Lesson Video: Surface Gravity Physics • 9th Grade

Video Transcript

In this video, our topic is surface gravity. This term refers to an acceleration due to gravity that all objects on or near the surface of a large spherical object experience. It’s surface gravity, for example, that pulls us as well as the objects around us toward the center of the Earth.

We can start to understand where surface gravity comes from by considering two other laws of physics. Both coming from Isaac Newton, one of these is the law of gravitation. This says that the gravitational force between two masses, one referred to as capital 𝑀 and the other as lowercase 𝑚, is equal to the product of these masses times the universal gravitational constant, big 𝐺, divided by the distance between the centers of mass of our masses squared. This is sometimes called the universal law of gravitation because these two masses, capital and lowercase 𝑚, could be anything. And, regardless of what they represent, this equation tells us the gravitational force between the masses.

The other of Newton’s laws we’ll consider is his second law of motion. This law tells us the net force on an object is equal to the object’s mass multiplied by its acceleration. One of the important conditions on Newton’s second law of motion is that this force here is the net force on our mass 𝑚. Knowing that, let’s imagine a scenario where we have some mass 𝑚. And the net force, indeed the only force acting on that object, is the gravitational force it experiences between itself and some other mass, capital 𝑀. In other words, let’s imagine that these two masses, lowercase 𝑚 and capital 𝑀, are the only masses in the universe.

In that case, the only force acting on this smaller mass is the gravitational force between it and the larger one. We can, therefore, say that the net force acting on our mass, lowercase 𝑚, is equal to the gravitational force on it. And that means that the right side of our law of gravitation equation is equal to the right side of our second law equation. And as we look at this new equation, notice what’s common to both sides. It’s the mass lowercase 𝑚. This means that if we divide both sides of the equation by that mass, it will cancel out completely.

And the equation we’re left with says that the universal gravitational constant times the mass of the larger object capital 𝑀 divided by the distance between the centers of mass of the two masses squared is equal to the acceleration experienced by the smaller mass, lowercase 𝑚. And notice something interesting about this equation. It doesn’t depend on the value of the smaller mass; that is, lowercase 𝑚 is nowhere in the equation. Now, that doesn’t mean that the position of this object with a mass lowercase 𝑚 is unimportant. Because we see in the denominator on the left-hand side this value 𝑟. Recall that that’s the distance between our two masses.

So, the acceleration of this object, which has a mass lowercase 𝑚, doesn’t depend on that mass, but it does depend on the object’s position with respect to the larger mass, capital 𝑀. Now just as a reminder, this gravitational force we’re talking about is a mutual force between these two masses. The magnitude of the force acting on this object is the same as the force magnitude acting on this one. If we had written the second law of motion a little bit differently, in this case, to highlight the larger mass capital 𝑀. Then when we combine this force equation with the law of gravitation, we would have found that it’s the larger of the two masses, the one represented by capital 𝑀, that cancels out in this equation. And then, the acceleration 𝑎, we will be referring to, is the acceleration experienced by that larger mass, capital 𝑀.

So just for completeness, we could come over here and specify that this acceleration is that experienced by the smaller mass, lowercase 𝑚. The point is, due to the mutual force of gravity between these two objects, each one will tend to accelerate toward the other. And the acceleration on these objects is not the same, even though the force magnitude acting on each one is. Each object’s acceleration depends on the other object’s mass and the distance between the centers of mass of the two masses.

Now, this relationship here, which is true in general for any two masses, is especially useful to us when our larger mass — we’ve called it capital 𝑀 — refers to the mass of a very large spherical body, like a planet or a moon. And our smaller object, with mass lowercase 𝑚, is on or near the surface of our larger object. A great example of this is when there’s a person, say it’s us, standing on the surface of the Earth. If we say that our mass is represented by lowercase 𝑚 and the mass of the Earth by uppercase 𝑀, then this equation here describes the acceleration we experience standing on Earth’s surface. In other words, this is Earth’s surface gravity.

And notice that, just like we said before, this acceleration doesn’t depend on our mass. So, then, this acceleration 𝑎 is actually the acceleration experienced by any object on or near Earth’s surface. Now, we need to be a bit careful when we say this. Because recall that this 𝑟 here represents the distance between the center of mass of the Earth, where all of its mass is effectively concentrated and our position on Earth’s surface.

