Chapter 5, Problem 35P

To determine

The distance of the Io and the Europa from Jupiter’s centre.

Answer to Problem 35P

The distance of the Io from Jupiter’s centre is $\underset{\_}{9.09\times {10}^{15}\text{m}}$ and the Europa from Jupiter’s centre is $\underset{\_}{1.44\times {10}^{16}\text{m}}$.

Explanation of Solution

Write the expression for the Newton’s gravitational force.

    $F_r=\frac{G{M}_{J}m}{r^2}$        (I)

Here, $F_r$ is the gravitational force, $G$ is the gravitational constant, $M_J$ is the mass of Jupiter, $m$ is the mass of satellite, $r$ is the separation between them.

Write the expression for the Newton’s second law of motion.

    $\begin{array}{c}{\displaystyle \sum F_r}=\frac{G{M}_{J}m}{r^2}\\ =m{a}_r\end{array}$        (II)

Here, $a_r$ is the radial acceleration of the satellite.

Write the expression for $a_r$.

    $a_r=\frac{v^2}{r}$        (III)

Here, $v$ is the velocity.

Write the expression for the $v$.

    $v=r\omega$        (IV)

Here, $\omega$ is the angular velocity of the satellite.

Write the expression for the $\omega$.

    $\omega =\frac{2\pi }{T}$        (V)

Here, $T$ is the period of the satellite.

Use equation (V) in (IV) to solve for $v$.

    $v=\frac{2\pi r}{T}$        (VI)

Use equation (VI) and (III) in (II) to solve for $r$.

    $\begin{array}{c}\frac{G{M}_{J}m}{r^2}=\frac{m}{r}{\left(\frac{2\pi r}{T}\right)}^2\\ r={\left(\frac{G{M}_J{T}^2}{4{\pi }^2}\right)}^{1/3}\end{array}$        (VII)

Conclusion:

Substitute $1.77\text{d}$ for $T$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $1.9\times {10}^{27}\text{kg}$ for $M_J$ in equation (VII) to find the distance of Io from Jupiter’s centre.

    $\begin{array}{c}{r}_{\text{Io}}={\left[\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(1.9\times {10}^{27}\text{kg}\right){\left(1.77\text{d}\times \frac{8.64\times {10}^4\text{s}}{1\text{d}}\right)}^2}{4{\pi }^2}\right]}^{1/3}\\ =9.09\times {10}^{15}\text{m}\end{array}$

Substitute $3.54\text{d}$ for $T$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $1.9\times {10}^{27}\text{kg}$ for $M_J$ in equation (VII) to find the distance of Europa from Jupiter’s centre.

    $\begin{array}{c}{r}_{\text{Europa}}={\left[\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(1.9\times {10}^{27}\text{kg}\right){\left(3.54\text{d}\times \frac{8.64\times {10}^4\text{s}}{1\text{d}}\right)}^2}{4{\pi }^2}\right]}^{1/3}\\ =1.44\times {10}^{16}\text{m}\end{array}$

Therefore, the distance of the Io from Jupiter’s centre is $\underset{\_}{9.09\times {10}^{15}\text{m}}$ and the Europa from Jupiter’s centre is $\underset{\_}{1.44\times {10}^{16}\text{m}}$.