# Seismic moment — optional lecture-formula practice

Subject: Disaster Management

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/dm/#tee/practice/calculations/moment

Canonical: https://deep-learning-study-room.vercel.app/read/dm/tee/practice/calculations--moment/

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 application · Same magnitude, different local effects

One earthquake has a source magnitude, but a nearby settlement with vulnerable buildings may experience stronger damaging effects than a distant settlement on firm ground. A one-unit magnitude difference means a tenfold amplitude ratio in the stated logarithmic comparison; it does not mean ten times the damage.

### 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 application · Same magnitude, different local effects

One earthquake has a source magnitude, but a nearby settlement with vulnerable buildings may experience stronger damaging effects than a distant settlement on firm ground. A one-unit magnitude difference means a tenfold amplitude ratio in the stated logarithmic comparison; it does not mean ten times the damage.

### 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.

#### Illustrative application · Same magnitude, different local effects

One earthquake has a source magnitude, but a nearby settlement with vulnerable buildings may experience stronger damaging effects than a distant settlement on firm ground. A one-unit magnitude difference means a tenfold amplitude ratio in the stated logarithmic comparison; it does not mean ten times the damage.

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