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Nuclear ignition of gas giant atmospheres

  1. Jun 19, 2003 #1
    Out of curiosity, can you produce a self perpetuating nuclear fusion chain reaction - in effect a new sun - by dropping a nuclear bomb deep into the atmosphere of a hydrogen rich gas giant?

    If a powerful enough weapon was fired on a non-hydrogen atmosphere like Earth's, bearing in mind that nitrogen and oxygen can still undergo nuclear fusion (although under much more extreme conditions with less nuclear binding energy released per nucleon), could the Earth's atmosphere, too, be ignited?
     
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  3. Jun 19, 2003 #2

    LURCH

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    Although a nuclear reaction could conceivably be ignited by a large energy released within a gas giant (in fact, I would be willing to bet half a peach that some hydrogen was fused when Schumacher Levy 9 hit Jupiter), such a reaction would not be self-perpetuating. You see, the energy state needed to fuse elements together requires very high temperatures. But as soon as fusion begins, the material being fused expands very rapidly, causing the temperature to drop. Once the temperature drops below a certain critical threshold (about 15 million degrees for hydrogen), there is no longer sufficient energy to sustain a reaction. The Sun has a strong enough gravitational influence to ritard this expansion enough so that sufficient energy is retained to perpetuate the reaction, but planets like Jupiter and Saturn do not.
     
    Last edited: Jun 19, 2003
  4. Jun 19, 2003 #3
    There seems to be two opposing factors here. One is that a fusion reaction on its own generates very high temperatures and pressures, another is that as expansion occurs, temperature and pressures drop. At a point, if the bomb is set off so that it generates a high enough starting temperature and pressure, it seems that even the expanding shockwaves would have high enough pressures and temperatures within them to generate a fusion reaction.

    Is there a relationship which describes how high pressure and high temperature affect the probability of a nuclear fusion reaction happening? Something like probability = f(p) + g(T)?
     
  5. Jun 19, 2003 #4
    The fusion reactions in the Sun

    are different than the fusion reactions in a weapon, in that they involve the weak interaction as well as the strong. Which is to say, in the Sun Neutrinos are emitted as part of the process, which makes it much slower than the fusion of Deuterium and Lithium6 in a weapon. This makes it not as easy to do as might seem at first glance.
     
    Last edited: Jun 19, 2003
  6. Jun 19, 2003 #5
    Re: The fusion reactions in the Sun

    Very interesting. Is the involvement of weak interactions in addition to strong interactions in solar fusion the result of gravity being the cause of the reaction, or is it just because its conditions are more extreme?

    For example, if inertia or magnetic confinement were used to create very extreme conditions, could terrestial scientists outside an intense gravity well like the Sun produce nuclear reactions which also involve the weak nuclear force?

    Does binding energy relate to the sum of the weak and strong nuclear potential energies corresponding to the weak and strong forces, or does it just relate to the strong nuclear force?

    It must seem that I am very much a tyro in this matter - where can I read up on the subject?
     
  7. Jun 19, 2003 #6

    chroot

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    Just for a sense of the difference between weapons and the Sun:

    The sun's core produces less energy per unit volume than does a candle flame. It just happens that the Sun is really, really huge!

    (I think marcus first brought this fact to my attention.)

    - Warren
     
  8. Jun 19, 2003 #7

    Hmmm....how did you or your firend come to that conclusion?

    I think that:
    1. The suns core produces more energy per unit volume than a candle flame.
    2. Calculations for the energy/vol of the sun assumed a uniform distribution of energy generation over its radius.
    3. The outer parts of the sun do not undergo fusion.
    4. The calculation would be biased if it included the outer "dead weight (or rather space)" of the sun.
    5. The calculation should try to do (energy output of sun)/(nuclear reacting volume) vs. (energy output of candle)/(reacting volume)....for a level playing field.
      [/list=1]
     
  9. Jun 19, 2003 #8

    chroot

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    Uh, Tyro. Duh. Do you think we're stupid?

    - Warren
     
  10. Jun 19, 2003 #9

    chroot

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    Let me know if I screw something up.

    The solar radius is 6.9 x 1010 cm.

    The core of the sun is generally taken to be 25% of its radius, for a radius of 1.73 x 1010 cm, or a volume of 2.1 x 1031 cm3.

    The solar luminosity is 3.9 x 1033 erg/s.

    Diving, the energy output of the core of the sun is 185 ergs/cm3-s, or 1.85 x 10-5 J/cm3-s.

    I'll let you figure out how small that is. (Hint: a snowflake strikes the earth with kinetic energy on the order of one erg.) In the future, maybe you should think more.

