ESCON fun theory:

How Long Would It Take to Fall Through the Earth?

I heard some people talking about the elevator scene. Here is what I gather from the plot (which could be wrong).

• There are essentially just two cities on Earth in the future.
• The only way to get from one city to the other is with a elevator that goes through the Earth.
• There is some plot point regarding the elevator – but I’m not sure what it is.
• I am pretty sure when the elevator gets to the half-way point, the people inside are weightless.

Ok, what about some physics. First, if you had a tunnel all the way through the Earth and you dropped an object, how long would it take to get to the other side? Yes, I understand that maybe this tunnel didn’t go straight through the center, but I am going to model it that way. How would you calculate this? Here (of course) is a diagram of an elevator going through the Earth (not to scale).

 If I assume there is no air for this elevator to fall through, then modeling the motion should be quite simple.

Modeling the Gravitational Force

Here are two options for the gravitational force that won’t work. First, I could use this expression for the force:

This says the gravitational force is some constant value. Of course this won’t work. Why? Well, for one thing, what would happen when you get to the center of the Earth? This says that there would still be a force. It should at least change directions after you pass through the center – I could make a modification to the expression, but that still wouldn’t be good enough. This expression for the gravitational force is an approximation for the case that an object is near the surface of the Earth. If you are in the center of the Earth, you clearly are not on the surface.

Another option would be to use the more universal expression for the gravitational force.

This says there is an attractive force between two objects that is inversely proportional to the square of the distance between their centers. We often use this force when dealing with planets and stuff. Does it work for the Earth-elevator (Earthvator)? Clearly, no. What would you use for the case when the elevator is at the center of the Earth? If you put in r = 0 meters, the above expression explodes. It literally explodes – so don’t do that.

In order to come up with a function for the gravitational force, let’s first look at a mass in the center of the Earth. What should the gravitational force be here? Well, in this case, there is mass all around it. All this mass does indeed exert a force on a separate mass in the center. If we like, we can break this Earth up into many many small spheres. Each sphere pulls on the mass in the middle, but in different directions. If the Earth’s mass is spherically symmetric, the net result would be a zero vector for the gravitational force.

This makes sense, if you place a mass in the center of the Earth (in an empty space), there shouldn’t be a gravitational force pulling it anywhere. It is already in the center.

Fine, neither of the above model works. We are just going to have to build our own model. To do that, I will start with a cheat. Let me state something and then give an example to demonstrate that it could possibly be true.

If a mass is inside a spherically symmetrical mass distribution, the net gravitational force due to this mass distribution is the zero vector. It doesn’t matter if you are in the center of this distribution or not.

Now let me demonstrate that this partially works. Suppose I have a series of small masses arranged in a circle. Since there are a finite number of masses, I can easily calculate the gravitational force at some point inside this circle. This works fairly nicely using Vpython. For my first run, I will show the forces on an object in the center of this circle.

 Here the red vector arrows represent gravitational forces from the masses in the circle that pull the center mass to the left and the yellow are for forces pulling to the right. If you added up all these gravitational forces, you would get something pretty close to the zero vector (but maybe not exactly zero since the masses aren’t perfectly spaced out).

Now, what if I move the location away from the center? Here is the same program and the same calculation for a mass off to the side a bit.

This might look like a non-zero vector force – but it is very close to zero. What you notice is the large magnitude of the yellow forces that are pulling to the right. This is because the location of the interior mass is closer to these masses on the right and thus have a greater force. However, for the forces pulling to the left (the red ones) they might be smaller in magnitude, but they are larger in quantity. If you counted, you would find 13 forces pulling to the right and 17 pulling to the left. I didn’t show an arrow for total force – it was just too small.

Yes, this calculation just shows the force on a mass due to 2-D distribution of masses in a circle. But what about a spherical distribution of masses? Well, the same concept still applies.

With that in mind, the gravitational force at some point in the center of the Earth only depends on the spherical distribution of mass that is closer to the center of the circle than the location of interest and for that mass, I can use the universal gravity model (the 1 over r squared). Here is a picture.

