10 ms·

The 6720000 joules number is misleading IMO. According to http://www.sal.wisc.edu/~khn/ast100/lectures/lecture16.pdf http://www.sal.wisc.edu/~khn/ast100/lectur

The 6720000 joules number is misleading IMO. According to http://www.sal.wisc.edu/~khn/ast100/lectures/lecture16.pdf http://www.sal.wisc.edu/~khn/ast100/lectures/lecture16.pdf the energy produced by a kilo of the sun is orders of magnitude higher. (something like 2.3×10e13 Joule/Kg). Of course it doesn't produce it in a measly 100 years (more like 100.000). At one time, only a very tiny fraction of the sun is being processed in it's fusion reactor at the core ( The core, inside 0.24 solar radius, generates 99% of the fusion power of the Sun. From http://en.wikipedia.org/wiki/Solar_core http://en.wikipedia.org/wiki/Solar_core ), and that's what people think about when talking about the sun's energy (the 2.3×10e13 Joule/Kg thing, not the 672000 one).

uvdiv 15y ago
yaakov34 is talking about power density, not energy density. The sun's core power density only about 0.3 watts/liter, compared with >100,000 W/L for a power-generating nuclear fission reactor. http://en.wikipedia.org/wiki/Sun#Core http://en.wikipedia.org/wiki/Sun#Core
- GFischer 15y ago
  Thanks for the clarification and putting it into perspective (I wasn't getting what yaakov wanted to convey). And yes, that does make things more difficult :) .
yaakov34 15y ago
Not 100,000 years, he is talking about the total energy produced over the Sun's lifetime of billions of years (his calculation is for the last 4 billion, actually). The number is still surprisingly small - 2.3e13J is only a few billion kcal, quite an imaginable number. Look at this another way: ask a layperson, who consumes 3000 kcal a day approximately, how much of the Sun's mass would be required to supply him with this energy input. The answer, unless I mislaid a zero, is 755 tons or so. Now, I will absolutely be a monkey's uncle if laymen give you an approximately correct answer. My guess is that people will say that it takes a tiny drop of solar material, or a microscopic amount. Can someone try this?
- scythe 15y ago
  This is mostly because the visible portion of the Sun is not currently undergoing nuclear fusion. Fusion occurs only in the core, not in the outer layers. Also, you're ignoring feedback, which isn't intuitive to ignore -- the blackbody radiation produced by a kilo of 5000 K plasma is huge, enough to supply a person's daily energy and then some -- it just doesn't get emitted from the Sun, which is why that energy will last for another ten billion years. A layperson would probably (try to) count all of the thermal energy which is emitted by such a plasma whenever it's not totally surrounded by more of the same plasma. In that sense, the layperson would be correct -- a kilo of Sun-matter, after being removed from the Sun -- produces a massive amount of power. The density of the Sun is approximately 2 * 10^-4 kg/m^3 in the photosphere ( http://en.wikipedia.org/wiki/Sun http://en.wikipedia.org/wiki/Sun ). We're ignoring the core for now, because that's at more like 15000000 K, so. A kilo thus takes up 5 * 10^3 m^3, which is a sphere with radius ~12 m, having a surface area of ~1.4 * 10^3 m^2. The power radiated by a kilo of plasma from the Sun's photosphere is thus is j' = area * sigma * temperature^4, where sigma = 5.67 * 10^-8 is the Stefan-Boltzmann constant ( http://en.wikipedia.org/wiki/Blackbody_radiation#Stefan.E2.80.93Boltzmann_law http://en.wikipedia.org/wiki/Blackbody_radiation#Stefan.E2.8... ), so j' = 1.4 * 10^3 * 5.67 * 10^-8 * (5 * 10^3)^4 = 4.96 * 10^10 watts, or well over a person's daily caloric intake every second (In fact, this is approximately the energy consumption of a small country).
  - uvdiv 15y ago
    The photosphere does not produce power, so attempts to calculate volumetric energy densities there are nonsensical. The power radiated from 1 square meter of the sun's surface (63 megawatts) comes from a cross section of the core, far below, that is 250,000 kilometers deep!
  - yaakov34 15y ago
    Thanks, I was not aware that the surface of the Sun is very bright. If we're going to analyze the performance of our "reactor", then we need to include the fusing fuel, the "confinement", which prevents us from being very quickly destroyed by rather nasty fusion products, and the photosphere, which is very beneficial to us. I continue to claim that if you ask laymen about the mass of their "personal portion" of this reactor, you will get answers that are about 8 orders of magnitude smaller than in reality. I actually tried this just now, with expected results ("some kind of a tiny pinhead thing"). There is no point to this exercise, other than to show that people have no intuitive vision of these scales; it doesn't make them stupid, the scales are just unintuitive and outside of their experience. I don't claim to have discovered anything about astrophysics.

