Universal Law of Gravitation

Concept: Universal Law of Gravitation

Video Transcript

Hey guys. So in this video I want to introduce you to the universal law of gravitation and we're basically going to look at how gravity really works in a more detailed left. Let's check it out. All right so the force of gravity we've been talking about this for a while. The force of gravity which is also referred to as weigh equals mg. That's what we've been doing the whole time. But it actually turns out that it's not just the Earth pulling down objects. In fact this is a mutual attraction because of Action-Reaction Newton's third law. It's a mutual attraction so that means that when you are here and the earth is pulling down you you are also pulling up on the earth. It just turns out that because the Earth is way more massive than you are you're going to accelerate down towards the earth and the earth isn't really going to move towards you though technically in theory it would. But the mass difference is just too great. And you are actually not going to have an effect on the earth. OK. So Newton's universal law of gravitation states that every two objects attract each other every two objects attract each other. OK. So again just as the earth attracts you Action-Reaction you pull back on the earth as well. OK the attraction force is wait's which is the same thing as mg that was a simpler version of the weight equation. And now we're going to learn is that the gravitational attraction is actually given by a more complicated equation and that equation is big G m1 m2 over a little r squared. Big G is called universal gravitational constant and that is given by six point six seven times ten to the negative 11 and then a bunch of units. These are the units. You probably don't have to memorize that. Ok 6.67 times ten to the negative 11. It's called the universal constant because this G is the same everywhere. You might remember little g equals 9.8. Well guess what that's the that's the acceleration due to gravity at the surface of the earth. So it's actually a very specific number one way to think about this is that in physics big letters like big G are constants and little letters like little g are variables. If you go to a different planet then g is going to be different but this G over here is always the same so everywhere everywhere you go G will be 6.67. So I have know two objects let's say that this is m1 and this is m2 and there's some distance from each other. First things first the force will happen in a line in a line or a long line between the two objects. So if this is the center of each object in a draw line connecting each other the forces will be like this. OK. Now I can say that this force here is the two poles on the one so you have two objects. This one pulls on this and then this pulls on this they are pulling on each other. So this guy here one is being pulled by two. So it's the force of two acting on one and this is the force of one acting on two. They have the same magnitude but they have different directions so I can see that the magnitudes are the same but obviously they have opposite directions because of action reaction. Instead of writing F21 or F12 you know the direction doesn't really matter. So you could just generally refer to this as F12 which would be the force between 1 and 2. OK. So that's G that's r and R, now little r is a distance not a radius, little r is the distance, it's distance between the object centers of mass. OK. So between centers of mass. So the idea is that the distance we're supposed to measure it's the distance between the center so if you have a perfect sphere that's going to be right on the middle. And this is the distance here. So just to be clear it's this distance between the center and not this distance here. OK not this distance but the one between the one between the centers. So one way that I have to think about this to remember this is a little perverted but if you look at this it looks kind of like two boobs right with nipples here and one way to remember this is that r is the distance between the nipples. Instead of this distance here which I guess would be the cleavage right. So nipples not cleavage. OK. So hopefully that works for you. And you remember so every two objects are attracted that way. I used to think of this attraction in terms of just mg. This is a more complicated version of that. And let's do some examples. Now I want to do one and then I want you guys to try the next one. So I have you the two identical solid spheres of mass 10. So there are let's just draw to their place side by side without any space between them so they're going to look like this. And they both have the same mass it's going to say m1 equals m2 equals m equals 10. Let's just call m and they both have the same diameter d1 equals d2 so I'm going to call this d equals 60 centimeters 4.6 meters you might remember that in physics we're never going to use diameter. We're always going to radius. So I'm just going to say the radius is point three meters. So I have using this information I want to find the attraction force, attraction force is an attraction due to gravity, gravitation to gravity. And it's given by the equation that I just showed you guys the attraction due to gravity or the gravitational attraction was given by this equation right here. Ok g is six point six seven times ten to the negative 11. The masses are 10 and 10 and a little r is a distance not a radius. And remember it's a distance from here to here. And how do we do this. Well the radius of this guy here is point three and the radius of this guy here is point three. So the distance between these two since they are touching side by side is actually twice the radius going the distance is twice the radius. So it's going to be point six. I can square this and I work this out. If you multiply everything. Be careful. But a lot of parentheses everywhere.

