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Rotation of earth and weight of an object on the earth

  1. May 1, 2008 #1
    We all know that g is more at the poles than at the equator because it is closer to the cenre. I think earth's rotation will not have any effect on the g because g depends only on the mass product and the distance only.

    But roration of the earth should be having some effect on the weight of a body as it may have a component in the direction of g and the net acceleration will be different from g. How does this net acceleration that determines the apparent of weight of an object on the earth change with the location (poles or equator) of the object on the earth and the angular velocity of the rotation?
  2. jcsd
  3. May 1, 2008 #2


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    Have you ever solved the standard kinematic problem of a rollercoaster at the top of a loop? If you have, it is the same problem here with what you are asking.

  4. May 2, 2008 #3

    Doc Al

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    Start by drawing a free body diagram for the object. (Imagine that it's suspended on a string and you want to find the tension in the string--that would be the apparent weight of the object.) Now apply Newton's 2nd law. If the earth didn't rotate, the acceleration in all directions would be zero. But if you include the rotation of the earth, you'll have an acceleration directed towards the axis of rotation (a centripetal acceleration).

    First analyze the cases where the object is on the equator or on a pole, then worry about the locations in between.
  5. May 2, 2008 #4


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    This is probably just a matter of terminology, but often it is the net acceleration (including Coriolis and centripetal effects) that is called 'g'.

    Furthermore, g does not depend on a mass product. There is only one mass involved in an expression for gravitational g, the mass of the earth.
  6. May 2, 2008 #5


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    FYI, there is a related thread at


    where the ellipticity of a planet due to rotation is calculated.

    The ratio of centrifugal to gravitational effects on an object's apparent weight is given by the ratio

    \frac{4 \pi^2 r^3}{GMT} \cos(\theta_{Lat})

    where r, M, and T are the planet's radius, mass, and rotation period, respectively. [tex]\theta_{Lat}[/tex] is the latitude.

    For the Earth this works out to 1/290 or 0.34% at the equator.
  7. Feb 1, 2009 #6
    Yes it does have an effect..
    This is due to the gravity's force is resolved into a centripetal force that acts parallel to the equator.

    Thus, At the Poles, apparent weight is the same as mg.

    At the equator, Apparent weight
    is g' = g - Rw^2
    Where w is the angular velocity

    At latitude 0 above the equator is
    g' = √(g2+(Rw2-2g )Rw2 cos2ѳ)
  8. Mar 1, 2010 #7
    ok, so I guess I'm not following this correctly, if I weighed 200 pounds at the equator, and the earth stopped it's rotation, how much would I weigh?
  9. Mar 2, 2010 #8

    Doc Al

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    Your apparent weight would increase by mRω². As Redbelly98 stated in post #5, that's an increase of about 0.34%. Thus your apparent weight would increase by about 0.7 lbs.
  10. Mar 2, 2010 #9
    So the take-away here is that supermodels should all be told these facts so that they migrate to mountaintop observatories, where astrophysicist and astronomers can have their way with the shivering lasses. :rofl:

    Or.... yes, your weight would increase. My point however is that the effect of gravity drops off so with distance from the mass, that distance from the surface of the earth is a much bigger issue than where on earth you are. Better yet, you ignore psuedo-forces this way, and the models...
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