The rings of Saturn are composed of chunks of ice that orbit the planet. The inner radius of the rings is 73,000 km, and the outer radius is 170,000 km. Find the period of an orabiting chunk of ice at the inner radius and the period of a chunk at the outer radius. Compare your numbers with Saturn’s own rotation period of 10 hours and 39 minutes. The mass of Saturn is \({\bf{5}}{\bf{.7 \times 1}}{{\bf{0}}^{{\bf{26}}}}\;{\bf{kg}}\).

The speed of an object moving in the orbit of a planet depends upon the radius of an orbital. This can be used to find the expression for the time period.

The mass of Saturn \({M_{planet}} = 5.7 \times 1{0^{26}}\;{\rm{kg}}\).

Satrun’s own rotation period is \(10\;h\) 39 minutes.

The inner radius of ring is \({r_{in}} = 73,000\;km\).

The outer radius of ring is \({r_{out}} = 170,000\;km\).

The velocity of rotation can be calculated as,

\(v = \frac{{2\pi r}}{T}\)…… (i)

Here, r is the radius and T is the orbitalperiod.

The expression for the orbital speed is given as

\(v = \sqrt {\frac{{G{M_{planet}}}}{r}} \)…… (ii)

Here, G is the universal gravitational constant and its value is \(6.67 \times {10^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}\).

On combining equation (i) and equation (ii),

\(\begin{aligned}\sqrt {\frac{{G{M_{{\rm{planet}}}}}}{r}} &= \frac{{2\pi r}}{T}\\T &= 2\pi \sqrt {\frac{{{r^3}}}{{G{M_{{\rm{planet}}}}}}} \end{aligned}\)

The period for the inner radius can be calculated as,

\(\begin{aligned}{T_{{\rm{inner}}}} &= 2\pi \sqrt {\frac{{{{\left( {73000\,{\rm{km}} \times \frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)}^3}}}{{\left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}}{{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right)\left( {5.7 \times {{10}^{26}}\;{\rm{kg}}} \right)}}} \\ &= 2.0 \times {10^4}\;{\rm{s}}\end{aligned}\)

The period for the outer radius can be calculated as,

\(\begin{aligned}{T_{{\rm{outer}}}} &= 2\pi \sqrt {\frac{{{{\left( {170000\,{\rm{km}} \times \frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)}^3}}}{{\left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}}{{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right)\left( {5.7 \times {{10}^{26}}\;{\rm{kg}}} \right)}}} \\ &= 7.1 \times {10^4}\;{\rm{s}}\end{aligned}\)

It is given that the time period is 10 h 39 minute that is, \(3.8 \times {10^4}\;{\rm{s}}\). Therefore, the upper ring will move faster than the Sun, however the outer ring will appear to move across the sky slower than the Sun.