Right. Make sure you try this out so that you can get good and confident that you were doing this correctly. I got 1.585 times 10 to the negative a. Notice how that's a tiny number. OK we'll talk a little bit about that a little later. And now what you can do is to just pause and try this out real quick. It's similar. And we're going to be playing with that equation again. OK. I'm going to jump right into it. The International Space Station ISS has a mass of this much let's write this now m is four point five times 10 to the fifth. And it is at this height above the earth. And I give you the mass of the earth. So this is the mass of the ISS, the mass of the earth is five ninety eight times ten to the 24th and it has a radius of roughly this much more than the radius of the earth radius is big R little r is the distance and that being this 6370 kilometers if you want to rewrite this in scientific notation you get six point three seven times ten to six meters. And there's a height here the International Space Station is at a height. Let's just put as an h of 370 kilometers or 3.7 times ten to the fifth meters everything in traditional standard units and we want to find the ISS's weight. So one way you could have done this you could have said well is a way just mg, Ok isn't which is mg. And actually it's not because what we're going to see and we'll talk about this more later. But as you move away from the earth as it moves the move away from the earth this g here is going to decrease. So g is only 9.8 if you are on the surface of the earth and that's because as you move away as two objects move away from each other the distance little r increases. This entire thing here will decrease. So the attraction between two objects gets smaller as they get farther and farther away. In fact if you're far enough from each other if two planets are far enough from each other they won't feel the attraction anymore because will be just too far away. OK so we wouldn't be able to do this because we don't know what g is. So the way to do this instead. To find the way is by using the Fg equation instead. So that's what we're going to do. OK. So I'm going to draw it over here. F g is g. m1 m2 over distance Square. So I have g, m1, m2 the sort of tricky part here is what is the distance. Well let me make a drawing. Remember it's the distance is the distance between the centers of mass So let's say this is the ISS over here. What matters is not this height here. It's not the radius here. The radius of the earth. But instead it's a combination of both. So I can say that little r and will see more of this later little r is big R plus h and I have to combine those two numbers. OK. So if I were to do this the numbers are six point thirty seven times ten to the six plus three point seven times ten to the fifth. And then when you add both you get six point seventy four times ten to the six meters. And that's the number that's going to go here. OK. So when I do this it's going to be six point six seven times ten to the negative 11 the two masses which are. This isn't gonna fit here. Let me just go. So that's the equation. But we're going to continue here. Six six seven times ten to the negative 11 the masses are in no particular order four point five times 10 to the fifth five ninety eight times 10 to the 24th. All of this divided by the distance between the two. The total distance here which is those two numbers added. And that is squared. So I hope you can see how this is going to be. It's a huge mass right. It's these all these huge numbers multiplied against each other or multiply each other. And if you do all of this you get that the weight is three point ninety five times ten to the six Newtons. The mass of this thing is four point five times into the fifth. If you want to multiply this by nine point eight you would get roughly you'd get roughly 4.4 let's say 10 percent of the six 4.4 times tend to the six. But instead we got a three point ninety five. And that's because as we're far away you're gravity is different. OK. So when we have these sort of problems with planets or satellites and whatnot and looking for the attraction of them you're going to use this equation right here. Let's get going. Let me do this example here and I have three objects that are fixed in place as shown. I want to find a net gravitational force gravitational force means of Fg. I wanna find the net force acting on m in terms of the given variables here G, m and L. So we can solve this not with numbers but with variables. Here's the idea I have three objects here. They each pull on each other. So this guy let's call this A, B and C, A is actually being pulled to the right by both of these objects I can call this the force between A and B. And this is the force between A and C, C is being pulled to the left right. Why because these guys are they make a straight line with each other they're all on the line. So C is being pulled this way by A towards A directed towards A. So I'm going to call this Fac is the force between the two. And then C is also being pulled towards the left by B, F between B and C ok. So you draw a line between them and you figure out that they're both going towards the middle they're both being pulled towards the middle. Now B is in the middle on the left this one last on purpose B is in the middle B's being pulled to the left by F by the mass over here. So you can call this. This is also the force between A and B and by the way this is the same number right just opposite directions and B is being pulled to the left by C and I'm going to call that the force BC again. Exactly like that. OK. Just to give a quick idea. I don't have to draw this but that's how many objects this guy is going to be pulled in every single direction. And what I want to do is find a net force for objects B. So this over here. So the net force on B is the same thing as the sum of all forces on B. And there are two forces. There is a force between A and B and there is a force between B and C. Now the going opposite directions so you have to put a sign here to indicate that this guy is to the left and then this guy is to the right. OK but before I can do this obviously I have to go calculate those two forces.

