Chapter 11, Problem 37QP

To determine

The tidal force of the Jupiter and Io with respect to that of the Earth and its Moon.

Answer to Problem 37QP

The tidal force experienced by the moon Io of Jupiter is $3.05\times {10}^5$ times stronger than the tidal force experienced by the Earth’s moon.

Explanation of Solution

The tidal force is the gravitational interaction felt by a celestial body towards and away from its centre of mass due to the presence of the other celestial body around it.

Write the expression for the tidal force acting on the moon.

  $F_{\text{tidal}}=\frac{2G{M}_p{M}_m{R}_m}{d_{p-m}^3}$        (I)

Here, $F_{\text{tidal}}$ is the tidal force on moon, $G$ is the gravitational constant, $M_p$ is the mass of the planet, $M_m$ is the mass of the moon, $R_m$ is the radius of the moon and $d_{p-m}$ is the distance between the moon and the planet.

Write the expression for the comparison of the tidal force on Earth-moon and Jupiter-Io system.

  $\frac{F_{\text{tidal-Io}}}{F_{\text{tidal-m}}}=\frac{\frac{M_{Jup}{M}_{Io}{R}_{Io}}{d_{Jup-Io}^3}}{\frac{M_E{M}_m{R}_m}{d_{E-m}^3}}$        (II)

Here, $M_{Jup}$ is the mass of the planet Jupiter, $M_{Io}$ is the mass of the moon Io, $R_{Io}$ is the radius of the moon Io, $d_{Jup-Io}$ is the distance between the moon Io and the planet Jupiter, $M_E$ is the mass of the planet Earth, $M_m$ is the mass of the Earth’s moon, $R_m$ is the radius of the Earth’s moon and $d_{E-m}$ is the distance between the Earth’s moon and the planet Earth.

The mass of the planet Jupiter is $1.898\times {10}^{27}\text{kg}$ whereas the mass of its moon Io is $8.9\times {10}^{22}\text{kg}$.

The mass of the Earth is $6\times {10}^{24}\text{kg}$ and the mass of the Earth’s moon is $7.34\times {10}^{22}\text{kg}$.

The radius of the Earth’s moon is $1737\text{km}$ and the radius of Io is $1820\text{km}$.

The orbital radius of the Earth’s moon is $385,000\text{km}$ and the orbital radius of Io is $422,000\text{km}$.

Conclusion:

Substitute $1.898\times {10}^{27}\text{kg}$ for $M_{Jup}$, $8.9\times {10}^{22}\text{kg}$ for $M_{Io}$, $1820\text{km}$ for $R_{Io}$ , $422,000\text{km}$ for $d_{Jup-Io}$, $6\times {10}^{24}\text{kg}$ for $M_E$, $7.34\times {10}^{22}\text{kg}$ for $M_m$, $1737\text{km}$ for $R_m$ and $385,000\text{km}$ for $d_{E-m}$ in equation (II).

  $\begin{array}{c}\frac{F_{\text{tidal-Io}}}{F_{\text{tidal-m}}}=\frac{\frac{\left(1.898\times {10}^{27}\text{kg}\right)\left(8.9\times {10}^{22}\text{kg}\right)\left(1820\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}{{\left(422,000\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}^3}}{\frac{\left(6\times {10}^{24}\text{kg}\right)\left(7.34\times {10}^{22}\text{kg}\right)\left(1737\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}{{\left(385,000\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}^3}}\\ =\frac{4.09\times {10}^{30}}{1.34\times {10}^{28}}\\ =3.05\times {10}^5\end{array}$

Thus, the tidal force experienced by the moon Io of Jupiter is $3.05\times {10}^5$ times stronger than the tidal force experienced by the Earth’s moon.