1

concept

## Universal Law of Gravitation

4m

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Hey, guys. So in this video, we're gonna talk about the universal law of gravitation. So the story is that Newton was sitting in a tree one day and apple fell and hit his head, and he realized that the force that causes the apple to fall to the earth is the same force that keeps the moon in orbit around the Earth. So basically, what Newton figured out was that all objects in the universe attract each other. They all exert gravitational forces on each other. So, for instance, in this diagram, I've got these two spheres, these two blue spheres here. And if they both are massive, so they have mass one and mass to then the gravitational force between them is known as Newton's gravitational law. And that equation is big G times, the product of both masses divided by r squared. Where G that capital G is just a number. It's known as the universal gravitational constant. It's 6.67 times 10 to the negative 11 here, the units for that number, you may or may not need to know those Just check your just check with your professor if you're unsure and that little are that's in the denominator. There is the distance between the objects centers of mass. So if you have these two massive objects here and they're separated by some distance, whoops they're separated by some. Distance are Newton's law of Gravitation says that mass to will pull on Mass one in this direction, but because of action reaction and because of everything attracts each other Mass one. Also a polls on mass to in that direction. The magnitude of those forces is given by this equation. And as for the direction, those things those forces air directed along a line that connects the two objects. So these forces always act on a line that connects the two objects centers, and that's basically it. That's the whole equation. The last thing I want to mention is that this capital G here is known as a universal constant, which means anywhere you go in the universe, that number is always going to be the same. Don't confuse that with little G, like we've seen in forces in telematics that 9.8 m per second squared. That's what's known as a local constant local constant means anywhere on Earth that you go, That number is going to be the same. But as soon as you go off of the earth or you go to different planet, that number is going to change. So just make sure that you know the difference between those two and you don't confuse them. Alright, guys, that's basically it. Let's go ahead and check out this example I've got here. So we've got to 30 kg fears and they're separated by 5 m. Suddenly get and draw that out. So I've got two spheres and there have a mass of 30 kg. So I've got mass. One equals 30 and I've got Mass two equals 30 and they're separated by a distance of 5 m. So they know that that is equal to five. And I'm asked to find out what the gravitational force is between them. So in other words, I need to find out what F G is equal to right? So that's not a six that's supposed to be a G Great. So we know the formula for Newton's law of gravitation. Thats f g equals capital G, the product of both of the masses divided by the distance between them squared. So I have all those variables. Well, let's check. I know that this capital G is just a constant. That's just a number. And I've got this mass here. That's 30 kg, the other mass. So I've got M one and M two, and I also have the distance between them. That's r equals 5 m. So I have all of those variables. I could go ahead and plug this stuff into my calculator, so just setting everything up. I've got f g equals. I've got 6.67 times, 10 to the minus 11. Then I've got the two masses 30 and 30 and I've got to divide them by the distance between them square that five. And so to do this, just make sure I've plugged in everything correctly, actually. Have little calculator that's gonna go and help us. So I'm gonna go ahead and plug all of this stuff in. So I've got 667 times. Whoops. Yep. Times 10 to the negative 11. And then I'm gonna multiply it by the two masses 30 and 30. And now we've just got a divided by in parentheses, the distance squared. Make sure that the nominators in parentheses because you don't want to mess up the order of operations. So if you go ahead and plug that in, we get the correct answer, which is that the force of gravity equals 2.4 times 10 to the minus nine, and the unit for that is in Newton's. So that's basically the gravitational force between these two spheres. Let me know if you guys have any questions, and if not, we're just gonna keep moving.

