 M2 their Center of masses are o1 and O2 respectively and the distance between the two Center of masses is R so this is you know the gravitational force is always attractive so Newton's law of gravitation so time the pass they ask to State the Newton's law of gravitation so the answer should be like this gravitational force is always attractive and this force is directly proportional to each mass and inversely proportional to square of the distance between the two Center of masses okay that's a uh Newton's law of gravitation so F proportional to M1 same thing like kum's law what we do for the Kum between two point charges same similar equation we'll get it f proportional to M1 there are q1 we put here M1 F proportional to M2 inversely proportional to r² so when I combine all these proportional to M1 M2 over r² so there must be a constant that is FAL G the constant is called Universal gravitational constant f = g M1 M2 over r² you can compare this with g f = k q1 Q2 r² but we used in electric field in unit 4 same similar type f = g M1 M2 r² where M1 M2 must be in kilogram R must be in meter so G is called Universal gravitational constant so the value of G normally given in the data 6.67 10 the^ minus 11 that's the reason I told the gravitational force between normally between two objects very small because the G value the index is very small so if g = 6.61 10^ - 11 Newton M squ uh kilog minus there in the Kum constant newon m s k minus 2 here kilog minus easy to remember so remember where R is the distance between the two Center of masses okay so we are going to get an equation for the gravitational field field strength created by a gravitational field strength of mass m so the mass m is a massive mass at a distance R from the center of mass of this object so this object or planet or whatever it is it has a mass capital M what will be the gravitational field strength of this Mass capital M at a distance R from it from the center of mass so at a distance R we imagine a point p and at that point a mass a small small Mass a is kept okay so the only force acting on the mass m the small M which is at Point P we are going to find the equation for the gravitational field strength at Point P so the gravitational force acting on the mass m is given by FAL g m m/ r² according to the definition of uh gravitational field uh according according to the Newton's law of gravitation but we know that gravitational field strength means gravitational force per unit Mass so the gravitational field strength at point B so gravitational field strength G at Point P that is g equal g means gravitation Fields I shortly wrote gravitation Fields I would have write it in words but I just put it in short symbol G at Point p is given by G is equal to gravitational force per unit Mass so know that the gravitational force acting on mass m is f = g M1 M r² so the gravitation field Str means Force per unit Mass I need to divide this one by the mass of the object which is kept at Point P so that is g m m/ r² / M so what will happen m m will get cancell GM r² so G = GM r² is a very important equation in gravitational field similar equation we got in electric field where e is equal to k q r² you can compare the similarities or the same pattern gal g m r² so this one next equation g = g m r² that means the gravitational field strength of a particular Mass capital M always inverse Square L that means G is constant m is constant that means G is inversely proportional to R squ okay so if the radius of this planet or radius of this uh moon or whatever it is capital r then the gravitational field strength on its surface what will it be use the same equation G equal the gravitational field strength on the surface gal G m/ r² where R is the radius of the planet normally the planet or moon or a star is spherical shape so if we want to find the gravitational field strength on the surface of it or the acceleration of freeall I can say Accel free fall on the surface of it GM over r² where R is the radius of the planet or Moon whatever we consider right so gal GM o r squ is a very important okay so if there's a planet of mass m which normally the planets are spherical shape uh and the radius of the planet if it is capital r the center of the planet is O the radius is capital r then the gravitational field strength on its surface is given by gal g m r² and at a distance are from the center of the planet which is outside the planet at Point P it's given by gal GM over r² where R is the distance of the point P from the center of the planet so that OB inverse square or if we plot a graph of G against R and if I start R from the center of sorry if I start R from the surface of the planet then the graph will be like this so here this is the gravitational field strength on the surface of the planet but if the graph is we plot a graph by cursing from the center of the planet so this is g0 and this is R from the center of the planet until the radius of the planet uh we will consider that now but before that so until the radius of the plan what could be the gravitational field Str I'll explain it but so far no questions based on it but on the surface of the planet if the gravitational field strength is G1 like here Beyond it it will obey inverse Square l so there if I say this is G we say the gravitational field STR on the surface of the planet if it is G not or G will leave it as G so at distance R1 if it is G1 at distance R2 if it is G2 then G1 R1 so inverse Square law is it so G1 R1 s equal to uh G2 R2 2 is equal to on the surface of FL G r² okay because it to is inverse the multiplication of g into r² should be the same constant okay so you can ask me what could be the gravitational field strength inside the planet when the distance from the planet say from the center of the planet is less than the radius of the planet inside the planet what could be the gravitational F set what could be the shape of the graph from 0 to capital r where capital r is the radius of the what could be the shape of it okay I can get that I can derive an equation for that by assuming G by taking the gaus Law which is not in your syllabus but as uh unit 5 A2 exam uh it's like a uh University entrance exam something like this so they could introduce it and asked to uh get an expression for the gravitational field strength inside the planet when the radius when the distance of the point from the center of the planet is less than the radius of the planet so there we need to use a gaus law so the gaus law says we need to consider where the m is the mass enclosed within the sphere of radius R so for example if I want to find the uh gravitational field strength at a distance R from the center this is the center r r r from the center at Point P I should consider the mass enclosed within the sphere of radius symbol r that is the mass m we should put we should substitute because outside the planet when I consider for example what I said g = GM r² outside the pl where m is the mass of the planet I said okay if you consider a sphere of radius symbol R here this R from the center to this point p where the mass enclosed within the sphere of radius simple R is the whole mass of the planet so that's G Law is correct if we consider outside the planet but when we consider inside the planet also where M must be the Gus law says Mass enclosed within the sphere of radius R where R is the distance at which we are finding the gravitational field strength so if I want to find the gravitational field State at a point B which is inside the planet I should know the dist distance of Point V from the center which is symbol R so there I should consider the mass enclosed within the sphere of radius R okay I can find it but only one assumption I should take the density of the planet is uniform if I take that assumption then I would be able to derive an expression for the gravitational field strength inside the planet okay we take the density of the planet is uniform uh by using it we'll find the by using that assumption uh we'll find the mass enclos with the radius R where R is the distance from the center of the planet so a the mass of the planet is equal to 4x3 Pi R that's the volume of the planet into its density row so row is going to be m/ 4x3 < R we assume the density is unit form everywhere on this inside the planet so the mass enclosed within the point P the spherical shape is M that is going to be 4x3 piun RQ into the density volume into density row so that is 4x3 piun R CU row inste of row I can substitute capital M over 4x3 pi r CU is it I can use even this one I don't need to convert this Mass I can use in terms of density row because density is uniform constant we are taking okay we can use in this equation so now the gravitational field the same equation g g r² g law says m is the mass enclosed within the sphere so the gravitational field strength inside the sphere at distance symbol R from the center of the planet at Point p g is equal to G the mass enclos is M which is 4x3 piun R Cub Row r² the distance of Point P from the uh center of the planet so here G is constant pi r so we'll solve it it will become 4x3 Pi G row 4x3 Pi G row R cu r² into R see these all constant 4x3 is a number Pi is a constant G is a constant row is a constant so gravitational field strength inside the planet is directly proportional to the