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Effective gravity, ge
Consider Newton's Second Law on the surface of the earth: In an inertial frame:
ΣF = ma = -mgr + other forces
(here -r is straight down to the center of the earth)
In the rotating frame:
ΣF - (2mv*) - (m(r)) - (md*/dtr) = ma*
(for the rotating earth, d*/dt = 0)
so:
other forces - mgr - (2mv*) - (m(r)) = ma* .
If we consider the (m(r)) term, we see that  for the earth is directed along the axis of the
spinning earth, or in the direction of the North star (which at Memphis, is about 35 above the
Northern horizon). The direction of r is straight up. Thus the direction of [r] will be {up and
to the North} cross {up} gives a direction of {East} with a magnitude that depends on  (which
is 2π/day), r (which is Rearth) and the sine of the angle between them which is the co-latitude [or
cosine of the latitude, which for Memphis would be cos(35)]. Further,  cross this vector would
be in the direction {North and up} cross {East} = {down and to the North}. The minus sign in
front means that the direction of the "fictitious force" would be {up and to the South}. Since this
will be essentially constant as long as r remains close to the surface of the earth, we can group
this term with the essentially constant -gr to make an effective ge that is a little less than the true
g due to gravity and also points a little bit South of straight down to the center of the earth.
ω
ω×r (East)
(E(East)
ω×(ω×r)
r
WRITTEN HOMEWORK PROBLEM #3:
a) Calculate the mass of the earth based upon the fact that the Moon orbits the earth with a
period of 27 days, 7 hours and 43 minutes at a distance of 384,400 km. (Since the moon orbits in
an ellipse rmin = 363,300 km and rmax = 405,500 km, and since the semimajor axis = ½ (rmin +
rmax), we get the 384,400 km figure to use. See the example under homework #27 at
http://www.cbu.edu/~jholmes/P380/Rrsystem.htm )
b) Using the above mass for the earth and given Rearth = 6,378 km, calculate the magnitude of the
true gravitational acceleration, g. (The 6,378 km is the equatorial radius; the polar radius is only
6,357 km.)
c) Using the fact that Memphis is close to sea level at latitude 35 North, and given that the
period for the earth's spin is 24 hours, calculate the magnitude and direction of the term
-((r)) at Memphis.
d) Using the results of parts b & c, calculate ge, and compare g and ge.