The Problem With Superman, and Other Physics Conundrums

The author of “Geek Physics” answers your pop culture physics questions.

This week, we asked you to submit pop culture physics questions. Now we’ve got some answers, thanks to Rhett Allain, author of the new book Geek Physics and a contributor

First, let me say that these are all great questions. Really, in some way, just asking the question is as much fun as answering it. But what do these outlandish queries have to do with real science? For me, thinking about fake things (like superheroes or video games) is very similar to looking at things in the real world. Let me offer an analogy.

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Geek Physics With Rhett Allain

Have you ever been rock climbing, either on real rocks or on a manmade climbing wall? In both situations, a human goes through similar motions. But some people really like the fake climbing walls because it gives them a chance to practice, and the facades are usually easier to get to than real mountains. The same is true for analyzing fake things—it’s good practice for scientific inquiry into real phenomena.

Now, on to your questions. I just realized that most of them deal with superheroes. I hope that doesn’t disappoint you.

@DuaneMieliwocki: Can Superman really fly? Or does the different gravity of Earth’s sun just allow him to jump really far?

I’m certain that Superman has been the topic of many scientific discussions. My answer to this question might not be unique, but it’s still fun. Originally, the idea was that Superman was so strong that he was “able to leap a tall building in a single bound.” Could jumping work as a form of flying?

Remember that there’s another superhero who likes to jump—the Hulk. Here’s a blog post that looks at a jumping superhero (there’s also a version in my book, Geek Physics). The problem with both the Hulk and Superman jumping is that it requires pushing on the ground. When a normal human jumps, the feet push on the ground with a force, and the human moves up in the air. In order to shoot off the ground at an even greater speed, you would need to push harder. In The Avengers movie, the Hulk jumps about 120 meters high—which would require a force of around 400,000 Newtons (remember that Hulk has a larger mass than a normal human). But this force is big enough that it would actually destroy the surface that the Hulk jumped from. The same would be true for a jumping Superman. As he applied the force necessary to jump to great heights, he would sink into the very ground he was trying to jump from. Just think about jumping in soft mud. That’s what this would be like.

If Superman doesn’t jump, then, how does he fly? Some people suggest that he has some sort of telekinetic power that allows him to push off the ground without touching it. However, that’s like saying he flies by magic. I think we can do better than that. There are a couple of ways that real objects fly. First, there’s a rocket, which stays off the ground by shooting gases out the bottom thruster. Could this work for Superman? Perhaps he could emit particles out of one side of his body. This would work, except that he would either need a large amount of particles (think of the massive fuel in a rocket), or those particles would have to travel at super speeds.

The second kind of flying object is an airplane, which stays off the ground by ramming into air and deflecting it downward (yes, that’s a very simplistic explanation of flying). But in order to crash into the air, the object must be moving forward—and now we’re back to the same problem: How does Superman move forward? I’m going to go out on a limb and say that he shoots particles out the bottom of his feet.

@alexziemianski: Why don’t the people whom The Flash rescues get whiplash from the sudden acceleration?

The short answer is that these humans actually would be injured. In the movie X-Men: Days of Future Past, Quicksilver (the Marvel version of a fast-running superhero) holds on to someone’s head in order to prevent whiplash, but whiplash is only a small part of the problem. The real problem is damage to internal organs.

When superheroes like The Flash or Quicksilver rescue someone, they must accelerate that person. And to accelerate someone, they must push on him in some way. If The Flash pushes on the back of a human, this will accelerate the human’s back. The back then pushes on the lungs, and the lungs push on the front chest. All this pushing is bad—it can cause all sorts of physical damage. A human might be able to survive acceleration up to 40 g’s for a very short time, but this is less than the acceleration for many superheroes.

So, how do they do it? In sum, I don’t know.

@DuaneMieliwocki: How was the Bionic Man able to lift so many heavy things without his spine or pelvis buckling?

This is a great point. Just because Steve Austin (his friends just call him Steve) has a bionic arm, this doesn’t mean he can do anything. If his arms lifted up some heavy mass, his back and legs would also have to support it.

This oversight didn’t just occur in The Six Million Dollar Man. In the movie Avengers: Age of Ultron, Tony Stark (as Iron Man) attempts to lift Thor’s hammer by using just the glove of the Iron Man suit.

