I have read recently in many websites dedicated to science that China has made an artificial sun hotter than the 'sun'. My question is how did they make a thing that can withstand such a high temperature and not burn the earth ?
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[Physics] Question about China’s Artificial Sun
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The reaction rate doesn't increase that quickly with temperature, but pressure does. If you perturb a solar model, making a zone near the core marginally hotter, the increased pressure will rapidly (at roughly the soundspeed divided by a characteristic length) cause it to expand. That lowers the pressure and temperature enough to substantially quench the reaction. So for normal stars during the long early phases of their evolution, the stability of thermonuclear burning is actually very high. There are some cases in stellar eveolution where pressure becomes insensitive to temperature, generally when the density is so high that electrons average kinetic energy is (almost) independent of temperature. In which case, fusion reactions can accelerate dramatically, but in most cases once the temperature is high enough to increase the pressure, the affected parts of the star rapidly expand and quench the reaction. When a sunlike star intitiates Helium fusion such a runaway reaction referred to as a Helium flash ocurrs, but it is easily contained. On the other hand, when a white dwarf accumulates a critical mass of material to intiate carbon burning, the reaction isn't contained and the star is destroyed in a class 1-a supernova. As dmkee points out, this differs from the core collapse supernova of massive stars.
A useful thing to remember is the virial theorem. If the star (or a portion of it) expands because of an increase in temperature, then once the system reaches hydrodynamic equilibrium the temperature must decrease, as the virial theorem determines that kinetic energy is proportional to gravitational binding energy. Thus unless the reaction=>heating=>reaction feedback is faster than the time to reach hydrostatic equilibrium, the star should be stable.
So a while ago I did a little project where I grabbed a "standard solar model" from this paper, which gives me some information that's useful for actually making an estimate. (Unsurprisingly the link given to download the data has changed in the last ten years; I haven't sleuthed to see whether the data is still publicly available.)
Only about 1.5% of the sun's mass is anything other than hydrogen and helium-4. This is true all the way out from the core to the surface. We'll assume that the sun contains only hydrogen and helium-4.
All but the outermost 0.2% of the sun's mass (out to 90% of the sun's radius) is at a temperature $kT>54\,\mathrm{eV}$, which is the energy needed to turn $\mathrm{He^+}$ into $\mathrm{He^{2+}}$. (This energy is four times the Rydberg energy.) So somewhere above 99% of the sun's mass is completely ionized.
The core temperature $kT\approx 1300\,\mathrm{eV}$ is much less than the electron mass, so the matter in the core is not relativistic.
I'm going to assume that the electrons aren't degenerate; this tool (via this question) makes me think that's a pretty safe assumption for matter at the core with density $\rho \approx 150\,\mathrm{g/cm^3}$ and temperature $T \approx 10^7\,\mathrm K$.
In that case we can treat the core of the sun as a mixture of three non-interacting ideal gases, $\mathrm H^+$, $\mathrm{He}^{2+}$, and $\mathrm e^-$. As
George Herold says, each ideal gas particle has mean kinetic energy $\frac32 kT$, so we'll want the number densities. The number density for hydrogen $n_\mathrm{H}$ is
$$
n_\mathrm{H} = \rho f_\mathrm{H}/{\mu_\mathrm{H} }
$$
where $\rho$ is the mass density, $f_\mathrm{H}$ is the hydrogen mass fraction, and $\mu_\mathrm{H} = 1\,\mathrm{gram/mole}$ is the atomic mass of hydrogen. You have a similar expression for helium (with $\mu_\mathrm{He} = 4\,\mathrm{gram/mole}$). The electron number density, thanks to complete ionization, is just
$$
n_\mathrm{e} = n_\mathrm{H} + 2n_\mathrm{He}.
$$
Here's a figure showing temperature, mass density, and composition from my source above and number density as computed here:
Note that the horizontal scale (radius) is mass-weighted: you find about half the mass of the sun between 0.1 and 0.3 solar radii, so that interval takes up about half the horizontal axis. This is purely a visualization technique, so that your eye isn't distracted by the (relatively) cool, diffuse outer layers of the sun.
To find the total thermal energy density, we have to integrate. We find the thermal energy density $$ \epsilon = (n_\mathrm{H} + n_\mathrm{He} + n_\mathrm{e})\frac32 kT $$ and the volume of a thin shell at radius $r$ is $$ dV = 4\pi r^2 dr $$ This integral $\int\epsilon\, dV$ gives me a total stored kinetic energy $E=3.09\times10^{41}\,\mathrm{J}$, of which about 95% is contained within half the sun's radius.
Now, if the sun had uniform density you could estimate its gravitational potential energy, the energy that was released when all the pieces fell together, as $$ U_\text{uniform sphere} = -\frac35 \frac{GM_\text{sphere}^2}{R_\text{sphere}} = - 2.3\times10^{41}\,\mathrm J \text{ (uniformly dense sun)}. $$ That's pretty close to our stored heat! We can do a little bit better since we actually know the density profile of the sun, by finding the potential energy released as you lay down each spherical shell, $$ U = - \int_0^{M_\text{sun}} \frac{G M_\text{enclosed}(r)}{r} dM = -6.15\times10^{41}\,\mathrm{J}. $$ This gravitational self-energy is roughly twice the stored kinetic energy --- which a real astronomer would have predicted as a consequence of the virial theorem.
Best Answer
I'm far from being an expert on this topic but I might trigger some more detailed answers so I will try my best to give a short intuitive explanation.
The general Idea in most nuclear fusion experiments is to produce a very hot plasma in which the fusion will take place and keep it trapped via a strong magnetic field so that it will not touch the walls of the fusion reactor.
In contrast to the fusion inside the sun where the fusion can take place mostly due to the high density of the plasma combined with a in comparison to the experiments on earth low temperature the density of the plasma inside such a reactor is very low. Therefore the total amount (mass) of plasma that is inside the reactor at the same time is very low (I remember a few grams) . This means that the energy density inside the reactor stays relatively low although the plasma is extremely hot. Due to this relatively low energy density it is possible to find materials that can withstand the generated heat.
So shortly speaking: A few gram of very hot plasma emit the same amount of energy as a lot of not so hot plasma. The temperature itself does not matter as long as the total amount of heat that is radiated towards the reactor material is not too high.