Rotation and Weight
Date: winter 2013-14
I was wandering about the effect of centrifugal force of earths rotation on the weight of objects. Since centrifugal force pulls an object away from the center of mass, should the weight of an object on Earth's equator be partially overcome by the effect of centrifugal force? I know that objects always retain the same mass regardless of the effects of gravity and other forces, but would the pull of gravity be weaker at the equator than at the poles of Earth? So for example, would a 999 gram weight on the equator weigh perhaps more like 1000 grams on the poles?
Yours is an excellent question, and it turns out that your guess is not so far off.
According to Newtonian physics, "centrifugal force" is actually radial acceleration. If you swing a tethered ball around your head while you are in space (away from gravity) the inertia of the moving ball is trying to make it travel in a straight line. It does not do so because the tension of the tether is radially accelerating it towards your head.
We will use the following in our calculations:
The earth is approximately round, and its circumference is about 25,000 miles.
The diameter of the earth is 25,000 miles/(pi = 3.14159) = 7958 miles
The radius of the earth = diameter/2 = 3979 miles = 6,403,000 meters
A radian is a measure of angle used in physics and mathematics. There are 2 pi radians in a circle.
The earth rotates once in 24 hours = 86,400 seconds. This is an "angular velocity" of 2*pi/86400 = 0.0000727 "radians" per second.
The radial acceleration is the angular velocity * angular velocity/ radius = (0.0339 meters per second) per second.
The acceleration of gravity, by comparison, is (9.8 meters per second) per second. Then the net acceleration on the equator is 9.8 - 0.0339 = 9.766 (meters per second)/per second So if we measure an object's "weight" on the North Pole (using a spring type scale) as 1000 grams, and then use the same spring scale to measure the same object on the equator, its "weight" should appear to be 1000*9.766/9.8 = 996.5 grams.
Your question is very perceptive; however, the effects are small but numerous when highly precise measurements are required. Even the changing magnetic field of the Earth can play a factor, as does, the longitude and latitude of the measurement, as well as the local changes in the Earth?s gravitational constant. The whole problem of these precise measurements are beyond the scope of this short format. If you want to follow up the definitive treatment of all these ?corrections? see:
"Handbook of Mass Measurement" by Frank E. Jones and Randall M. Schoonerover; CRC Press 2002 ISBN 0-8493-2531-5.
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Update: November 2011