 ## Physics

Learn the toughest concepts covered in Physics with step-by-step video tutorials and practice problems by world-class tutors

8. Centripetal Forces & Gravitation

# Newton's Law of Gravity

1
concept

## Universal Law of Gravitation 4m
Play a video:
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?

4
concept

## Center of Mass Distance 7m
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5
example

## Find force between spheres with radii 1m
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