SSS3 Physics: Gravitational Field

[Music] come back from the break we take your questions and then we close session so for today's end topic the objectives include we are going to learn to understand the concept of the gravitational field then we shall state the relationship between the gravitational constant g and acceleration due to gravity we will learn to calculate the escape velocity of the earth and then we shall state some effects and applications of the gravitational field but before we start let's um define force a force is a parameter that changes the states of an object let's say we have an object at rest this object is at rest this object will continue to be at rest until a force acts on this object so we say that a force is a quantity that causes motion because without the force there won't be motion and at the other end so if this ball is in motion if the ball is in motion we need to apply a force to stop the movement of the wall of the ball so a force is a parameter or quantity that changes the states of an object we have two types of forces we have them contact foxes and then we have um force fields contact forces require a direct contact with the objects for example now i'm holding this marker now i can apply a force to this marker now because the marker is in direct contact with my hand another example is i'm holding this um duster now i can push this duster like this and push it back like this that's contact forces examples of contact forces are we have pull we have push we have tension and we have a frictional forces so the other type of force which is our interest for today are the force fields for force fields we don't need any contacts in as much as we are in the radius of the field you experience the force examples of that are if we have a magnet we have a magnet this is the north and the south pole of the magnet if we bring a metallic object very close to this magnet we are going to notice that them these magnets will attract this metallic on objects that means that the object doesn't need to work to be in contact with the magnet before it experience the magnetic force of the magnets so that's what we are trying to say that um for force fields we don't need a direct contact for the object to experience the force another example of force field is them electric forces which acts the same way as the magnetic forces and then the third one is the gravitational field which is our main interest today so what is the gravitational field a gravitational field is a region or space where gravitational forces are filled every object is believed to be in its own gravitational field what i mean by that or what i mean by that is just that every object in space will attract another object in space with a force that acts in opposite directions in the same way the earth would also attract the moon and the moon would also attract the eggs it is believed that these forces are what keeps all objects in the universe in their position a newton helped us in calculating the force of attraction between these two objects if this object is mass m1 and the other object is mass as a mass of m2 nielsen was able to deduce that the force of attraction between the two bodies is directly proportional to the product of their masses and inversely proportional to the square of their distance a path let's say this is the center of this mass and this is the center of the other mass then their distance our paths will be will be represented as r so the newton's law of investor gravitation states that the force of attraction between two masses is directly proportional to the products of their masses and inversely proportional to the square of their distance apart so that means that if two light masses are in contact with themselves the force is going to be low that means that a high mass with what would result to a i force and then a i radius would result to a low force what we are just trying to say scientifically just that um as the mass increases the force of attraction increases and as the distance apart increases the force of attraction reduces if we introduce our constants here we are going to have that um f is equal to g m1 m2 all divided by r squared where g is referred to as the universal gravitational gravitational constants and it has a value of 6.67 times 10 to the power of minus 11 newton meter squared per kg so let's move on from here so um mathematically we are able to deduce that f is equal to g m1 m2 over r squared where m1 and m2 are the masses of the two particles measured in kilogram g is the universal gravitational constant and then r is their distance apart measured in so let's take a look at the same question so five kg spherical balls are placed are placed so that their center are 53.0 centimeters apart so the mass m1 is 5.0 kg and the mass m2 is also 5.0 kg because they have the same masses and then their distance apart half is 50.0 centimeters and then we are told to take the gravitational constant g as 6.67 times 10 to the power of minus 11 newton per meter squared per kg so from here newton was able to tell us that um the relationship between the masses the force of attraction and their distance apart is represented as f is equal to g m 1 m 2 over r squared so the force of attraction f will be equal to 6.67 times 10 to the power of minus 11 which is the value of g times m1 which is 5 kg times m2 which is also 5 kg all divided by their distance apart which is r which is a 50.0 centimeters but we have a problem here the problem is that centimeter is not the s i need forward for distance so that means we need to convert our centimeters to meters which is the s i need for distance and we know that um one meter is equal to hundred centimeter so to convert these two meters we have to divide by what by hundred so fifty fifty centimeter will be equal to