### Video Transcript

In this video, our topic is surface
gravity. This term refers to an acceleration
due to gravity that all objects on or near the surface of a large spherical object
experience. It’s surface gravity, for example,
that pulls us as well as the objects around us toward the center of the Earth.

We can start to understand where
surface gravity comes from by considering two other laws of physics. Both coming from Isaac Newton, one
of these is the law of gravitation. This says that the gravitational
force between two masses, one referred to as capital 𝑀 and the other as lowercase
𝑚, is equal to the product of these masses times the universal gravitational
constant, big 𝐺, divided by the distance between the centers of mass of our masses
squared. This is sometimes called the
universal law of gravitation because these two masses, capital and lowercase 𝑚,
could be anything. And, regardless of what they
represent, this equation tells us the gravitational force between the masses.

The other of Newton’s laws we’ll
consider is his second law of motion. This law tells us the net force on
an object is equal to the object’s mass multiplied by its acceleration. One of the important conditions on
Newton’s second law of motion is that this force here is the net force on our mass
𝑚. Knowing that, let’s imagine a
scenario where we have some mass 𝑚. And the net force, indeed the only
force acting on that object, is the gravitational force it experiences between
itself and some other mass, capital 𝑀. In other words, let’s imagine that
these two masses, lowercase 𝑚 and capital 𝑀, are the only masses in the
universe.

In that case, the only force acting
on this smaller mass is the gravitational force between it and the larger one. We can, therefore, say that the net
force acting on our mass, lowercase 𝑚, is equal to the gravitational force on
it. And that means that the right side
of our law of gravitation equation is equal to the right side of our second law
equation. And as we look at this new
equation, notice what’s common to both sides. It’s the mass lowercase 𝑚. This means that if we divide both
sides of the equation by that mass, it will cancel out completely.

And the equation we’re left with
says that the universal gravitational constant times the mass of the larger object
capital 𝑀 divided by the distance between the centers of mass of the two masses
squared is equal to the acceleration experienced by the smaller mass, lowercase
𝑚. And notice something interesting
about this equation. It doesn’t depend on the value of
the smaller mass; that is, lowercase 𝑚 is nowhere in the equation. Now, that doesn’t mean that the
position of this object with a mass lowercase 𝑚 is unimportant. Because we see in the denominator
on the left-hand side this value 𝑟. Recall that that’s the distance
between our two masses.

So, the acceleration of this
object, which has a mass lowercase 𝑚, doesn’t depend on that mass, but it does
depend on the object’s position with respect to the larger mass, capital 𝑀. Now just as a reminder, this
gravitational force we’re talking about is a mutual force between these two
masses. The magnitude of the force acting
on this object is the same as the force magnitude acting on this one. If we had written the second law of
motion a little bit differently, in this case, to highlight the larger mass capital
𝑀. Then when we combine this force
equation with the law of gravitation, we would have found that it’s the larger of
the two masses, the one represented by capital 𝑀, that cancels out in this
equation. And then, the acceleration 𝑎, we
will be referring to, is the acceleration experienced by that larger mass, capital
𝑀.

So just for completeness, we could
come over here and specify that this acceleration is that experienced by the smaller
mass, lowercase 𝑚. The point is, due to the mutual
force of gravity between these two objects, each one will tend to accelerate toward
the other. And the acceleration on these
objects is not the same, even though the force magnitude acting on each one is. Each object’s acceleration depends
on the other object’s mass and the distance between the centers of mass of the two
masses.

Now, this relationship here, which
is true in general for any two masses, is especially useful to us when our larger
mass — we’ve called it capital 𝑀 — refers to the mass of a very large spherical
body, like a planet or a moon. And our smaller object, with mass
lowercase 𝑚, is on or near the surface of our larger object. A great example of this is when
there’s a person, say it’s us, standing on the surface of the Earth. If we say that our mass is
represented by lowercase 𝑚 and the mass of the Earth by uppercase 𝑀, then this
equation here describes the acceleration we experience standing on Earth’s
surface. In other words, this is Earth’s
surface gravity.

And notice that, just like we said
before, this acceleration doesn’t depend on our mass. So, then, this acceleration 𝑎 is
actually the acceleration experienced by any object on or near Earth’s surface. Now, we need to be a bit careful
when we say this. Because recall that this 𝑟 here
represents the distance between the center of mass of the Earth, where all of its
mass is effectively concentrated and our position on Earth’s surface.

