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-   -   Circular Motion & Gravitation (https://www.askmehelpdesk.com/showthread.php?t=109464)

  • Jul 12, 2007, 10:22 PM
    Gostronge
    Circular Motion & Gravitation
    This question is driving me crazy. I'm using the formula v=2(pi)(R) / T to find the velocity and then a=v^2 / R to find the final acceleration, but have no luck in getting the correct answer. What I don't understand is how does the period (T) is used in the first formula? Please help.


    The radius of the earth at the equator is approximately 6.37 * 10^6m
    The centripetal acceleration of the person standing at the equator is?

    a 3.4 * 10^-2 m/s^2
    b 7.27 * 10^-5 m/s^2
    c 8.5 * 10^-4 m/s^2
    d 9.79 m/s^2
  • Jul 12, 2007, 10:44 PM
    Capuchin
    The period of motion is 1 day = 24*60*60 seconds. Using this I do get one of your answers.

    The result you get here is quite interesting, it's part of the reason why people of that same mass at the poles are heavier than on the equator. They have to overcome centripetal forces by using the gravitational force. But as you will see the difference from this effect is very small. The other part is that at the poles you are 13 miles closer to the center of gravity of the Earth than at the equator, because the Earth is not a perfect sphere. These two effects add and there's nearly a 0.5% difference in gravitationaly force between poles and equator.

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