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High school physics - NGSS
Course: High school physics - NGSS > Unit 2
Lesson 1: Newton's law of universal gravitationViewing g as the value of Earth's gravitational field near the surface
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There are two ways to interpret the constant g = (-)9.81 m/s². The first interpretation is the acceleration due to gravity of an object in free fall near Earth's surface. The second interpretation is the average gravitational field at Earth's surface, which can be used to calculate the force of gravity on an object. Created by Sal Khan.
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This video got me thinking. Suppose that my friend and I are tunneling downwards towards the Earth's center. As we proceed, I reason out that we should start to weigh less, since some of the Earth's mass that used to be beneath us is now "above" us (but not directly above us, of course) and is exerting a force that starts to contradict the force pulling us towards the Earth's center. Yet my friend reasons out that we should start to weigh more since according to Newton's law of universal gravitation, which states that F = G(m1)(m2)/r^2, the closer we get to Earth's center, the smaller the distance between the centers of our masses and of the Earth, which translates to a greater force.
We then think about what would happen to our weights when we reach the Earth's center. I say that we should become weightless, since there are equal amounts of mass pulling us in all directions, and therefore all forces should cancel out (we assumed that the Earth is a perfect sphere). My friend says, "No. Since the distance between the centers of our masses and of the Earth is zero, we would experience... infinite force?"
Which of us is correct? And what are your views on this?(217 votes)
The law of gravitation treats the two objects as particles, meaning all the mass is located at one central point (this is Gauss's law, and only applies when the mass being influenced is outside of the sphere). However, in reality, the mass of the earth is distributed all around this blue sphere we call home. Although conceptually you are correct and your friend is wrong (forces of gravity influence you from all directions, ultimately canceling out), the pressure and heat at the center of the earth would compress/melt you to death before you could enjoy such a feeling of weightlessness.(145 votes)
What is the cause of gravitation?
Is it even known yet?
Does the theory of general relativity explain it or is it a fundamental interaction carried by a hypothetical graviton particle?(17 votes)
From a quantum field theory perspective all forces are carried by particles referred to as gauge bosons, for gravity it would be the Graviton but we do not really have a quantum description of gravity that is mathematically consistent.
The best description we have of gravity is General Relativity and it depicts gravity as curvature of space-time.(14 votes)
I don't understand what (kg*m)/s^2 means. How can you have kilogram meters? And how do seconds matter for the force experienced by an object on the ground?(20 votes)
lets say we have a point mass(something as small as a point with some mass) now lets say u tie a string to it and pull it, to pull it you have to apply a force, and how would you define the amount of force you have you put in to flex your strength? like sal says pause and think, now sumin u have had a go at it, we work this out together, you will say, i pulled a mass of 'x' kilograms, 'x' meters away, in 'x^2' amount of time(aka the measure of the relative changes in your environment while you were at work). and thats the description of force. mass, distance, time, they are the dimensions, the descriptions we made to describe things in our real life. atleast thats my understanding of it.(2 votes)
How were scientists able to determine that g is equal to 9.81 m/s^2? And how are they able to determine the force of gravity on other planets?(18 votes)
By Newton's law of universal gravitation F1 = F2 = G* (m1*m2)/r^2
we multiply the Gravitational constant G = 6.673X10^-11 by the earth's mass divided by the earth's radius which will give us F/m2 = acceleration derived from the formula F = ma.(5 votes)
- So, every time I trip over, I fall at an acceleration of 9.81 m/s^2 ?(2 votes)
It actually depends upon exactly where on Earth you fall. g is not truly constant; it varies from location to location. 9.8 m/s^2 is just an average. The true value varies with your latitude and longitude (mainly due to surface variations in Earth's density). g also varies with altitude according to the formula: g/g' = (r'/r)^2, where g' equals g at the Earth's (or other celestial body's surface) and r' equals distance from the center of gravity of Earth (or other celestial body). For example, if you are at an altitude of 2.2x10^7 m (22000 km), the value of g drops to 0.512 m/s!(27 votes)
At3:00, it talks about the average gravitational field on the Earth's surface. I wanted to know how far does an object's gravity effect on matter? If it has no limit, even though it is negligible, does combing the distance and mass of objects near by give a combined for of gravity? (For example would the Earth's gravitational pull on the other planets be greater than 9.8 m/s^2, would it be the same, or based on the distance that other objects are in space it would decrease)
...and if theoretically there is no limit to how far matter's gravity is affecting other matter, than wouldn't that mean that the whole universe is pulling on itself?(3 votes)- To answer your last question first, yes, there is no limit to the effect of gravity, and this does indeed mean that the whole universe is 'pulling' on itself. However, the strength of a gravitational pull gets weaker depending on the square of the distance between two objects, so at interstellar distances the gravitational effect of, say, our Sun would rapidly become very weak.
