Io, one of the satellites of Jupiter, has an orbital period of $1.769$ days and the radius of the orbit is $4.22 \times 10^{8} \; m$. Show that the mass of Jupiter is about one-thousandth that of the Sun.

(N/A) The orbital period of $Io$, $T_{Io} = 1.769 \; \text{days} = 1.769 \times 24 \times 3600 \; s$.
The orbital radius of $Io$, $R_{Io} = 4.22 \times 10^{8} \; m$.
The mass of Jupiter $(M_J)$ is given by Kepler's third law: $M_J = \frac{4 \pi^{2} R_{Io}^{3}}{G T_{Io}^{2}} \quad ... (i)$.
Similarly, for the Earth orbiting the Sun, the mass of the Sun $(M_S)$ is: $M_S = \frac{4 \pi^{2} R_e^{3}}{G T_e^{2}} \quad ... (ii)$, where $R_e = 1.496 \times 10^{11} \; m$ and $T_e = 365.25 \; \text{days}$.
Dividing equation $(ii)$ by $(i)$: $\frac{M_S}{M_J} = \left( \frac{R_e}{R_{Io}} \right)^{3} \times \left( \frac{T_{Io}}{T_e} \right)^{2}$.
Substituting the values: $\frac{M_S}{M_J} = \left( \frac{1.496 \times 10^{11}}{4.22 \times 10^{8}} \right)^{3} \times \left( \frac{1.769}{365.25} \right)^{2}$.
Calculating the ratio: $\frac{M_S}{M_J} \approx (354.5)^{3} \times (0.00484)^{2} \approx 44558000 \times 0.0000234 \approx 1045$.
Thus, $\frac{M_S}{M_J} \approx 1000$, which implies $M_J \approx \frac{M_S}{1000}$.
Hence, the mass of Jupiter is about one-thousandth that of the Sun.