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Acceleration vs acceleration due to gravity

  1. Apr 30, 2012 #1
    Are these the same thing? but let's say that there is an object on a table that is not moving - I would say it is not accelerating (it has no change in velocity), but it does have an acceleration due to gravity (it has weight). I'm conflicted with the two. Can anyone clear up my confusion?
     
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  3. Apr 30, 2012 #2

    phinds

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    Einstein showed (or rather I guess it was postulated) that if you are in a closed room, there is absolutely nothing you can do to distinguish between weight that is due to being near a massive body and weight that is due to movement-type acceleration, so no you can't really say they are different, in effect at least.
     
  4. Apr 30, 2012 #3

    tiny-tim

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    hi kjamha! :smile:
    it does not have an acceleration due to gravity, it has a force due to gravity (the weight)

    that force is equal and opposite to the reaction force from the table, so the net force (and the acceleration) is 0 :wink:
     
  5. Apr 30, 2012 #4
    There are different "types" of acceleration.

    1) Co-ordinate acceleration: this is the dv/dt type acceleration.
    2) 4-acceleration: the invariant acceleration used in relativity (co-ordinate independent).
    3) Proper acceleration: 4-acceleration projected onto co-moving co-ordinates.

    Your confusion is in mixing up these different definitions. In newtonian gravity the first definition is used, so a change in velocity is required. In GR gravity is curvature in spacetime. Objects follow geodesics (equivalent of straight lines in flat spacetime), if they are prevented from doing so by a table they are not in freefall and therefor not following a geodesic and therefor are being accelerated, i.e. there 4-acceleration is not zero.
     
  6. May 1, 2012 #5
    So is it accurate to say that an object at rest on a table is not accelerating, but is experiencing an acceleration (acc due to gravity)
     
  7. May 1, 2012 #6

    tiny-tim

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    it is experiencing an acceleration of zero

    it is experiencing a force of mg
     
  8. May 1, 2012 #7
    The acceleration is 0 in Newtonian gravity, but in general relativity, the state of no acceleration is considered to be the state of free fall, so the object sitting on the table is accelerating upward. It doesn't look like it's moving because you are also accelerating upward, propelled by the floor. It seems kind of silly to think this way, but it works.
     
  9. May 1, 2012 #8
    I'm going to say the same thing for you a different way: In GR, if you feel a force you are accelerating. So Einstein takes the view that jumping out of a plane and free falling is inertial motion....no acceleration because there is no force [neglecting air resistance] you feel. A Newtonian observer on the ground sees you coming toward earth faster and faster so he would describe you as accelerating.

    Lots of stuff in math and physics has different descriptions analogous to this: plus and minus axis directions are purely arbitrary....convention has positive increasing to the right and up.....potential is another concept which is arbitrary...we pick a reference and work out differences....your positive might be my negative....convention has zero potential energy when considering gravity at infinity.....when objects are widely separated....

    In math, if you multiply both sides of any equation by [-1] and in so doing change all the signs, the equation is still valid.
     
    Last edited: May 1, 2012
  10. May 1, 2012 #9
    In a pendulum, isn't acceleration a maximum at the amplitude - where the velocity is zero? Likewise, an object that is tossed upward experiences (although I'm not sure that is the correct way to say this based on your quote) the same acceleration on the way up, down, and at the apex (velocity = 0). I'm still a bit confused.
     
  11. May 1, 2012 #10

    tiny-tim

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    yes :smile:
    yes, a projectile always has acceleration g downwards

    but an object stationary on a stationary table has zero velocity and zero acceleration :wink:
     
  12. May 1, 2012 #11

    HallsofIvy

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    Are you clear on what acceleration is? If an object is sitting still its speed is 0 and stays 0- its speed is not changing- and acceleration is the rate of change of speed.
     
  13. May 3, 2012 #12
    There are a couple of things that are vague in this. "Sitting still" is one. The other is defining acceleration as rate of change of speed. If you mean speed to be the magnitude of the velocity then surely this is not correct. In relativity for example the velocity has unit magnitude and it is the angle that is changing producing a 4-acceleration. Objects in circular orbit are also a common 3D example.

    I'm sure you know this so I'm a little confused by your post.
     
  14. May 3, 2012 #13
    In the Newtonian model yes, not in GR.
     
  15. May 3, 2012 #14
    Yes, but I don't think the OP is necessarily ready for GR.
     
  16. May 6, 2012 #15
    Sort of. He does mention that an object can have an acceleration due to gravity when sitting on a table. This says to me that he has read a GR interpretation and is confusing it with the Newtonian one. Perhaps not. :-)
     
  17. May 7, 2012 #16
    In every theory, "stationary" implies that you have chosen a reference system in which the object's acceleration is zero. In GR one may choose instead a reference system in which the object is not stationary but accelerating. Hopefully it will become clear to the OP that here two points of view are described that in GR are equivalent (but equivalent is not "the same thing").
     
    Last edited: May 7, 2012
  18. May 7, 2012 #17
    I just feel like pointing out one of the finer points.

    In Newtonian relativity, it is experiencing a force of 0 N. This is because the normal force cancels it out.

    Now if we rewrote that to contain "a gravitational force," then, from a Newtonian viewpoint, yes, you'd be right.

    And from an Einsteinian viewpoint, the quote, as is, is completely correct. The observer is also experiencing a net force of mg. An object in freefall will be experiencing no force.
     
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