Chapter 5, Problem 68P

(a)

To determine

The acceleration due to gravity on the surface of planet Tungsten.

Answer to Problem 68P

The acceleration due to gravity on the surface of planet Tungsten is $\underset{\_}{4g}$.

Explanation of Solution

Write the expression for the acceleration due to gravity on Earth.

    $g=\frac{G{M}_{\text{Earth}}}{r_{\text{Earth}}^2}$        (I)

Here, $g$ is the acceleration due to gravity on Earth, $G$ is the gravitational constant, $M_{\text{Earth}}$ is the mass of Earth, $r_{\text{Earth}}$ is the radius of Earth.

The Planet twice the radius of the Earth and twice its density.

    $\begin{array}{c}r=2r_{\text{Earth}}\\ \rho =2{\rho }_{\text{Earth}}\end{array}$        (II)

Here, $\rho$ is the density of the Planet, ${\rho }_{\text{Earth}}$ is the density of the Planet, $r$ is the radius of the planet.

Use equation (II) and write the expression for the volume of the Planet.

    $\begin{array}{c}{V}_{\text{Planet}}=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\pi {\left(2r_{\text{Earth}}\right)}^3\\ =\left(8\right)\left(\frac{4}{3}\pi r_{\text{Earth}}^3\right)\end{array}$        (III)

Here, $V_{\text{Planet}}$ is the volume of the planet.

Write the expression for the density of the Planet.

    $\rho =\frac{M_{\text{Planet}}}{V_{\text{Planet}}}$        (IV)

Here, $M_{\text{Planet}}$ is the mass of the Planet.

Use equation (II) in (IV) to solve for $M_{\text{Planet}}$.

    $\begin{array}{c}{M}_{\text{Planet}}=\rho V_{\text{Planet}}\\ =\left(2{\rho }_{\text{Earth}}\right)\left(\frac{32}{3}\pi r_{\text{Earth}}^3\right)\\ =2\left({\rho }_{\text{Earth}}\right)\left(8\right)\left(\frac{4}{3}\pi r_{\text{Earth}}^3\right)\\ =16\left({\rho }_{\text{Earth}}\right)\left(\frac{4}{3}\pi r_{\text{Earth}}^3\right)\\ =16M_{\text{Earth}}\end{array}$        (V)

Write the expression for the $g$ value for the Planet.

    $g_{\text{Planet}}=\frac{G{M}_{\text{Planet}}}{r_{\text{Planet}}^2}$        (VI)

Here, $g_{\text{Planet}}$ is the force of gravity of the Planet.

Use equation (II) and (V) in (VI) to compare the $g$ value of the Planet.

    $\begin{array}{c}{g}_{\text{Panet}}=\frac{G\left(16M_{\text{Earth}}\right)}{{\left(2r_{\text{Earth}}\right)}^2}\\ =4\left(\frac{G{M}_{\text{Earth}}}{r_{\text{Earth}}^2}\right)\\ =4g\end{array}$        (VII)

Conclusion:

Therefore, the acceleration due to gravity on the surface of planet Tungsten is $\underset{\_}{4g}$.

(b)

To determine

The period of rotation of the Planet.

Answer to Problem 68P

The period of rotation of the Planet is $\underset{\_}{69\text{min}}$.

Explanation of Solution

A person standing at the pole experiences the full $4g$ of gravitational acceleration

Write the expression for the force acting on the person standing on the poles.

    $\begin{array}{c}{\displaystyle \sum F}=N-m{g}_{\text{Planet}}\\ =0\\ N=m{g}_{\text{Planet}}\end{array}$        (VIII)

Here, $N$ is the normal force.

Write the expression for the force acting on the person standing on the equator.

    $\begin{array}{c}{\displaystyle \sum F}=N-m{g}_{\text{Planet}}\\ =-\frac{m{v}^2}{r}\\ N=m{g}_{\text{Planet}}-\frac{m{v}^2}{r_{\text{Planet}}}\end{array}$        (IX)

Here, $v$ is the velocity.

Write the expression for $v$.

    $v=\frac{2\pi r_{\text{Planet}}}{T}$        (X)

Here, $T$ is the period of rotation.

Use equation (VIII) and (VII) in (IX) to solve for $v$.

    $\begin{array}{c}mg=m{g}_{\text{Planet}}-\frac{m{v}^2}{r_{\text{Planet}}}\\ mg=4mg-\frac{m{v}^2}{r_{\text{Planet}}}\\ v=\sqrt{3g{r}_{\text{Planet}}}\end{array}$        (XI)

Use equation (X) and (I) in (XI) to solve for $T$.

    $\begin{array}{c}\frac{2\pi r_{\text{Planet}}}{T}=\sqrt{3g{r}_{\text{Planet}}}\\ T=2\pi \sqrt{\frac{r_{\text{Planet}}}{3g}}\\ =2\pi \sqrt{\frac{2r_{\text{Earth}}}{3g}}\end{array}$        (XII)

Conclusion:

Substitute $6.37\times {10}^6\text{m}$ for $r_{\text{Earth}}$, $9.8\text{m}/{\text{s}}^2$ for $g$ in equation (XII) to find $T$.

    $\begin{array}{c}T=2\pi \sqrt{\frac{2\left(6.37\times {10}^6\text{m}\right)}{3\left(9.8\text{m}/{\text{s}}^2\right)}}\\ =4136\text{s}\\ =69\text{min}\end{array}$

Therefore, the period of rotation of the Planet is $\underset{\_}{69\text{min}}$.