weighing the earth

Essentially Cavendish used one mass to weigh another and get $G$. But we weigh masses every time we go to buy fish. We're using the earth to weigh the fish. Could we use this to get the mass of the earth? But wait, there are a couple of things I should say to make this clearer.

It is by no means obvious that the gravity that we feel sitting quite confortably in our chairs is related to the forces that drive the planets around the sun. It was quite bold for Newton to assume it was the same force. All the particles in the earth are pulling on you, keeping you in your seat. Atoms that make up the core of the earth, a sea slug off the coast of Tasmania, an old smelly sock sitting in some dorm room in Amherst. All these atoms add up using eqn. 1.1 to give you total gravitational force acting on you. But what will that be? Clearly the earth is not a point particle, and the law of gravity stated above only applies to point particles.

Using some math, one can show, as Newton did, a quite amazing result. The force between a uniform sphere and a point object is the same as if all the mass of the sphere was concentrated at its center! The easiest way to show this is to use something called ``Gauss's Law" which you'll learn about when you study electricity and magnetism. You can either wait until then, or show it the brute force way, by doing a some multiple integration.

So let's get back to the mass of the earth. We know that when you weigh a fish of mass $m$, the force the earth exerts on the fish is $mg$. But from what we've just said, the law of gravity gives us the same thing a different way

$$F ~=~ G {M_em\over R^2}$$ (1.3)

Where $M_e$ is the mass of the earth, that we're concentrating at the earth's center. Then $R$, is the distance between the fish and the center of the earth. Unless your fish is horribly confused, $R$ will be the radius of the earth. So now we can solve for the mass of the earth

$$M_e ~=~ {F R^2\over G m} ~=~ {mg R^2\over G m} ~=~ {g R^2\over G }$$ (1.4)

But we know from various means what the radius of the earth is. It's about $6.37\times 10^{6}m$. Taking $g ~=~ 9.8 m/s^2$, we get the mass of the earth is about $6 \times 10^{24} kg$. That's a big number even in a state like Texas.

From this we can calculate the average density of the earth since the volume of the earth is $V ~=~ {4\over 3}\pi R^3$. It's

$$\rho ~=~ {M_e\over V} \approx 5.5 {g\over cm^3}$$ (1.5)

That's about twice the density of rock. So that gives you the idea that things underneath us are under an awful lot of pressure!

Now let's see if we can understand Kepler's laws