So technically, if we were to do
something, like climb up on a stepladder, this distance represented by 𝑟 would
increase, and that would cause the acceleration we experience, the surface gravity
of the Earth, to decrease. It turns out, though, that for
small elevation changes like this, this equation is still so nearly correct that we
consider it essentially accurate. In other words, the difference in
distance between the radius of the Earth and the radius of the Earth plus this small
distance we’ve climbed up is so very small that it doesn’t really have an effect on
the acceleration we experience.

And so, we can say that this
acceleration equals the acceleration due to gravity experienced by any object that’s
on or near Earth’s surface. This is why all objects near that
surface fall at the same rate regardless of their mass. By the way, for the specific case
where we are talking about Earth’s gravitational attraction, this acceleration, what
we’ve called 𝑎 sub 𝑀, is given a specific symbol. It’s represented by a lowercase
𝑔. The value of lowercase 𝑔, we can
recall, is approximately 9.8 meters per second squared.

This is what we get if we take the
mass of the Earth, multiply it by the universal gravitational constant, and then
divide that product by the radius of the Earth squared. Try it for yourself and see. Knowing all this about acceleration
due to gravity, let’s get some practice with these ideas through an example.

Two objects, object A with a mass
of five kilograms and object B with a mass of 100 kilograms, are near an even larger
object with a mass of 10 to the 20th kilograms. Object A and object B are at an
equal distance away, 100 kilometers, from the center of mass of the very large
object. What is the magnitude of the
gravitational force experienced by object A due to the very large object? Give your answer to three
significant figures.

Okay, looking over at our sketch,
we see a five-kilogram object. That must be object A. And as well, we see a 100-kilogram
mass. This must be object B. Object A and object B interact
gravitationally with this very large spherical object. Now, this object is represented
here using a rectangle that’s not very much bigger than A and B. But we can imagine that actually
this object is so large it couldn’t fit on the screen at this scale. We have an indication of this
thanks to the very large mass of this object.

So, we could almost think of these
three objects, A and B and our large spherical one, as though our large spherical
object is the Earth, object B, say, is a person, and object A is a book the person
is holding on to. That’s roughly the scale we’re
talking about as we consider these three objects. Our first question related to this
scenario says, what is the magnitude of the gravitational force that object A
experiences due to the very large object?

As we get started figuring this
out, let’s record some of the information we’ve been given. First, we’ve been told that our
large spherical object has a mass, we’ll call it capital 𝑀, of 10 to the 20th
kilograms. And we’re also told that the
distance between the center of mass of this large spherical object, that is, where
all of its mass is effectively concentrated, and our two other objects, object A and
object B, is the same. It’s 100 kilometers. We’ll call that distance 𝑟. And as we see, it’s the same for
object A and object B in relation to our large spherical object. With these values written down,
let’s clear some space and consider again our question.

We want to know the magnitude of
the gravitational force experienced by object A, that’s this object here, due to our
very large spherical object. So, let’s recall the general
equation, called Newton’s law of gravitation, that describes the force of gravity
between two objects that have mass. This equation tells us that that
force is equal to the product of those masses. We’ll call them capital and
lowercase 𝑚, respectively. Multiplied by the universal
gravitational constant, capital 𝐺, divided by the distance between our two masses
squared.

Now, the gravitational constant,
big 𝐺, is approximately equal to 6.67 times 10 to the negative 11th cubic meters
per kilogram second squared. So, if we want to know the
gravitational force on object A caused by the very large object, we can call this
force 𝐹 sub A, then we’ll take this constant and we’ll multiply it by the mass of
our large object. And multiply that by the mass of
object A, we’ll call it 𝑚 sub A, and divide all this by the distance between the
centers of mass of object A and our large spherical object squared.

When we substitute in the values
for all these variables, we’re just about ready to calculate this force magnitude,
except for one thing. Notice that the distance in our
denominator is in units of kilometers. In order to agree with the distance
units in the rest of our expression, we’d like to convert this into meters. We can recall that one kilometer is
equal to 1000 meters and therefore 100 kilometers is equal to 100,000 meters. With that all figured out, when we
calculate this fraction, to three significant figures, we find a result of 3.34
newtons. That’s the magnitude of the
gravitational force experienced by object A due to the very large object.

Now, let’s consider the next
question in this exercise.

What is the acceleration of object
A toward the very large object? Give your answer to three
significant figures.

Okay, so, we’ve calculated the
gravitational force between these objects. And now, we want to calculate the
acceleration object A experiences. We can start doing this by
recalling Newton’s second law of motion. This law has it that the net force
on an object is equal to that object’s mass times its acceleration. Acceleration, in the case of object
A, is just what we want to solve for. And here’s how we can do it. As we saw, Newton’s second law says
that the net force on an object is equal to that object’s mass times its
acceleration.

