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Introduction to Newton's law of gravitation

Gravity is a force of mutual attraction between two objects that both have mass or energy. Newton's universal law of gravitation can be used to approximate the strength of gravitation forces between two objects as a function of the objects' masses and the distance between them. Created by Sal Khan.

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  • blobby green style avatar for user Arslan Raza
    Actually andrew sir there is a problem which i want to discuss with you sir because one of my collegue said to me that gravity of earth is directly proportional to the radius he further says to me
    that he prove it by his geophysics fellow but my thinking is that gravity is inversly proportional to the radius kindly sir help me from this issue or give me a valid example i am very thankful to you if you help me.
    (11 votes)
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    • male robot hal style avatar for user Reinhard Grünwald
      The gravitational force is F=G*m*M/d^2 (look at the video at )
      But the mass M of a planet is its density [rho] times its volume and the volume V=4/3pi d^3 (I took the volume of a sphere). So if you plug this into the equation for force you get F=4/3 pi*G*m*[rho]*d
      So in this representation the gravity of a planet is indeed proportional to its radius.
      (8 votes)
  • purple pi purple style avatar for user Adam Staples
    Why would G not be constant everywhere in the universe? Would it be because the density of matter is different in different parts of space?
    (5 votes)
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    • old spice man green style avatar for user Dan Surerus
      G is the universal constant for the gravitational force. It never changes. The units for G are m^3/(kg*s^2)

      g is the local acceleration due to gravity between 2 objects. The unit for g is m/s^2 an acceleration.

      The 9.8 m/s^2 is the acceleration of an object due to gravity at sea level on earth. You get this value from the Law of Universal Gravitation.
      Force = m*a = G(M*m)/r^2

      Here you use the radius of the earth for r, the distance to sea level from the center of the earth, and M is the mass of the earth. Notice that little m cancels out on both sides of the equation.


      If you put in the mass of the earth and the radius to sea level you will get 9.8 m/s^2 for a. This is what we call little g.

      Notice if you change your radius that the acceleration(g) will fall off as 1/r^2. If you are twice as far out 2*r, you will have 4 times less gravitational acceleration.

      That is why g can change from place to place on earth. If you are on Mt. Everest your radius will be larger than if you were in Death Valley.. for example.

      If you want to know how G came about, read on the "Cavendish experiment"

      Hope that clears it up.
      (9 votes)
  • duskpin tree style avatar for user Saradha S
    I've read that the acceleration due to gravity in the center of the earth is zero. I believe this is b'cos the mass beneath the body at the center of the earth is practically nothing; and so the force experienced downward is zero (so acceleration is zero). But, in that case, the mass above the body should be huge, and the gravitational force should be acting upward from the center of mass of the body. Hence, the body should be attracted upward, shouldn't it?
    (1 vote)
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    • leaf green style avatar for user Mark Zwald
      If you are at the center of a sphere of mass, there is equal amounts of mass in all directions around you. This means that the gravitational pull of that mass is also equal in all directions which results in zero net gravitational force since all the forces have a component which is equal and opposite.
      (5 votes)
  • female robot grace style avatar for user stutidash
    What is the difference between g and G.?? Except the difference in values
    (2 votes)
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  • blobby green style avatar for user jacob borne
    how was the formula derived in the first place?
    (3 votes)
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  • aqualine tree style avatar for user Devika
    What is the value of G (gravitational constant) ?
    (2 votes)
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  • male robot johnny style avatar for user sairahul
    in which cases G is not constant? which sal said at .
    (3 votes)
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  • hopper jumping style avatar for user Mishri
    If small earth has 1/2 mass of Earth, then what would be its density?
    (2 votes)
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  • aqualine ultimate style avatar for user raghav.228.2020
    what is the difference between the big G and the small g
    (1 vote)
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    • male robot hal style avatar for user Andrew M
      They are not the same thing at all.
      G is the universal gravitational constant. It's the constant in the equation F = GMm/r^2.
      It's value is 6.67E-11 N*m^2/kg^2. Note the units.

      g is the acceleration due to gravity, which on the surface of the earth is 9.8 m/s^2. See how these units are nothing like the other ones?

      On the surface of earth, the force of gravity is m*g, and it is also GMm/r^2, so if we set those two things equal to each other we find that g = GM/r^2 where M is mass of earth and r is radius of earth.
      (3 votes)
  • male robot hal style avatar for user Karthikk Kumaresan
    where can be a situation where is no gravity present??
    (0 votes)
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    • male robot hal style avatar for user Jacob Elessar
      There is nowhere in the universe with no gravity. Every single object extends it's gravitational field infinitely into space. However the strength of the field also becomes infinitely weaker the further it gets from it's source, causing objects at a considerable distance to be virtually unaffected.
      (3 votes)

