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Weight and shape of earth

  1. Nov 26, 2007 #1
    Hi,

    Which formula may help to compute the effect to weight due to geoid shape of earth? I assume earth is not rotating.

    Thanks
     
  2. jcsd
  3. Dec 3, 2007 #2
    Maybe i couldn't explain well. In other words; rotating of earth and shape(geoid) both effects the accelerate(g).

    1. The rotating is decreasing weight and i can understand this.

    2. Shape of earth. I am not sure to understand the effect of this.

    Say, fill in the blanks.

    -* The shape of earth effects the weight. So, weight is bigger than equator because ....................

    it is closer? It is far? ... What??

    ps: is my question too easy to ask(stupid) or hard or unknown or else could not explain my question.
     
  4. Dec 3, 2007 #3
    1. The rotating decreases the weight of an object on earth surface. The effect of rotating is smaller if you are closer to the poles.
    2. The weight is slightly smaller in equator because the earth radius there is somewhat larger than that at the poles.

    The two effects do not cancel each other.
     
  5. Dec 4, 2007 #4

    LURCH

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    It was difficult to picture the Earth being oblong if it's not rotating. However, if you were on a non-rotating planet of the same size and mass as the Earth, and this planet was not rotating, but for some reason still kept its geoid shape (and the density variations that go with it), then the basic rule is that gravity (and therefore weight) decreases by the square of the distance. In this case, the square of the distance between the center of the Earth and an object on the surface.

    If you take a person as an example, then let us suppose that we use some unit of measure that fits exeactly 1,000 times in between that person and the center of the planet, when that person stands at the poles. Assume that when the person moves to the Equator, the distance between their feet and the center of the planet increases to 1,002 (same units of measure). Then their distance has increased by 2/1,000 (.2%). If we also use some unit of measure for weight so that the person weighs exactly 100 something-or-others at the pole, then they will weigh .4% less (.2%2), or 999.6 s/o at the equator.

    BTW, these numbers are generated ex posteriori, and do not reflect the actual eccentricity of the Earth's shape.
     
    Last edited: Dec 4, 2007
  6. Dec 4, 2007 #5

    D H

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    Staff Emeritus
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    That "basic rule" is only true for spherical objects. For non-spherical objects you need to use a more complex form, typically a spherical harmonic representation of the Earth's gravity field.
     
  7. Dec 4, 2007 #6
    Thanks to everybody,

    I read that weight is changing with rotating velocity of earth and shape of it too. Rotating effect is more understandable for me. But the shape is not easy to understand for me.

    As i know, the gravity field(g) is increasing linear untill earth's surface but then decreases by the square of the distance. Now, pole and equator both on surface but both has different diameters. Then i ask myself, which physical reality can tell me the effect of shape(different radius) to gravity field. Honestly, non of approach what i know, didn't satisfied me. Then thought that the time to ask physicsforums.

    In this case, D_H's opinion is more close my thought. I mean,

    Code (Text):
    g = G [tex]\frac{ M  m }{R^{2}}[/tex]
    Code (Text):
    g = K d R  (d: dencity, R: Radius, K: constant )
    both equates above will not explain the effect to accelerate the shape of world.

    If i understand wrong, please warn me.

    Thank you very much
     
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