2

example

## Where is Net Force on middle object zero?

5m

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Alright, guys, let's check out this problem. It is a classic one in gravitation, so we're gonna work it out together. So we've got two spheres and we're gonna position them along the line like this. So I've got two spheres like this. I know what the masses are, and I know the distance between them. So I'm gonna call the left Mass the 10 kg mass mass A on the 25 1 mass. B. And I know what the distance between them is. I'm told that the distance between them is 5 m. That's little are. So the whole point of this problem is that I'm gonna position a mass somewhere along the line between them, and I know what the I'm just gonna call that mass mass. See, I don't actually know what that Massey is, but I want to figure out where I have to place it so that the gravitational forces of the Net gravitational force is equal to zero. So it means I need to find out what this distance is. I'm gonna call that are a So that's my target. Variable. Really? That's what I'm trying to solve to solve four. So I've got are a Is my target variable and I want the Net. Gravitational force Thio equal zero What We know that Newton's Law says Newton's gravitational law says that there are forces between any two objects. So there is a force from this guy, which I'm gonna call F A. C. And there's a force from that guy over there on the right, and that's fbc. And I know with the I know how to express that as the equation. And so we want these things to basically cancel out, which means that the magnitude of F A F A C needs to be the magnitude of fbc, but they have to be equal and opposite in direction. So I'm gonna actually gonna write out the gravitational forces for those We're gonna start out with Newton's Law of Gravity. So I've got G m A M C divided by R A squared. And then over here you got G M B EMC divided by r b squared where that r B is just the distance right here between, um m a r m c and N b. So that's our be all right off course. We could take a look at this equation, we could actually cancel some terms out. I noticed the G pops up in both sides and then also, the emcee also pops up in both sides. So it's actually good. We didn't need to know that the mass of that center thing in there and then the Ara is my target. Variable. Now I know what Mass A and Mass B both are. So all I need to do is just figure out what this RB is. That's my only unknown variable not told what that distance is, so I actually have to go here and figure it out. So what is the distance between this thing and M B? Well, we don't know what the individual distances are between our A and R B. We don't know what those things individually are, but we do know that the whole entire distance between the two spheres is five. So we could actually come up with another equation for this. We know that our A plus RB is equal to five. So now what happens is in this equation where we've come up with two unknowns. We've got this other equation that we're gonna use to help solve it. So if I could figure out what RB is that I can plug it back into that equation. RB is this If I move this guy over, it's just gonna be five minus r a So I'm just gonna substitute this equation back into that guy right there and now we're only gonna have one variable. So now we've got mass A divided by r A squared equals mass B divided by five minus R a squared Notice how we've gone from two unknown variables to now on Lee one and that's the one that they need to find out. So because I need to figure out what that variable is, I want to start getting everything over to one side. So I'm gonna do is I'm gonna take this expression right here and move it up to the other side on the left on the M B is gonna come down and trade places with it. So when I do that, I get five minus r a squared over a squared equals mass be over mass a. Now I actually know what these two masses are, right. Uh, this mass B was equal to 25. 25. I've got this mass A was equal to 10. So how do we get rid of this whole like squared thing? Right? We've got this, like expression here that squared you might be tempted to, like, foil it out and start like, basically multiplying everything out. But that's actually gonna be way too complicated. Notice how both of these things on the top and bottom or both squared so we can actually take the square root of both sides. So don't go. Don't don't go ahead and foil it because it's actually gonna be, like, way more complicated. So instead, we could do is basically square roots both sides of this thing, and we do that. We get that the squares cancel five minus ra over Ara. And then this is just a number, right? The square root of 25 divided by 10 is just a number, and that number is 1.58 So cool. How do we get now? This are a by itself. I've got a subtraction in the numerator, so I can't just, like, go ahead and start splitting this thing off. But what I can do is I could bring the r a over to the other side. And when I do that, I guess it makes more room. So I've got five minus r A equals 1.58 r a. So now the last step one of the last steps is I've got to move this other are a to the other side So I've got five equals and I've got 1.5 are a plus R a. So this is actually sort of like a grace common factor. This are a between both of these terms, So I've got five equals R a and then I've got one plus 1.58 and that's just 2.58 Right? So if I do five divided by 2.58 and I basically move this thing down to the other side, then I've got our A and that is just gonna be 1.94 m and that is our final answer. So this thing needs to be 1.94 m away, So let's go back to the diagram away from the left mass in order Thio workout. So this has to be 1.94 so it actually has to be a little bit closer. Because if this is 1.94 then this is gonna be This is gonna be, like, three or something like that, right? It actually has to be closer to the left Mass because the mass on the right is a little bit bigger, so it's gonna pull on it harder. Our guys, let's say for this one, let me know if you have any.

3

Problem

Two spheres of mass 300 kg and 500 kg are placed in a line 20 cm apart. If another sphere of mass 200 kg is placed between them, 8 cm from the 300 kg sphere, what is the net gravitational force on the 200 kg sphere?

