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Q: Centrifugal Force and Gravity ( Answered,   1 Comment )
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 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?```
 ```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?" http://curious.astro.cornell.edu/question.php?number=310 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 smaller: G x M1 x M2 Fg = ------------------ (Newton's Law of Gravitation). R^2 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: http://www.ursa.fi/~mpi/earth/ 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 articles: "The Bulging Earth" from Math Pages: http://www.mathpages.com/home/kmath182.htm "Weight Changes with Position On/In Earth" by J. D. Jones from M. Casco Associates: http://mcasco.com/QA22.html 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...": http://mathforum.org/library/drmath/view/56342.html Another article: "Gravity variation from the equator to the poles" by Ramin Amirmardfar: http://www.geocities.com/ramin1102000/chap2-2page.html 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 ://www.google.com/search?q=earth+gravity+centrifugal+ratio&btnG=Google+Search&hl=en&lr=&ie=ISO-8859-1 weight poles centrifugal ://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&q=weight+poles+centrifugal&btnG=Google+Search I hope this helps you, but if you need some clarification, please post a request for it before rate my answer.```
 ```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: http://curious.astro.cornell.edu/question.php?number=310 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: http://mathforum.org/library/drmath/view/56342.html Agreement that there will be less force: http://newton.dep.anl.gov:70/askasci/phy00/phy00083.htm . Some barometers have to take this fact into account in their design: see "Latitude Correction" at http://www.mid-c.com/manmar/Latitude.htm . 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: http://www.mid-c.com/manmar/Latitude.htm radius of the earth: 6.371 x 10^6 m (http://whatis.techtarget.com/definition/0,,sid9_gci816253,00.html ----- general discussion of centripetal acceleration: http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html keywords: centrifugal centripetal centrifugal eart centrifugal earth "weigh less" "centripetal acceleratin" earth weigh less "centripetal acceleration" earth```