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How much does a person weigh on Mount Everest? 
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#1
Nov710, 01:24 PM

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1. The problem statement, all variables and given/known data
If a person has a mass of 68kg, how much does he weigh on the top of mount everest (8488km above sea level? Given: m=68kg, d=8488m 2. Relevant equations F = (G(m1)(m2)) / r^2 3. The attempt at a solution I tried figuring out the new radius. Once i get the radius i enter all the data and rearrange the formula to get m2 by itself. Fr^2/Gm1 = m2 My problem is getting the new radius. 


#2
Nov710, 01:26 PM

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What's the "old" radius?



#3
Nov710, 01:30 PM

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#4
Nov710, 01:33 PM

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How much does a person weigh on Mount Everest?
You didn't understand my question. I'll take another tack. How much does the person weigh at sea level? Why?



#5
Nov710, 01:37 PM

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#6
Nov710, 01:38 PM

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What about Newton's law of gravity? Does that not apply at sea level?



#7
Nov710, 01:46 PM

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#8
Nov710, 01:51 PM

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Newton's universal law of gravitation, F=GMm/r^{2}, is not called universal just on a whim.
In other words, it applies at sea level as well as atop Mt. Everest. 


#9
Nov710, 01:56 PM

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#10
Nov710, 02:10 PM

P: 6

Whoops, was i not supposed to read that xD?



#11
Nov710, 02:42 PM

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No, you were not supposed to see that.



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