# The Science of *Total Recall*

Like many people, I went to see *Total Recall* recently. Much of the story for this science fiction thriller (based loosely on a short story by Philip K. Dick) concerns “The Fall.” As explained in the introduction for the movie, “The Fall” is a tunnel bored through the center of the Earth that connects the fictional United Federation of Britain and The Colony, which is located in present day Australia. These are the last two habitable locations on a world destroyed by biochemical warfare. Workers commute between the two locations via this tunnel in a trip that takes 17 minutes, a commute many people today would find short.

So what can Wolfram|Alpha tell us about such a scenario? First off, we can find that the diameter of the Earth is 7,913.1 miles. 7,913.1 miles in 17 minutes indicates an average speed of 27,929 miles per hour, which is slightly more than the escape velocity for Earth.

The point of the tunnel is that gravity does the hard work of accelerating passengers from one end to the other. But what is the force of gravity as one descends into the center of the Earth? If we approximate the Earth as a solid uniform sphere of material, we derive this by starting with Newton’s law of gravitation:

Where *r* is the distance from the center of the Earth and *m* is the mass of the object falling. As we can see, the effective mass of the Earth *M* (*r*) varies with depth. From the volume of a sphere, we can determine it as:

We can approximate ρ based on our knowledge of the total mass of the Earth and its volume:

Putting this all together we find:

where *g* is the gravity felt at the surface. Thus in this approximation as one descends into the center of the earth the acceleration drops linearly rather than some sudden drop in the force felt.

We can also solve for the time it takes for one to travel through the center of the Earth. First we can look at the gravitational attraction inside a massive ball. For the inside of a homogeneous sphere, the differential equation is r”(t)=-G M/R^3 r(t). We can include the initial conditions that we start at the surface of the Earth with zero velocity (r”(t) = -G M/R^3 r(t), r(0) = R, r’(0)=0):

From this we can determine the period from the frequency of the solution, which is Sqrt[*G**M*_{earth} / *R*_{earth}^{3}]. The period is 2pi / sqrt(G Earth mass / Earth radius^3):

So this takes significantly longer than the movie version of “The Fall.”

Of course the Earth is not a homogenous ball of rock, but it is made up of several layers of differing density. This will affect the gravitational acceleration significantly. We can ask Wolfram|Alpha about these layers in general:

We see that the core is significantly denser than the outer layers. We can also look at gravity specifically at a given depth such as 3,000 miles below the surface of the Earth. From the last graph for this result, we can see that gravity, rather than falling linearly as our calculations suggest, actually remains steady and even rises until one hits the outer core.

One consequence of this is that since you accelerate faster for longer, the maximum velocity is higher than one would expect from a uniform sphere. But how long does it take?

By extracting the data points from the plot using *Mathematica*, we can find out.

After interpolating over the points, we can construct an equation based on the relationship between acceleration and velocity.

We integrate over both sides, and from that we can extract the time.

But this is the time to fall from the surface to the center of the Earth. For the full fall we need to double this, since our upward velocity is slowed down by the increasing pull of gravity. Two times 1384.12 seconds is just over 46 minutes, or a bit shorter than an average college lecture. So it is not the worst commute in the world, but definitely not *Total Recall*‘s 17 minutes.

So, if the time doesn’t work out for our trip from Great Britain to Australia, how about the tunnel’s route? For this we need to check that they are antipodes, that is, that they are directly on the opposite sides of the Earth. Unfortunately the antipode for the United Kingdom is on the opposite side of New Zealand from Australia and the anitpode for Australia is in the mid Atlantic. Thus the tunnel must deviate significantly from the center of the planet.

Because of this, the passengers would never be subjected to zero acceleration from gravity. On the other hand, as the crew of the International Space Station knows, just because you are subject to gravity doesn’t mean you can’t experience weightlessness. The astronauts are subject to gravity just slightly less than us on the surface, but because they and their surroundings are constantly falling (since they are in orbit), they feel weightless. Our hypothetical commuters on “The Fall” would feel the same effects for their entire trip, as long as the device they were riding wasn’t accelerated by outside forces such as the clamps that brought them to a stop. You see the same effect on the steep drop of a roller coaster until something (in this case, the rails at the bottom of the drop) returns the sensation of having weight.

Cool!

Hope you will discuss the next Bond movie too!!!

Interesting! I would assume the 17 minutes could be reached if there is an engine that would increase speed on the way down and reduce it on the way up. Say 1g both ways. The capsule could start off in free fall, rotate the passengers with their head down, start power – the passengers would then feel normal gravity. As the capsule approaches the core, stop power and rotate passengers. After passing the core start slowing down with 1g and when close to the exit stop power, rotate passengers for a smooth “landing”. This is however not consistent with what is shown in the movie.

Two other problems though: the tunnel would have to be emptied of air (the pressure would be higher in the deep, creating huge friction at those speeds) and the Coriolis effect.

Hey thanks for this analysis. I left the movie annoyed that there was apparently normal gravity the whole time except 0 for what seemed like 30-60 seconds in the middle. I just wish movies would get the physics right! It is in fact quite interesting to deal with reality. Assuming you could build such a tunnel, they would be nearly weightless the whole time, so it’d be like an action scene on the moon or something.

Jason, I don’t understand the connection between the figures 3188.02 and 1384.12 in your post. Earlier, you showed that free fall through an Earth of constant density would take about 42 mins. But you then say that the time to free fall through an Earth with theoretical density distribution would take about 46 mins, whereas I calculate it to be about 38 mins. Using 5g, it is easy to show that the time is about 17 mins, using simple kinematic formulae, or by integration, as per the movie (see the postings on the IMDB website under Total Recall 2012, “The Fall”).

Hi, I’m late to the party but just watched the film today.

I believe you’re right about the weightlessness during the first half of the trip (you’re in free-fall), but I’m pretty sure you’d experience weight on the way up, as a passenger and carriage would be travelling upwards but the gravity pulling downwards.

I assume/believe this would be a continually rising resultant g from zero (weightless at the center) to 1g once you reach the surface. (again, no sudden change)

Also, what the heck is up with the magnets in that film??? :-D

Alex – I think you’re right about gravity, but not the perception of weight. Since you and the whole train are always accelerating due to the same forc of gravity, you and the train are always moving the same speed, and are actually ‘falling’ the whole way; the very definition of free fall. This is analogous to the astrounaut trainer at Nasa (google ‘Vomit Comet’) and would only feel weight if another force besides gravitational acceleration was acting on you. The movie is actually more believable after coming here – the only way they could reduce the time to 17 min would be to be constantly be accelerating the train (in addition to gravity) towards the core – that would give the conditions seen here (always normal gravity except when passing the core, as there the engine providing the extra acceleration would be turned off.

I also notice that the path of the fall is unlikely to go through the center of the earth. Australia is too far to the East. The clostest possible land to land diameter might be from Ireland to New Zealand.

How is it possible for the acceleration to be everywhere larger than in the uniform case, but the travel time to be longer? I ran a calculation myself using a realistic earth model and came up with just over 38 minutes.

Is it really true that v = {dr \over dt} = t g{r) ? The first part is trivially true, but the last strikes me as very dubious.

My calculations and reasoning agree with Steve B and John Thorstensen that the travel time through an Earth with realistic density is about 38 minutes (the calculation that leads to 46 minutes is hard to follow) and the travel time through an Earth of constant average density would be about 42 minutes. Your 84 minutes assume a full cycle of your cosine function wheres only half a cycle is needed to go from one end (+R) of the Earth to the other (-R).