Chapter 1, Problem 1P

(a)

To determine

The average density of the Earth.

Answer to Problem 1P

The average density of the Earth is $5.52\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Given information:

The average mass of the Earth is $5.97\times {10}^{24}\text{kg}$ and the mean radius of the Earth is $6.37\times {10}^6\text{m}$.

Formula to calculate the average density of the Earth is,

$\rho =\frac{m}{V}$ (I)

• $\rho$ is the average density of the Earth.
• $m$ is the average mass of the Earth.
• $V$ is the average volume of the Earth.

Formula to calculate the average volume of the Earth is,

$V=\frac{4}{3}\pi r^3$

• $r$ is the mean radius of the Earth.

Substitute $\frac{4}{3}\pi r^3$ for $V$ in equation (I).

$\begin{array}{c}\rho =\frac{m}{\frac{4}{3}\pi r^3}\\ =\frac{3m}{4\pi r^3}\end{array}$

Substitute $5.97\times {10}^{24}\text{kg}$ for $m$ and $6.37\times {10}^6\text{m}$ for $r$ to find $\rho$.

$\begin{array}{c}\rho =\frac{3\left(5.97\times {10}^{24}\text{kg}\right)}{4\pi {\left(6.37\times {10}^6\text{m}\right)}^3}\\ =\frac{1.791\times {10}^{25}\text{kg}}{3.24\times {10}^{21}{\text{m}}^{\text{3}}}\\ =5.52\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the average density of the Earth is $5.52\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$.

(b)

To determine

The position of the density of Earth fit among those listed in Table 14.1 and compare the density of surface rocks like granite to the density of Earth.

Answer to Problem 1P

The average density of Earth lies in between the densities of aluminum and iron and the average density of granite is less than the density of the Earth.

Explanation of Solution

Section 1:

To find the position of the density of Earth fit among those listed in Table 14.1:

The average density of the Earth is $5.52\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$. From the table 14.1, it lies in between the density of aluminum $2.70\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ and the density of tin $7.30\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$.

Conclusion:

Therefore, the average density of Earth lies in between the densities of aluminum and tin.

Section 2:

To determine the comparison of density of a typical surface rock like granite and the average density of Earth:

Introduction: The density of any substance is defined as the mass of the substance per unit volume.

The average density of the Earth is $5.52\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ and the average density of the typical surface rocks like granite is $2.65\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$

Granite is a type rock that is mostly found on the surface of the earth.

Since the value $2.65\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ is less than $5.52\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$. Hence the density of granite type surface rocks is less than the average density of the Earth.

Conclusion:

Therefore, the average density of granite type rocks is less than the average density of the Earth.