distance from the center of the planet so now if I plot the graph gravitational field strength from the center of the plan how it varies up to the radius of the planet the gra will be a straight line so graph will be a straight line because G proportional to R this is the gravitational field strength on the surface of the planet then beyond the surface of the planet it will obey inverse Square this would be the uh graph we should plot so the variation of gravitational field inside the planet is G proportional to R so nothing new actually it only the G Law we need to use and we take an assumption uh the density of the plant is uniform at all points so in the exam they might state that and ask you to get an expression okay so most of a satellite around a planet in circular path the same derivation what I'm going to do could be used for motion of a planet around the sun we take it as a circular path there also or we can use that motion of a uh Moon around a planet everywhere the same derivation okay so here motion of a satellite around a planet we take the path is circular path around the planet mass of the planet is capital M mass of the satellite is simple M uh so since it's moving on a circular path it needs a cental force the C the required cental force is provided by the gravitational force exerted by the planet on the satellite required Gra Cal force is provided by the gravitational pull by the planet on the uh satellite so that is given by the Newton's law of gravitation so the required Cal force is provided by the gravitational force which is g m m/ r² so that is the Cal Force that's going to be m v² / r i can get the uh expression for the speed of the satellite so make the V subject Sol r² R get cancel capital M cap sorry simple M simple M will get cancel m is the mass of the satellite simple M so a m of the satellite and M will get cancel so V is going to be square root of g m/ r better remember this equation you must be able to derive this also not a derivation just simple thing but remember that the speed of the satellite around a particular planet is inversely proportional to square root of r inversely proportional to square root of R better remember that b inversely proportional to square root of R better know that uh and you should be able to derive it okay we need this equation to explain or how did they identify uh Dark Matter the same thing I'll do it but same idea is used there also to the experimental evidence where they identify the dark matter right okay in under cosmology okay so same thing I can continue if I want I can say v r Omega and I can get an expression or I can substitute here R Omega squ anything I can do I can substitute here m r Omega squ and do it again or even I can say V = to squ root of GM r or V squ equal GM r i take the square of it GM R but we equal R Omega we know that v r Omega means r² Omega s we know that v r om GA we learned that in circular motion so r² Omega 2 = g m/ r Omega we know that Omega = 2 pi/ T So substitute here cross multiply RQ into Omega 2 Pi / T all things squ = GN so t² will be the subject 4 Pi sare so from this t² is equal the t² subject is equal to 4² over GM * R so for a given planet around it satellites can move at different radius different distance from the center of Planet everything should be measured from the center so the period of the circular motion so m is constant G is constant 4 Pi s is constant so t² proportional to R important conclusion t² proportional to R Cu uh you should be able to derive this whenever there's a question related to time period of circular path around circular path of a satellite around a planet or motion of a moon around a planet or motion of a planet around around the star Sun the same way we have to derive it so even if it is a numerical calculation you have to derive it and get it don't memorize this equation and don't use this directly without derivation you will lose lot of marks so you have to derive simple the derivation remember the derivation this is called K plus law the derivation is called the outcome the final conclusion t² proportional to RQ is uh one of the K plus law we call in your syllabus just Kus law okay so first question based on what we learned so far it's an explanation type question a satellite is orbiting a planet of mass m at a speed v a constant speed V at distance R so it will satisfy this equation equal square root of GM o r you can't use it directly you have to derive and get it not a derivation just one line you can get it okay so uh the motion of the satellite around the planet of mass m it has spe V right the question is due to some reason or maybe collision with something the speed of the satellite suddenly drops the speed of the satellite suddenly drops what will happen to its following motion okay so the speed suddenly drops due to cision with something else the speed of the satellite suddenly drops what will happen there's no change in the direction immediately suddenly drops when it is at particular height H at a particular distance are from the center of the planet the speed only suddenly drops what will happen to its motion next okay so for that uh stents mostly answer like this they use this equational root of GM over r m is the mass of the planet R is the distance of the satellite from the uh center of the planet so the speed of the satellite suddenly drops so V becomes lower so when V decreases G and M constant so R should increase St normally say when B decreases R should increase so the satellite will move away outward spiral path it will move away it will spiral out and move away from the planet that's the answer most of the students they give okay careful that is not the answer actually you know we call square root of GM how we obtain we derive this equation we said that when the satellite is in circular path it it needs a cental force so the required centripetal force is given by m m v^ 2/ r according to circular motion it needs a cal force that is equal to MV ² R that's a required Cal force to move on a circular path when it has this Cal Force it's provided by physically it's given by the gravitational force between the planet and the satellite that provides the required Cal Force so this is the required Cal Force when it is at a particular distance R from the center of planet that is provided by the gravitational force it's a type of force gra gravitational force is a type of force that Force provides the required Cal Force now what happens the speed suddenly drops when the speed suddenly drops the required Cal force is decreasing the required Cal force is suddenly dropping so when the required Cal Force suddenly drops at that position the gravitational P given by the Earth on the satellite the planet on the satellite will remain the same but the required Cal force is dropping that means the gravitational pull now becomes larger than the required sentimental Force so the gravitational pool becomes larger than required Cal Force means what happen the satellite will move in an inward spiral path and fall on the uh Planet finally because the required Cal force is dropping the gravitational pool remains same as earlier now the gravitational force has become greater than the required Cal Force so the satellite will have a force which is more than the requir Imagine something like a small sphere a marble or a bob is attached to a string and you are rotating on circular part suddenly you pull a small pull towards it you rotate in a circular part the string provides the tension provides the required Cal Force suddenly give a small rate of force rate of tension just pull it when you pull it just the marble will come towards you so like that the satellite will start to move in an inward spiral path because it won't just fall because it has velocity or speed along the tangent at that moment so it will move on an inward spiral path and finally fall on the surface of the planet that's the answer you should say Okay so this type of questions normally come uh mass of Planet X is M and its radius is 2 R gravitational field strength on its surface is G mass of planet Y is 2 m and radius is 3 R find its gravitational field on its surface means we need to give it in terms of G okay so for X we can write gal GM r² I'm going to use this equation so g = g the mass is m/ r² that is 2 R all things squ so that will be g m over 4 r² first equation for Y for the planet y I need to find G1 I'm going to use the same thing G in of mass I should put 2 m the mass is 2 m t * 2 m over 3 R Allin squared so that will be 2 g m over 9 r² second equation now nothing divide second equation by first equation you'll get the G1 in terms of G so second equation divided by first equation G1 over G equal uh 2 gm/ 9 r² into this will go up 4 r² over GM so this will be r² r² will get cancel uh GM G will get cancel 8 over 9 so G1 is going to be 8 G over 9 okay so that is G1 is going to be 8/ 9 G okay so the third question wait weight of an object on the surface of a planet is 120 newton find the weight of this object at a height of 60 km from the surface of the planet radius of the planet is 7,200 km so the weight means that's a gravitational pull gravitational force between the object and the planet so that could be given as F equal g m m/ r² okay R must be measured from the center of the planet okay on the surface of the