If the Earth Stopped Rotating...Date: 05/11/2000 at 00:15:43 From: Scott J. Bryan Subject: Effects of the Earth's rotation on objects Dear Dr. Math, I teach an Inertial Navigation Theory and Analysis course for the United States Navy. We discuss gyro theory, velocity meter theory, and effects of gravity and Earth rotation. Here is a question that was asked of me: If the Earth were to stop rotating (we'll assume that we wouldn't have to worry about weather, or flying off or anything) would objects weigh less or would they weigh more? Here's the answer we came up with: We know that we are experiencing centrifugal force due to the rotation. We also know that for every action, there is an equal and opposite reaction. This would mean that we may be under the effects of centripetal force as well. However, we believe that centripetal force is no match for centrifugal force. So if the Earth were to stop rotating, we would no longer be feeling the centrifugal force trying to throw us off, and so gravity would have no competition. Therefore, it would be able to be felt at its fullest capacity. This leads us to believe that we would weigh more. Well, there it is. I hope you get a chance to read and respond to this. It's quite a good little brain teaser, if you ask me. But then, I may be considered a "geek" in certain circles. I can live with that ;-) Thanks a ton! Scott Bryan Date: 05/11/2000 at 10:45:57 From: Doctor Rick Subject: Re: effects of the Earths rotation on the objects. Hi, Scott. You're basically correct. The total force that you feel as you stand on the ground is zero; otherwise you would be falling (accelerating in some direction). This force is the result of three forces: (1) gravity (acting downward), (2) centrifugal "force" (acting outward, perpendicular to the axis of the earth), and (3) the force of the ground pushing up on you (acting upward). It is the last that we measure with a scale and call weight. It is opposite to the result of (1) and (2). The centripetal force is gravity. It isn't a reaction to the centrifugal force, but an independent force that makes us move in a circle in the first place, rather than a straight line. Centrifugal "force" is not a real force, but an artifact of our choosing to work with a coordinate system that is rotating, rather than fixed in space. If gravity (the centripetal force) were not acting to hold us on the surface, we would fly off at a tangent to the surface - in a straight line, because no force would be acting on us. From the point of view of someone on the surface, however, we would appear to accelerate upward, as if a force were acting; this apparent force is centrifugal force. If the centrifugal force is removed, the gravitational force will not change. Since the centrifugal force acted in a direction generally opposite to that of gravity, it reduced the resultant force; in its absence, the resultant force increases and you will weigh more. Note that centrifugal force does not act straight up, and that its strength depends on one's latitude. At the poles, centrifugal force has no effect on one's weight. The only contrary effect that occurs to me is that the earth's shape is slightly flattened at the poles and fattened at the equator because the earth itself is pulled out by centrifugal force. If the earth stopped spinning, it would revert to a more spherical shape - but VERY slowly. Over eons, objects at the equator would gradually increase in weight as they got closer to the center of gravity; this would be in addition to the immediate increase in weight due to the removal of the centrifugal force. On the other hand, objects at the poles would have no immediate change in weight since they would experience no centrifugal force, but they would gradually decrease in weight as the earth rebounded and they got farther from the center. It would take some work to figure out how great each of these changes would be. Would you like to try? - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 05/13/2002 at 06:53:18 From: Steve Subject: If the Earth Stopped Rotating... Dr. Math, in your response you suggest Centrifugal Force is a fictitious force, and that I certainly can see. But how can you then talk in subsequent paragraphs about it as though it is perfectly real? Surely if the force is non-existent, it is unnecessary to use it as a means of determining the effect the rotation of the Earth has on one's apparent weight. In order to explain why I tend to feel an outward (ie. centrifugal) force when I travel around a curve in my car, I can dispense with the concept entirely and use only the concepts of Newton's First Law and Centripetal Force. Why should standing on the Earth be any different? However let me clarify one thing: I agree with both of you that a person would feel lighter when the Earth rotates than if it were not to rotate! I just am not satisfied with the explanation given. I saw a solution to a problem recently that had a Free-Body Diagram of forces acting on a person at the equator of the rotating Earth; two forces were shown: (1) gravity acting DOWNWARD and (2) the force - calculated as a CENTRIPETAL FORCE, mysteriously and inexplicably acting UPWARD! How can a centre-seeking force be interpreted as acting UPWARD? And if this depicted force is in fact centrifugal force, why are we allowed to show it on a Free-Body diagram? After all, centrifugal force is a non-existent force isn't it? I sincerely hope you can solve this conundrum for me! Date: 05/13/2002 at 09:22:59 From: Doctor Rick Subject: Re: If the Earth Stopped Rotating... Hi, Steve. The free-body diagram you describe is incorrect. As I stated in my answer, it is gravity that is the centripetal ("center-seeking") force -- pulling the person toward the center of the earth. The (fictitious) centrifugal ("center-fleeing") force acts away from the center of rotation. Scott's apparent misunderstanding was about the nature of the centripetal force: he stated that it arose as a reaction to centrifugal force. I suppose this common misunderstanding arises because of the similarity of the words, which makes them appear to be a set of twins, whereas in fact they arise from unrelated sources: gravity (or tension in a string, etc. depending on the situation) and rotation. The fact that centrifugal force is fictitious, or (as I stated) "an artifact of our choosing to work with a coordinate system that is rotating", does not mean that it is non-existent. It means that we must be careful to be consistent: we either choose to work in a rotating coordinate system and use centrifugal force, or we choose to work in an inertial coordinate system and realize that there is no centrifugal force. Both approaches work, as long as we do not mix them. In other words, if you wish to work with a rotating coordinate system (such as latitude and longitude on a rotating earth), then you MUST include the "fictitious" centrifugal force. It is wrong to omit it. If you wish to avoid fictitious forces, you MUST use a stationary (inertial) coordinate system. Consider the problem at hand, of a person standing on the equator. In the coordinate system rotating with the earth, the man is stationary; his acceleration is zero. The forces acting on the man, as I described (and in contrast to the free-body diagram you saw), are: (1) gravity, acting downward with force -mg (where m is the man's mass); (2) centrifugal force, acting upward with force mRw^2 where R is the radius of the earth and w (for omega) is the angular velocity of the earth; (3) the reaction force F of the ground pushing up on the man, which is the man's apparent weight. The net force equals m times the man's acceleration (in the rotating reference frame), which is zero as I said; so F - mg + mRw^2 = 0 F = mg - mRw^2 Thus we can calculate the man's apparent weight (and express it in mass units, in kilograms say, as F/g). You are correct that we can solve the same problem in an inertial coordinate system and forget about "fictitious forces". Looking at it this way, we just have two forces, (1) and (3) above; but the man is moving, and in particular he has an acceleration: a = -Rw^2 The acceleration is downward because a straight-line horizontal trajectory would be a tangent to the earth's surface, eventually rising above the surface; in order to stay "in one place", the trajectory must be bent downward -- it must have a downward acceleration. Now, with a non-zero acceleration Newton's second law takes the form F - mg = -Rw^2 F = mg - Rw^2 The resulting equation is just the same as before. It is a matter of preference. If you can easily picture the fact that this man, just minding his own business standing in "one place", is really orbiting the earth at some 1000+ miles per hour and accelerating toward the earth at 250 miles per hour per hour, then you'll get along fine with the second view. If it's easier to picture the man as stationary, and you're willing to use a fictitious force as the cost of using a rotating coordinate system, then you'll want to take the first view. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/