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In this video, we’re going to learn about Newton’s law of universal gravitation. We’ll see what this law says, how it was developed, and how we can apply it practically. To start out imagine that you lived in the time of Isaac Newton in the 1600s. During that time, scientists were facing a difficult challenge. On the one hand, thanks to astronomical observations, they had clear evidence that planets, celestial bodies they called them, moved around one another in regular orbits.

Based on this, it seemed clear that there was some sort of attraction between these masses as they moved. But on the other hand, on planet Earth with regular everyday size masses, there was no sense that those masses would attract one another even if they were held very close. For the time being then, it seemed as though the physical laws governing massive bodies, celestial bodies the size of planets, might have a different form or just be fundamentally different than the laws governing everyday size more tangible objects.

To see how these two worlds were brought together, it will be helpful to learn about Newton’s law of universal gravitation. As the story goes, one day Isaac Newton was resting under an apple tree pondering all these mysteries. Then, it said, an apple falls from the tree, hits him on the head, and he has a grand idea. There’s a good chance that the story is not accurate. But nonetheless, it shows how Newton’s law connects everyday objects we’re familiar with with much larger objects on the scale of the size of our planet or other planets.

One reason this law is so meaningful is because it truly is universal. It applies to any masses no matter how big or small and no matter where they’re located in the universe. Considering the scope of this development, it’s remarkable that in one simple statement it’s possible to sum up the gravitational force of attraction between any two masses. But that’s just what Newton along with help from his contemporaries did. This universal law says that the gravitational force between two objects, mass one and mass two, is equal to their product divided by the square of the distance between them.

This is the backbone of Newton’s law from a physical perspective. And this value is then multiplied by a constant value called big 𝐺. This value, called the universal gravitational constant, was developed in order to make the units in the overall expression work out. It has its own interesting story of development. But before telling that story, let’s consider the rest of this universal law of gravitation. Sometimes this law, which is one of the more recognizable equations in physics, can become so familiar to us that we lose sight of what makes it special. At the time of its development, this law was not at all obvious though.

For example, consider the denominator where we see an 𝑟 squared term. This means that the universal law of gravitation is an inverse square law. Though we see these laws across physics, it’s still a striking result. Why one over 𝑟 squared? Why not one over 𝑟 cubed or one over 𝑟 to the 1.99? Or even why should the force of gravity decrease with increasing distance? All of the alternatives we can think of for a one over 𝑟 squared relationship remind us of just how special this law is and how it helps establish the structure of our universe.

Now, let’s move to considering the mass values 𝑚 one and 𝑚 two. This law tells us that if we have two objects of any shape, as long as they have mass, there’s a gravitational force of attraction between them. Our masses could be spheres or blocks or crocodiles or atoms. Any objects that have mass meet the standard and therefore have a gravitational force between them. Regardless of the shapes our two masses have, when we talk about the distance between them, we’re speaking of the distance between their centers of mass wherever those centers are located within the overall mass itself.

Now, let’s say we try an experiment. What if we get two masses and we let their masses both equal exactly one kilogram and we separate these masses by exactly one meter? So we can see when we look at this law of gravitation that we’ll have an equation that reads the gravitational force of attraction between these two one-kilogram masses is equal to one kilogram times one kilogram, which is one kilogram squared, all over one meter squared times 𝐺, this gravitational constant.

In order to show why this gravitational constant is needed, let’s imagine for a second that it’s not there. In other words, let’s let it equal one and see what we get as a result of this calculation. If big 𝐺 had this value with no units, that would mean that the gravitational force of attraction between these two masses is one kilogram squared meter squared. But wait a second, we know that force is measured in units of newtons and that a newton has base units of a kilogram meter per second squared. That means the two sides of our equation don’t add up when we let big 𝐺 equal simply one.

We’re now getting into the whole reason for the existence of the gravitational constant in the first place. We’re seeing that if it’s not there — that is, if it’s just equal to one — then our whole expression for the gravitational force of attraction doesn’t make sense. Isaac Newton postulated this constant in order to give gravitational forces the right magnitude as well as the right units. Gravity as the weakest of the four fundamental forces creates an attractional force between two masses of one kilogram separated by one meter of much much less than one newton. So 𝐺 does serve double duty. It gives us the right units for our expression as well as a magnitude that agrees with experiment.

We’ve said that gravity is a weak force compared to the other four fundamental forces of electromagnetism and the strong and weak nuclear forces. One way we can see that weakness in action is to take any two household objects we might come across, say a pencil and a water glass. If we hold those two objects together, we can’t feel the gravitational force of attraction between them. It’s just too weak. On the other hand, if we had two magnets, one in each hand, we could definitely feel the force of attraction or repulsion between them as they got close.

