Thursday, October 22, 2015

How long would it take to reach the other side of the world?

joão pestana

Imagine that there was a straight tunnel passing right through the center of the Earth and it was large enough for you to fit in. For how long would you fall if you jumped right into it? The answer I'll provide might surprise you and I'll try to keep it as simpler as possible, but not simpler.

What are the ingredients we need to cook up our answer? The most important one is imagination followed closely by intuition and, at a close third, some mathematics. Let us assume there is no air resistance, just to make things easier. The more real situation just implies a terminal velocity that would make the trip longer.

It might seem plausible to expect that one accelerates until reaching the core of the planet and then decelerates for the rest of the journey, coming to a halt at the very edge of the other side of the planet. In this scenario, the highest value for the velocity would be achieved at the very center when the acceleration drops to zero. Doesn't this description resemble that of an oscillating spring?

We're going to need a few tools from the trade such as Sir Isaac Newton's law of universal gravitation which states the following:
Any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
What does this mean? It just says that if any object has a mass, than that object attracts other objects that have mass as well and that the closer they are to each other, the greater their attraction becomes. For example, when our moon was formed it used to be much, much closer to the Earth than it is now. Not only it appeared bigger and brighter in the sky, but the tides that it caused on our planet were enormous. Since then it has been slowly drifting away. The tides run shorter, the days grow larger and the skies become dimmer. This is all due to the fact that distance plays a very important role in the force of gravity — even bigger that the mass of the objects!

If the mass of one object doubles then the gravitational force that that object exerts over others doubles as well — this means it's directly proportional —, but if the distance between both objects doubles then the force of attraction does not fall to a half... It falls to a quarter of what it was initially! This is what inversely proportional to the square of the distance means.

So far, we've been describing a physics problem without recurring to mathematics. Unfortunately, we cannot go on like that ad aeternum. Let's us start by simply writing down Newton's law as many will easily recognize:


The force `F` increases as any of the masses `M` (the Earth) or `m` (you) increases and decreases as the distance `r` increases. `G` is just some number that makes everything come out right in the units that we use daily to measure things such as distances and masses. Of course — and this is important — if we can write down the mass of an object, we can also define it's density `rho` as the ration between its mass and its volume `V` such that `M = rho V`.

Why is this important? Because one of our biggest assumptions in order to solve this problem is that the density of Earth is constant — which is not. Can you remember how to calculate the Volume of a sphere? We're also assuming the Earth to be perfectly spherical — which again it's not.

`V_(sphere) = 4/3 pi r^3`

The next ingredient that we need also comes from Newton and is his second law of motion which was translated in 1729 by Andrew Motte as:
If a force generates a motion, a double force will generate double the motion, a triple force triple the motion, whether that force be impressed altogether and at once, or gradually and successively. And this motion (being always directed the same way with the generating force), if the body moved before, is added to or subtracted from the former motion, according as they directly conspire with or are directly contrary to each other; or obliquely joined, when they are oblique, so as to produce a new motion compounded from the determination of both.
Imagine that you and your friends form a circle around a table and there is a blanket on top of it. It's very cold and everyone wants the blanket for oneself so every person starts to pull it their way. Nobody moves, because everyone is pulling in opposing directions and cancel each other out. Imagine now that your car won't start and you and your friends have to push it. You can only do this if everyone pushes in the same direction. Essentially, this is what Newton's second law of motion states — a change in the motion of an object is the result of all the combined `N` forces that act upon it within a given interval of time.

`sum_(i=1)^N F_i Delta t = Delta p`

A change in motion can also be stated as a change `Delta` in the momentum `p` of an object. That is, the will of an object to keep moving. Imagine a big truck that is at rest (not moving). Its will to move is none. Unless something acts upon it, it will remain at rest — Newton's first law of motion. If, on the other hand, it's going at a great velocity, then its will to keep moving is very large and it'll be very hard for you to force it to come to a halt. This is not only due to its velocity `v`, but also to its incredible mass `m`.

