GATE-2018 (24) :Wyllie – Time average equation

 

54 and 55) The seismic slip of a fault after an earthquake is measured to be 0.5 m and the fault area is estimated to be 250 km². The rigidity of the medium surrounded the fault is 30 Gpa.

Then calculate

1)      The seismic moment(in Nm) of the earthquake

2)      The moment magnitude of the earthquake.

(Thanks to Chandrasekhar, ANU)

Solution:

 Given that

Seismic slip(S) = 0.5 m

Rigidity of Modules (µ) = 30 GPa = 30*10⁹ Pa

Ares of the Fault (D) = 250 km² = 250*10⁶ m²

 Seismic moment (M₀) =?

Moment magnitude (M_w) =?

 

The Seismic moment can be defined as

$M_{0}= \mu \times S\times D$

 

$M_{0}= (30\times 10^{9}) \times (0.5)\times (250\times 10^{6})$

 

$M_{0}= 3750\times 10^{15}Nm$

 

$M_{0}= 3.750\times 10^{18}Nm$

                                                           

 The Moment magnitude of the earthquake can be defined as

$M_{w}= \frac{2}{3}(\log(m_{0})-9.1)$

 

$M_{w}= \frac{2}{3}(\log(3.75\times 10^{18})-9.1)$

 

$M_{w}= \frac{2}{3}(18.5740-9.1)$

 

$M_{w}= \frac{2}{3}(9.474)$

 

$M_{w}= 6.316$