1. Schematic core temperature
T_core = 1.570e+07 × (1.0788)^0.7 = 1.656e+07 K = T6 16.56.
| Criterion | Points |
|---|---|
| Correct exponent application | 10 |
| Correct T6 conversion | 10 |
Worked Solutions
2. Gamow peak energy
E0 = 1.22(1·1·0.5·16.56²)^(1/3) = 6.29 keV, essentially the same order of magnitude as the Sun's own value (6.07 keV), since Alpha Centauri A's core temperature is close to the Sun's.
| Criterion | Points |
|---|---|
| Correct Gamow-peak arithmetic | 8 |
| Correct comparison to the Sun | 7 |
3. pp/CNO classification
Alpha Centauri A's T6 (16.56) is well below the 24.2 crossover, so it is pp-dominated, consistent with its near-solar mass and G2V spectral type (very similar to the Sun's own pp-dominated core).
| Criterion | Points |
|---|---|
| Correct classification | 8 |
| Correct connection to mass/spectral type | 7 |
4. Comparative reasoning
Because ε_CNO ∝ T6²⁰, an extremely steep power law, even the modest 1.91× mass ratio (via the T ∝ M^0.7 scaling) produces a large relative change in core temperature ratio (1.57), and raising that ratio to the 20th power amplifies it enormously -- explaining why such a modest mass difference flips the dominant mechanism entirely.
| Criterion | Points |
|---|---|
| Correct quantitative use of the T6²⁰ sensitivity | 12 |
| Correct connection to the modest mass/temperature ratio | 8 |
Grading Notes
Award full marks for correct method, units, and a numeric result consistent with the arithmetic shown (allow reasonable rounding). Verify that the student re-derives the number for this problem's specific inputs rather than quoting a lecture example without adaptation. Watch specifically for unit-conversion errors (km/AU/m, years/seconds, kg/M⊙) and for comparative-magnitude statements stated in the wrong direction (e.g., "X times greater" versus "a small fraction of") -- these are the most common sources of order-of-magnitude mistakes in this course.