Chapter 20, Problem 1SOP

To determine

The fraction of Mercury’s surface that is covered by Caloris Basin and the fraction of Moon’s surface that is covered by The South-Pole Aitken Basin.

Answer to Problem 1SOP

The fraction of Mercury’s surface that is covered by Caloris Basin is $0.025$ and the fraction of Moon’s surface that is covered by South-Pole Aitken Basin is $0.13$.

Explanation of Solution

Write the expression for the area of the basin.

    $A_{\text{basin}}=\pi r^2$        (I)

Here, $r$ is the radius of the basin.

Write the expression for the area of the heavenly body.

    $A_{\text{body}}=4\pi {\left(\frac{d}{2}\right)}^2$        (II)

Here, $d$ is the diameter of the heavenly body.

Write the expression for the fraction of the area of the heavenly body covered by the basin.    $\text{Fraction}\text{of}\text{Area}=\frac{A_{\text{basin}}}{A_{\text{body}}}$        (III)

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

Refer to Celestial profile of Moon to obtain the value of the diameter of Moon as $3.48\times {10}^3\text{km}$.

Calculation:

Substitute $775\text{km}$ for $r$ in equation (I) to find the area of The Caloris Basin, $A_{\text{basin,Caloris}}$.

    $\begin{array}{c}{A}_{\text{basin,Caloris}}=\pi {\left(775\text{km}\right)}^2\\ =\pi \left(600625\right){\text{km}}^{\text{2}}\end{array}$

Substitute $4.88\times {10}^3\text{km}$ for $d$ in equation (II) to find the area of Mercury, $A_{\text{body,Mercury}}$.

    $\begin{array}{c}{A}_{\text{body,Mercury}}=4\pi {\left(\frac{4.88\times {10}^3\text{km}}{2}\right)}^2\\ =4\pi {\left(2.44\times {10}^3\right)}^2\\ =4\pi \left(5.95\times {10}^6\right){\text{km}}^{\text{2}}\end{array}$

Substitute $\pi \left(600625\right){\text{km}}^{\text{2}}$ for $A_{\text{basin,Caloris}}$ and $4\pi \left(5.95\times {10}^6\right){\text{km}}^{\text{2}}$ for $A_{\text{body,Mercury}}$ in equation (III) to find the fraction of the area of the Mercury covered by The Caloris Basin.

    $\begin{array}{c}\text{Fraction}\text{of}\text{Area}=\frac{\pi \left(600625\right){\text{km}}^{\text{2}}}{4\pi \left(5.95\times {10}^6\right){\text{km}}^{\text{2}}}\\ =\frac{1}{4}\left(100945.37\times {10}^{-6}\right)\\ =0.025\end{array}$

Substitute $1250\text{km}$ for $r$ in equation (I) to find the area of The South-Pole Aitken Basin, $A_{\text{basin,Aitken}}$.

    $\begin{array}{c}{A}_{\text{basin,Aitken}}=\pi {\left(1250\text{km}\right)}^2\\ =\pi \left(1562500\right){\text{km}}^{\text{2}}\end{array}$

Substitute $3.48\times {10}^3\text{km}$ for $d$ in equation (II) to find the area of Moon, $A_{\text{body,Moon}}$.

    $\begin{array}{c}{A}_{\text{body,Moon}}=4\pi {\left(\frac{3.48\times {10}^3\text{km}}{2}\right)}^2\\ =4\pi {\left(1.74\times {10}^3\right)}^2\\ =4\pi \left(3.02\times {10}^6\right){\text{km}}^{\text{2}}\end{array}$

Substitute $\pi \left(1562500\right){\text{km}}^{\text{2}}$ for $A_{\text{basin,Aitken}}$ and $4\pi \left(3.02\times {10}^6\right){\text{km}}^{\text{2}}$ for $A_{\text{body,Moon}}$ in equation (III) to find the fraction of the area of the Moon covered by The South-Pole Aitken Basin.

    $\begin{array}{c}\text{Fraction}\text{of}\text{Area}=\frac{\pi \left(1562500\right){\text{km}}^{\text{2}}}{4\pi \left(3.02\times {10}^6\right){\text{km}}^{\text{2}}}\\ =\frac{1}{4}\left(517384.105\times {10}^{-6}\right)\\ =0.13\end{array}$

Therefore, the fraction of Mercury’s surface that is covered by Caloris Basin is $0.025$ and the fraction of Moon’s surface that is covered by South-Pole Aitken Basin is $0.13$.