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Re: A weighty subject



Suddenly I'm worrying about the definition of g! Is it the rate at which
objects fall, or GM/R? It can't be both!

Using the best data I could find,
M = 5.9736 x 10^24
R = 6371.0 (this is the average (volumetric) radius)
G = 6.6726 x 10-11

So g = GM/R = 9.8201 m/s^2. I'll grant you that this assumes a spherical
earth, so 9.81 could be the correction due to oblateness.

However, this ignors the rotation of the earth. At the equator this
amounts to a centripetal acceleration of
a(c) = (464 m/s)^2 / 6378000 m = 0.034 m/s^2

So the rate at which objects dropped at the equator would fall is
determined by
a = g - a(c) = 9.81 - 0.03 = 9.78 m/s^2

From these calculations it looks like g = GM/R, but g is not the
acceleration of falling objects. (The other possibility is that GM/R would
really be closer to 9.84 and the spinning of the earth makes objects fall
at 9.81 m/s^2)

Thus our freshmen labs should apparently aim for 9.78 m/s^2, not 9.81
m/s^2! Indeed, the NASA website I checked for the mass and size of the
earth listed the "surface gravity" as 9.78 m/s^2. Only if you are moving
westward at several hundred km/hr would a dropped object fall at 9.81 m/s^2


Tim Folkerts