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Terminal Velocity of a Bomb

  1. Nov 11, 2007 #1
    Recently, the New York times ran the obituary of Paul Tibbets who
    was the pilot of the plane that dropped an atom bomb on Hiroshima.
    At one point the article says:

    "At 8:15 a.m. local time, the bomb known to its creators as Little Boy
    dropped free at an altitude of 31,000 feet. Forty-three seconds later,
    at 1,890 feet above ground zero, it exploded..."

    If the above is true, then a little work with a calculator shows it
    must have been traveling at about 930 miles per hour if we ignore air
    resistance.

    This seems quite high to me. It's over 20% faster than the speed of sound
    and it seems to me that the shock of going through the sound barrier
    might have set off the bomb before it reached it's designated height.

    So... Do the above numbers seem reasonable? There is a formula for
    terminal speed in Knight's Physics text that indicates the maximum speed
    for an object like this to be around 600 miles per hour. Does the air
    resistance calculation seem reasonable? (The formula says that terminal
    speed is sqrt(4mg/A) where A is the cross sectional area of the object.
    Using information on the internet, I found the mass was 4000 kg and I was
    able to estimate the area as about 2 square meters. If you want to find
    the same information, go into Google and look up "Little Boy", the bomb's
    code name.)

    Just curious.

    --PatF
     
  2. jcsd
  3. Nov 11, 2007 #2

    DaveC426913

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    Not that this helps you towards an anwer but:
    AFAIK atomic bombs cannot be accidentally set off. They're not volatile the way that conventional bombs are; they need a fairly advanced and well-coordinated set of events to happen in order to work.
     
  4. Nov 12, 2007 #3

    rbj

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    since A-bombs are triggered by a conventional explosive (whether it's the gun-type or implosion mechanism), i would think the safety mechanism of a conventional bomb could be made just as secure as in an A-bomb if they wanted to do such.

    if i recall correctly, for the A-bombs used in Japan, they had three different mechanisms with criteria that all had to be satified for the bomb to go off. one was a timer, and then there were two altimeters, one that was atmospheric pressure based and another proximity based (radar?). i thought there was the arming the device with the neutron initiator by hand, some minutes before it was dropped, was another safety mechanism.
     
  5. Nov 12, 2007 #4

    mathman

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    I have not made a quantitative study. But I suspect the answer is simply a result of the fact that air density is far lower at high altitiude than at ground level. As a result, the bomb could fall much faster there.
     
  6. Nov 12, 2007 #5
    Patf, according to wikipedia you overestimated the cross section more than enough to account for the discrepancy [!], but you should have already been satisfied that your estimates had the same order of magnitude (without you giving the context of your formula, there is no reason to expect it to work at all in this regime).

    As for the shock wave, conventional explosives are routinely unaffected by supersonic flight, so [even aside from the measures rbj described] there is no reason to expect premature ignition (in fact less so [as Dave was saying] since, unlike conventional explosives, nuclear explosives are not triggered just by a smaller explosion, in fact [if I may add something further for those that haven't watched "broken arrow"] such is just likely to disable the weapon; specially synchronised detonations are often required to bring about a critical configuration).
     
    Last edited: Nov 12, 2007
  7. Nov 12, 2007 #6

    DaveC426913

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    Dis is wot I'm sayin'. Isn't dis wot I'm sayin'? *gestures with both hands*
     
  8. Nov 13, 2007 #7
    mind you these were not conventional bombs these were atomic bombs.
    again a formula of terminal velocity
    v=2r2(the second2 stands for square)(p1-p2)g/(divided by)9n(eta)
    where p1 and p2 stands for the density of the fluid it is travelling and p2 the density of the material which is travelling.
    eta is the coefficient of viscosity
     
  9. Nov 13, 2007 #8

    Andrew Mason

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    I suspect that the author of the article simply calculated the drop time without taking into account air resistance and terminal velocity. Your terminal velocity calculation shows that it would not reach supersonic speed. So, in actual fact, the descent must have taken longer than 43 seconds.

    AM
     
  10. Nov 13, 2007 #9
    The OP didn't cite their source (how terrible for a man's final legacy to be nuking Hiroshima.. all I can say is at least it wasn't Nagasaki), but yes, 43 seconds and 930mph assumes free-fall between those heights. According to wikipedia, it actually took 57 seconds.
     
    Last edited: Nov 13, 2007
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