Video Transcript
In this video, we’re talking about
gravitational potential energy in a radial field. We’re going to learn how to define
gravitational potential energy in such a field mathematically. And we’ll also see how changes in
this potential energy relate to changes in kinetic energy.
As we get started, we can think
about gravitational potential energy not in a radial field but rather in a uniform
gravitational field. Now, a uniform gravitational field
is one where the spacing between the gravitational field lines is constant all
throughout. In other words, we could say that,
for a uniform field, the density of field lines is constant. In such a field, if we were to toss
an object, say a rock with a mass 𝑚, up into the air such that the rock at one
moment in time was a height ℎ above ground level, then we could say that the
gravitational potential energy between the rock and the Earth that attracts it is
equal to the rock’s mass multiplied by the local acceleration due to gravity times
its height above ground level. This relationship, gravitational
potential energy is equal to 𝑚 times 𝑔 times ℎ, is true for a uniform
gravitational field.
But if we were to zoom way out and
look over larger space scales, say we were looking now at the entire planet Earth,
then here we can see that the spacing between our gravitational field lines is no
longer constant. That is, the density of these field
lines increases as we get closer to the surface of the planet. This means we no longer have a
uniform field. And we can’t use this equation to
solve for gravitational potential energy anymore. But then if we had some mass 𝑚 a
distance 𝑟 away from the center of this planet, how would we calculate the
gravitational potential energy now?
We’re going to learn how to do this
with a field that’s no longer uniform, but the name we give to it instead is a
radial field. A radial gravitational field is
created by any object that is spherically symmetric. For example, a sphere of uniform
density makes a field like this. In a radial field, if we trace all
the field lines, we see that they meet at the center of the object creating the
field. In a situation like this, the
gravitational potential energy between a smaller mass — we can call it a test mass,
lowercase 𝑚 — and a relatively larger mass that’s creating this radial field — we
can call that capital 𝑀 — is given by this equation. 𝐺𝑃𝐸 is equal to negative 𝐺, the
universal gravitational constant, times the product of these two masses involved
divided by the distance between their centers of mass.
As we look at this equation, the
first thing that may catch our attention is this negative sign here. According to this equation, the
gravitational potential energy between a test mass and a larger mass creating a
radial field is always negative. This in fact is accurate. And to see why, let’s clear a bit
of space on screen. And then let’s imagine moving our
test mass, lowercase 𝑚, so that it’s at a distance infinitely far away from the
larger mass, capital 𝑀.
If we do this, moving our test mass
an infinite distance away, we can see that, at this location, it no longer
experiences a gravitational force created by the larger mass, capital 𝑀. This is because the gravitational
force of attraction between two masses is proportional to one over the distance
between their centers squared. With our two masses separated by an
infinite distance, 𝑟 becomes ∞. And so one over ∞ squared is
zero. So there’s no gravitational force
acting on our test mass. And therefore, there is no
gravitational potential energy shared between this system of two masses. We can see that this is so because,
once again, our denominator is infinite, making the fraction equal to zero. So with our masses infinitely
separated, the gravitational potential energy of the system is zero.
But now let’s imagine this. Let’s say that we give our mass a
tiny little nudge in the direction of the planet. By doing this, we can say that
we’re moving the test mass into the gravitational field of the planet. Once that happens, we know the test
mass will be attracted gravitationally to the planet. And it will begin to move toward it
with higher and higher speed. This is so because our test mass is
essentially in free fall toward the planet.
As our test mass picks up speed, we
can say that it’s also gaining in kinetic energy. Recall that an object’s kinetic
energy is equal to one-half its mass times its speed squared. And so our test mass, which started
out with a speed of zero, that is, it was at rest, infinitely far away from the
planet, now that it’s moving at some nonzero speed, here we could say, has picked up
kinetic energy. And that change in kinetic energy
the mass has experienced is equal to one-half its mass multiplied by its speed at
this moment in time — we can call that speed 𝑣 — squared.
At this point, we can realize that
the system that we’re working with here consists of two objects. One is the smaller mass, our test
mass lowercase 𝑚, and the other is this planet, with its mass capital 𝑀. This is our system, and we can see
that no energy is coming in from outside that system and none is leaving the
system. Moreover, the energy of our larger
mass, capital 𝑀, hasn’t changed at all as the test mass moves closer to it, which
means that all the energy exchange in our system is confined to our test mass as it
moves from here, its original position, to now here. But then, what was the overall
energy of our test mass in its original location?
