Chapter 8, Problem 1P

By what factor would the Sun be shrunk to be the size of a large beach ball, 1 meter in diameter? (b) Calculate the distances and diameters of Mercury, Earth, Ceres, Jupiter, Neptune, and Pluto if the whole Solar System were shrunk by the same amount. (b) Find their masses if their densities stayed the same.

To determine

The factor that the sun has to shrink to be the size of a large beach ball, one meter in diameter.

Answer to Problem 1P

The sun has to shrink by a factor of $\underset{\_}{1.4\text{billion}}$ times to be the size of large beach ball of $1\text{m}$ in diameter.

Explanation of Solution

Given that the diameter of the sun is $1.39\times {10}^9\text{m}$ and the diameter of the ball is $1\text{m}$.

Write the expression for the ratio of the diameter of the ball to the sun’s diameter.

  $R=\frac{d_{\text{ball}}}{d_{\text{sun}}}$        (I)

Here, $R$ is the ratio of the diameter of the ball to the sun’s diameter, $d_{\text{ball}}$ is the diameter of the ball, and $d_{\text{sun}}$ is the diameter of the sun.

Conclusion:

Substitute $1\text{m}$ for $d_{\text{ball}}$ and $1.39\times {10}^9\text{m}$ for $d_{\text{sun}}$ in Equation (I) to find the factor that the sun has to shrink to be the size of a large beach ball.

  $\begin{array}{c}\frac{d_{\text{ball}}}{d_{\text{sun}}}=\frac{1\text{m}}{1.39\times {10}^9\text{m}}\\ =\frac{1\text{m}}{1.4\times {10}^9\text{m}}\end{array}$

Therefore, the sun has to shrink by a factor of $\underset{\_}{1.4\text{billion}}$ times to be the size of large beach ball of $1\text{m}$ in diameter.

To determine

The diameters of Mercury, Earth, Ceres, Jupiter, Neptune, and Pluto, distance from the sun of each planets, and masses if the densities of the planets stayed same.

Answer to Problem 1P

The relative masses and model distance of each planet from the sun is $5.27\times {10}^{-8},7.70\times {10}^{-7},3.44\times {10}^{-10},10.90\times {10}^{-4},4.53\times {10}^{-5},4.66\times {10}^{-9}$ and $1.72\times {10}^{-2},6.68\times {10}^{-3},2.42\times {10}^{-3},1.28\times {10}^{-3},2.22\times {10}^{-4},1.69\times {10}^{-4}$ respectively.

Explanation of Solution

Given from appendix A in the Table A.3, that the radius of mercury, earth, Ceres, Jupiter, Neptune, and Pluto is $2440\text{km}$, $\text{6378}\text{km}$, $\text{487}\text{km}$, $\text{71,492}\text{km}$, $\text{24764}\text{km}$, and $\text{1160}\text{km}$ respectively.

Write the formula for the diameter.

  $d=2r$        (I)

Here, $d$ is the diameter, and $r$ is the radius.

Write the expression for the density.

  $\rho =\frac{M}{V}$        (II)

Here, $\rho$ is the density, $M$ is the mass, and $V$ is the volume.

Write the expression for the volume.

  $V=\frac{4\pi r^3}{3}$

Here, $V$ is the volume.

From Equation (I) the radius is equal to half of the diameter. So the above relation becomes,

  $\begin{array}{c}V=\frac{4\pi {\left(\frac{d}{2}\right)}^3}{3}\\ =\frac{\pi d^3}{6}\end{array}$        (III)

Write the expression for the model diameter.

  $\text{Model diamter}=\frac{d_{\text{planet}}}{d_{\text{sun}}}$        (IV)

Here, $d_{\text{planet}}$ is the diameter of each planet.

Conclusion:

For the case of mercury:

Substitute $2440\text{km}$ for $r$ in Equation (I) to find the diameter of mercury.

  $\begin{array}{c}{d}_{\text{m}}=2\left(2440\text{km}\right)\\ =4880\text{km}\end{array}$

Here, $d_{\text{m}}$ is the diameter of mercury.

