Lesson Video: Newton’s Law of Gravitation Physics • First Year of Secondary School

Video Transcript

In this video, our topic is Newton’s law of gravitation. This is one of the foundational laws of classical physics, and in it, we discover a universal principle for how objects with mass interact.

Now, back in Newton’s time, just as it’s true for us today, it was common everyday-type events — walking down the road, moving things from one place to another — that people noticed and paid attention to most often. But, so the story goes, it was just such an everyday occurrence that helped Newton conceive of a universal law of gravitation. Upon observing a gravitational event on Earth’s surface, Newton connected this in his mind with the motion of stars and planets very far away from the Earth.

His remarkable claim was that the same force responsible for making objects near Earth’s surface fall to the ground was also responsible for the motion of stars and planets in space. That is, Newton held that all masses attract all other masses. So, whether that mass is an apple or the Earth or a person or another planet, the same principle applies.

Specifically, Newton said that if we have one point mass over here, say we call this mass 𝑚 one, and that we have another point mass over here. We can call this 𝑚 two. And if these point masses are separated from one another by a distance 𝑑. Then, Newton said, there’s a gravitational force of attraction between these two masses that’s proportional to the product of the two masses divided by the distance between them squared.

This is the basic form of Newton’s law of gravitation, sometimes also called the universal law of gravitation. The reason it’s called that is because these masses could be anything. We could be talking about any two masses in the universe and the same law would apply. Newton developed the form of this law by looking at experimental measurements made by other scientists. By careful analysis of those results, he came to see that this proportionality relationship was true. Still, as it’s written, this law of gravitation isn’t quite complete. To see why that so, let’s take a look at the units on either side of this relationship.

We know that today, the SI base unit of force is the newton and that a newton can be written as a kilogram meter per second squared. So, those are the units on the left-hand side. But on the right, we see, first of all, a mass that will have units of kilograms, then a second mass having the same units. And all of that is divided by a distance. Let’s say it’s in units of meters squared. On the left-hand side then, we have units of kilograms meters per second squared, while on the right, we have units of kilograms squared per meter squared. We can see that these units clearly do not agree.

The way to resolve this discrepancy is to introduce another factor into this relationship for force. That factor is symbolized using a capital 𝐺. It’s called the universal gravitational constant or just gravitational constant for short. The units of this constant, big 𝐺, are such that when we multiply them by kilograms squared per meter squared. We do end up with units that are equal to the units on the left-hand side, kilograms meters per second squared or newtons.

Despite being notoriously hard to measure, the constant, capital 𝐺, has been determined to be approximately 6.674 times 10 to the negative 11th cubic meters per kilogram second squared. So, we see that inserting 𝐺 into our equation doesn’t only affect the units, it also affects the overall value. This has a very important implication for how we experience gravitational attraction. But before we get to that, let’s write out our new and improved law of universal gravitation.

This law, formulated by Newton, says that the gravitational force of attraction between two masses, 𝑚 one and 𝑚 two, is equal to the product of their masses divided by the square of the distance between them all multiplied by the universal gravitational constant. Just as this law is universal because it applies to any masses anywhere in the universe, so this constant is sometimes called universal because we believe its value to be the same everywhere.

Now, let’s take this equation for a test drive. And we’ll do it by using our masses 𝑚 one and 𝑚 two separated by a distance 𝑑. Let’s imagine, though, that we have very specific values for these masses and this distance. Let’s say that each one of the masses is one kilogram and the distance separating them is one meter. Well, as a first guess, we might expect this to lead to a force of attraction between these masses of one newton. But this is where the particular value of our gravitational constant comes into play.

The gravitational force of attraction between 𝑚 one and 𝑚 two is equal to 𝐺 times one kilogram times one kilogram divided by one meter quantity squared. We can see that all of this part of the right side of our equation, if we were to calculate it separately, we’d come out to one kilogram squared per meter squared. When we multiply all this, though, by the value of 𝐺, we end up with a force of this value in between these two one-kilogram masses. This force is clearly very small, much too small for us to be sensitive to if we experienced it in our everyday life.

But then, let’s say that instead of two one-kilogram masses separated by a distance of one meter. We had two people walking along separated by this distance. And that each person, we’ll say, has a mass of 100 kilograms. When we plug these new masses into our equation for force, all that we’d do is shift the decimal place four spots to the right. So, the force of attraction between these two people would still be a very small value. The relative smallness of this force at everyday distances and everyday masses is one of the reasons this law of Newton’s was such a special insight.

Now, another insight of Newton’s has to do with the direction that this gravitational force acts. As we saw, this force is always attractive. No masses ever repel one another due to gravity. More than that, for massive objects that take up space, such as our two walkers here, the force of attraction can be thought to act on the center of mass of each object. The spot where all of an object’s mass can effectively be concentrated. So, say, our two walkers have centers of masses here and here. In that case, we could say that the walker on the right exerts a force like this on the walker on the left. And an equal but opposite force is exerted on the walker on the right.

Note that these two force vectors act on a line between these two centers of mass. This explains why it is that regardless of where we stand on the surface of the Earth. If we release an object from our hand, that object will fall down toward the Earth’s surface. These objects, as they fall, are tending towards the center of mass of the Earth. Now, here’s a very strange thought. These objects that have been dropped are tending towards the center of the Earth. We know that based on their motion, that motion is due to the gravitational attraction between these objects and the Earth.