A

-7.32×10

^{-4}N (toward 300kg sphere)B

-1.62×10

^{-4}N (toward 300kg sphere)C

7.32×10

^{-4}N (toward 500kg sphere)D

1.62×10

^{-4}N (toward 500kg sphere)4

concept

## Center of Mass Distance

7m

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Hey guys. So we saw from the universal law of gravitation that we could calculate the force between two objects. Now, if these objects are relatively small, like we have here in this diagram, we call those objects point masses because we can treat them as points. So in this case, if these point masses have mass one and mass to and they're separated by some distance are the universal law of gravitation says that we could calculate the gravitational force between these two things, and that's given by this equation up here g m one m two over our square. But let's say we have a different situation in which we have something huge like a planet. And then we're calculating the force on a person that's orbiting some distance above that. So the universal law of gravitation still says that we need G times, the product of both of the masses. But in this situation, we're working with large objects instead of using em ones and M twos. The big object. It's a capital M, and you'll see the small object has a lower case M. So it's still the product of the two masses. Little big M and little em over R squared. This is really the same equation. It's the same thing. I'm not telling you anything. I'm not giving a different equation. It's just for a separate, um, situation. Okay, so what is this little our distance between the centers of Mass for point masses? It was just this distance between the two objects. But in this situation, in this diagram here, it's a little bit more complicated because now, if we want to get from the center of mass of the earth to the center of mass of the astronaut, because this little are is the center of mass distance between the two objects. First we have to go from the core or the center of the earth, out to the surface, and then we have to go some extra distance here above that surface. So, really, the whole center of mass length is this. Little are here, and that's made up of two lengths. The first length from the core out to the surface is called the radius. If you think about the earth is just a giant sphere in the court of the surface is capital are, which is the radius of the earth. And this extra distance here above the surface is called the height. So that little H is called the height. And we can see from this diagram that this little are is really just the sum of both of the radius and the height. So for large objects, the little R is gonna be capital R plus H. Now I want to point out that I have two different kinds of variables. Here are two different kinds of letters. I haven't uppercase letter or capital letter and a lower case letter. So in physics, a capital letter is always used to represent Constance. So, for instance, the radius of the earth a constant never changes. Whereas this lower case letter H here represents a variable because you could go any distance away from the surface of the Earth. But the radius always stays the same. So that's basically it, guys. So let's go ahead and start working out this example. So this example we have the height above the earth. So we're asked for at what height above the Earth is the gravitational force on a satellite equal 2000 Newtons. Let's go ahead and draw a little diagram here and figure out what's going on. So we have the earth representatives, that sphere, and I'm just gonna go ahead and represent the satellite is dot Don't want to show you my bad drawing skills. Okay, so we have, um Let's see, we've got the force of gravity. So we got the force. The gravitational force is equal to 1000 Newtons, and we have the mass of the satellite is equal to, ah 1000 kg. And we're because we're working with the gravitational force here. I want to go ahead and write out that equation. So I've got f g equals G times the big M little em over r squared. Remember, Because we're working with a planet and a small mass, we're gonna use that. And really, what am I looking for while I'm actually looking for the height above the surface. So I'm looking for a church, but I'm gonna use this gravitational force in order to solve for that. Okay, So what you might be tempted to dio is you might be tempted to replace this formula with our plus H capital R plus H squared and then use that to solve for little H, but I wanna warn you against doing that because you're actually gonna make the math a little bit more complicated than it needs to be. So instead of doing this here, I've got a pro tip for you guys. If you ever have a problem that asks you to solve for capital are or H first, go ahead and solve for little. Our first using Newton's law of gravity and then using this equation, you consult for whatever you want, so don't do this instead, What you're gonna do is we're gonna solve for f g equals G m m over little r squared. Go ahead and sulfur little r and then we can use this equation r equals big R plus h in order to figure out the variable that we're looking for And this problem We're looking for this h variable here. So let's go ahead and solve for that gravitational force and see if we can find the little our distance. Okay, so let me go ahead and write out all of my knowns here. Um right, So we've got the in this diagram. We've got the center of the earth. And then if I wanted to find a little our distance. That's gonna be two things. I've got the radius of the earth and that I've got a height above the center so that height is really what I'm looking for. And I've got this Little are here That is that distance we're gonna be solving for that first. Okay, so I've got that the, um the mass of the satellite is equal to 1000. What about Capital M? Because we need to know what capital M is. Well, I've got I've got G, which is the gravitational constant. I've got capital M, which is actually given over here. The mass of the earth. Remember? That's a capital letter. So gets a constant. I have the mass of the satellite, and I also have the gravitational force between them, so I could go ahead and use on soft for little are let me go ahead and write all that stuff out. So I've got that capital M is equal to 5 97 times 10 of the 24th, and I know G is and then yeah, that's basically awesome. So if I go ahead and rearrange this equation right here, so this this gravitation equation. I could come over this expression r squared. If I move that to the other side and then move the F G down is equal to G times capital, M lower case M over the force of whoops, the force of gravity. So because this is a square, I could take the square root of both sides, and I'm gonna get that r equals the square root of GM little em over the force of gravity. I'm gonna go ahead and start moving this over here so I can actually just go ahead and start plugging in values for this. So I've got 6.67 times 10 of the mice, 11 that I've got the mass of the earth 5 times 10 to the 24th. And then I've got the mass of the satellite, which is 1000. And then I've got the force of gravity, which is also 1000. If you go ahead and plug all of this stuff into your calculator, you should notice that we should get two times 10 to the seventh meters. So we're done, right? Well, no, because remember, this number here on Lee represents the full center of mass distance, not the H, which is what we're really looking for. So our last step is we're basically just gonna have to solve using the r equals R plus H equation. So if I wanted to figure out what H is going and use this equation and figure out that H is equal to little ar minus big are which is the radius of the earth. But what is that value? What is that capital are we haven't been given a value for that yet. Well, if I look here in my gravitational constants that capital r is just represents the radius of the earth which is given right here as this number. So I've got that from my final answer. I've got h is equal to little are, which is two times 10 to the seventh minus big are, which is 6.37 times 10 to the six, and that's gonna be in meters. So if you go ahead and work this out for the finance so you get 1 times 10 to the seventh, and that's gonna be in meters, so that's about 13,600 kilometers above the surface. So that is the answer for this. Let me know if you guys have any questions with this

