Chapter 22, Problem 9P

What is the orbital velocity and period of a ring particle at the outer edge of Saturn’s A ring? (Hint: Use the formula for circular velocity, Eq. 5-1a. The formula requires input quantities in kg and m.) (Note: The radius of the outer edge of the A ring is 136,500 km.)

To determine

The orbital velocity and period of a ring particle at the outer edge of Saturn’s A ring.

Answer to Problem 9P

The orbital velocity of a ring particle at the outer edge of Saturn’s A ring is $1.67\times {10}^4\text{m}/\text{s}$ and orbit period of the ring particle is $14.3\text{h}$.

Explanation of Solution

Write the expression for the orbital velocity.

    $V_{\text{s}}=\sqrt{\frac{G{M}_{\text{s}}}{R_{\text{s}}}}$        (I)

Here, $V_{\text{s}}$ is the orbital velocity of Saturn’s A ring, $G$ is the universal gravitational constant, $M_{\text{s}}$ is the mass of Saturn and $R_{\text{s}}$ is the radius of the edge of A ring.

Write the expression for the time period.

    $T_{\text{s}}=\frac{2\pi R_{\text{s}}}{V_{\text{s}}}$        (II)

Here, $T_{\text{s}}$ is the time period of a ring particle at the outer edge of Saturn’s A ring.

From the celestial profile, the mass of the Saturn is $5.68\times {10}^{26}\text{kg}$.

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $5.68\times {10}^{26}\text{kg}$ for $M_{\text{s}}$ and $136500\text{km}$ for $R_{\text{s}}$ in equation (I) to find $V_{\text{s}}$.

    $\begin{array}{c}{V}_{\text{s}}=\sqrt{\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(5.68\times {10}^{26}\text{kg}\right)}{\left(136500\text{km}\right)}}\\ =\sqrt{\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(5.68\times {10}^{26}\text{kg}\right)}{\left(136500\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}}\\ =\sqrt{\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(5.68\times {10}^{26}\text{kg}\right)}{\left(136500\times {10}^3\text{m}\right)}}\\ =1.67\times {10}^4\text{m}/\text{s}\end{array}$

Substitute $136500\text{km}$ for $R_{\text{s}}$ and $1.67\times {10}^4\text{m}/\text{s}$ for $V_{\text{S}}$ in equation (II) to find $T_{\text{s}}$.

    $\begin{array}{c}{T}_{\text{s}}=\frac{2\pi \left(136500\text{km}\right)}{\left(1.67\times {10}^4\text{m}/\text{s}\right)}\\ =\frac{2\pi \left(136500\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}{\left(1.67\times {10}^4\text{m}/\text{s}\right)}\\ =\left(5.14\times {10}^4\text{s}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\\ =14.3\text{h}\end{array}$

Therefore, the orbital velocity of a ring particle at the outer edge of Saturn’s A ring is $1.67\times {10}^4\text{m}/\text{s}$ and orbit period of the ring particle is $14.3\text{h}$.