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$M_J = 1.9 \times 10^{27} kg$

Giancoli 6th Edition, Chapter 5, Problem 61

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Chapter 5, Problem 61 is solved.

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Transcript for this Giancoli solution

In part a we’re going to find the mass of Jupiter based on knowing the radius and period of orbit of its moon Io and let’s recognize that centripetal force on the moon Io is provided by gravity due to Jupiter. So we’ll have the gravity formula: ‘G’ times mass of Jupiter ‘mJ’ times that mass of the moon ‘mm’ divided by the radius of the moon’s orbit ‘rm’ squared equals centripetal acceleration which is mass of the moon ‘mm’ times the centripetal acceleration four pi squared 'r' over ‘T’ squared, that involves the period which we know. Doing some algebra, we’re going to multiply both sides by ‘rm’ squared and divide both sides by ‘G’ and we’re left with the mass of Jupiter: ‘mJ’ equals four pi squared times radius of its orbit ‘rm’ cubed divided by the period of the moon’s orbit ‘T’ squared and putting in some numbers our answer for part a is: four times three point one four squared time four hundred and twenty two times ten raised to power six meters cubed divided by six point six seven times ten raised to power minus eleven which is ‘G’ times one point seven seven days times twenty four hours per day times three thousand six hundred second per hour and we’ll square that period, and this gives us one point nine zero zero zero times ten raised to power twenty seven kilograms. We kept a lot of decimals on that because we’re going to be comparing our answer for part b to that number. In part b we calculate the mass of Jupiter based on data about Europa, Ganymede and Callisto and we’ll see if we get the same answer or not. Mass of Jupiter according to data about Europa is four times pi squared times six hundred and seventy one times ten raised to power six meters cubed divide by six point six seven times ten raised to power eleven times three point five five days times twenty four hours per day times tree thousand six hundred seconds per hour and square that and we get one point eight nine eight eight times ten raised to power twenty seven kilograms. Using the same process you can see the data in the table, we have mass of Jupiter according to Ganymede that’s going to give us one point eight nine two seven times ten raised to power twenty seven kilograms and then according to the data about Europa the mass of Jupiter is one point eight nine six two times ten raised to power twenty seven kilograms. If you wanted to do a sophisticated comparison you could find the percent differences between each of these results but they're really close so for now we’ll just say they're roughly consistent.