"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

#1**+2 **

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

CPhill
Jan 30, 2018

#2**+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

edited by Guest Jan 30, 2018