1. Schwarzschild radius of a stellar-mass black hole
R_s = 2GM/c² = 2 × 6.674e-11 × 9.945e+30 / (2.998e+08)² = 14769.2 m ≈ 14.8 km.
| Criterion | Points |
|---|---|
| Correct mass conversion to kg | 5 |
| Correct formula application | 5 |
| Correct final answer in km | 5 |
Worked Solutions
2. Earth's Schwarzschild radius
R_s = 2 × 6.674e-11 × 5.97e24 / (2.998e+08)² = 8.9 mm ≈ 0.9 cm. Earth's actual radius (6,371 km) is about 7.19e+08 times larger than its Schwarzschild radius, so Earth would need to be compressed by roughly that factor to become a black hole.
| Criterion | Points |
|---|---|
| Correct Schwarzschild radius (order of a few cm) | 8 |
| Correct compression-factor comparison | 7 |
3. Enclosed mass from the rotation curve
R = 14 kpc = 4.320e+20 m, V = 224 km/s = 2.240e+05 m/s. M = (2.240e+05)² × 4.320e+20 / 6.674e-11 = 3.248e+41 kg = 1.633e+11 M☉.
| Criterion | Points |
|---|---|
| Correct unit conversions | 8 |
| Correct final enclosed mass | 7 |
4. Textbook problem: dark matter fraction
Dark mass at 14 kpc ≈ 1.633e+11 − 6.000e+10 = 1.033e+11 M☉. Dark-to-visible ratio ≈ 1.72, i.e. roughly 1.7 times as much dark mass as visible mass is enclosed within 14 kpc.
| Criterion | Points |
|---|---|
| Correct dark-mass subtraction | 8 |
| Correct ratio and interpretation | 7 |
Grading Notes
Award full marks for correct method, units, and a numeric result consistent with the arithmetic shown (allow reasonable rounding). Verify the student re-derives the number rather than quoting it from the lecture without adapting it to the problem's specific inputs.