At left is a force diagram for both the hammer and the glove as Tony tries to lift it.

No matter how hard the glove pulls on the hammer, Tony (the mere human) would have to pull on the glove. It’s just a bad idea for the rest of his body. But maybe Tony never intended to really lift the hammer…

Fahad Uddin Siddiqui: How realistic is it to dream about antimatter as a plausible power source in the next century?

Ah ha! A non-superhero question. First, what is antimatter? In short, antimatter is just like normal matter, but with the opposite charge (that’s not the full story, but it’s good enough). If you have a positively charged proton, for instance, there is an antimatter proton (called the antiproton) with the same mass but a negative charge. There is also an anti-electron (also called the positron) that is just like an electron, but with a positive electrical charge.

What happens when you get a positron and an electron near each other? Since they are opposite charges, they attract. There is nothing to keep these particles apart, so they just annihilate each other—and in the process, they create energy. How much energy? Well, you’ve probably seen the famous equation E = mc2.  That can tell you how much energy you’d need for a given amount of mass (m). And guess what? In this case, it’s a lot. If you had just 1 kg of matter and 1 kg of antimatter, this would create 1.8 x 1019 Joules. That’s close to the amount of energy used in the entire U.S. in one year.

This all said—and I hate to burst your bubble—antimatter won’t be a source of energy. Why not? Well, to use this as an energy source, you must have antimatter. It turns out that our universe has much more matter than antimatter (at least as far as we know). So, you can’t really use antimatter to make energy unless you either find a bunch of it or create it yourself—but it takes a lot of energy to make antimatter. So, you see the problem.

Ok, that’s it. I hope you enjoyed the answers. I sure enjoyed answering them.

The following is an excerpt from Allain’s book, Geek Physics:

Geek Physics: Surprising Answers to the Planet's Most Interesting Questions (Wiley Pop Culture and History Series)



How much bubble wrap do you need to survive jumping out of the sixth floor of a building? Let me roughly say it would be a height of 20 meters. Where would you start with a question like this? Well, first we need some bubble wrap. What properties can I even measure from bubble wrap? First, I can measure the thickness of a sheet of bubble wrap. Yes, there are all different kinds of bubble wrap but I’m using mine (you have to start somewhere). Instead of just measuring the thickness of one sheet, let me make a plot of total thickness as I stack several sheets.

I measured the stack every time I added a layer and plotted a graph with height on one side and the number of sheets on the other. The slope of this linear fitting equation is 0.432 cm/sheet and this will be a good estimate for the thickness of one sheet.

Next, I need to see how “springy” the bubble wrap behaves. Is it like a spring? If so, how stiff is it? If it were an actual spring, I would just add some weight to it and see how much it compressed. Let me do exactly that. If you plot this, it looks fairly linear—proportionally more force leads to proportionally more compression. So, I would have to say the bubble wrap does behave like a spring. I can model the force the bubble wrap pushes on other stuff (like a person) as though it were a spring. The force would be proportional to the amount the bubble wrap is compressed.

From this data, I have found an effective spring constant for bubble wrap of this size. But what about other sizes? Suppose I have two sheets of bubble wrap stacked on top of each other instead of just one. If a mass is placed on top, each sheet will be compressed the same amount as just one sheet, since they both have the same force pushing down. But two sheets compressing gives an overall larger compression than one.

What if I compare a small single sheet and a larger sheet of bubble wrap? This would be like two sheets of bubble wrap next to each other. When a mass is placed on top, they both push up on the mass so that each would only have half of the force compressing it. So, two sheets side by side wouldn’t compress as much as one single sheet.

In short, the bigger the area of bubble wrap, the more the bubble wrap acts like a stiffer (higher spring constant) spring. The thicker the stack of bubble wrap, the smaller the effective spring constant. The property of a material that shows its stiffness, independent of the actual dimensions of that material, is called Young’s modulus. Since I know the size of my sheet, I get a Young’s modulus of 4,319 N/m2 for this particular bubble wrap. What about the jumping part? It isn’t the jumping that is dangerous, it’s the landing. The best way to estimate the safety of a landing is to look at the acceleration. Fortunately, I don’t need to collect experimental data on the maximum acceleration a body can take, NASA already did this. Here is essentially what they came up with (from the Wikipedia page on g-tolerance):

From this, you can see a normal body can withstand the greatest accelerations in the “eyeballs in” position. This is the orientation such that the acceleration would “push” the eyeballs into the head. In the case of jumping, this means landing on your back.

