Find the Richter scale magnitude of an earthquake that releases energy of 1000E

[kjeanch]

Hello. I need some help with a question on my final exam review sheet. Any help would be greatly appreciated. Thanks in advance.

Find the Richter scale magnitude of an earthquake that releases energy of 1000E. Then find the energy released by an earthquake that measures 5.0 on the Richter scale given that that Eo= 10^4.40 joules. Finally find the ratio in energy released between an Earthquake that measures 8.1 on the Richter scale and an aftershock measuring 5.4 on the scale. Use the formula R = 2/3 log E/Eo


A) R = 2, E = 7.94 x 10^11 joules and the ratio E1/E2 = 10200/1
B) R = 2, E = 7.94 X 10^10 joules and the ratio E1/E2 = 11200/1
C) R = 2, E = 7.94 X 10^11 joules and the ratio E1/E2 = 11200/1
D) R = 3, E = 7.94 X 10^11 joules and the ratio E1/E2 11200/1
E) R = 2, E= 5.94 X 10^11 joules and the ratio E1/E2 = 11200/1
F) None

 

[HallsofIvy]

Hello. I need some help with a question on my final exam review sheet. Any help would be greatly appreciated. Thanks in advance.

Find the Richter scale magnitude of an earthquake that releases energy of 1000E.

Do you mean that \(\displaystyle E= 1000 E_0\) or just that \(\displaystyle E= 1000\) Joules? Apparently the former since then (2/3) log(E/E0)= (2/3)log(1000)= (2/3)(3)= 2 which is the answer most of the choices below give.

Then find the energy released by an earthquake that measures 5.0 on the Richter scale given that that Eo= 10^4.40 joules.

\(\displaystyle 10^{4.40}\) Joules is 25,119 Joules. Is that what you mean? Or do you mean \(\displaystyle 40 \times 10^4\) which is 400000 Joules?
In either case, \(\displaystyle (2/3)log(E/E_0)= 5\) gives \(\displaystyle log(E/E_0)= 15/2\) so \(\displaystyle E/E0= 10^{15/2}= 31,622,777\)

Finally find the ratio in energy released between an Earthquake that measures 8.1 on the Richter scale and an aftershock measuring 5.4 on the scale. Use the formula R = 2/3 log E/Eo

Since \(\displaystyle R= (2/3)log(E/E_0)\), \(\displaystyle log(E/E_0)= 3R/2\) and then \(\displaystyle E/E_0= 10^{3R/2}\). Taking \(\displaystyle E_1\) and \(\displaystyle R_1\) for the first event, \(\displaystyle E_2\) and \(\displaystyle R_2\) for the second, the ratio is \(\displaystyle \frac{E_2}{E_1}= \frac{E_010^{3R_2/2}}{E_010^{3R_1/2}}= \frac{10^{3R_2/2}}{10^{3R_1/2}}= 10^{3R_2/2- 3R_1/2}=a 10^{3(R_2- R_1)/2}\)

A) R = 2, E = 7.94 x 10^11 joules and the ratio E1/E2 = 10200/1
B) R = 2, E = 7.94 X 10^10 joules and the ratio E1/E2 = 11200/1
C) R = 2, E = 7.94 X 10^11 joules and the ratio E1/E2 = 11200/1
D) R = 3, E = 7.94 X 10^11 joules and the ratio E1/E2 11200/1
E) R = 2, E= 5.94 X 10^11 joules and the ratio E1/E2 = 11200/1
F) None

 

[kjeanch]

Do you mean that \(\displaystyle E= 1000 E_0\) or just that \(\displaystyle E= 1000\) Joules? Apparently the former since then (2/3) log(E/E0)= (2/3)log(1000)= (2/3)(3)= 2 which is the answer most of the choices below give.


\(\displaystyle 10^{4.40}\) Joules is 25,119 Joules. Is that what you mean? Or do you mean \(\displaystyle 40 \times 10^4\) which is 400000 Joules?
In either case, \(\displaystyle (2/3)log(E/E_0)= 5\) gives \(\displaystyle log(E/E_0)= 15/2\) so \(\displaystyle E/E0= 10^{15/2}= 31,622,777\)


Since \(\displaystyle R= (2/3)log(E/E_0)\), \(\displaystyle log(E/E_0)= 3R/2\) and then \(\displaystyle E/E_0= 10^{3R/2}\). Taking \(\displaystyle E_1\) and \(\displaystyle R_1\) for the first event, \(\displaystyle E_2\) and \(\displaystyle R_2\) for the second, the ratio is \(\displaystyle \frac{E_2}{E_1}= \frac{E_010^{3R_2/2}}{E_010^{3R_1/2}}= \frac{10^{3R_2/2}}{10^{3R_1/2}}= 10^{3R_2/2- 3R_1/2}=a 10^{3(R_2- R_1)/2}\)

Yes \(\displaystyle E= 1000 E_0\) and \(\displaystyle 10^{4.40}\) is what I meant. I just didn't know how to write it that way. I'm still not sure which answer out of the 6 provided would be the correct one given the information you provided me with though.