The Physics of Super Mario Galaxy

For the last week I’ve been playing this game called “Super Mario Galaxy”, and if it isn’t the best game ever made, then it’s in the top three. It has fanservice by the buckets, and a soundtrack that soars as if the game itself is just happy it is being played, and little blue tractor beam stars that sing you sad little theremin songs, and evil hats. Things happen for no reason, like they used to in old NES video games, and you don’t care why, you just love it. I don’t even know the words to express how much fun I’ve been having with this game.

So instead of trying, I’m going to instead write about the physics problems the game poses.

The gimmick in Super Mario Galaxy is that where previous Mario games had you jumping between platforms floating in the air in the Mushroom Kingdom, the new one has you jumping between little bitty planetoids floating out in space. It works like this:

You get the idea pretty quickly. So here’s what I wonder:

Most of Mario Galaxy is spent running around on the surfaces of those little asteroids, like you see on the video. The asteroids vary in size and shape, although most are roughly spherical and on average each asteroid is about the size of a Starbucks franchise. Every single one of them, regardless of size or shape, has completely normal earth gravity. And it’s hard not to think, when you see Mario taking an especially long jump on a small asteroid and landing about halfway around the planet, that he kinda looks like he’s having a lot of fun. So what I want to know is, is any of this possible in real life?

Would it be physically possible– assuming that you have more or less unlimited resources and futuristic engineering, but are still bound by the laws of known physics– to actually build a bunch of little asteroids like this, with compact volume but earth gravity? Say, if you’re a ringworld engineer or a giant turtle with magic powers. Could you build the Mario Galaxy universe?

Here’s what I worked out:

So first off, clearly we’re not going to be able to do most of the things in the Mario Galaxy universe, since SMG is of course a game and obviously they weren’t trying to be realistic or imitate anything like real physics. I’m going to just ignore these things– for example, I’m ignoring anywhere where there’s just a vanilla gravity field pointing in some direction, as if generated by a machine. As far as we know, that’s just not possible. In the physics we know, the only way to create a gravity field is to put a bunch of mass in the place that you want the gravity to point toward.

This isn’t so bad, since a lot of the planetoids in Galaxy look like this might well be what they are– just a big lump of mass creating a gravity well. In fact, some of the planets– for example the one at the start of the video above– are actually shown to just be thin hollow shells with a black hole at the center. So, in the post that follows, I show what you’d have to do to build one of these planets. I’m going to consider just a single example planet, of about the size of the one from the video above; you’d have to use different masses and densities for each planet in the game, since each has a different size but the exact same amount of gravity at the surface.

ANALYSIS

The first question to ask here is, how much mass do we need? Well, the planet in the video, eyeballing it, looks like it’s about as wide as a 747 is long. Wikipedia says a 747 is 230 feet long.

Newton’s formula for gravity is m_1*a = G * m_1 * m_2 / r^2, where I take m_1 as the mass of Mario and m_2 is the mass of the planetoid. Solving for m_2, I get:

m_2 = a * r^2 / G

r is 115 feet (half a 747), and a is earth gravity, 9.8 m/s/s.

Plugging in to google calculator, I get:

((9.8 m (s^(-2))) * ((115 ft)^2)) / G = 1.8043906 × 10^14 kilograms

So, if you want to get earth-like gravity on the surface of a 230-foot-diameter sphere, you’re going to need about 1.8 * 10^14 kg of mass. This is not that much! Checking Wikipedia I find even the smallest moons and dwarf planets in the solar system get up to about 10^20 kg. in fact, 2 * 10^14 kg is just about exactly the mass of Halley’s Comet. This sounds attainable; other characters are shown hijacking comets for various purposes elsewhere in Mario Galaxy, so no one will notice a few missing.

Meanwhile, Wikipedia’s formula for escape velocity for a planet ( sqrt( 2GM / r ) ) tells us that escape velocity from this planetoid will only be about:

sqrt((2 * G * (1.8043906 × (10^14) kilograms)) / (115 feet)) = 26.2110511 m / s

Which is about 60 MPH. So the cannon stars Mario uses in the video to move from planet to planet should work just fine. (Although platforms might not work so well, at least not tall ones; since the planetoid is so small, you’d only have to get about 45 or 50 feet up off the ground before the amplitude of gravity is halved. Actually gravity will fall off so quickly on the planetoid that there would be a noticeable gravity differential between your feet and your head: a six-foot-tall person standing on it would experience about 1 m/s^2 more acceleration on their feet than their head. So expect some mild discomfort.)

