Question

The 1906 earthquake in San Francisco had a magnitude of 8.3 on the richter scale. At the same time in Japan an earthquake with magnitude of 4.9 caused minor damage. How many times more was the San Francisco earthquake that the Japan earthquake?

Answer 1

In 1906 in San Francisco the magnitude of earthquake on richter scale was 8.3.

In the same year the magnitude of earthquake on richter scale in Japan was 4.9.

Now we have to find how many times more was the San Francisco earthquake that the Japan earthquake.

The amount of energy released in an earthquake is very large, so a logarithmic scale avoids the use of large numbers.

The formula used for these calculations is:

$M=log_{10}(\frac{I}{I_{0}})$

Where M is the magnitude on the richter scale,  I is the intensity of the earthquake being measured and I₀ is the intensity of a reference earthquake.

So because the magnitude is a base 10 log, the Richter number is actually the exponent that 10 is raised to in order to calculate the intensity of the earthquake.

So the difference in magnitudes of the earthquakes can be calculated as follows:

$M=log_{10}(\frac{10^{8.3}}{10^{4.9})}$

M=3.4

Answer: The San Francisco earthquake was 3.4 times more than Japan earthquake.