Physics
posted by Larry on .
I've been stuck on this equation for a while and I don't know if i'm missing a key part to the equation.. i just need help!
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

You 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
Rr = 0.0014316r = 9134m
That seems a bit high. Better check my math. 
The 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).