Video: Gravitational Potential Energy in Radial Gravitational Fields

In this video we use the law of gravity to solve for gravitational potential energy, and also apply this law to find the potential energy of an object near a planetary surface.

12:51

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.

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