At that point, the gravitational
potential energy it shared in was zero. And it was also stationary, which
means it had no kinetic energy. So let’s say this. The initial energy — we’ll
represent it 𝐸 sub i, of our system, where our system consists of these two masses,
is equal to zero. But then by energy conservation,
this is also equal to the final energy of our system, that is, the energy when our
test mass is right here. That final energy is equal to the
change in kinetic energy our system underwent plus the change in gravitational
potential energy. So 𝐸 sub f is equal to one-half
𝑚𝑣 squared plus 𝛥𝐺𝑃𝐸. And according to the line above,
this must be equal to zero. That’s what conservation of energy
requires.
So if a positive value one-half
𝑚𝑣 squared plus 𝛥𝐺𝑃𝐸 is equal to zero, that tells us that this term in our
equation must be less than zero. It must be negative. That’s the only way that energy can
be conserved in our system. And that is why we have this
negative sign here in our equation.
Another interesting thing about
this equation for gravitational potential energy in a radial field is that once we
have the two masses we’re working with, everything in the equation is a fixed value
or a constant except for this distance 𝑟. We could say that’s the only
variable leading to a change in 𝐺𝑃𝐸. So here’s what that means
practically. Say we sketch in a couple of
circular paths that are concentric to the surface of our planet. The fact that this equation here
varies only with the radius 𝑟 means that if our test mass was located, say, here,
then the gravitational potential energy of our system would be the same as if the
mass were here or here or anywhere else along this outer ring. And the same is true for our inner
ring or really any ring that we could draw concentric to the planet’s surface. Once the masses in our system are
specified, it’s only the radial distance between the centers of these masses that
affects the gravitational potential energy of the system.
Knowing all this, let’s now get a
bit of practice with these ideas through an example.
A planet has a mass of 8.08 times
10 to the 24th kilograms. A small asteroid with a mass of
1,420 kilograms is in space near the planet. The asteroid is 23,900 kilometers
away from the planet’s center of mass. What is the magnitude of the
gravitational potential energy of the asteroid-and-planet system? Give your answer to three
significant figures.
Drawing a sketch of this scenario,
let’s say that this here is our planet. We’ll say it has a mass 𝑚 sub p,
with that mass value given to us. And let’s say further that out here
is our asteroid, and we’ll label the asteroid’s mass 𝑚 sub a. The centers of these two masses are
separated by a distance we’ll call 𝑟. And that distance is given to us as
23,900 kilometers. Knowing all this, we want to
calculate the magnitude of the gravitational potential energy of the
asteroid-and-planet system.
At this point, we must be very
careful because we may think that the gravitational potential energy between these
two objects is equal to the mass of the asteroid times 𝑔 times its height above the
surface of the planet. But this representation of GPE is
only true when the gravitational field is uniform. In the case of our planet and
asteroid though, if we were to sketch in some of the gravitational field lines
around the planet, we would see that the field is not uniform, but rather it’s
radial. That is, the field lines move out
from the center of the planet, like spokes on a bicycle wheel.
In a situation like this, we can no
longer use this equation to calculate gravitational potential energy. Instead, we need to recall the form
of this equation that applies to this kind of field, a radial field. When a field is structured this
way, then the gravitational potential energy between two masses, big 𝑀 and little
𝑚, where big 𝑀 is the mass creating the field, is equal to negative the product of
those masses times the universal gravitational constant divided by the distance
between the centers of mass of the masses.
To get started using this equation,
we can recall that the universal gravitational constant, to three significant
figures, is equal to this value here. And so applied to our scenario, the
gravitational potential energy of our asteroid-and-planet system equals negative big
𝐺 times the mass of the planet — this is the mass creating the radial field — times
the mass of the asteroid divided by the distance between their centers.
One important point is that our
question wants us to solve for the magnitude of the gravitational potential energy
of this system. To calculate that, we would take
the absolute value of a negative number. And so to simplify things, we could
drop both the absolute value bars and the negative sign.
We’re now ready to plug in for
these values and solve for the magnitude of 𝐺𝑃𝐸. With these values substituted in,
notice that our radial distance has units of kilometers, whereas in our numerator,
we have distance in units of meters. We’ll want these units to agree
before we calculate the magnitude of 𝐺𝑃𝐸. So recalling that 1,000 meters is
equal to one kilometer, we can recognize that 23,900 kilometers is equal to that
number with three zeros added on the end meters, in other words, 23,900,000
meters.