planet uh the weight is given as 120 newton so 120 is equal g m m r² is on the surface means that should be that must be the radius of the planet 7,200 into 10^ 3 all things squar let it be first equation now the question is what will be the weight of the same object at a height of 60 km from that planet so weight I need to find that be W is equal to g m mass of the planet M over it's from the at a height of 60 kilm from the surface of the planet so at a height of 60 km from the surface of Planet means from the center the equation must be used then we use so that's going to be radius plus 60 km radius is 7,200 so 7,200 + 60 7,260 into 10 the^ 3 all things squar okay I need to find the uh W so it's not a big issue divide the second equation by the first equation so divide the first second equation by the first equation that is W over 120 equal to GMM over 7,6 sorry 7,260 10^ 3 all things squared into this will go up when I divide it that is 7,200 into 10^ 3 all squared over G mm Okay g mm will get canc 10^ 3 10^ 3 inside the square will get cancel so w is going to be W is equal to be 7,200 over 7,260 all the squar into 12 find the W so you'll get 118 Newton 118 Newton okay so the next question question number four uh period of the Moon around the earth is 28 days find the distance or height of the Moon from the surface of the Earth which is uh radius of the Earth is 6,370 KM mass is given I how to write it mass of the earth is uh also given uh mass of the Earth uh 6.02 10 the^ 24 kilog okay so by using this we need to find the height of the mo above the uh surface of the Earth okay so for this type of question uh we should start from the basic principles that is the required Cal force that is m r Omega s is provided by the gravitational pole g m M2 r² G mass of the Earth mass of the Moon Over r r² so M mass of the Moon and the mass of the moon will get cancell so r² into r r Omega means we know that Omega = 2 pi/ T in circular motion so into 2 pi T all things squ is equal to GM so solve it and make the T sub t² = 4² / GM into r or you can make the r Cub subject because we are going to take the r the height we are going to find so we'll make the r Cub subject so R cube is going to be R cube is equal to GM over 4 piun 2 * T 2 GM 4 Pi sare t s okay so now we need to find the uh radius of the orbit of the moon from the center of the Earth from that we'll subtract the radius of the Earth okay so we'll substitute that is GC 6.67 10^ -1 mass of the earth is given 6.02 10 the^ 24 the mass of the Moon got cancel Syle m is the mass of the Moon that got cancelled so over 4 Pi 4 Pi s into t² t² means the square of the period that's 28 days we had to give it in seconds so 28 days we need to convert to our so 24 hours per day then 60 minutes into 60 seconds all squar okay solve it and find the R so you will get R Cube the value you will get when you solve the right side that will be the value of the r Cube take cube root of it you'll get the r the value of R will be equal to uh 6 sorry 3.90 10^ 8 m in kilom gu it will be 3.90 10^ 5 kilom so that's the radius of the orbit of the moon from the center of the earth the height of the moon is the question so height of the Moon from the surface of Earth is radius of the orbit minus radius of the Earth is it so this is the Earth this is the orbit so we know the radius of the orbit radius of the orbit from that we subtract the radius of the Earth we get the height okay so the height we need to find it that's radius of the orbit 3.90 10^ 5 minus radius of the Earth is given which is uh 6,370 that also in kilomet so Solve IT subtract it you will get finally 3.84 10^ 5 kilm that's the height of the moon or distance of the Moon from the surface of Earth check it with internet type it in the Google uh height of the Moon above the surface of Earth you'll get exactly the same answer okay so the fifth question mass of the earth is 6.0 into 10^ 24 kg and its radius 6370 km normally uh sometimes in the past paper they say 6,400 km this is 6,370 K I think this more accurate value find the height of the geost satellite above the surface of Earth geostationary satellite geost satellite means the satellite that has period equal to 24 hours it always moves of it Orit is above the equator of the earth so for the for the certain region of the earth it looks like stationary related to Earth so the period of the earth is 24 hours same as this period of spin of the earth that also 24 hours so this satellite also has the same period 24 hours probally the satellite the geost cite is parked above the equator of the earth okay so they as to find the first part uh the height of the geost satellite which has period 24 hours so geost satellite if they say the period is 24 hours so same equation what I used earlier uh first part F equal that is M Omega Square that's a required sentimental Force which is pointed by the gravitational force G mm over r² capital M is the mass of the Earth symbol is the mass of the geost satellite mass of the geost satellite will get cancel r² here R into Omega = 2 Pi / T so that is 2 pi/ T alling squ is equal to GM so find the r cu make the r CU subject R is going to be GM over 4² into t² so substitute the values and find the r 6.67 10^ -1 mass of the earth is 6 into 10^ 24 over 4 piun 2 into t² 24 hours so 24 into hours to minute minute to second and all the squar that is R Cub find the r take solve it on the right side and take cube root of it so you'll get R is equal to we'll get r that is equal to uh 42.3 10^ 6 m in kilom if you find that is going to be uh 4,200 4,200 uh sorry 42,00 297 km so this approximate value 42,2 197 km so the height of the satellite is equal to radius of orbit minus radius of Earth as I explained same as earlier for the moon so R we know 42299 minus radius of the Earth is uh 6,370 kilom so that is going to be 35,000 the height is 35,9 127 uh kilom that's the height of the geost station satellite check it in the Google height of the geost station satellite it says 35786 slightly different because mass of the satellite is uh sorry mass of the earth is not 6 10 about 24 I think something 5.97 or something anyway we use 6 10 before any the the internet the standard value it says 35,7 186 around 120 km different doesn't matter we are getting a good answer for this calculation okay second part find the speed of it so that we can simply find v r Omega or we can use this equation derive and get it derive and get it you can't directly use this equation this is not a standard equation you have to derive it and get it okay so we R Omega Omega equal 2 pi/ T we can use that equation so R we found it here uh 40 42 uh 3 10^ 6 Omega 2 pi/ t t is 24 hours so 24 into 60 into 60 so this much of m per second solve it you'll get the speed which is uh 30, sorry 376 m/s or if you're using this one you must derive it from basic the initial the uh you have to derive it and get it so again I'll show the derivation uh it's like this we have to derive it means like let start with We R the gravitational force the required Cal force is m v² / R which is given by a gravitational pull G mm r² so mass of the set it cancels so v² is = to R GM over R so V is = to root of GM R okay so substitute the values square root of G that is 6.67 10^ -1 mass of the Earth given 6 into 10^ 24 over R we found it in the previous part that is radius 42.3 10^ 6 M solve it you'll get exactly same answer it's 30 375 you will get 3,75 M either way you can do okay okay so uh sixth question by deriving an appropriate equation show that the orbital speed of a satellite decreases as the radius of the Orit increases so that is actually they asking to derive this one anyway I don't need to derive again again I already derive it so first part just we leave it okay second part by deriving an appropriate equation show that the orbital period of a satellite increases as the orbital speed decreases okay so for that we can show like this okay so from this one first derivation second part I'm it from the first deriv that we can derive already dered it so B is equal to square root of GM R so from that b² = g m r okay so by deriving an appropriate equation show that the orbital period increases as the orbital speed decrease so I need to get a relationship between period and the speed okay so there we can see speed B is equal to 2i R over T 2 pi r is the circumference over T is the time taken for one complete uh orbit so here we can see uh from this I'll substitute uh r by V this equation so from this I would eliminate the r because second one is something relation between speed and the time period so from this I'll substitute R in terms of B so R is going to be GM over b² substitute that here so V is going to be 2 pi/ T into inste of R GM over p² okay nothing so they are saying uh period of so make the T subject so T is going to be make the T subject 2 pi GM over V CU so that is T proportional to inversely proportional to VQ so when the speed increases uh the orbital time period will decrease okay so the seventh question first part there are three parts first part A PPE of question radius of the Earth is 6,400 KM and the gravitational field strength on its surface is 9.81 Kil Newton per kog or m/s Square okay whatever it is say uh find the gravitational field strength at a height of 40 km from the surface of Earth so we are going to