Question

The intensity levels I of two earthquakes measured on a seismograph can be compared by the formula log I₁/I₂ = M₁ – M_2, where M is the magnitude given by the Richter Scale. In August 2009, an earthquake of magnitude 6.1 hit Honshu, Japan. In March 2011, that same region experienced yet another, more devastating earthquake, this time with a magnitude of 9.0. How many times greater was the intensity of the 2011 earthquake? Round to the nearest whole number.

Answer 1

The intensity levels I of two earthquakes measured on a seismograph can be compared by the formula

logI1/I2 = M1 – M2

 

Where M is the magnitude given by Richter scale.

 

The magnitude of earthquake hit in August, 2009 is M2 = 6.1.

The magnitude of earthquake hit in March, 2011 is M1 = 9.0.

Substituting numerical values,

logI1/I2 = 9.0 – 6.1

logI1/I2 = 2.9

 

So,

I1/I2 = 102.9

I1/I2 = 794.13

I1/I2 ≈ 794

 

Therefore, the intensity of 2011 earthquake was 794 times greater than the intensity of 2009 earthquake.

Therefore, the intensity of 2011 earthquake was 794 times greater than the intensity of 2009 earthquake.