That force, though, the gravitational force between a dropped object and the Earth, acts on both masses, the object and the Earth, with equal magnitude. Which means that this pink object here is exerting a gravitational attractive force on the center of the Earth and that force is nonzero. Likewise, the orange object here is also exerting a gravitational attractive force. This is part of what it means for all masses to attract all other masses.

By the way, realizing this fact helped Newton understand that the star at the center of our solar system, our Sun, actually isn’t entirely stationary like it was thought to be. This is due to the mutual gravitational attraction between the Sun and the masses orbiting it. Knowing all this about Newton’s law of gravitation, now let’s get some practice with these ideas through an example.

Each of the following figures shows two rocks in outer space. Which figure correctly shows the direction of the gravitational force exerted on each rock? (a), (b), (c), (d), or (e).

Okay, so here we’re told that we have two rocks in outer space. That means there are no other masses anywhere nearby these rocks. Therefore, we can assume that the gravitational force exerted on each rock is due only to the other of the two rocks. In order to figure out which of these five diagrams correctly shows the direction of the forces acting on each rock, we can recall a principle about gravitational force. That is, gravitational force is always attractive.

This means that if we have two masses, say one mass here and the other mass here, the force of gravity will cause this mass to be attracted to the other one. And this second mass to be attracted to the first. That is, the force vectors we could show on each mass lie along a line between the two masses. This is always true when objects exert gravitational forces on one another. Those attractional forces lie along a line between the two objects’ respective centers of mass.

Looking at our answer options, we see option (a) agreeing with the rule we just described. If the center of mass of the reddish-brown rock is here and the center of mass of the yellowish rock is right here, then we can see that the two gravitational force vectors lie along the line between these points. And what’s more, they point in such a way that indicates these forces as attractive. That is, the rocks will tend to move toward one another.

Before we confirm that option (a) is the correct answer, let’s look at the remaining choices. For option (b), if we draw a straight line between the two centers of mass, we see the forces don’t lie along this line. So, that means option (b) won’t be our choice. Then, looking at option (c), here the force vectors do lie along this line. But we notice that the force on the reddish-brown rock is pointed in the wrong direction. The implication here is that the golden-colored rock is somehow repelling the reddish-brown one. But we know that gravity doesn’t act that way. We’ll cross off option (c) then as well.

Looking at option (d), this fails for the same reason option (b) did. The force vectors do not lie along the line connecting these rocks’ centers of mass. And lastly, for option (e), the vectors are along this line. But now, they both imply a repulsive force rather than an attractive one. But gravity is always attractive. So, we’ll cross out option (e) too.

This confirms our earlier assessment that it’s figure (a) that correctly shows the direction of the gravitational force exerted on each rock.

Let’s now look at a second example.

Two objects, A and B, are in deep space. The distance between the centers of mass of the two objects is 20 meters. Object A has a mass of 30,000 kilograms and object B has a mass of 55,000 kilograms. What is the magnitude of the gravitational force between them? Use a value of 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, in this scenario, we have these two objects, called A and B. So, let’s say here we have our objects A and B. And we’re told that the distance between the centers of mass of these two objects is 20 meters. If the center of mass of object A is here and the center of mass of object B is here, that tells us that this distance here is 20 meters. Along with this, we’re told the mass of object A and the mass of object B as well as the fact that these two objects are in deep space.

Being in deep space means that A and B are the only objects nearby. When we compute the magnitude of the gravitational force between them, we can ignore or neglect any other masses or objects. Continuing on then, we can represent the mass of object A as 𝑚 sub A and that of object B as 𝑚 sub B. Now that we know our object masses as well as the distance that separates the centers of mass of these two objects, let’s recall Newton’s law of gravitation.

This law says that the gravitational force between two objects, we’ll call it 𝐹, is equal to the universal gravitational constant, big 𝐺, times the mass of each one of these objects. We’ll call them 𝑚 one and 𝑚 two. Divided by the distance between the objects’ centers of mass, we’ll call that 𝑟, squared. Looking at this equation, we can see that for our scenario with objects A and B, we know their masses. And we also know the distance separating their centers of mass. And along with that, we’re told in the problem statement a particular value to use for the universal gravitational constant.

At this point then, we can begin to calculate the magnitude of the gravitational force between objects A and B. It’s equal to the value we’re given to use as capital 𝐺 times the mass of object A, 30,000 kilograms, times the mass of object B, 55,000 kilograms. All divided by 20 meters quantity squared. Note that all the units in this expression are already in SI base unit form. We have meters and kilograms and seconds. To three significant figures, this force is 2.75 times 10 to the negative fourth newtons. That’s the magnitude of the gravitational force between objects A and B.

Let’s summarize now what we’ve learned about Newton’s law of gravitation. We began this lesson by noting one of the realizations of Isaac Newton that all masses attract all other masses. Given two masses 𝑚 one and 𝑚 two, the magnitude of the force of attraction between them is given by Newton’s law of gravitation.

If the centers of mass of these two massive objects are separated by a distance 𝑟. Then the gravitational force of attraction between the masses is equal to the universal gravitational constant, capital 𝐺, times the two mass values themselves, 𝑚 one and 𝑚 two. Divided by the distance between the centers of mass of these masses squared. And lastly, we saw that the gravitational force between two objects: one, always attracts and two, always acts along the line connecting the objects’ respective centers of mass. This is a summary of Newton’s law of gravitation.