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$r_E= 6.711 \times 10^5km$
$r_G=1.0714 \times 10^6km$
$r_C = 1.884 \times 10^6km$

Giancoli 6th Edition, Chapter 5, Problem 63

(4:25)

Chapter 5, Problem 63 is solved.

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Transcript for this Giancoli solution

We’re going to use data about the moon Io in order to figure out the radius of orbit of the other moon’s orbiting about Jupiter. So we have Kepler’s Law saying: the radius of one of the moon’s orbit cubed divided by the radius of Io’s orbit cubed equals the period of the other moon’s orbit squared divided by the period of Io’s orbit squared. We’ll solve this for the radius of the moon’s orbit, that is the other moon besides Io and so we’ll multiply both sides by Io’s orbit cubed and then we’ll take the cube root of both sides and solve for the radius of the moon’s orbit, which is the cube root of Io’s orbit radius cubed times the moon’s orbital period squared divided by Io’s orbital period squared. We could write this as: the radius of Io’s orbit times the cube root of the moon’s orbit squared divided by Io’s orbital period squared. Plugging in numbers three times we get three answers for the radius of Europa’s orbit, Ganymede’s orbit and Callisto’s orbit. Let’s do one at a time. So we have radius of Europa’s orbit: four hundred and twenty two times ten raised to power three kilometers, that’s Io’s orbit, and we’ll multiply this by the cube root of three point five five days squared divided by one point seven seven days squared giving six point seven one one times ten raised to power five kilometers, which is in agreement with the book where it says Europa has a radius of six point seven one times ten raised to power three kilometers. Ganymede’s orbital radius is four hundred and twenty two times ten raised to power three kilometers times the cube root of seven point one six days squared divided by one point seven seven days squared which results in one point zero seven one four times ten raised to power six kilometers and that also is in agreement with the book which says one thousand and seventy one times ten raised to power three kilometers. And then for Callisto: four hundred and twenty two times ten raised to power three kilometers times the cube root of sixteen point seven days squared divided by one point seven seven days squared and we get one point eight eight four times ten raised to power six kilometers which is approximately equal to what the book says, one thousand eight hundred and eighty three times ten raised to power three kilometers.