use the g = g m/ r² so G = GM r² so on the surface of Earth 9.81 = GM over r² 6,400 I don't need to convert to meter because I'm going to divide two equations the 10^ 3 will get cancelled but anyway to avoid unmounted confusion I included 10^ 3 okay so GM r² the question is find the gravitational F at 40 km from the surface of Earth at a height means from the surface of Earth so you know the equation is applicable are always from the center of the earth so G1 I need to find so G1 is equal to GM over 6,440 into 10^ 3 all the s so that's a second equation divide second equation by the first equation G1 over 9.81 is equal to GM over 64 440 into 10^ 3 all things squared into 6,400 into 10^ 3 all squ over GM GM GM get cancel 10^ 3 can 3 get cancel so 6,400 over 6,440 all things squared so G1 is going to be uh 6,400 over 6, 440 all things squared into 9.81 so that will be uh 9.69 m/s square or Newton per kog okay so the Earth is represented by a circle in the diagram below add to the diagram to indicate the gravitational field so you know how do we identify the gravitational field the direction of the force or the gravitational force acting on a test Mass indicates the direction of the gravitational field so if you keep a test Mass a tiny Mass it will be pulled by the Earth so the force will be towards so always the gravitational field lines are towards the center of the planet draw a few lines symmetrical uh here also the gravitational field lines indicates the strength of the field number of lines per unit area indicates a strength the gravitation field lines never cross each other don't draw gravitation field inside normally we don't draw inside the planet outside the planet but they should Point towards the center of mass so if I extend these gravitational field lines they'll meet at the center of mass of the planet so gravitational field lines never cross each other the number of lines for unit area indicates the strength of the field same as the electric field but here it's like a radial field arrows are towards the center of the right center of the planet okay so explain why the gravitational can be thought of as approximately uniform over a smaller region of the earth even if the Earth is sphere spherical shape if you think about a region for example a football ground it looks perfectly flat so the gravitational field lines are always radial means they will be perpendicular to the surface at any point because they meet at the center of the Earth or center of the planet so if you consider smaller region of the earth smaller region of the Earth at every point the gravitation field lines are perpendicular to the surface and if you draw the smaller portion looks like a flat line there all the gravitational field lines are perpendicular means the gravitational field lines appear parallel among them therefore the strength of the field is the same in a smaller region when you consider uh above the the surface of Earth okay so the eighth question this also a pass paper question in binary system of black holes two black holes are at a separation of 3.2 10^ 15 M and their masses are given the black holes orbit about a 7.7 10^ 13 M from the larger Mass black hole that means they loit about the Common Center of mass of the black hole Common Center of mass that means imagine if there's a larger sphere a steel sphere massive Mass not larger not in size a massive Mass M1 there is another sphere the same size you can imagine M2 M2 is lighter both are connected by a light Rod like this a light Rod like this okay so when they are connected by a light Rod the center of mass of the system will be closer to the massive sphere that's a center of mass now about the center of mass you rotate it so this will rotate about this one so the radius of the orbit of this one will be this one this will rotate about this way this will rotate with the radius of this is the radius of orbit for the M1 this is R1 this p is the center of mass of the system of two spheres the the rod which connects them much lighter so we can neglect its mass so if you use the center of mass equation Sigma zma so you don't need it but I'm just telling so it's they orbiting about Point p means this mass will Orit in a circular path of radius R1 this will Orit at radius of R2 from P to this one is the R2 so from P2 the mass M2 is R2 that will Orit in a circular part of radius this one so like this so that means imagine on this PA on two sides spheres are attached massive sphere lighter sphere by holding it at the center of mass rotate it so you rotate like this both will have the same period of circular motion is it when this is here this is here when this comes to this point this will be here like that they will have the same period of circular motion but they'll orbit about the Common Center of mass same way here the two black holes they are forming binary system binary system means combination of two uh massive objects like stars or black holes uh they together form a binary system they orbiting about Point P Point p is the center of M calculate and see you'll get 7.83 I don't know why they give 7.7 anyway it's okay we use the given data uh so they Orting about Point P right show that the orbital period of the binary system is both will have the same period This Will Orit about Point P this also will Orit about point B both will have the same period of uh circular motion so if I know know that I need to find the time period how can I find the time period the idea so here first we'll find the uh gravitational force because when the massive Mass orbit about P it needs a centripetal force so wherever it stays this will be in opposite to its position so the force from acting on the massing mass towards p is the gravitational force or gravit pull exerted by lighter Mass on the massive Mass same way when lighter mass or lighter or low mass black hole also orbiting about p means the low mass black hole also needs a centripetal force the Cal force is the gravitational pull exerted by massive black hole on the lighter black hole so we need to find the gravitational pull or gravitational force between the two uh black HS that is going to be the centripetal force because the gravitational pull on this one will act towards the uh lighter black hole means that will act towards the center of mass a same way the gravitational force on lighter uh black hole also will act towards point B because it will act towards the massive uh black hole so we'll find the uh gravitational force between the two uh black holes F equal G M1 M2 r² 6.67 10 power - 11 Mass M1 is 1.6 into 10 power 39 Mass M2 is 4 into 10^ 37 over r² is the distance between the two black holes that is 3.2 10 the power of 15 all squ so find the gravitational force that will be equal to 4.17 10^ 35 Newton so that will be the this force is a cental force for the circular motion of massive black hole about p and this is the same uh force is the Cal Force for the lighter black hole for it circular motion about P so I can use now a m r Omega s f I know R I know for the massive black hole R I know because it's given 7.7 13 F = Mr R Omega s we'll find the Omega Omega = 2 pi/ T we'll find the T so if you find the time period of this one that is same as for this one also if you want you can use f Mr Omega s for this one then you have to use the distance of this one from the P so that you have to subtract 3.2 10^ 15 - 7.7 10^ 13 you will get this one but anyway that also you can do but since I know this distance I'll use for the massive black hole so FAL m r Omega s for massive black hole if we know 4. 17 10 power 35 its mass is 1.6 10^ 39 is it R the radius of Orit 7.7 10^ 13 in Omega squ Omega = 2 piun / T all squ so 2 piun / T all square I can say 4² over t² solve it and get the find the T so you'll get T is equal to uh 3.41 5 into 10^ 9 seconds you will get it but they also to show that 100 years so I need to convert this into years okay I'll raise this part now okay so T is equal to 3.41 5 10^ 9 seconds I to convert years means seconds to minutes divide by 60 Minutes to hour then hour to day 24 into day two years 365 years so it you will get 108 years so approximately 100 years okay so the ninth question it is claimed that a satellite in orbit at a height of 8.50 10^ 5 m would make 15 orits of of the Earth every 24 hours this 15 look like is it's 15 15 orbits of the Earth every 24 hours assess the validity of this claim m of the earth is 5.98 10^ 24 kg okay so here I'm going to use the equation but we use so far t² = 4² GM r g GM into R Cub any if you you must derive that equation or they might ask you to derive in the previous part anyway I'll use uh FAL the required Cal Force = m r Omega s = g m m R 2 m m will get cancell so R CU into Omega means 2 piun T all things squ equal GM so make the t² subject t² isal to uh 4 piun 2/ GM into R CU okay so uh at a height of 8.50 m above the surface of Earth so is orbiting around the earth so the mass of the earth is given H sorry the radius uh of the earth is also given sorry I I forgot to write it radius of Earth radius of Earth is uh 6,3 60 kilom that's the way it's given 6,360 uh kilomet okay that's the radius of the Earth so radius of the Earth also given so this is the height so you know if this is the Earth the radius of the Earth we know that that is 6, uh 6,360 km but the satellite is