Weight in an Accelerated Frame of Reference
Date: Spring 2013
Since spinning of Earth does not affect its gravity (gravitational field intensity), then why is the magnitude of a gravitational force between the Earth and an object on the surface of the Earth is the same as the weight of the object? The weight of the object should be less than the gravitational force. If I have an object near equator of Earth then centripetal force due to rotation of Earth reduces the weight of the object compared to the situation where Earth does not rotate at all; or is it now called an apparent weight or does the weight remain unchanged. My question is, how does weight and the magnitude of the gravitational attraction depend on the rotation of the Earth. And are these two equal in magnitude? Why?
First, we must realize that the reading on a scale is not the weight of an object. This is how hard the scale pushes upward on the object. This is actually the ?normal force? from the scale.
If there were no spinning of the Earth (zero acceleration), and if the only two forces on the object are gravity and the scale, then the normal force on the scale (the reading) would equal the gravitational force (the weight). When accelerating up or down, as in an elevator or as with Earth rotation, the reading on the scale will not be the same as the weight.
Dr. Ken Mellendorf
Illinois Central College
You have entered the dark forest of high precision measurements – in this case high accuracy mass measurements. First, “weight” is determined by “gravity”. This is different than the “mass” of an object. The definition of mass is: “The property of a body by which it requires force to change its state of motion is called inertia, and mass is the numerical measure of this property.” One of the outstanding problems in modern physics is a method for measuring mass in terms of other physical constants. To date that has not been solved. “The present embodiment of the mass of a kilogram is based on the French platinum kilogram of the Archives constructed in 1792.” Weight and mass are different. The “true weight” of an object varies depending upon many small corrections. Some are barometric pressure, humidity (which alters the density of air), location on Earth, and several others. This is not an easy topic, despite its common usage.
The book “Handbook of Mass Measurement” by Frank E. Jones and Randall M. Schoonover published by CRC Press. Goes into all these issues in detail, but it is not easy reading.
It sounds like you understand what is going on pretty well. Imagine if the gravitational force were so weak that it was just barely sufficient to keep you in orbit - say one inch above the surface of the Earth and moving at the same speed as the land beneath you. The acceleration required to accomplish that is v^2/r, where v is your orbital speed and r is the Earth's radius. If you work the numbers out, the acceleration is around .03 m/s^2 - a very small fraction of the 9.8 m/s^2 total gravitational acceleration. So, if the Earth were a perfect sphere, and you weighed 100 lb at the equator, you would weigh about five ounces more at the pole.
Thanks for the question. Let us ignore the spinning of the Earth. The weight of an object that one measure (on a scale) is the normal force or apparent weight. The normal force is the force of the ground pushing pack on the object.
The gravitational attraction does not depend on the spinning of Earth. However, the weight (or normal force) that one measures is dependent on the spinning of Earth. A 100 kg object has a slightly smaller weight (normal force) on the equator than at the 45 degree latitude line.
Often, folks will confuse the gravitational force with the normal force or weight. They are different when one considers the effect of Earth's rotation.
I hope this helps. Please let me know if you have more questions.
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Update: November 2011