What is the force between A and B. Well the gravitational force is always going to be g mass one mass two over distance squared. So between B and B it's g, A and B tells me that it's between mass A and mass B. So Mass A is 2m mass B is m and the distance between them is zero this is 3l. So the gap between these two guys is 3l the gap between these two over here is 2l because it's to five and obviously the gap between A and C we don't need that but the gap between A and C is 5 L . So this gap here between A and B is three L and that's squared. OK. So I do this I get 2g. Mass times mass is mass square. This is square so I have a 9 L squared k. If you want to you can we can kind of put this in a calculator 2 over 9 and you'll see that this is 0.222 gm squared over l square. Let's leave it like this for now. Let's find the force between. And by the way obviously when we play this and we're going to play it as a negative because it's going to the left let's now find F. Fbc so it's g the masses and the radius squared. So what are the masses. Well B and C, B is m, C is 3m and the distance between them is 2l. So I end up with 3gm squared over 4l squared. So it's three over 4 which is if you want the decimal version. Is this OK. And the reason I get a decimal version is because in this case it's going to be easier to work this out. And I can just plug in these forces here. Fab will be negative it's negative point 222gm squared over l squared plus positive point 75gm squared over l squared. I hope you see right away that this part here is the same which means I can just combine this with this one of them is negative. So when you do this again very carefully I gets point fifty three ok. So this is point fifty three gm square over l square. This is force total on B and that's part a. I know we don't have a lot of space but we don't need a lot of space for B and C are for part B. We're going to carve out a little bit of room here for part B says Suppose 2m comes loose so they're fixed in place I forgot to mention is there fixed in place. So we pulled each other but not really moving, let's say 2m comes loose in what direction in which direction would accelerate. Well I hope you agree that it would accelerate to the rights. And that's because the force that's pulling into the rights this one is positive to the right is greater than the forces that's pulling it to the left. This is a point 75 this is point 22 now if you look at it is that your net force came out as a positive which means a positive force outweighs the negative force or the force of the right out with the force of the left. So this is going to be to the rights and that's because sum of our forces on B positive are the stronger force on B to the right. So with this resulting acceleration B constants with the resulting acceleration B constant. Well remember F equals ma. So the acceleration depends on the mass and the force where's the force coming from, this force here is a combination of these two forces. So A is is A has to do with I can say that A is proportional to the gravitational force. But remember the gravitational force is (gm1m2) r square. And this is not going to be constant because as you release and the middle one starts moving to the right. The r between these two guys will be closer. They r between these two guys is going to be getting farther and farther. Right. So because the r's will be changing the forces will change as well. And if the forces change of the acceleration changes OK. So it's constantly changing. In fact what's going to happen is that they will be going be getting faster and faster as they get closer to each other and the acceleration itself will be growing. OK. So not only are they getting towards each other they're getting faster and faster and the acceleration is also getting greater and greater as the force gets greater as the distance gets smaller. OK. So will it be constant not. Constant, ok, the last one I have to make here and I briefly mentioned this earlier is notice how gravitational forces are very weak for non planetary stuff. We did it here and we got one point eight five times in the negative eight that's nothing. So basically the idea is that we're going to ignore these forces unless otherwise stated. What does that mean. Well if you have if you have a little. A pulley with two objects. You're not really going to start thinking that all but they're pulling on each other and all that kind of stuff right. So this doesn't mean that you're not going to look at everything a different way for things that are sort of human sized. The gravitational force doesn't really effectively do much or anything really. So we can just ignore it. We're going to consider gravitational force when we're dealing with planetary stuff. So huge masses ok.