fifty divided by hundred so that we have we now have our f to be equal to 6.67 times 10 to the power of minus 11 times 5 times 5 divided by 0.5 times 0.5 and if we do this we are going to have f to be equal to 6.67 times 10 to the power of minus 9 newton 6.67 times 10 to the power of minus 19. so if we have two masses m1 5kg and m2 5kg separated by 50 centimeters their force of attraction will be equal to 6.67 times 10 to the power of minus 9 newton so let's move on to calculate the mass of the watts the mass of the earth but before we do that let's take a look at this example too determine the force of attraction between the sensor of the sun the center of the sun which is represented as ms ms which has a value of 1.99 times 10 to the power of 30 kg and earth mass of the earth is giving us 5.98 times 10 to the power of 24 kg the distance apart is equal to 1.5 times 10 to the power of 8 kilometers so we are told to calculate the force of attraction f newton told us that the the force of attraction f is equal to g m1 f is equal to gm1 m2 all divided by half squared so if we impute our the values of our parameters here we are going to find out that f is equal to g which has a value it's a constant it has a value of 6.67 times 10 to the power of minus 11 newton meters squared per kg squared times m1 the value of m1 let's take m1 as the mass of the sun and let's take m2 as the mass of the earth so our m1 is 1.99 times 10 to the power of 30 times the mass of the earth is 5.98 times 10 to the power of 24 kg all divided by their distance apart is 1.5 times 10 to the power of 8 but we have to convert that to meters because i told you in the la in the other example that um the si unit for distance is what is meters so if we make use of our kilometers here we are going to get the wrong answer so we have to convert that towards two meters and how do we convert kilometers to meters we know that a thousand meters is equal to one kilometer so therefore one point five times ten to the power of eight kilometers will be equal to one point five times ten to the power of eight times 1000 since a thousand meter is equal to one kilometer then one point five times times the power of eight kilometers will be equal to one point five times ten to the power of eight times one thousand meters so this value is what we are going to use here now as our distance so we have 1.5 times 10 to the power of 8 times 10 to the power of 3 don't forget the square because in our equation what we have is r squared so if we push for that we are going to get our f to be equal to 3.53 times 10 to the power of 22 newton if we do that we are going to get 3.53 times 10 to the power of 22 newton so let's move forward so as i said now we want to state the relationship between our universal gravitational constants and our isolation due to gravity and in this um sub-topic we are going to learn to calculate our small g which is this relation due to gravity of the earth we are also going to learn to calculate the mass of the earth and then we are also going to learn to calculate the distance of every object from the center of the earth so from the newton's law of universal gravitation which was given as f is equal to g m1 m2 over r squared f is equal to g m1 m2 over r squared we also know that every object in space falling down would move with an acceleration equal to the isolation due to gravity meaning that f which is equal to m a and since the object is falling down the value of a will be equal to the acceleration due to gravity because every object falling to the surface of the earth must fall with an acceleration equal to the isolation of this gravity and from here we can deduce that f is equal to what mg let's represent this as equation one and this has equation two so what we are going to do now is from equation one we know that f is equal to g m one m two all divided by r squared and from equation two we know that f is equal to what mg so since the values of f are both equal to this and this what we want to do now is to equate those two values together so that we have g m1 m2 all over r squared must be equal to mg so one assumption we are going to make here is just that then our m2 is equal to n so that we now have g m small m all divided by r squared must be equal to us to mg and from here we have m at both ends of the walls of the equation and m can cancel m here so that we have um g m over r squared is equal to g and this equation shows the relationship between the big g which is universal gravitational constant and our small g which is the acceleration of this gravity and then from experiments m is the mass of the earth and we can deduce that term g is equal to g m e over r e squared where m e is equal to the mass of the earth mass of the earth and re is equal to the radius of the earth which shows the distance of every object in space from the earth so let's move further to this example three so i want to determine the mass of the s if the radius of the s is approximately 6.38 times 10 to the power of 6 meters g is taken as 6.67 times 10 to the power of minus 11 newton meter square per kg squared and then small g is taken as 9.8 ms squared the truth is that from this equation g is equal to g m e over r squared knowing the values of g and this big g the values which are not g is equal to 9.8 ms minus 2 which is a constant g is the isolation due to gravity of the x and every object falling to the surface of the earth must move with this value and big g which was defined as the universal gravitational constant is equal to 6.67 times 10 to the power of minus 11. knowing these values we can calculate the mass of the earth and the radius of the earth so for this question the question wants us to calculate the mass of the earth so if we cross multiply yeah we are going to have g r squared is equal to g m e