Question

The average density of Earth’s crust $\mathbf{10}\mathbf{\text{\hspace{0.17em}km}}$beneath the continents is $\mathbf{2.7}\mathbf{\text{\hspace{0.17em}g}}\mathbf{/}{\mathbf{\text{cm}}}^{\mathbf{\text{3}}}$. The speed of longitudinal seismic waves at that depth, found by timing their arrival from distant earthquakes, is role="math" localid="1661499978175" $\mathbf{5.4}\mathbf{\text{\hspace{0.17em}km}}\mathbf{/}\mathbf{\text{s}}$. Use this information to find the bulk modulus of Earth’s crust at that depth. For comparison, the bulk modulus of steel is about role="math" localid="1661499995851" $\mathbf{16}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{\text{\hspace{0.17em}Pa}}$.

Short answer

Bulk Modulus of earth’s crust at depth is $7.9\times {10}^{10}\text{\hspace{0.17em}Pa}$.

Step-by-step solution

The given data

1. Density of earth, $\mathrm{\rho }=2.7\text{\hspace{0.17em}g}/{\text{cm}}^{\text{3}}\text{or}2.7\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$
2. Speed of longitudinal seismic wave, $\mathrm{v}=5.4\text{\hspace{0.17em}km}/\text{sor}5400\text{\hspace{0.17em}m}/\text{s}$

Understanding the concept of bulk modulus

Here, we have to use the relation between velocity, the density of the earth, and bulk modulus to find bulk modulus.

Formula:

The velocity of the wave,

$\mathbf{v}\mathbf{=}\sqrt{\frac{\mathbf{B}}{\mathbf{\rho }\mathbf{}}}$ …(i)

Calculation of bulk modulus of Earth’s crust at depth

Using equation (i) and the given values, we get the value of bulk modulus of Earth’s crust at depth as:

$\begin{array}{c}5.4\times {10}^3\text{\hspace{0.17em}m}/\text{s}=\sqrt{\frac{\mathrm{B}}{2.7\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}}}\\ \mathrm{B}=7.87\times {10}^{10}\text{\hspace{0.17em}Pa}\\ \mathrm{B}\simeq 7.9\times {10}^{10}\text{\hspace{0.17em}Pa}\end{array}$

Hence, the value of bulk modulus is $7.9\times {10}^{10}\text{\hspace{0.17em}Pa}$.