When we think about the forces
acting on object A, we can see that there are two. One is the gravitational force due
to object B and another is the gravitational force due to the large spherical
object. But here’s the thing. The force between A and B is much,
much, much smaller than the force between A and the large spherical object. This is because the mass of our
large object absolutely dwarfs the mass of object B. We can say then that the
gravitational force on object A due to the very large object is effectively the only
force on object A.

And this means we can say that this
equation is equal to the mass of object A multiplied by the acceleration of object
A. We’ll call it 𝑎 sub A. And note that we got this
expression here from Newton’s second law. But now, as we consider this
equation, notice that the mass of object A appears on both sides, and therefore we
can cancel it out. If we do that, this is the equation
that results. The acceleration of object A is
equal to the universal gravitational constant times the mass of our very large
object divided by the distance between the centers of mass of object A and our very
large object squared.

Now, as we consider this part of
our expression here, we can see that it’s actually equal to this expression if we
drop out the mass of our object, in this case, object A. With that mass gone, see that we
have the gravitational constant, uppercase 𝐺, multiplied by the mass of our large
object divided by 𝑟 squared. Which are identical to the factors
we see here on the right-hand side of this equation. So then, if we calculate all this,
we’ll solve for the acceleration of object A due to the very large object. To three significant figures, it’s
0.667 meters per second squared. That’s the acceleration of object A
toward the very large object.

Now, let’s consider the next part
of our exercise.

What is the magnitude of the
gravitational force experienced by object B due to the very large object? Give your answer to three
significant figures.

Okay, whereas before, we were
considering the gravitational force between object A and the large spherical object,
now we’re considering that force between the large object and object B. Once again, we’ll use Newton’s law
of gravitation to solve for this force. But this time, instead of solving
for the force experienced by object A, we’ll solve for that experienced by object
B. And we’ll say that this object has
a mass 𝑚 sub B. Because both objects A and B are
the same distance away from the center of mass of our large spherical object, we
won’t change 𝑟, the distance between the centers of mass of the two masses in our
equation. Which means that when we go to
calculate this force, 𝐹 sub B, using an expression like this, all we’ll need to do
is substitute the mass of object B in where we used to have the mass of object
A.

We see that this mass is 100
kilograms. And so, now, we have an expression
which, when we solve it, will give us this force experienced by object B due to the
gravitational attraction from the large spherical object. To three significant figures, this
force is 66.7 newtons. Note that this is larger than the
force magnitude experienced by object A. In fact, it’s 20 times larger
because object B has a mass 20 times larger than object A.

Now that we found this out, let’s
consider the last part of our question.

What is the acceleration of object
B toward the very large object? Give your answer to three
significant figures.

To answer this question, once
again, we’ll use Newton’s second law of motion. And just like we did for object A,
we’ll also assume that the gravitational force between objects A and B is negligibly
small compared to that force between object B and the very large object. This means we can say that the mass
of object B multiplied by its acceleration is equal to the gravitational force
experienced by object B due to the very large object. So then, we can go to our equation
for that force, and we can equate it to the mass of object B times the acceleration
of object B. We’ll call it 𝑎 sub B.

And then, like we saw earlier, the
mass of our object, in this case, object B, is common to both sides. And that means that the
acceleration experienced by object B is equal to capital 𝐺 times the mass of our
very large object divided by the distance between B and our very large object
squared. And this expression here is equal
to this expression here if we ignore this mass of our object B. And so, to solve for 𝑎 sub B,
we’ll multiply the universal gravitational constant by the mass of our large object
divided by the distance between that object and B squared.

To three significant figures, this
is 0.667 meters per second squared. Note that this is the same as the
acceleration experienced by object A. And this comes back to the fact
that our equation for object acceleration does not depend on the mass of the object
whose acceleration we’re considering. So, this is how object B and object
A, and any other object, will accelerate toward the very large object.

Let’s recall now some key points
about surface gravity.

Starting off with Newton’s law of
gravitation and second law of motion, we saw that these laws can be combined to
yield an expression for the acceleration one of the masses involved would
experience. We noted that if this is an
acceleration of the mass indicated by a lowercase 𝑚. Then equivalently the other mass,
indicated by a capital 𝑀 in Newton’s law of gravitation, also experiences an
acceleration due in this case to the smaller mass, lowercase 𝑚. Related to this, we saw that the
acceleration an object experiences does not depend on its mass. And lastly, we saw that surface
gravity describes an object’s acceleration when it is on or near a much larger
spherical object.