Video Transcript
In this video, our topic is
mechanical energy conservation in orbits. A foreign object that’s in orbit
around some other body, we’re going to learn how to calculate its mechanical energy
and also see how that energy changes over time.
The first thing we can consider
here is that an object’s mechanical energy, we’ll refer to it as capital ME, is
equal to the sum of kinetic and potential energy. We know that kinetic energy is
energy due to motion. And while there are many different
kinds of potential energy, in this case gravity will always be its source. That is, for an object in orbit,
its mechanical energy is equal to the sum of kinetic and gravitational potential
energy.
If we consider what energy due to
motion is, we can recall that mathematically, it’s equal to one-half an object’s
mass times its speed squared. So we can imagine that our orbiting
body has some mass, we’ll call it 𝑚, as well as some speed 𝑣 so that there is
kinetic energy for this body in motion. And what’s more, we can see that
this body in orbit is specifically moving in a circle around the center of the mass
that it orbits.
Whenever a mass is moving in a
circular path, we know it’s subject to a force called the centripetal force and that
this force is always directed towards the center of the circle that the mass moves
around. Now, a center-seeking or
centripetal force always has some physical mechanism causing it. In this case, that force is the
force of gravity, the attractional force between the mass being orbited and the mass
in orbit. Mathematically, then, we can write
that the centripetal force, 𝐹 sub 𝑐, this center-seeking force acting on our mass
in orbit, is equal to the gravitational force acting on that mass, we’ll call it 𝐹
sub 𝑔. In other words, the physical cause
of a center-seeking force on this mass is the force of gravity.
We can recall that a foreign object
in circular motion around an arc that has a radius we’ll call lowercase 𝑟, the
centripetal force on that object equals its mass times its speed squared divided by
𝑟. Looking back to our sketch, we can
say that this distance here is 𝑟. Once we know 𝑟 and 𝑚 and 𝑣, then
we’re able to calculate the center-seeking force acting on our mass. As we’ve said, though, this is
equal to the gravitational force acting on the mass. And that, written as an equation,
is equal to the universal gravitational constant multiplied by the larger mass being
orbited, we’ll call that capital 𝑀, and the smaller mass, lowercase 𝑚, all divided
by the distance between the centers of mass squared. And in the case of our circular
orbit, as we’ve seen, we can think of this distance as the radius of the circle that
the smaller mass moves in.
So the centripetal force acting on
our object is equal to the gravitational force on it. And looking at this equation, we
see that a couple of terms cancel. First, and interestingly, all this
is independent of the smaller mass 𝑚. That mass disappears completely
from this equation. We also see that a factor of 𝑟
cancels out in the denominator of these two fractions. And so we find this: the speed of
our smaller mass squared is equal to the universal gravitational constant times the
mass of the object being orbited, this is the larger mass, all divided by the
distance between the centers of mass of that larger and our smaller mass.
Now that we have this expression
for 𝑣 squared, let’s go back to our equation for kinetic energy, which we see
involves a 𝑣-squared factor.
This means that we can take this
expression for 𝑣 squared and substitute it in for 𝑣 squared here in our equation
for kinetic energy and see that when we do this, we arrive at an equation for the
kinetic energy of a mass, lowercase 𝑚, that’s in circular orbit around a larger
mass, capital 𝑀. It’s important to keep in mind this
limit, that this equation applies only to circular orbits. Remember that we derived it using
an assumption that our object was moving in a circle and, therefore, experienced a
centripetal force.
For these types of orbits, though,
we now have an expression for kinetic energy. And now, let’s consider the
potential energy, specifically gravitational potential energy. We can refer to this energy using
GPE and recall that it’s equal to negative the universal gravitational constant
times the product of the two masses involved divided by the distance between their
centers. So then if we want to write an
expression for the mechanical energy, that is, kinetic plus potential, of a mass in
circular orbit, then that’s equal to kinetic energy plus gravitational potential
energy or this term here minus this term here.
And we can see there’s a lot in
common between these two terms, capital 𝐺, capital 𝑀, the larger mass that’s being
orbited, lowercase 𝑚, the smaller mass that’s doing the orbiting, and one over
𝑟. We can factor all of these values
out from both terms, which gives us 𝐺 times big 𝑀 times little 𝑚 divided by 𝑟
all multiplied by one-half minus one. But then one-half minus one is
negative one-half. And now, we have a simplified
expression for the overall mechanical energy of an object in circular orbit.