The momentum of an object is proportional to both such that `p=mv`. Since the mass of an object can be considered constant, this implies that a change in momentum is due to a change in velocity or `Delta p = m Delta v`. The change in the velocity is what we usually call the acceleration `a`. Since there is only one force on our problem — that of Earth's gravity — we can put it simply as:

`F_(gravity) = G(Mm)/r^2 = ma`

Notice that the force of Earth's gravity is acting upon you and that force is the only source of change in your velocity — your acceleration. If you recall basic equation mathematics, you can cancel all the common terms that appear on both sides. In this case, we can cancel your mass — yes, the result is independent of it! This means that the travel time from one side of the planet to the other is the same for every single person. We're left with an expression for the acceleration that the Earth causes upon you:

`a = G M_(Earth)/r^2`

I'm now going to move on with the mathematics and feel free to skip this part if you're not inclined to follow through. It's actually not very important. We just have to replace Earth's mass with the statement that defines it as the product of its density and its volume.

`a = G (rho V_(Earth))/r^2 = G 4/3 (rho pi r^3)/r^2 = 4/3 G rho pi r`

I should mention in case you're wondering that `r` is the distance between you and the center of the planet and is our variable. It changes as you fall through the tunnel. We can do this, because it was also shown by Newton that any object with mass can be modelled as a single point with the same mass, placed at its center of gravity.

We shall also consider that same center as our point of reference which means that when you fall towards it, you go from a higher point to a lower one — implying that the change in your position is negative. That is, your velocity is negative and thus your acceleration is negative as well. Don't worry if you don't understand and just ignore the minus sign I'm going to use from now on.

The rest follows from basic calculus — we need to solve a differential equation. From basic physics, we know that acceleration is a change in velocity which is itself a change in motion. If we consider a straight line along the tunnel as the `x` axis, we can write the above expression in terms of the position along that line as:

`a = (Delta v ) / (Delta t) = (Delta ( Delta x ) ) / ( Delta ( Delta t ) ) = - 4/3 G rho pi x`

Let us just call `k^2` to the product of all that is constant and we make the expression much nicer. If `k^2=4/3 G rho pi` we can call upon our differential language and write everything simply as:

`(d^2x)/dt^2 = - k^2 x`

Which you might recognize as the equation that describes the harmonic oscillator. Your intuition that you would just bounce back and forth along the tunnel was right after all! The solution for that equation is an oscillating one — any sine, cosine or a combination of the two will do. Since we considered the center as our reference, we must use the cosine, because it starts at 1 and halfway through its period intercepts the zero point. Your position as you fall through the tunnel is given by:

`x(t) = R cos(kt)`

Where `R` is the radius of the Earth — your starting point relative to its center. How do we get to the travel time? We've all that we need right now. We just have to find the time `t` for which `x(t)` is zero and double that, because from `R` to `0` is just halfway through your epic journey!

If you don't recall your trigonometry classes, a cosine function is null when its angle is a quarter of a circle — that is, `90` degrees or `pi/2` radians. We simply need to make sure that its argument is equal to that value:

`kt = pi/2`

After we solve for `t` and properly replace `k` we get what we wanted all along. And remember that `k^2` was the product of the constants, so `k` is the square root of that.

`t = pi/(2k) = pi/2 sqrt(3/(4 pi rho G )) = sqrt((3pi)/(16 rho G))`

Since you're asking for actual values, we must lookup the experimental vales for the Earth's density `rho_("average")=5514   kgm^(-3)` and the gravitational constant `G=6.67408(31)xx10^-11   Nm^2kg^(-2)`. The answer is `t=1265` seconds which translates to 21 minutes and some milliseconds. It would take you that long to reach the core of our planet. If we double that amount, we know that it would take you 42 minutes to reach the other side of the world.

On the American Journal of Physics 83, 231 (2015), Alexander R. Klotz wrote an article entitled The gravity tunnel in a non-uniform Earth. He uses a more realistic model which takes into consideration the non-constant density of the Earth as well as the fact that it's not perfectly spherical. His result is an interval between 38 and 42 minutes for any given point on the surface of our planet.

Now I know that I quoted Douglas Adams in my previous post and that the final answer here is 42, but there is no correlation whatsoever. Or is it?