On August 23, 2011, an earthquake of magnitude 5.8 on the Richter scale shook the East Coast of the United States. (Round your answers to two decimal places.)
What would be the magnitude of a quake four times as powerful as the Veracruz quake?
On August 23, 2011, an earthquake of magnitude 5.8 on the Richter scale shook the East Coast of the United States. (Round your answers to two decimal places.)
What would be the magnitude of a quake four times as powerful as the Veracruz quake?
Richter scale magnitude are generally logarithmic in scale, to the base 10 in amplitude. But the energy release is to the scale of 10^1.5 per 1 logarithm. Therefore, an earthquake that is 4 times as powerful as 5.8, in amplitude, would read on the Richter scale: 5.8 + Log 4=.60, or 5.8 + .60=6.4. So, an earthquake of magnitude 6.4 on the Richter scale would be 4 times as powerful as 5.8 magnitude quake. The total amount of energy released by a 6.4 magnitude quake would be (10^6.4)^1.5=10^9.6 Joules.
On August 23, 2011, an earthquake of magnitude 5.8 on the Richter scale shook the East Coast of the United States. (Round your answers to two decimal places.)
What would be the magnitude of a quake four times as powerful as the Veracruz quake?
Richter scale magnitude are generally logarithmic in scale, to the base 10 in amplitude. But the energy release is to the scale of 10^1.5 per 1 logarithm. Therefore, an earthquake that is 4 times as powerful as 5.8, in amplitude, would read on the Richter scale: 5.8 + Log 4=.60, or 5.8 + .60=6.4. So, an earthquake of magnitude 6.4 on the Richter scale would be 4 times as powerful as 5.8 magnitude quake. The total amount of energy released by a 6.4 magnitude quake would be (10^6.4)^1.5=10^9.6 Joules.
+130553 We can solve this as follows:
10^(M - 5.8) = 4 where M is the magnitude we are looking for
Take the log of each side
log10 ^(M - 5.8) = log 4 and we can write
M - 5.8 log10 = log 4 and log 10 = 1, so we can ignore it
M - 5.8 = log 4 add 5.8 to both sides
M = log4 + 5.8 = about 6.4