    - Warren
     
  11. Jun 19, 2003 #10

    marcus

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    I dont remember making that comparison but it sounds like a very interesting one and I do know the wattage of the sun.

    It is somewhere around 3.8 x 1026 watts.

    The core is often thought of as the inner 10 percent. We could do a quick back of envelope and get the watts per cubic centimeter.
    The sun radius is 70 billion cm, so the core radius is 7 billion cm.
    So the core volume is 1440 x 1027 cubic cm

    so the wattage per cc is a quarter of a milliwatt per cc

    incredibly small!

    but thats what you get if you divide 3.8 by 14400

    I will recheck after posting, seems so small. I never did this
    calculation, must have been someone else.
     
  12. Jun 19, 2003 #11

    marcus

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    If I'd known you were working it out I wouldnt have, but I think my result confirms yours since your figure for the core is some 15 or 16 times larger than mine, if I take my quarter of a milliwatt and divide by 15, I get 0.017 milliwatt per cc
    which I think is about what you got.

    Candle flame is raging inferno compared with inside of sun:wink:

    BTW I liked the "slow roll" question in Astronomy gamethread and am holding off answering so other player(s) can get in.
     
    Last edited: Jun 19, 2003
  13. Jun 19, 2003 #12

    chroot

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    Well, you chose to use a much more conservative volume for the core and still got a result which confirms that candle flame is MUCH more luminous, so yes, I'd say it's in agreement.

    In any event, the reason the sun's core is so boringly quiet is because of the weak interaction -- as has been pointed out here. Were it not for the weak interaction, the Sun would not fuse at all. The weak interaction allows just a trickle of power to escape, rather than an all-out flood.

    - Warren
     
  14. Jun 19, 2003 #13
    I seem to recall that the energy producing

    part of the Sun's core is only about 60-100% larger than the Earth.
     
  15. Jun 19, 2003 #14

    chroot

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    Tyger,

    Two times the size of the Earth would be 17% of the Sun's radius, which is right in the middle of my lenient calculation and marcus' conservative one.

    - Warren
     
  16. Jun 19, 2003 #15
    Hm,. let 's see

    Earth diam.×2=26,000Km, Solar diam. 1,400,000KM, I get a factor of approx 55 or 1.8%.
     
  17. Jun 19, 2003 #16

    chroot

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    Oops! I missed a decimal point -- 1.7%. In any event, I don't think the reacting core of the Sun is so small.

    - Warren
     
  18. Jun 20, 2003 #17
    I said no such thing. Feeling a bit insecure are we? Don't jump to conclusions.

    I brought that up because the post below of yours was a little ambiguous. Corrections to remove ambiguity outlined in red.


     
  19. Jun 20, 2003 #18
    I don't like this comment. In the future, maybe you should reserve unconstructive comments to yourself.

    There is a difference between not knowing and not thinking. You, Mr. Great Thinker, should know this by now.
     
  20. Jun 20, 2003 #19
    I also read off another source, solar energy outputs which agree with those stated in this thread.

    But if I work backwards from the solar constant and the Earth's radius from the sun, I get a bigger value. Why is this wrong?

    Earth's avg. orbital radius = 2.87e12m
    Solar constant = 1370 W/m

    Total solar ouput = 3.142*1370*(2.87e12)2
    or 3.55e28W

    Not that it makes much of a difference.

    The energy output of the sun seems to take into account the radiant energy. My above calculation took into account only radiant energy and already it is larger than the stated value (here or officially).

    Does anyone know the amount of gravitational potential energy being fuelled by solar fusion as a result of particles escaping the sun or non-returning solar flares?

    Incidentally, do photons have gravitational potential energy? On one hand, they should not, because they have zero rest mass and also because they are partly waves. On the other hand, they should, because light gets attracted by black holes too.

    Also, can the frequency shift of light due to gravity (e.g. red shift) be explained by energy conservation; i.e. if it gains gravitational PE and its speed must be at c, then it has to lose energy by changing (lowering) its frequency. Same if it loses gravitation PE.
     
  21. Jun 20, 2003 #20

    marcus

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    BTW Warren you were right about the size of the reacting core---at least Allen's handbook "Astrophysical Quantities" agrees with you. Almost all the fusion goes on in the inner 25 percent.

    I finally looked it up.

    Inside a slightly larger ball, radius = 0.3 of solar radius, Allen's model shows 0.998 of the fusion happening. Inside a ball of 0.25 radius, your figure, essentially all the fusion happens.

    I was just guesstimating order of magnitude when I said 10 percent, remembering onion-diagrams in textbooks and vaguely recalling lectures on stellar structure.

    In fact inside 0.1 radius, according to Allen's model, only 42 percent of the fusion happens.

    Your figure of 0.25 radius was realistic.
     
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