Nuclear ignition of gas giant atmospheres

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  • #1
Tyro
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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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  • #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.
 
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  • #3
Tyro
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Originally posted by LURCH
You see, the energy state needed to fuse elements together requires very high temperature. But as soon as fusion begins, the material being fused expands very rapidly, causing the temperature to drop.

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)?
 
  • #4
Tyger
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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.
 
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  • #5
Tyro
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Originally posted by Tyger
are different than the fusion reactions in a weapon, in they 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.

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?
 
  • #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
 
  • #7
Tyro
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Originally posted by chroot
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


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]
 
  • #8
chroot
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Uh, Tyro. Duh. Do you think we're stupid?

- Warren
 
  • #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
 
  • #10
marcus
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Originally posted by chroot
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

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.
 
  • #11
marcus
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Originally posted by chroot
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

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.
 
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  • #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
 
  • #13
Tyger
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I seem to recall that the energy producing

part of the Sun's core is only about 60-100% larger than the Earth.
 
  • #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
 
  • #15
Tyger
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Hm,. let 's see

Earth diam.×2=26,000Km, Solar diam. 1,400,000KM, I get a factor of approx 55 or 1.8%.
 
  • #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
 
  • #17
Tyro
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Originally posted by chroot
Uh, Tyro. Duh. Do you think we're stupid?

- Warren

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.


Originally posted by chroot
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's core is really, really huge!

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

- Warren
 
  • #18
Tyro
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Originally posted by chroot
In the future, maybe you should think more.

- Warren

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.
 
  • #19
Tyro
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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.
 
  • #20
marcus
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Originally posted by chroot
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

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.
 
  • #21
marcus
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Originally posted by Tyro
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.


It makes two orders of magnitude difference----a factor of 100.
All life would be burnt up if your figure for the wattage of the sun were correct, Tyro.

Allen's handbook figure is 3.8 E26 watts. It is a standard reference in Astrophysics. But also one can easily calculate it from the solar constant.

In your calculation please verify your figure of 2.87E12 meters for the radius of the earth's orbit
and the formula used for the area of a sphere (should be 4piR2 and not simply piR2

Your figure for the radius of the earth's orbit might be too large by a factor of 19.
 
  • #22
chroot
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Originally posted by marcus
figure of 0.25 radius was realistic.
Thanks for the citation, I appreciate it. I was going to look up the core radius in Carroll and Ostlie last night, but ended up rock climbing somehow.

- Warren
 
  • #23
chroot
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Originally posted by Tyro
I brought that up because the post below of yours was a little ambiguous. Corrections to remove ambiguity outlined in red.
The first sentence was precise, and used the word 'core' appropriately. I assumed that before reading the second sentence, you would have read the first.

In any event, your orignal objection was "I think that: The suns core produces more energy per unit volume than a candle flame." The bottom line is: no one cares what you think. I hate to be so blunt, but you're new here and you're already ticking me off. You continued by explaining to me how my calculation was flawed, without ever actually doing any work to verify it. This is just not a scientific attitude, and will have to change if you're going to be a welcome member here. Capiche?
I don't like this comment. In the future, maybe you should reserve unconstructive comments to yourself.
If you'd close your mouth and spend a bit more time thinking, no such unconstructive comments would be forthcoming. It's cause-and-effect.

Now, enough personal flub.
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?
The figures I've seen indicate the sun loses about 10-14 solar masses per year due to the solar wind. I have no idea what you mean by "gravitational potential energy being fuelled by solar fusion," though.
Incidentally, do photons have gravitational potential energy?
The concept of gravitational potential energy is only valid in Newtonian physics. General relativity does not recognize any kind of gravitational field, and thus does not really recognize any form of gravitational potential energy. However, the concepts are still pretty pervasive.

As you are probably aware, Newtonian physics predicts that photons do not couple gravitationally (since they have zero mass). So if you're going to talk about photons, you really need to use general relativity.

Now, photons DO get redshifted and blueshifted while moving through curved spactime. For example, if you fire a photon off the surface of the earth straight upwards, the photon will be redshifted. You can think of this in a couple of ways -- as the photon "working against gravity," and trading energy for potential energy (the Newtonian picture which you described in your last paragraph), or as time running faster as you get further out of the gravitational influence of the earth.
and also because they are partly waves
I'm growing to hate the phrase 'wave-particle duality.' Quantum mechanics does not make a distinction between waves and particles. At the macroscopic scale, there is only one kind of thing. You can ascribe some classical particle properties or some classical wave properties to the things, depending upon the experiment, but the things themselves are neither waves nor particles -- nor some kind of mish-mash of waves + particles. The things just are what they are.

