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On Randall Munroe’s new blog “what if”, he answers the question:

“What would happen if you tried to hit a baseball pitched at 90% the speed of light?”

http://what-if.xkcd.com/1/

He concludes:

… the air molecules in front of this ball don’t have time to be jostled out of the way. The ball smacks into them hard that the atoms in the air molecules actually fuse with the atoms in the ball’s surface. Each collision releases a burst of gamma rays and scattered particles.

These gamma rays and debris expand outward in a bubble centered on the pitcher’s mound. They start to tear apart the molecules in the air, ripping the electrons from the nuclei and turning the air in the stadium into an expanding bubble of incandescent plasma.

… Suppose you’re watching from a hilltop outside the city. The first thing you see is a blinding light, far outshining the sun. This gradually fades over the course of a few seconds, and a growing fireball rises into a mushroom cloud. Then, with a great roar, the blast wave arrives, tearing up trees and shredding houses.

Everything within roughly a mile of the park is leveled, and a firestorm engulfs the surrounding city. The baseball diamond is now a sizable crater, centered a few hundred feet behind the former location of the backstop.

Is that all actually reasonable, or is it all just hyperbole?

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i think its not reasonable at all; i mean, to be allowed to advance only to first base after being hit by this, does not make any sense to me. I think this should qualify as forfeit, in which case the home team should lose: en.wikipedia.org/wiki/Forfeit_(baseball) –  lurscher Jul 10 '12 at 21:07
    
uhh. cool! If an explosion would happen from an object moving that fast, why doesn't an explosion happens when a fighter jet flies at the speed of sound in the air? –  user10749 Jul 21 '12 at 23:38
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Speed of sound in air (~340m/s depending on the weather/environment) is much much slower than the speed of light (300000000m/s) –  DouglasHeriot Jul 22 '12 at 5:16
    
I have a question about this. Due to relativity wouldn't the baseball have nearly infinite mass at .9c? –  user14430 Oct 27 '12 at 1:08

3 Answers 3

up vote 6 down vote accepted

I found the narrative to be consistent with my view of physics. Let me address this major point about the integrity of the ball as it travels through the air.

The distance from the pitcher's mound to home plate is $18.39 m$, and the diameter of the ball is about $7.4 cm$. Since the ball is sufficiently fast, we are comfortably out of normal fluid mechanics and know that collisions will happen based entirely on the trajectory of the ball. Let me be clear that the ball holds together mostly due to the fact that all the atoms in the ball have the same momentum vector - NOT because molecular forces are making much of a difference.

The density of air is about $1.2 kg/m^3$. Multiply this by the area of the ball times the distance it travels to get that the ball collides with $92.4 g$ in its path to home plate. The weight of a major league baseball is about $145 g$. In the absence of fusion, it would appear correct to say that the speed of the ball changes very little through that distance, due to the simple reasoning that the ball is heavier than the air.

At the point that the ball hits the bat, it really shouldn't matter much to us what happens because the conservation of energy mandates absolutely that such an explosion happens. I would maintain, however, that the center could be considerably past home plate, since the nuclei in the ball will have to go through a good number of collisions before the momentum is dissipated. Keep in mind, the Earth is only normally supporting the full weight of the stadium directly downward. The weight of Wembly Stadium is in the neighborhood of $(2.2 g/cm^3) \times 90,000 m^3 + 23,000 \text{tonnes} = 2.2 \times 10^{8} kg$, which leads to a normal downward force of about 2 billion Newtons. The momentum of the ball is about 89 million Newton-seconds.

I think a fraction of the stadium itself will head down the street a little bit while exploding. In fact, it might travel quite a good distance, since the shine from the ball plasma is highly directional and will hit the bleachers which have less of the high mass concrete (mostly foundations). The ball has the momentum of about 90 fully loaded 18-wheelers traveling at 60 mph. That is a lot of momentum, and it would likely break through a few walls, but it may still dissipate sufficiently (the momentum, not the energy) within the stadium but not the field. That's the one major point where I take issue with Munroe.