The Earth's pull at its surface is 9.8 m/s^2, but an object at its surface is only about 6400 km from the centre. The Moon is 384000 km away, which is 60 times as far, so the Earth's gravitational pull on the Moon is 60^2 (which is 3600) times as weak -- only 2.7 millimeters per second squared. Going out to other planets, Jupiter is 2000 times further away than the Moon, so the Earth's pull on Jupiter is a whopping 15 billion times smaller than the Earth's pull at its own surface. So the numbers do get very small very quickly.(11 votes)
The acceleration in free fall is -9.81 m/s2 , but while skydiving you would be in equilibrium, but shouldn't it still be 9.8 m/s2 ?(5 votes)
that's true because when you are sitting in a chair you are not free falling, and that chair is supporting your weight.(2 votes)
- For an object at the poles and the earth is rotating , why is there no centripetal acceleration for the object . Also what if the earth is stationary(3 votes)
- If you are at the pole, the radius from the axis of rotation is zero, so your centripetal acceleration from the Earth's rotation will also be zero.
a = v²/r = rω²
If r=0, then a=0.
If you are at the equator, you will experience the largest centripetal acceleration since the distance from you to the axis of rotation is at a maximum (the full radius of the Earth).
The distance from the axis of rotation ends up being R*cos(Θ) where Θ is your latitude on the Earth's surface and R is the Earth's radius. The centripetal acceleration from the Earth's rotation then becomes:
a = R*cos(Θ)*ω²(4 votes)
- what would happen to g if earth stopped rotating(3 votes)
- The acceleration due to gravity is not related to the rotation of the earth. It is the gravitational acceleration caused by the mass of the earth.
If you are asking about the decrease in an objects measured weight because of the centrifugal force because of the rotation lets take a look at that.
The speed of a point on a circle is the radius times the angular velocity. The angular velocity of the earth's rotation is 0.00007272 radian/second the radius of the earth is 6,371,000 meters so you have 0.00007272 * 6,371,000 = 463 m/s at the equator. For centripetal acceleration you have a = (v^2)/radius so you have 463^2/6,371,000 = 214,369/6,371,000 = 0.034 m/s^2.
So for the value of g being 9.8 and the the centripetal acceleration being 0.034 which is about 0.34% of the value of g and depending on where you are on the earth the actual gravitational acceleration can vary by up to 0.7% so using 9.8 m/s^2 for problems like this is fine. If you end up doing actual engineering work where more precision is needed you will have to take the local gravitational acceleration and all other factors into your calculations.(2 votes)
3:00
Video transcript
What I want to do in this
video is think about the two different ways of
interpreting lowercase g. Which as we've talked about
before, many textbooks will give you as either 9.81
meters per second squared downward or towards
the Earth's center. Or sometimes it's given with
a negative quantity that signifies the direction, which
is essentially downwards, negative 9.81 meters
per second squared. And probably the
most typical way to interpret this value, as
the acceleration due to gravity near Earth's surface for
an object in free fall. And this is what we're going
to focus on this video. And the reason why I'm
stressing this last part is because we know
of many objects that are near the surface
of the Earth that are not in free fall. For example, I am near the
surface of the Earth right now, and I am not in free fall. What's happening to me right
now is I'm sitting in a chair. And so this is my chair--
draw a little stick drawing on my chair,
and this is me. And let's just
say that the chair is supporting all my weight. So I have-- my legs
are flying in the air. So this is me. And so what's
happening right now? If I were in free fall,
I would be accelerating towards the center of
the Earth at 9.81 meters per second squared. But what's happening is, all
of the force due to gravity is being completely
offset by the normal force from the surface of the
chair onto my pants, and so this is normal force. And now I'll make
them both as vectors. So the net force in my
situation-- the net force is equal to 0, especially
in this vertical direction. And because the net
force is equal to 0, I am not accelerating towards
the center of the Earth. I am not in free fall. And because this 9.81
meters per second squared still seems relevant
to my situation-- I'll talk about
that in a second. But I'm not an
object in free fall. Another way to interpret this