Quantum mechanically, all particles (even photons) are regarded as being the same things. Neutrons diffract just as well as photons, and are no less wavey.

- Warren
 
  • #24
Tyro
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Originally posted by chroot
The figures I've seen indicate the sun loses about 10-14 solar masses per year due to the solar wind. I have no idea what you mean by "gravitational potential energy being fuelled by solar fusion," though.

The only source of energy in the sun is nuclear fusion. For mass to leave the immediate vicinity of the sun, it has to gain gravitational PE. So GPE gain is sourced, or fuelled, from fusion.


Originally posted by chroot
The concept of gravitational potential energy is only valid in Newtonian physics. General relativity does not recognize any kind of gravitational field, and thus does not really recognize any form of gravitational potential energy. However, the concepts are still pretty pervasive.

I see. Thanks for the info.


Originally posted by chroot
As you are probably aware, Newtonian physics predicts that photons do not couple gravitationally (since they have zero mass). So if you're going to talk about photons, you really need to use general relativity.

Now, photons DO get redshifted and blueshifted while moving through curved spactime. For example, if you fire a photon off the surface of the earth straight upwards, the photon will be redshifted. You can think of this in a couple of ways -- as the photon "working against gravity," and trading energy for potential energy (the Newtonian picture which you described in your last paragraph), or as time running faster as you get further out of the gravitational influence of the earth.

Is there a way to reconcile the views of either, or are they exclusive with the former being somewhat inaccurate because it is too classical?

I don't know much about redshifting and blueshifting because my most advanced physics course before university did not over it in detail apart from alluding to it. What I meant by the reconciliation of views or the views being exclusive is...If you find the change in GPE, will it correspond exactly to the change in energy due to a frequency change, or would it almost correspond with the frequency change, and the deficit being made up by what you are talking about - time running faster further out of the gravitational field?




chroot, about your post section on 'personal flub'...

Originally posted by chroot
The first sentence was precise, and used the word 'core' appropriately. I assumed that before reading the second sentence, you would have read the first.

Already your reply has ambiguity.

Originally posted by chroot
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!

You mean to say, "The second sentence was precise, and used the word 'core' appropriately. I assumed that before reading the third sentence, you would have read the second."

I wouldn't use the word precise. Let me put it this way - as you pointed out on this forum, I am new here. I have no idea who the knowledgeable and otherwise, posters are. Someone claims to have done some calculation, I question their assumptions. THAT IS ALL. Don't get your panties in a bind over me just being thorough and checking to see if the right assumptions have been made.

Your sentence would be clear if someone knew you know what you are talking about, for someone who doesn't and is checking assumptions, the slight ambiguity there may point to a mistaken assumption.


Originally posted by chroot

In any event, your orignal objection was "I think that: The suns core produces more energy per unit volume than a candle flame." The bottom line is: no one cares what you think. I hate to be so blunt, but you're new here and you're already ticking me off. [/b]

Really, is that why you replied to the post, and are replying again? I think - again - and I know you are going to care - that you are just being "blunt" because you are angry and you can not make a direct insult without moderator intervention. So you make an oblique one.


Originally posted by chroot
You continued by explaining to me how my calculation was flawed, without ever actually doing any work to verify it. This is just not a scientific attitude, and will have to change if you're going to be a welcome member here. Capiche?[/b]

I COULD NOT DO ANY CALCULATIONS TO VERIFY IT BECAUSE I DO NOT KNOW YOUR ORIGINAL ASSUMPTIONS. So I do have to make, in order to check -- quite exactly as you say, in fact -- an assumption on your assumptions. And get flamed for it. Capiche?


Originally posted by chroot
If you'd close your mouth and spend a bit more time thinking, no such unconstructive comments would be forthcoming. It's cause-and-effect.[/b]

If you'd remove that rude tone in your posts I wouldn't have a problem with them. Even though I am new here I don't have a problem with the other posters, except you.

Bottom line chroot: From what you post, you seem to know what you are talking about. This is the observation of a new person here. So mistakes about the ambiguity of your post will less likely happen by virtue of the reliability of the source - something inherently difficult to quantify online, I might add. I just think your posts could do with a little less caustic side comments. I've added a few to my responses to your responses. Do you like them? Would you want other people to feel that same way?

Originally posted by chroot
Now, enough personal flub.

Of course.
 
  • #25
chroot
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Originally posted by Tyro
You mean to say, "The second sentence was precise, and used the word 'core' appropriately. I assumed that before reading the third sentence, you would have read the second."
For ****'s sake, you really need to get laid.

