Superman is so strong, he can do anything, right? Could he punch someone so hard that they ended up in space? Let’s do this.

## How High Is Space?

When I say space, you might say “outer space.” But how high is that? The Earth’s atmosphere doesn’t just stop at some height. No, instead the density of air gets lower and lower until you can’t even really detect it. But for this problem, we have to pick a height. I am going to pick 420 km above the surface of the Earth as “space.” Why? Why not. That is about the height of the International Space Station’s orbit, so I think it is a good choice.

## How Fast Would the Person Have to Go?

I am talking about after the punch from Superman. Let’s just look at a person moving up at some initial speed v0. If this were a problem in an introductory physics course, I would hope you would think of the work-energy principle.

Let’s say that Superman is punching a clone of himself (called Superman-b) – just as an example. If I take Superman-b and the Earth as my system, then after the punch from Superman there is no external work done on the system. There will be two types of change in energy – kinetic and gravitational potential.

I know the values of these variables. If I plug in what I know, I get a “launch” speed of 2778 m/s (6214 mph). Yes, that is fast – but actually Superman-b would have to be going even faster than that. Why? Air resistance, that’s why.

## Launch Speed With Air Resistance

Here is a diagram of Superman-b shortly after he was hit by Superman.

I will use the two following models for the magnitude of the gravitational force and the air resistance force.

For the gravitational force, the two masses are the mass of the Earth and the mass of Superman-b and r is the distance between Superman-b and the center of the Earth. This force will decrease somewhat as Superman-b rises to space.

In the model for air resistance, A is the cross sectional area of the object and C is some drag coefficient that depends on the shape of the object. The ρ is the density of the air. As you get higher in the atmosphere, this will decrease. So, you see this air resistance force changes with both the speed and the altitude. Actually, the drag coefficient can depend on speed too – but I will pretend like it is constant. So, this isn’t such an easy problem.

Let me get some estimates for some of these values. I am going to assume Superman-b is the same size and shape as a normal human. Maybe he has a mass of 70 kg. For the product of AC, let me estimate this based on the terminal speed of a sky diver. If a sky diver falls at 120 mph (54 m/s) then the air resistance would be equal to the weight of the sky diver. This means that AC would be:

The terminal speed of a sky diver is near the surface of the Earth. That is why I can use the mg for the weight. Also, I can use a value of 1.2 kg/m

^{3}for the density of air. Putting in my values gives a product of AC at about 0.392 m^{2}. I will use an AC value of just 0.05 m^{2}. Why? Because the previous calculation was for a sky diver in a typical skydiver position. If Superman-b is “launched” in a head-up position, he will have a much lower cross sectional area. This is probably WAY too low, but that’s ok.
The other problem is dealing with a changing density of air. Fortunately, I have looked at air resistance at high altitudes before. Yes, the Red Bull Stratos Space Jump started at a point where the density of air was much lower than it is on the surface of the Earth. In the calculation of his falling speed, I used this model for the density of air.

That model isn’t really valid for super high altitudes. So, I will just use it up to around 100 km and then assume the density of air is negligible after that. Yes, I know this is wrong – but it will still work. First, I am trying to show that the starting speed of Superman-b is super large. Cutting off the density of air at high altitudes will just lower the starting speed. Also, when Superman-b reaches these high altitudes, he won’t be going so fast such that the air resistance force will be small even if I did have some air up there.

What now? I can’t directly calculate the required starting speed. However, I can pick some starting speed and create a numerical model to determine how high Superman-b will go. Then I can keep increasing the starting speed until I get the height that I want. For each starting speed, I will break the motion into tiny steps of time. During each of these steps, I will do the following (these are the basic principles of a numerical calculation).

- Calculate the density of air based on the height.
- Using the height, density of air and the speed – calculate the sum of the gravitational and air resistance forces.
- With this net force, calculate the change in momentum during this time step.
- Based on the momentum, determine the change in height during this time step.
- Repeat the above.