5

example

## Find force between spheres with radii

1m

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Okay, guys, let's check this one out. So we have two identical spheres and we go and draw out these fears right here. And we're told the masses of those two spheres and were told that their surfaces were just barely in contact with each other. So we didn't write the stuff out. So I've got m equals 10 kg or M one and M two also equals 10 kg, right? I'm also told that these fears have a physical size thes physical. These seers have a physical size, which the diameter is equal to 60 centimeters, which is 0.6 m, right. Just be careful. That's in centimeters there. Okay, So in order to figure out what the gravitational attraction is, let's start off with Newton's Law of gravitation. We have f G equals G times, the mass of both objects divided by the center of mass distance between them. But if these two things have a physical size, what's that center of mass distance? Or remember, if you have two objects here that the gravitational force always acts between the centers of mass and it's a proportional or it's it's divided by the center of mass distance between them. So the center of mass between this thing is this thing has to go all the way out to the surface and that has to go inside of the other one. That's the center of mass distance between them. So we're told that this whole thing, the whole distance is the diameter, Which means that this guy is just capital are, which is the radius. So in other words, that the center of mass distance between these two objects little are is actually just two times the radius of each individual sphere and they're identical, right? So that's just the two times two times the radius, which is the diameter of the sphere, which is just 0.6 m. So now we have little g. Sorry, we have big G. We have the both masses and we have little are So let's go ahead. Just plug everything in. Right. So we've got f G equals 6.67 times 10 of minus 11. Then we've got both spheres Mass 10 10 and then we've got the center of mass distance 0.6, and that's gonna be squared. Go ahead and plug it in and you should get 1 85 times 10 to the minus eight. And that's Newton's. Alright, guys, let me know if you have any questions about this.

6

Problem

A 2,000-kg spacecraft is blasting away from the surface of an unknown planet the same size as the Earth. At 1500km above the surface, an instrument onboard reads the gravitational force to be 18000 N. What is the planet’s mass?

A

8.36×10

^{24}kgB

2.89×10

^{12}kgC

1.67×10

^{28}kgD

5.78×10

^{15}kgAdditional resources for Newton's Law of Gravity

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