For the case of Earth:

Substitute $\text{6378}\text{km}$ for $r$ in Equation (I) to find the diameter of Earth.

  $\begin{array}{c}{d}_{\text{e}}=2\left(\text{6378}\text{km}\right)\\ =12756\text{km}\end{array}$

Here, $d_{\text{e}}$ is the diameter of Earth.

For the case of Ceres:

Substitute $\text{487}\text{km}$ for $r$ in Equation (I) to find the diameter of Ceres.

  $\begin{array}{c}{d}_{\text{C}}=2\left(\text{487}\text{km}\right)\\ =974\text{km}\end{array}$

Here, $d_{\text{C}}$ is the diameter of Ceres.

For the case of Jupiter:

Substitute $\text{71,492}\text{km}$ for $r$ in Equation (I) to find the diameter of Jupiter.

  $\begin{array}{c}{d}_{\text{J}}=2\left(\text{71,492}\text{km}\right)\\ =142984\text{km}\end{array}$

Here, $d_{\text{J}}$ is the diameter of Jupiter.

For the case of Neptune:

Substitute $\text{24764}\text{km}$ for $r$ in Equation (I) to find the diameter of Neptune.

  $\begin{array}{c}{d}_{\text{N}}=2\left(\text{24764}\text{km}\right)\\ =49528\text{km}\end{array}$

Here, $d_{\text{N}}$ is the diameter of Neptune.

For the case of Pluto:

Substitute $\text{1160}\text{km}$ for $r$ in Equation (I) to find the diameter of Pluto.

  $\begin{array}{c}{d}_{\text{P}}=2\left(\text{1160}\text{km}\right)\\ =2320\text{km}\end{array}$

Here, $d_{\text{P}}$ is the diameter of Pluto.

For the case of mercury:

Substitute $4880\text{km}$ for $d_{\text{m}}$ and $1.39\times {10}^6\text{km}$ for $d_{\text{sun}}$ in Equation (IV) to find the model diameter of the mercury.

  $\begin{array}{c}\text{Model diamter}=\frac{4880\text{km}}{1.39\times {10}^9\text{m}}\\ =3.75\times {10}^{-3}\end{array}$

For the case of Earth:

Substitute $12756\text{km}$ for $d_{\text{e}}$ and $1.39\times {10}^6\text{km}$ for $d_{\text{sun}}$ in Equation (IV) to find the model diameter of the Earth.

  $\begin{array}{c}\text{Model diamter}=\frac{12756\text{km}}{1.39\times {10}^6\text{km}}\\ =9.17\times {10}^{-3}\end{array}$

For the case of Ceres:

Substitute $974\text{km}$ for $d_{\text{C}}$ and $1.39\times {10}^6\text{km}$ for $d_{\text{sun}}$ in Equation (IV) to find the model diameter of the Ceres.

  $\begin{array}{c}\text{Model diamter}=\frac{974\text{km}}{1.39\times {10}^6\text{km}}\\ =0.7\times {10}^{-3}\end{array}$

For the case of Jupiter:

Substitute $142984\text{km}$ for $d_{\text{J}}$ and $1.39\times {10}^6\text{km}$ for $d_{\text{sun}}$ in Equation (IV) to find the model diameter of the Jupiter.

  $\begin{array}{c}\text{Model diamter}=\frac{142984\text{km}}{1.39\times {10}^6\text{km}}\\ =102.86\times {10}^{-3}\end{array}$

For the case of Neptune:

Substitute $49528\text{km}$ for $d_{\text{N}}$ and $1.39\times {10}^6\text{km}$ for $d_{\text{sun}}$ in Equation (IV) to find the model diameter of the Neptune.

  $\begin{array}{c}\text{Model diamter}=\frac{49528\text{km}}{1.39\times {10}^6\text{km}}\\ =35.63\times {10}^{-3}\end{array}$

For the case of Pluto:

Substitute $2320\text{km}$ for $d_{\text{P}}$ and $1.39\times {10}^6\text{km}$ for $d_{\text{sun}}$ in Equation (IV) to find the model diameter of the Pluto.