Looking at the units in this
expression, notice now what happens. One factor of meters cancels from
numerator and denominator. And then the kilogram unit in both
of our masses cancels out with one over kilogram squared here. We’re left with newtons times
meters, and a newton times a meter is one joule, the SI base unit of energy. When we compute this fraction, to
three significant figures, we find a result of 3.20 times 10 to the 10th joules. But then, if we recall that 10 to
the ninth or one billion joules is equal to what’s called a gigajoule, then we can
equivalently express our answer as 32.0 gigajoules. This is the magnitude of the
gravitational potential energy of the asteroid-and-planet system.
Let’s look now at a second example
exercise.
A spacecraft heading toward the
Moon to deploy a rover on the surface has a mass of 870 kilograms. The magnitude of the gravitational
potential energy of the spacecraft-and-Moon system is 427 megajoules. How far away from the Moon’s center
of mass is the spacecraft? Use a value of 7.35 times 10 to the
22nd kilograms for the mass of the Moon and 6.67 times 10 to the negative 11th cubic
meters per kilogram second squared for the universal gravitational constant. Give your answer to three
significant figures.
If we say that this pink circle is
the Moon and that this is our spacecraft heading toward it, our problem statement
tells us the mass of the spacecraft, let’s call that 𝑚 sub s; the mass of the moon,
we’ll call that 𝑚 sub m. And it also tells us the
gravitational potential energy of this system of two masses, the spacecraft and the
Moon. We’ll call that magnitude
𝐺𝑃𝐸. Knowing all this, what we want to
solve for is the distance between the spacecraft and the center of the Moon. And we can call that distance
𝑟.
To get started, we can recall the
equation for gravitational potential energy in a radial gravitational field. This is the kind of field created
by a spherically symmetric object, such as the Moon. For this kind of field, the
gravitational potential energy shared between two masses, the larger one creating
the field and the smaller one experiencing it, is equal to negative 𝐺, the
universal gravitational constant, times those masses divided by the distance between
their centers. In our scenario, it’s not 𝐺𝑃𝐸 we
want to solve for though, but rather the distance 𝑟.
If we multiply both sides of this
equation by the fraction 𝑟 divided by 𝐺𝑃𝐸, then on the left-hand side, that
gravitational potential energy cancels out. And on the right, the distance 𝑟
cancels. And that leaves us with this
equation. 𝑟 is equal to negative big 𝐺
times big 𝑀 times little 𝑚 divided by 𝐺𝑃𝐸.
One important thing to remember
here is that this term, the gravitational potential energy in this equation, is
negative. And so if we were to plug in actual
values in this equation, we would have a negative multiplied by a negative, which is
a positive, giving us then an overall positive value for 𝑟. We can apply this equation to our
scenario like this. The distance between the center of
the Moon and the spacecraft is equal to the universal gravitational constant times
the mass of the Moon times the mass of the spacecraft all divided by the magnitude
of the gravitational potential energy of this two-mass system. And notice that we’re given all of
the values that appear on the right-hand side of this equation.
Our next step then is to substitute
them in. So here we have the value for the
universal gravitational constant. Here’s the mass of the Moon, here’s
the mass of our spacecraft, and here’s the magnitude of the gravitational potential
energy between these masses. As a last step, before we calculate
𝑟, we’ll want to convert this gravitational potential energy from units of
megajoules into units of joules. That way, all the units in this
expression will be on the same footing. Now, one megajoule is equal to 10
to the sixth or a million joules. So we can rewrite our denominator
as 427 times 10 to the sixth joules. This number isn’t in scientific
notation, but it is the correct number of joules of energy. When we calculate 𝑟, to three
significant figures, we get a result of 9.99 times 10 to the sixth meters. We can equivalently express this as
9,990 kilometers. That’s the distance from the center
of mass of the Moon to the spacecraft.
Let’s summarize now what we’ve
learned about gravitational potential energy in a radial field. In this lesson, we first recalled
that, in a uniform gravitational field, the gravitational potential energy shared by
some object is equal to that object’s mass times the local acceleration due to
gravity times its height above some larger surface. For a radial field, however, such
as one created by a large spherically symmetric object, like a planet, the
gravitational potential energy between that planet and some smaller mass, lowercase
𝑚, is given by negative the universal gravitational constant 𝐺 times the product
of the two masses divided by the distance between their centers of mass. And lastly, we saw that, for a test
mass moving about in a radial gravitational field, the magnitude of its change in
kinetic energy is equal to the magnitude of the change in gravitational potential
energy. This is a summary of gravitational
potential energy in a radial field.