And notice that this value doesn’t
depend on the speed of the object. That doesn’t appear anywhere in the
equation. The reason for that, as we saw a
moment ago, is that the object’s speed squared can be expressed in terms of other
parameters. We can say then that in the special
case of a circular orbit, there’s a particular relation between an object’s kinetic
and its gravitational potential energy. We can see that by dividing one by
the other. When we do this, many of the
factors cancel, big 𝐺, big 𝑀, little 𝑚, and 𝑟, the radius. And we end up simply with negative
one-half. This relationship here is the
specific one that KE and GPE will always share for a circular orbit.
At this point, though, we might
recall that circular orbits aren’t the only ones there are. It’s also possible for a mass to
orbit another in the shape of what’s called an ellipse. This is like a circle that’s been
smushed in one direction. We can see right away that not
everything we’ve said so far will also apply to an elliptical orbit. Mostly, that’s because our object
is no longer moving in a circular arc and, therefore, is not experiencing a
centripetal force.
So for an elliptical orbit, we can
no longer say that the orbiting object’s kinetic energy is given by this
relationship. But that doesn’t mean that
everything is different for an elliptical orbit. For example, this equation here
still applies for the gravitational potential energy between our orbiting mass and
the mass being orbited. And it’s still true that the
mechanical energy of our object is the sum of its kinetic and gravitational
potential energies.
So to write this object’s
mechanical energy mathematically, once again, that’s kinetic energy plus
gravitational potential energy. But now, we’ll write out kinetic
energy as simply one-half the object’s mass times its speed squared. And expressed this way, we can see
that there’s only one factor common to both of these terms. It’s the orbiting object’s mass,
lowercase 𝑚. And so we can write the mechanical
energy for an object in elliptical orbit this way. And here we see the object’s speed
doesn’t cancel out of the expression like it did for our object in circular
orbit.
Despite the differences between
these two equations, one thing that’s true for both is that the orbiting object’s
mechanical energy is constant all throughout its orbit. That is, the mechanical energy of
an object in orbit, whether that orbit is circular or elliptical, stays the
same. Note that we’re not saying that all
orbiting objects have the same mechanical energy, only that for a given object in
orbit, that amount is maintained all through its orbit. This fact may seem more clearly
true in the case of an object in circular orbit. In this case, everything on the
right-hand side of the expression is either a constant or a fixed value.
But what about for an elliptical
orbit, where that mechanical energy is given by this equation? After all, we can see that 𝑟, the
distance between the center of mass of our orbiting body and the body being orbited,
clearly changes throughout the orbit. When our mass, for example, is out
here, that value is relatively large, whereas when our mass is closer in, that value
shrinks. Despite this, it is true that the
mechanical energy of this orbiting body is constant. And that’s because as the distance
𝑟 varies, so does the speed of our orbiting object 𝑣.
It turns out that the bigger 𝑟 is,
that is, the farther our orbiting body is away from the body that it’s orbiting, the
smaller that object’s speed 𝑣 gets. This means, for example, that when
our orbiting mass is way out here, its speed 𝑣 is relatively small, whereas when 𝑟
is relatively small, say, when our mass is here, it has a greater speed. The speed of the object, 𝑣, and
its distance from the center of mass of the object it’s orbiting, 𝑟, balance one
another out so that the difference between these two terms is always the same. And this leads to the mechanical
energy of our orbiting body being constant all throughout its motion.
Knowing all this about how
mechanical energy is conserved for objects in circular or elliptical orbit, let’s
get some practice now with these ideas through an example.
A spacecraft that has been launched
into space is moving along a circular path around Earth. The radius of the orbit is 7,000
kilometers, and the spacecraft has a mass of 2,200 kilograms. What is the kinetic energy of the
spacecraft? Use a value of 5.97 times 10 to the
24th kilograms for the mass of Earth 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.