is not as the acceleration due to gravity near
Earth's surface for an object in free
fall, although it is that-- a maybe more
general way to interpret this is the gravitational-- or
Earth's gravitational field. Or it's really the average
acceleration, or the average, because it actually
changes slightly throughout the
surface of the Earth. But another way to view this, as
the average gravitational field at Earth's surface. Let me write it
that way in pink. So the average gravitational
field-- and we'll talk about what a field
means in the physics context in a second-- the
average gravitational field at Earth's surface. And this is a little bit
more of an abstract thing-- we'll talk about
that in a second-- but it does help
us think about how g is related to
this scenario where I am not an object in free fall. A field, when you think of
it in the physics context-- slightly more
abstract notion when you start thinking about it
in the mathematics context-- but in the physics
context, a field is just something that
associates a quantity with every point in space. So this is just a quantity
with every point in space. And it can actually be a
scalar quantity, in which case we call it a scalar
field, and in which case it would just be a value. Or it could be a
vector quantity, which would be a magnitude
and a direction associated with every point in space. In which case you are
dealing with a vector field. And the reason why
this is called a field is, because at near
Earth's surface, if you give me a mass--
so for example-- actually, I don't know what my
mass is in kilograms. But if you're near
Earth's surface and you give me a
mass-- so let's say that mass right over
there is 10 kilograms-- you can use g to figure out
the actual force of gravity on that object at
that point in space. So for example, if this
has a mass of 10 kilograms, then we know-- and
this right over here is the surface of the Earth, so
that's the center of the Earth. So it actually associates
a vector quantity whose magnitude,--
so its direction is towards the center of the
Earth, and the magnitude of this vector quantity is
going to be the mass times g. And you could take--
since we're already specifying the
direction, we could say 9.81 meters
per second squared towards the center of the Earth. And so in this
situation, it would be 10 kilograms times 9.81
meters per second squared. Which is 98.1. And even this I've
rounded a little bit, so it's actually
approximate number. 98.1 kilogram meters
per second squared which is the unit of
force, or 98.1 newtons. And this thing might
not be in free fall, so this is why g
is relevant even in a situation where the
object isn't in free fall. g has given us the
force per unit mass-- the force per mass of
gravity on an object near the surface of the Earth. Another way to think
about it-- so this is the average gravitational
field, and what it's giving is force per mass. So you give me a mass
near Earth's surface-- whether it's an
object in free fall or not-- you multiply
that mass times g, because it's giving
you force per mass, and it will give you the
force of gravity acting on that object near the
surface of the Earth, whether or not
it's in free fall. So I just want to make
this little distinction, because although g
tends to be referred to this way right over here. Sometimes you might
encounter a stickler who says oh no, no, no, no,
no but g is relevant even when an object is not in free fall. You obviously can't say
that my acceleration when I'm sitting in my chair is
9.81 meters per second squared towards the center of the Earth. I am not accelerating towards
the center of the Earth. And so they'll say,
oh no, no, no, no, you can't just call
this acceleration. It is true, it is
the acceleration when an object is in free fall
near the surface of the Earth-- if you don't have really air
resistance, if the net force really is the force of
gravity-- then this really would be the object's
acceleration. But it becomes
relevant, and we know most objects that we know
of aren't in free fall. Obviously, an object in free
fall doesn't stay in free fall for long. It eventually hits something. But we know that
now g is actually relevant to all objects. It tells us the force
per mass And it's tempting to call it
always acceleration-- because the units
are acceleration-- but even when you
talk about in terms of the gravitational field,
it's still the same quantity. It still has the exact same
units, the same magnitude, and the same
direction-- it's just a different way of viewing it. Here, acceleration for
an object in free fall. Here, something to
multiply by mass to figure out the
force due to gravity.