All this is to say that the universal gravitational constant big 𝐺 is a very small value. It’s not one. In fact, it’s much much less than one. It’s so small in fact that it’s very hard to measure big 𝐺. One of the most accurate and ingenious measurements of this gravitational constant came under the watch of a gentleman named Henry Cavendish. In his experiment, Cavendish suspended a very very thin metal wire from a solid frame. From the end of this wire, he hung a small piece of metal that had masses of carefully measured values on either end. Once this system had stabilized and wasn’t moving or twisting in anyway, Cavendish brought two relatively large heavy masses close to opposite sides of the smaller suspended masses.

In response to the gravitational attraction, the smaller suspended masses moved ever so slightly towards the larger ones causing a twist in the wire. Cavendish was able to measure that twist with high precision. And since he knew all the masses involved as well as the distances separating them, he had values for 𝑚 one, 𝑚 two, 𝑟 as well as the force, 𝐹, the torsional force acting on the wire. In other words, he had all the ingredients needed to arrive at a value for 𝐺. The value for 𝐺 that Cavendish found is very close to the value that we’ll often use today, which is that 𝐺 is approximately equal to 6.67 times 10 to the negative 11th cubic meters per kilogram second squared.

Looking at this value, two things perhaps stand out: one, 𝐺 is indeed small much smaller than one and, two, it has a strange set of units attached to it. But, remember, the units of 𝐺 are designed to make the rest of the universal law of gravitation have consistent units. Measurements for increasingly accurate values of 𝐺 still go on today. For our purposes though, we’ll be well served to use this given value for 𝐺, very close to the one that Cavendish found experimentally. Let’s get some practice working with Newton’s law of universal gravitation through an example.

An asteroid has a mass of 4.7 times 10 to the 13th kilograms. The asteroid passes near Earth, and at its closest approach, the separation of the centers of mass of the asteroid and Earth is four times the average orbital radius of the Moon. What force does the asteroid exert on Earth when at its minimum distance from Earth? Use a value of 384400 kilometers for the average orbital radius of the Moon.

We’ll label this force, we want to solve for, capital 𝐹 and start off by drawing a sketch of the situation. In this situation, our asteroid labelled 𝑎 passes by the Earth labelled 𝐸 at a minimum distance of four times the orbital radius of the moon around the Earth. Given the mass of the asteroid 𝑚 sub 𝑎 and the orbital radius of the moon around the Earth 𝑂𝑅 sub 𝑚, we want to solve for the gravitational force of attraction between the asteroid and Earth when they’re closest. To solve for this force, we recall that the gravitational force of attraction between any two masses, 𝑚 one and 𝑚 two, is equal to their product divided by the square of the distance between their centers of mass all multiplied by the universal gravitational constant capital 𝐺.

We’ll let that constant 𝐺 be exactly 6.67 times 10 to the negative 11th cubic meters per kilogram second squared. When we apply the mathematical relationship for gravitational force to our scenario, we can say that 𝐹, the force we wanna solve for, is equal to 𝐺 times the mass of the Earth times the mass of the asteroid all divided by four times the orbital radius of the Moon quantity squared. We know the value in the denominator given to us. And we also know the mass of the asteroid, as well as the constant 𝐺. All that remains is to solve for the mass of the Earth. And we can do that by looking up this value.

A commonly accepted value for the mass of the Earth is 5.95 times 10 to the 24th kilograms. Knowing that, we’re ready to plug in and solve for 𝐹. When we do plug these values in, we’re careful to convert the orbital radius of the Moon into units of meters so that it’s consistent with the units in the rest of our expression. Speaking of units, let’s take a second to consider what the final units of this calculation will be. Looking in the numerator of this overall expression, we see that a factor of kilograms will cancel out. Looking then at the units of meters that appear in our expression, we see that the meter squared in the denominator will take away two meter factors in our numerator so that overall we’ll simply have meters to the first.

The units of second squared will remain in our overall denominator. So we expect to get final units of kilograms meters per second squared, which agrees with the units we’d expect for a force, that is, units of newtons. When we calculate this result, we find, to two significant figures, that it’s 7.9 times 10 to the ninth newtons. That’s the gravitational force of attraction between the asteroid and the Earth. Let’s summarize what we’ve learnt so far about Newton’s law of universal gravitation.

In this video, we’ve seen that Newton’s law of universal gravitation specifies the gravity force between any two masses separated by a distance. As an equation, it says that that gravitational force is equal to the product of the masses divided by the square of the distance between their centers of mass all multiplied by this universal gravitational constant called big 𝐺. We’ve also seen that gravity is the weakest of the four fundamental forces of gravity electromagnetism and the strong and weak nuclear force. And because gravity is such a weak force, we’ve seen that the gravitational constant 𝐺 is difficult to measure. However, a working value for 𝐺 has been determined to be 6.67 times 10 to the negative 11th cubic meters per kilograms second squared.