How many times stronger is an earthquake with a magnitude of 8 than an earthquake with a magnitude of 6?

AngelRay
Feb 12, 2018

#3**+2 **

Since the question only mentions magnitude and intensity, I guessed that a "stronger" earthquake meant that it has a higher intensity.

Let I_{1} be the intensity of the earthquake with magnitude 8 .

Using the formula given in your question...

8 = log( I_{1}/S ) Solve for I_{1} .

Take the inverse of the log of both sides.

10^{8} = l_{1}/S

Multiply both sides by S .

10^{8}S = l_{1}

Let I_{2} be the intensity of the earthquake with magnitude 6 .

6 = log( l_{2}/S )

10^{6}S = l_{2}

The question is... l_{1} equals *what* times l_{2} ?

l_{1} = x * l_{2}

10^{8}S = x * 10^{6}S

[ 10^{8}S ] / [ 10^{6}S ] = x

10^{8} / 10^{6} = x

100 = x

So...

l_{1} = 100 * l_{2}

An earthquake with a magnitude 8 is 100 times more intense than an earthquake with magnitude 6 .

hectictar
Feb 13, 2018

#1**+1 **

8/6=1.3

magnitude 8 earthquake is 1.3 times stronger than magnitude 6 earthquake.

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lynx7
Feb 12, 2018

#2**0 **

I don't quite understand the formula given in the question for "standard earthquake" !!

This is the most modern updated formula used by the USGS(United States Geological Survey):

Energy release in Megatons =10^((1.5 * M)+4.8) / (4.184E15), where M = Magnitude.

So, 6 Magnitude earthquake will release: 10^((1.5*6) +4.8) /4.184E15 =15 Kilotons - or the size of the atomic bomb dropped on Hiroshima in 1945.

But, 8 Magnitude earthquake will release: 10^((1.5*8 +4.8) /4.184E15 =15 Megatons !!!

15 Megatons / 15 Kilotons =15,000,000 / 15,000 =1,000x more poweful is an 8 Magnitude over 6 Magnitude earthquake. The old "Richter Scale" was simply a power of 10. So, using that old scale, the difference between 8 Magnitude and 6 Magnitude earthquake would have been: 10^(8 - 6) =10^2, or 100 times more poweful. But, that is no longer used to measure the Magnitudes of earhquakes. You may go to the USGS website and learn about the moderm formulas used to measure earthquakes.

Note: You should refer this to your teacher.

Guest Feb 13, 2018

edited by
Guest
Feb 13, 2018

#3**+2 **

Best Answer

Since the question only mentions magnitude and intensity, I guessed that a "stronger" earthquake meant that it has a higher intensity.

Let I_{1} be the intensity of the earthquake with magnitude 8 .

Using the formula given in your question...

8 = log( I_{1}/S ) Solve for I_{1} .

Take the inverse of the log of both sides.

10^{8} = l_{1}/S

Multiply both sides by S .

10^{8}S = l_{1}

Let I_{2} be the intensity of the earthquake with magnitude 6 .

6 = log( l_{2}/S )

10^{6}S = l_{2}

The question is... l_{1} equals *what* times l_{2} ?

l_{1} = x * l_{2}

10^{8}S = x * 10^{6}S

[ 10^{8}S ] / [ 10^{6}S ] = x

10^{8} / 10^{6} = x

100 = x

So...

l_{1} = 100 * l_{2}

An earthquake with a magnitude 8 is 100 times more intense than an earthquake with magnitude 6 .

hectictar
Feb 13, 2018