Problem: (a) How hard does the Earth (5.98 x 1024 kg) pull on the Moon (7.35 x 1022 kg) if they are 3.85 x 108 m apart? (b) How hard does the Sun (1.99 x 1030 kg) pull on the Moon if they are 1.50 x 1011 m apart?


Problem: Calculate the net force acting on the Moon when it is aligned with the Sun and the Earth, as shown below. Use the values given and forces found in EXAMPLE 1 (above). 

EXTRA: What acceleration (magnitude and direction) does the Moon have as a result of this net force?


Example: Gravitational Force

Video Transcript

Alright so this question is pretty tricky but it's a big classic in physics. So I wanted to show how to work it out. We have three objects lined up as shown the masses 2m and 3m are 10 l apart. So this is some sort of length. So these two guys here are a distance of 10 l apart. I want to know how far from 2m in terms of l what m have to be. How far from 2m this m have to be. So I want to know what is this distance here. I'm going to call this y. Because it's the distance in the y axis could have called D, x would be kind of weird right but. So how far in terms of l would m have to be in order to experience no net force experience no net force means that the sum of all forces on m equals zero. OK. So let's call this guy A, B and C and I want the sum of all forces and B to be zero. I hope you see that B is being pulled up by C I can call that Fbc or just Fc and it's being pulled down by Fab the idea is that if the net force is zero these two forces cancel. So if the sum of all force zero it means that these two forces will cancel. They are going to have the same magnitude. So Fbc is going to have the same magnitude as Fab. To solve this we're now going to replace Fac 'mand Fab with their equations so remember F is Gm1m2 r. So in this case will be G MbMc divided by the radius between I'm sorry the distance between B and C. Remember little r is the distance not the radius. And then this is G ma mb over the distance between A and B squared. Right away I hope you realize that G's cancel and Mb cancels as well. So I have this ratio over here. If you try to solve this you're going to be kind of stuck you're gonna get stuck because you don't know all these distances. Now I know A.

And B the distance between A and B is what I'm looking for and I call this y. However I don't know this distance here. OK so the best you can do here is say well if the whole thing is 10 and this is y then this is just 10 minus y. This is just 10 l minus y and that's what I'm going to do I'm going to replace that here and then try to solve. Let's clear this up a little bit, Mc divided by 10l minus y squared equals ma divided by y squared and we're looking to solve for is y. Now. I have these numbers here. Ok in the answer. At least in terms of m and I have l here. So I'm supposed to solve for. I'm supposed to solve for y. The easiest way to do this is going to be to actually take the square root of both sides to get y out of here. There's a few other ways you might want to try this. I can for example Cross multiply and distribute the y's and bring everything together but that's going to be harder because we're going to have to foil this thing so I'm going to say I'm going to take the square root of both sides. And that's going to get rid of the squares here so I'm going to get the square root of mc, mc by the way it's 3m so the square of three m. Equals the square root of ma is 2m. And then this is going to be 10l minus y and this is just y. By the way here I can cancel out the m's because I have two m's that aren't a square root. This would've been the same thing as saying if we had something like this. Three m equals two m. And then some stuff in the bottom here. This is same thing as saying 3m equals 2m so the m's will cancel just to be clear. We can do that even though it's sort of a it. And now I'm going to cross multiply this so I'm going to have you on the left. This is going to multiply with that so I gonna have square root of 2 10 L minus y equals y squared of three. OK and this is just algebra it's kind of annoying but I have to get the y's by themselves so that these two y's have to be by by themselves on the left side or the right side doesn't matter so that I can solve for y.

