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Effect of earth's rotation on an object on the surface

  1. Jan 20, 2013 #1
    Hi,


    This is one of the things that confuses me.

    Assume an object on the surface of the earth has a mass m and F=m*g is the force on it due to the gravity. But also the earth is rotating and although the radius is extremely large compared to the size of the object but it must be affected by this rotation as well, isn't it?

    This is the part that I couldn't figure out. What is the total force on the body?

    Is it:

    (G*m*M/R^2) - (m*v^2/R); obviously first one gravity component and the second one angular acceleration.


    Thanks in advance.
     
  2. jcsd
  3. Jan 20, 2013 #2

    D H

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    No.

    Look at the problem kinematically:
    • What kind of motion does an object on the surface of the Earth undergo from the perspective of an inertial observer?
      Ignore the Earth's orbit about the Sun. (e.g., use a frame in which the Earth is rotating but the center of mass is stationary.)
    • What is the acceleration of the object from the perspective of this inertial observer?
    • What does Newton's second law say that the total force must be?
     
  4. Jan 20, 2013 #3
    Thanks a lot for the response D H. I will try to answer in the same order;

    OK, it's only earth and the object and the observer. The observer looking from the top.
    The object is undergoing a uniform angular motion.

    The acceleration is a=v^2/R

    F = m*a

    And the only acceleration is the angular one (v2/R); so the earth pulls the object due to gravity and the object pulls earth equally (as if the earth is stationary - Newton #3). But then there is the rotational acceleration which means there is some unbalanced force. Earth is pulling the object more than the object pulls her.

    Am I correct in thinking that;

    N - m*M*G/R^2 = m*v^2/R

    Even if it's correct, still not clear on the N though.
     
  5. Jan 20, 2013 #4

    rude man

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    That would be the weight of the mass as measured by a scale, on the equator. But not the total force.

    Remeber that a mass resting on the Equator has net force applied to it close, but not equal to, zero (gravity pulling it in, the Earth pushing it out almost with the G force, the difference being your mv^2/R.
     
  6. Jan 20, 2013 #5

    D H

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    Correct.

    Correct.

    Correct. You know m, you know a, so what does that mean that the net force F must be? This is kinematics, not dynamics. You don't need to know *anything* about causal agents such as gravity. Just use F=ma.

    Exactly. Gravitation obviously is not the only force acting on an object sitting at rest (with respect to the rotating Earth) on the surface of the Earth. There's also a force that keeps the object object from sinking into the Earth. This is the same force that keeps your hand from penetrating into a wall when you lean against a wall, and that keeps your hand from penetrating into a book when you hold the book from beneath. It's called the normal force.

    If you know the acceleration of some object, you do not need to know the individual contributors to the net force. Simply apply Newton's second law and the net force just pops out. In fact, you need to compute this net force kinematically to determine the normal force. You know the net force, you know the gravitational force, and assuming that the only other force, you can compute the normal force based on the fact that forces are subject to the superposition principle (net force is the vector sum of the individual component forces).
     
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