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Chapter 5: Q5-62P (page 109)

Table 5鈥�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

\({\bf{1883 \times 1}}{{\bf{0}}^{\bf{3}}}\)

Short Answer

Expert verified

(a) The mass of the Jupiter using the data of Io鈥檚 is \(1.901 \times {10^{27}}\;{\rm{kg}}\)

(b) 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.

Step by step solution

01

Concept

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.

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

02

Calculation of mass of the Jupiter using the data of Io

(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 \(1.902 \times {10^{27}}\;{\rm{kg}}\).

03

Calculation of mass of the Jupiter using the data of Ganymede

(b)

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

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

04

Calculation of mass of the Jupiter using the data of Europa

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

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

05

Calculation of mass of the Jupiter using the data of Callisto

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.

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