Okay, so here we have the Earth,
and we’re told that there is a spacecraft, we’ll say that craft is right here, in
circular orbit around it. We’re told that the radius of the
orbit, which is the distance from the center of the spacecraft to the center of the
Earth, is 7,000 kilometers. And we’ll call that distance
𝑟. We’re also told that this orbiting
spacecraft has a mass of 2,200 kilograms. And we’ll call that mass lowercase
𝑚. We want to know what is the kinetic
energy of the spacecraft. And we’re given values to use for
the mass of the Earth and the universal gravitational constant.
Now, if we recall that, in general,
an object’s kinetic energy is given by one-half its mass times its speed squared, we
might be confused about how to solve for the spacecraft’s kinetic energy because we
don’t know its speed. However, we do know that this
spacecraft is moving in a circular orbit, and any object moving in a circular arc is
subject to what is called a centripetal or center-seeking force. We can call it 𝐹 sub 𝑐.
This centripetal force acting on an
object as it moves in a circle of radius 𝑟 is given by the object’s mass times its
speed squared divided by that radius. Now, any center-seeking force needs
to have some physical mechanism that causes it. In this case, our spacecraft is
moving in a circle around the Earth thanks to the gravitational attraction between
the two masses. In general, the force of gravity
between two masses, we’ll call them uppercase and lowercase 𝑚, is equal to the
product of those masses times the universal gravitational constant divided by the
distance between their centers of mass squared.
What we’re saying is that in the
case of our satellite, the gravitational force experienced by it is equal to the
center-seeking force that makes it move in a circle. In other words, the centripetal
force 𝑚𝑣 squared over 𝑟 is equal to the gravitational force, where we’re saying
that lowercase 𝑚 is the mass of our satellite and uppercase 𝑀 is the mass of the
body being orbited, in this case, Earth.
In this equation, the mass of our
satellite cancels out, as does one factor of the radius 𝑟. And so we find this expression: 𝑣
squared, the speed of our satellite squared as it orbits the Earth, is equal to 𝐺
capital 𝑀 over 𝑟. And this now solves our problem of
not knowing the speed of our satellite in order to calculate its kinetic energy. Now that we know its speed squared
and can express that in terms of these factors, we can say that the kinetic energy
of our orbiting spacecraft is equal to one-half its mass times 𝐺, the universal
gravitational constant, multiplied by the mass of, in this case, the Earth, divided
by the distance between our satellite and the center of the Earth.
And we can now notice that we’re
given the masses of these two bodies as well as the universal gravitational constant
and this distance 𝑟. All that remains, then, is for us
to substitute in these values and then calculate the kinetic energy. So here we have our satellite mass,
2,200 kilograms; our value for the universal gravitational constant; our mass of the
Earth, 5.97 times 10 to the 24th kilograms; and 𝑟, 7,000 kilometers.
There’s just one change we’ll want
to make before calculating kinetic energy. And it comes down to what is right
now a disagreement in units. Notice that in our value for the
universal gravitational constant, we have units of distance given in meters, whereas
our radius is currently in kilometers. If we recall, though, that 1,000
meters is equal to one kilometer, then that tells us that 7,000 kilometers is equal
to 7,000 with three zeros added to the end meters. That is seven million meters. Now, the units all throughout our
expression do agree, and we can go ahead and compute the kinetic energy of the
spacecraft.
To three significant figures, we
find a result of 6.26 times 10 to the 10th joules. And if we recall that 10 to the
ninth, or a billion joules, is equal to one gigajoule, then we can express our
answer as 62.6 gigajoules. This is the spacecraft’s kinetic
energy, which, notice, is constant all throughout its circular orbit.
Let’s summarize now what we’ve
learned about mechanical energy conservation in orbits. In this lesson, we first recall
that an object’s mechanical energy, we refer to it using capital ME, equals the sum
of that object’s kinetic and potential energies. We saw further that when an object
is in circular orbit, we can express its kinetic energy this way, where all the
factors in this equation are either constants or fixed values, whereas when an
object is in elliptical orbit, we express its kinetic energy this way. And we note that it changes in
time.
Along with this, we recall that for
all orbits, the gravitational potential energy shared between the two masses is
equal to negative 𝐺, the universal gravitational constant, times the product of the
two masses divided by the distance between their centers. All this led us to expressions for
the mechanical energy of an object in both circular and elliptical orbit and,
finally, the observation that these mechanical energies are constant in time. That is, these energies are
conserved. This is a summary of mechanical
energy conservation in orbits.