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Then does the pitcher get a new contract? –  Mike Dunlavey Jul 10 '12 at 20:06

Starting with the conservation of energy and Einstein's formula for relativistic kinetic energy, I determine worst case scenario that a rigid .9c 0.148834996kg baseball (0.0373934539m radius) will go no further than between 37.6572765763794m and 75.3145531527588m through Earth air with a molecular density of 1.1644kg per cubic meter at sea level (80F/30C).

I understand Einstein's relativistic kinetic energy formula to be K = (γ-1)mc^2 using a Lorentz factor γ = 2.2941573387056 for .9c, and a rest mass energy of 1.33766223432274E16 J for the 0.148834996kg mass baseball (see: http://goo.gl/oXZkN for the kinetic energy formula and γ).

I determine that the total energy for the baseball alone is 1.7311453972581E16 J kinetic + 1.33766223432274E16 J rest mass energy for a total of 3.06880763158084E16 J or about 7.3346 megaton TNT. This does not count the rest mass energy of the air which would also be released.

This total energy is nearly the energy released by the Krakatoa eruption. Or, the impact energy that created Meteor Crater, Arizona (10 megaton).

Details of my analysis of this thought problem, and comments are on two Google+ posts originating from my re-share: https://plus.google.com/109667384864782087641/posts/S9xLmLBj8ZV

I'm still working on determining what cross-section of mass-energy will interact with the bat at a distance of 18.4m. I am also still trying to understand the forces at work, & generally trying to find any mistakes in my thinking.

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Nice calculation, but if start with one significant digit (0.9), you shouldn't use so many digits for the rest. One more at most. Interesting comparison to the Meteor crater instead of atomic bombs by the way! –  Tobias Kienzler Jul 14 '12 at 7:48

The rest mass of a baseball is around 150g (i.e., http://hypertextbook.com/facts/1999/ChristinaLee.shtml). The relativistic kinetic energy is

$ E = \sqrt{p^2c^2+m_0^2c^4} - m_0c^2 $

where $p = m_0u/\sqrt{1-v^2/c^2}$. Given that $u=0.9c$, $m_0 = 0.15$kg, you find that

$E \approx 17 $ PJ

which is in the thermonuclear explosion regime (i.e., http://hypertextbook.com/facts/2000/MuhammadKaleem.shtml).

The Hollywood-esque part about "molecules disintegrating" and "atoms fusing" etc. is actually true -- you already see this happening at a mere Mach 30 or so (meteors).

Also, supposing that each collision with an air molecule knocks off a molecule from the baseball, it would be disintegrated completely after

$ \frac{0.150}{\pi \left(\frac{0.074}{2}\right)^2 \cdot 1.225} \approx 30 $m

where 0.074 = 7.4cm is the diameter of the ball, 1.225kg/m$^3$ is the average air density @ sealevel. So all of this energy would be released quite locally, indeed very much like a nuclear explosion.

I see no argument against very hard X-ray/gamma shining in the forward direction -- wouldn't want to be the catcher there :)

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What could cause the energy to "spread out", other than the emission of emission of particles and EM radiation that Monroe already describes? –  Nathaniel Jul 10 '12 at 15:17
    
Come to think of it...Supposing that each collision with an air molecule knocks off one molecule from the baseball, it would only travel a few meters at most. So it would not spread much, you're right –  Rody Oldenhuis Jul 10 '12 at 15:28
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What you have given is the total relativistic energy. If you want the kinetic energy, you need to subtract mc^2 from your result. This cuts the value nearly in half. Also, think you may be off by a factor of 1000. I get 30 PJ (3e16) for total energy and 17 PJ for kinetic. (But, I haven't had any coffee yet...) –  kbeta Jul 10 '12 at 16:43
    
@rody_o Using that logic it would make it to home plate, as it only intercepts about 10% worth of its mass, as per my answer. Do the fusion products have speeds greater than 0.9c? The spread arguments seem to objectively address momentum conservation well enough. –  Alan Rominger Jul 10 '12 at 17:04
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@AlanSE right so you are...I half-assumed a factor of 1000 would bring it way out of anything we ever produced un-peacefully. Shocking to find out I was wrong. –  Rody Oldenhuis Jul 10 '12 at 21:01

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