The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and assuming the density of seawater is constant all the way down.

Solution

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Here given that the depth of the trench is

\begin{align*} h_{\text{T}} = 11\,\mathrm{km} = 11\times 10^3\,\mathrm{m} \end{align*}

From table 11.1, the density of the sea water is

\begin{align*} \rho_{\text{W}} = 1.025\times 10^3\,\mathrm{kg\,m^{-3}} \end{align*}

Now we know that the pressure is given by

\begin{align*} P_{\text{W}} & = \rho_{\text{W}} \,g \,h_{\text{T}}\\ & = 1.025\times 10^3\,\mathrm{kg\,m^{-3}} \times \,9.8\,\mathrm{m\,s^{-2}} \times \,11\times 10^3\,\mathrm{m}\\ & = 1.10\times 10^5\,\mathrm{kg\,m^{-1}\,s^{-2}} \end{align*}

Now as we know

\begin{align*} 1 \mathrm{N} & = \mathrm{kg\,m\,s^{-2}} \tag{3} \end{align*}

Therefore

\begin{align*} P_{\text{W}} & = 1.10\times 10^5\,\mathrm{kg\,m^{-1}\,s^{-2}} \times \frac{\mathrm{N}}{\mathrm{kg\,m\,s^{-2}} }\\ & = 1.10\times 10^8\,\mathrm{N\,m^{-2}}\\ & = 1.10\times 10^8\,\mathrm{Pa} \end{align*}

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