Associated Topics || Dr. Math Home || Search Dr. Math

### 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

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/
```
Associated Topics:
High School Physics/Chemistry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search