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Answer: Velocity of Jupiter Satellite is 6.1 times that of the velocity of the Earth Satellite.
Since you are looking for qualitative, we need to determine the relationships between Earth and Jupiter. Before we can do that, we need to review the science of circular orbits.
For an object to orbit, two conditions must be met. The only force acting on the satellite will be the force due to gravity from the planet, and the force must be equal to the centripetal force required to keep the object in a circular orbit
`F_g=G*(Mm)/R^2` (where `GM/R^2` is the acceleration due to gravity at the distance from the center of the planet), and `F_c = (mv^2)/R` Where R is the distance to the center of the planet. v is the orbital speed we are trying to compare, so let's make some sense out of this.
`F_g = F_c` so `G(Mm)/R^2=(mv^2)/R`
which gives us that `v_(text(orbit)) = sqrt(GM/R)`
G is a constant that will be the same for both Jupiter and Earth, so that doesn't affect our answer qualitatively.
From this equation we need to determine the relative mass of Jupiter to Earth, and the Relative radii of the satelite orbits.
Jupiter is equal to 317.83 Earth masses (see NASA link)
Since we are dealing with the same altitude, we need to determine R for the satellite above Jupiter and Earth.
Jupiter has a radius of 69,911 km, so `R_J = 71,911text(km)`
Earth has a radius of 6,371 km, so `R_E = 8,371text(km)`
so `R_J/R_E ~~ 8.59`
Now we set up the ratio:
`v_J/v_E = (sqrt(GM_J/R_J))/(sqrt(GM_E/R_E))`
Which simplifies to
We know those numbers so we plug in `M_J/M_E = 317.83` and `R_J/R_E = 8.59`
ANSWER: This means that the orbital velocity of the satellite around Jupiter will be about 6.1 times that of the Earthly satellite.
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