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"What is the Richter Scale measurement for an earthquake that is 5 times as intense as an earthquake measuring 8.2 on the Richter Scale?"

I understand that the measurement of magnitude is exponential but I don't know how to apply that information and I don't have a clue as to how to go about solving this. 

Aleguan  Jan 30, 2018
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2+0 Answers

 #1
avatar+85757 
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Not an expert on this....but...I think we can use this formula to find the result :

 

Magnitude  =  log 5  + 8.2  =

 

.699  + 8.2   ≈

 

8.9

 

Here's a website that goes into more detail :

 

http://www.sosmath.com/algebra/logs/log5/log56/log56.html

 

 

cool cool cool

CPhill  Jan 30, 2018
 #2
avatar
+2

Here is the formula used by USGS(United States Geological Survey) to determine modern Magnitude of Eartquakes:

10^((1.5*M) + 4.8) / (4.184E15) =Energy Release in Megatons eqivalent of TNT. M=Magnitude.

10^((1.5*8.2 + 4.8) / (4.184E15)=~30 Megatons of TNT

10^((1.5*M + 4.8) / (4.184E15)=30*5, solve for M

M = 8.67, so:

8.67 - 8.2 =0.47 - difference in Magitude which will release 5 times as much as 8.2 Magnitude.

In other words, 8.67 Magnitude will release 5 times as much energy as 8.2 Magnitude.

Note: Wolfram/Alpha will give you the same answer if you input 8.67 Earthquake Manitude. See here:

https://www.wolframalpha.com/input/?i=Earthquake+magnitude+8.67

Guest Jan 30, 2018
edited by Guest  Jan 30, 2018
edited by Guest  Jan 30, 2018

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