orbiting at a height of at a height of 8.50 10 the power 5 m so when we use this equation R must be the radius Rus of the orbit R must be radius of the orbit means height Plus radius of the Earth that's a very common mistake stents make they don't add the Earth of the radius of the Earth when the height is given so you have to add the uh radius of the Earth with the height of the satellite above the surface of Earth then only you'll get the radius of the orbit of the satellite okay so we'll continue we'll find the time period here so from this t² is equal to 4 y^ 2 / G uh 6.67 10 power -1 mass is given mass of the planet that is mass of the Earth 5.98 10 the^ 24 into R Cub so radius means we have to add both that is U 8.5 10 the^ 5 that is in me this is in kilom convert to meter 6360 into 10^ 3 m all in Cube okay so after that find the T squ and take square root of it you'll get T is equal to Sol it you'll get t equal to 691 or 90 second 691 second you will get it okay so the field of the earth is 24 hours is it it they are given it uh uh every 24 hours so during 24 hours how many times it will it make complete orits we know the time period of the Orit so how many times will it Orit during 24 hours divide the 24 hours first convert the 24 hours into second divided by the time for one Orit you'll get the number of orits during 24 hours equal 24 hours 60 Minute into 60 second divided by uh 691 so you will get almost 14.2 oit approximately 15 orits it should be 14 but more most closest uh value we take it 15 here it's okay okay so this is May 20122 May June 2022 question number 16 uh part A nothing we several time came across we have to show t² proportional to R CU that's a cap plus law so I'm not doing it a part first part that's nothing A Part second part it says when planets align as they orbit the sun they are said to be in opposite uh in opposition the diagram shows the Earth and Jupiter in opposition so that means this is Sun this is Earth this is Jupiter this diagram is given in the question this is Sun so the opposition means they are on the same line sun earth Jupiter they are the same light then it's called opposition the website states that the Earth and Jupiter are in opposition every 13 months okay so deduce whether this statement is correct mean distance from Earth to Sun so this distance Earth to Sun this radius of orbit of the earth is given that is radius of orbit of the earth is given as 1.5 10^ 11 M radius of Orit of the Jupiter is given that is 7.8 10 the power 11 M okay so Earth orbit the time period is also given we know that 12 months 1 year so it's given 12 months so we need to show that the statement is correct every 13 months the sun earth Jupiter becomes in opposition means they come on the same line okay so the idea is this one they are in same line like this now later you know according to the equation t² proportional to R Cub means the uh planet that has smaller radius of or bit that means which is closer to sun will have shorter time period which has which is closer to Sun of smaller radius of Orit will have smaller time period and also we know that they are going to Orit in different time periods they will have different time periods we know that Omega equal 2 pi over T and also we know that Omega equal Theta over T Omega equal Theta over T because Theta is angle of uh as central angle change in central angle over time taken T okay so we know that the plan the planet that has larger time p period that means Which is far away from the Sun or which has larger radius of orbit will have larger time period larger time period means when we consider particular time during the time the angle in the change in central angle will be smaller change in central angle will be smaller so when we consider what will be their position from this time they are in opposition okay later at time T from this position we take this position as at time tal zero later at time T what will be their position means the Earth which has smaller radius Orit that has smaller time period smaller time period means angular speed is larger angular speed is larger means change in central angle will be large angular speed larger mean change in central angle will be large of for Earth at a time T later so Earth might be from this position Earth might be somewhere here at time T which it would have gone through angle Theta 1 but Jupiter that has larger radius of orbit means larger time period larger time period means at a particular time the change in central angle will be smaller for Jupiter the Jupiter might be somewhere here that might makes an angle Theta 2 which is smaller with the center of the Orit so that is going to be Theta 2 okay so the condition mathematically when they are in opposition again Theta 1 minus Theta 2 that's change in central angle Theta 1us Theta 2 should be zero that happened now Theta 1 minus Theta 2 the difference between the center angle Z means that happen now when t equal 0 later when they become again opposition Theta 1 minus Theta 2 could be 2 pi we don't know they will come and meet at the same position we can't see maybe this would have completed 5 orits this would have completed uh four orits means Theta uh the change in central angle of the earth minus change in centrer angle of the Jupiter when it is 2 pi there it will happen again they will be in opposition somewhere here we don't know at this position again when the change in central angle becomes 4 Pi also so they will be again in opposition so like that when the change in central angle that is when Theta 1- Theta 2 is equal to 0er that happen now R 2 pi or 4 Pi or do dot 6 Pi 8 Pi like that 2 N Pi General format 2 N Pi they will be in opposition where n is an integer n = 0 1 2 and so on it's an integer so during that time then when Theta 1 - Theta = 0 2 pi 4 Pi 2 and Pi then they will be or these planets will be in opposition okay so we need to find Theta 1- Theta 2 from this we can get Theta is equal to generally 2 Pi / t that's a period into T so I want to use this one here so Omega = 2 pi/ T that is changing central angle during time T so that time T when will it happen we had to show during that time T if Theta 1 minus Theta 2 for the two planets the difference between the change in central angle becomes 0 or 2 pi or 4 Pi that is 2 * uh 2 pi times into n n is an integer then the planets will be in opposition they will be in the along the same line okay so I need to find the time the time period for the Jupiter because time period for the Earth is given I need to find the time period for the Jupiter okay that's nothing I'm going to use the first part A Part first part they has to derive t² proportional to R CU I'm going to use that t² proportional to RQ we derived in the a part first part so from that t² is equal to a constant time R the constant is 4 S GM any I'm not going to substitute anything so for Earth we know t² is 12 months 12 s = k into the earth radius is 1.5 10^ 11 alling Cube let it be first equation for the Jupiter I need to find the time period that it be TJ that is going to be K into its o is 7.8 10^ 11 all CU second equation divide the second equation by the first equation I'll continue here second equation divide the second equation by the first equation you will get TJ sorry this T squ is it t² J so t² j/ 12 2 is = to 7.8 10^ 11 10^ 11/ 1.5 10 the^ 11 all in Cube I can take the cube out then solve it and find the t uh square root of it and find the time period of the Jupiter you will get 142 3 months okay so I know the condition if they want to be in opposition again the condition uh Theta 1- Theta 2 should be either 0 2 pi 4 pi and 2 and Pi generally I can say n is equal to 0 1 integer right so Theta 1- Theta 2 Theta = 2 Pi / T the period into T So Theta 1 - Theta 2 means 2 pi/ T Jupiter yeah the actually sorry uh uh the largest angle for um Earth so T earth - 2 pi over T Jupiter into T because here t comes 2 pi T into t - 2 pi T into T the T I took it common is equal to 2 and Pi because they ask every 3 months they asking every 13 months so they didn't say after 13 months every 13 months so that's the reason I did it in generally with n mostly students ask why are you doing it including n why can't you put just 2 pi okay if the question is uh website states that the Earth and Jupiter are in opposition next three during next 13 months then I I don't need to put n equal uh I don't need to include n here just I can put 2 pi but every 13 months means yes 13 months then again 26 months again 39 months and so on so that's the reason I put in here General format okay 2 and Pi okay we'll substitute and find the T So 2 pi 2 pi 2 and pi means I can take I now 2 pi into n the 2 pi here this 2 pi 2 pi will get cancel so 1 over t e time period of Earth is I'll keep it in months that is 12 1 / 12 - 1/ 42.3 into T is equal to 2 pi 2 pi 2 pi got cancel n what is n n = 0 1 2 3 integer okay so Solve IT find the T Now 1 you can use the calc 1 12 - 1 42.3 into t = n so find the T you'll get at T is equal to 131 n where Nal 0 already over because that happened now next 0 over 1 2 3 and so on so that means if I substitute 0 t = 0 that happened now then if I substitute n = 1 n = 1 means after 13 months n = 2 2 after 26 months n = 3 after 39 months so General format 13.1 n where n = 0 1 2 3 therefore they will be in opposition when T = 13 13 