Chapter 9, Problem 37QAP

(a)

To determine

The surface area and volume of moon Io.

Answer to Problem 37QAP

The surface area of moon Io is $4.16\times {10}^7{\text{km}}^2$ and volume of moon Io is $2.52\times {10}^{10}{\text{km}}^3$  .

Explanation of Solution

Write the expression for surface area of sphere.

$A=4\pi r^2$        (I)

Here, $r$ is the radius of sphere and $A$ is the surface area of sphere.

Write the expression for volume of sphere.

$V=\frac{4}{3}\pi r^3$        (II)

Here, $V$ is the volume of sphere.

Conclusion:

Substitute $1820\text{km}$ for $r$ in equation (I).

$\begin{array}{c}A=4\pi {\left(1820\text{km}\right)}^2\\ =4.16\times {10}^7{\text{km}}^2\end{array}$

Substitute $1820\text{km}$ for $r$ in equation (II).

$\begin{array}{c}V=\frac{4}{3}\pi {\left(1820\text{km}\right)}^3\\ =2.52\times {10}^{10}{\text{km}}^3\end{array}$

Thus, surface area of moon Io is $4.16\times {10}^7{\text{km}}^2$ and volume of moon Io is $2.52\times {10}^{10}{\text{km}}^3$  .

(b)

To determine

The volume of volcanic material deposited on Io’s surface each year.

Answer to Problem 37QAP

The volume of volcanic material deposited on Io’s surface each year is $1.25\times {10}^{11}{\text{m}}^3/\text{year}$ .

Explanation of Solution

Write the expression for volume of deposited material.

$V_d=Ad$        (III)

Here, $d$ is the depth of material deposited and $V_d$ is the volume of material deposited.

Conclusion:

Substitute $3\text{mm}$ for $d$ and $4.16\times {10}^7{\text{km}}^2$ for $A$ in equation (III).

$\begin{array}{c}{V}_d=\left(4.16\times {10}^7{\text{km}}^2\right){\left(\frac{1000\text{m}}{1\text{km}}\right)}^2\left(3\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ =1.25\times {10}^{11}{\text{m}}^3\end{array}$

Thus, the volume of volcanic material deposited on Io’s surface is $1.25\times {10}^{11}{\text{m}}^3/\text{year}$ .

(c)

To determine

The time required by volcanism to deposit the total volume of Io on its surface.

Answer to Problem 37QAP

The time required by volcanism to deposit the total volume of Io on its surface is $2.02\times {10}^8\text{years}$  .

Explanation of Solution

Write the expression for required time.

$t=\frac{V}{V_d}$        (IV)

Here, $t$ is the required time.

Conclusion:

Substitute $2.52\times {10}^{10}{\text{km}}^3$ for $V$ and $1.25\times {10}^{11}{\text{m}}^3/\text{year}$ for $V_d$ in equation (IV).

$\begin{array}{c}t=\frac{2.52\times {10}^{10}{\text{km}}^3{\left(\frac{1000\text{m}}{1\text{km}}\right)}^3}{1.25\times {10}^{11}{\text{m}}^3/\text{year}}\\ =2.02\times {10}^8\text{years}\end{array}$

Thus, the time required by volcanism to deposit the total volume of Io on its surface is $2.02\times {10}^8\text{years}$.

(d)

To determine

The number of times Io would turn inside-out during life of Solar System.

Answer to Problem 37QAP

The Io would turn inside-out during life of Solar System is $22\text{times}$ .

Explanation of Solution

The life of the solar system is $4.5\text{billion years}$ . The time required by moon to turn inside-out once is $2.02\times {10}^8\text{years}$ .

So, the number of times the Io would turn inside-out during life of solar system.

$\begin{array}{c}n=\frac{4.5\text{billion years}\left(\frac{{10}^9\text{years}}{\text{1}\text{billion years}}\right)}{2.02\times {10}^8\text{years}}\\ \approx 22\end{array}$

Conclusion:

Thus, the Io would turn inside-out during life of Solar System is $22\text{times}$ .