Question

(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

(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

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.

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 ρ_o =11100kg/m³.

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

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

The radius of the mantle is r_M = 6380 Km.

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

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

The average density of the Earth is calculated as;

$$\begin{array}{r}\rho =\frac{{\rho }_{\mathrm{I}}{V}_{\mathrm{I}}+{\rho }_{\mathrm{O}}{V}_{\mathrm{O}}+{\rho }_{\mathrm{M}}{V}_{\mathrm{M}}}{V_{\mathrm{I}}+V_{\mathrm{O}}+V_{\mathrm{M}}}\\ =\frac{{\rho }_{\mathrm{I}}\left(\frac{4}{3}\pi r_{\mathrm{I}}^3\right)+{\rho }_{\mathrm{O}}\left(\frac{4}{3}\pi \left(r_{\mathrm{O}}^3-r_{\mathrm{I}}^3\right)\right)+{\rho }_{\mathrm{M}}\left(\frac{4}{3}\pi \left(r_{\mathrm{M}}^3-r_{\mathrm{O}}^3\right)\right)}{\left(\frac{4}{3}pi{r}_{\mathrm{I}}^3\right)+\left(\frac{4}{3}\pi \left(r_{\mathrm{O}}^3-r_{\mathrm{I}}^3\right)\right)+\left(\frac{4}{3}\pi \left(r_{\mathrm{M}}^3-r_{\mathrm{O}}^3\right)\right)}\end{array}$$

Here V_I is the volume of the inner core, V_O is the volume of the outer core, and V_M 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.

 Calculation of the actual density of the Earth

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

$$\begin{array}{r}{\rho }_{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}}=\frac{m_{\mathrm{E}}}{V_{\mathrm{E}}}\\ {\rho }_{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}}=\frac{5.98\times {10}^{24}\mathrm{k}\mathrm{g}}{\frac{4}{3}\pi {\left(6380\mathrm{k}\mathrm{m}\times \frac{1000\mathrm{m}}{1\mathrm{k}\mathrm{m}}\right)}^3}\\ {\rho }_{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}}=5497.3\mathrm{k}\mathrm{g}/{\mathrm{m}}^3\end{array}$$

Here, V_E 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,

$$\begin{array}{r}\left(\frac{{\rho }_{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}}-\rho }{\rho }\right)\times 100=\frac{\left(5497.3\mathrm{k}\mathrm{g}/{\mathrm{m}}^3-5500.5\mathrm{k}\mathrm{g}/{\mathrm{m}}^3\right)}{5500.5\mathrm{k}\mathrm{g}/{\mathrm{m}}^3}\times 100\\ =-0.058\mathrm{\%}\end{array}$$

The percentage difference in density is -0.058%.