Chapter 20, Problem 2SOP

To determine

The fraction of Mercury’s volume occupied by its core and the fraction of Earth’s volume occupied by its core.

Answer to Problem 2SOP

The fraction of Mercury’s volume occupied by its core is $0.40$ and the fraction of Earth’s volume occupied by its core is $0.15$.

Explanation of Solution

Write the expression for the volume occupied by the core.

    $V_{\text{core}}=\frac{4}{3}\pi r^3$        (I)

Here, $r$ is the radius of the core of the planet.

Write the expression for the volume occupied by the planet.

    $V_{\text{planet}}=\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$        (II)

Here, $d$ is the diameter of the planet.

Write the expression for the fraction of planet’s volume occupied by the core.

    $\text{Fraction}\text{of}\text{volume}=\frac{V_{\text{core}}}{V_{\text{planet}}}$        (III)

Refer to Celestial profile of Mercury to obtain the value of the diameter of Mercury as $4.88\times {10}^3\text{km}$.

Calculation:

Substitute $1800\text{km}$ for $r$ in equation (I) to find volume of core of mercury, $V_{\text{core,M}}$.

    $\begin{array}{c}{V}_{\text{core,M}}=\frac{4}{3}\pi {\left(1800\text{km}\right)}^3\\ =\frac{4}{3}\pi \left(5832\times {10}^6\right){\text{km}}^{\text{3}}\end{array}$

Substitute $4.88\times {10}^3\text{km}$ for $d$ in equation (II) to find the volume of Mercury, $V_{\text{planet,M}}$.

    $\begin{array}{c}{V}_{\text{planet,M}}=\frac{4}{3}\pi {\left(\frac{4.88\times {10}^3}{2}\right)}^3\\ =\frac{4}{3}\pi \left(\frac{116.21\times {10}^9}{8}\right)\\ \text{=}\frac{4}{3}\pi \left(14.52\times {10}^9\right){\text{km}}^{\text{3}}\end{array}$

Substitute $\frac{4}{3}\pi \left(5832\times {10}^6\right){\text{km}}^{\text{3}}$ for $V_{\text{core}}$ and $\frac{4}{3}\pi \left(14.52\times {10}^9\right){\text{km}}^{\text{3}}$ for $V_{\text{planet}}$ in equation (III) to find the fraction of Mercury’s volume occupied by the core.

    $\begin{array}{c}\text{Fraction}\text{of}\text{volume}=\frac{\frac{4}{3}\pi \left(5832\times {10}^6\right){\text{km}}^{\text{3}}}{\frac{4}{3}\pi \left(14.52\times {10}^9\right){\text{km}}^{\text{3}}}\\ =\frac{\left(5832\times {10}^6\right)}{\left(14.52\times {10}^9\right)}\\ =401.65\times {10}^{-3}\\ =0.401\end{array}$

Substitute $3400\text{km}$ for $r$ in equation (I) to find volume of core of Earth, $V_{\text{core,E}}$.

    $\begin{array}{c}{V}_{\text{core,E}}=\frac{4}{3}\pi {\left(3400\text{km}\right)}^3\\ =\frac{4}{3}\pi \left(39304\times {10}^6\right){\text{km}}^{\text{3}}\end{array}$

Substitute $12742\text{km}$ for $d$ in equation (II) to find the volume of Earth, $V_{\text{planet,E}}$.

    $\begin{array}{c}{V}_{\text{planet,E}}=\frac{4}{3}\pi {\left(\frac{12472\text{km}}{2}\right)}^3\\ =\frac{4}{3}\pi \left(258596602811\right){\text{km}}^{\text{3}}\end{array}$

Substitute $\frac{4}{3}\pi \left(39304\times {10}^6\right){\text{km}}^{\text{3}}$ for $V_{\text{core}}$ and $\frac{4}{3}\pi \left(258596602811\right){\text{km}}^{\text{3}}$ for $V_{\text{planet}}$ in equation (III) to find the fraction of Earth’s volume occupied by the core.

    $\begin{array}{c}\text{Fraction}\text{of}\text{volume}=\frac{\frac{4}{3}\pi \left(39304\times {10}^6\right){\text{km}}^{\text{3}}}{\frac{4}{3}\pi \left(258596602811\right){\text{km}}^{\text{3}}}\\ =\frac{\left(39304\times {10}^6\right){\text{km}}^{\text{3}}}{\left(258596602811\right){\text{km}}^{\text{3}}}\\ =1.52\times {10}^{-7}\times {10}^6\\ =0.152\end{array}$

Therefore, the fraction of Mercury’s volume occupied by its core is $0.40$ and the fraction of Earth’s volume occupied by its core is $0.15$.