Answer to Question #156782 in Calculus for Hector A Zepeda

Question

One earthquake has MMS magnitude 4.4. If a second earthquake has 730 times as much energy (earth movement) as the first, find the magnitude of the second quake.

Answer 1

For an earthquake with seismic moment S, a measurement of earth movement, the MMS value, or magnitude of the earthquake, is

                                                                               $M = \large\frac{2}{3}log(\large\frac{S}{S_o})$

Where $S_o = 10^{16}$ is a baseline measure for the seismic moment.

Let's solve we have following quantities:

            $M_1 = 4.4$           n = 730 = $\large\frac{S_2}{S_1}$                 $M_2 - ?$

$S_1 = S_o 10^{1.5M_1}$

$S_2 = S_o 10^{1.5M_2}$

$\frac{S_2}{S_1} = 10^{1.5(M_2-M_1)} = 730$

$1.5(M_2-M_1) = log730$    $\to M_2 = \frac{2}{3}log730 +M_1 = 6.3$

Answer:   the magnitude of the second quake is 6.3