Chapter 13, Problem 9Q

To determine

The average density of Moon, if it contains solid rock of density $3500{\text{ kg/m}}^3$ in its inner portion comprising 50% of its radius and water for the rest of its volume. Consider the density of water to be $1000{\text{ kg/m}}^3$.

Answer to Problem 9Q

Solution:

$1313{\text{ kg/m}}^3$.

Explanation of Solution

Given data:

The density of soil rock is $3500{\text{ kg/m}}^3$.

The density of water is $1000{\text{ kg/m}}^3$.

Formula used:

The expression for volume of a sphere is written as,

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

Here, r is the radius of the sphere.

The expression for density of an object is written as,

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

Here, M is the mass of the object.

Explanation:

Recall the expression for volume of a sphere to determine the volume of Moon (considering Moon to be a perfect sphere).

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

Here, r is the radius of Moon.

The volume of the sphere varies with $r^3$. The inner portion of Moon which consists of rock has a radius which is 50% $\left(\frac{1}{2}\right)$ of the total radius of Moon. So,

$r_i=\frac{r}{2}$

Here, subscript i represents the corresponding quantities for the inner portions of Moon.

The volume of the inner portion of Moon is,

$V_i=\frac{4}{3}\pi r_i^3$

Substitute $\frac{r}{2}$ for $r_i$.

$\begin{array}{c}{V}_i=\frac{4}{3}\pi {\left(\frac{r}{2}\right)}^3\\ =\frac{1}{8}\left(\frac{4}{3}\pi r^3\right)\end{array}$

Substitute V for $\frac{4}{3}\pi r^3$.

$V_i=\frac{V}{8}$

So, $\frac{1}{8}$ of the total volume of Moon is contained in the inner part of the Moon, the rest will be contained in the outer portion of the Moon. Therefore,

$V_o=V-V_i$

Here, the subscript o represents the corresponding quantity for the outer portion of Moon.

Substitute $\frac{V}{8}$ for $V_i$.

$\begin{array}{c}{V}_o=V-\frac{V}{8}\\ =\frac{7V}{8}\end{array}$

The mass of the inner potion can be written as,

$M_i={\rho }_i{V}_i$

The mass of the outer portion is written as,

$M_o={\rho }_o{V}_o$

The total mass of Moon is written as,

$M=M_i+M_o$

Substitute ${\rho }_i{V}_i$ for $M_i$ and ${\rho }_o{V}_o$ for $M_o$.

$M={\rho }_i{V}_i+{\rho }_o{V}_o$

Substitute $\frac{V}{8}$ for $V_i$ and $\frac{7V}{8}$ for $V_o$.

$\begin{array}{c}M={\rho }_i\left(\frac{V}{8}\right)+{\rho }_o\left(\frac{7V}{8}\right)\\ =V\left(\frac{{\rho }_i}{8}+\frac{7{\rho }_o}{8}\right)\end{array}$

Write the expression for density of Moon as,

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

Substitute $V\left(\frac{{\rho }_i}{8}+\frac{7{\rho }_o}{8}\right)$ for M.

$\begin{array}{c}\rho =\frac{V\left(\frac{{\rho }_i}{8}+\frac{7{\rho }_o}{8}\right)}{V}\\ =\frac{{\rho }_i}{8}+\frac{7{\rho }_o}{8}\end{array}$

Substitute $3500{\text{ kg/m}}^3$ for ${\rho }_i$ and $1000{\text{ kg/m}}^3$ for ${\rho }_o$.

$\begin{array}{c}\rho =\frac{3500{\text{ kg/m}}^3}{8}+\frac{7\left(1000{\text{ kg/m}}^3\right)}{8}\\ =1313{\text{ kg/m}}^3\end{array}$

Conclusion:

The density of Moon is $1313{\text{ kg/m}}^3$.