1. Central pressure
M = 2.1457e+30 kg, R = 8.5084e+08 m. P_c = (3/8π)GM²/R⁴ = 6.999e+13 Pa.
| Criterion | Points |
|---|---|
| Correct unit conversions | 8 |
| Correct final pressure | 7 |
Worked Solutions
2. Comparison to the Sun
Alpha Centauri A's estimated central pressure (6.999e+13 Pa) is 0.52 times the Sun's (1.345e+14 Pa). Because P_c ∝ M²/R⁴, and Alpha Centauri A has both larger mass (raising P_c) and larger radius (lowering P_c, more strongly since R is raised to the fourth power), the net effect and its sign depend on which scaling dominates; the computed ratio above shows which one wins for these specific values.
| Criterion | Points |
|---|---|
| Correct computed ratio | 8 |
| Correct qualitative M vs. R⁴ scaling argument | 7 |
3. Virial mass of 47 Tucanae
M ≈ 5σ²R_h/G = 5.841e+05 M⊙.
| Criterion | Points |
|---|---|
| Correct unit conversions (km/s to m/s, pc to m) | 12 |
| Correct final virial mass | 8 |
4. Agreement check
Percent agreement = |5.84e+05 − 6.40e+05| / 6.40e+05 × 100% = 8.7%, a similar order-of-magnitude level of agreement to the M15 case in Lab 03, consistent with the single-mass virial estimator being a genuine but approximate (not exact) mass estimator for real, non-uniform clusters.
| Criterion | Points |
|---|---|
| Correct percent-agreement arithmetic | 8 |
| Correct comparison to the M15 precedent | 7 |
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 (AU/m, years/seconds, km/Mpc, Msun/kg) -- these are the most common source of order-of-magnitude mistakes in this course.