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Giancoli 6th Edition, Chapter 5, Problem 43

(2:11)

Chapter 5, Problem 43 is solved.

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Transcript for this Giancoli solution

For a satellite orbiting the earth we have the centripetal force provided by gravity so we say: ‘Fg’ equals ‘Fc’. The gravity formula is: ‘G’ times mass of the satellite ‘msat’ times mass of the earth ‘mE’ divided by distance from the earth’s centre ‘r’ squared and that equals mass of the satellite times the centripetal acceleration ‘v2’ over ‘r’ and solving for ‘v’: the satellite mass cancels and we multiply both sides but ‘r’ and take the square root of both sides so we have ‘v’ is equal to the square root of ‘G’ times the mass of the earth divided by ‘r’. The thing to be cautious about in this question is that ‘r’ isn’t the height given in the question but instead that height plus the radius of the earth because that height is the distance above the earth’s surface. So we’ll say 'r' equals radius of the earth plus the height given, three thousand six hundred kilometers, so that’s the radius of the earth six point three eight times ten raised to power six meters plus three thousand six hundred times ten raised to power three meters so we have nine point nine eight times ten raised to power six meters is what we’ll substitute for 'r'. Putting in numbers: ‘v’ equals the square root of six point six seven times ten raised to power minus eleven times five point nine eight times ten raised to power twenty four kilograms all divided by nine point nine eight times ten raised to power six meters which gives us six point three two times ten raised to power three meters per second.

Comments

Hey why doesn't the equation for velocity (2*pi*r)/T work for this question as well?