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/cm
3
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.
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