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Acceleration vs acceleration due to gravity |
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| Apr30-12, 03:53 PM | #1 |
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Acceleration vs acceleration due to gravity
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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| Apr30-12, 03:58 PM | #2 |
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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.
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| Apr30-12, 03:58 PM | #3 |
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hi kjamha!
![]() that force is equal and opposite to the reaction force from the table, so the net force (and the acceleration) is 0
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| Apr30-12, 04:47 PM | #4 |
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Acceleration vs acceleration due to gravity
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. |
| May1-12, 05:13 AM | #5 |
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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)
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| May1-12, 05:22 AM | #6 |
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it is experiencing an acceleration of zero
it is experiencing a force of mg |
| May1-12, 11:41 AM | #7 |
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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.
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| May1-12, 03:33 PM | #8 |
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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. |
| May1-12, 05:14 PM | #9 |
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| May1-12, 05:52 PM | #10 |
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![]() but an object stationary on a stationary table has zero velocity and zero acceleration
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| May1-12, 06:27 PM | #11 |
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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.
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| May3-12, 05:17 PM | #12 |
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I'm sure you know this so I'm a little confused by your post. |
| May3-12, 05:19 PM | #13 |
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| May3-12, 05:46 PM | #14 |
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| May6-12, 04:54 PM | #15 |
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| May7-12, 03:28 AM | #16 |
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| May7-12, 08:51 AM | #17 |
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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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