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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# In analyzing certain geological features, it is often appropriate to assume that the pressure at some horizontal level of compensation, deep inside Earth, is the same over a large region and is equal to the pressure due to the gravitational force on the overlying material. Thus, the pressure on the level of compensation is given by the fluid pressure formula. This model requires, for one thing, that mountains have roots of continental rock extending into the denser mantle (Figure). Consider a mountain of height ${\mathbit{H}}{\mathbf{=}}{\mathbf{6}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{km}}$ km on a continent of thickness ${\mathbit{T}}{\mathbf{=}}{\mathbf{32}}{\mathbf{}}{\mathbit{k}}{\mathbit{m}}$. The continental rock has a density of${\mathbf{2}}{\mathbf{.}}{\mathbf{9}}{\mathbf{}}{\mathbit{g}}{\mathbf{/}}{\mathbit{c}}{{\mathbit{m}}}^{{\mathbf{3}}}$ , and beneath this rock the mantle has a density of ${\mathbf{3}}{\mathbf{.}}{\mathbf{3}}{\mathbf{}}{\mathbit{g}}{\mathbf{/}}{\mathbit{c}}{{\mathbit{m}}}^{{\mathbf{3}}}$ . Calculate the depth of the root. (Hint: Set the pressure at points a and b equal; the depth y of the level of compensation will cancel out.)

The depth of the root is $44km$

See the step by step solution

## Step 1: The given data

1. The height of the mountain, $H=6.0km\mathrm{or}6000\mathrm{m}$
2. Thickness of a continent, $T=32kmor\mathit{}32000m$
3. Density of continental rock, ${\mathrm{\rho }}_{\mathrm{c}}=2.9g/c{m}^{3}\mathrm{or}2.9×{10}^{3}kg/{m}^{3}$
4. Density of mantle, ${\mathrm{\rho }}_{\mathrm{m}}=3.3g/c{m}^{3}\mathrm{or}3.3×{10}^{3}kg/{m}^{3}$

## Step 2: Understanding the concept of pressure

The pressure at some horizontal level of compression, deep inside the earth, is the same over a large region and is equal to the pressure exerted by the gravitational force on the overlying material. Therefore, using the formula of gauge pressure, we can find the depth of the root.

Formula:

Pressure applied on a body, $p=\mathrm{\rho }gh$ (i)

## Step 3: Calculation of depth of the root

$As{p}_{a}={p}_{b},\phantom{\rule{0ex}{0ex}}Fromequation\left(i\right)andthegivenvalues,weget\phantom{\rule{0ex}{0ex}}But,{p}_{a}={\mathrm{\rho }}_{c}g\left(6.0km+32km+D\right)+{\mathrm{\rho }}_{\mathrm{m}}\left(y\mathit{-}D\right).......\left(a\right)and\phantom{\rule{0ex}{0ex}}{p}_{b}={\mathrm{\rho }}_{\mathrm{c}}g\mathit{\left(}\mathit{32}km\mathit{\right)}\mathit{+}{p}_{\mathit{m}}y\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(b\right)\phantom{\rule{0ex}{0ex}}Therefore,equatingequations\left(a\right)and\left(b\right),weget\phantom{\rule{0ex}{0ex}}{\mathrm{\rho }}_{\mathrm{c}}g\left(6.0\mathrm{km}+32\mathrm{km}+D\right)+{\mathrm{\rho }}_{\mathrm{m}}\left(y\mathit{-}D\right)={\mathrm{\rho }}_{\mathrm{c}}g\left(32km\right)+{\mathrm{\rho }}_{\mathrm{m}}y\phantom{\rule{0ex}{0ex}}{\mathrm{\rho }}_{\mathrm{c}}\left(38.0km+D\right)+{\mathrm{\rho }}_{\mathrm{m}}\left(y\mathit{-}D\right)={\mathrm{\rho }}_{\mathrm{c}}\left(32km\right)+{\mathrm{\rho }}_{\mathrm{m}}y\phantom{\rule{0ex}{0ex}}{\mathrm{\rho }}_{\mathrm{c}}\left(6.0km\right)+{\mathrm{\rho }}_{\mathrm{c}}D+{\mathrm{\rho }}_{\mathrm{m}}y\mathit{-}{\mathrm{\rho }}_{\mathrm{m}}D-{\mathrm{\rho }}_{\mathrm{m}}y=0\phantom{\rule{0ex}{0ex}}{\mathrm{\rho }}_{\mathrm{c}}\left(6.0km\right)+{\mathrm{\rho }}_{\mathrm{c}}D\mathit{-}{\mathrm{\rho }}_{\mathrm{m}}D\mathit{=}0\phantom{\rule{0ex}{0ex}}D\mathit{=}\frac{{\mathit{\rho }}_{\mathit{c}}\mathit{\left(}\mathit{6}\mathit{.}\mathit{0}\mathit{k}\mathit{m}\mathit{\right)}}{{\mathit{\rho }}_{\mathit{m}}\mathit{-}{\mathit{\rho }}_{\mathit{c}}}\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}D\mathit{=}\frac{\left(\mathrm{2}\mathrm{.}\mathrm{9}g/c{m}^{3}\right)\left(\mathrm{6}\mathrm{.}\mathrm{0}km\right)}{\mathit{3}\mathit{.}\mathit{3}\mathit{g}\mathit{/}\mathit{c}{\mathit{m}}^{\mathit{3}}\mathit{-}\mathit{2}\mathit{.}\mathit{9}\mathit{g}\mathit{/}\mathit{c}{\mathit{m}}^{\mathit{3}}}\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}D=43.5km$

Therefore, the depth of the root is $43.5km$