Video transcript

We're now going to learn a little bit about gravity. And just so you know, gravity is something that, especially in introductory physics or even reasonably advanced physics, we can learn how to calculate it, we can learn how to realize what are the important variables in it, but it's something that's really not well understood. Even once you learn general relativity, if you do get there, I have to say, you can kind of say, oh, well, it's the warping of space time and all of this, but it's hard to get an intuition of why two objects, just because they have this thing called mass, they are attracted to each other. It's really, at least to me, a little bit mystical. But with that said, let's learn to deal with gravity. And we'll do that learning Newton's Law of Gravity, and this works for most purposes. So Newton's Law of Gravity says that the force between two masses, and that's the gravitational force, is equal to the gravitational constant G times the mass of the first object times the mass of the second object divided by the distance between the two objects squared. So that's simple enough. So let's play around with this, and see if we can get some results that look reasonably familiar to us. So let's use this formula to figure out what the acceleration, the gravitational acceleration, is at the surface of the Earth. So let's draw the Earth, just so we know what we're talking about. So that's my Earth. And let's say we want to figure out the gravitational acceleration on Sal. That's me. And so how do we apply this equation to figure out how much I'm accelerating down towards the center of Earth or the Earth's center of mass? The force is equal to-- so what's this big G thing? The G is the universal gravitational constant. Although, as far as I know, and I'm not an expert on this, I actually think its measurement can change. It's not truly, truly a constant, or I guess when on different scales, it can be a little bit different. But for our purposes, it is a constant, and the constant in most physics classes, is this: 6.67 times 10 to the negative 11th meters cubed per kilogram seconds squared. I know these units are crazy, but all you have to realize is these are just the units needed, that when you multiply it times a mass and a mass divided by a distance squared, you get Newtons, or kilogram meters per second squared. So we won't worry so much about the units right now. Just realize that you're going to have to work with meters in kilograms seconds. So let's just write that number down. I'll change colors to keep it interesting. 6.67 times 10 to the negative 11th, and we want to know the acceleration on Sal, so m1 is the mass of Sal. And I don't feel like revealing my mass in this video, so I'll just leave it as a variable. And then what's the mass 2? It's the mass of Earth. And I wrote that here. I looked it up on Wikipedia. This is the mass of Earth. So I multiply it times the mass of Earth, times 5.97 times 10 to the 24th kilograms-- weighs a little bit, not weighs, is a little bit more massive than Sal-- divided by the distance squared. Now, you might say, well, what's the distance between someone standing on the Earth and the Earth? Well, it's zero because they're touching the Earth. But it's important to realize that the distance between the two objects, especially when we're talking about the universal law of gravitation, is the distance between their center of masses. For all general purposes, my center of mass, maybe it's like three feet above the ground, because I'm not that tall. It's probably a little bit lower than that, actually. Anyway, my center of mass might be three feet above the ground, and where's Earth's center of mass? Well, it's at the center of Earth, so we have to know the radius of Earth, right? So the radius of Earth is-- I also looked it up on Wikipedia-- 6,371 kilometers. How many meters is that? It's 6 million meters, right? And then, you know, the extra meter to get to my center of mass, we can ignore for now, because it would be .001, so we'll ignore that for now. So it's 6-- and soon. I'll write it in scientific notation since everything else is in scientific notation-- 6.371 times 10 to the sixth meters, right? 6,000 kilometers is 6 million meters. So let's write that down. So the distance is going to be 6.37 times 10 to the sixth meters. We have to square that. Remember, it's distance squared. So let's see if we can simplify this a little bit. Let's just multiply those top numbers first. Force is equal to-- let's bring the variable out. Mass of Sal times-- let's do this top part. So we have 6.67 times 5.97 is equal to 39.82. And I just multiplied this times this, so now I have to multiply the 10's. So 10 to the negative 11th times 10 to the negative 24th. We can just add the exponents. They have the same base. So what's 24 minus 11? It's 10 to the 13th, right? And then what does the denominator look like? It's going to be the 6.37 squared times 10 to the sixth squared. So it's going to be-- whatever this is is going to be like 37 or something-- times-- what's 10 to the sixth squared? It's 10 to the 12th, right? 10 to the 12th. So let's figure out what 6.37 squared is. This little calculator I have doesn't have squared, so I have to-- so it's 40.58. And so simplifying it, the force is equal to the mass of Sal times-- let's divide, 39.82 divided by 40.58 is equal to 9.81. That's just this divided by this. And then 10 to the 13th divided by 10 to the 12th. Actually no, this isn't 9.81. Sorry, it's 0.981. 0.981, and then 10 to the 13th divided by 10 to the 12th is just 10, right? 10 to the first, times 10, so what's 0.981 times 10? Well, the force is equal to 9.81 times the mass of Sal. And where does this get us? How can we figure out the acceleration right now? Well, force is just mass times acceleration, right? So that's also going to just be equal to the acceleration of gravity-- that's supposed to be a small g there-- times the mass of Sal, right? So we know the gravitational force is 9.81 times the mass of Sal, and we also know that that's the same thing as the acceleration of gravity times the mass of Sal. We can divide both sides by the mass of Sal, and we have the acceleration of gravity. And if we had used the units the whole way, you would have seen that it is kilograms meters per second squared. And we have just shown that, at least based on the numbers that they've given in Wikipedia, the acceleration of gravity on the surface of the Earth is almost exactly what we've been using in all the projectile motion problems. It's 9.8 meters per second squared. That's exciting. So let's do another quick problem with gravity, because I've got two minutes. Let's say there's another planet called the planet Small Earth. And let's say the radius of Small Earth is equal to 1/2 the radius of Earth and the mass of Small Earth is equal to 1/2 the mass of Earth. So what's the pull of gravity on any object, say same object, on this? How much smaller would it be on this planet? Well, actually let me save that to the next video, because I hate being rushed. So I'll see you