Chapter 11, Problem 35QP

(a)

To determine

The surface are and the volume of the moon Io.

Answer to Problem 35QP

The surface are of the moon is $4.16\times {10}^{13}{\text{m}}^{\text{2}}$ and the volume of the moon is $2.52\times {10}^{19}{\text{m}}^{\text{3}}$.

Explanation of Solution

The surface of the moon Io is considered to be perfectly spherical.

Write the expression for the surface area for the spherical body.

  $A=4\pi R^2$        (I)

Here, $A$ is the surface are of the moon Io and $R$ is the radius of the moon.

Write the expression for the volume of the spherical body.

  $V=\frac{4}{3}\pi R^3$        (II)

Here, $V$ is the volume of the moon.

Conclusion:

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

  $\begin{array}{c}A=4\pi {\left(1820\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}^2\\ =12.56\left(3.31\times {10}^{12}\right){\text{m}}^{\text{2}}\\ =4.16\times {10}^{13}{\text{m}}^{\text{2}}\end{array}$

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

  $\begin{array}{c}V=\frac{4}{3}\pi {\left(1820\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}^3\\ =4.188\left(6.028\times {10}^{18}\right){\text{m}}^{\text{3}}\\ =2.52\times {10}^{19}{\text{m}}^{\text{3}}\end{array}$

Thus, the surface are of the moon is $4.16\times {10}^{13}{\text{m}}^{\text{2}}$ and the volume of the moon is $2.52\times {10}^{19}{\text{m}}^{\text{3}}$.

(b)

To determine

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

Answer to Problem 35QP

The volume of volcanic material deposited on the moon is one year is $1.24\times {10}^{11}{\text{m}}^{\text{3}}$.

Explanation of Solution

The thickness of the layer of lava and ash on the surface of the Io per year is about $3\text{mm}$.

Write the expression for the volume of the volcanic material deposited on the moon.

  $V^{\prime }=A\times d$        (III)

Here, $V^{\prime }$ is the volume of volcanic material deposited and $d$ is the average depth of deposition per year.

Conclusion:

Substitute $4.16\times {10}^{13}{\text{m}}^{\text{2}}$ for $A$ and $3\text{mm}$ for $d$ in equation (III).

  $\begin{array}{c}{V}^{\prime }=\left(4.16\times {10}^{13}{\text{m}}^{\text{2}}\right)\left(3\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)\\ =1.24\times {10}^{11}{\text{m}}^{\text{3}}\end{array}$

Thus, the volume of volcanic material deposited on the moon is one year is $1.24\times {10}^{11}{\text{m}}^{\text{3}}$.

(c)

To determine

The time taken for the volcanic deposition equal to that of the  volume of the moon.

Answer to Problem 35QP

The time taken for the volcanic eruption to be equal to the volume of the moon is $2.03\times {10}^8\text{years}$.

Explanation of Solution

Write the expression for the time taken for volcanic deposition equal to mass of moon Io.

  $t=\frac{V}{V^{\prime }}$        (IV)

Here, $t$ is the time taken for the deposition of the volcanic material.

Conclusion:

Substitute $2.52\times {10}^{19}{\text{m}}^{\text{3}}$ for $V$ and $1.24\times {10}^{11}{\text{m}}^{\text{3}}$ for $V^{\prime }$ in equation (IV).

  $\begin{array}{c}t=\frac{2.52\times {10}^{19}{\text{m}}^{\text{3}}}{1.24\times {10}^{11}{\text{m}}^{\text{3}}/\text{year}}\\ =2.03\times {10}^8\text{years}\end{array}$

Thus, the time taken for the volcanic eruption to be equal to the volume of the moon is $2.03\times {10}^8\text{years}$.

(d)

To determine

The number of times the moon had turned inside out all over the age of the solar system.

Answer to Problem 35QP

The moon could be turned inside out for $20$ times for all over the age of the solar system.

Explanation of Solution

The moon takes $2.03\times {10}^8\text{years}$ for the total volume of the volcanic eruption to be equal to its mass. Therefore, the number of times it had turned inside out all through the age of the solar system will be the ratio of the total age to the time taken to turn out once.

The age of the solar system is $4.0\text{billion}\text{years}$ or $4\times {10}^9\text{years}$.

Write the expression for the number of times the moon turned inside out over the age of solar system.

  $n=\frac{T}{t}$        (V)

Here, $n$ is the number of times the moon turned inside out and $T$ is the total age of the solar system.

Conclusion:

Substitute $4\times {10}^9\text{years}$ for $T$  and $2.03\times {10}^8\text{years}$ for $t$ in equation (V).

  $\begin{array}{c}n=\frac{4\times {10}^9\text{years}}{2.03\times {10}^8\text{years}}\\ \approx 20\end{array}$

Thus, the moon could be turned inside out for $20$ times for all over the age of the solar system.