4. Forward problem — worked example
A circular rupture, A=πR2, on a crustal fault.
Given: R=5 km, sˉ=0.4 m, μ=30 GPa.
M0=(3×1010)⋅π(5×103)2⋅0.4≈9.4×1017N⋅m
MW=32log10(9.4×1017)−6.03≈5.93
→ Realistic for a small subduction event.
Read more → Lecture 15 §4