Question

On the Richter scale, the magnitude \(R\) of an earthquake of intensity \(I\) is \(R=\frac{\ln I-\ln I_{0}}{\ln 10}\) where \(I_{0}\) is the minimum intensity used for comparison. Assume that \(I_{0}=1\) (a) Find the intensity of the 1906 San Francisco earthquake \((R=8.3)\) (b) Find the factor by which the intensity is increased if the Richter scale measurement is doubled. (c) Find \(d R / d I\).

Short answer

The intensity associated with a \(R = 8.3\) earthquake on the Richter scale is \(I = 10^{8.3}\). The intensity is increased by a factor of \(10^{8.3}\) if the Richter scale measurement is doubled. The derivative of the Richter scale with respect to Intensity is given by \(d R / d I = \frac{1}{I \cdot \ln 10}\)

Step-by-step solution

Calculating the intensity of the earthquake

Substitute \(R = 8.3\) and \(I_{0} = 1\) in the formula \(R=\frac{\ln I-\ln I_{0}}{\ln 10}\) to calculate the intensity. The equation becomes\(8.3=\frac{\ln I-\ln 1}{\ln 10}\). Now, solve for I. So, \(I = 10^{8.3}\)

Calculation of intensity change

To find how much the intensity increases when the Richter scale measurement is doubled, substitute \(R = 2R\) into the formula and find the new \(I'\). \(2R = \frac{\ln I' - \ln I_{0}}{\ln 10}\), which simplifies to \(2 \cdot 8.3 = \frac{\ln I'}{\ln 10}\). Solving for \(I'\) gives \(I' = 10^{2 \cdot 8.3}\). The factor by which intensity is increased is then \(I' / I = 10^{2 \cdot 8.3} / 10^{8.3} = 10^{8.3}\)

Finding the derivative of Richter scale with respect to intensity

The derivative of R with respect to I, \(dR/dI\), should be found using logarithm differentiation rules. Starting from the base relation \(R=\frac{\ln I - \ln I_{0}}{\ln 10}\). Because \(\ln I_{0}\) is a constant, its derivative will be 0. Therefore, \(dR / dI = \frac{1}{I \cdot \ln 10}\)