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(III) The Earth is not a uniform sphere, but has regions of varying density. Consider a simple model of the Earth divided into three regions—inner core, outer core, and mantle. Each region is taken to have a unique constant density (the average density of that region in the real Earth):





(a) Use this model to predict the average density of the entire Earth. (b) If the radius of the Earth is 6380 km and its mass is 5.98 × 1024Kg, determine the actual average density of the Earth and compare it (as a percent difference) with the one you determined in (a).

Short Answer

Expert verified

(a) The average density of the earth from the densities of three regions is 5500.5 kg/m3.

(b) The actual density of the earth is 5497.3 kg/m3 and the percentage difference in density is -0.058%.

Step by step solution

01

Step-1: Density of Earth

The density of the entire earth is the total mass of the earth divided by the total volume of the earth.

02

Given the data

The density of the inner core is ρ1 = 13000 kg/m3.

The radius of the inner core is r1 =1220 km.

The density of the outer core is ρo =11100kg/m3.

The radius of the outer core is ro = 3480km.

The density of the mantle is ρM = 4400 kg/m3.

The radius of the mantle is rM = 6380 Km.

The mass of the earth is mE = 5.98 × 1024kg.

03

Step-3- Calculation of the average density of the earth

The average density of the Earth is calculated as;

\begin{aligned}\rho=\frac{{{\rho_{\rm{I}}}{V_{\rm{I}}}+{\rho_{\rm{O}}}{V_{\rm{O}}}+{\rho_{\rm{M}}}{V_{\rm{M}}}}}{{{V_{\rm{I}}}+{V_{\rm{O}}}+{V_{\rm{M}}}}}\\=\frac{{{\rho_{\rm{I}}}\left({\frac{4}{3}\pi r_{\rm{I}}^3}\right)+{\rho_{\rm{O}}}\left({\frac{4}{3}\pi\left({r_{\rm{O}}^3-r_{\rm{I}}^3}\right)}\right)+{\rho_{\rm{M}}}\left({\frac{4}{3}\pi\left({r_{\rm{M}}^3-r_{\rm{O}}^3}\right)}\right)}}{{\left({\frac{4}{3}pi r_{\rm{I}}^3}\right)+\left({\frac{4}{3}\pi\left({r_{\rm{O}}^3-r_{\rm{I}}^3}\right)}\right)+\left({\frac{4}{3}\pi\left({r_{\rm{M}}^3-r_{\rm{O}}^3}\right)}\right)}}\end{aligned}

Here VI is the volume of the inner core, VO is the volume of the outer core, and VM is the volume of the mantle.

On rearranging and substituting the known values of the above equation:


The average density of the Earth is 5500.5 kg/m3.

04

 Calculation of the actual density of the Earth

The density of the Earth from the given mass of the Earth is calculated as;

\begin{aligned}{\rho_{{\rm{actual}}}}=\frac{{{m_{\rm{E}}}}}{{{V_{\rm{E}}}}}\\{\rho_{{\rm{actual}}}}=\frac{{5.98\times{{10}^{24}}{\rm{kg}}}}{{\frac{4}{3}\pi{{\left({6380\;{\rm{km}}\times\frac{{1000{\rm{m}}}}{{1{\rm{km}}}}}\right)}^3}}}\\{\rho_{{\rm{actual}}}}=5497.3{\rm{kg/}}{{\rm{m}}^3}\end{aligned}

Here, VE is the volume of the Earth.

The actual density of the Earth is 5497.3 kg/m3.

The percentage of the difference between the actual density and the calculated density is,

\begin{aligned}\left({\frac{{{\rho_{{\rm{actual}}}}-\rho}}{\rho}}\right)\times100=\frac{{\left({5497.3{\rm{kg/}}{{\rm{m}}^3}-5500.5{\rm{kg/}}{{\rm{m}}^3}}\right)}}{{5500.5{\rm{ kg/}}{{\rm{m}}^3}}}\times100\\=-0.058\%\end{aligned}

The percentage difference in density is -0.058%.

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Most popular questions from this chapter

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

Explain how the tube in Fig. 10–45, known as a siphon, can transfer liquid from one container to a lower one even though the liquid must flow uphill for part of its journey. (Note that the tube must be filled with liquid to start with.)


A barge filled high with sand approaches a low bridge over the river and cannot quite pass under it. Should sand be added to, or removed from, the barge? [Hint: Consider Archimedes’ principle.]

A 3.2-N force is applied to the plunger of a hypodermic needle. If the diameter of the plunger is 1.3 cm and that of the needle is 0.20 mm, (a) with what force does the fluid leave the needle? (b) What force on the plunger would be needed to push fluid into a vein where the gauge pressure is 75 mm-Hg? Answer for the instant just before the fluid starts to move.

(II) The surface tension of a liquid can be determined by measuring the force \(F\) needed to just lift a circular platinum ring of radius \(r\) from the surface of the liquid. (a) Find a formula for \(\gamma \) in terms of \(F\) and \(r\). (b) At \(30^\circ C\), if \(F = 6.20 \times {10^{ - 3}}\;{\rm{N}}\) and \(r = 2.9\;{\rm{cm}}\), calculate \(\gamma \) for the tested liquid.

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