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.

Recall from the previous question** that a photon created a the center of the Sun takes 3R2S/(lc) amount of time to randomly walk out. How many years does a photon’s random walk take?

** Previous question**
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 × 10−15erg cm−3 K−4). 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.

Answer 1

FIrst, Considering your previous question, we have,

Pc ( central pressure) = 2.16024*10^17 Barye
E-thermal = 3Pc/2 = 3.24036*10^17 Barye
E-rad = aT^4 = 7.56*10^-15 * (1.5*10^7)^4 = 3.8275*10^14 Barye
E-thermal/E-radiation = 846.654

Now Comes to Your Question that you have asked, . How many years does a photon’s random walk take?

   Here, time = 3*Rs^2 / lc = 3*(6.96*10^10 )^2 / 0.5*3*10^10 = 9.68832*10^11 s = 30721.46119 years

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