### How Dense?

In the last post I talked about the physics and considerations behind the gravity train. A crucial part of the calculations involves the average Earth density. This is what determines the period, or how fast the train will go. It also determines the pressure our tunnel must sustain, and the amount of work we must perform in order to mine out rocks. As stated, the average Earth density is about 5515 [kg/m3], an-
Stop right there! How do you know that the average density is 5515 [kg/m3]?
You know, you could have at least waited until I finished the sentence.
Yes, but I want to know now.
Well, finding Earth’s density can be either difficult, or easy, depending on what you want to find. Creating a 3d map which tells us for each point within Earth its density is no easy feat at all – we have no way to directly measure the material that is underground. What’s more, the core and mantle are part liquid and tend to move with time. But we can do is speculate on the nature of the materials there, record seismic activity, and measure micrograviational changes on the tug of a satellite. All these give us a good approximation to the density of the matter beneath us.
Of course, we might want just the average density, disregarding internal changes, and then we can take Earth’s mass and divide it by its volume.

$\rho = \dfrac{M}{V}$

Earth’s mass is 5.9722 x 1024 kilograms. Its volume is 1.08321 x 1021 meters cubed. Dividing, we get 5513 [kg/m3], which we rounded to 5515 in our calculations.
But how do you know that Earth’s volume is 1.08321 x 1021 meters cubed?
In everyday life, the classical way of measuring an object’s volume is to submerge it in water and see by how much the water level rises. We’d have a hard time doing this for Earth, for many, many reasons. So we have to use a theoretical model. The easiest way is to look at Earth as a sphere. In this case, we have the formula:

$V = \dfrac{4}{3} \pi R ^ 3$

If we want something more accurate, we notice that Earth is more of an ellipsoid: the spin around its axis causes it to bulge at the equator a bit. In this case,

$V = \dfrac{4}{3} \pi a b c$

Where a,b,c are the axes of the ellipsoid. We know that the radius at the equator is Re = 6378.1 km, while the polar radius is Rp = 6356.8 km. Putting a = Re, b = Re, c = Rp, we get the proper result.
But how do you know that Re is 6478.1 km, and Rp is 6356.8 km?
There are many ways of measuring Earth’s radius. The earliest one was well over 2000 years ago by the Greek Eratosthenes. He assumed that Earth is a sphere, and used the the differences in shadow angle in two different places at the same time. Similar inaccurate measurements can be done using only a stopwatch.
More accurate measurements can be made with satellites – you send a signal from the surface to a satellite, and see how long it takes for it to get there (or to get back). Aka , GPS. Do this for two opposite sides of Earth, and you can get the distance between two opposite points, which gives you the radius.
You can also measure the time delays between measurements in different parts of the Earth of distant quasars (VLBI).
But how do you know that Earth’s mass is 5.9722 x 1024 kg?
We can approximate the mass using two of Newton’s laws. The first is that every two objects attract each other gravitationally, according to:

$F = G \dfrac{mM}{r^2}$

Where G is the gravitational constant, M and m are the masses of the two objects, and r is the distance between them. The second links between the forces acting on a body and the resulting acceleration:

$F = ma$

We take the two of them together for, say, a cat that we drop and let fall to the ground. We replace the F’s:

$G \dfrac{mM}{r^2} = ma$

Here, m is the mass of the cat, M is Earth’s mass, and r is the distance between Earth to cat (equivalent to Earth’s radius). Rewriting:

$M = \dfrac{ar^2}{G}$

We already found out what r is – let’s suppose we drop the cat at the north pole, so we get r = 6356.8 km (and a very cold cat). G is about 6.673 x 10-11, and the acceleration a is 9.832 [m/s2]. Putting all that together, we get… M = 5.953 x 1024 kg. Note that this is not the mass I cited (which I got from a NASA fact sheet), although it’s quite close to it. I suspect that the difference comes from the treatment of Earth as a sphere, while in fact it is much more of an an ellipsoid. The mass we got in the calculation goes larger as the radius gets larger; measuring the pole in a spherical model “undershoots” and uses too small a radius: we are farther away from the parts that bulge at the equator, than we would have been had those parts not bulged, and thus require more mass to compensate. This seems like a plausible explanation, especially since if we take the radius and acceleration that are proper for the equator, we get a mass higher than the one cited (the opposite bulging effect). I haven’t done the calculations to see if this is indeed the case, though. As a funny side note, it seems that even NASA got a bit confused, as the mass data in one fact sheet contradicts the data in another fact sheet
But how do you know that G is 6.673 x 10-11?
This is a fundamental property of the universe. There is currently no theory which gives us G’s value in advance. Thus, we have to measure it. If we place two known masses at a known distance away from each other, we can measure the force between them. The problem is, as you can see, G is very small, so we have to think of clever ways to measure such a minute effect. Henry Cavendish did this in 1798, by looking at how balls of lead turn on a torsion balance. You can find a good physical explanation here. We have progressed quite a bit in terms of technology and precision since 1798, but even today the methods used don’t always deviate far from the original experiment.
But how do you know that the acceleration is 9.832 [m/s2] at the north pole?
This is not a fundamental property of the universe, but you still have to go and measure it. One way would be to drop a cat from a known height and measure when it hits the ground. If it moves due to gravity alone, then its height as a function of time is:

$h = \dfrac{1}{2}gt^2$

And therefore:

$g = \dfrac{2h}{t^2}$

Of course, that’s neglecting air resistance, but: for small distances and velocities, it doesn’t make a lot of difference; we know how drag affects motion, and can put that in the equation as well; we can drop the cat in a vacuum.
Note that if we were to do this in the equator, or any other region apart from the poles, we would also have to consider the centrifugal force caused by Earth’s spinning: the acceleration is caused both by gravity and the rotational force. But that’s not so hard: we already know Earth’s radius, and figuring out its angular velocity is only a matter of looking up at the stars and seeing how often they pass overhead.
Is this the only way to discover Earth’s mass?
Of course not! While G and R are essential, there are other things you could do. If you didn’t want to use the acceleration of falling objects, you could, for example, put a satellite into orbit around Earth. According to Kepler and Newton, there is a relationship between the distance of the satellite from Earth’s center and its orbit period:

$T^2 = \dfrac{4\pi}{GM}a^3$

Where a is the semi-major axis of the ellipse in which the satellite orbits Earth. We already know Earth’s radius, so the semi-major axis can be found by measuring how far away the satellite is (you can use simple trigonometry or radar for this).
Putting putting a satellite into orbit without knowing g in advance is not recommended; you’re going to have to guess how much fuel you need to put in… Luckily, Earth has a natural satellite of its own – the Moon…

What I like about all this reasoning, is that it shows how science is built upon itself. Everything eventually relies on measurements and observations, and the mathematical theories which glue all the pieces together. In this case, the density of Earth lies on very solid foundations.
Further, throughout history, we have always advanced our methods of obtaining these measurements and increased our precision. This allows us to better use them in engineering applications.