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Centrifugal force? Why does the Earth bulge at the equator?

  1. Mar 24, 2014 #1
    I'm trying to understand why a superdeformed nucleus may be represented as bulging perpendicular to the axis of rotation, and I'm guessing this is akin to why the Earth does so too. I've gone through secondary school and 3 years of University to have professors/teachers snigger every time they hear the word centrifugal force. But whenever I look up the explanation for bulging, centrifugal force is always mentioned.

    I honestly can't recall a situation where I've had to consider anything called a centrifugal force and I can't with any confidence even say what such a force even is. If I were to guess, in the non-inertial rotating frame of reference it's a force that appears to exist due to the Coriolis effect - throw a ball and it will seem to move from its original trajectory, as if a force was being applied. However to a stationary observer outside the rotating frame we of course just see the ball being thrown in a straight line - no force is being applied to anything.

    Is there any way to understand why the Earth bulges at the equator due to its rotational motion, without having to delve in to the concept of centrifugal forces?
     
  2. jcsd
  3. Mar 24, 2014 #2

    A.T.

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    On the poles all the gravity acts to compress the planet. On the equator some of the gravity is used up for centripetal acceleration.
     
  4. Mar 24, 2014 #3
    I don't know how, but I'll try. A point on the equator is inert. It wants to keep moving forward forever, but the rest of the planet is gravitationally curving it. At the same time, the point is also pulling the planet and trying to make it go forward with it, that's why the Earth bulges.
    I do hope someone can explain this better

    cb
     
  5. Mar 24, 2014 #4
    There is nothing wrong with using the centrifugal force and in this case it is the easiest way to understand what's going on. just keep in mind that the centrifugal is an inertial force that appears because the rotating frame isn't an inertial frame. You also should know that the centrifugal force and the Coriolis force are two completely separate phenomena. Both appear in rotating reference frames but one does not create the other.
     
    Last edited: Mar 24, 2014
  6. Mar 24, 2014 #5

    FactChecker

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    Centrifugal force and Coriolis force are not directly related. Either can be 0 while the other is not. Both are "pseudo" forces to allow F=mA in a rotating frame but the Coriolis force is at right angles to the centrifugal force and is caused by a different velocity. It is zero if there is no change in the distance from the object of interest to the axis of rotation, whereas centrifugal force can be nonzero.

    Yes. The bulge is because objects want to continue in a straight line tangent to the circle they are on. But you might as well give it a name and introduce centrifugal force.
     
  7. Mar 24, 2014 #6

    AlephZero

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    Almost, but not quite. If the non-inertial frame is rotating at constant angular velocity, the centrifugal force on a particle depends on its position in the rotating frame, and is a fictitious force of ## m r \omega^2## acting radially outwards.

    The Coriolis force depends on the velocity of the particle in the rotating frame (which presumably is zero for your question).

    For a rotating object in an inertial frame, there is a real force ##m r \omega^2## acting radially inwards on a parrticle, creating the centripetal acceleration.

    I'm not sure why you are your professors are making such a big deal about one formulation rather than the other.
     
  8. Mar 25, 2014 #7

    A.T.

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    Coriolis force is perpendicular to the velocity (in the rotating frame), not to the centrifugal force. If the velocity is purely tangential the Coriolis force is radial, not zero.
     
    Last edited: Mar 25, 2014
  9. Mar 25, 2014 #8

    FactChecker

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    I stand corrected. Coriolis has more to it than I visualized. Thanks.
     
  10. Mar 25, 2014 #9

    jtbell

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    Imagine whirling an object around over your head at the end of a string. At a fixed angular speed (radians per second or revolutions per minute or whatever), the larger the radius of the object's path, the more centripetal force you have to exert on the string (and the string has to exert on the object).

    All parts of the Earth rotate at the same angular speed. As the latitude decreases, moving towards the equator, the radius of revolution (at the Earth's surface) increases. Therefore the parts closer to the equator need larger centripetal forces to keep them revolving. To get those larger forces, the material of the Earth must "stretch out" more, similar to a spring whose tension force increases as you stretch it.
     
  11. Mar 25, 2014 #10

    bcrowell

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    BTW, superdeformation doesn't always have to be stabilized by rotation. Superdeformation was discovered way back in the 60's, in nuclei in the uranium/plutonium region, where it exists at zero spin. The term to google for is "fission isomer." What *is* needed in all cases in order to stabilize superdeformed shapes is a gap in the single-particle states at superdeformed shape, i.e., a superdeformed shell closure.
     
  12. Mar 26, 2014 #11
    As an (hopefully) interesting side note, relativity predicts that if the Earth was a perfect rotating sphere, clocks at the Equator would tick slower than clocks at the poles due to time dilation caused by the tangential velocity at the equator. Relativity also predicts that time dilation if a function of distance from the gravitational mass and for a bulging Earth, the reduction in time dilation due the increased distance from the centre of the Earth at the equator cancels out the increased time dilation due to tangential velocity at the equator. This means that all clocks tick at the same rate at sea level anywhere on the Earth.

    This is more than just a coincidence. The time dilation due to the combined factors of gravitational potential and tangential velocity create an effective potential and matter flows from a high effective potential to low effective potential and this shaped the Earth into its bulged shape. There is of course much more to it all than mentioned here, but this is just an aside (and definitely not a classical physics explanation).
     
    Last edited: Mar 26, 2014
  13. Mar 26, 2014 #12
    Can you give an example of an inertial frame where these forces do not appear and the bulge can still be explained?
     
  14. Mar 26, 2014 #13

    A.T.

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    See post #2
     
  15. Mar 26, 2014 #14

    D H

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    From the perspective of a frame that rotates with the Earth, the surface of the Earth is very well approximated as a surface of constant potential energy, where the potential from both the gravitation and centrifugal forces contribute to the total.

    From the perspective of an inertial (non-rotating) frame, the surface of the Earth is very well approximated as a surface of constant total energy, where both the gravitation force and kinetic energy contribute to the total.

    These two perspectives yield the same result.
     
  16. Mar 26, 2014 #15
    And what exactly is the inertial reference frame you are considering? The rotating earth is not an inertial reference frame.

    Even if you pick the "fixed" stars as your inertial frame, you will still measure a bulge in the earth. So where did the force to cause this deformation come from? It can't be a fictitious force because they supposedly do not exist in an inertial frame.

    Suppose you had two planets of equal mass and density but you only could measure their diameters. You would have to infer that one of them was rotating if you measured a bulge on one of them. Yet what external real force was causing the bulge?
     
    Last edited: Mar 26, 2014
  17. Mar 26, 2014 #16

    A.T.

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    How exactly do you compute that total energy, which is constant over the surface in the inertial frame?
     
  18. Mar 26, 2014 #17

    A.T.

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    See post #2

    There are no external forces, just gravity versus pressure.
     
  19. Mar 26, 2014 #18

    A.T.

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    The Earth is under radial tension?
     
  20. Mar 26, 2014 #19
    See post #15.
     
  21. Mar 26, 2014 #20

    A.T.

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    I did, and answered it in post #17.
     
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