  $\begin{array}{c}\text{Model diamter}=\frac{2320\text{km}}{1.39\times {10}^6\text{km}}\\ =1.67\times {10}^{-3}\end{array}$

From Equation (III) the model volume of the sun is,

  $V_{\text{sun}}=\frac{\pi D_{\text{sun}}^3}{6}$

Here, $D_{\text{sun}}$ is the model diameter of sun and $V_{\text{sun}}$ is the model volume of sun.

Substitute $1$ for $D_{\text{sun}}$.

  $\begin{array}{c}{V}_{\text{sun}}=\frac{\pi {\left(1\right)}^3}{6}\\ =0.523\end{array}$

Similarly,

From Equation (III) the model volume of the Mercury is,

  $V_{\text{Mercury}}=\frac{\pi D_{\text{Mercury}}^3}{6}$

Here, $D_{\text{mercury}}$ is the model diameter of sun and $V_{\text{Mercury}}$ is the model volume of the mercury.

Substitute $3.75\times {10}^{-3}$ for $D_{\text{mercury}}$.

  $\begin{array}{c}{V}_{\text{mercury}}=\frac{\pi {\left(3.75\times {10}^{-3}\right)}^3}{6}\\ =2.76\times {10}^{-8}\end{array}$

From Equation (III) the model volume of the Earth is,

  $V_{\text{Earth}}=\frac{\pi D_{\text{Earth}}^3}{6}$

Here, $D_{\text{Earth}}$ is the model diameter of Earth and $V_{\text{Earth}}$ is the model volume of the Earth.

Substitute $9.17\times {10}^{-3}$ for $D_{\text{Earth}}$.

  $\begin{array}{c}{V}_{\text{Earth}}=\frac{\pi {\left(9.17\times {10}^{-3}\right)}^3}{6}\\ =4.03\times {10}^{-7}\end{array}$

From Equation (III) the model volume of the Ceres is,

  $V_{\text{Ceres}}=\frac{\pi D_{\text{Ceres}}^3}{6}$

Here, $D_{\text{Ceres}}$ is the model diameter of Ceres and $V_{\text{Ceres}}$ is the model volume of the Ceres.

Substitute $0.7\times {10}^{-3}$ for $D_{\text{Ceres}}$.

  $\begin{array}{c}{V}_{\text{Ceres}}=\frac{\pi {\left(0.7\times {10}^{-3}\right)}^3}{6}\\ =1.80\times {10}^{-10}\end{array}$

From Equation (III) the model volume of the Jupiter is,

  $V_{\text{Jupiter}}=\frac{\pi D_{\text{Jupiter}}^3}{6}$

Here, $D_{\text{Jupiter}}$ is the model diameter of Jupiter and $V_{\text{Jupiter}}$ is the model volume of the Jupiter.

Substitute $102.86\times {10}^{-3}$ for $D_{\text{Jupiter}}$.

  $\begin{array}{c}{V}_{\text{Jupiter}}=\frac{\pi {\left(102.86\times {10}^{-3}\right)}^3}{6}\\ =5.70\times {10}^{-4}\end{array}$

From Equation (III) the model volume of the Neptune is,

  $V_{\text{Neptune}}=\frac{\pi D_{\text{Neptune}}^3}{6}$

Here, $D_{\text{Neptune}}$ is the model diameter of Neptune and $V_{\text{Neptune}}$ is the model volume of the Neptune.

Substitute $35.63\times {10}^{-3}$ for $D_{\text{Neptune}}$.

  $\begin{array}{c}{V}_{\text{Neptune}}=\frac{\pi {\left(35.63\times {10}^{-3}\right)}^3}{6}\\ =2.37\times {10}^{-5}\end{array}$

From Equation (III) the model volume of the Pluto is,

  $V_{\text{Pluto}}=\frac{\pi D_{\text{Pluto}}^3}{6}$

Here, $D_{\text{Pluto}}$ is the model diameter of Pluto and $V_{\text{Pluto}}$ is the model volume of the Pluto.

Substitute $1.67\times {10}^{-3}$ for $D_{\text{Pluto}}$.