Ok when we continue over here so I'm going to. Distribute this. So to me 10 l square of 2 minus square of 2y equals square of three Y and I'm going to move this over here and I'm going to get 10 squared of two L equals squared of two y plus squared of three y. And the way to solve for y here. Is going to be to factor out y. And then send it over to the other side. So the final answer is that y equals 10 squared of 2l divided by a square of two plus square of three. OK. And that's your. Sort of final version of this. If you wanted to you could calculate this entire numbers here. Let me show you real quick. You could calculate this entire number here. Time's l And if you did this you would get to 4.50 l it runs pretty nicely actually. So if this was a sort of a multiple choice test question maybe the answer would be presented this way maybe the answer will be presented this way. Again you've got to be ready for both. So anyway the big idea here is there is no net force. So you set them equal to each other. And what's different about this question is that this little set up here I'm going to ask you how far one of these two have to be from each other and what you have to do is make one the variable and the other one. You write in terms of the variable . So this was why I knew this was 10. So I have to rewrite this which is the one I won't know in terms of the other two. So that way you can just you just use algebra to finish the rest. OK. That's it for this one I mean if you guys have any questions.

Universal Law of Gravitation Additional Practice Problems

Let there be two planets X and Y with masses M X and MY . The radius of planet X is twice that of planet Y. If an object weighs 4 times as much on the surface of planet X as it does on the surface of planet Y, then MX/MY is given by

A. 1/8 

B. 16 

C. 1/4 

D. 8 

E. 1 

F. 1/16 

G. 1/2 

H. 2 

I. 4

Watch Solution

Consider the six equally-spaced objects shown in the diagram below. The points A, B, C and D represent locations in empty space and the large circles represent planets with masses m and 2m. 

1. At which location(s) would the magnitude of the net gravitational force on an object be a maximum?

2. At which location(s) would the magnitude of the net gravitational force on an object be a minimum?

Separate your answers by a semicolon.

1. A and D; B and C

2. C; A

3. B and C; A

4. B; D

5. D; B

6. A; C

7. B and C; A and D

8. B; A and D

9. A; B and C

10. D; B and C


Watch Solution

Two masses, separated by a distance d, experience a gravitational force F. In order to drop the force to F/3, the distance has to be changed to

a) d√(3)

b) 3d

c) / 3

d) d / √(3)

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One day a human being will land on Mars. Mars has a diameter of 6786 km and a mass of 6.42×1023 kg. If you weighed 834 N on Earth, how much would you weigh on Mars?

A. 1270 N

B. 316 N

C. 224 N

D. 834 N

E. None of the above

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A certain object weighs 22.2 N on the surface of Earth. If the radius of the moon is 0.276 times the radius of Earth and the mass of the moon is 0.0123 times the mass of Earth, the object's weight on the surface of the moon is approximately

A. 9.79 N

B. 133 N

C. 3.58 N

D. 22.2 N

E. 0.365 N

Watch Solution

What is the difference in the force of gravity on a 1.0-kg mass at the bottom of the deepest ocean trench and that at the top of the highest mountain? Assume that g = 9.8 m/s2 at sea level. The radius of Earth at sea level is 6.37 × 106 m. The deepest trench is the Marianas Trench, south of Guam, which has a depth d = 1.103 × 104 m below sea level. The highest mountain is Everest in Nepal, which has a height of h = 8.847 × 103 m above sea level. The difference is

A. 0.0062 N

B. 0.061 N

C. 0.034 N

D. 0.0067 N

E. 0.027 N

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If you have a mass of 60 kg, answer the following questions:

a. What is your weight on Earth?

b. What is your mass on the Moon?

c. What is your weight on the Moon? Note that the mass of the moon is 7.34 x 10  22 kg and the radius is 1.74 x 103 km. 

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A 15 kg mass lies at rest on a table with a coefficient of static friction of 0.4. If you were to place a second mass 5 cm away from the first, what would that mass have to be to cause the 15 kg mass to move towards it?

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A 15 kg mass and a 7 kg mass are separated by 10 cm. Where should you place a 4 kg mass in between the two masses so that the net gravitational force on it is zero?

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Which of the following is true about the gravitational force?

a. The closer two object get, the larger the force gets but the smaller the speed gets

b. The closer two objects get, the smaller the force gets but the larger the acceleration gets

c. The further two objects get, the smaller the force gets and the smaller the acceleration gets

d. The further two objects get, the larger the force gets and the larger the speed gets

Watch Solution