dividing both sides by g because our interest is in mass of the earth dividing both sides by g then we have the mass of the earth to be equal to g r squared over g and that is equal to g is 9.8 times r is we are told to take r as um six points six we are thought to take our six point three eighths times times the power of 6 meters so from here we have 6.38 times 10 to the power of 6 all squared all divided by the value of g which is 6.6 times 10 to the power of minus 11 and from there the mass of the x will be 6 times 10 to the power of 24 k g which is the approximate value for the mass of the earth and then every object on the surface of the earth has a distance of 6.38 times 10 to the power of 6 meters from the earth escape velocity so sk velocity is defined sk velocity is defined as the minimum velocity required for an object to just escape the gravitational influence of the earth as we all know that every object that goes up must come down yes but there's a certain velocity at which objects can be what can be launched and then they are going to escape the influence of the words gravitational force of the head and that is the principle rocket satellite that is the principle that are all applied towards to them so they are launched at a value um at a value of velocity which is equal to the escape velocity if they are launched at this value we are going to see that then they are going to escape the influence of the gravitational force of the wall of the head so our main focus today is to calculate that velocity so um but first we must learn to derive the relationship between them between the velocity and the walls and the other parameters we've mentioned earlier so we can do that from we stated earlier that f is equals to g m m over half squared but before we move further i would like to write all the equations of um we've proved at this at this point here the first one we said was that f i'll just write it at this little corner f is equals to g m m over r squared we also proved that g is equal to g m over r squared where this is the mass of the earth and this is the radius of the earth so let's go back to this and proof so we have that f is equals to gm times m over r squared and from here our kinetic energy is equal towards force times distance r recall that the energy is a product of volts of force times what's distance so the distance in this question is what is the value of volt of her and from here kinetic energy is equal to us half mv squared which is equal to our force from here our force is giving us what g m m over r squared times small r all i just did there is to equate my kinetic energy to be equal to force times distance the force has a value of g m m over what over r squared times r and from here we're going to see that um this r is going to work cancel one of the arrows here so that we have f mv squared is equal to g m m and then this m here goes with this m here and then we have v squared is equal to us su g n but g m bottom gmr is equal to g r g m r is equal to what g r how do i get that from this part of this equation here where we say that g m over r squared is equal to g from here we can say that um we can say that gm over r times one over r we've not changed anything here since it's r squared out times hours to give us half squared is equal to g and from here we can cross multiply and take this r to this side so that we have um gm over r is equal to gr so that is what we have here that gm over r is synonymous towards g her the area was just represented as capital letter r since we are dealing with the words with the s which is and this r is represented as the radius of the earth so from here now we can now say that then our b squared which is equal to su g m r is equal to 2 g r and finally we can say that um v squared which is equal to 2g r taking the square root of both sides our v is equal to the square root of 2g r so using this um equation now we can calculate the escape velocity for x but before we do that i would like to use um a different method to prove this equation here we can also equate the um the potential energy to be equal to the kinetic energy what we are just trying to do is what is to prove using a different method that v is equal to square root of 2g r so our potential energy as a formula of mgh our hs represented as what as r which is equal to still f mv squared so that we have mgr is equal to half mv squared and goes with m and then we have gr is equal to half v squared when we cross multiply we have v squared to be equal to 2g r and when we take the square root of both sides we are going to have that what v is equal to the square root of m 2 g r so you can see that um with a simpler proof we were also able to calculate the um the escape velocity of the s although this is the method they are going to find in most textbooks this method is also very correct so let's move further to now calculate the numerical value of the escape velocity of the x don't forget that our formula is v e is equal to the square root of 2gr so calculate the escape velocity of the satellites from the edge gravitational field if g is equal to 9.8 ms minus 2 the x radius re is equal to 6.4 times 10 to the power of 6 meters so we want to calculate the sk velocity of the earth our formula is giving us v is equal to 2gr and from here ve is equal to the square root of 2 times g has an american value of 9.8 times the radius of the earth that is the distance of every object from the surface of the earth as a value of 6.4 times 10 to the power of 6 meters and if we calculate that we are going to get the value of 11 kilometers per seconds which is also represented as 11 000 meters per second so that means that if i want to launch a satellite or a rocket into space i must give it a velocity greater than the escape velocity because the escape velocity is