The Missing Source of the Sun’s Energy: As photons from the Sun’s interior leak

Question

The Missing Source of the Sun’s Energy: As photons from the Sun’s interior leak away, they are replaced with new ones. But what is the source of those new photons? In the core of the Sun, the gas is hot and thermodynamics tells us that the thermal energy per unit volume (the thermal energy density) stored in the motion of the gas particles is Ethermal = 3P/2.

Using the Sun’s central pressure that you calculated in question 3, what is the thermal energy density, Ethermal = 3Pc/2? In the previous question we saw that the radiation energy density is Erad = aT4 (recall that a = 7.56 × 1015erg cm3 K4). What is the ratio of the thermal energy density to the radiation energy density, Ethermal/Erad, assuming a central temperature of Tc = 1.5 × 107 K? (Notice that the thermal energy density is much larger than the radiation density.) The thermal energy in the Sun acts as a reservoir for new photons. Let us imagine that the thermal reservoir is the only source of new photons which leak out of the Sun. Let’s find out how long the Sun could shine.

** suns central pressure is believed to be 2.1388537 X 10^11
   Value = in previous question:
      The Luminosity of the Sun from Random Walking: The conditions inside the Sun imply that a photon has a mean free path length of about l = 0.5 cm; that means on average a photon “flies” about half of a centimetre before interacting or bumping into matter. Imagine a photon emitted at the center of the Sun, but only in one dimension—it can move right (+0.5 cm) or left (0.5 cm) at each time step with equal probability. The average position is zero, the photon has no net drift pushing it to the left or to right. However, the square of the displacements does have a net drift—even though on average photons are just as likely to be to the left or to the right, over a long period of time they will have had a chance to wander a long way from the centre, some far to the left and some far to the right (and no, they are not political photons!). The mean square distance after N steps is R 2 = Nl2. In three dimensions, the result becomes 3R 2 = Nl2. Therefore, the number of steps that a photon needs to take on average to get to the photosphere from the center is N = 3R2S/l2. The random walk time, t, is just the total distance covered divided by the speed of light (c = 3.0 × 1010 cm/s) , t = Nl/c = 3R2S/(lc). A hot gas at temperature T (in Kelvin) has radiation energy per unit volume of aT4, where a = 7.56 × 1015erg cm3 K4. The “leakage” of photons from the Sun—which is the Sun’s luminosity—must be:

LS = (volume) × (radiation per unit volume) random walk time to cover distance RS

= (4R3S/3)(aT4). 3 R 2S / ( l c )

(a) Using the Sun’s core temperature that you calculated in the previous question, what does the above argument give for the Suns luminosity. Compare your result with the observed value of LS = 3.90 × 1033 erg/sec. How does the comparison work when using a temperature of T = 4.5 × 106 K? What does this result tell you about the average temperature of the Sun compared to the core temperature?

Explanation / Answer

Accoridng to the given problem,

1.) Using the relations,

Pc ( central pressure) = 2.1388537*1011 Barye

E-thermal = 3Pc/2 = 3.2083*1011 Barye

E-rad = aT4 = 7.56*10-15 * (1.5*107)4 = 3.8275*1014 Barye

Then the ration is,

E-thermal/E-radiation = 1192.93.

As per chegg guidlines only 1Q per chance, please post the other question separately.

  I hope you understood the problem, If yes rate me!! or else comment for a better solution.

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