Callisto and io are two of jupiter's satellites. the distance from callisto to the center of jupiter is approximately 4.5 times farther than the distance from io to the center of jupiter. how does callisto's orbital period, tc, compare to that of io, ti?

Respuesta :

To solve the problem, we can use Kepler's third law, which states that the ratio between the square of the orbital period and the cube of the radius of the orbit is constant. So, we can rewrite it as follows:
[tex] \frac{T_c^2}{R_c^3} = \frac{T_i^2}{R_i^3} [/tex]
where T are the periods, R the radius of the orbits (so, the distance of the satellites from Jupiter), and where the label c refers to Callisto and i to Io.

The problem says that the distance of Callisto from Jupiter is 4.5 times the distance of Io from Jupiter, i.e.:
[tex]R_c = 4.5 R_i[/tex]
If we substitute this into the previous equation, we can find a relationship between the two orbital periods Tc and Ti:
[tex] \frac{T_c^2}{T_i^2} = \frac{R_c^3}{R_i^3} = (4.5)^3 \frac{ R_i^3}{R_i^3} [/tex]
and so we have
[tex] \frac{T_c}{T_i}=(4.5)^{3/2} [/tex]
that we can rewrite as
[tex]T_c = 9.5 T_i[/tex]
so, the orbital period of Callisto is 9.5 times the orbital period of Io.
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