Putting this together with the expression for the gravitational force, I get (I am just writing the magnitude of the force):

There are two things to check with this model. First, what is the force at the center of the Earth? According to this model, it would be zero – so that is good. Second, what about at the surface of the Earth, I should get back to the m*g expression. If you put in the density and radius of the Earth into that model, you do get 9.8*m – good.

What about the density of the Earth? I could use an average density of 5.52 g/cm³ and that will probably be good enough. Really, the density of the material in the Earth increases as you get closer to the center. Wikipedia has a nice graph showing the density of the Earth as a function of radius.

You could easily make this a step type function and use that to find the mass of the “interior” part of the Earth. Maybe I will save that for a homework problem.

Modeling the Motion of a Falling Elevator

Now that I have an expression for the force, I can model the motion. One trick to do this is to notice that the gravitational force is linear. What other forces look like this? Oh, the force from a spring. This means that the “spring constant” for this case would be:

The motion of a mass on a spring is already a solved problem. We know that the period of oscillation is:

For the Earthevator, I don’t want the period of oscillation. I just want to get there – not there and back. Putting in my value for the “gravitational spring constant”, I get:

The mass of the elevator cancels – which one would kind of expect. If I put in values for G and the density, I get 2529 seconds or 42 minutes. BOOM. You knew the answer was 42, you just didn’t know the question.

Numerical Model

Now for a better answer. If I want to take into account the changing density of the Earth, I need to use a numerical model. I will use python to break the calculation into a whole bunch of small time steps. During each step, I will calculate the force based on the location of the elevator. Note: you can’t just use the same formula as the constant density calculation. Why? Because what you really need is the total mass inside of a sphere at the location of the elevator. This depends not only on the density at that location but the density all the way to the center.

Ok, here is a plot of position from the center of the Earth as a function of time for both the constant density case and a more realistic Earth-density.

From this, the constant density case gives a time of 42 minutes. With the changing density, I get a time of 32.6 minutes. Why is this one greater? Well, for the more realistic density the mass of the Earth that is still closer to the center than the elevator is much larger. That core volume with a 12,000 kg/m² density is still there for the first parts of the fall. This gives a much larger force earlier on to give a much larger increase in speed.

Here is a comparison of elevator speeds for both cases.

The first thing I noticed was the maximum speed. Even in the case of constant density, the elevator gets up to 8,000 m/s. That is super fast. Really, it is crazy to go this fast. What about air resistance? Oh sure, you could pump all of the air out of this giant elevator shaft. But what if there was air? The first question would be to get a model for the density of air. On the surface of the Earth, the density is about 1.2 kg/m³. As you know, the density of air decreases as you get higher. Of course it would have to increase as you get deeper in the Earth. It has to increase in density in order to support all of the air above it. The density would really depend on the weight of the air above it which would depend on the value of the gravitational field. Hmmmmm… an interesting homework problem. I suppose you would get a good estimate if you just used a density of 1.2 kg/m³. It would be better than nothing.

Yeah. Just turn that calculation in for homework. If you wait too long, I will probably do it myself.

Would They Be Weightless in the Middle?

Here is another scene from the movie (that I have not seen). When the elevator gets halfway on its trip to the other side of the Earth, the people become weightless and float around.  From a story line perspective, this makes sense.  If the people start on one side of the Earth, they have their feet towards the center of the Earth (we call this “down”).  Once they get to the other side of the Earth, they have to spin around to have their feet towards the center again.  There has to be some “spin around” part.  There should be some part where the gravitational force is zero and they float around.

Yes, there is a location where the gravitational force is zero (the zero vector).  However, we humans don’t really feel the gravitational force since it pulls on all parts of our bodies the same.  Instead, we feel the force of something else pushing on us.  We call this our apparent weight.  If you want more details about apparent weight, this probably goes over it in more detail than you asked for.

The correct answer is that the people in the elevator would feel weightless during the entire trip since they are in an elevator that is accelerating due to just the gravitational force.  It is interesting that this idea that they would be weightless at the “gravity flip over point” is the same idea that Jules Verne used in his novel From the Earth to the Moon.