Lesson Video: Mechanical Energy Conversation in Orbits Physics

In this video, we will learn how to calculate the kinetic energy required to maintain circular orbit around a star or planet.

14:54

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.

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