  $\begin{array}{c}{V}_{\text{pluto}}=\frac{\pi {\left(1.67\times {10}^{-3}\right)}^3}{6}\\ =2.44\times {10}^{-9}\end{array}$

From Equation (II),

The expression for the mass is,

  $M=\rho V$

The ratio of the mass of the planet to the mass of the sun is,

  $\begin{array}{c}{M}_{\text{r}}=\frac{\rho V_{\text{planet}}}{\rho V_{\text{sun}}}\\ =\frac{V_{\text{planet}}}{V_{\text{sun}}}\end{array}$        (V)

Here, $V_{\text{planet}}$ is the model volume of the planet and $V_{\text{sun}}$ is the model volume of sun.

For the case of mercury:

From Equation (V) the relative mass of the mercury is,

  $M_{\text{r}}=\frac{V_{\text{mercury}}}{V_{\text{sun}}}$

Substitute $0.523$ for $V_{\text{sun}}$ and $2.76\times {10}^{-8}$ for $V_{\text{mercury}}$.

  $\begin{array}{c}{M}_{\text{r}}=\frac{2.76\times {10}^{-8}}{0.523}\\ =5.27\times {10}^{-8}\end{array}$

For the case of Earth:

From Equation (V) the relative mass of the Earth is,

  $M_{\text{r}}=\frac{V_{\text{Earth}}}{V_{\text{sun}}}$

Substitute $0.523$ for $V_{\text{sun}}$ and $4.03\times {10}^{-7}$ for $V_{\text{Earth}}$.

  $\begin{array}{c}{M}_{\text{r}}=\frac{4.03\times {10}^{-7}}{0.523}\\ =7.70\times {10}^{-7}\end{array}$

For the case of Ceres:

From Equation (V) the relative mass of the Ceres is,

  $M_{\text{r}}=\frac{V_{\text{Ceres}}}{V_{\text{sun}}}$

Substitute $0.523$ for $V_{\text{sun}}$ and $1.80\times {10}^{-10}$ for $V_{\text{Ceres}}$.

  $\begin{array}{c}{M}_{\text{r}}=\frac{1.80\times {10}^{-10}}{0.523}\\ =3.44\times {10}^{-10}\end{array}$

For the case of Jupiter:

From Equation (V) the relative mass of the Jupiter is,

  $M_{\text{r}}=\frac{V_{\text{Jupiter}}}{V_{\text{sun}}}$

Substitute $0.523$ for $V_{\text{sun}}$ and $5.70\times {10}^{-4}$ for $V_{\text{Jupiter}}$.

  $\begin{array}{c}{M}_{\text{r}}=\frac{5.70\times {10}^{-4}}{0.523}\\ =10.90\times {10}^{-4}\end{array}$

For the case of Neptune:

From Equation (V) the relative mass of the Neptune is,

  $M_{\text{r}}=\frac{V_{\text{neptune}}}{V_{\text{sun}}}$

Substitute $0.523$ for $V_{\text{sun}}$ and $2.37\times {10}^{-5}$ for $V_{\text{neptune}}$.

  $\begin{array}{c}{M}_{\text{r}}=\frac{2.37\times {10}^{-5}}{0.523}\\ =4.53\times {10}^{-5}\end{array}$

For the case of Pluto:

From Equation (V) the relative mass of the Pluto is,

  $M_{\text{r}}=\frac{V_{\text{Pluto}}}{V_{\text{sun}}}$

Substitute $0.523$ for $V_{\text{sun}}$ and $2.44\times {10}^{-9}$ for $V_{\text{Pluto}}$.

  $\begin{array}{c}{M}_{\text{r}}=\frac{2.44\times {10}^{-9}}{0.523}\\ =4.66\times {10}^{-9}\end{array}$

The distance between the sun and mercury is $57.91\times {10}^6\text{km}$, distance between the sun and earth is $\text{149}\text{.6}\times {10}^6\text{km}$, distance between the sun to Ceres is $413.60\times {10}^6\text{km}$, distance between the sun and Jupiter is $778.50\times {10}^6\text{km}$, distance between the sun and Neptune is $4498\times {10}^6\text{km}$, and the distance between the sun and Pluto is $5909\times {10}^6\text{km}$.

