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August 28, 2014

August 28, 2014

Posted by **Larry** on Monday, January 21, 2013 at 10:51am.

If the gravitational field strength at the top of Mount Everest is 9.772 N/kg, approximately how tall (in feet) is the mountain?

F= g ( m1 x m2)/ radius^2

f= gravitational force

g= 6.67x10^-11

m= mass

d= radius^2

earth mass= 5.98x10^24

earth radius = 6.38x10^6

- Physics -
**Steve**, Monday, January 21, 2013 at 11:29amYou know that at sea level

GM/r^2 = 9.8

Since G and M don't change, at radius R at the mountain top,

GM/R^2 = 9.772

so, R^2/r^2 = 9.8/9.772 = 1.0028653

R = 1.0014316r

R-r = 0.0014316r = 9134m

That seems a bit high. Better check my math.

- Physics -
**drwls**, Monday, January 21, 2013 at 11:33amThe actual height of Everest is 8848 m. Your number is close, and you used the right method. The difference is probably due to the use of too few significant figures for G(or GM/r^2).

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