The Earth–Moon distance is about R_(ZM) = 3.85 · 10^8 m and…

The Earth–Moon distance is about $R_{ZM} = 3.85 \cdot 10^8\ \mathrm{m}$ and the Moon's period is $T_M = 27.3$ days. Saturn's moon Dione orbits Saturn at a radius $R_{SD} = 3.78 \cdot 10^5\ \mathrm{m}$ with a period of $T_D = 2.7$ days. From these data, find the ratio of the masses of the Earth and Saturn.

Step 1: Idea

The orbit of any moon reveals the mass of its planet (as in Problem 16 for the Sun). Writing this for both systems and dividing, $\gamma$ and $4\pi^2$ cancel, and the periods can even stay in days.

Step 2: Mass from a moon's orbit

\[ \gamma\frac{Mm}{R^2} = m\frac{4\pi^2}{T^2}R, \]

\[ M = \frac{4\pi^2R^3}{\gamma T^2}. \]

\[ \frac{M_Z}{M_S} = \frac{R_{ZM}^3/T_M^2}{R_{SD}^3/T_D^2} = \left(\frac{R_{ZM}}{R_{SD}}\right)^3\left(\frac{T_D}{T_M}\right)^2. \]

Step 3: Checking the data

With $R_{SD} = 3.78 \cdot 10^5\ \mathrm{m}$ = 378 km the ratio would come out at about $10^7$: the Earth would be ten million times heavier than Saturn, which is impossible. Also, 378 km is far less than Saturn's own radius (58 000 km), so a moon cannot orbit there. The radius must be $3.78 \cdot 10^8\ \mathrm{m}$ (Dione's real orbit is 377 400 km); the exponent in the data is a misprint.

Step 4: Numbers

\[ \frac{M_Z}{M_S} = \left(\frac{3.85}{3.78}\right)^3\left(\frac{2.7}{27.3}\right)^2 = 1.0566 \cdot 0.00978. \]

\[\begin{gathered} \frac{M_Z}{M_S} \approx 0.0103 \\ M_S \approx 97\,M_Z \end{gathered}\]

Dione's orbit is about the same size as the Moon's, but Dione goes round ten times faster. A ten-times-shorter period at the same distance needs a $10^2 = 100$ times stronger pull, hence a planet about 100 times more massive.

Check

Accepted value: Saturn has 95 Earth masses $\checkmark$