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$m_M=6.5 \times 10^{23}kg$

Giancoli 6th Edition, Chapter 5, Problem 41

(1:32)

Chapter 5, Problem 41 is solved.

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Transcript for this Giancoli solution

The acceleration due to gravity on the moon is going to be zero point three eight times acceleration due to gravity on the earth and using our acceleration due to gravity formula we have ‘G’ times mass of the moon ‘mM’ divided by the radius of the moon ‘rM’ squared which is going to equal zero point three eight times ‘G’ times mass of the earth ‘mE’ divided by the radius of the earth ‘rE’ squared. Solving for the mass of the moon: the ‘G’s cancel, multiplying both sides by the radius of the moon squared we’ll end up with mass of the moon is zero point three eight times mass of the earth times the radius of the moon squared divided by the radius of the earth squared. Putting in numbers: zero point three eight times five point nine eight times ten raised to power twenty four kilograms times radius of the moon three thousand four hundred kilometers times ten raised to power three to get meters squared divided by six point three eight times ten raised to power six which is the radius of the earth squared getting six point five times ten raised to power twenty three is the mass of the moon.