Chapter 9, Problem 18Q

(a)

To determine

To analyze: That the average density of Earth is $5500\text{}{\text{kg/m}}^3$, if the diameter of Earth is $12,756\text{ km}$ and the mass of Earth is $5.974\times {10}^{24}\text{ kg}$.

Answer to Problem 18Q

Solution:

$5500{\text{kg/m}}^3$.

Explanation of Solution

Given data:

Diameter of Earth is $12,756\text{km}$.

Mass of Earth is $5.974\times {10}^{24}\text{kg}$.

Formula used:

Write the expression for finding the volume of a spherical object.

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

Here, $r$ is the radius of the sphere and $V$ is the volume of the sphere.

Write the expression for finding the density of an object.

$\rho =\frac{M}{V}$

Here, $\rho$ is the density of the object, $M$ is the mass of the object and $V$ is the volume of the object.

Explanation:

Radius of Earth ($r$) is half of its diameter. That is,

$\begin{array}{c}r=\frac{12,756\text{km}}{2}\\ =6,378\text{km}\\ \text{=}\text{6}\text{.378}\times {\text{10}}^6\text{m}\end{array}$

To find the volume of Earth, refer to the expression for finding the volume of a spherical object.

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

Upon substituting $\text{6}\text{.378}\times {\text{10}}^3\text{ m}$ for $r$ in the formula,

$\begin{array}{c}V=\frac{4}{3}\pi {\left(6.378\times {10}^6\text{ m}\right)}^3\\ =1.087\times {10}^{21}{\text{ m}}^3\end{array}$

Mass of Earth ($M$) is given in the table, that is, $5.974\times {10}^{24}\text{kg}$.

Refer to the expression for finding the density of an object.

$\rho =\frac{M}{V}$

Upon substituting $5.974\times {10}^{24}\text{ kg}$ for $M$ and $1.087\times {10}^{21}{\text{ m}}^3$ for $V$,

$\begin{array}{c}\rho =\frac{5.974\times {10}^{24}\text{kg}}{1.087\times {10}^{21}{\text{m}}^3}\\ =5495.8{\text{ kg/m}}^3\\ \simeq 5500{\text{ kg/m}}^3\end{array}$

Conclusion:

Hence, the density of Earth is $5500{\text{ kg/m}}^3$.

(b)

To determine

The expected average density of the core assuming that the average density of material in the Earth’s mantle is about $3500{\text{kg/m}}^3$. Also check whether evaluated answer is consistent with the value given in the table.

	Earth’s Internal Structure
Region	Depth Below surface (km)	Distance from center (km)	Average density (kg/m3)
Crust (solid)	0-5 (under oceans) 0-35 (under continent)	6343-6378	3500
Mantie (plastic, solid)	From bottom of crust to 2900	3500-6343	3500-5500
Outer core (liquid)	2900-5100	1300-3500	10,000-12,000
Inner core (solid)	5100-6400	0-1300	13,000

Answer to Problem 18Q

Solution:

Average density of the core is calculated to be $15,265{\text{ kg/m}}^3$ and it is consistent with value given in the table.

Explanation of Solution

Given data:

Average density of material in the Earth’s mantle is $3500{\text{kg/m}}^3$.

Formula used:

Write the expression for finding the density of an object.

$\rho =\frac{M}{V}$

Here, $\rho$ is the density of the object, $M$ is the mass of the object and $V$ is the volume of the object.

Explanation:

Total mass of Earth is composed of core, mantle and crust. The mass of any object is the product of its density and its volume. Therefore, mass of crust and mantle is the product of their density $\left({\rho }_m=3500{\text{kg/m}}^3\right)$ and volume $\left(V_{\text{m}}\right)$. Similarly, mass of the core is the product of its density $\left({\rho }_c\right)$ and volume $\left(V_c\right)$.

The volume of core is 17% of that of Earth. Hence, the volume of core in terms of the volume of Earth is given as,

$V_{\text{c}}=0.17V_{\text{e}}$

Here, $V_c$ is the volume of core and $V_{\text{e}}$ is the volume of Earth.

The volume of mantle is 83% of that of Earth. Hence, the volume of mantle in terms of the volume of Earth is given as,

$V_{\text{m}}=0.83V_{\text{e}}$

Here, $V_{\text{m}}$ is the volume of the mantle and $V_{\text{e}}$ is the volume of Earth.

Write the expression for the mass of Earth in terms of the mass of mantle and the mass of core.

$M_{\text{e}}=M_{\text{m}}+M_{\text{c}}$

Here, $M_{\text{m}}$ is the mass of mantle and crust, $M_{\text{e}}$ is the mass of Earth and $M_{\text{c}}$ is the mass of core.

It is known that mass is the product of density and volume. Hence, in terms of density, the above expression can be rewritten as,

${\rho }_{\text{e}}{V}_{\text{e}}={\rho }_{\text{m}}{V}_{\text{m}}+{\rho }_{\text{c}}{V}_{\text{c}}$

Upon substituting $0.17V_{\text{e }}$ for $V_{\text{m}}$, $0.83V_{\text{e}}$ for $V_{\text{c}}$ and dividing the whole equation by $V_{\text{e}}$,

$\begin{array}{c}{\rho }_{\text{e}}{V}_{\text{e}}={\rho }_{\text{m}}\left(0.17V_{\text{m}}\right)+{\rho }_{\text{c}}\left(0.83V_{\text{c}}\right)\\ {\rho }_e=0.17{\rho }_{\text{m}}+0.83{\rho }_{\text{c}}\end{array}$

Substitute $5500{\text{ kg/m}}^3$ for ${\rho }_{\text{e}}$ and $3500{\text{ kg/m}}^3$ for ${\rho }_{\text{m}}$ and solve for ${\rho }_{\text{c}}$.

$\begin{array}{c}5500{\text{ kg/m}}^3=0.17\left(3500{\text{ kg/m}}^3\right)+0.83{\rho }_{\text{c}}\\ {\rho }_{\text{c}}=15,265{\text{ kg/m}}^3\end{array}$

Conclusion:

Hence, the density of core is $15,265{\text{ kg/m}}^3$ and it is approximately four times denser than the crust. Thus, the calculated value is consistent with the value given in the table.