So technically, if we were to do something, like climb up on a stepladder, this distance represented by 𝑟 would increase, and that would cause the acceleration we experience, the surface gravity of the Earth, to decrease. It turns out, though, that for small elevation changes like this, this equation is still so nearly correct that we consider it essentially accurate. In other words, the difference in distance between the radius of the Earth and the radius of the Earth plus this small distance we’ve climbed up is so very small that it doesn’t really have an effect on the acceleration we experience.

And so, we can say that this acceleration equals the acceleration due to gravity experienced by any object that’s on or near Earth’s surface. This is why all objects near that surface fall at the same rate regardless of their mass. By the way, for the specific case where we are talking about Earth’s gravitational attraction, this acceleration, what we’ve called 𝑎 sub 𝑀, is given a specific symbol. It’s represented by a lowercase 𝑔. The value of lowercase 𝑔, we can recall, is approximately 9.8 meters per second squared.

This is what we get if we take the mass of the Earth, multiply it by the universal gravitational constant, and then divide that product by the radius of the Earth squared. Try it for yourself and see. Knowing all this about acceleration due to gravity, let’s get some practice with these ideas through an example.

Two objects, object A with a mass of five kilograms and object B with a mass of 100 kilograms, are near an even larger object with a mass of 10 to the 20th kilograms. Object A and object B are at an equal distance away, 100 kilometers, from the center of mass of the very large object. What is the magnitude of the gravitational force experienced by object A due to the very large object? Give your answer to three significant figures.

Okay, looking over at our sketch, we see a five-kilogram object. That must be object A. And as well, we see a 100-kilogram mass. This must be object B. Object A and object B interact gravitationally with this very large spherical object. Now, this object is represented here using a rectangle that’s not very much bigger than A and B. But we can imagine that actually this object is so large it couldn’t fit on the screen at this scale. We have an indication of this thanks to the very large mass of this object.

So, we could almost think of these three objects, A and B and our large spherical one, as though our large spherical object is the Earth, object B, say, is a person, and object A is a book the person is holding on to. That’s roughly the scale we’re talking about as we consider these three objects. Our first question related to this scenario says, what is the magnitude of the gravitational force that object A experiences due to the very large object?

As we get started figuring this out, let’s record some of the information we’ve been given. First, we’ve been told that our large spherical object has a mass, we’ll call it capital 𝑀, of 10 to the 20th kilograms. And we’re also told that the distance between the center of mass of this large spherical object, that is, where all of its mass is effectively concentrated, and our two other objects, object A and object B, is the same. It’s 100 kilometers. We’ll call that distance 𝑟. And as we see, it’s the same for object A and object B in relation to our large spherical object. With these values written down, let’s clear some space and consider again our question.

We want to know the magnitude of the gravitational force experienced by object A, that’s this object here, due to our very large spherical object. So, let’s recall the general equation, called Newton’s law of gravitation, that describes the force of gravity between two objects that have mass. This equation tells us that that force is equal to the product of those masses. We’ll call them capital and lowercase 𝑚, respectively. Multiplied by the universal gravitational constant, capital 𝐺, divided by the distance between our two masses squared.

Now, the gravitational constant, big 𝐺, is approximately equal to 6.67 times 10 to the negative 11th cubic meters per kilogram second squared. So, if we want to know the gravitational force on object A caused by the very large object, we can call this force 𝐹 sub A, then we’ll take this constant and we’ll multiply it by the mass of our large object. And multiply that by the mass of object A, we’ll call it 𝑚 sub A, and divide all this by the distance between the centers of mass of object A and our large spherical object squared.

When we substitute in the values for all these variables, we’re just about ready to calculate this force magnitude, except for one thing. Notice that the distance in our denominator is in units of kilometers. In order to agree with the distance units in the rest of our expression, we’d like to convert this into meters. We can recall that one kilometer is equal to 1000 meters and therefore 100 kilometers is equal to 100,000 meters. With that all figured out, when we calculate this fraction, to three significant figures, we find a result of 3.34 newtons. That’s the magnitude of the gravitational force experienced by object A due to the very large object.

Now, let’s consider the next question in this exercise.

What is the acceleration of object A toward the very large object? Give your answer to three significant figures.

Okay, so, we’ve calculated the gravitational force between these objects. And now, we want to calculate the acceleration object A experiences. We can start doing this by recalling Newton’s second law of motion. This law has it that the net force on an object is equal to that object’s mass times its acceleration. Acceleration, in the case of object A, is just what we want to solve for. And here’s how we can do it. As we saw, Newton’s second law says that the net force on an object is equal to that object’s mass times its acceleration.

