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Relativistic Baseball

  1. Jul 10, 2012 #1

    Simon Bridge

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    Relativistic Baseball
    When it reaches the batter, the center of the cloud is still moving at an appreciable fraction of the speed of light. It hits the bat first, but then the batter, plate, and catcher are all scooped up and carried backward through the backstop as they disintegrate.
    That's interesting since I'd have guessed the energy of the x-rays and other light to have already incinerated the batter?! Perhaps the distance is too short?
     
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  3. Jul 10, 2012 #2

    BobG

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    Don't know about the fusion, x-rays, etc, but the way objects behave during high speed collisions is pretty interesting. Going fast enough (merely as fast as a satellite for example), a collision between a baseball and a garbage bag is over before the far ends of the garbage bag even know there's been a collision (information that will move at the speed of sound for the materials involved). As a result, the garbage bag is even more brittle than a plate glass window!

    That makes for some strange results when satellites collide with each other in space. You get ghosting - at the speeds involved, the material is so brittle the satellites appear to pass right through one another, except on one side of the collision they're a whole satellite and on the other they're a debris cloud.
     
  4. Jul 10, 2012 #3

    collinsmark

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    I don't know the details about the fusion, x-rays, etc. either (although it sounds perfectly reasonable to me), but in terms of raw energy I don't think it matters that much in the end, as I'll indirectly allude to below.

    There's an easy way to approximate how big the explosion ends up, by just assuming/approximating that the final net energy of the explosion ultimately comes the baseball's original kinetic energy.

    Mass of baseball, m: ~0.145 [kg]
    speed of light c: ~3 x 108 [m/s]
    Velocity of ball, v: (0.9)c

    [tex] \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}= \frac{1}{\sqrt{1 - 0.9^2}} = ~2.294 [/tex]

    Kinetic energy, T = [itex] (\gamma -1) mc^2 [/itex] = (2.294 - 1)(0.145 [kg])(3 x 108 [m/s])2 = 16.9 x 1015 [J].

    1 megaton of TNT is 4.184 x 1015 [J]

    Explosion due to baseball is roughly (16.9 x 1015 [J])/(4.184 x 1015 [J/Mt]) ≈ 4 Megaton of TNT.
     
    Last edited: Jul 10, 2012
  5. Jul 11, 2012 #4

    Simon Bridge

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    Assuming that all the KE goes into the explosion ... which amounts to stopping the ball at the bat. I suspect the ball would simply go right through the bat, snapping it, reducing the KE contribution to the collision.

    The ball is going fast enough for nuclear reactions with the air ... so we want to compare the likly nuclear contribution.
     
  6. Jul 11, 2012 #5

    Drakkith

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    Hmmm. I'm trying to find the equivalent temperature of a plasma with the same KE as an atom of carbon moving at 0.9c, which is about 15 GeV. Anyone help me out a bit? I'm getting lost trying to find all the right equations.
     
  7. Jul 11, 2012 #6

    collinsmark

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    Not quite so fast. :smile: There is conservation of energy involved here. The breaking of the bat heats up the bat. The energy is not gone, it has merely changed form.

    But you have a point that the assumption may involve a restriction such as perhaps the ball was thrown with a slight downward angle, such that it hits the ground or maybe a nearby hill, within a kilometer or two of the pitcher's mound.
    Which leads to another assumption/approximation. But first, something which needs no approximation is that the material of a the baseball, or anything in the ballpark, is certainly not fissile -- not particularly fissionable either. But there might be some nuclear fusion going on anyway, to a very small extent, which then might immediately decay again quickly releasing that energy. The assumption is that when comparing the end/final products of any nuclear reactions that take place and the beginning/initial elements (involving the atoms originally in the baseball/air/earth), the difference is negligible compared to the overall energy released in the entire pitching event. I think that's a fair approximation.
     
    Last edited: Jul 11, 2012
  8. Jul 11, 2012 #7

    Drakkith

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    Not sure if I've done this right, but my calculations tell me that at 0.9c the collisions between the atoms in the baseball and the air is like a plasma at 180 trillion kelvin. I don't think there would be fusion. I think the atoms would be smashed together like in a particle collider. End result: TOTAL ANNIHILATION!!!
     
  9. Jul 12, 2012 #8

    Simon Bridge

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    Of course I know that - I'm saying that not all the kinetic energy is stopped/converted at the bat. Some of it (I'm suggesting: lots) continues with whatever the ball looks like at this stage past the bat.
    I didn't say "fission" I said "nuclear reactions" ... if I shoot 4MeV protons at a solid target, the protons interact with the nucleus. The energies we are dealing with here are much much higher.

    As Drakkith says, it's smashing nuclei together. The sort of process that happens in particle accelerators on a big scale.

    Thinking about it though - would we get nuclear energy released in this kind thing or would the energy out be limited to the available kinetic energy.

    But any way I look at it - there is a very big explosion.
     
  10. Jul 12, 2012 #9

    collinsmark

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    I guess what I was saying is that assuming the ball doesn't exit the Earth's atmosphere and head out into space, it's a good assumption/approximation that the final, net thermal energy generated by the pitching event is approximately the same as the ball's initial kinetic energy.

    Small differences can be attributed to (a) differences in the mass energies of any atoms that did undergo nuclear reactions, and (b) differences in the chemical energies of any molecules that underwent chemical reactions (such that the initial and final states have a net difference). Note that in both cases, the individual reactions that make up the net energy, may contain either positive or negative energy differences. (Edit: And then there is a little electromagnetic radiation energy which escapes into space, which I neglected to mention.)

    But the vast majority will end up as thermal energy, even if there are intermediate steps in between (one example of an intermediate step could be
    Kinetic Energy → electromagnetic potential energy (plasma) + some thermal energy → radiated electromagnetic energy + chemical potential energy + some more thermal energy → some additional radiated electromagnetic energy + even yet more thermal energy → thermal energy.)
    Oh, yes, I agree. It will continue way past the bat (as BobG points out). But assuming the ball (or what's left of it) enters the ground or the side of a nearby hill* (within a roughly a kilometer or so), it still won't make it out of the blast area. A 4 megaton explosion is enough to devastate a modestly sized city.

    *(And the hill/earth won't stop the energy transformation either, although it might introduce other intermediate steps such as
    Kinetic energy → gravitational potential energy → back to kinetic energy → thermal energy. :smile:)
     
    Last edited: Jul 12, 2012
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