A gravity train is such a (theoretical) form of transportation, that takes advantage of the Earth's natural gravitational pull downward to accelerate into the planet, and that also takes advantage of that same pull to slow down on approach to the destination. Ideally, the train would move through a frictionless environment, so that it doesn't have to provide its own thrust. While such a means of transportation is very likely not feasible because of the amount of energy needed to maintain a several thousand kilometer-long vacuum tube (and electromagnetic rail for levitating, so on), we can still model how it would behave if it existed.
I'll walk through the equations governing the train's motion and how I attempted to model said motion using the math-based coding language MatLab, and also explore some possible density distributions for a given planet and how that affects the travel time for the train.
The math
A gravity train will follow a path through a planet at an angle relative to the planet's centerline (a line through the planet that intersects with the center of the planet. For simplicity's sake, we will assume that the planet that we're working with has a mass and size similar to Earth's (mass of 5.974*10^24 kilograms, radius of 6371 kilometers), is a sphere, and has a density distribution that is more or less constant across large layers of the planet. For a brief sketch of what our problem looks like:
The path that the train takes along the main diagonal, at an angle theta from the centerline, we will call length D. The distance from the train to the center of the planet will be r, and the radius of the planet as a whole will be R.
If we want to use a program to solve for the time that it takes to travel through the planet at a given angle theta, we can use an Euler step method to estimate either the distance after each time step; or the velocity after each time step; or the acceleration after each time step, etc. What I found useful was to use an Euler step to estimate the velocity at each time step; since an Euler step is defined as
I could use the fact that v' is equivalent to acceleration to solve the equation at each point. The force experienced by a pair of masses due to gravity is:
and force is also equal to mass*acceleration, we can simplify the expression and find that acceleration is a function of the mass of the planet and the distance from the center of the planet:
But, what happens as the train moves through the planet? How does the effective mass of the planet appear to change? Going with the assumption that the planet effectively has a uniform density distribution for a given layer, we can use the pair of facts that outside of a uniform shell, the gravitational force exerted by the shell is equivalent to the gravitational force exerted by a point mass – of the same mass as the shell – whose location is the same as the shell's center; and that inside such a uniform shell, the gravitational force exerted by all of the matter in the shell cancels to net zero. So, when the train moves through the planet, it does not feel any force on it from the mass in the planet that lies outside of the train's radius toward the planet center. The amount of mass that gets ignored is a function of the distance traveled so far (this helps determine the volume of the exterior shell) and the density at each shell passed (which is a function of radius).
Since we will determine the density distribution ourselves though, and since the density could vary from one time step to another, we will not explicitly include it in our equation, which to this point has become:
The next trick is to figure out what the distance from the train to the center of the planet, r, is. For this, we'll need to know how far the train has traveled so far, which we will call d, and will have to use the law of cosines with the uppermost triangle in the first figure, whose vertices are the train, the starting point of the train, and the center of the planet:
What more, we need to take into account the fact that the planet's pull toward its center is not impacting the train in full. Because the train is only moving along an angle to the center of the planet, it will feel a component of the gravitational force that points in the direction of the track it sits on/levitates over. We just solved for r, and can use the law of sines this time to solve for the component of the gravitational force that points along the direction of the track. To keep the equation condense, I'll keep to using r instead of its expanded equivalent:
And – finally! – we need to modify d after each step. Since velocity*time is equivalent to distance, we can estimate the total distance traveled with a similar Euler step, but with a slight change:
Testing the model
The simplest setup for our planet is to have a uniform density throughout. For a planet similar in mass and size to Earth, this would be a mean density of about 5150 kg/m^3.
There are a couple other simplifications that can be made as well that do not depend on density distribution. One is that since the density distribution will be even throughout any given layer, there is rotational symmetry from any point on the planet, such that an angle of descent of, say, 20˚ will result in a travel time the same as a descent of the same angle from any other point on the planet. So, if we check angles between 0˚ and 90˚, we do not need to check angles outside of that range.
The second is due again to the evenness of the density distribution: the trip that a train takes to the center of its path is analogous to the trip it takes out toward its destination. That is, the second half of its journey is a reflection of its first half, in time and in speed and in distance. So, stopping the train after it crosses the halfway point (and appropriately taking into account any factor of 2 that needs to be applied) will cut down on the computations required without losing information.
With these simplifications, the model run: Wikipedia says that the time that it takes for a gravity train to travel through such a planet described above is a little over 42 minutes, no matter the angle of descent. Here is the result from my model (this graph is in polar coordinates):
The exact values may be difficult to see, but the time to travel though the close-to-centerline path is 42.27 minutes, which matches what Wikipedia gave; the times for the rest of the angles are appreciably higher though, and converge to almost exactly 50 minutes for small angles of descent. I do not know if this means that Wikipedia is wrong, or if this means that my model is not taking into account some other factors at low angles. It does not appear to be an issue with choice of time step either, since the results do not very noticeably change across about 3 orders of magnitude (0.1 second time steps to 10 second time steps). The similarities, however, are somewhat encouraging.