months or 26 months or I can put on 13 months and I can put because many solutions 13 months uh uh okay 26 months I put just comma 39 months dot dot dot you know every 13 months yes so the difference between this is 13 26 - 13 13 39 - 26 13 and so on that's the answer for this question okay so the next topic is gravitational potential something like uh electric potential but we learned in unit four here gravitational potential when there is mass m for example when there is planet or Moon it will create gravitational field around it everywhere the same way it will create gravitational potential around it everywhere so what is the gravitational potential of a mass m means mostly in your syllabus we take a spherical shape object cical shape Mass planet or moon or star uh so if there is a planet or star or Moon of mass capital M at distance R from its Center of mass at Point P what is the gravitational potential of this Mass capital M at Point e means how do we Define okay definition careful work done in moving 1 kilg Mass from Infinity to point p is defined as the gravitational potential at Point P again I say work done in moving 1 kg Mass from infinity or infinite distance to point p is defined as the gravitational potential at Point P you know gravitational force is always attractive so gravitational force is attractive means actually when a 1 kog mass is moved from infinite distance to point P the gravitational pull exerted by the mass capital M or Planet will do work on the object of mass symbol m sorry 1 kilogram Mass cap 1 kilogram and this mass will gain kinetic energy so when it gains kinetic energy it lose its gravitational potential energy since the force is attractive the gravitational potential will be always a negative quantity because its gravitational potential will decrease when the object move from Infinity to towards the object of mass m when 1 kilog Mass moves the gravitational potenti of this created by this mass m will be will decrease when the distance of it from the mass decreases so the gravitational potential is always negative because of the at ractive force uh between the uh mass m and the 1 kg Mass so we can say why the gravitational potential is negative means because of attractive Force the gravitational potential decreases when the distance from the mass decrease okay so the gravitational potential at Point p is given by V = minus g m over R you must know that the gravitational potential is given by minus g m over R okay so when a mass m is kept at Point P when a mass m is kept at Point D the gravitational potential energy of the mass m i told that in unit one we never in unit one work Energy power we never measure the gravitational potential energy in unit one we actually we measure the change in gravitational potential enery if the mass moves up we write MGH which is the gain in GP when the mass moves down we write loss in GP so we never measure the gravitational potential energy in unit one uh when we do questions related to gravitational potential energy but here we can measure the gravitational we can calculate the gravitational potential energy of mass m when it is kept in the gravitational field created by mass capital M so the gravitational potential energy of object it could be a small object object a satellite or whatever it is the gravitational potential energy of this object of mass symbol m when it is in the gravitational field of gravitational field created by the mass capital M at distance R from it the gravitational potential energy is given by gravitational potential into the mass of the object so the gravitational potential is - g m / R into M so the gravitational potential energy of the mass is always negative the reason here also when the distance of the mass m from the massive mass m decreases its gravitational potential energy will decrease so that's the reason it's given by a negative quantity so the gravitational potential energy is given byus g m / R into M that is V gravitational potential into Mass okay so the unit of gravitational potential is Jewel per kilogram because we Define work done in moving 1 kilog charge so that means work run per kilogram jwel per kilogram is the uh unit of uh gravitational potential uh the unit of gravitational potential energy is Jews okay so the next topic is uh change in gravitational potential energy okay for example there's a planet the radius of the planet is R the center is O center of the planet is so so an object is moved object of mass a is moved from point P1 to point P2 so we know know that uh when the distance from the planet decreases the numerical value will be with larger number with negative sign that means when object moves closer and closer to the planet or moon or whatever it is the gravitational potential energy will decrease I told already kinetic energy will increase of the moving object so for example gravitational potential energy decrease means it will have a negative larger value say for example -1 J is less than -4 je when we consider the gravitational potential ener this is smaller value uh negative sign okay so uh when object moves away from the planet what will happen it will gain gravitational uh potential energy uh so imagine a point p is at a distance R from the center of the planet so this distance is r from the center of the planet uh the R1 will make it as R1 so it's moving from P1 to P2 through a height H or distance H from it uh so what will happen to the gravitational potentialy so it will move away from it so it's going to the object of mass m is going to move from P1 to P2 so gain in gravitational potential energy gravitational potential energy of the mass m when it moves from P1 to P2 so gain means final minus initial so the gravitational potential energy at Point P2 is V2 M minus V1 M V2 is the gravitational potential of the planet which has mass capital M uh at a distance uh of P2 that is R1 + H is the distance of P2 from the center of the Planet so the gravitational potential energy of the mass m at Point P2 is V2 into M that a gravitational potential of the planet at Point P2 into Mass that's I told that yeah and the gravitational potential energy of the same mass at Point P1 is V1 M okay so we'll continue this so V2 is equal to - g m over distance of Point P2 is R1 + h minus again there'll be minus so minus GM R1 into M okay so Solve IT minus into minus plus so it's going to be g m m/ R1 minus g m m over R1 + H H is the distance or height through which the object of mass m is moved from P1 to P2 such that away from the planet okay we's find the gain in gravitational potential energy so I put e sign because I can say Delta gravitational potential energy Delta potential energy we'll see Delta p uh yeah Delta p e so that's going to be uh solid this one uh g m m you can take common within the bracket 1/ R1 - 1 / R1 + H so solve it g m m R1 R1 + H is the LCM R1 + H - R1 okay so it's going to be g m m R1 / R1 into R1 + H okay now this is the uh increase in gravitational potential energy okay now we'll see we'll analyze it if R1 is very small comp sorry if H the height through which it moved the height through which the object is moved H is much smaller than R1 say the point P1 is at a distance R1 from the center of the earth the distance through which the object of mass m is moved is H if H is very small compared to R1 so for example uh this may be around uh 7,000 kilm the distance moved may be around 200 M something like that so if r h is very small if H is much smaller than R1 then what can we say then R1 + H is equal to approximately equal to R1 is it R1 + H will be equal to almost equal to R1 so then what will happen the change in gravitational potential Delta potential energy will become g m m into R1 over uh sorry here that's a mistake I made yeah this is H sorry there's a mistake this is H yeah R1 R1 get cancel H over so this is H H over uh H becomes R1 + H is almost equal to R1 means it's going to be R1 s so this I can write g m R1 s into uh H times this m will be there so GM o R1 s into m h g o GM o R1 squ what is GM o r s R1 sare GM R1 squ is the gravitational field strength at Point P1 so that I can see G G1 so G1 that is the gravitational field strength of the planet at Point P1 is G1 so G1 m h so that is going to be M1 sorry that's going to be M G1 H so that is the change in gravitational potential energy M g1h where G1 is the gravitational field strength at Point P1 so this format we used in uh unit one MGH the gain or loss because the distance moved if it is very small compared to the uh distance of the point from where to where it's moving so this height through which it's moving is very small compared to the initial height from the center of mass uh if it is very small then we can say the change in gravitational potential energy is equal to MGH okay so what we learned in unit one the same format an object is uh projected from the surface of a planet say from the surface of Earth uh at a velocity initially away from the surface at speed U uh it moved through a height H and the speed becomes a drop to be so there when it moves away it's going to gain gravitational potential energy and lose uh kinetic energy so gaining gravitational potentialy um we learned that in unit one MGH but that is true only when the height through which it's moved which is very small compared to the initial