What is the magnitude of an earthquake 1410 times as intense as a standard earthquake?

Question

BlackTom AI · Accepted Answer

Begin by identifying what the question is asking: the magnitude change corresponding to a certain factor of intensity relative to a standard earthquake.
The magnitude scale used in seismology is logarithmic, meaning that a change in magnitude ΔM corresponds to a ratio of intensities I2/I1 = 10^(ΔM).
Here, the intensity is 1410 times greater than the standard, so the ratio I2/I1 = 1410 corresponds to a magnitude increase ΔM where 1410 = 10^(ΔM).
To find ΔM, compute the base-10 logarithm of 1410: ΔM = log10(1410).
Calculating or estimating: 1410 = 1.41 × 10^3, so log10(1410) = log10(1.41) + log10(1000) ≈ 0.1492 + 3 = 3.1492, which rounds to approximately 3.15.
Therefore, the magnitude increase relative to the standard earthquake is about 3.15 magnitude units. 
If one interprets the problem as asking for the absolute magnitude of the earthquake assuming the standard earthquake has magnitude M0, then the new magnitude would be M0 + 3.15. However, since the question presents only the factor 1410 without giving M0, the key takeaway is the magnitude change of approximately 3.15 units.
Now evaluate the given options:
- 2.96: This is too small to correspond to a 1410× increase in intensity, since 10^3 would give a 1000× increase and 2.96 is less than the calculated 3.15.
- 2.61: Even smaller; not consistent with a 1410× rise in intensity.
- 3.15: This matches the computed ΔM = log10(1410) ≈ 3.15, aligning with the logarithmic magnitude relation.
- 3.74: This would imply a larger intensity ratio than 1410 (since 10^(3.74) ≈ 5500), which is inconsistent with the given factor.
Thus, the option that aligns with the logarithmic relationship between magnitude and intensity for a 1410× increase is the one around 3.15.

MHF4U - Quattrociocchi - S1 Unit 6 Online Quiz

What is the magnitude of an earthquake 1410 times as intense as a standard earthquake?

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