Question

Io and Europa are two of Jupiter's many moons. The mean distance of Europa from Jupiter is about twice as far as that for Io and Jupiter. By what factor is the period of Europa's orbit longer than that of Io's?
TEu/TIo =

Answer 1

The universal gravitation law and Newton's second law allow us to find that the answer for the relation of the rotation periods of the satellites is:

        $\frac{T_{Eu}}{T_{Io}}$ = 2.83

The universal gravitation law states that the force between two bodies is proportional to their masses and inversely proportional to their distance squared

           $F = G \frac{Mm}{r^2}$

Where G is the universal gravitational constant (G = 6.67 10⁻¹¹ $\frac{N m^2 }{kg^2}$), F the force, m and m the masses of the bodies and r the distance between them

Newton's second law states that force is proportional to the mass and acceleration of bodies

          F = m a

Where F is the force, m the mass and the acceleration

In this case the body is the satellites of Jupiter and the planet,

            $G \frac{Mm}{r^2} = m a$

Suppose the motion of the satellites is circular, then the acceleration is centripetal

           a = $\frac{v^2}{r}$r

Where v is the speed of the satellite and r the distance to the center of the planet

     

we substitute

      $G \frac{Mm}{r^2} = m \frac{v^2}{r} \\G \frac{M}{r} = v^2$

Since the speed is constant, we can use the uniform motion ratio

      v = $\frac{\Delta x}{t}$

In the case of a complete orbit, the time is called the period.

The distance traveled is the length of the orbit circle

           Δx = 2π r

We substitute

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

           

Let's write this expression for each satellite

Io satellite

Let's call the distance from Jupiter is  

            r = $r_{Io}$  

           $T_{Io}^2 = (\frac{4 \pi ^2}{GM} ) \ r_{Io}^3$TIo² = (4pi² / GM) rIo³

Europe satellite

Distance from Jupiter  is

         $r_{Eu} = 2 \ r_{Io}$

We calculate

         $T_{Eu} = ( \frac{4\pi ^2 }{GM} (2 \ r_{Io})^3\\T_{Eu} = ( \frac{4 \pi ^2 }{GM}) r_{Io} \ 8$

         

         $T_{Eu}^2 = 8 T_{Io}^2$            

         

         $\frac{ T_{Eu}}{T_{Io}} = \sqrt{8} = 2.83$

           

In conclusion, using the universal gravitation law and Newton's second law, we find that the answer for the relationship of the relation periods of the satellites is:

        $\frac{T_{Eu}}{T_{Io}} = 2.83$

Learn more about universal gravitation law and Newton's second law here:

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