For the case of mercury the model distance is,

  $y=\frac{y_{\text{sun}}}{y_{\text{mercury}}}$

Here, $y$ is the distance, $y_{\text{sun}}$ is the distance of the sun, and $y_{\text{mercury}}$ is the distance between the sun and mercury.

Substitute $57.91\times {10}^6\text{km}$ for $y_{\text{mercury}}$ and ${10}^6\text{km}$ for $y_{\text{sun}}$.

  $\begin{array}{c}y=\frac{{10}^6\text{km}}{57.91\times {10}^6\text{km}}\\ =1.72\times {10}^{-2}\end{array}$

For the case of earth the model distance is,

  $y=\frac{y_{\text{sun}}}{y_{\text{earth}}}$

Here, $y_{\text{sun}}$ is the distance of the sun, and $y_{\text{earth}}$ is the distance between the sun and earth.

Substitute $\text{149}\text{.6}\times {10}^6\text{km}$ for $y_{\text{earth}}$ and ${10}^6\text{km}$ for $y_{\text{sun}}$.

  $\begin{array}{c}y=\frac{{10}^6\text{km}}{\text{149}\text{.6}\times {10}^6\text{km}}\\ =6.68\times {10}^{-3}\end{array}$

For the case of Ceres the model distance is,

  $y=\frac{y_{\text{sun}}}{y_{\text{Ceres}}}$

Here, $y_{\text{sun}}$ is the distance of the sun, and $y_{\text{Ceres}}$ is the distance between the sun and Ceres.

Substitute $413.60\times {10}^6\text{km}$ for $y_{\text{Ceres}}$ and ${10}^6\text{km}$ for $y_{\text{sun}}$.

  $\begin{array}{c}y=\frac{{10}^6\text{km}}{413.60\times {10}^6\text{km}}\\ =2.42\times {10}^{-3}\end{array}$

For the case of Jupiter the model distance is,

  $y=\frac{y_{\text{sun}}}{y_{\text{Jupiter}}}$

Here, $y_{\text{sun}}$ is the distance of the sun, and $y_{\text{Jupiter}}$ is the distance between the sun and Jupiter.

Substitute $778.50\times {10}^6\text{km}$ for $y_{\text{Jupiter}}$ and ${10}^6\text{km}$ for $y_{\text{sun}}$.

  $\begin{array}{c}y=\frac{{10}^6\text{km}}{778.50\times {10}^6\text{km}}\\ =1.28\times {10}^{-3}\end{array}$

For the case of Neptune the model distance is,

  $y=\frac{y_{\text{sun}}}{y_{\text{Neptune}}}$

Here, $y_{\text{sun}}$ is the distance of the sun, and $y_{\text{Neptune}}$ is the distance between the sun and Neptune.

Substitute $4498\times {10}^6\text{km}$ for $y_{\text{Neptune}}$ and ${10}^6\text{km}$ for $y_{\text{sun}}$.

  $\begin{array}{c}y=\frac{{10}^6\text{km}}{4498\times {10}^6\text{km}}\\ =2.22\times {10}^{-4}\end{array}$

For the case of Pluto the model distance is,

  $y=\frac{y_{\text{sun}}}{y_{\text{Neptune}}}$

Here, $y_{\text{sun}}$ is the distance of the sun, and $y_{\text{Pluto}}$ is the distance between the sun and Pluto.

Substitute $5909\times {10}^6\text{km}$ for $y_{\text{Pluto}}$ and ${10}^6\text{km}$ for $y_{\text{sun}}$.

  $\begin{array}{c}y=\frac{{10}^6\text{km}}{5909\times {10}^6\text{km}}\\ =1.69\times {10}^{-4}\end{array}$

Therefore, relative masses and model distance of each planet from the sun is $5.27\times {10}^{-8},7.70\times {10}^{-7},3.44\times {10}^{-10},10.90\times {10}^{-4},4.53\times {10}^{-5},4.66\times {10}^{-9}$ and $1.72\times {10}^{-2},6.68\times {10}^{-3},2.42\times {10}^{-3},1.28\times {10}^{-3},2.22\times {10}^{-4},1.69\times {10}^{-4}$ respectively.