University Physics V · topic UPV-16.07
The proton-proton chain and stellar energy generation
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
The proton–proton chain and stellar fusion: predict a finite change
The proton–proton chain is the sequence p + p → d + e⁺ + ν, d + p → ³He + γ and ³He + ³He → ⁴He + 2p that turns four protons into one ⁴He nucleus, its overall rate fixed by the weak first step and its temperature dependence by that step's Gamow peak. Use the supplied topic model Y_out = 100/(1+10^(log(xh/xc)-21.25+51.45/t^(1/3))), where Y_out is CNO SHARE OF ε (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. t is Core temperature T_6, measured in MK. Keep xh (Hydrogen mass fraction X) = 0.35; xc (CNO mass fraction X_CNO) = 0.004 fixed. For this exercise use only 7.825 MK ≤ t ≤ 36.175 MK on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at t = 14.062 MK and 18.031 MK. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CNO SHARE OF ε against t
Predictions from the supplied The proton–proton chain and stellar fusion model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CNO SHARE OF ε / %
Horizontal axis (x): t: Core temperature T_6 / MK
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Core temperature T_6 / MK | CNO SHARE OF ε / % |
|---|---|
| 7.824999999999999 | 0.000024696237545724304 |
| 14.912499999999998 | 2.4797646402253717 |
| 22 | 89.83557033140933 |
| 29.0875 | 99.73674643752807 |
| 36.175 | 99.98227552555367 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 14.062 MK, substitute into Y_out = 100/(1+10^(log(xh/xc)-21.25+51.45/t^(1/3))): Y₀ = 0.972479 %.
- At t = 18.031 MK, the same fixed-parameter model gives Y₁ = 32.744 %.
- Signed change: ΔY_out = Y₁ − Y₀ = 31.7715 %; the input change is Δt = 3.969 MK.
- Average sensitivity is ΔY_out/Δt = 8.00491 % per MK. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.972479 %, Y₁ = 32.744 %; ΔY_out = 31.7715 %; average sensitivity = 8.00491 % per MK.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
The proton–proton chain and stellar fusion: infer an input with uncertainty
The proton–proton chain is the sequence p + p → d + e⁺ + ν, d + p → ³He + γ and ³He + ³He → ⁴He + 2p that turns four protons into one ⁴He nucleus, its overall rate fixed by the weak first step and its temperature dependence by that step's Gamow peak. Use the supplied topic model Y_out = 100/(1+10^(log(xh/xc)-21.25+51.45/t^(1/3))), where Y_out is CNO SHARE OF ε (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. t is Core temperature T_6, measured in MK. Keep xh (Hydrogen mass fraction X) = 0.35; xc (CNO mass fraction X_CNO) = 0.004 fixed. For this exercise use only 7.825 MK ≤ t ≤ 36.175 MK on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 98.213035 %, with a bounded reading uncertainty ±0.26539 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.002835 MK, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
CNO SHARE OF ε against t
Predictions from the supplied The proton–proton chain and stellar fusion model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CNO SHARE OF ε / %
Horizontal axis (x): t: Core temperature T_6 / MK
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Core temperature T_6 / MK | CNO SHARE OF ε / % |
|---|---|
| 7.824999999999999 | 0.000024696237545724304 |
| 14.912499999999998 | 2.4797646402253717 |
| 22 | 89.83557033140933 |
| 29.0875 | 99.73674643752807 |
| 36.175 | 99.98227552555367 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(t) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(7.825) = 0.0000246962 % and Y_out(36.175) = 99.9823 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 7.825 + (98.213 − 0.0000246962)(36.175 − 7.825)/(99.9823 − 0.0000246962) = 35.6733 MK. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives t = 25.1185 MK; substituting back gives Y_out = 98.213035 %.
- The interpolation error in the input is |t_line − t| = 10.5548 MK. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 97.947645 % and 98.478425 %. Sort the roots even if the response decreases: 24.8573 MK ≤ t ≤ 25.4255 MK.
- The lower and upper input deviations are 0.261163 MK and 0.307001 MK. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer t ≈ 25.1185 MK; compatible interval [24.8573, 25.4255] MK. Endpoint interpolation gives 35.6733 MK.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.