- Warren
 
  • #26
chroot
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Oh, and by the way, my claim was:
The sun's core produces less energy per unit volume than does a candle flame.
to which you replied
I COULD NOT DO ANY CALCULATIONS TO VERIFY IT
You COULD NOT DO ANY CALCULATIONS? Why not? Are the solar luminosity and core radius known only by me or something? *snicker* You're a funny one.

- Warren
 
  • #27
Tyro
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Originally posted by marcus
It makes two orders of magnitude difference----a factor of 100.
All life would be burnt up if your figure for the wattage of the sun were correct, Tyro.

Allen's handbook figure is 3.8 E26 watts. It is a standard reference in Astrophysics. But also one can easily calculate it from the solar constant.

In your calculation please verify your figure of 2.87E12 meters for the radius of the earth's orbit
and the formula used for the area of a sphere (should be 4piR2 and not simply piR2

Your figure for the radius of the earth's orbit might be too large by a factor of 19.

When I talked about it not making much of a difference, check the low energy/volume figures you are getting. It is not going to make much of a difference in the conclusion that the energy/volume of the sun < energy/volume of a candle.

My calculations were based on using the solar constant, which was constant throughout the calculations and is what would determine whether we get burnt up or not. So, all life would not get burnt up in one sense.

marcus - I am not saying your calculations are wrong. I am asking why mine are wrong. So quoting your reference is unnecessary - I know when there is an error.

As for verifying the dimensions of Earth's orbit, I got it by pulling off the first figure I could Google out (link below):

here

Edit:
Whoops. I found the mistake. I was reading the info off the first link too fast and didn't read the header. I am getting 3.874e26 now - so that is fine now. Earth's orbital radius should be 1.5e11m. The one I quoted was for Uranus.



Whoops @ the factor of '4'. Summer heat quite literally cooks my brain . But the conclusion is still the same and still out by a magnitude of 100.


Basically, what are the assumptions made in calculating the sun's power output?
  • Part of the sun's energy goes to radiant energy, the rest to gravitational energy to the outbound mass flux. Is this included in the power output calculations?
  • If the sun's power output is calculated using the solar constant, does the solar constant itself involve energies within the visible and near visible wavelengths, or beyond it?

Both of these factors could potentially increase the actual value of the sun's power output, although it seems that the conclusion made previously regarding a candle/sun would be the same because of the sheer minuteness of the figure.

*Goes to look up the definition of solar constant to check for its assumptions.*

Note: From these calculations, it seems that the solar energy output does not take into account the energy lost to GPE from outwards mass flux.
 
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  • #28
Tyro
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Originally posted by chroot
Oh, and by the way, my claim was:

to which you replied

You COULD NOT DO ANY CALCULATIONS? Why not? Are the solar luminosity and core radius known only by me or something? *snicker* You're a funny one.

- Warren

*snicker* You did not provide me with any of YOUR assumptions before you started. If I started making assumptions and my assumptions were different from yours, of course our results will differ.

Your assumptions only became clearer in your NEXT post so from there only can I start doing calculations. Capiche?
 
  • #29
chroot
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I'm going to ask you a second time.

What assumptions are necessary to verify or disprove the following statement:
The sun's core produces less energy per unit volume than does a candle flame.

- Warren
 
  • #30
Tyro
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Originally posted by chroot
I'm going to ask you a second time.

What assumptions are necessary to verify or disprove the following statement:
Originally posted by chroot

The sun's core produces less energy per unit volume than does a candle flame.

- Warren

Unless I have been reading too fast again or you edited a previous post, this is the first time you asked me what assumptions are necessary.

Assumptions/info needed:
  1. Assumed radius of region actively undergoing nuclear fusion.
  2. Whether this is the same as your definition of 'core'.
  3. Your value for the power output of the sun.
  4. Actually, power outputs for a candle you are using too.
  5. Your definition of a candle flame - to determine the volume of the reacting area for comparison.
  6. Finally, checking that the assumptions first made are relevant to the problem. (E.g. the solar power output seems dubious because it seems to have been derived purely from radiant energy calculations)
    [/list=1]

    I have to check that I am starting off with the right assumptions as you and that you are starting off with the right assumptions, before barreling of to do a calculation. It is just that. Different assumptions, different results.
 
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  • #31
chroot
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Originally posted by Tyro
Assumptions/info needed:
You are misusing the term 'assumption.' What you really mean is prior, a result of a previous experiment. An assumption has no experimental basis. I made no assumptions, but I did make use of priors.

- Warren
 

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