Disaster Management · TEE · practice

Seismic moment — optional lecture-formula practice

Optional calculation practice

Supplementary worked examples · no numerical question found in the available 12 TEE sittings

Source of the calculations. M2 PDF p 55 teaches seismic moment and moment magnitude. The three sets of values below were authored for practice, not taken from a recovered question paper. SI conversion and explicit unit conditions clarify the lecture formula.

Definition. Seismic moment measures source strength using rock rigidity, slipped fault area and average slip. Moment magnitude converts this physical measure into a logarithmic magnitude.

Simple meaning. Pehle moment ke units dekho, tab magnitude wali constant choose karo.

M_0=\mu A D=\mu LWD

Symbol	Meaning	SI unit
μ	Shear modulus / rigidity of rock	N/m² (Pa)
A = L × W	Area of the fault that slipped	m²
D	Average slip	m
M₀	Seismic moment	N·m

M_w=\frac{2}{3}\left(\log_{10}M_0-9.1\right)\quad(M_0\text{ in N m})

M_w\approx\frac{2}{3}\log_{10}M_0-10.7\quad(M_0\text{ in dyn cm})

The second convention is printed in the lecture without its unit condition. Do not insert N·m into the −10.7 formula. Since 1 N = 10⁵ dyn and 1 m = 100 cm, 1 N·m = 10⁷ dyn·cm. The constants are rounded conventions, so the two formula results may differ slightly at the last decimal.

Illustrative calculation (not a PYQ). A fault slips over an area 10 km long and 5 km wide with average slip 1 m. Rock rigidity is 3 × 10¹⁰ N/m². Calculate seismic moment and moment magnitude using SI units.

Answer

Given

L=10 km; W=5 km; D=1 m; μ=3×10¹⁰ N/m².

To find

M₀ and M𝑤.

Formula and why

M_0=\mu LWD;\quad M_w=\tfrac23(\log_{10}M_0-9.1)

Convert lengths to metres before forming area so the moment is in N·m.

Step-by-step solution

Step 1. Convert lengths

L=10\times1000=10000\text{ m};\quad W=5000\text{ m}

Step 2. Find the slipped area

A=10000\times5000=5\times10^7\text{ m}^2

Step 3. Multiply rigidity, area and slip

M_0=(3\times10^{10})(5\times10^7)(1)=1.5\times10^{18}\text{ N m}

Step 4. Take the logarithm

\log_{10}(1.5\times10^{18})=18.176091

Step 5. Convert to magnitude

M_w=\tfrac23(18.176091-9.1)=6.050727

Final answer: M₀ = 1.5 × 10¹⁸ N·m; M𝑤 ≈ 6.05.

Illustrative calculation (not a PYQ). The seismic moment is 10²⁵ dyn·cm. Calculate moment magnitude using the lecture convention.

Answer

Given

M₀=10²⁵ dyn·cm.

To find

M𝑤.

Formula and why

M_w=\tfrac23\log_{10}M_0-10.7

The moment is already in dyn·cm, so this constant is appropriate.

Step-by-step solution

Step 1. Take log

\log_{10}(10^{25})=25

Step 2. Multiply then subtract

M_w=\tfrac23\times25-10.7=16.666667-10.7=5.966667

Final answer: M𝑤 ≈ 5.97. Do not compute ⅔ × (25 − 10.7); that is a different expression.

Illustrative calculation (not a PYQ). If area and rigidity remain fixed but slip doubles, how do moment and moment magnitude change?

Answer

Given

μ and A fixed; D₂ = 2 D₁.

To find

M₀ ratio and ΔM𝑤.

Formula and why

M_0=\mu AD;\quad \Delta M_w=\tfrac23\log_{10}(M_{02}/M_{01})

Multiplication makes moment proportional to slip; subtracting the magnitude equations cancels the common constant.

Step-by-step solution

Step 1. Find moment ratio

M_{02}/M_{01}=2

Step 2. Take log

\log_{10}(2)=0.301030

Step 3. Find magnitude change

\Delta M_w=\tfrac23\times0.301030=0.200687

Final answer: Moment doubles; magnitude increases by about 0.201, not by a factor of two.

Yaad rakho: Moment = rigidity × area × slip; magnitude = log of moment.