There is a slight problem. If a jumper is wrapped in bubble warp, the acceleration during the collision with the ground wouldn’t be constant. Here is a diagram showing a person covered in bubble wrap while impacting the ground:

So, there are two main forces on the jumper during this time: the force from the bubble wrap (which is like a spring) and the gravitational force. In order for the jumper to stop, the acceleration must be in the upwards direction and the bubble wrap force has to be greater than the gravitational force.

The acceleration depends on the value of the spring constant as well as the distance the spring is compressed. I don’t know either of those values. However, if I use the work-energy principle then I can look at the entire fall. At both the start and finish of this fall, the kinetic energy is zero. The gravitational potential energy will decrease during the fall, and the spring potential energy (in the bubble wrap) will increase during the impact. Since there is no other work done on the system, I can create a relationship between the height of the jump and the needed spring constant (for the acceptable acceleration).

In order to get a value for k, I need some other values. Here are my assumptions:

• Mass of the jumper plus bubble wrap is 70 kg. Here, I am assuming that the mass of the bubble wrap is small compared to the jumper.

• Maximum acceleration of 300 m/s2 and also an impact that lasts less than one second.

• Starting height of 20 meters.

With this, I need a bubble wrap spring constant of 1.7 x 104 n/m in order to have an acceleration during the landing that won’t exceed the NASA recommendations for g-tolerance.

Now that I know the spring constant needed to stop the jumper, I am one step closer to determining how many layers of bubble wrap would be needed. There is one thing I need to estimate first: the area of contact between the ground and the bubble wrap. I know this area should actually change during the collision, so I am just going to estimate it. Suppose the contact makes a square about 0.75 meters on a side. This would give an area of 0.56 m2.

Because I know Young’s modulus for bubble wrap, I can calculate the thickness (which I have oddly called L) to be 0.142 meters. Since each sheet is 0.432 centimeters thick, I would need thirty-nine sheets.

Maybe thirty-nine sheets seems a bit low. Let me calculate the mass of this bubble wrap and how big it would look. If I assume the bubble wrap is wrapped cylindrically around the jumper, it would look something like this:

When looking down on a person, that person roughly appears as a cylinder with a radius of 0.3 meters (just a guess). If the bubble wrap cylinder extends another 0.142 meters, then what is the volume of bubble wrap? Oh, I guess I also need to have a person with a height of about 1.6 meters (another guess). This would give a bubble wrap volume of 0.53 m3.

I can use the thickness of bubble wrap along with mass data and find the density of bubble wrap. With this volume of bubble wrap, it would have a mass of 9 kg. Not too bad, but technically this would change the amount of bubble wrap needed to land. Maybe I could just be on the safe side and add a couple of layers to compensate for the added weight of the bubble wrap. Now that I know the size of this bubble-wrapped person, I can consider the air resistance on the fall. The typical model for air resistance shows a force proportional to both the cross-sectional area and the square of the speed of the object.

Because the force changes with speed, the problem isn’t so simple to solve on paper. However, it is pretty straightforward to calculate by breaking the problem into many short time intervals with a computer. Doing this numerical calculation, I get the following plots for position and vertical speed of the falling object.

This shows that the falling object, with air resistance accounted for, ends up with a slightly lower speed before impact (17.8 m/s instead of about 20 m/s). I could redo all the calculations, but I won’t. Instead, you could just consider this lower speed part of the safety margin (although I would never consider doing something like this “safe”).

What about a follow up question (for your homework)? How much bubble wrap would you need to survive a jump out of a plane? I suspect it wouldn’t be that much more. If you add some layers to the bubble wrap, you will decrease the terminal speed of this falling object.

Should we really use bubble wrap for human safety situations, though? Clearly, that would be a bad idea.

From Geek Physics by Rhett Allain. Copyright © 2015 by Rhett Allain. By permission from Turner Publishing Company. ALL RIGHTS RESERVED.

Meet the Writer

About Rhett Allain

Rhett Allain is the author of Geek Physics and an associate physics professor at Southeastern Louisiana University. He also writes WIRED’s Dot Physics blog.