When we think about the forces acting on object A, we can see that there are two. One is the gravitational force due to object B and another is the gravitational force due to the large spherical object. But here’s the thing. The force between A and B is much, much, much smaller than the force between A and the large spherical object. This is because the mass of our large object absolutely dwarfs the mass of object B. We can say then that the gravitational force on object A due to the very large object is effectively the only force on object A.

And this means we can say that this equation is equal to the mass of object A multiplied by the acceleration of object A. We’ll call it 𝑎 sub A. And note that we got this expression here from Newton’s second law. But now, as we consider this equation, notice that the mass of object A appears on both sides, and therefore we can cancel it out. If we do that, this is the equation that results. The acceleration of object A is equal to the universal gravitational constant times the mass of our very large object divided by the distance between the centers of mass of object A and our very large object squared.

Now, as we consider this part of our expression here, we can see that it’s actually equal to this expression if we drop out the mass of our object, in this case, object A. With that mass gone, see that we have the gravitational constant, uppercase 𝐺, multiplied by the mass of our large object divided by 𝑟 squared. Which are identical to the factors we see here on the right-hand side of this equation. So then, if we calculate all this, we’ll solve for the acceleration of object A due to the very large object. To three significant figures, it’s 0.667 meters per second squared. That’s the acceleration of object A toward the very large object.

Now, let’s consider the next part of our exercise.

What is the magnitude of the gravitational force experienced by object B due to the very large object? Give your answer to three significant figures.

Okay, whereas before, we were considering the gravitational force between object A and the large spherical object, now we’re considering that force between the large object and object B. Once again, we’ll use Newton’s law of gravitation to solve for this force. But this time, instead of solving for the force experienced by object A, we’ll solve for that experienced by object B. And we’ll say that this object has a mass 𝑚 sub B. Because both objects A and B are the same distance away from the center of mass of our large spherical object, we won’t change 𝑟, the distance between the centers of mass of the two masses in our equation. Which means that when we go to calculate this force, 𝐹 sub B, using an expression like this, all we’ll need to do is substitute the mass of object B in where we used to have the mass of object A.

We see that this mass is 100 kilograms. And so, now, we have an expression which, when we solve it, will give us this force experienced by object B due to the gravitational attraction from the large spherical object. To three significant figures, this force is 66.7 newtons. Note that this is larger than the force magnitude experienced by object A. In fact, it’s 20 times larger because object B has a mass 20 times larger than object A.

Now that we found this out, let’s consider the last part of our question.

What is the acceleration of object B toward the very large object? Give your answer to three significant figures.

To answer this question, once again, we’ll use Newton’s second law of motion. And just like we did for object A, we’ll also assume that the gravitational force between objects A and B is negligibly small compared to that force between object B and the very large object. This means we can say that the mass of object B multiplied by its acceleration is equal to the gravitational force experienced by object B due to the very large object. So then, we can go to our equation for that force, and we can equate it to the mass of object B times the acceleration of object B. We’ll call it 𝑎 sub B.

And then, like we saw earlier, the mass of our object, in this case, object B, is common to both sides. And that means that the acceleration experienced by object B is equal to capital 𝐺 times the mass of our very large object divided by the distance between B and our very large object squared. And this expression here is equal to this expression here if we ignore this mass of our object B. And so, to solve for 𝑎 sub B, we’ll multiply the universal gravitational constant by the mass of our large object divided by the distance between that object and B squared.

To three significant figures, this is 0.667 meters per second squared. Note that this is the same as the acceleration experienced by object A. And this comes back to the fact that our equation for object acceleration does not depend on the mass of the object whose acceleration we’re considering. So, this is how object B and object A, and any other object, will accelerate toward the very large object.

Let’s recall now some key points about surface gravity.

Starting off with Newton’s law of gravitation and second law of motion, we saw that these laws can be combined to yield an expression for the acceleration one of the masses involved would experience. We noted that if this is an acceleration of the mass indicated by a lowercase 𝑚. Then equivalently the other mass, indicated by a capital 𝑀 in Newton’s law of gravitation, also experiences an acceleration due in this case to the smaller mass, lowercase 𝑚. Related to this, we saw that the acceleration an object experiences does not depend on its mass. And lastly, we saw that surface gravity describes an object’s acceleration when it is on or near a much larger spherical object.