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$T_i=2.0 \times 10^4s$
$T_o=7.1 \times 10^4 s$
$T_s=3.8 \times 10^4s$

Giancoli 6th Edition, Chapter 5, Problem 49

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Chapter 5, Problem 49 is solved.

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Transcript for this Giancoli solution

The centripetal acceleration of the ice will be the same as acceleration due to gravity at the location of the ice. So we’ll have centripetal acceleration is gravity due to Saturn. Let’s consider the inner ice first: we have that four times pi squared times the radius of the inner ice divided by the period squared of the inner ice revolution equals the acceleration due to gravity of Saturn at that position, so the radius of the inner ice squared. We solve this for the period and multiply both sides by period of inner ice squared and also multiply by ‘ri’ squared divide by ‘G’ times mass of Saturn squared on both sides, simplifying and taking the square root of both sides and we’re left with the period of the inner ice is: four times pi squared times the radius of the inner ice cubed divided by ‘G’ times the mass of Saturn. So the time it takes for the inner ice to make one complete orbit around Saturn is going to be: the square root of four times three point one four squared times seventy three thousand times ten raised to power three meters cubed divided by six point six seven times ten raised to power minus eleven times five point seven times ten raised to power twenty six kilograms, the mass of Saturn. This works out to twenty thousand and eighty eight seconds which is better expressed as two point zero times ten raised to power four seconds. For the outer ice we have ‘To’ period of the outer ice is the same formula so we plug in the numbers right away: the square root of four times three point one four squared times one hundred and seventy thousand times ten raised to power three meters cubed divided by six point six seven times ten raised to power minus eleven times five point seven times ten raised to power twenty six kilograms, the mass of Saturn. This works out to seven point one times ten raised to power four seconds. The period of Saturn’s own rotation is ten hours times three thousand six hundred seconds per hour plus thirty nine minutes times sixty seconds per minute which works put to three point eight times ten raised to power four seconds. What's interesting to think about is what would the sky look like on Saturn? Let’s draw the surface of the planet and let’s suppose you’re standing on the surface of the planet and the planet is rotating right, that means the sun will appear to go left across the sky and the inner ice since it’s going faster than the rotation of the planet the inner ice will appear to go right across the sky because it’s going faster than the rotation of the planet. The outer ice on the other hand since it’s going slower than the rotation of the planet will appear to go the same direction as the sun but it won't go across the sky as fast as the sun because it has its velocity in the same direction as the rotation of the planet so the outer ice is in fact moving in the same direction as the planet but since it’s going slower than the planet its apparent motion will be to the left, the same as the sun, but it’ll go slower than the sun. So it’ll be an interesting sky to see.

Comments

How did you get the first equation with 4 pi