3 ms·

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, actual

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.
  - scythe 15y ago
    >Thanks, I was not aware that the surface of the Sun is very bright. Well, you also clearly weren't aware of the point I was making: if you ask a misleading question, you get a misleading answer. >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. The 'laypeople' are basically right, though: if you actually extract the energy from a bit of thermal plasma it only takes a "pinhead" to supply energy for an entire human life. Obviously the energy being actually emitted is much smaller, just as there is no light emitted from a box with a bright light in it, but that doesn't mean that the light in the box couldn't easily light up a room. Like, the amount of 15000000 degree plasma at the core of the Sun that could destroy a small city if it were released is probably no larger than my fist. >rather nasty fusion products There really aren't any seriously nasty fusion products -- tritium and positrons, the latter being instantly destroyed and the former already present in the atmosphere.

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.
- scythe 15y ago
  >Thanks, I was not aware that the surface of the Sun is very bright. Well, you also clearly weren't aware of the point I was making: if you ask a misleading question, you get a misleading answer. >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. The 'laypeople' are basically right, though: if you actually extract the energy from a bit of thermal plasma it only takes a "pinhead" to supply energy for an entire human life. Obviously the energy being actually emitted is much smaller, just as there is no light emitted from a box with a bright light in it, but that doesn't mean that the light in the box couldn't easily light up a room. Like, the amount of 15000000 degree plasma at the core of the Sun that could destroy a small city if it were released is probably no larger than my fist. >rather nasty fusion products There really aren't any seriously nasty fusion products -- tritium and positrons, the latter being instantly destroyed and the former already present in the atmosphere.