At this point the only question is: If one were to attempt to fit Halley’s Comet into a 230-foot-diameter sphere, would this turn out to be impossible for any reason? In answering, I’m going to consider two cases: One where the mass is in a black hole at the center of the sphere; and one where the mass is distributed through the sphere evenly. In both I’ll try to see if anything really bad happens.

So, first, the black hole case. Looking here I’m told that the event horizon of a mature black hole should be equal to G * M / c^2, and Wikipedia tells me that the Schwarzschild radius (the radius you have to pack a given mass into before the black hole starts to form on its own) is about twice that. So plugging this in, I find that the radius of this black hole at the center of the sphere is going to be:

(G * (1.8043906 × (10^14) kilograms)) / (c^2) = 1.33970838 × 10^-13 meters

… uh oh! Now we seem to be running into trouble. If someone wanted to construct a black hole with the mass of Haley’s comet, they’d somehow have to pack the mass of the whole thing into… well, google claims the diameter of a gold atom is 0.288 nanometers, so… about one-hundredth of the diameter of a gold atom?! That doesn’t sound very feasible.

On the other hand, considering the case where the sphere is solid, one finds that the required density is going to be about:

(1.8043906 × (10^14) * kilograms) / ((4 / 3) * pi * ((115 feet)^3)) = 1.00023843 × 10^9 kg / m^3

As remarkable coincidence would have it, 10^9 kg/m^3 is, according to wikipedia, exactly the density of a white dwarf star, or the crust of a neutron star! So this is sounding WAY easier than the black hole plan: All you have to do is go in and chip off 180,000 cubic feet of the crust of a neutron star, and you’ve got your planetoid right there.

Of course, there’s still one more step. First off, I’m not sure what the temperature of a block of white dwarf matter that size would be, but you might not want to actually walk on it. For another thing, the reason why a white dwarf or the crust of a neutron star has that much density in the first place is that all that matter is being held in place by the incredible gravitational weight of the star the matter is attached to– your average white dwarf is about the size of Earth, which by star standards is tiny, but which compared to our little Mario Galaxy planet is rather large. So if you tore off a 747-sized chunk of one of these stars and just dumped it in space, it would almost certainly explode or expand enormously or something, because the chunk’s gravitational pull on itself would probably not be sufficient to hold it together at the white dwarf density. So we’re going to need some kind of a shell, to hold the white dwarf matter in place and (one assumes) trap things like heat and radiation inside.

When you get to the density a white dwarf is at, the main thing you have to worry about is what’s called degeneracy pressure. Usually, when you try to compress matter, the resistance you meet is due to some force or other– the force holding atoms in a solid apart, for example, or the accumulated force of marauding gas molecules striking the edge of their container. When you get to anything as dense as a white dwarf, though, the particles are all basically touching (“degenerate”), and the main thing preventing you from compressing any further is literally just the physical law that prevents any two particles from being in the same place at the same time, the Pauli Exclusion Principle. If you try to compress past that point, a loophole in the exclusion principle comes into play: it’s actually possible for two particles to share the same chunk of space as long as they’re in some way “different”, for example if one of them is in a higher energy state than the other (say, it’s moving faster). So in order to compress two particles “on top” of each other, you have to apply enough force that you’re basically pushing one of the particles up to a higher energy. For large numbers of particles, this can get hard.

Different kinds of matter degeneracy happen at different pressures. At the center of a neutron star, the pressure is so high that neutrons become degenerate and start stacking up on top of each other. The matter in white dwarfs and neutron star crusts, on the other hand, exhibits only electron degeneracy. Whew! So, how bad is this going to be? Well, looking here, we find the formula for electron degeneracy pressure to be:

P = ( (pi^2*((planck’s constant)/(2*pi))^2)/(5*(mass of an electron)*(mass of a proton)^(5/3)) ) * (3/pi)^(2/3) * (density/(ratio of electrons to protons))^(5/3)

(I’m assuming that our little white-dwarf-chunk will not be so warm that the particles will be moving at relativistic speeds; if not, we have to switch to a different formula, found here.)

