Question

Question: A simple model (Fig. 10–56) considers a continent as a block (density\( \approx {\bf{2800}}\;{\bf{kg/}}{{\bf{m}}^{\bf{3}}}\)) floating in the mantle rock around it (density\( \approx {\bf{3300}}\;{\bf{kg/}}{{\bf{m}}^{\bf{3}}}\)). Assuming the continent is 35 km thick (the average thickness of the Earth’s continental crust), estimate the height of the continent above the surrounding mantle rock.

Figure 10-56

Short answer

The height of the continent is \(5250\;{\rm{m}}\).

Step-by-step solution

Given Data

The density of continent is\({\rho _1} = 2800\;{\rm{kg/}}{{\rm{m}}^3}\).

The density of mantle rock is\({\rho _2} = 3300\;{\rm{kg/}}{{\rm{m}}^3}\).

The thickness of the Earth’s crust is \(d = 35\;{\rm{km}}\).

Understanding the weight of the continent

In this problem, the weight of the mantle displaced will be equivalent to the weight of the continent. Consider x as the height of the continent.

Estimating the height of the continent

The relation ofweight of the continentis given by,

\(\begin{array}{c}{W_1} = {W_2}\\{\rho _1}gd = {\rho _2}g\left( {d - x} \right)\\{\rho _1}d = {\rho _2}\left( {d - x} \right)\\x = d\left( {1 - \frac{{{\rho _1}}}{{{\rho _2}}}} \right)\end{array}\)

Here,\({W_1}\)and\({W_2}\)are the weight of the continent and mantle respectively,\(g\)is the gravitational acceleration and\(x\)is the height of the continent.

On plugging the values in the above relation,

\(\begin{array}{l}x = \left( {35\;{\rm{km}} \times \frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\left( {1 - \frac{{2800\;{\rm{kg/}}{{\rm{m}}^3}}}{{3300\;{\rm{kg/}}{{\rm{m}}^3}}}} \right)\\x = 5250\;{\rm{m}}\end{array}\)

Thus, the height of the continent is \(5250\;{\rm{m}}\).