said to be the minimum velocity required to launch an object into space so if the velocity of the satellite is less than this then it's going to come back to the surface of the earth but if we give it a velocity greater than this that means that it's going to escape the influence of the x gravitational force so let's take an example um as gravitational force so let's take a look at them the effects of the x gravitational field so what are the effects of the x gravitational field so the first one we calculated is that term we said that then the s gravitational field gives us a clue about the mass of the walls the mass of the earth so from the x gravitational field we can calculate the mass of the earth and that is at the value of 6 times 10 to the power of 24 kg without that formula it should be very impossible to know that the value and then we can also calculate the radius of the s which has a value of um 1.5 times 10 to the power of 8 meters and from there to we are able to calculate the value of g which has a value of 9.8 ms minus 2. so um the edge gravitational field helps us to know these n values and aside that it also keeps every object in what in space in their position when we started this class i told you that every object is in its own gravitational field that means that every object would attract the other objects closest to him with the same force and creating was able to us was able to give us the value of the force using this formula so in that sense we can say that um the earth and the sun and the moon are all in their position due to the x as in this the gravitational field of the universe without the gravitational field of the universe then the sun the moon and the sun the sun the moon and the earth might not be able to us to keep orbiting in their worlds in their position so the gravitational field of the universe helps them to be what to be put in their words in their place and then another effect is that then another effect is that them we were able to calculate the escape velocity of the world of the earth without this value then we won't be able to like um and study um some of these astronomical bodies like pluto like other astronomical bodies won't be able to what to have a quest study about them because we make use of this satellite and rockets what we send them into space so that we can get data about what about them and before we can do this we need to what we need to launch them with a watch with a velocity greater than the watts than the escape velocity so you can see that the gravitational field has a lot of votes it has a lot of benefit to us so let's um move studies and practice questions so the first practice question is telling us to state the newton's law of universal out to the product of the analysis and inversely proportional to the square of their distance apart and we can write that mathematically as f is proportional to the product of their masses and is inversely proportional to the words so the square of their distance apart where m1 and m2 are the masses of the body of the bodies and are is their distance apart and from there if we are bringing our equality sign then our constant comes into place and then we have that then f is equals to what g m one m su over r squared so the newton's law of inverse of gravitation states that the force of attraction between two masses is proportional to the to their products and inversely proportional to their square and mathematically it is written as this and then from there we can say that f is equals to g m 1 m 2 over r squared where g is equal to the gravitational constants to the gravitational constants so the b part of the question says that we should define the gravitational field of the earth the gravitational field of the earth is just a space or region where the effects of gravitational forces are words are fails simply push it is just a region where gravitational force is what is felt so let's move to the next practice question so derive the equation relating the universal gravitational constant g and the acceleration of free fall small g at the x surface don't forget that we proved this equation and then i say that all we need to do is what is to equate since we know that f is equals to us g m over r squared is equal to us mg where this is the force of attraction between two bodies on the s and this is the what this is the force on any object experiencing free fall from here m cancels m and then we have that g is equal to us g m e over r squared where this is the relationship between what's between the inversal gravitational constant and the acceleration of free fall g then the b part of the question is saying that we should state two assumptions for which the relationship a holds the assumptions are one the x is assumed to be spherical without the x being spherical then this equation might not hold another assumption is that the x is assumed to be what to be uniform to be uniformly dense another assumption is that the mass of the earth is assumed to be concentrated at the walls at the center of the earth the mass of the earth is assumed to be concentrated as it was at the center of the earth so moving to the next stem practice question define escape velocity we define escape velocity as the minimum velocity required to launch an object into space so sk velocity is the minimum velocity required to launch an object into space so the b part of the question wants us to derive the equation for escape velocity and obtain its numerical value for the hex so i'm going to go with the same problem i thought those two methods to to doing this so i the second one i said then let pe be equal to what to ke and we know that our potential energy which has a formula of mass times isolation due to gravity times distance r is equal to kinetic energy which has a formula of f times m times