Thermonuclear Fusion in the Sun and Other Stars — Question 6

Section 43-6 · Physics pages 7262–7263

Question 6 A steeper temperature dependence lets CNO burning overtake the pp chain At fixed density and composition, use supplied local laws ϵpp ∝T 4 and ϵCNO ∝T 17. At T0, their ratio is .0100. Treat the laws as valid over the interval considered. T/T0 ϵCNO/ϵpp equal Schematic; labels, not drawing lengths, specify the data. (a) Find the ratio at 1.10T0. (b) Find the temperature ratio at which the contributions become equal and identify the key extrapolation assumption.
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Question 6 - Solution A steeper temperature dependence lets CNO burning overtake the pp chain T/T0 ϵCNO/ϵpp equal (a) Method and calculation ϵCNO ϵpp = .0100(T/T0)13 . ϵCNO/ϵpp|1.10T0 = .0100(1.10)13 = .0345227 . (b) Interpretation and further calculation .0100(T/T0)13 = 1 ⇒T/T0 = 1001/13 = 1.42510 . d ln(ϵCNO/ϵpp) d ln T = 17 −4 = 13. Physical check The powers are local approximations and composition includes catalyst abundance. A real crossover requires temperature-dependent reaction rates and stellar structure.

Question 6

A steeper temperature dependence lets CNO burning overtake the pp chain

At fixed density and composition, use supplied local laws ϵpp∝T4\epsilon_{pp}\propto T^4 and ϵCNO∝T17\epsilon_{\mathrm{CNO}}\propto T^{17}. At T0T_0, their ratio is .0100. Treat the laws as valid over the interval considered.

Diagram shown in the original SVG worksheet.

  1. Find the ratio at 1.10T01.10T_0.

  2. Find the temperature ratio at which the contributions become equal and identify the key extrapolation assumption.

Question 6 - Solution

A steeper temperature dependence lets CNO burning overtake the pp chain

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ϵCNOϵpp=.0100(T/T0)13.\boxed{\frac{\epsilon_{\mathrm{CNO}}}{\epsilon_{pp}}=.0100(T/T_0)^{13}}.ϵCNO/ϵpp|1.10T0=.0100(1.10)13=.0345227.\boxed{\left.\epsilon_{\mathrm{CNO}}/\epsilon_{pp}\right|_{1.10T_0}=.0100(1.10)^{13}=.0345227}.

(b) Interpretation and further calculation

.0100(T/T0)13=1⇒T/T0=1001/13=1.42510..0100(T/T_0)^{13}=1\Rightarrow\boxed{T/T_0=100^{1/13}=1.42510}.dln⁡(ϵCNO/ϵpp)dln⁡T=17−4=13.\frac{d\ln(\epsilon_{\mathrm{CNO}}/\epsilon_{pp})}{d\ln T}=17-4=13.

Physical check

The powers are local approximations and composition includes catalyst abundance. A real crossover requires temperature-dependent reaction rates and stellar structure.