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a) see video
b) $\rho=5.4 \times 10^4 kg/m^3$

Giancoli 6th Edition, Chapter 5, Problem 55


Chapter 5, Problem 55 is solved.

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Transcript for this Giancoli solution

We have to show that density, defined as mass divided by volume, of this planet that has a satellite orbiting around it near its surface works out to three pi over ‘G’ times period ‘T’ squared. The best way to approach this is let’s figure out what is the volume ‘V’ and then hopefully substitute that into the expression and it’ll work out to be that. So we have to find ‘V’ in terms of period and we’ll start by saying ‘V’ equals four thirds pi times 'r' cubed, which is the definition of the volume of a sphere, what is 'r'? We know that the force of gravity on this satellite that’s orbiting the planet is going to be equal to the centripetal force being provided only by gravity and so we’ll substitute in our formulas for each of these. For gravity we have ‘G times mass of the planet ‘mp’ times the mass of the satellite ‘ms’ divided by the radius of the planet 'r' squared. Since the satellite is basically located at the surface I can say its distance from the centre is equal to the radius of the planet which equals mass of the satellite times four times pi squared times 'r' over ‘T’ squared, four times pi squared times 'r' over ‘T’ squared being the centripetal acceleration of the satellite and I chose to write it this way instead of ‘v’ squared over 'r' because in this form we have the period which the problem in the textbook gives us a clue that’s the way we want it, we want to have a ‘T’ in the solution so we approach the solution using ‘T’ whenever possible. We’ll solve this for 'r' cubed which we can then substitute directly into the volume formula. So multiplying both sides by 'r' squared and simplifying we end up with: 'r' cubed is equal to ‘G’ times ‘mp’ times ‘T’ squared divided by four pi squared. Substituting that into our volume formula ‘V’ is equals four thirds times pi times 'r' cubed: ‘V’ equals four thirds times pi times ‘G’ times ‘mp’ times ‘T’ squared over four pi squared and this reduces to: ‘V’ is ‘G’ times ‘mp’ times ‘T’ squared divided by three pi. So for part a we can substitute that back into the density formula, density is the mass of the planet divided by the volume: ‘mp’ multiplied by three pi over ‘G’ times ‘mp’ times ‘T’ squared which is three pi over ‘G’ times ‘T’ squared which is what we had to show. For part b we’ll substitute in some numbers into this formula to get the density of earth. So: density is three times pi divided by six point six seven times ten raised to power minus eleven times the period, the time it takes the satellite to go around the earth which is eighty five minutes which we’ll have to convert into seconds, square that and we get an answer of five point four times ten raised to power three kilograms per meters cubed. For the 5th Edition we have ninety minutes instead of eighty five minutes and for the 5th Edition the answer would be four point eight times ten raised to power three kilograms per meters cubed.