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### Course: Physics library > Unit 4

Lesson 3: Newton's law of gravitation- Introduction to gravity
- Mass and weight clarification
- Gravity for astronauts in orbit
- Would a brick or feather fall faster?
- Acceleration due to gravity at the space station
- Space station speed in orbit
- Introduction to Newton's law of gravitation
- Gravitation (part 2)

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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.

## Want to join the conversation?

- 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)- The gravitational force is F=G*m*M/d^2 (look at the video at0:50)
**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)

- 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)
- 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.

m*a=G(M*m)/r^2

a=G*M/r^2

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.(7 votes)

- What is the difference between g and G.?? Except the difference in values(2 votes)
- g = 9.8 m/s^2 (acceleration on Earth) whereas G = 6.67430(15)×10^-11 m^3⋅kg^–1⋅s^–2. Hope this helps!(1 vote)

- 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)
- 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.(4 votes)

- how was the formula derived in the first place?(3 votes)
- What is the value of G (gravitational constant) ?(2 votes)
- G ≈ (6.7±0.6)×10−11 m3⋅kg–1⋅s−2(1 vote)

- in which cases G is not constant? which sal said at2:07.(3 votes)
- As far as we know, G is constant everywhere, but we don't know that to be true for certain. g = GM/r^2, so it varies depending on the mass and radius of a planet.(0 votes)

- If small earth has 1/2 mass of Earth, then what would be its density?(2 votes)
- what is the difference between the big G and the small g(1 vote)
- 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)

- where can be a situation where is no gravity present??(0 votes)
- 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