### Video Transcript

In this video, we’re going to learn
about gravitational potential energy in radial gravitational fields. We’ll see the form of this
potential energy both at large and at small distances. To get started, imagine that you’re
making plans to visit the red planet, Mars. As part of your planning, you wanna
figure out how much energy it will take your rocket ship to make it to Mars for the
purpose of figuring out how much fuel you’ll need.

To help with your planning, it will
be good to know about gravitational potential energy in radial gravitational
fields. We’ve seen that whenever we have a
mass of object, that object creates a field around itself. We can call this the gravitational
field that surrounds the object. And as the mass of this object gets
bigger, the strength of its field increases too. Even though the field lines of this
field have only been drawn with a certain length, we know that in fact they extend
theoretically to infinity; that is, there is no finite limit to the reach of this
mass’s gravitational field.

It’s this field, the gravitational
field, that mediates the gravitational force between the object creating the field
and another massive object. If the distance 𝑟 is the distance
from the center of mass of one mass to the other, then we’ve seen by Newton’s law of
universal gravitation that the gravitational force between these two masses is given
by the product of their masses divided by the distance between their centers squared
multiplied by the universal gravitational constant big 𝐺.

This expression is also known
simply as the law of gravity. Looking at this law, one thing to
notice about it is that it has nothing to do with the direction our smaller, we
could call it test, mass is from the larger mass capital 𝑀 that’s creating the
field we’re talking of. In other words, if the mass 𝑚 was
over here or over here, as long as the same distance 𝑟 separated its center from
the center of the mass capital 𝑀, then the gravitational force between these two
masses will be the same.

This means that we’re speaking of a
radially oriented gravitational field. The only factor that changes the
strength of the field is our radius or linear distance from its center. Direction of approach doesn’t make
a difference. All that matters is our distance 𝑟
from its centre. Here’s a question: we’ve said that
these gravitational field lines extend out indefinitely. Well what if we took our small mass
lower case 𝑚 and moved it out to infinity, infinitely far away from capital 𝑀, the
larger mass?

If we did that, then in calculating
the gravitational force between these two masses, we would replace the symbol 𝑟
with infinity. And infinity squared divided into
some finite number would be zero. That’s an important result because
it means that if our smaller test mass was infinitely far away from the mass
creating the field, then there will be no attraction between those masses. Without any force of attraction,
our smaller test mass 𝑚 would stay motionless. It wouldn’t move towards the larger
mass at all; that is, when our smaller test mass is infinitely far away from the
mass that creates the radial field that could draw it in, it has zero gravitational
potential energy. It has no likelihood of starting to
move.

Often we symbolize gravitational
potential energy with the symbol capital 𝑈. And we’ll use that symbol from here
on. So far we’ve seen that if we put
our mass at infinity, then the gravitational potential energy of that mass is
zero. What if, after starting with our
test mass at infinity, we move it in towards the larger mass capital 𝑀 until the
centers of the two are separated by our original distance 𝑟? What, we wonder, is the change in
gravitational potential energy from moving the test mass over this distance?

At this point, there is a
definition that can help us: the change in an object’s gravitational potential
energy is equal to the work done to bring that object from infinity, that is
infinitely far away, to a distance 𝑟 from the mass that creates the field. Just to be clear, any and all
masses create their own gravitational field. So when we say the mass creating
the field, we’re speaking of the one who’s primarily drawing in our smaller test
mass. This definition tells us then that
the mass’s change in gravitational potential energy is equal to the work we need to
do to move it from infinity to its final location a distance 𝑟 from the mass
capital 𝑀.

We recall that work is equal to
force times distance. And in this case, the force we’re
speaking out of is the gravitational force 𝐹 sub 𝐺, defined by Newton’s law of
gravity. And the distance 𝑑 is equal to the
distance our test mass started from the larger mass minus the distance between that
point and its final location; that is. minus what we can call capital 𝑅. Infinity minus this value is equal
simply to lower case 𝑟. And one last thing about this
equation, as we move the test mass closer in further and further, we’re working with
the direction of the gravitational field.

This means that rather than
positive work being required to move the test mass this way, our work will be
negative; that is to say, less than zero work or negative work is required to move
the mass this way since we’re going with the direction of the gravitational
field. When we plug in for the general
equation of Newton’s law of universal gravitation, we see that one factor of 𝑟
cancels out and our final result is that the change in the object’s gravitational
potential energy is equal to negative 𝐺 times the product of the two masses
involved all divided by the final distance between their centers of mass.

This relationship is one we’ll want
to remember when we’re working with large bodies separated by large spaces. So far we’ve seen that Newton’s law
of universal gravitation paired with an assumption of zero gravitational potential
energy for a test mass at infinity leads to this relationship for the change in an
object’s gravitational potential energy. Incidentally, this is the integral
with respect to distance of 𝐹 sub 𝐺, the gravitational force. When we look at this expression for
gravitational potential energy though, it may not be the one most familiar to
us.