It looks complicated, but it really isn’t too bad. Here is a plot of height as a function of time for the case that Superman hits Superman-b with an initial speed of 2778 m/s (from above).

You can see that in this case, Superman-b doesn’t get to an altitude of 420 km. Not even close. Now we just need to keep increasing the launch speed until we get to the speed we want. Here is a plot of the maximum altitude as a function of starting speeds up to a speed of 10

^{5}m/s.
Even at 10

^{5}m/s, Superman-b would only get to a height around 13 km. I am a little disappointed. I thought I would get Superman-b higher than that. What would happen if I started this problem from the top of Mt. Everest at an altitude of 8.5 km? That way, the density of air would be less and maybe I could get much higher.
That’s better, but still not in space. Ok, let’s just say Superman punches Superman-b such that he has a starting speed (after the punch of 10

^{5}m/s) but Superman-b doesn’t actually go into space. He just goes really high. Could he get into space? Not with my air resistance model. Maybe there is a way, but not this way.## What About the Punch?

Ok. Say that Superman hits Superman-b really hard. So hard that he has a speed of 10

^{5}m/s. What would happen? Let’s say that The punch is right on the chin – an upper cut. Here is a diagram of Superman-b during that hit.
Here, Superman-b goes from a speed of zero to a speed of 10

^{5}m/s over a distance of Δy. What kind of force from Superman would this take? I will ignore gravity (really, its effect will be small in this case) and use the work-energy principle. If Superman-b is my object then only Superman will do work.
This is the average force the fist exerts on Superman-b. The only number I have not estimated is the distance over which the punch is exerted on Superman-b. I think 0.75 meters would be a generous estimation. With that, I get an average force of 4.67 x 10

^{11}Newtons. Yup.
Suppose that Superman’s fist makes contact with a surface area of 70 cm

^{2}(I measured the front of my fist as an estimate – of course I made Superman’s larger). What kind of pressure would this punch produce on Superman-b’s skin?
That’s a high pressure. I typical scuba tank has 3,000 psi inside it and the steel tanks have a wall thickness of 1/4 inch. What am I trying to say? I am thinking that if Superman could hit Superman-b this hard, I think his fist would just push right through his head. Gross, I know.

What about the pressure between Superman’s feet and the ground? The force of Superman pushing on the ground would be somewhere around the same magnitude as the force he pushes on Superman-b. Of course, the contact area of his feet are probably higher, but the pressure would still be HUGE. I’m sure he would get pushed into the ground by his own punch.

## What About Superman-b?

If Superman-b has a mass of 70 kg, then I can get a value for his average acceleration during the punch. This would just be the force divided by the mass (again, the gravitational force is small in comparison). His average acceleration would be 6.67 x 10

^{9}m/s^{2}.
What if I pretend like Superman-b is made of two parts. His head with a mass of 7 kg and the rest of his body with a mass of 63 kg. Superman pushes just on the head of Superman-b. Then why does the rest of his body also accelerate? Well, of course the head is connected to the body. This means that Superman-b’s head pulls up on the body through the neck. In order for the body to have the same acceleration as the head, it would have to have a force of 4.2 x 10

^{11}Newtons.
A Nimitz class aircraft carrier has a mass of about 9 x 10

^{7}kg. In order to produce the same force on Superman’s neck, you could hang him upside down and then have 4,500 aircraft carriers hang from his head. I don’t know about you, but I think his head would come off (also there aren’t 4,500 aircraft carriers in the whole world).
Back to the original question. Could Superman punch someone into space? No. Here’s why.

- If you account for air resistance, the faster you start the greater the air resistance force. It’s just not going to happen.
- Even if Superman did hit someone super-hard, Superman’s fist would probably just go through the target’s head.
- By pushing so hard on someone, Superman’s feet would crush the ground under him.
- The acceleration of the victim would be so great that his head would come off.

Here is a homework question. What would be the power needed for Superman to punch someone this hard? If he got all of this energy from the Sun, how long would it take to “charge up”?

Oh, I know. Superman isn’t real. I look forward to the epic internet battle that will follow this post.

Source: Wired