It’s fun to think about how we know stuff. For instance, the sun has a mass of about 2 x 10 ^{30} kilograms. That is such an enormous mass that it’s difficult to comprehend. And if it’s so difficult for us to even imagine numbers that large, how would we go about finding those values? Well, the original method was to use some small masses, a stick, and a string. Yes, this is one of the important steps in determining masses of both the sun and all the planets in our solar system. It’s called the Cavendish experiment—first performed by Henry Cavendish in 1798. It’s really cool, so I’m going to explain how it works.

Objects with mass have a gravitational attraction between them. A basketball has a gravitational interaction with the Earth (since they both have mass). It is this gravitational interaction that makes the basketball speed up as it falls toward the ground if you let go of it. But of course everyone has always known that if you let go of an object it will fall. However, it was around the time of Newton that people realized that this interaction also worked with astronomical objects like Earth, moon, and sun. That gives us this force model—it’s often called Newton’s Law of Universal Gravity, but like most big ideas it likely had a lot of contributors.

Let’s go over this gravitational force model. First, the magnitude of this force depends on the product of the two masses interacting (m_{1} and m_{2}). Second, the magnitude decreases with the square of the distance between the two objects (r). Finally, there’s that G. This is the universal gravitational constant. It’s the key to finding the mass of the Earth.

So, just step back for a moment. When we measure stuff, we always have to make some type of choice. If we want to have a mass in kilograms, then we have to decide how to specify the value of 1 kg. One way would be to say that a kilogram is the mass of 1 liter of water. Of course, that’s not the best definition (we have better methods now). OK, what about measuring force? We use a unit called the Newton where 1 Newton is the force required to accelerate 1 kilogram at 1 meter per second per second. Yes, things are getting out of control—but the key is that you can make these definitions and build one unit on another unit.

Now imagine this experiment. Suppose I take my 1 liter of water (that I know is 1 kilogram) and measure the gravitational force exerted by the Earth. If I know the radius of the Earth (the Greeks did a pretty nice job figuring this out) and the gravitational constant G, then I can solve the gravitational force equation above for the mass of the Earth. But what is the gravitational constant? That’s the tough part and this is how you can find the value of G.

It turns out that this gravitational constant is super small. That means that the interaction between two ordinary objects like bottles of water is ridiculously tiny. The only way to get a noticeable gravitational force is if one of the interacting masses is huge (like the Earth). However, there’s a way to figure it out—using a torsion balance.

Let’s start with a simple physics demo that you can try at home. Take a pencil and place it on the edge of a table so that about half of the pencil is hanging over the edge and it’s almost about to fall over (but it doesn’t). At this point, the pencil is mostly balancing right on the edge of the table. With only this tiny contact point supporting the pencil, the frictional force can’t really exert any torque to stop it from rotating. Even a super tiny force pushing the end of the pencil will get it to rotate. Try a tiny puff of air from your mouth to get it to rotate.

I like to put my fingers near the pencil so that I can pretend like I am using my superhero powers to move it. Now let’s replace the pencil with a longer stick and instead of placing it on a table, I could hang it from a string. Since it’s supported from the middle, tiny forces can get it to rotate just like the pencil. Instead of blowing with air, we could get a small gravitational force to move it. Here’s how this works.

There are two smaller masses (labeled m_{1}) at the end of the rotating horizontal rod. These masses interact with the larger masses (m_{2}) that are a distance (r) away. The horizontal rod will eventually reach some equilibrium position since there is a tiny amount of torque from the twisting of the cable that supports the rod. The cable acts like a rotational spring. The more it twists, the greater the torque. If you know the relationship between rotation angle (θ) and torque then you can figure out the gravitational force pulling the mass on the end of the stick and the bigger stationary mass together. In the configuration shown in the diagram above, the large masses would make the stick rotate clockwise (as seen from above). If you move the bigger masses to the other side of the stick, the gravitational forces would cause it to rotate counterclockwise. This shows that the rotation is due to the gravitational interaction between paired masses. Once the stick settles into a stable position, it’s just a matter of measuring the masses and the distance between them to get the gravitational constant.

In this case we get a gravitational constant of G = 6.67 x 10^{-11} N*m^{2}kg^{2}. You can see that this constant is indeed tiny. As an example, we can do a sample calculation. Suppose that you are a human standing 1 meter away from another human of the same mass (about 75 kilograms). What magnitude of force would pull on you due to the gravitational interaction? Putting these values (along with the constant) into the force equation, we get:

But this is meaningless. No one can get a good feeling for a force this small. Let’s try to imagine a situation with a force comparable to the gravitational attraction between two humans. How about this? Suppose you put a small object in your hand. You can then feel the gravitational force from the Earth on this object because your hand has to push up on it to balance the gravitational force. What mass of an object would produce an Earth-caused gravitational force that’s equal to the force between two humans? For the surface of the Earth, some of these values are always the same (the gravitational constant, the mass of the Earth and the distance to the center of the Earth). We can group all of these values into one single number.

We can call this the local-Earth gravitational constant. All you need to do is take a mass and multiply by “g” (we use lower-case “g” so that it’s not confused with the other gravitational constant “G”) and you get the gravitational force (the weight). In this instance you would need an object with a mass around 4 x 10^{-11} grams to have a weight equal to the force between two people. That’s still too small to understand. How about this? Human hair can have a linear mass density of 6.5 grams per kilometer (from this publication). That means with a piece of hair just 6 x 10^{-6} millimeters long, you would have a weight equal to the attraction between two people. That’s so crazy.

Bonus, here are my calculations if you want to change the values.

Oh, you could repeat this exact same calculation but use a known mass and solve for the mass of the Earth. This gives a value of about 5.97 x 10^{24} kilograms. But why stop there? You can also use the value of G to find the mass of the sun. I’m going to give you the short version of how this calculation works.

So, you have a planet like Mercury that orbits the sun. If you assume a circular orbit, then there is a a gravitational force on Mercury exerted by the sun.

The gravitational force makes the planet accelerate and move around in a circle (centripetal acceleration). But this centripetal acceleration depends on both the angular velocity (ω) and the orbital distance (R). Since there is only one force on the planet (the gravitational force), this will be equal to mass multiplied by acceleration to give the following relationship.

Notice that this assumes the sun is stationary—which is mostly true. The mass of the sun is ginormous compared the mass of the Mercury such that the mass of Mercury is basically irrelevant. So, solving for the sun’s mass:

Now you just need to find the orbital distance for Mercury. You can do this by starting with the radius of the Earth. Then you need to find the angular velocity—you can get this by looking at how long it takes Mercury to complete an orbit. After that, you are finished. You have the gravitational constant and you can calculate the mass of the sun. It’s amazing to think this would all start with some masses on a horizontally rotating stick—but it’s true.

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