We may have seen an expression for
gravitational potential energy equal to an object’s mass times the acceleration due
to gravity times its height above some reference level. Is there a connection between this
expression we found and 𝑚 times 𝑔 times ℎ? Well, it turns out there is. Let’s imagine we have some solid
object. We’ll give it a mass value 𝑚, and
we’ll set it on a table top. Then imagine that we lift this mass
up a height ℎ above its original elevation. Let’s consider what the change in
that object’s gravitational potential energy would be using the equation be derived
from the law of gravity.

We can say that the change of mass
𝑚’s gravitational potential energy is equal to its final GPE minus its initial
GPE. Using a relationship for
gravitational potential energy, we can say that the final GPE of this object is
equal to capital 𝐺 times the mass of the Earth times the mass of the object all
divided by the radius of the Earth plus ℎ, the height it’s been elevated. Then we subtract from that the
gravitational potential energy of the mass 𝑚 when it started out; that is, a
distance of the radius of the Earth from the Earth’s center.

The only difference between these
two terms then is the fact that we’ve raised the object an extra height ℎ in the
first term. If we now take this denominator
term with our height ℎ in it, if we take that term by itself, we can factor out a
factor of the radius of the Earth and rewrite the denominator this way. Then taking this rewritten version
of our denominator, we apply a theorem known as Taylor’s theorem, which is a
mathematical theorem for expressing a function. This theorem tells us that we can
rewrite our original expression as one over 𝑟 sub 𝐸, the radius of the Earth,
multiplied by the quantity one minus ℎ over 𝑟 sub 𝐸.

Our next step is to take this
result we found from Taylor’s theorem and substitute it in to our original
expression. With that substituted in, we’re in
the home stretch for solving for Δ𝑈. The first thing we’ll do to
simplify this is to multiply through this fraction, one over 𝑟 sub 𝐸 times one
minus ℎ over 𝑟 sub 𝐸. That leaves us with this
expression. Now as we look over to the last
term in this equation, we see that the minus minus creates a plus and we see that
this last term will cancel out with the first term in our expression because they’re
opposites one of another.

This means Δ𝑈 reduces to capital
𝐺, the gravitational constant, times the mass of the Earth times the mass of our
test object multiplied by its height differential ℎ all divided by the radius of the
Earth squared. But there’s one last step. Remember that 𝐺 times the mass of
the Earth all divided by the radius of the Earth squared has a symbol that
represents it; it’s the acceleration due to gravity represented as a lowercase
𝑔. This means that for an object near
Earth’s surface raised a height ℎ above its initial elevation, the change in its
gravitational potential energy is indeed 𝑚 times 𝑔 times ℎ.

We can think of this gravitational
potential energy form as the form useful when we’re near a planetary body. And the original form we found as a
form it’s useful in general whether or not we’re close to another large mass. Let’s get some practice now with
gravitational potential energy through an example.

What is the gravitational potential
energy between two spheres, each of mass 9.5 kilograms, separated by a
center-to-center distance of 20.0 centimeters? We can call this gravitational
potential energy between the two spheres capital 𝑈.

And to start on our solution, we’ll
recall the mathematical forms of gravitational potential energy. When it comes to gravitational
potential energy, there are two relationships we can recall. The first relationship is the
general equation, that gravitational potential energy between two masses 𝑚 one and
𝑚 two equals the product of their masses times big 𝐺 all divided by the distance
between their centers. And this potential energy is the
negative of this fraction. There’s also a special formulation
of GPE, which is when a relatively smaller mass lowercase 𝑚 is near a planetary
body whose acceleration due to gravity, we’ve called it lowercase 𝑔, is known.

In that case, the GPE of the
smaller mass is equal to 𝑚 times 𝑔 times the height it moves through. In our case, because our two masses
have the same value, we’ll use the more general expression for gravitational
potential energy. If each mass has a value we
symbolize with lowercase 𝑚 and the center-to-center distance between the masses we
represent with 𝑟, and if we let the universal gravitational constant equal exactly
6.67 times 10 to the negative 11 to cubic meters per kilogram second squared, then
given the values of 9.5 kilograms for 𝑚 and 20.0 centimeters for 𝑟, we can plug
all these values into our equation being careful to convert our distance from units
of centimeters to units of meters.

Entering this expression on our
calculator, to two significant figures, we find a result of negative 3.0 times 10 to
the negative eighth joules. That’s the gravitational potential
energy between these two spheres.

Now let’s summarize what we’ve
learned about gravitational potential energy in radial gravitational fields. In this video, we’ve seen that in a
radial gravity field, the force of gravity between two masses capital 𝑀 and
lowercase 𝑚 equals their product times the universal gravitational constant divided
by the distance between their centers squared.

We’ve also seen that gravitational
potential energy, often symbolized using the letter capital 𝑈, equals the work done
to bring a mass from infinity to a distance 𝑟 from another mass. And finally we’ve seen that as an
equation, gravitational potential energy in general is equal to negative capital 𝐺
multiplied by the product of the two masses involved divided by the distance between
their centers of mass. We’ve also seen that in a special
case of GPE when a relatively small mass is near the surface of a large planet, the
gravitational potential energy of that smaller mass is equal to the mass times the
strength of the acceleration due to gravity near that planet multiplied by the
height through which that mass has moved.