This is not Earth.
The Earth is not uniformly distributed, however. We can look to estimates of the Earth's density profile from Jordan, T.H. (1979):
0–35 kilometers: ~2550 kg/m^3 (crust)
35–2890 km: ~4500 kg/m^3 (mantle)
2890–5100 km: ~11,000 kg/m^3 (outer core)
5100–6371 km: ~13,000 kg/m^3 (inner core)
Using such a distribution, our travel-time graph changes to:
The time that it takes to travel along a path at a low angle of descent (here shown as an angle of ~90˚!) is not appreciably changed, which is what we expect if, like with the previous scenario, we're choosing a path that does not cut through (and thus exclude gravitationally) a lot of mass. In general however, there appears to be a shift toward less travel-time for all angles than in the constant-density scenario. The time it takes to travel through the core is about 39.2 minutes, 3 minutes less than before.
Varying density
An Earth-like distribution of mass allowed the gravity train to fall through the planet at a faster rate than a constant-density mass distribution did. What if density were concentrated more toward the crust? Toward the mantle? Even more to the core?
Let's break up the planet into 4 layers of constant thickness: we'll have a "crust" of some density that extends for about 1580 kilometers, a "mantle" that extends for the next 1580 kilometers, and an inner core of the same thickness and a core with a radius of 1580 kilometers. Each layer will in its turn contain 90% of the mass of the planet, while the remaining layers split the remaining mass evenly (this means they will not have the same density).
Graphing them as above, with 90% of the mass being with:
dark blue = inner core
red = outer core
green = mantle
cyan = crust
we obtain:
From this it seems that the greater the mass distribution toward the core, the faster that the gravity train will travel; and, apparently if mass is concentrated to the surface, it will actually take more time to travel through the planet at a relatively steep angle of descent than it would take to travel near the surface.
The first observation is obvious given what we know about the gravitational pull from shells of constant density: if we stay outside of the shells of mass, we will still be accelerated by them. The more mass closer to the core, the longer we will stay outside of the interior of that mass, and so the longer we are pulled by its gravity.
The second observation is not so obvious. If, for instance, we put all of the mass in a small (but not arbitrarily thin) layer close to the surface, and had a hollow interior, we would accelerate while we are within that layer but then continue at a constant pace afterward, while in the interior. There would be no gravity to give us a push in any direction.
Two opposing effects can be posited, one being that if you enter the planet at a shallower angle, you will take longer to accelerate, and take longer to get into the interior of the planet. On the other hand, if you enter the planet at a steep angle of descent, you would be traveling at a constant speed (not accelerating) for a much longer distance. Which contributes to a longer travel time is not clear to me from first principles, though the model appears to show that the longer distance traveled with a steep descent contributes more. I am a bit skeptical to conclude that's absolutely the case, mainly because my model shows a non-constant travel-time for a constant density throughout the planet, but the way my model's "bias" ran was in adding more time to shallow angles of descent; so, it seems reasonable to conclude the model isn't biasing this particular result, since it's the opposite observation.
(It is entirely possible the apparent decrease in travel time for the cyan graph as the angle of descent becomes very steep is due to this potential bias.)
Refining, more details soon
I somewhat glossed over the trigonometry in describing the component forces and radial distance from the train to the center of the planet, and will follow up with another post detailing those in a few days when I have more time to sit down and draw out those sketches (similar to the first one).
There are potentially some errors in how I'm treating the forces on the train or in how I'm telling the model to stop running, and while the results make intuitive comparative sense between all of the individual runs, the nominal values might best be taken with a grain of salt. I'll be seeing if some of my peers who are slightly more familiar with the fundamental gravity train problem can correct any mistakes I made.
But at this point: given the odds of a gravity train ever being built, this is probably not a very practical problem to solve! However, there are some interesting observations here, namely from the final graph, namely that while from a distance two planets may interact the same with an object on a static track (our train, for example), inside the planets the interactions can be very different, to the point of either increasing or decreasing said object's net travel time, relative to travel times at other angles of descent.
It seems not too early to also lend support in a numerical sense to the estimate given that to travel through a planet of constant density – of Earth's mass and size – would take roughly 42 minutes is true. It certainly appears to be true for steep angles of descent, whether or not it is true for shallow angles; and whereas the 42 minute estimate is derived analytically, its match with the numerical solution is in part a testament to the physics of this model.
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