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Gravitation question

  1. Jun 8, 2015 #1
    1. The problem statement, all variables and given/known data
    There's a sphere with the mass of the earth and same radius.It rotates at constant speed. It hhas the period of the earth. a mass of a kg sits on the surface.
    1) find grav force on mass
    2) find centripetal force on mass
    3) find the difference in magnitude between them
    4) suggest a value of acceleration of free fall for the mass.
    2. Relevant equations


    3. The attempt at a solution
    Cant do last part; the way I see it is that the centripetal force does not exist separately; it is criteria for circular motion at constant speed. So clearly the gravitational force over-provides, but I dont understand how this excess is related to the acceleration.
     
  2. jcsd
  3. Jun 8, 2015 #2

    Simon Bridge

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    What is the definition of "free fall"?

    Note for (2): latitude is not given.
     
    Last edited: Jun 8, 2015
  4. Jun 8, 2015 #3
    Motion where the only force acting is weight
     
  5. Jun 8, 2015 #4

    Simon Bridge

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    So what is the weight of the object close to the surface?
     
  6. Jun 8, 2015 #5
    mg, where g is the GFS at that point, so is it the grav force?
     
  7. Jun 8, 2015 #6
    When you say latitude isn't given, what is the significance of that?
     
  8. Jun 8, 2015 #7

    Simon Bridge

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    No - you are asked to find the acceleration.
    Acceleration and force are different physical concepts - important not to mix them up.
    What the question is basically asking is what the acceleration would be if there was no ground to stop it falling.
    (Note: the centripetal force is the difference between the object's weight and the normal force to the surface.)

    Compare the centripetal force sitting on the equator to that when sitting on a pole.
    What about latitudes in between?
     
  9. Jun 8, 2015 #8

    haruspex

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    How far is the mass from the axis of rotation.
    By the way, I agree with you that it is not right to ask what is the centripetal force on the mass. It would be better to ask what centripetal force is required by the mass in order to stay on the surface.
     
  10. Jun 8, 2015 #9
    Haruspex - its a whole radius away isnt it?
    Mr.Bridge- so if we removed the surface the reaction force would disappear. The only acting force would be the grav force?
     
  11. Jun 8, 2015 #10
  12. Jun 8, 2015 #11

    haruspex

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    How far are you away from the Earth's axis of rotation now?
     
  13. Jun 8, 2015 #12
    Pretty far , im english so certainly on a smaller radius than the mass in the question, which is on the equator.
     
  14. Jun 8, 2015 #13
    The acceleration due to gravity and centripetal force are in the same direction if you are on the equator. I would use the equator as your teacher probably expected. Then it's easy to add the two acceleration vectors together. And g is easy if we use the standard r at the equator. And then Ac is easy if we use the equator. Note you will have to find v by looking up r. I'm thinking this is what your teacher wants you to do.

    And the gravitational force idea just states that any two masses attract each other with the same force. But one can also see that an object on the surface is also changing direction( if not exactly on a rotational pole) so there must be centripetal force as well. And Fc Is supplied by the force of gravity.
     
    Last edited: Jun 8, 2015
  15. Jun 8, 2015 #14
    The actual question: 1433777769338.jpg
     
  16. Jun 8, 2015 #15

    haruspex

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    This is the first time you have stated it is on the equator. This is why Simon asked for the latitude.
     
  17. Jun 8, 2015 #16

    haruspex

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    This is backwards.
    Centripetal force is not some extra force that you can add on to gravity. It is a required component of the resultant force, namely, that necessary to keep the object travelling on the arc. See https://www.physicsforums.com/insights/frequently-made-errors-pseudo-resultant-forces/
    The two forces acting on the mass are gravity and the normal force from the surface. The resultant of the two is the centripetal force. The apparent weight of the mass is equal and opposite to the normal force.
     
  18. Jun 9, 2015 #17
    Ah

    Because the normal is a 3rd law reaction to what is the weight pushing down
     
  19. Jun 9, 2015 #18

    haruspex

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    To the apparent weight, yes.
     
  20. Jun 9, 2015 #19

    Simon Bridge

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    Technically the 3rd law reaction to the weight of the object due to the Earth is the weight of the Earth due to the object. This force acts at the center of mass of the Earth and points towards the object.

    The normal force to the ground is the reaction to the force of the object pressing into the ground. It can be a tad confusing because "how hard we press against the ground" is what we commonly think of as "weight". In physics, "weight" is a technical term for the force of gravity on an object, so it should be used carefully.
     
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