v squared since we have m at both ends of the equation we can cancel them out so that we now have g half to be equal to us f times v squared so while we cross multiply we haven't two g r is equal to what v squared and since we are calculating the escape velocity of the f then um this r can be represented as the radius of the s and v as the word as the sk velocity of the s taking the square root of both sides now we finally have that v e is equal to the square root of 2 g r the question also said we should calculate the numerical value calculating the numerical value we are going to get the escape velocity of the x as um 11 000 meters per seconds meaning that any object that should be launched into space must have a velocity greater than 11 000 meter per second so question c says we should state two differences between the universal gravitational constants and three four we want to state two differences between them universal gravitational constant and three four the first difference is just that them they have different units let's start from there they have a different units then from our discussion today we should know that the g has a unit of n m squared kg minus 2 and then g as a unit of volts of ms minus two and g is also referred to as the word as the first per unit mass of the word of the body so they have difference in it then another difference is datum another difference is just that then g is a universal constant g is a universal constant is a universal constant while g is the acceleration due to gravity of the x isolation due to gravity of the earth so those are the two differences between them the big g which is universal gravitational constant and the small g which is that relation due to gravity so g has a unit of nm squared kg minus 2 and small g has unit of mx minus two and then the other one is that big g is an inverse of constant that means it applies to the universe while small g is related to the words to the heads that means when we get to other parts of the world of the universe then we are going to have a different value for what for g so a round down of everything we've talked about today before we go for break we say that then every object in space attracts themselves with the value with a force of attraction that has a value of f is equals to g m1 m2 over half squared that means that you can calculate the force of attraction between you and then and the person near you by what by knowing the masses of us of your mass the mass of the person near you and knowing the distance the past knowing your distance the paths so the f is equal to what g which has a value of it's a constant 6.67 times 10 to the power of minus 11 so use that to multiply the mass your mass with the mass of the person next to you and define divide by the words by the square of your distance apart from the person and you are going to calculate the force of attraction between you and then the person and from there we also derive the formula relating g which says that g is equal to g m over what over half squared and g is also referred as as to the force by unit mass acting on the body and from there we talk about escape velocity which is the minimum velocity required to launch an object into space and then then we close them if by what by learning some effects of the graph of gravitational fields and then we will we were made to know that them the moon the earth and the sun all stay in that position due to what due to the gravitational field of the walls of the universe so um thank you for we are going to go on a short break now when we come back we are going to take your um questions thank you very much okay so i'm emilio are we got your question the escape velocity is not energy is not energy it's the minimum velocity required to launch an object into space ana monera thank you for answering them that question for us that's correct just that the force of attraction is directly proportional to the product of the masses yes you got that right but um it is also inversely proportional to the square of their distance apart so let me take that again the escape velocity is the minimum velocity required to launch an object into space and then the force of an attraction is what is directly proportional to the product of their masses and inversely proportional to the square of their distance apart so to round up let them take this assessment question now so and calculate the force per unit mass on the earth surface m e the mass of the earth has a value of 5.98 times 10 to the power of 24 kg 5.98 times 10 to the power of 24 kg and then and then the radius of the s has a value of 6.38 times 10 to the power of 6 meters and g is giving us 6.67 times 10 to the power of minus 11 newton per meter squared per kg squared so the force by unit mass is also given as what as g our small g which is isolation due to gravity is also known as the force by unit mass and this is equal to what's to g m over r squared so if we put in these values then we are going to have that then our g is 6.67 times ten to the power of minus eleven multiplies and the mass of the earth which is five point nine eight times ten to the power of 24 kg all divided by r which has a value of 6.38 times 10 to the power of 6 times the power of 6 meters all squared and if we do this we are going to get a value for g which is 9.8 ms minus 2 which is the acceleration due to gravity of the s and is synonymous to the watts to the force by units mass axis on the edge surface so thank you for attending today's session for further reading and reference please select um gravitational field lesson content available on learnathon.ng so thank you for your time please join us next week thank you [Music] oh [Music] you

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Length: 47min 8sec (2828 seconds)

Published: Sat Apr 24 2021