So, we know the density; the ratio of electrons to protons has to be 1 (Otherwise the white dwarf would have an electric charge. Of course, if you can somehow find a positively charged white dwarf somewhere, you can reduce your required pressure noticeably!); and everything else here is a constant. Plugging this in we get:

P = ( (pi^2*((6.626068 * 10^-34 m^2 kg / s)/(2*pi))^2)/(5*(9.10938188 * 10^-31 kilograms)*(1.67262158 * 10^-27 kilograms)^(5/3)) ) * (3/pi)^(2/3) * (1.00023843 * 10^9 kg/m^3)^(5/3) = 9.91935718 × 10^21 kg m^-1 s^-2

Or in other words, if we neglect the assistance that the white dwarf material will be providing in holding itself together in terms of gravitational pull, the shell for our planetoid would need to be able to withstand 9.91935718 × 10^21 Pascals of pressure in order to keep all of that degenerate matter in. That’s a bit of a problem. Actually, it’s more than a bit of a problem. It’s most likely impossible. The strongest currently known material in the entire universe is the carbon nanotube, and it has a theoretical maximum tensile strength (I think tensile strength is what we want to be looking at here) of more like 10^11 Pascals. So you’d have to find a material that could do better than that by a power of like 10^10. But, hey, that’s just an engineering problem, right?

CONCLUSION

So what the above basically tells us is this. As long as you can do one of the following things:

  1. Hijack Halley’s Comet and collapse all of its mass down into a volume with diameter 5.35 milliangstroms, to create a small black hole
  2. Build a pressure vessel capable of withstanding 1022 Pascals of stress, then trap inside a big chunk torn out of a white dwarf

Then you, too, can have a Super Mario Galaxy style planetoid in your very own space station! Now, given, these things aren’t easy, and possibly not even possible. But isn’t it worth it?

DISCLAIMERS

The above analysis has some limitations which should be kept in mind.

  • In the case of the black hole strategy, you’d have to somehow stabilize the position of the black hole relative to the shell, so that the surface of the shell always stayed exactly 115 feet away from the black hole– otherwise random drift would cause one side or other of the shell to gradually drift toward the black hole and eventually cause the whole thing to fall in. Which probably would be kind of fun to watch, but isn’t what you want if you’re standing on it at the time.
  • In the case of the solid-body/white dwarf strategy, I am assuming that the final planetoid can be treated like a point mass. This is almost certainly wrong. I’m not sure how to go about figuring out exactly the gravitational force from a chunk of matter which rather than being treated like a point is distributed through a sphere immediately underfoot.
  • On the other hand, it might be interesting to find out, because if you knew how to do that you’d probably know also how to figure out the gravitational force from matter distributed through an irregular body. Most of the planets in Mario Galaxy are not spheres! There’s also planetoids shaped like barbells, and avocados, and cubes. Most interestingly, there are a handful of planetoids in certain places in Mario Galaxy shaped like toruses (doughnuts). I’m curious but not sure how to figure out, if you actually were to construct such a thing in real life, what would the gravity when walking on it be like? What would happen if you tried to walk onto the inside rim?
  • None of these planetoids would be able to maintain an atmosphere. The escape velocity is just too stupidly low. (Puzzlingly, this does not seem to matter in Mario Galaxy, since Mario seems to be able to breathe even when floating out in deep space. One would be tempted to simply conclude that Mario, being a cartoon character, does not need air, but no, there are sections in the game with water and Mario suffocates if he stays underwater too long. Apparently the Great Galaxy is permeated with some kind of breathable aether?)
  • Even the gravity and breathable aether aside, many the elements of Super Mario Galaxy do not seem to be possible to replicate under normal physics. For one thing no matter where Mario walks on any structure anywhere in the game, his shadow is always cast “down” underneath his feet, as if light in Mario’s universe always falls directly toward the nearest source of gravity, or if the shadows weren’t cast by light at all but were simply visual markers in a video game allowing the player to tell where they are about to land.
  • I am not a licensed structural engineer, space architect or astrophysicist. The data above is provided on an “as is” basis without any representations or warranties and should not be used in the construction of any actual space vessel or celestial object. John Carmack, this means you.

Special thanks to pervect at Physicsforums for help with the degenerate matter stuff, and Treedub for pointing out what the electron/proton ratio of a white dwarf would be.

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