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Q: Centrifugal Force and Gravity ( Answered,   1 Comment )
Subject: Centrifugal Force and Gravity
Category: Science > Physics
Asked by: manofwar-ga
List Price: $10.00
Posted: 02 Nov 2002 10:08 PST
Expires: 02 Dec 2002 10:08 PST
Question ID: 96574
Centrifugal force from the spinning earth tends to throw off objects
on the surface. Gravity wins out over centrifugal force and objects
stay on the earth. Everything else being equal, if the earth was not
spinning, would a person weigh
more? (Ignore the complication arisng from the fact that the earth
would then not bulge at the equator which would then make the earth
denser and result in increased gravitational force.)  If so, by how
much? Does a person weigh more at the North Pole than at the equator
where centrifugal force is greater? Maybe the greater distance from
the center of the earth at the equator cancels the difference out?
Subject: Re: Centrifugal Force and Gravity
Answered By: livioflores-ga on 03 Nov 2002 00:57 PST
Hi manofwar!!

If the earth was not spinning, would a person weigh more?
The answer is yes on the Equator, where the centrifugal acceleration
due by the Earth rotation rise the maximum. But in the poles the
effects of the Earth´s rotation are null, so the weight without
rotation will be the same.

If so, by how much?
In the Equator the weight will be 0.3468% greater, and in the poles
will be the same.
To see how to obtain, aproximately, this data I you must visit the
following page: "Curious About Astronomy - Does your weight change
between the poles and the equator?"

Does a person weigh more at the North Pole than at the equator where
centrifugal force is greater?
The answer is yes, the centrifugal force pull the person out of the
Earth and the gravity pull the person to the center of the Earth, then
weight is:
W = Fg - Fc  where Fg is the gravity force and the Fc is the
centrifugal force.
In the North Pole Fc is null, then the weight is greater here.

Maybe the greater distance from the center of the earth at the equator
cancels the difference out?
Here you have a little confusion: The answer is NO, because if the
distance from the center of the Earth is greater, the weight will be

      G x M1 x M2
Fg = ------------------ (Newton's Law of Gravitation).

Then the person weigh more in the Pole because here he is closer to
the center of the Earth than in the Equator.
I addition we have the fact that the person weigh more in the North
Pole than in the Equator due the Earth rotation.
Then the weight difference considering the closer distance of the
North Pole to the center of the Earth is greater.

I found the exact values of related measures at the "MikaP Astro -
Earth" page:

From this page we have the following data of the Earth:
-Mass: 5.9763E27 g
-Mean equatorial radius: 6378.245 km (A.A.Izotov,1950); 
                         6378.077 km (I.D.Zhongolovich,1956)
-Difference in equatorial and polar semi-axes:
                           21.382 km (A.A.Izotonov,1950)
                           21.500 km (I.D.Zhongolovich,1956)
-Mean radius:            6370.949 km
-Mean acceleration of gravity at equator:
                            9.780573 m/s^2 (I.D.Zhongolovich,1952)
-Mean acceleration of gravity at poles:
                            9.832251 m/s^2
-Mean acceleration of gravity for entire surface of terrestial :
                            9.797830 m/s^2
-Ratio of centrifugal force to force of gravity at equator:
                            0.0034677 (0.34677%)
And a lot more.

For more reference and nice info related you can read the following
"The Bulging Earth" from Math Pages:

"Weight Changes with Position On/In Earth" by J. D. Jones from M.
Casco Associates:

At "The Math Forum - Ask Dr. Math" you can read a very interesting
discussion about the "Effects of the Earth's rotation on objects" and
the nature of the centrifugal force. The discussion is entitled: "If
the Earth Stopped Rotating...":

Another article: "Gravity variation from the equator to the poles" by
Ramin Amirmardfar:

I answer this question based in my own knowledge and using the
following search strategy:

Search engine: Google

Keywords and results pages:

earth gravity centrifugal ratio

weight poles centrifugal

I hope this helps you, but if you need some clarification, please post
a request for it before rate my answer.
Subject: Re: Centrifugal Force and Gravity
From: rbnn-ga on 02 Nov 2002 11:41 PST
I am able to answer your questions assuming a spherical earth, but I
do not know how to account precisely for the effect of bulge at the
equator, so I am posting what I have as a comment. If it's OK I can
repost it as an answer.

A discussion of the effect of the effect of the earth spinning on on
the weight of a person at the equator is given here:

The conclusion is:
"So some of the force of gravity is being used to make you go round in
a circle at the equator (instead of flying of into space) while at the
pole this is not needed. The centripetal acceleration at the equator
is given by 4 times pi squared times the radius of the Earth divided
by the period of rotation squared (4*pi2*r/T2). The period of rotation
is 24 hours (or 86400 seconds) and the radius of the Earth is about
6400 km. This means that the centripetal acceletation at the equator
is about 0.03 m/s2 (metres per seconds squared). Compare this to the
acceleration due to gravity which is about 10 m/s2 and you can see how
tiny an effect this is - you would weigh about 0.3% less at the
equator than at the poles! "

A discussion of centrifugal and centripetal forces:

Agreement that there will be less force: .

Some barometers have to take this fact into account in their design:
see "Latitude Correction" at

The earth would have to rotate with a period of 1 hour and 24 minutes
in order for there to be no gravity at the poles:

radius of the earth: 6.371 x 10^6 m

general discussion of centripetal acceleration:
centrifugal centripetal
centrifugal eart
centrifugal earth "weigh less"
"centripetal acceleratin" earth weigh less
"centripetal acceleration" earth

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