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Gravity and Acceleration

  1. Jul 29, 2010 #1
    Here's a weird, theoretical question that I couldn't quite decide the answer to by myself. Let's say you have a guy standing on a planetoid of some sort. The planetoid has the same gravity at its surface as Earth does: 1g (just for simplicity). Now, if this planetoid accelerates directly away from the person (i.e. in the direction they'd fall if the ground gave out) at 1g itself, would the gravity of the planet cancel out the feeling of acceleration for our guy? Would he be weightless as he's dragged along by the planet's gravity, constantly falling toward it as long as the rate of acceleration and everything else remain perfect? I have a feeling I might be missing other factors, and I'm also not sure how Newton's third law would factor in here.

    Thanks for any thoughts.
     
  2. jcsd
  3. Jul 29, 2010 #2
    Hi there,

    From your hypothetical question, and if I understand it right, your guy on his planetoid would be in the same situation as if you would take the elevator that falls at 1g.

    The inside of the elevator remains the same, just the apparent gravity that changes. Your guy could not be considered in an inertial frame. His reference frame would be accelerating, therefore Newton's laws of motion would have to be adapted.

    Cheers
     
  4. Jul 29, 2010 #3

    Ich

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    The point of the elevator is to illustrate that the guy in free fall has every right to see himself as inertially moving. Same with the guy on the planetoid.
    Strictly defined, that means zero proper acceleration, which in turn means "feeling weightless".
     
  5. Jul 29, 2010 #4
    Like the guy in the elevator, if the cable breaks, he would feel weightless.
     
  6. Jul 29, 2010 #5

    Ich

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    Yep. In fact, the elevator example may not be the best analogy for the planetoid.
    However, in GR, everything accelerating under the influence of gravity alone may be justifiably deemed not accelerating at all.
     
  7. Jul 29, 2010 #6
    As soon as the planet "takes off" he will be free falling.
     
  8. Jul 29, 2010 #7
    And what does this "zero proper acceleration" mean?

    AB
     
  9. Jul 29, 2010 #8

    Ich

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    I didn't join that discussion for a reason. Just let me say: As far as I've read, I completely agree with DaleSpam's position, while I have no idea what you're arguing about there.
    BTW, I explicitly said what I mean by "zero proper acceleration" in #3.
     
  10. Jul 29, 2010 #9

    Jonathan Scott

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    The term "proper" in relativity effectively means "as seen from its own point of view".

    This use of the word "proper" as relating to "its own", similar to the French "propre", is mostly archaic in English, but it is often used as an English translation of scientific and mathematical terms which include "Eigen" in German.
     
  11. Jul 30, 2010 #10
    Thanks for the replies, guys. So if you put the guy in a box that was bolted down to the ground, he wouldn't feel any g-forces in any direction? That was the conclusion I came to, and I partly compared it to the old elevator-in-free-fall explanation too, but it's still pretty weird to think about.
     
  12. Jul 31, 2010 #11
    That doesn't make sense to say "feeling weightless means zero proper acceleration" until you clarify what such terms mean in the language of GR. If you have Dale's idea as well, then I'd guess you take "zero proper acceleration" as meaning "moving along a geodesic". Okay but how do you relate this to "feeling weightless"? I'm asking a simple question and if you don't have any answer, then try not to use such terms together.

    AB
     
  13. Aug 1, 2010 #12

    Ich

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    An uncharged point particle in geodesic motion feels weightless.

    That said, I don't like being forced into discussions that I avoid for good reasons. Please respect that, and please respect my right to use terms and phrases as I see fit even without your approval.
     
  14. Aug 1, 2010 #13
    Well if you're cool with using it, then there is no problem. I respect your idea!

    AB
     
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