height of the or initial distance of the object from the center of the mass of the planet otherwise you can't use MGH H format anyway again I'll drive it derive it so gaining GP here final GP that is at this point uh we can say potential at that point final Point BF into M minus VI initially on the surface into M so the final gravitational uh potential is equal to minus GM over the radius of the planet is r r + h r + H into M minus - g m/ r r on the surface of the planet into M so it's going to be minus into minus plus g m m/ r r is the radius of the planet minus g m m/ r + H same thing what I did I'm doing it again from the surface earlier I took two different points outside now from the surface but we are using in uh unit one I'm going to show that so here the change in gravitational potential is going to be so take the LCM R into r + H so that will be R uh r + H minus r into we can take G mm common okay so here it will become r r get cancel H over R into r + H into g m m okay here also if the height through which the object moved that is H is very small compared to the radius of the planet that's a way in unit one we do a ball is thrown vertically upward so the distance it mov maybe maximum 10 kilm even that you can't throw but a bullet is fired the this height it went is around uh 500 M or 1,000 M 4,000 M anyway even if it is 5,000 m compared to radius of the Earth 6,400 km it's negligible much small is it so when I can say when H is uh H is very much smaller compared to the radius of the planet then r + H approximately equal to R so if I use that equation here again so the Delta potential energy change in gravitation potential energy so GMM H over r + H is almost R so it's going to be r² so that's going to be g m r² time M this m again into h g m r² is G MH then we write as m g h so that's we wrote in unit one because the height it moves is very small negligible but actually if the height is not small through which it's moving is not small this is the final equation you have to substitute the given numerical values when the height is not small compared to the radius of the planet or not compared to the uh not smaller than the initial distance from the center of the planet so here what's happened it's mov thrown away so loss in GP you can use either this one if the HED is not much smaller if this is not true if this is not true H much smaller than R is not true then you have to use this as a uh gain in gravitational potential energy that is equal to loss in kinetic energy loss in kinetic energ means initial ktic energy half M u² minus half M v² but if this is true the height through which it moved is very small compared to radius of the planet then the loss in uh sorry the gaining gravitational potential energy will be equal to MGH so that is equal to loss in ktic half m u s minus half M v² that's the way we did in unit one okay so a satellite of mass symbol m is Mo moving on a circular path around a planet which has radius capital r and mass capital M so the satellite is at a distance R from it so the total energy of the satellite is the kinetic energy of the satellite plus the gravitational potential energy those are the two energies it possess so in unit one we don't see uh gravitational potential that is change in GP in unit one a ball is thrown upwards uh it loses ktic energy gains GP when a ball or stone moves down it uh loss in GP equal gain in case so here we are not considering change in GP so here we are calculating the GP gravitational po potential energy of the object okay not change in GPA there we consider change in GPA okay so because here everything are measured from the center of the planet center of mass of the planet so we are calculating the GP itself not the change in GP in this uh topic okay so kinetic energy of the satellite is half M the mass of the satellite M the speed is V half M v² the gravitational potential energy is going to be minus g m m over the radius of toit is R so this is the gravitational potential energy okay we learned that when object moves on a circular path uh around under the influence of uh gravitational force the speed should satisfy when it is moving a circular partal to square root of GM I derive it several times earlier in this lesson so in so v² we substitute this one it's going to be half m in so v² is going to be g m over R minus g m m/ r so that means it's going to be g m m / 2 R minus g m m/ r so 2 R is the LCM so 2 R is going to be uh t m m- 2 g m m so that will be uh minus g m m over 2 R that is the total energy of the satellite okay so STS normally ask the total energy becomes negative quantity what's the meaning of why is it negative okay how to explain it we'll explain what's the meaning of if the total energy of the satellite equal to 0 or total energ of the satell equal to positive quantity what the meaning of it then we can come to the conclusion why does it have a negative total energy okay we know that when the r becomes Infinity that is when it is at far distance the gravitational potential energy will be equal to Zer when R becomes zero the gravitational potential energy equal Z Z and we know that when object moves away from a planet it's gaining gravitational potential energy and when it reaches infinite distance the amount of gravitational potential energy possessed by the object will become zero so at infinite distance the maximum gravitational potential enery it could have due to the uh gravitational field of the planet is equal to zero so when it is in the influence of the gravitation field of the planet not infinite distance somewhere closer to the planet it will have a value gravitational potential energy lower than zero so it has a negative GP okay that's not an issue that is about the meaning why the GP is negative right but why the total energy is negative what's the meaning of it so there we can see first we'll see what's the meaning of total energy equal to Z the total energy equal to Z means the object has has reached infinite R becomes Infinity the total energy will become zero that means when the object reaches the infinite or the satellite reaches the infinite distance the gravitational potential will become zero and if the object is at rest just reaches means when the satellite just reaches the infinite distance if the speed also become zero gravitational potential energy will become zero when R becomes infinity and the speed becomes zero means the total energy will be equal to zero that means the satellite is at infinite distance without the influence of the planet without the influence of the gravitational effect of the planet it has reached infinite distance the potential ener becomes zero it could be addressed so the total energy is equal to zero that's the meaning of total energy equal to zero right total energy equal to positive quantity what's the meaning of it total equal to positive quantity means it has reached infinite distance the satellite has reached infinite distance so R becomes Infinity the gravitational potential energy equals zero but it has ktic energy so it means the satellite is not going to Orit o it might move in a random path with a particular speed so it poses kinetic energy so it poses kinetic energy no gravitational potential energy due to the gravitational effect of the planet so it will have total energy a positive quantity so you know the meaning of both if the total energy is positive means the satellite has reached infinite distance without the influence of the gravitational effect of the planet and it has a speed so it has a total uh energy which is positive quantity if the total energy equal to zero means it has reached infinite distance so no influence of the satellite uh Earth Planet so the gravitational potential energy has become zero also the satellite might be at rest at infinite distance so kinetic energy is zero so in these two uh situations the total energy become zero and totaler become in positive quantities the satellite is not under the influence or effect of the gravitational field so now we'll come back if the satalite is under the influence of the gravitational field it can't have a zero value it can't have a positive value the total energy then what could be the possible it must have a negative value that's the only way we can explain I hope you understood again I'm explain when it is at infinite distance gravitational potential energy is zero it could be at rest so infinite distance means the satellite is not under the influence of the uh gravitational effect of the planet so it has come out of the gravitational effect so its gravitational potential energy is zero when r equal Infinity it might be at r so no kinetic energy so total energy equal zero same way if the total energy equal positive quantity means it is at infinite distance gravitational potential energy has become uh zero but the satellite might be moving at random path at a particular speed so it has kinetic so total energy become positive in these two conditions situations the satellite is out of the gravitational influence of the planet so if it wants to be under the influence of the grav under the influence of the gravitational effect of the planet the satellite can't have zero value satellite cannot have positive value so what could be the value it should have a negative value so negative value means the satellite