Table 5–3 gives the mean distance, period, and mass for the four largest moons of Jupiter (those discovered by Galileo in 1609). Determine the mass of Jupiter: (a) using the data for Io; (b) using data for each of the other three moons. Are the results consistent?

Table 5-3 Principal Moons of Jupiter
Moon	Mass(kg)	Period (Earth days)	Mean distance from Jupiter (km)
Io	\({\bf{8}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{22}}}}\)	1.77	\({\bf{422 \times 1}}{{\bf{0}}^{\bf{3}}}\)
Europe	\({\bf{4}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{22}}}}\)	3.55	\({\bf{671 \times 1}}{{\bf{0}}^{\bf{3}}}\)
Ganymede	\({\bf{15 \times 1}}{{\bf{0}}^{{\bf{22}}}}\)	7.16	\({\bf{1070 \times 1}}{{\bf{0}}^{\bf{3}}}\)
Callisto	\({\bf{11 \times 1}}{{\bf{0}}^{{\bf{22}}}}\)	16.7	\({\bf{1883 \times 1}}{{\bf{0}}^{\bf{3}}}\)

Centripetal force and Gravitational force acts on the satellite during the orbital motion.

On equating these two forces the mass of the planet can be calculated.

\({\bf{M = }}\frac{{{\bf{4}}{{\bf{\pi }}^{\bf{2}}}{{\bf{r}}^{\bf{3}}}}}{{{\bf{G}}{{\bf{T}}^{\bf{2}}}}}\)

(a)

The expression for the mass of the planet is given as,

\(\begin{aligned}M &= \frac{{4{\pi ^2}{r^3}}}{{G{T^2}}}\\ &= \frac{{4{\pi ^2} \times {{\left( {4.22 \times {{10}^8}\;{\rm{m}}} \right)}^3}}}{{6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2} \times {{(1.77\;{\rm{days}} \times 24 \times 60 \times 60\;\frac{{\rm{s}}}{{{\rm{days}}}}{\rm{ }})}^2}}}\\ &= 1.902 \times {10^{27}}\;{\rm{kg}}\end{aligned}\)

Thus, the mass of the Jupiter is \(1.902 \times {10^{27}}\;{\rm{kg}}\).

(b)

Substituting the values in the mass of the planet is given as,

\(\begin{aligned}{M_{{\rm{Ganymede }}}} &= \frac{{4{\pi ^2}{r^3}}}{{G{T^2}}}\\ &= \frac{{4{\pi ^2} \times {{\left( {1.07 \times {{10}^9}\;{\rm{m}}} \right)}^3}}}{{6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2} \times {{(7.16\;{\rm{days}} \times \left( {24 \times 60 \times 60} \right)\;\frac{{\rm{s}}}{{{\rm{days}}}}{\rm{ }})}^2}}}\\ &= 1.895 \times {10^{27}}\;{\rm{kg}}\end{aligned}\)

Substituting the values in the mass of the planet is given as,

\(\begin{aligned}{M_{{\rm{Europa }}}} &= \frac{{4{\pi ^2}{r^3}}}{{G{T^2}}}\\ &= \frac{{4{\pi ^2} \times {{\left( {6.71 \times {{10}^8}\;{\rm{m}}} \right)}^3}}}{{6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2} \times {{(3.55\;{\rm{days}} \times \left( {24 \times 60 \times 60} \right)\;\frac{{\rm{s}}}{{{\rm{days}}}})}^2}}}\\ &= 1.901 \times {10^{27}}\;{\rm{kg}}\end{aligned}\)

Substituting the values in the mass of the planet is given as,

\(\begin{aligned}{M_{{\rm{Callisto }}}} &= \frac{{4{\pi ^2}{r^3}}}{{G{T^2}}}\\ &= \frac{{4{\pi ^2} \times {{\left( {1.883 \times {{10}^9}\;{\rm{m}}} \right)}^3}}}{{6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2} \times {{(16.7\;{\rm{days}} \times \left( {24 \times 60 \times 60} \right)\;\frac{{\rm{s}}}{{{\rm{days}}}})}^2}}}\\ &= 1.898 \times {10^{27}}\;{\rm{kg}}\end{aligned}\)

Thus, the mass of the Jupiter using the data for Europa is \(1.901 \times {10^{27}}\;{\rm{kg}}\), using the data of Ganymede is \(1.895 \times {10^{27}}\;{\rm{kg}}\) and that using the data of Calisto is \(1.898 \times {10^{27}}\;{\rm{kg}}\) and the results are consistent.