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

 

31) An earthquake causes an average of 25 m strike slip displacement over a 50 km long, 25 km deep portion of a transform fault. Assuming that the rock rigidity is 3 ×10¹⁰ Nm^-2, the Moment magnitude (Mw) of the earthquake is__________

 (Thanks to Chandrasekhar, ANU)

Solution:

 Given that Rock Rigidity (µ) =3 ×10¹⁰ Nm^-2

Strike slip(D) = 25 m

Length = 50km

Depth (width)= 25 km

The Area of the fault(S) = Length × Width

                                    = 50×25=1250 $km^2$

        =$1250\times 10^{6} m^2$

The Seismic Moment$M_{0}=\mu\times S\times D$

$M_{0}=(3\times 10^{10})\times (1250\times 10^{6})\times 25$

$M_{0}= 93750 \times10^{16}$

$M_{0}= 9.3750 \times10^{20}$

The Moment Magnitude (Mw)

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

$M_{w}=\frac{2}{3}(\log 9.375 \times 10^{20}-9.1)$

$M_{w}=\frac{2}{3}(20.9719-9.1)$

$M_{w}=\frac{2}{3}(11.8719)$

$M_{w}=7.9146$