is moving under the influence of the gravitation effect of the planet that's the meaning of it okay so the first question under gravitation potential radius of the Earth is 6.4 10^ 6 M and its mass 6 10 24 kg geost satellite of mass 650 kg is launched from the equator and reach its Orit of radius 4.2 10^ 7 m uh calculate the increase in gravitational potential energy of the satellite during its laun from the Earth's surface okay so increasing gravitational potential energy so there the increase in gravitational potential energy Delta GP is equal to final GP you know the GP = B into M minus g m m R2 - g m m/ R2 R2 is the radius of the uh geous satellites or bit minus Min - G mm over R1 R1 is the initial radius that is from the surf of Planet so R1 is the uh radius of the Earth okay so that's going to be minus minus plus so it's going to be uh g m m we can take it out 1 / R1 - 1 / R2 so that's going to be GMM R1 minus GM R2 I took GMM common so 6.67 10^ - 11 uh the mass of the earth is given capital M is the mass of the Earth 6 uh 6 into 10^ 24 m is the mass of the satellite which is given as 650 yeah 1/ R1 so R1 is the uh radius uh of the earth initial radius so that is going to be uh 6.4 10^ 6 minus R2 is the radius of the Geary path that is 4.2 10 the power 7 okay that much of jewels that's the increase in GP so solve it uh you will get 3.44 10 the power 10 JW that's the increase in GPA okay so the rocket is launched from the surface of a planet the following table gives the uh data for the speed of the rocket at two heights above the planet surface after the rocket engine has been switched off so so at height H1 which is 19.9 10^ 6 Speed is 5370 m/s at a height of 22.7 so it's moving away from the planet the speed is decreasing uh the radius of the planet is given find the mass of the planet is the question so here the satellite is moving away from the planet so it's going and speed is decreasing means the gravitational potential of the satellite is increasing so gain in GP equal to loss in uh kinetic energy so we can say gain in GPA in gravitational potential energy equal to loss in kinetic energy we are writing this for this the satellite which is in motion gaining gravitational po final GP minus initial GP so that is minus G mm R2 the R2 is height H2 uh plus the radius of the uh Planet H + r r + H2 that is the radius of the orbit where it reaches the height H2 because all the distances should be measured from the center so these are the heights above the surface of Earth so minus GM over R2 means r + H2 - - G mm over r + H1 that is equal to loss in ktic energy half M uh initial ktic energy V1 s that is given here minus half M V2 s okay we'll substitute okay so here minus into minus plus so we'll write that one first so g m uh the symbol m symbol m symbol m all will get cancelled that is the m mass of the uh satellite so M simple M simple M these all M will get cancel they are the mass of the satellite so GM into uh minus into minus plus 1 / r + H1 - 1 / r + H2 is going to be half V1 s - E2 2 okay so we know everything except capital M so G we know 6.67 10^ -1 into m 1/ r + H1 so R is given as that is given as 6.38 10^ 6 H1 is 19.9 10^ 6 so we'll add it and put it that is easy so r + H1 6.38 10^ 6 + H1 19.9 10^ 6 if you add both uh we will get 26.2 so that's easy rather than making too much of writing too much of numbers here we'll find that first that is 6.38 10^ 6 + 19.9 10^ 6 that will be uh 26.2 10^ 6 uh that is r + H1 - 1 / r + H2 that also 6.38 10^ 6 + 22.7 into 10^ 6 uh better do it separately and substitute that will be uh 29.8 29.8 into 10^ 6 = half into V2 V1 squ 5,370 squ minus 590 s okay so we know everything all the numbers we know only thing solving is a bit careful so you will get when you solve the left side you will get 2.44 10^ -9 M it will Sol everything you'll get this much on the right side when you solve this one you will get 14 uh 64 into 10^ 5 so find them you'll get 5.99 10^ 24 that is 6 into 10^ 24 kog okay so uh third question mass of a planet is M and its radius capital r so this is like this is the planet its mass is M this is radius is this is the center radius is R uh a rocket is launched from the surface of Planet such that rocket moves radially along okay so like this uh away from the planet the rocket engines are stopped when the rocket is at a height of R above the surface so R above the surface mean uh from the center it's at a height of distance of 2 R so this is r at that point the engines are switched off then it's moving from R above the surface uh height of R above the surface of the planet so that the change in GPA of the rocket when it moves from height R to 2 R so it's moving from R to end the point which is 2 R so that means we can say it's moving from X to y when it move from X to Y we need write get an expression this one okay so again it's not a big issue so it's moving away so it's losing G uh gaining GPA so gaining GPA that is Del equal find the GP at point Y so the radius the distance from the center of the earth will be 2 r + r 3 r- G mm over 3 r - - g mm/ 2 R that's all nothing just solve it so that's going to beus into minus plus g m m/ 2 r- G mm/ 3 R the LCM is 6 R so that's going to be 3 G mm minus 2 G mm so that will be g m M over 6r that's a change in or increase in gravitational potential energy okay the same question second part when the rocket moves from the height R to 2 hour its speed changes from 7,600 m/s to 7,200 so gaining GP so you can see the speed decreasing means it's losing the kinetic energy we got an expression for the loss gaining GP we can write gaining GP gravitational potential energy equal to loss in kinetic energy so ging GP we got the expression for that already we derived it that is GMM 6r equal to half M Los in K initial k at a height R which is 7,6 s - 7,320 squ okay so here m m will get cancel and the radius of the planet is given 3.40 10^ 6 that is uh R so we know everything only thing we need to find the M these two M the mass of the satellite sorry the rocket will get cancell so we need to find the capital M uh so m is equal to 6 * R we cross multiply 6 * r that is uh okay so I'll write the M subject here again from this m is equal to when 6 R goes here half so that will become three 3 R into 7,600 s - 7,320 s over divided by uh capital G okay so we'll solve it m is = to 3 R so 3 into the radius is given 3.40 10 the^ 6 6 7,600 s - 7,3 and 20 s/ G the capital G 6 67 10^ - 11 okay solve it you'll get the answer so mass m = 6.39 10 the^ 23 kilog that's the mass of the planet okay so the last topic is escape speed Escape speed means speed at which an object is projected from the surface of a planet so what's the minimum speed required to escape completely from the uh gravitational pull of the planet so that means if it wants to completely escape from the gravitational p uh it should not experience any gravitational force when it reaches a point where that's the maximum distance from the point that should be the infinite distance because the gravitational force could exert up to infinite distance so infinite distance means very large distance yeah so the object should be launched from the surface of a planet at a speed U so the minimum speed is called Escape speed such that it never comes back when it is projected not a rocket motion just one throw uh after that it should not come back so it should reach an infinite point where just reaches so minimum speed so just reaches where its speed becomes equal to zero so the radius of the planet we take it as R so uh the initial when it moves from this point to this point it's losing the kinetic energy and gaining gravitational potential energy so loss in kinetic energy is equal to gaining gravitational potential energy so at infinite distance the gravitational potential energy is equal to zero because it has reached infinite GMM r r is infinity also we are calculate the minimum speed so minimum speed means it just reaches so at this point the kinetic energy also becomes equal to zero okay so loss in kinetic energy equal gain in gravitational potential energy so loss in kinetic energy means initially it's ktic energy half M u² we are finding the minimum ktic energy such that it completely escapes from the gravitational effect of the planet means it should reach the infinite distance half m u s at infinite distance the ktic energy equal to zero just reaches so the speed becomes zero gaining gravitational potential energy that means the fining gravitational potential energ is zero because at infinite distance it should reach because otherwise there will be gravitational pull on it so the final gravitational potential energ is 0 minus- g m m r is the initial gravitational potential energy on the surface of the planet so half m u² is equal to g m m r m m get cancel so u² = 2 g m is the mass of the Planet 2 g m over R so we can write this as 2 g m/ r² into r r is the radius of the plan GM o r² is the gravitational field strength so that means u² is equal to 2 g r so the minimum speed what we found is the minimum speed so that is the escape Velocity U Escape is going to be square root of 2 g r g is the gravitational field strength on the surface of the planet R is the radius of the planet okay so that's all this uh gravitation field I hope you understood uh okay we'll meet during the next lesson bye