uvdiv 15y ago

yaakov34 is talking about power density, not energy density. The sun's core power density only about 0.3 watts/liter, compared with >100,000 W/L for a power-generating nuclear fission reactor. http://en.wikipedia.org/wiki/Sun#Core http://en.wikipedia.org/wiki/Sun#Core

GFischer 15y ago
Thanks for the clarification and putting it into perspective (I wasn't getting what yaakov wanted to convey). And yes, that does make things more difficult :) .

yaakov34 15y ago

Not 100,000 years, he is talking about the total energy produced over the Sun's lifetime of billions of years (his calculation is for the last 4 billion, actually). The number is still surprisingly small - 2.3e13J is only a few billion kcal, quite an imaginable number. Look at this another way: ask a layperson, who consumes 3000 kcal a day approximately, how much of the Sun's mass would be required to supply him with this energy input. The answer, unless I mislaid a zero, is 755 tons or so. Now, I will absolutely be a monkey's uncle if laymen give you an approximately correct answer. My guess is that people will say that it takes a tiny drop of solar material, or a microscopic amount. Can someone try this?

scythe 15y ago
This is mostly because the visible portion of the Sun is not currently undergoing nuclear fusion. Fusion occurs only in the core, not in the outer layers. Also, you're ignoring feedback, which isn't intuitive to ignore -- the blackbody radiation produced by a kilo of 5000 K plasma is huge, enough to supply a person's daily energy and then some -- it just doesn't get emitted from the Sun, which is why that energy will last for another ten billion years. A layperson would probably (try to) count all of the thermal energy which is emitted by such a plasma whenever it's not totally surrounded by more of the same plasma. In that sense, the layperson would be correct -- a kilo of Sun-matter, after being removed from the Sun -- produces a massive amount of power. The density of the Sun is approximately 2 * 10^-4 kg/m^3 in the photosphere ( http://en.wikipedia.org/wiki/Sun http://en.wikipedia.org/wiki/Sun ). We're ignoring the core for now, because that's at more like 15000000 K, so. A kilo thus takes up 5 * 10^3 m^3, which is a sphere with radius ~12 m, having a surface area of ~1.4 * 10^3 m^2. The power radiated by a kilo of plasma from the Sun's photosphere is thus is j' = area * sigma * temperature^4, where sigma = 5.67 * 10^-8 is the Stefan-Boltzmann constant ( http://en.wikipedia.org/wiki/Blackbody_radiation#Stefan.E2.80.93Boltzmann_law http://en.wikipedia.org/wiki/Blackbody_radiation#Stefan.E2.8... ), so j' = 1.4 * 10^3 * 5.67 * 10^-8 * (5 * 10^3)^4 = 4.96 * 10^10 watts, or well over a person's daily caloric intake every second (In fact, this is approximately the energy consumption of a small country).
- uvdiv 15y ago
  The photosphere does not produce power, so attempts to calculate volumetric energy densities there are nonsensical. The power radiated from 1 square meter of the sun's surface (63 megawatts) comes from a cross section of the core, far below, that is 250,000 kilometers deep!
- yaakov34 15y ago
  Thanks, I was not aware that the surface of the Sun is very bright. If we're going to analyze the performance of our "reactor", then we need to include the fusing fuel, the "confinement", which prevents us from being very quickly destroyed by rather nasty fusion products, and the photosphere, which is very beneficial to us. I continue to claim that if you ask laymen about the mass of their "personal portion" of this reactor, you will get answers that are about 8 orders of magnitude smaller than in reality. I actually tried this just now, with expected results ("some kind of a tiny pinhead thing"). There is no point to this exercise, other than to show that people have no intuitive vision of these scales; it doesn't make them stupid, the scales are just unintuitive and outside of their experience. I don't claim to have discovered anything about astrophysics.