Chapter 5, Problem 61P

(a)

To determine

The mass of the Jupiter using the given IO.

Answer to Problem 61P

Solution:

The mass of Jupiter was found using the given data ${\text{I}}_{\text{0}}$ , ${\text{M}}_{\text{J}}\text{=1}{\text{.90×10}}^{\text{27}}\text{kg}$ .

Explanation of Solution

Given:

  $\begin{array}{l}{\text{R=422×10}}^{\text{3}}\text{km}\\ {\text{T}}_{\text{1}}\text{=1}\text{.77}\text{Earth}\text{days}\end{array}$

Formula Used:

FromKepler’s third law of planetary motion:

  $M_J=\frac{4{\pi }^2{R}^3}{G.T^2}$

Where, R -Mean distance from Jupiter, $T$ -periodof revolution.

  $\text{G=6}{\text{.67×10}}^{\text{11}}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$ (G is gravitational constant)

  $\text{T=(1}\text{.77 days)}\left(\frac{\text{86400}\text{s}}{\text{days}}\right)$

Calculation:

  $M_J\text{=}\frac{{\text{4×π}}^{\text{2}}{\text{×(422×10}}^{\text{6}}{\text{)}}^{\text{3}}}{\text{6}{\text{.67×10}}^{\text{-11}}\text{×(1}{\text{.77×86400)}}^{\text{2}}}\text{kg}$

  $M_J\text{=1}{\text{.90×10}}^{\text{27}}\text{kg}$

Conclusion:The mass of Jupiter was found using the given data ${\text{I}}_{\text{0}}$ , ${\text{M}}_{\text{J}}\text{=1}{\text{.90×10}}^{\text{27}}\text{kg}$ .

(b)

To determine

The mass of the Jupiter using the data of other three moons i.e. Europa, Ganymede and Calisto and find if all the resultants are consistent.

Answer to Problem 61P

Solution:

The mass of Jupiter was found using the given data ${\text{M}}_{\text{J}}\text{=1}{\text{.90×10}}^{\text{27}}\text{kg}$ and the results are consistent with each other.

Explanation of Solution

Given:

Period of Europa, T= 3.55 Earth days

Period of Ganymede, T= 7.16 Earth days

Period of Callisto, T=16.7 Earth days

Mean Distance of Europa from Jupiter $=671\times {10}^3\text{km}$

Mean Distance of Ganymede from Jupiter $\text{=1070}\times {\text{10}}^3\text{km}$

Mean Distance of Callisto from Jupiter $\text{=1883}\times {\text{10}}^3\text{km}$

Formula Used:

FromKepler’sthird law of planetary motion:

  $M_J=\frac{4{\pi }^2{R}^3}{G.T^2}$

Where, R -Radius of Jupiter $T$ -period

  $\text{G=6}{\text{.67×10}}^{\text{11}}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$ (G is gravitational constant)

Calculation:

By using the moon Europa data is

  $M_J\text{=}\frac{{\text{4×π}}^{\text{2}}{\text{×(671×10}}^{\text{6}}{\text{)}}^{\text{3}}}{\text{6}{\text{.67×10}}^{\text{-11}}\text{×(3}{\text{.55×86400)}}^{\text{2}}}\text{kg}$

  $M_J\text{=1}{\text{.90×10}}^{\text{27}}\text{kg}$ .

By using the data for the moon Ganymede

  $M_J\text{=}\frac{{\text{4×π}}^{\text{2}}{\text{×(1070×10}}^{\text{6}}{\text{)}}^{\text{3}}}{\text{6}{\text{.67×10}}^{\text{-11}}\text{×(7}{\text{.16×86400)}}^{\text{2}}}\text{kg}$

  $M_J\text{=1}{\text{.89×10}}^{\text{27}}\text{kg}$ .

By using the data for the moon Calisto

  $M_J\text{=}\frac{{\text{4×π}}^{\text{2}}{\text{×(1883×10}}^{\text{6}}{\text{)}}^{\text{3}}}{\text{6}{\text{.67×10}}^{\text{-11}}\text{×(16}{\text{.7×86400)}}^{\text{2}}}\text{kg}$

  $M_J\text{=1}{\text{.89×10}}^{\text{27}}\text{kg}$

Conclusion:The mass of Jupiter that was found using the given data was consistent.