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OCW Scholar - Physics I: Classical Mechanics 8 Online
OpenStudy (anonymous):

Problem Solving and Estimation: Challenge Problem 1: How does the solver get g(eur) to relate to g(earth) (eqn 1.10)? In particular, what am I missing between 1.9 and 1.10? Thanks!

OpenStudy (jh3power):

Generally, \[\frac{ GM }{ R^2 }=g\] Look at the solution. Look at 1.9 and 1.10. Well, obvioulsy, F=mg, so the mass on the right becomes "destroyed" and the F becomes g. but I do not understand why g eur=Reur/Rearth *g earth. It should be the opposite(Rearth/Reur). I hope it helped.

OpenStudy (anonymous):

I am sure that as soon as I understand what is going on, Ill have to smack my head pretty hard, but for now, I guess I just don't get it. I have no problems arriving at 1.9, and I also see how \[g=\frac{ GM }{R^2}\]. However, I can get as far as \[\frac{ F _{g, earth} }{F _{g, eur}}=\frac{ R _{eur}^2 M _{earth} }{R _{earth}^2 M _{eur}}=\frac{ g _{earth} }{g _{eur}}\]but I cannot make \[g=f(R)\]Regardless of what I do, \[g=f(M,R^2)\] Working 1.10 backwards, \[g _{eur}=\frac{ R _{eur} }{R _{earth} }g _{earth}\] \[mg _{eur}=\frac{ R _{eur} }{R _{earth} }mg _{earth}\] or \[F_{eur}=\frac{ R_{eur} }{R_{earth} }F_{earth}\] Rearranging: \[\frac{ F_{earth} }{ F_{eur} }=\frac{ R_{earth} }{ R_{eur} }\] which is not derivable from Newtons Law of Gravitation. Even if this were a typo, and the inverse relationship is true, we have still lost factors of mass and radius from our relationship in 1.9. I think I am digging myself even deeper into the muck on this one, so I'm going to move on, but if anyone can help pull me out, I'd appreciate it! Thanks for your help, jh3power!

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